Heat and Mass transfer(NPTEL).pdf

Motivation and HighlightsMotivation: In this course, we are primarily interested in heat, which is the form of energy than can be transferred from one system to another as a result of temperature difference. The science that deals with the rates of such energy transfers is called Heat Transfer. Why do we need to undertake a detailed study on heat transfer? After all, we can determine the amount of heat transfer for any system undergoing any process using thermodynamic analysis alone. The reason is that thermodynamics is concerned with the amount of heat transfer as a system undergoes a process from one equilibrium state to another, and it gives no indication about how long the process should take or what is the mode of heat transfer. But engineers are more concerned with the rate of heat transfer than the amount. For example, it is relatively simple to calculate, using principle of thermodynamics, the amount of heat that is transferred when water in a thermos flask cools from 90 ºC to 60 ºC. But a typical user is more interested in knowing how quickly the liquid in the thermos bottle cools, and an engineer who designs the thermos tries to understand the mode of heat transfer and then designs the bottle appropriately for low rates of heat transfer. Relevance of heat transfer: In the vernacular of the time, heat transfer is indeed a relevant subject, not to mention an inherently fascinating part of engineering sciences. We will devote much time to acquire an understanding of heat transfer effects and to developing the skills needed to predict heat transfer rates. What is the value of this knowledge and to what kinds of problems may be applied? Heat transfer phenomenon plays an important role in many industrial and environmental problems. As an example, consider the vital area of energy production and conversion. There is not a single application in this area that does not involve heat transfer effects in some way. In the generation of electrical power, whether it is through nuclear fission or fusion, the combustion of fossil fuels, magneto hydrodynamic processes, or the use of geothermal energy sources, there are numerous heat transfer problems that must be solved. These problems involve conduction, convection, and radiation processes and relate to the design of systems such as boilers, condensers, and turbines. One is often confronted with the need to maximize heat transfer rates and to maintain the integrity of material in high temperature environments. On a smaller scale there are many heat transfer problems related to the development of solar energy conversion systems for space heating, as well as for electric power production. Heat transfer processes also affect the performance of propulsion systems, such as the internal combustion, gas turbine, and rocket engines. Heat transfer problem arise in the design of conventional space and water heating systems, in the design of conventional space and water heating systems, in the design of incinerators and cryogenic storage equipment, in the cooling of electronic equipment, in the design of refrigeration and air conditioning systems, and in many manufacturing processes. Heat transfer is also relevant to air and water pollution and strongly influences local and global climate. Highlights: Classification of heat transfer problems: The heat transfer problems typically encountered in practive can be broadly classified into two groups: (1) rating and (2) sizing problems. The rating problems deal with the determination of heat transfer rate for an existing system at a specified temperature difference. The sizing problems deal with the determination of the size of a system in order to transfer heat at a specified rate for a specified temperature difference. In other words, rating problems deal with an analysis of a given system, while sizing problems deal with the design of a new system for a specified performance. Experimental vs. analytical studies: A heat transfer process or equipment can be studied either experimentally (testing and taking measurements) or analytically (by analysis or calculations). The experimental approach has the advantage that we deal with the actual physical system (or a scaled down model), and the desired quantity is obtained by measurement, within the limits of experimental error. However, this approach is time consuming, expensive and often impossible. For example, the system we are analyzing may not be existing (at the design stage) and hence measurement approach will not be practical at all. The analytical approach (including computational modeling) has the advantage that it is fast and inexpensive, but the results obtained are subject to numerical accuracy and the validity of the assumptions and idealization made in the analysis. Modeling in heat transfer: Over the years, modeling has gradually evolved to become a costeffective alternative to experimentation with respect to engineering design in many cases. The development of advanced computational tools in heat transfer and the increase in computing power which roughly doubles every two years has contributed immensely to the feasibility of solving realistic engineering problems. With modeling approach, the lead time in design and development of equipment can be considerably reduced. How does modeling work? The description of most scientific problems involve expressions that relate the changes in some key variables to each other. Usually, the smaller the increment chosen in the changing variables, the more accurate is the description. In the limiting case of infinitesimal or differential changes in variables, we obtain differential equations that provide precise mathematical formulations for the physical principles and laws by representing the rates of changes as derivatives. Hence, the differential equations are used Solving Heat Transfer Problems: General Guidelines A major objective of this learning material is to prepare you to solve engineering problems that involve heat transfer processes. In view of this, several problems are provided and worked out at the end of each module. In working these problems you will gain a deeper appreciation of the fundamentals of the subject, and you will gain confidence in your ability to apply these fundamentals to the solution of engineering problems. It is strongly recommended that the student tries to work out the problems first by himself/herself, before looking at the solutions provided. In solving heat transfer problems, use of a systematic procedure is advocated. Our experience shows that most heat transfer problems can be tackled by a solution procedure characterized by a prescribed format. This procedure has been consistently employed in all the solutions provided, and we require the students to use it in their problem solutions in the future. The general solution procedure consists of the following steps: 1. Known: After carefully reading the problem, state briefly and concisely what is known about the problem. Do not repeat the problem statement verbatim. 2. Find: State briefly and concisely what must be found. 3. Schematic: Draw a schematic of the physical system. If application of the conservation laws is anticipated, represent the required control surface by dashed lines on the schematic. Identify relevant heat transfer processes by appropriately labeled arrows on the schematic. Treat this entire step seriously, as a good picture paints a thousand words! 4. Assumptions: List all pertinent simplifying assumptions. 5. Properties: Compile property values need for calculations, and identify sources from where you obtain the values. Make sure that the source is reliable. 6. Analysis: Begin your analysis by applying appropriate conservation laws, and introduce rate equations as needed. Develop the analysis as completely as possible before substituting numerical values. Check the units before substituting any numerical value. Perform the calculations needed to obtain the desired results. 7. Comments: Discuss your results (even if you are not asked do to so, specifically). Such a discussion may include a summary of key conclusions, a critique of the original assumptions, and an inference of trends obtained. This step will give you additional insights. Module 1: Learning objectives • • • Overview: Although much of the material of this module will be discussed in greater detail, the objective of this module is to give you a reasonable overview of heat transfer. Heat transfer modes: You should be aware of the several modes of transfer modes of transfer and their physical origins. Physical insight: Given a physical situation, you should be able to perceive the relevant transport phenomena. The importance of developing this insight must not be underestimated. You will be investing a lot of time to acquire the tools needed to calculate heat transfer phenomena. However, before you can begin to use these tools to solve practical problems, you must have the intuition to determine what is happening physically. In short, you must be able to look at a problem and identify the pertinent transport phenomenon. The example and problems at the end of this module should help you to begin developing this intuition. Rate equations and conservation laws: You should also appreciate the significance of the rate equations and feel comfortable in using them to compute transport rates. You must also recognize the importance of the conservation laws and the need to carefully identify control volumes. With the rate equations, the conservation laws may be used to solve numerous heat transfer problems. • DR.PRADIP DUTTA Department of Mechanical Engineering Indian Institute of Science Bangalore What is Heat Transfer? “Energy in transit due to temperature difference.” Thermodynamics tells us: How much heat is transferred (δQ) How much work is done (δW) Final state of the system Heat transfer tells us: How (with what modes) δQ is transferred At what rate δQ is transferred Temperature distribution inside the body Heat transfer complementary Thermodynamics MODES: Conduction - needs matter - molecular phenomenon (diffusion process) - without bulk motion of matter Convection - heat carried away by bulk motion of fluid - needs fluid matter Radiation - does not needs matter - transmission of energy by electromagnetic waves APPLICATIONS OF HEAT TRANSFER Energy production and conversion - steam power plant, solar energy conversion etc. Refrigeration and air-conditioning Domestic applications - ovens, stoves, toaster Cooling of electronic equipment Manufacturing / materials processing - welding,casting, soldering, laser machining Automobiles / aircraft design Nature (weather, climate etc) (Needs medium, Temperature gradient) T1 . . . . . . .. . . . . . . . . . . . . .. . . . .. . .. . . . .. . .. . .. . . . .. . .. . . . . . . . . .. . . . . . . .. . .. . .. . .. . . . . solid or stationary fluid ………….. . . ... . . . . . T1>T2 T2 q’’ RATE: q(W) or (J/s) (heat flow per unit time) Conduction (contd…) x Rate equations (1D conduction): A Differential Form q = - k A dT/dx, W k = Thermal Conductivity, W/mK A = Cross-sectional Area, m2 T = Temperature, K or oC x = Heat flow path, m Difference Form q = k A (T1 - T2) / (x1 - x2) T1 q T2 k Heat flux: q” = q / A = - kdT/dx (W/m2) (negative sign denotes heat transfer in the direction of decreasing temperature) Conduction (contd…) Example: The wall of an industrial furnace is constructed from 0.15 m thick fireclay brick having a thermal conductivity of 1.7 W/mK. Measurements made during steady state operation reveal temperatures of 1400 and 1150 K at the inner and outer surfaces, respectively. What is the rate of heat loss through a wall which is 0.5 m by 3 m on a side ? moving fluid Ts>T∞ T∞ q” Ts Energy transferred by diffusion + bulk motion of fluid Rate equation (convection) U∞ y U∞ u(y) q” y T∞ T(y) Ts Heat transfer rate q = hA( Ts-T ∞ ) W Heat flux q” = h( Ts-T ∞ ) W / m2 h=heat transfer co-efficient (W /m2K) (not a property) depends on geometry ,nature of flow, thermodynamics properties etc. Convection (contd…) Free or natural convection (induced by buoyancy forces) Convection Convection Forced convection (induced by external means) May occur with phase change (boiling, condensation) Convection (contd…) Typical values of h (W/m2K) Free convection gases: 2 - 25 liquid: 50 - 100 Forced convection gases: 25 - 250 liquid: 50 - 20,000 Boiling/Condensation 2500 -100,000 q1” q2” T1 T2 RATE: q(W) or (J/s) Flux : Heat flow per unit time. q” (W/m2) Rate equations (Radiation) RADIATION: Heat Transfer by electro-magnetic waves or photons( no medium required. ) Emissive power of a surface (energy released per unit area): E=εσTs4 (W/ m2) ε= emissivity (property)……… σ=Stefan-Boltzmann constant Rate equations (Contd….) Ts u r q”r a d. Ts q”co n v. Area = A Radiation exchange between a large surface and surrounding Q”r a d = εσ(Ts4-Tsur4) W/ m2 Radiation (contd…) Example: An uninsulated steam pipe passes through a room in which the air and walls are at 25 °C. The outside diameter of pipe is 70 mm, and its surface temperature and emissivity are 200 °C and 0.8, respectively. If the coefficient associated with free convection heat transfer from the surface to the air is 5 W/m2K, what is the rate of heat loss from the surface per unit length of pipe ? Module 1: Short questions 1. What is the driving force for (a) heat transfer (b) electric current flow and (c) fluid flow? 2. Which one of the following is not a property of the material ? A. thermal conductivity B. heat transfer coefficient C. emissivity 3. What is the order of magnitude of thermal conductivity for (a) metals (b) solid insulating materials (c) liquids (d) gases? 4. What is the order of magnitude for the convection heat transfer coefficient in free convection? Forced convection? Boiling? 5. When may one expect radiation heat transfer to be important? 6. An ideal gas is heated from 50 ºC to 70 ºC (a) at constant volume and (b) at constant pressure. For which case do you think the energy required will be greater? Why? 7. A person claims that heat cannot be transferred in a vacuum. How do you respond to this claim? 8. Discuss the mechanism of thermal conduction in gases, liquids and solids. 9. Name some good conductors of heat; some poor conductors. 10. Show that heat flow lines must be normal to isotherms in conduction heat transfer. Will it be true for convection heat transfer? Module 1: Worked out problems Problem 1: A freezer compartment consists of a cubical cavity that is 2 m on a side. Assume the bottom to be perfectly insulated. What is the minimum thickness of Styrofoam insulation (k=0.030W/m.K) which must be applied to the top and side walls to ensure a heat load less than 500 W, when the inner and outer surfaces are -10 ºC and 350C? Solution: Known: Dimensions of freezer component, inner and outer surfaces temperatures. Find: value. Thickness of Styrofoam insulation needed to maintain heat load below prescribed Schematic: Assumptions: (1) perfectly insulted bottom, (2) one-dimensional conduction through five walls of areas A=4m2, (3) steady-state conditions Analysis: Using Fourier’s law, the heat rate is given by ΔT A total L Solving for L and recognizing that A total =5*W2 q = q '' .A = k L= L= 5kΔTW 2 q 5 * 0.03W / m.k * 45 0 C * 4m 2 500W L = 0.054m = 54mm Comments: The corners will cause local departures from one–dimensional conduction and, for a prescribed value of L, a slightly larger heat loss. Problem 2: A square silicon chip (k=150W/m.k) is of width W=5mm on a side and of thickness t=1mm. the chip is mounted in a substrate such that its side and back surfaces are insulated, while the front surface is exposed to a coolant. If 4W are being dissipated in circuits mounted to the back surface of the chip, what is the steady-state temperature difference between back and front surfaces? Known: Dimensions and thermal conductivity of a chip. Power dissipated on one surface. Find: temperature drop across the chip Schematic: Assumptions: (1) steady-state conditions, (2) constant properties, (3) uniform dissipation, (4) negligible heat loss from back and sides, (5) one-dimensional conduction in chip. Analysis: All of the electrical power dissipated at the back surface of the chip is transferred by conduction through the chip. Hence, Fourier’s law, P = q = kA ΔT = ΔT t t.P 0.001m * 4 W = 2 kW 150 W / m.K (0.005m 2 ) ΔT = 1.10 C Comments: for fixed P, the temperature drop across the chip decreases with increasing k and W, as well as with decreasing t. Problem 3: Air at 3000C flows over a plate of dimensions 0.50 m, by 0.25 m. if the convection heat transfer coefficient is 250 W/m2.K; determine the heat transfer rate from the air to one side of the plate when the plate is maintained at 400C. Known: air flow over a plate with prescribed air and surface temperature and convection heat transfer coefficient. Find: heat transfer rate from the air to the plate Schematic: Assumptions: (1) temperature is uniform over plate area, (2) heat transfer coefficient is uniform over plate area Analysis: the heat transfer coefficient rate by convection from the airstreams to the plate can be determined from Newton’s law of cooling written in the form, q = q '' .A = hA(T∞ − Ts ) where A is the area of the plate. Substituting numerical values, q = 250W / m 2 .K * (0.25 * 0.50)m 2 (300 − 40) 0 C q = 8125W Comments: recognize that Newtown’s law of cooling implies a direction for the convection heat transfer rate. Written in the form above, the heat rate is from the air to plate. Problem 4 : A water cooled spherical object of diameter 10 mm and emissivity 0.9 is maintained at 4000C. What is the net transfer rate from the oven walls to the object? Known: spherical object maintained at a prescribed temperature within a oven. Find: heat transfer rate from the oven walls to the object Schematic: Assumptions: (1) oven walls completely surround spherical object, (2) steady-state condition, (3) uniform temperature for areas of sphere and oven walls, (4) oven enclosure is evacuated and large compared to sphere. Analysis: heat transfer rate will be only due to the radiation mode. The rate equation is q rad = εA s σ(Tsur 4 − Ts 4 ) Where As=πD2, the area of the sphere, substituting numerical values, q rad = 0.9 * π(10 * 10 −3 ) 2 m 2 * 5.67 *10 −8 W / m 2 .K[(400 + 273) 4 − (80 + 273) 4 ]K 4 q rad = 3.04W Comments: (1) this rate equation is useful for calculating the net heat exchange between a small object and larger surface completely surrounds the smaller one . this is an essential, restrictive condition. (2) Recognize that the direction of the net heat exchange depends upon the manner in which T sur and Ts are written. (3) When performing radiant heat transfer calculations, it is always necessary to have temperatures in Kelvin (K) units. Problem 5: A surface of area 0.5m2, emissivity 0.8 and temperature 1500C is placed in a large, evacuated chamber whose walls are maintained at 25 0C. What is the rate at which radiation is emitted by the surface? What is the net rate at which radiation is exchanged between the surface and the chamber walls? Known: Area, emissivity and temperature of a surface placed in a large, evacuated chamber of prescribed temperature. Find: (a) rate of surface radiation emission, (b) net rate of radiation exchange between the surface and chamber walls. Schematic: Assumptions: (1) area of the enclosed surface is much less than that of chamber walls. Analysis (a) the rate at which radiation is emitted by the surface is emitted q emit = q emit. A = εAσTs 4 qemit = 0.8(0.5m 2 )5.67 * 10 −8 W / m 2 .K 4 [(150 + 273)K ] 4 qemit = 726 W (b) The net rate at which radiation is transferred from the surface to the chamber walls is q = εAσ(Ts 4 − Tsurr 4) q = 0.8(0.5m 2 )5.67 *10 −8 W / m 2 .K 4 [(423K ) 4 − (298K ) 4 q = 547 W Comments: the foregoing result gives the net heat loss from the surface which occurs at the instant the surface is placed in the chamber. The surface would, of course, cool due to this heat loss and its temperature, as well as the heat loss, would decrease with increasing time. Steady-state conditions would eventually be achieved when the temperature of the surface reached that of the surroundings. Problem 6: A solid aluminium sphere of emissivityε is initially at an elevated temperature and is cooled by placing it in chamber. The walls of the chamber are maintained at a lower temperature and a cold gas is circulated through the chamber. Obtain an equation that could be used to predict the variation of the aluminium temperature with time during the cooling process. Do not attempt to solve. Known: Initial temperature, diameter and surface emissivity of a solid aluminium sphere placed in a chamber whose walls are maintained at lower temperature. Temperature and convection coefficient associated with gas flow over the sphere. Find: equation which could be used to determine the aluminium temperature as a function of time during the cooling process. Schematic: Assumptions: (1) at any time t, the temperature T of the sphere is uniform, (2) constant properties; (3) chamber walls are large relative to sphere. Analysis: applying an energy balance at an instant of time to a control volume about the sphere, it follows that E st = − E out Identifying the heat rates out of the CV due to convection and radiation, the energy balance has the form d (ρVcT) = −(q conv + q rad ) dt dT A 4 =− [h (T − T∞ ) + εσ(T 4 − Tsurr )] dt ρVc dT 6 4 = [h (T − T∞ ) + εσ(T 4 − Tsurr )] dt ρcD Where A=πD2, V=πD3/6 and A/V=6/D for the sphere. . . Comments: (1) knowing T=Ti at t =0, the foregoing equation could be solved by numerical integration to obtain T (t). (2) The validity of assuming a uniform sphere temperature depends upon h, D. and the thermal conductivity of the solid (k). The validity of the assumption improves with increasing k and decreasing h and D. Problem 7: In an orbiting space station, an electronic package is housed in a compartment having surface area As = 1m2 which is exposed to space. Under normal operating conditions, the electronics dissipate 1 kW, all of which must be transferred from the exposed surface to space. If the surface emissivity is 1.0 and the surface is not exposed to the sun, what is its steady- state temperature? If the surface is exposed to a solar flux of 750W/m2 and its absorptivity to solar radiation is 0.25, what is its steady –state temperature? Known: surface area of electronic package and power dissipation by the electronics. Surface emissivity and absorptivity to solar radiation. Solar flux. Find: surface temperature without and with incident solar radiation. Schematic: Assumptions: steady state condition Analysis: applying conservation of energy to a control surface about the compartment, at any instant E in − E out + E g = 0 It follows that, with the solar input, α S A s q '' S − A s q '' emit + P = 0 α S A s q '' S − A s εσTs4 + P = 0 ⎛ α A q '' S + P ⎞ 4 ⎟ Ts = ⎜ S s ⎜ ⎟ A s εσ ⎝ ⎠ '' In the shade ( q = 0 ) 1000W ⎞4 ⎛ Ts = ⎜ 2 ⎟ = 364K 2 4 −8 ⎝ 1m *1 * 5.67 *10 W / m .K ⎠ In the sun, ⎛ 0.25 *1m 2 * 750W / m 2 + 1000W ⎞ 4 ⎟ Ts = ⎜ ⎜ 1m 2 *1 * 5.67 *10 −8 W / m 2 .K 4 ⎟ = 380K ⎝ ⎠ 1 1 1 . . . Comments: in orbit, the space station would be continuously cycling between shade, and a steady- state condition would not exist. Problem 8: The back side of a metallic plate is perfectly insulated while the front side absorbs a solar radiant flux of 800 W/m2. The convection coefficient between the plate and the ambient air is 112 w /m2. K. (a) Neglecting radiation exchange with the surroundings, calculate the temperature of the plate under steady-state conditions if the ambient air temperature is 200C. (b) For the same ambient air temperature, calculate the temperature of the plate if its surface emissivity is 0.8 and the temperature of the surroundings is also200C. Known: front surface of insulated plate absorbs solar flux, qs” and experiences for case (a) Convection process with air at T and for case (b): the same convection process and radiation exchange with surroundings at Tsur Find: temperature of the plate, Ts, for the two cases. Schematic: Assumptions: (1) steady state conditions, (2) no heat loss out backside of plate, (3) surroundings large in comparison plate. Analysis: (a) apply a surface energy balance, identifying the control surface as shown on the schematic. For an instant of time the conversation requirement is E in − E out =0. The relevant processes are convection between the plate and the air, qconv, and absorbed solar flux, qs”. Considering the plate to have an area As solve for Ts and substitute numerical values to find q '' s .A s − hA s (Ts − T∞ ) = 0 Ts = T∞ + q " / h s Ts = 20 0 C + 800 W / m 2 = 20 0 C + 66.7 0 C = 87 0 C 2 12 W / m .K . . (b) Considering now the radiation exchange between the surface and its surroundings, the surface energy balance has the form E in − E out =0. . . q '' s .A s − qconv − qrad = 0 4 qs" A s − hA s (Ts − T∞ ) − εA s (Ts − T∞ ) = 0 4 800 W W W − 12 2 (Ts − [20 + 273]K ) − 0.8 * 5.67 *10 − 8 2 4 (Ts4 − [20 + 273]4K 4 ) = 0 2 m m .K m .K 12Ts + 4.536 *10 − 8Ts 4 = 4650.3 By trial and error method, find that Ts=338K=650C. Comments: note that by considering radiation exchange, Ts decreases as expected. Note the manner in which qconv is formulated using qconv is formulated using Newton’s law of cooling: since qconv is shown leaving the control surface, the rate equation must be h (Ts − T∞ ) and not h ( T∞ -Ts). MODULE I BASICS OF HEAT TRANSFER 1.1 Difference between heat and temperature In describing heat transfer problems, we often make the mistake of interchangeably using the terms heat and temperature. Actually, there is a distinct difference between the two. Temperature is a measure of the amount of energy possessed by the molecules of a substance. It is a relative measure of how hot or cold a substance is and can be used to predict the direction of heat transfer. The usual symbol for temperature is T. The scales for measuring temperature in SI units are the Celsius and Kelvin temperature scales. On the other hand, heat is energy in transit. The transfer of energy as heat occurs at the molecular level as a result of a temperature difference. The usual symbol for heat is Q. Common units for measuring heat are the Joule and calorie in the SI system. What is Heat Transfer? “Energy in transit due to temperature difference.” 1.2 Difference between thermodynamics and heat transfer Thermodynamics tells us: • how much heat is transferred (δQ) • how much work is done (δW) • final state of the system Heat transfer tells us: • how (with what modes) δQ is transferred • at what rate δQ is transferred • temperature distribution inside the body Heat transfer complementary Thermodynamics 1.3 Modes of Heat Transfer • Conduction: An energy transfer across a system boundary due to a temperature difference by the mechanism of inter-molecular interactions. Conduction needs matter and does not require any bulk motion of matter. x A T1 q T2 k Conduction rate equation is described by the Fourier Law: r q = − kA∇T where: q = heat flow vector, (W) k = thermal conductivity, a thermodynamic property of the material. (W/m K) A = Cross sectional area in direction of heat flow. (m2) ∇T = Gradient of temperature (K/m) = ∂T/∂x i + ∂T/∂y j + ∂T/∂z k Note: Since this is a vector equation, it is often convenient to work with one component of the vector. For example, in the x direction: qx = - k Ax dT/dx In circular coordinates it may convenient to work in the radial direction: qr = - k Ar dT/dr • Convection: An energy transfer across a system boundary due to a temperature difference by the combined mechanisms of intermolecular interactions and bulk transport. Convection needs fluid matter. moving fluid T∞ Ts>T∞ q Ts Newton’s Law of Cooling: q = h As ΔT where: q = heat flow from surface, a scalar, (W) h = heat transfer coefficient (which is not a thermodynamic property of the material, but may depend on geometry of surface, flow characteristics, thermodynamic properties of the fluid, etc. (W/m2 K) As = Surface area from which convection is occurring. (m2) ΔT = TS − T∞ = Temperature Difference between surface and coolant. (K) Free or natural convection (induced by buoyancy forces) Convection Forced convection (induced by external means) May occur with phase change (boiling, condensation) Table 1. Typical values of h (W/m2K) Free convection Forced convection Boiling/Condensation gases: 2 - 25 liquid: 50 – 100 gases: 25 - 250 liquid: 50 - 20,000 2500 -100,000 • Radiation: Radiation heat transfer involves the transfer of heat by electromagnetic radiation that arises due to the temperature of the body. Radiation does not need matter. Emissive power of a surface: where: E=σεTs4 (W/ m2) ε = emissivity, which is a surface property (ε = 1 is black body) σ = Steffan Boltzman constant = 5.67 x 10-8 W/m2 K4. Ts = Absolute temperature of the surface (K) The above equation is derived from Stefan Boltzman law, which describes a gross heat emission rather than heat transfer. The expression for the actual radiation heat transfer rate between surfaces having arbitrary orientations can be quite complex, and will be dealt with in Module 9. However, the rate of radiation heat exchange between a small surface and a large surrounding is given by the following expression: Ts u r q”r a d. Ts q”co n v. Area = A q = ε·σ·A·(Ts4 – Tsur4) where: ε = Surface Emissivity A= Surface Area Ts = Absolute temperature of surface. (K) Tsur = Absolute temperature of surroundings.(K) 1.4 Thermal Conductivity, k As noted previously, thermal conductivity is a thermodynamic property of a material. From the State Postulate given in thermodynamics, it may be recalled that thermodynamic properties of pure substances are functions of two independent thermodynamic intensive properties, say temperature and pressure. Thermal conductivity of real gases is largely independent of pressure and may be considered a function of temperature alone. For solids and liquids, properties are largely independent of pressure and depend on temperature alone. k = k (T) Table 2 gives the values of thermal conductivity for a variety of materials. Table 2. Thermal Conductivities of Selected Materials at Room Temperature. Material Copper Silver Gold Aluminum Steel Limestone Bakelite Water Air Thermal Conductivity, W/m K 401 429 317 237 60.5 2.15 1.4 0.613 0.0263 It is important that the student gain a basic perspective of the magnitude of thermal conductivity for various materials. The background for this comes from the introductory Chemistry courses. Molecules of various materials gain energy through various mechanisms. Gases exhibit energy through the kinetic energy of the molecule. Energy is gained or lost through collusions of gaseous molecules as they travel through the medium. Kinetic energy transfer between gaseous molecules. Lattice vibration may be transferred between molecules as nuclei attract/repel each other. Solids, being are much more stationary, cannot effectively transfer energy through these same mechanisms. Instead, solids may exhibit energy through vibration or rotation of the nucleus. Another important mechanism in which materials maintain energy is by shifting electrons into higher orbital rings. In the case of electrical conductors the electrons are weakly bonded to the molecule and can drift from one molecule to another transporting their energy with them. This is an especially effective transport mechanism, so that materials which are excellent electrical conductors are excellent thermal conductors. Module 2: Learning objectives • The primary purpose of this chapter is to improve your understanding of the conduction rate equation (Fourier’s law) and to familiarize you with heat equation. You should know the origin and implication of Fourier’s law, and you should understand the key thermal properties and how they vary for different substances. You should also know the physical meaning of each term appearing in the heat equation. The student should understand to what form does the heat equation reduce for simplified conditions, and what kinds of boundary conditions may be used for its solution? The student should learn to evaluate the heat flow through a 1-D, SS system with no heat sources for rectangular and cylindrical geometries. Many other geometries exist in nature or in common engineering designs. The student, using a similar development, should be able to develop an appropriate equation to describe systems of arbitrary, simple geometry. The student should be comfortable with the use of equivalent thermal circuits and with the expressions for the conduction resistances that pertain to each of the three common geometric. Composite thermal resistances for 1-D, Steady state heat transfer with no heat sources placed in parallel or in series may be evaluated in a manner similar to electrical resistances placed in parallel or in series. The student should learn to evaluate the heat flow through a 1-D, SS system with no heat sources for rectangular and cylindrical geometries. In short, by the end of the module, the student should have a fundamental understanding of the conduction process and its mathematical description. • • • • • • Objectives of conduction analysis To determine the temperature field, T(x,y,z,t), in a body (i.e. how temperature varies with position within the body) T(x,y,z,t) depends on: T(x,y,z) - boundary conditions - initial condition - material properties (k, cp, ρ …) - geometry of the body (shape, size) Why we need T(x,y,z,t) ? - to compute heat flux at any location (using Fourier’s eqn.) - compute thermal stresses, expansion, deflection due to temp. etc. - design insulation thickness - chip temperature calculation - heat treatment of metals Unidirectional heat conduction (1D) Area = A 0 x x x+Δx Solid bar, insulated on all long sides (1D heat conduction) qx A qx+Δx & q = Internal heat generation per unit vol. (W/m3) Unidirectional heat conduction (1D) First Law (energy balance) & & & & ( E in − E out ) + E gen = E st q x − q x + Δx ∂E + A(Δx)q = & ∂t ∂T ∂x x E = ( ρ AΔx)u ∂E ∂t = ρ AΔx ∂u ∂t = ρAΔxc ∂T ∂t q q x = − kA = q x + Δ x + ∂q ∂x x Δ x Unidirectional heat conduction (1D)(contd…) ∂T ∂T ∂ ⎛ ∂T − kA + kA + A ⎜k ∂x ∂x ∂x ⎝ ∂x ∂ ⎛ ∂T ⎞ ∂T ⎜k ⎟+q = ρ c & ∂x ⎝ ∂x ⎠ ∂t Longitudinal conduction Internal heat generation ∂T ⎞ ⎟ Δx + AΔxq = ρ AcΔx & ∂t ⎠ Thermal inertia If k is a constant ∂ 2T q ρ c ∂T 1 ∂T = + &= 2 ∂x k k ∂t α ∂t Unidirectional heat conduction (1D)(contd…) For T to rise, LHS must be positive (heat input is positive) For a fixed heat input, T rises faster for higher α In this special case, heat flow is 1D. If sides were not insulated, heat flow could be 2D, 3D. Boundary and Initial conditions: The objective of deriving the heat diffusion equation is to determine the temperature distribution within the conducting body. We have set up a differential equation, with T as the dependent variable. The solution will give us T(x,y,z). Solution depends on boundary conditions (BC) and initial conditions (IC). Boundary and Initial conditions (contd…) How many BC’s and IC’s ? - Heat equation is second order in spatial coordinate. Hence, 2 BC’s needed for each coordinate. * 1D problem: 2 BC in x-direction * 2D problem: 2 BC in x-direction, 2 in y-direction * 3D problem: 2 in x-dir., 2 in y-dir., and 2 in z-dir. - Heat equation is first order in time. Hence one IC needed 1- Dimensional Heat Conduction The Plane Wall : Hot fluid …. .. .. . … .. .. .. . . .. .. k . Ts,1 ........... ....... . . . .... .. .. .. .... ...... .. .. .. .. .. . . . . ..... . . x=0 d dx dT ⎛ ⎜ k dx ⎝ Ts,2 x=L ⎞ ⎟ = 0 ⎠ Cold fluid T∞,2 Const. K; solution is: q x = − kA dT kA (T s ,1 − T s , 2 ) = T s ,1 − T s , 2 = dx L L / kA Thermal resistance (electrical analogy) OHM’s LAW :Flow of Electricity V=IR elect Voltage Drop = Current flow×Resistance Thermal Analogy to Ohm’s Law : Δ T = qR therm Temp Drop=Heat Flow×Resistance 1 D Heat Conduction through a Plane Wall T∞,1 …. ..... … .. .. .. . . .. .. k . Ts,1 ........... ....... . . . .... .. .. .. .... ...... .. .. .. .. .. . . . . ..... . . x=0 Ts,1 Ts,2 Ts,2 x=L Cold fluid T∞,2 T∞,2 Hot fluid qx T∞,1 1 h1 A L k A 1 h1 A + L kA + 1 1 h2 A h2 A ∑ R t = (Thermal Resistance ) Resistance expressions THERMAL RESISTANCES • Conduction • Convection Rconv = (hA) • Fins Rcond = Δx/kA -1 Rfin = (hηΑ)−1 • Radiation(aprox) 1.5 -1 Rrad = [4AσF(T1T2) ] Composite Walls : T∞,1 h1 A B C KA KB KC h2 T∞,2 T∞,1 qx LA LB LB kB A LC LC kC A T∞,2 1 h1 A T∞ LA kAA 1 h2 A q x = ∑ ,1 − T∞ R t ,2 = 1 T∞ 1 h1 A + LA kA − T ∞ ,2 = UA Δ T LC LB 1 + + + kB kC h2 A ,1 where, U = Rtot A = Overall heat transfer coefficient Overall Heat transfer Coefficient U = 1 R total A = 1 h 1 1 L + Σ k + 1 h 2 Contact Resistance : TA TB ΔT A R t , c B = Δ T q x U = L 1 + h1 k A A 1 LC LB 1 + + + k B kC h2 Series-Parallel : A T1 KA B KB C Kc D KD T2 AB+AC=AA=AD LB=LC Series-Parallel (contd…) LA kA A LB kB A LC kC A LD kD A T1 T2 Assumptions : (1) Face between B and C is insulated. (2) Uniform temperature at any face normal to X. Example: Consider a composite plane wall as shown: kI = 20 W/mk qx T1 = 0°C AI = 1 m2, L = 1m Tf = 100°C kII = 10 W/mk AII = 1 m2, L = 1m h = 1000 W/ m2 k Develop an approximate solution for th rate of heat transfer through the wall. 1 D Conduction(Radial conduction in a composite cylinder) h1 r1 h2 r2 k1 T∞,1 T∞,2 r k 3 2 qr = T∞,1 1 ( h 1 )( 2 π r1 L ) T ∞ , 2 − T ∞ ,1 ∑R t T∞,2 1 ( h 2 )( 2π r2 L ) r1 r2 1 ln 2 π Lk ln r2 r3 2 2 π Lk Critical Insulation Thickness : T∞ h Insulation Thickness : r o-r i ri r0 Ti R tot 1 = + 2 π kL ( 2 π r0 L ) h ln( r0 ri ) Objective : decrease q , increases R tot Vary r0 ; as r0 increases ,first term increases, second term decreases. Critical Insulation Thickness (contd…) Maximum – Minimum problem Set dR tot = 0 dr 0 1 1 − 2 π kr 0 L 2 π hLr 2 = 0 0 r0 = Max or Min. ? k h Take : d 2 R tot = 0 dr 2 0 at r0 = k h d 2 R tot −1 1 = + 2 π kr 2 0 L dr 2 0 π r 2 0 hL h2 = 2 π Lk 0 r0 = k h 3 Critical Insulation Thickness (contd…) Minimum q at r0 =(k/h)=r c r (critical radius) R tot good for electrical cables good for steam pipes etc. R c r=k/h r0 1D Conduction in Sphere 1D Conduction in Sphere r2 r1 T∞,2 k Inside Solid: Ts,2 Ts,1 T∞,1 1 d ⎛ ⎜ kr 2 r dr ⎝ 2 dT ⎞ ⎟ = 0 dr ⎠ → T ( r ) = T s ,1 → q r = − kA → R t , cond {T s ,1 − T s , 2 } dT 4 π k (T s ,1 − T s , 2 ) = (1 / r1 − 1 / r 2 ) dr − ⎡ 1 − (r / r ) ⎤ 1 ⎢ ⎥ ⎢ 1 − ( r1 / r 2 ) ⎥ ⎣ ⎦ 1 / r1 − 1 / r 2 = 4π k Conduction with Thermal Energy Generation & E q= & = Energy generation per unit volume V Applications: * current carrying conductors * chemically reacting systems * nuclear reactors Conduction with Thermal Energy Generation The Plane Wall : k Ts,1 T∞,1 Hot fluid x= -L x=0 x=+L q & Ts,2 T∞,2 Cold fluid Assumptions: 1D, steady state, constant k, & uniform q Conduction With Thermal Energy Generation (contd…) d T dx 2 2 + q& k =0 cond . : x = −L, x = +L, T = Ts , 1 T = Ts , 2 Boundary Solution : T = − q& 2k x 2 +C x +C 1 2 Conduction with Thermal Energy Generation (cont..) Use boundary conditions to find C1 and C2 qL2 ⎛ − x2 ⎞ + Ts , 2 −Ts ,1 x + Ts , 2 +Ts ,1 & ⎜ Final solution : T = ⎜1 L2 ⎟ ⎟ 2k ⎝ 2 L 2 ⎠ No more linear Heat flux : qx′′ = −k dT dx Derive the expression and show that it is no more independent of x Hence thermal resistance concept is not correct to use when there is internal heat generation Cylinder with heat source T∞ h ro r Ts Assumptions: 1D, steady state, constant & k, uniform q Start with 1D heat equation in cylindrical co-ordinates: q & 1 d r dr ⎛ dT ⎞ q & ⎜r ⎟ + =0 ⎝ dr ⎠ k Cylinder With Heat Source Boundary cond. : r = r0 , r = 0, T = Ts dT =0 dr q 2 ⎛ r2 ⎞ & Solution : T (r ) = r0 ⎜1 − 2 ⎟ +Ts 4k ⎜ r0 ⎟ ⎝ ⎠ Ts may not be known. Instead, T∝ and h may be specified. Exercise: Eliminate Ts, using T∝ and h. Cylinder with heat source (contd…) Example: A current of 200A is passed through a stainless steel wire having a thermal conductivity K=19W/mK, diameter 3mm, and electrical resistivity R = 0.99 Ω. The length of the wire is 1m. The wire is submerged in a liquid at 110°C, and the heat transfer coefficient is 4W/m2K. Calculate the centre temperature of the wire at steady state condition. Solution: to be worked out in class Module 2: Short questions 1. How does transient heat transfer differ from steady state heat transfer? 2. What is meant by the term “one-dimensional” when applied to conduction heat transfer? 3. What is meant by thermal resistance? Under what assumptions can the concept of thermal resistance be applied in a straightforward manner? 4. For heat transfer through a single cylindrical shell with convection on the outside, there is a value for the shell radius for a nonzero shell thickness at which the heat flux is maximized. This value is (A) k/h (B) h/k (C) h/r (D) r/h 5. The steady temperature profile in a one-dimensional heat transfer across a plane & slab of thickness L and with uniform heat generation, q , has one maximum. If the slab is cooled by convection at x = 0 and insulated at x = L, the maximum occurs at a value of x given by & q x x=0 (A) 0 L (B) 2 & q (C) k (D) L x=L 6. Consider a cold canned drink left on a dinner table. Would you model the heat transfer to the drink as one-, two-, or three-dimensional? Would the heat transfer be steady or transient? Also, which coordinate system would you use to analyse this heat transfer problem, and where would you place the origin? 7. Consider a round potato being baked in an oven. Would you model the heat transfer to the potato as one-, two-, or three-dimensional? Would the heat transfer be steady or transient? Also, which coordinate system would you use to analyse this heat transfer problem, and where would you place the origin? 8. Consider an egg being cooked in boiling water in a pan? Would you model the heat transfer to the egg as one-, two-, or three-dimensional? Would the heat transfer be steady or transient? Also, which coordinate system would you use to analyse this heat transfer problem, and where would you place the origin? MODULE 2 ONE DIMENSIONAL STEADY STATE HEAT CONDUCTION 2.1 Objectives of conduction analysis: The primary objective is to determine the temperature field, T(x,y,z,t), in a body (i.e. how temperature varies with position within the body) T(x,y,z,t) depends on: - Boundary conditions - Initial condition - Material properties (k, cp, ρ) - Geometry of the body (shape, size) Why we need T (x, y, z, t)? - To compute heat flux at any location (using Fourier’s eqn.) - Compute thermal stresses, expansion, deflection due to temp. Etc. - Design insulation thickness - Chip temperature calculation - Heat treatment of metals 2.2 General Conduction Equation Recognize that heat transfer involves an energy transfer across a system boundary. A logical place to begin studying such process is from Conservation of Energy (1st Law of Thermodynamics) for a closed system: dE dt dE dt & & = Qin − Wout system The sign convention on work is such that negative work out is positive work in. & & = Qin + Win system The work in term could describe an electric current flow across the system boundary and through a resistance inside the system. Alternatively it could describe a shaft turning across the system boundary and overcoming friction within the system. The net effect in either case would cause the internal energy of the system to rise. In heat transfer we generalize all such terms as “heat sources”. dE dt & & = Qin + Qgen system The energy of the system will in general include internal energy, U, potential energy, ½ mgz, or kinetic energy, ½ m 2. In case of heat transfer problems, the latter two terms could often be neglected. In this case, E = U = m ⋅ u = m ⋅ c p ⋅ T − Tref = ρ ⋅ V ⋅ c p ⋅ T − Tref ( ) ( ) where Tref is the reference temperature at which the energy of the system is defined as zero. When we differentiate the above expression with respect to time, the reference temperature, being constant, disappears: ρ ⋅ cp ⋅ V ⋅ dT dt & & = Qin + Qgen system Consider the differential control element shown below. Heat is assumed to flow through the element in the positive directions as shown by the 6 heat vectors. qz+Δz qx qy qy+Δy qx+Δx z y x qz In the equation above we substitute the 6 heat inflows/outflows using the appropriate sign: ρ ⋅ cp ⋅ (Δ x ⋅ Δ y ⋅ Δ z) ⋅ ρ ⋅ c p ⋅ (Δx ⋅ Δy ⋅ Δz ) ⋅ ∂T ∂t dT dt & = q x − q x + Δx + q y − q y + Δy + qz − qz + Δz + Qgen system Substitute for each of the conduction terms using the Fourier Law: system ⎧ ⎤⎫ ∂T ⎡ ∂T ∂ ⎛ ∂T ⎞ = ⎨− k ⋅ (Δy ⋅ Δz ) ⋅ − ⎢− k ⋅ (Δy ⋅ Δz ) ⋅ + ⎜ − k ⋅ (Δy ⋅ Δz ) ⋅ ⎟ ⋅ Δx ⎥ ⎬ ∂x ⎣ ∂x ∂x ⎝ ∂x ⎠ ⎦⎭ ⎩ ⎧ ⎤⎫ ∂T ⎡ ∂T ∂ ⎛ ∂T ⎞ ⎪ ⎪ + ⎨ − k ⋅ ( Δ x ⋅ Δ z) ⋅ − ⎢ − k ⋅ ( Δ x ⋅ Δ z) ⋅ + ⎜ − k ⋅ ( Δ x ⋅ Δ z) ⋅ ⎟ ⋅ Δ y ⎥ ⎬ ∂y ⎣ ∂y ∂y ⎝ ∂y ⎠ ⎪ ⎦⎪ ⎩ ⎭ ⎧ ∂T ⎡ ∂T ∂ ⎛ ∂T ⎞ ⎤⎫ + ⎨ − k ⋅ (Δ x ⋅ Δ y) ⋅ + ⎢ − k ⋅ (Δ x ⋅ Δ y) ⋅ + ⎜ − k ⋅ (Δ x ⋅ Δ y) ⋅ ⎟ ⋅ Δ z⎥ ⎬ ∂z ⎣ ∂z ∂z ⎝ ∂z ⎠ ⎦⎭ ⎩ + &&& ⋅ ( Δ x ⋅ Δ y ⋅ Δ z ) q q where &&& is defined as the internal heat generation per unit volume. The above equation reduces to: ρ ⋅ cp ⋅ (Δ x ⋅ Δ y ⋅ Δ z) ⋅ dT dt system ⎧ ⎡∂ ⎛ ⎫ ∂T ⎞ ⎤ = ⎨ − ⎢ ⎜ − k ⋅ (Δ y ⋅ Δ z) ⋅ ⎟ ⎥ ⋅ Δ x ⎬ ∂x ⎠ ⎦ ⎩ ⎣ ∂x ⎝ ⎭ ⎧ ⎡∂ ⎛ ⎤⎫ ∂T ⎞ ⎪ ⎪ + ⎨ − ⎢ ⎜ − k ⋅ ( Δ x ⋅ Δ z) ⋅ ⎟ ⋅ Δ y ⎥ ⎬ ∂y ⎠ ⎪ ⎣ ∂y ⎝ ⎦⎪ ⎭ ⎩ ⎧⎡ ∂ ⎛ ∂T ⎞ ⎤⎫ + ⎨ ⎢ ⎜ − k ⋅ ( Δ x ⋅ Δ y ) ⋅ ⎟ ⋅ Δ z ⎥ ⎬ + &&& ⋅ ( Δ x ⋅ Δ y ⋅ Δ z) q ∂z ⎠ ⎦⎭ ⎩ ⎣ ∂z ⎝ Dividing by the volume (Δx⋅Δy⋅Δz), ρ ⋅ cp ⋅ dT dt =− system ∂ ⎛ ∂T ⎞ ∂ ⎛ ∂T ⎞ ∂ ⎛ ∂T ⎞ q ⎜−k⋅ ⎟ − ⎜ − k ⋅ ⎟ − ⎜ − k ⋅ ⎟ + &&& ∂x ⎝ ∂x ⎠ ∂y ⎝ ∂y ⎠ ∂z ⎝ ∂z ⎠ which is the general conduction equation in three dimensions. In the case where k is independent of x, y and z then ρ ⋅ c p dT ⋅ k dt system ∂ 2 T ∂ 2 T ∂ 2 T &&& q = 2 + 2 + 2 + ∂x ∂y ∂z k Define the thermodynamic property, α, the thermal diffusivity: k α ≡ ρ ⋅ cp Then 1 dT ⋅ α dt or, : = system q ∂ 2 T ∂ 2 T ∂ 2 T &&& 2 + 2 + 2 + k ∂x ∂y ∂z 1 dT ⋅ α dt = ∇ 2T + system &&& q k The vector form of this equation is quite compact and is the most general form. However, we often find it convenient to expand the del-squared term in specific coordinate systems: Cartesian Coordinates 1 ∂T ∂ 2T ∂ 2T ∂ 2T q ⋅ = + + + a ∂τ ∂x 2 ∂y 2 ∂z 2 k Circular Coordinates L Z R y Θ x 1 ∂T 1 ∂ ⎛ ∂T ⎞ 1 ∂ 2T ∂ 2T q ⋅ = + + ⎜r ⋅ ⎟+ ⋅ a ∂τ r ∂r ⎝ ∂r ⎠ r 2 ∂θ 2 ∂z 2 k L Spherical Coordinates Z φ r y ϕ x L 1 ∂T 1 ∂ ⎛ 2 ∂T ⎞ 1 1 ∂T ⎞ q ∂ 2T ∂ ⎛ + ⋅ ⋅ ⋅ = ⎜ sin θ ⋅ ⎟+ ⎜r ⋅ ⎟+ 2 2 2 2 2 a ∂τ r ∂r ⎝ r sin θ ∂θ ⎝ ∂z ⎠ k ∂r ⎠ r ⋅ sin θ ∂φ In each equation the dependent variable, T, is a function of 4 independent variables, (x,y,z,τ); (r, θ ,z,τ); (r,φ,θ,τ) and is a 2nd order, partial differential equation. The solution of such equations will normally require a numerical solution. For the present, we shall simply look at the simplifications that can be made to the equations to describe specific problems. • • • Steady State: Steady state solutions imply that the system conditions are not changing with time. Thus ∂T / ∂τ = 0 . One dimensional: If heat is flowing in only one coordinate direction, then it follows that there is no temperature gradient in the other two directions. Thus the two partials associated with these directions are equal to zero. Two dimensional: If heat is flowing in only two coordinate directions, then it follows that there is no temperature gradient in the third direction. Thus the partial derivative associated with this third direction is equal to zero. No Sources: If there are no heat sources within the system then the term, q = 0 . L • Note that the equation is 2nd order in each coordinate direction so that integration will result in 2 constants of integration. To evaluate these constants two additional equations must be written for each coordinate direction based on the physical conditions of the problem. Such equations are termed “boundary conditions’. 2.3 Boundary and Initial Conditions • The objective of deriving the heat diffusion equation is to determine the temperature distribution within the conducting body. • • We have set up a differential equation, with T as the dependent variable. The solution will give us T(x,y,z). Solution depends on boundary conditions (BC) and initial conditions (IC). How many BC’s and IC’s ? - Heat equation is second order in spatial coordinate. Hence, 2 BC’s needed for each coordinate. * 1D problem: 2 BC in x-direction * 2D problem: 2 BC in x-direction, 2 in y-direction * 3D problem: 2 in x-dir., 2 in y-dir., and 2 in z-dir. - Heat equation is first order in time. Hence one IC needed. 2.4 Heat Diffusion Equation for a One Dimensional System & q T1 q Ax L x T2 y z Consider the system shown above. The top, bottom, front and back of the cube are insulated, so that heat can be conducted through the cube only in the x direction. The internal heat & generation per unit volume is q (W/m3). Consider the heat flow through an arbitrary differential element of the cube. qx qx+Δx From the 1st Law we write for the element: & & & & ( E in − E out ) + E gen = E st & q x − q x + Δx + Ax (Δx )q = q x = − kA x ∂T ∂x (2.1) (2.2) ∂E ∂t (2.3) q x + Δx = q x + − kA ∂q x Δx ∂x (2.4) ∂T ∂T ∂ ⎛ ∂T ⎞ ∂T & + kA + A ⎜ k ⎟ Δ x + A Δ x q = ρ Ac Δ x ∂x ∂x ∂x ⎝ ∂x ⎠ ∂t (2.5) ∂ ⎛ ∂T ⎞ ∂T & ⎜ k ⎟ + q = ρ cΔ x ∂x ⎝ ∂x ⎠ ∂t Longitudinal conduction Internal heat generation Thermal inertia (2.6) If k is a constant, then & ∂ 2T q ρ c ∂ T 1 ∂T + = = 2 ∂x k k ∂t α ∂t (2.7) • • • For T to rise, LHS must be positive (heat input is positive) For a fixed heat input, T rises faster for higher α In this special case, heat flow is 1D. If sides were not insulated, heat flow could be 2D, 3D. 2.5 One Dimensional Steady State Heat Conduction The plane wall: The differential equation governing heat diffusion is: d ⎛ dT ⎞ ⎜k ⎟=0 dx ⎝ dx ⎠ With constant k, the above equation may be integrated twice to obtain the general solution: T ( x ) = C1 x + C 2 where C1 and C2 are constants of integration. To obtain the constants of integration, we apply the boundary conditions at x = 0 and x = L, in which case T (0) = Ts ,1 T ( L) = Ts , 2 and Once the constants of integration are substituted into the general equation, the temperature distribution is obtained: x T ( x ) = (Ts , 2 − Ts ,1 ) + Ts ,1 L The heat flow rate across the wall is given by: dT kA (Ts ,1 − Ts ,2 ) = Ts ,1 − Ts ,2 = qx = − kA dx L L / kA Thermal resistance (electrical analogy): Physical systems are said to be analogous if that obey the same mathematical equation. The above relations can be put into the form of Ohm’s law: V=IRelec Using this terminology it is common to speak of a thermal resistance: ΔT = qRtherm A thermal resistance may also be associated with heat transfer by convection at a surface. From Newton’s law of cooling, q = hA(Ts − T∞ ) the thermal resistance for convection is then T − T∞ 1 Rt ,conv = s = q hA Applying thermal resistance concept to the plane wall, the equivalent thermal circuit for the plane wall with convection boundary conditions is shown in the figure below The heat transfer rate may be determined from separate consideration of each element in the network. Since qx is constant throughout the network, it follows that T − Ts ,1 Ts ,1 − Ts , 2 Ts , 2 − T∞ , 2 = = q x = ∞ ,1 1 / h1 A L / kA 1 / h2 A In terms of the overall temperature difference T∞ ,1 − T∞ , 2 , and the total thermal resistance Rtot, the heat transfer rate may also be expressed as T − T∞ , 2 q x = ∞ ,1 Rtot Since the resistance are in series, it follows that 1 L 1 + + Rtot = ∑ Rt = h1 A kA h2 A Composite walls: Thermal Resistances in Series: Consider three blocks, A, B and C, as shown. They are insulated on top, bottom, front and back. Since the energy will flow first through block A and then through blocks B and C, we say that these blocks are thermally in a series arrangement. The steady state heat flow rate through the walls is given by: qx = T∞ ,1 − T∞ , 2 ∑ Rt = T∞ ,1 − T∞ , 2 = UAΔT L A LB LC 1 1 + + + + h1 A k A k B k C h2 A where U = 1 Rtot A is the overall heat transfer coefficient. In the above case, U is expressed as U= 1 1 L A LB LC 1 + + + + h1 k A k B k C h2 Series-parallel arrangement: The following assumptions are made with regard to the above thermal resistance model: 1) Face between B and C is insulated. 2) Uniform temperature at any face normal to X. 1-D radial conduction through a cylinder: One frequently encountered problem is that of heat flow through the walls of a pipe or through the insulation placed around a pipe. Consider the cylinder shown. The pipe is either insulated on the ends or is of sufficient length, L, that heat losses through the ends is negligible. Assume no heat sources within the wall of the tube. If T1>T2, heat will flow outward, radially, from the inside radius, R1, to the outside radius, R2. The process will be described by the Fourier Law. T1 R1 R2 T2 L The differential equation governing heat diffusion is: 1 d ⎛ dT ⎞ ⎜r ⎟=0 r dr ⎝ dr ⎠ With constant k, the solution is The heat flow rate across the wall is given by: dT kA (Ts,1 − Ts,2 ) = Ts ,1 − Ts ,2 = qx = − kA dx L L / kA Hence, the thermal resistance in this case can be expressed as: ln rr1 2 2πkL Composite cylindrical walls: Critical Insulation Thickness : Rtot = ln( rr0i ) 2πkL + 1 ( 2πr0 L)h Insulation thickness : ro-ri Objective : decrease q , increase Rtot Vary ro ; as ro increases, first term increases, second term decreases. This is a maximum – minimum problem. The point of extrema can be found by setting dRtot =0 dr0 or, or, 1 1 − =0 2πkr0 L 2πhLro2 r0 = k h In order to determine if it is a maxima or a minima, we make the second derivative zero: d 2 Rtot = 0 at dro2 r0 = k h = h2 0 2πLk 3 d 2 Rtot 1 −1 = + 2 2 2 dro 2πkro L πro hL r = k 0 h Minimum q at ro =(k/h) = rcr (critical radius) 1-D radial conduction in a sphere: 1 d ⎛ 2 dT ⎞ ⎜ kr ⎟=0 dr ⎠ r 2 dr ⎝ 1− → T ( r ) = Ts ,1 − {Ts ,1 − Ts , 2}⎢ 1−((rr1 //rr )) ⎥ ⎢ ⎥ ⎡ ⎣ ⎤ ⎦ 1 2 → qr = − kA → Rt ,cond = dT 4π k (Ts ,1 − Ts , 2 ) = (1 / r1 − 1 / r2 ) dr 1 / r1 − 1 / r2 4π k 2.6 Summary of Electrical Analogy System Electrical Cartesian Conduction Cylindrical Conduction Current I Resistance R L kA ln r2 r1 2πkL 1 / r1 − 1 / r2 4π k 1 h ⋅ As Potential Difference ΔV q q q q ΔT ΔT ΔT ΔT Conduction through sphere Convection 2.7 One-Dimensional Steady State Conduction with Internal Heat Generation Applications: current carrying conductor, chemically reacting systems, nuclear reactors. & E & Energy generated per unit volume is given by q = V & Plane wall with heat source: Assumptions: 1D, steady state, constant k, uniform q & d 2T q + =0 2 dx k Boundary cond. : x = − L, x = + L, Solution : T = − T = Ts ,1 T = Ts ,2 & q 2 x + C1 x + C2 2k Use boundary conditions to find C1 and C2 & qL2 Final solution : T = 2k ⎛ x 2 ⎞ Ts , 2 − Ts ,1 x Ts , 2 + Ts ,1 ⎜1 − 2 ⎟ + + ⎜ 2 2 L ⎟ L ⎝ ⎠ dT dx Heat flux : q′′ = −k x Note: From the above expressions, it may be observed that the solution for temperature is no longer linear. As an exercise, show that the expression for heat flux is no longer independent of x. Hence thermal resistance concept is not correct to use when there is internal heat generation. & Cylinder with heat source: Assumptions: 1D, steady state, constant k, uniform q Start with 1D heat equation in cylindrical co-ordinates & 1 d ⎛ dT ⎞ q ⎜r ⎟+ =0 r dr ⎝ dr ⎠ k Boundary cond. : r = r0 , T = Ts dT =0 dr & q 2⎛ r2 ⎞ Solution : T ( r ) = r0 ⎜1 − 2 ⎟ + Ts 4k ⎜ r0 ⎟ ⎠ ⎝ r = 0, Exercise: Ts may not be known. Instead, T∞ and h may be specified. Eliminate Ts, using T∞ and h. Module 3, Learning Objectives: • Students should recognize the fin equation. • Students should know the 2 general solutions to the fin equation. • Students should be able to write boundary conditions for (a) very long fins, (b) insulated tip fins, (c) convective tip fins and (d) fins with a specified tip temperature. • Students should be able to apply the boundary conditions to the fin equation and obtain a temperature profile. • Students should be able to apply the temperature profile to the Fourier Law to obtain a heat flow through the fin. • Students should be able to apply the concept of fin efficiency to define an equivalent thermal resistance for a fin. • Students should be able to incorporate fins into an overall electrical network to solve 1-D, SS problems with no sources. EXTENDED SURFACES / FINS Convection: Heat transfer between a solid surface and a moving fluid is governed by the Newton’s cooling law: q = hA(Ts-T∞). Therefore, to increase the convective heat transfer, one can Increase the temperature difference (Ts-T∞) between the surface and the fluid. Increase the convection coefficient h. This can be accomplished by increasing the fluid flow over the surface since h is a function of the flow velocity and the higher the velocity, the higher the h. Example: a cooling fan. Increase the contact surface area A. Example: a heat sink with fins. Extended Surface Analysis Tb x dT q x = − kAC dx AC is the cross-sectional area P: the fin perimeter Ac: the fin cross-sectional area q x + dx = q x + dq x dx dx dq conv = h( dAS )(T − T∞ ), where dA S is the surface area of the element dq Energy Balance: q x = q x + dx + dq conv = q x + x dx + hdAS (T − T∞ ) dx d 2T − kAC dx + hP(T − T∞ )dx = 0, if k, A C are all constants. 2 dx Extended Surface Analysis (contd….) d 2 T hP − (T − T∞ ) = 0, A second - order, ordinary differential equation 2 dx kAC Define a new variable θ ( x ) = T ( x ) − T∞ , so that d 2θ hP m 2θ = 0, where m 2 = − , ( D 2 − m 2 )θ = 0 dx 2 kAC Characteristics equation with two real roots: + m & - m The general solution is of the form θ ( x ) = C1e mx + C2 e − mx To evaluate the two constants C 1 and C 2 , we need to specify two boundary conditions: The first one is obvious: the base temperature is known as T(0) = Tb The second condition will depend on the end condition of the tip Extended Surface Analysis (contd...) For example: assume the tip is insulated and no heat transfer dθ/dx(x=L)=0 The temperature distribution is given by T ( x ) - T∞ θ cosh m( L − x ) = = θb Tb − T∞ cosh mL The fin heat transfer rate is q f = − kAC dT ( x = 0) = hPkAC tanh mL = M tanh mL dx These results and other solutions using different end conditions are tabulated in Table 3.4 in HT textbook, p. 118. Temperature distribution for fins of different configurations Case Tip Condition A Convection heat transfer: hθ(L)=-k(dθ/dx)x=L B C Adiabatic (dθ/dx)x=L=0 Given temperature: θ(L)= θL Temp. Distribution cosh m( L − x ) + ( h cosh mL + ( h Fin heat transfer mk mk ) sinh m( L − x ) ) sinh mL Mθo sinh mL + ( h ) cosh mL mk cosh mL + ( h ) sinh mL mk cosh m( L − x ) cosh mL (θ L Mθ 0 tanh mL (cosh mL − θ L sinh mL θb ) sinh m( L − x ) + sinh m( L − x ) sinh mL Mθ 0 θb ) D Infinitely long fin θ(L)=0 e − mx M θ0 θ ≡ T − T∞ , m2 ≡ hP kAC M = hPkAC θ b θ b = θ (0) = Tb − T∞ , Example An Aluminum pot is used to boil water as shown below. The handle of the pot is 20-cm long, 3-cm wide, and 0.5-cm thick. The pot is exposed to room air at 25°C, and the convection coefficient is 5 W/m2 °C. Question: can you touch the handle when the water is boiling? (k for aluminum is 237 W/m °C) T∞ = 25 °C h = 5 W/ m2 °C x 100 °C Example (contd...) We can model the pot handle as an extended surface. Assume that there is no heat transfer at the free end of the handle. The condition matches that specified in the fins Table, case B. h=5 W/ m2 °C, P=2W+2t=2(0.03+0.005)=0.07(m), k=237 W/m °C, AC=Wt=0.00015(m2), L=0.2(m) Therefore, m=(hP/kAC)1/2=3.138, M=√(hPkAC)(Tb-T∞)=0.111θb=0.111(100-25)=8.325(W) T ( x ) - T∞ θ cosh m( L − x ) = = Tb − T∞ θb cosh mL T − 25 cosh[3138(0.2 − x )] . = , 100 − 25 cosh(3138 * 0.2) . T ( x ) = 25 + 62.32 * cosh[3138(0.2 − x )] . Example (contd…) Plot the temperature distribution along the pot handle 100 95 T( x ) 90 85 0 0.05 0.1 x 0.15 0.2 As shown, temperature drops off very quickly. At the midpoint T(0.1)=90.4°C. At the end T(0.2)=87.3°C. Therefore, it should not be safe to touch the end of the handle Example (contd...) The total heat transfer through the handle can be calculated also. qf=Mtanh(mL)=8.325*tanh(3.138*0.2)=4.632(W) Very small amount: latent heat of evaporation for water: 2257 kJ/kg. Therefore, the amount of heat loss is just enough to vaporize 0.007 kg of water in one hour. If a stainless steel handle is used instead, what will happen: For a stainless steel, the thermal conductivity k=15 W/m°C. Use the same parameter as before: ⎛ hP ⎞ m=⎜ ⎟ ⎜ kA ⎟ ⎝ C⎠ 1/ 2 = 12.47, M = hPkAC = 0.0281 Example (contd...) T ( x ) − T∞ cosh m( L − x ) = Tb − T∞ cosh mL T ( x ) = 25 + 12.3 cosh[12.47( L − x )] 100 75 T( x) 50 25 0 0 0.05 0.1 x 0.15 0.2 Temperature at the handle (x=0.2 m) is only 37.3 °C, not hot at all. This example illustrates the important role played by the thermal conductivity of the material in terms of conductive heat transfer. Fins-2 If the pot from previous lecture is made of other materials other than the aluminum, what will be the temperature distribution? Try stainless steel (k=15 W/m.K) and copper (385 W/m.K). Recall: h=5W/m2°C, P=2W+2t=2(0.03+0.005)=0.07(m) AC=Wt=0.00015(m2), L=0.2(m) Therefore, mss=(hP/kAC)1/2=12.47, mcu=2.46 Mss=√(hPkssAC) (Tb-T∞)=0.028(100-25)=2.1(W) Mcu= √(hPkssAC) θb=0.142(100-25)=10.66(W) Tss ( x ) - T∞ cosh m( L − x ) θ For stainless steel, = = Tb − T∞ θb cosh mL Tss − 25 cosh[12.47(0.2 − x )] = , 100 − 25 cosh(12.47 * 0.2) Tss ( x ) = 25 + 12.3 * cosh[12.47(0.2 − x )] Fins-2 (contd....) Tcu ( x ) - T∞ θ cosh m( L − x ) For copper, = = Tb − T∞ θb cosh mL Tcu − 25 cosh[2.46(0.2 − x )] , = cosh(2.46 * 0.2) 100 − 25 Tcu ( x ) = 25 + 66.76 * cosh[2.46(0.2 − x )] 100 95 90 85 T cu( x ) 80 75 0 T( x ) T ss( x ) copper aluminum stainless steel 0.04 0.08 x 0.12 0.16 0.2 Fins-2 (contd...) Inside the handle of the stainless steel pot, temperature drops quickly. Temperature at the end of the handle is 37.3°C. This is because the stainless steel has low thermal conductivity and heat can not penetrate easily into the handle. Copper has the highest k and, correspondingly, the temperature inside the copper handle distributes more uniformly. Heat easily transfers into the copper handle. Question? Which material is most suitable to be used in a heat sink? Fins-2 (contd...) How do we know the adiabatic tip assumption is good? Try using the convection heat transfer condition at the tip (case A in fins table) We will use the aluminum pot as the example. h=5 W/m2.K, k=237 W/m.K, m=3.138, M=8.325W Long equation cosh[ m( L − x )] + (h / mk )sinh[ m( L − x )] θ = θb cosh mL + (h / mk )sinh mL T ( x ) - T∞ Tb − T∞ = cosh[3138(0.2 − x )] + 0.00672 sinh[3138(0.2 − x )] . . = cosh(0.6276) + 0.00672 sinh(0.6276) T ( x ) = 25 + 62.09{cosh(0.6276 − 3138 x ) + 0.00672 sinh(0.6276 − 3138 x )} . . Fins-2 (contd….) 100 96.25 T( x ) 92.5 T c( x ) 88.75 85 T: adiabatic tip Tc: convective tip T(0.2)=87.32 °C Tc(0.2)=87.09 °C 0 0.04 0.08 x 0.12 0.16 0.2 Note 1: Convective tip case has a slightly lower tip temperature as expected since there is additional heat transfer at the tip. Note 2: There is no significant difference between these two solutions, therefore, correct choice of boundary condition is not that important here. However, sometimes correction might be needed to compensate the effect of convective heat transfer at the end. (especially for thick fins) Fins-2 (contd...) In some situations, it might be necessary to include the convective heat transfer at the tip. However, one would like to avoid using the long equation as described in case A, fins table. The alternative is to use case B instead and accounts for the convective heat transfer at the tip by extending the fin length L to LC=L+(t/2). Insulation LC=L+t/2 With convection t t/2 L Then apply the adiabatic condition at the tip of the extended fin as shown above. Original fin length L Fins-2 (contd...) Use the same example: aluminum pot handle, m=3.138, the length will need to be corrected to LC=l+(t/2)=0.2+0.0025=0.2025(m) Tcorr ( x ) - T∞ cosh m( Lc − x ) θ = = Tb − T∞ θb cosh mLc Tcorr − 25 cosh[3138(0.2025 − x )] . = , 100 − 25 cosh(3138 * 0.2025) . Tcorr ( x ) = 25 + 62.05 * cosh[3138(0.2025 − x )] . Fins-2 (contd...) 100 96.25 92.5 T( x) T c( x) T(0.2)=87.32 °C Tc(0.2)=87.09 °C Tcorr(0.2025)=87.05 °C slight improvement over the uncorrected solution T corr x) ( 88.75 85 0 0.04 0.08 0.12 0.16 0.2 Correction Length The correction length can be determined by using the formula: Lc=L+(Ac/P), where Ac is the cross-sectional area and P is the perimeter of the fin at the tip. Thin rectangular fin: Ac=Wt, P=2(W+t)≈2W, since t << W Lc=L+(Ac/P)=L+(Wt/2W)=L+(t/2) Cylindrical fin: Ac=(π/4)D2, P= πD, Lc=L+(Ac/P)=L+(D/4) Square fin: Ac=W2, P=4W, Lc=L+(Ac/P)=L+(W2/4W)=L+(W/4) Optimal Length of a Fin In general, the longer the fin, the higher the heat transfer. However, a long fin means more material and increased size and cost. Question: how do we determine the optimal fin length? Use the rectangular fin as an example: q f = M tanh mL, for an adiabatic tip fin 1 0.8 0.6 R( mL ) 0.4 0.2 ( q f ) ∞ = M , for an infinitely long fin Their ratio: R(mL)= qf ( q f )∞ = tanh mL Note: heat transfer increases with mL as expected. Initially the rate of 0 0 1 2 3 4 change is large and slows down mL drastically when mL> 2. R(1)=0.762, means any increase beyond mL=1 will increase no more than 23.8% of the fin heat transfer. Temperature Distribution For an adiabatic tip fin case: T − T∞ cosh m( L − x ) Rθ = = Tb − T∞ cosh mL High ΔT, good fin heat transfer 1 1 Use m=5, and L=0.2 as an example: Low ΔT, poor fin heat transfer R θ ( x ) 0.8 0.648054 0.6 0 0 0.05 0.1 x 0.15 0.2 0.2 Correction Length for a Fin with a Non-adiabatic Tip The correction length can be determined by using the formula: Lc=L+(Ac/P), where Ac is the cross-sectional area and P is the perimeter of the fin at the tip. Thin rectangular fin: Ac=Wt, P=2(W+t)≈2W, since t << W Lc=L+(Ac/P)=L+(Wt/2W)=L+(t/2) Cylindrical fin: Ac=(π/4)D2, P= πD, Lc=L+(Ac/P)=L+(D/4) Square fin: Ac=W2, P=4W, Lc=L+(Ac/P)=L+(W2/4W)=L+(W/4) Fin Design T∞ Tb Total heat loss: qf=Mtanh(mL) for an adiabatic fin, or qf=Mtanh(mLC) if there is convective heat transfer at the tip hP where m= , and M= hPkA Cθ b = hPkA C (Tb − T∞ ) kAc Use the thermal resistance concept: (Tb − T∞ ) q f = hPkA C tanh( mL)(Tb − T∞ ) = Rt , f where Rt , f is the thermal resistance of the fin. For a fin with an adiabatic tip, the fin resistance can be expressed as (Tb − T∞ ) 1 Rt , f = = qf hPkA C [tanh(mL)] Fin Effectiveness How effective a fin can enhance heat transfer is characterized by the fin effectiveness εf: Ratio of fin heat transfer and the heat transfer without the fin. For an adiabatic fin: εf = qf q = qf hAC (Tb − T∞ ) = hPkA C tanh(mL) kP = tanh( mL) hAC hAC If the fin is long enough, mL>2, tanh(mL) → 1, it can be considered an infinite fin (case D of table3.4) kP k⎛ P ⎞ εf → = ⎜ ⎟ hAC h ⎝ AC ⎠ In order to enhance heat transfer, ε f > 1. However, ε f ≥ 2 will be considered justifiable If ε f <1 then we have an insulator instead of a heat fin Fin Effectiveness (contd...) εf → kP k⎛ P ⎞ = ⎜ ⎟ hAC h ⎝ AC ⎠ To increase εf, the fin’s material should have higher thermal conductivity, k. It seems to be counterintuitive that the lower convection coefficient, h, the higher εf. But it is not because if h is very high, it is not necessary to enhance heat transfer by adding heat fins. Therefore, heat fins are more effective if h is low. Observation: If fins are to be used on surfaces separating gas and liquid. Fins are usually placed on the gas side. (Why?) Fin Effectiveness (contd...) P/AC should be as high as possible. Use a square fin with a dimension of W by W as an example: P=4W, AC=W2, P/AC=(4/W). The smaller W, the higher the P/AC, and the higher εf. Conclusion: It is preferred to use thin and closely spaced (to increase the total number) fins. Fin Effectiveness (contd...) The effectiveness of a fin can also be characterized as εf = qf q = qf hAC (Tb − T∞ ) = (Tb − T∞ ) / Rt , f (Tb − T∞ ) / Rt ,h Rt ,h = Rt , f It is a ratio of the thermal resistance due to convection to the thermal resistance of a fin. In order to enhance heat transfer, the fin's resistance should be lower than that of the resistance due only to convection. Fin Efficiency Define Fin efficiency: η f = qf q max where q max represents an idealized situation such that the fin is made up of material with infinite thermal conductivity. Therefore, the fin should be at the same temperature as the temperature of the base. q max = hA f (Tb − T∞ ) Fin Efficiency (contd…) T(x)<Tb for heat transfer to take place For infinite k T(x)=Tb, the heat transfer is maximum Tb x Total fin heat transfer qf Real situation x Ideal heat transfer qmax Ideal situation Fin Efficiency (cont.) Use an adiabatic rectangular fin as an example: η = f = qf q m ax = M ta n h m L = h A f (Tb − T∞ ) h P k A c ( T b − T ∞ ) ta n h m L h P L (Tb − T∞ ) ta n h m L ta n h m L = ( s e e T a b le 3 .5 f o r η f o f c o m m o n f in s ) mL hP L k Ac 1 Tb − T∞ T − T∞ = b , w h ere Rt, f = η f hA f Rt, f 1 /(η f h A f ) 1 h Ab T h e f in h e a t tr a n s f e r : q f = η f q m a x = η f h A f ( T b − T ∞ ) qf = T h e r m a l r e s is ta n c e f o r a s in g le f in . A s c o m p a r e d to c o n v e c tiv e h e a t tr a n s f e r : R t , b = In o r d e r to h a v e a lo w e r r e s is ta n c e a s th a t is r e q u ir e d to e n h a n c e h e a t tr a n s f e r : R t , b > R t , f o r A b < η f A f Overall Fin Efficiency Overall fin efficiency for an array of fins: qf qb Define terms: Ab: base area exposed to coolant Af: surface area of a single fin At: total area including base area and total finned surface, At=Ab+NAf N: total number of fins Overall Fin Efficiency (contd…) qt = qb + Nqf = hAb (Tb − T∞ ) + Nη f hAf (Tb − T∞ ) = h[( At − NAf ) + Nη f Af ](Tb − T∞ ) = h[ At − NAf (1 −η f )](Tb − T∞ ) = hAt [1 − NAf At (1 −η f )](Tb − T∞ ) = ηOhAt (Tb − T∞ ) NAf At (1 −η f ) Define overall fin efficiency: ηO = 1 − Heat Transfer from a Fin Array Tb − T∞ 1 where Rt ,O = Rt ,O hAtηO qt = hAtηO (Tb − T∞ ) = Compare to heat transfer without fins 1 q = hA(Tb − T∞ ) = h( Ab + NAb, f )(Tb − T∞ ) = hA where Ab,f is the base area (unexposed) for the fin To enhance heat transfer AtηO >> A That is, to increase the effective area ηO At . Thermal Resistance Concept L1 t A=Ab+NAb,f Rb=t/(kbA) T1 T∞ T1 T2 Tb T∞ R1=L1/(k1A) Rt ,O = 1 /( hAtηO ) T2 Tb T1 − T∞ T1 − T∞ q= = ∑ R R1 + Rb + Rt ,O Problem 1: A long, circular aluminium rod attached at one end to the heated wall and transfers heat through convection to a cold fluid. (a) If the diameter of the rod is triples, by how much would the rate of heat removal change? (b) If a copper rod of the same diameter is used in place of aluminium, by how much would the rate of heat removal change? Known: long, aluminum cylinder acts as an extended surface. Find: (a) increase in heat transfer if diameter is tripled and (b) increase in heat transfer if copper is used in place of aluminum. Schematic: T∞,h Aluminum Or copper q D Assumptions: (1) steady-state conditions, (2) one-dimensional conduction, (3) constant properties, (4) uniform convection coefficient, (5)rod is infinitely long. Properties: aluminum (pure): k=240W/m. K; copper (pure): k=400W/m. K Analysis: (a) for an infinitely long fin, the fin rate is qr = M = (hpkAc ) 2 θ b q f = (hπDkπD 2 / 4) 2 θ b = 1 1 π 2 (hk ) 2 D 2 θ b 1 3 Where P=πD and Ac=πD2/4 for the circular cross-section. Note that qf≈D3/2. Hence, if the diameter is tripled, q f (3D ) q f ( D) = 3 = 5.2 3 2 And there is a 520 % increase in heat transfer. (b) in changing from aluminum to copper, since qf ≈ k1/2, it follows that q f (Cu ) ⎛k = ⎜ Cu q f ( Al ) ⎜ k Al ⎝ ⎞ 2 ⎛ 400 ⎞ 2 ⎟ =⎜ ⎟ = 1.29 ⎟ ⎝ 240 ⎠ ⎠ 1 1 And there is a 29 % increase in the heat transfer rate. Comments: (1) because fin effectiveness is enhanced by maximum P/Ac = 4/D. the use of a larger number of small diameter fins is preferred to a single large diameter fin. (2) From the standpoint of cost and weight, aluminum is preferred over copper. Problem 2: Two long copper rods of diameter D=1 cm are soldered together end to end, with solder having melting point of 6500C. The rods are in the air at 250C with a convection coefficient of 10W/m2. K. What is the minimum power input needed to effect the soldering? Known: Melting point of solder used to join two long copper rods. Find: Minimum power needed to solder the rods. Schematic: Copper ∞ ∞ D=1cm Junction Tb=6500C Air T∞ = 250C H=10W/m2. K Assumptions: (1) steady-state conditions, (2) one-dimensional conduction along the rods, (3) constant properties, (4) no internal heat generation, (5) negligible radiation exchange with surroundings, (6) uniform, h and (7) infinitely long rods. Properties: copper T = (650+25)0C ≈ 600K: k=379 W/m.K Analysis: the junction must be maintained at 6500C while energy is transferred by conduction from the junction (along both rods). The minimum power is twice the fin heat rate for an infinitely long fin, 1 = 2q = 2(hpkA ) 2 (T − T ) q b ∞ min f c substituting numerical values, 1 ⎛ ⎞2 W W ⎞π ⎛ 2 0 q = 2⎜10 (π × 0.01m)⎜ 379 ⎟ (0.01m) ⎟ (650 − 25) C. ⎜ ⎟ min m.K ⎠ 4 ⎝ ⎝ m 2 .K ⎠ therefore, q min = 120.9W Comments: radiation losses from the rod are significant, particularly near the junction, thereby requiring a larger power input to maintain the junction at 6500C. Problem 3: Determine the percentage increase in heat transfer associated with attaching aluminium fins of rectangular profile to a plane wall. The fins are 50mm long, 0.5mm thick, and are equally spaced at a distance of 4mm (250fins/m). The convection coefficient affected associated with the bare wall is 40W/m2. K, while that resulting from attachment of the fins is 30W/m2. K. Known: Dimensions and number of rectangular aluminum fins. Convection coefficient with and without fins. Find: percentage increase in heat transfer resulting from use of fins. Schematic: L=50mm N=250m-1 W=width hw = 30W/m2.K(with fins) hwo=40W/m2.K(without fins) Aluminium Assumptions: (1) steady-state conditions, (2) one-dimensional conduction, (3) constant properties, (4) negligible fin contact resistance, (6) uniform convection coefficient. Properties: Aluminum, pure: k≈240W/m. K Analysis: evaluate the fin parameters Lc = L + t = 0.0505m 2 A p = L c t = 0.0505m × 0.5 × 10 −3 m = 25.25 × 10 −6 m 2 / KA P ) 1/ 2 L3c/ 2 (h w (0.0505m) 3/ 2 ⎛ 30 W / m 2 .K ⎜ ⎜ 240 W / m.K × 25.25 × 10 −6 m 2 ⎝ ⎞ ⎟ ⎟ ⎠ 1/ 2 L3c/ 2 (h w / KA P )1 / 2 = 0.80 it follows that, η f = 0.72. hence q f = ηf q max = 0.72h w 2 wLθ b q f = 0.72 × 30 W / m 2 .K × 2 × 0.05m × ( wθ b ) = 2.16 W / m / K ( wθ b ) With the fins, the heat transfer from the walls is q w = Nq f + (1 − Nt ) wh w θ b W ( wθ b ) + (1m − 250 × 5 × 10 − 4 m) × 30 W / m 2 .K ( wθ b ) m.K W ( wθ b ) = 566 wθ b q w = (540 + 26.63) m.K Without the fins, q wo = h wo 1m × wθ b = 40wθ b . q w = 250 × 2.16 Hence the percentage increases in heat transfer is qw 566wθ b = 14.16 = 1416% = q wo 40wθ b Comments: If the finite fin approximation is made, it follows that qf θ b = [hw2wkwt] overestimated. (hPkAc) 1/2 1/2 θ b = (30*2*240*5*10 ) -4 1/2 = w θ b =2.68w θ b . Hence qf is Module 3: Short questions 1. Is the heat flow through a fin truly one-dimensional? Justify the one-dimensional assumption made in the analysis of fins. 2. Insulated tip condition is often used in fin analysis because (choose one answer) A. Fins are usually deliberately insulated at their tips B. There is no heat loss from the tip C. The heat loss from the tip is usually insignificant compared to the rest of the fin, and the insulated tip condition makes the problem mathematically simple 3. What is the difference between fin effectiveness and fin efficiency? 4. The fins attached to a surface are determined to have an effectiveness of 0.9. Do you think the rate of heat transfer from the surface has increased or decreased as a result of addition of fins? 5. Fins are normally meant to enhance heat transfer. Under what circumstances the addition of fins may actually decrease heat transfer? 6. Hot water is to be cooled as it flows through the tubes exposed to atmospheric air. Fins are to be attached in order to enhance heat transfer. Would you recommend adding fins to the inside or outside the tubes? Why? 7. Hot air is to be cooled as it is forced through the tubes exposed to atmospheric air. Fins are to be attached in order to enhance heat transfer. Would you recommend adding fins to the inside or outside the tubes? Why? When would you recommend adding fins both inside and outside the tubes? 8. Consider two finned surface which are identical except that the fins on the fist surface are formed by casting or extrusion, whereas they are attached to the second surface afterwards by welding or tight fitting. For which case do you think the fins will provide greater enhancement in heat transfer? Explain. 9. Does the (a) efficiency and (b) effectiveness of a fin increase or decrease as the fin length is increased? 10. Two pin fins are identical, except that the diameter of one of them is twice the diameter of the other? For which fin will the a) efficiency and (b) effectiveness be higher? Explain. 11. Two plate fins of constant rectangular cross section are identical, except that the thickness of one of them is twice the thickness of the other? For which fin will the a) efficiency and (b) effectiveness be higher? Explain. 12. Two finned surface are identical, except that the convection heat transfer coefficient of one of them is twice that of the other? For which fin will the a) efficiency and (b) effectiveness be higher? Explain. MODULE 3 Extended Surface Heat Transfer 3.1 Introduction: Convection: Heat transfer between a solid surface and a moving fluid is governed by the Newton’s cooling law: q = hA(Ts-T∞), where Ts is the surface temperature and T∞ is the fluid temperature. Therefore, to increase the convective heat transfer, one can • • Increase the temperature difference (Ts-T∞) between the surface and the fluid. Increase the convection coefficient h. This can be accomplished by increasing the fluid flow over the surface since h is a function of the flow velocity and the higher the velocity, the higher the h. Example: a cooling fan. Increase the contact surface area A. Example: a heat sink with fins. • Many times, when the first option is not in our control and the second option (i.e. increasing h) is already stretched to its limit, we are left with the only alternative of increasing the effective surface area by using fins or extended surfaces. Fins are protrusions from the base surface into the cooling fluid, so that the extra surface of the protrusions is also in contact with the fluid. Most of you have encountered cooling fins on air-cooled engines (motorcycles, portable generators, etc.), electronic equipment (CPUs), automobile radiators, air conditioning equipment (condensers) and elsewhere. 3.2 Extended surface analysis: In this module, consideration will be limited to steady state analysis of rectangular or pin fins of constant cross sectional area. Annular fins or fins involving a tapered cross section may be analyzed by similar methods, but will involve solution of more complicated equations which result. Numerical methods of integration or computer programs can be used to advantage in such cases. We start with the General Conduction Equation: 1 dT ⋅ system α dτ = ∇ 2T + &&& q k (1) After making the assumptions of Steady State, One-Dimensional Conduction, this equation reduces to the form: d 2 T &&& q =0 2 + dx k (2) This is a second order, ordinary differential equation and will require 2 boundary conditions to evaluate the two constants of integration that will arise. 1 Consider the cooling fin shown below: h, T∞ Ts, q L Ac y The fin is situated on the surface of a hot surface at Ts and surrounded by a coolant at temperature T∞, which cools with convective coefficient, h. The fin has a cross sectional area, Ac, (This is the area through with heat is conducted.) and an overall length, L. Note that as energy is conducted down the length of the fin, some portion is lost, by convection, from the sides. Thus the heat flow varies along the length of the fin. We further note that the arrows indicating the direction of heat flow point in both the x and y directions. This is an indication that this is truly a two- or three-dimensional heat flow, depending on the geometry of the fin. However, quite often, it is convenient to analyse a fin by examining an equivalent one–dimensional system. The equivalent system will involve the introduction of heat sinks (negative heat sources), which remove an amount of energy equivalent to what would be lost through the sides by convection. Consider a differential length of the fin. h, T∞ T0, q Δx Across this segment the heat loss will be h⋅(P⋅Δx)⋅(T-T∞), where P is the perimeter around the fin. The equivalent heat sink would be &&& ⋅ ( Ac ⋅ Δ x ) . q 2 Equating the heat source to the convective loss: &&& = q − h ⋅ P ⋅ (T − T∞ ) Ac (3) Substitute this value into the General Conduction Equation as simplified for One-Dimension, Steady State Conduction with Sources: d 2T h ⋅ P − ⋅ (T − T∞ ) = 0 dx 2 k ⋅ Ac (4) which is the equation for a fin with a constant cross sectional area. This is the Second Order Differential Equation that we will solve for each fin analysis. Prior to solving, a couple of simplifications should be noted. First, we see that h, P, k and Ac are all independent of x in the defined system (They may not be constant if a more general analysis is desired.). We replace this ratio with a constant. Let m2 = h⋅ P k ⋅ Ac (5) then: d 2T 2 2 − m ⋅ (T − T ) = 0 ∞ dx (6) Next we notice that the equation is non-homogeneous (due to the T∞ term). Recall that nonhomogeneous differential equations require both a general and a particular solution. We can make this equation homogeneous by introducing the temperature relative to the surroundings: θ ≡ T - T∞ Differentiating this equation we find: dθ dT = +0 dx dx (7) (8) Differentiate a second time: d 2θ d 2 T = dx 2 dx 2 Substitute into the Fin Equation: d 2θ − m2 ⋅ θ = 0 dx 2 (9) (10) This equation is a Second Order, Homogeneous Differential Equation. 3.3 Solution of the Fin Equation 3 We apply a standard technique for solving a second order homogeneous linear differential equation. Try θ = eα⋅x. Differentiate this expression twice: dθ = α ⋅ eα ⋅ x dx (11) d 2θ = α 2 ⋅ eα ⋅ x dx 2 Substitute this trial solution into the differential equation: α2⋅eα⋅x – m2⋅eα⋅x = 0 Equation (13) provides the following relation: α=± m (12) (13) (14) We now have two solutions to the equation. The general solution to the above differential equation will be a linear combination of each of the independent solutions. Then: θ = A⋅em⋅x + B⋅ e-m⋅x (15) where A and B are arbitrary constants which need to be determined from the boundary conditions. Note that it is a 2nd order differential equation, and hence we need two boundary conditions to determine the two constants of integration. An alternative solution can be obtained as follows: Note that the hyperbolic sin, sinh, the hyperbolic cosine, cosh, are defined as: sinh(m ⋅ x ) = e m⋅x − e − m⋅ x 2 cosh(m ⋅ x ) = e m⋅x + e − m⋅x 2 (16) We may write: C ⋅ cosh(m ⋅ x ) + D ⋅ sinh(m ⋅ x ) = C ⋅ e m⋅ x + e − m÷ x e m⋅ x − e − m÷ x C + D m⋅ x C − D − m⋅ x + D⋅ = ⋅e + ⋅ e (17) 2 2 2 2 We see that if (C+D)/2 replaces A and (C-D)/2 replaces B then the two solutions are equivalent. θ = C ⋅ cosh(m ⋅ x ) + D ⋅ sinh(m ⋅ x ) (18) Generally the exponential solution is used for very long fins, the hyperbolic solutions for other cases. 4 Boundary Conditions: Since the solution results in 2 constants of integration we require 2 boundary conditions. The first one is obvious, as one end of the fin will be attached to a hot surface and will come into thermal equilibrium with that surface. Hence, at the fin base, θ(0) = T0 - T∞ ≡ θ0 (19) The second boundary condition depends on the condition imposed at the other end of the fin. There are various possibilities, as described below. Very long fins: For very long fins, the end located a long distance from the heat source will approach the temperature of the surroundings. Hence, θ(∞) = 0 Substitute the second condition into the exponential solution of the fin equation: ∞ 0 (21) (20) θ(∞) = 0 = A⋅em⋅∞ + B⋅ e-m⋅∞ The first exponential term is infinite and the second is equal to zero. The only way that this equation can be valid is if A = 0. Now apply the second boundary condition. θ(0) = θ0 = B⋅ e-m⋅0 ⇒ B = θ0 The general temperature profile for a very long fin is then: θ(x) = θ0 ⋅ e-m⋅x If we wish to find the heat flow through the fin, we may apply Fourier Law: q = − k ⋅ Ac ⋅ dT dθ = − k ⋅ Ac ⋅ dx dx (22) (23) (24) Differentiate the temperature profile: dθ = − θ o ⋅ m ⋅ e − m⋅ x dx (25) So that: ⎡ h ⋅ P ⎤ 2 − m⋅ x q = k ⋅ Ac ⋅ θ 0 ⋅ ⎢ ⎥ ⋅e = ⎣ k ⋅ Ac ⎦ where M = hPkAc . 1 h ⋅ P ⋅ k ⋅ Ac ⋅ e − m⋅ x ⋅ θ 0 = Mθ 0 e − mx (26) Often we wish to know the total heat flow through the fin, i.e. the heat flow entering at the base (x=0). 5 q= h ⋅ P ⋅ k ⋅ Ac ⋅ θ 0 = Mθ 0 (27) The insulated tip fin Assume that the tip is insulated and hence there is no heat transfer: dθ dx The solution to the fin equation is known to be: =0 x =L (28) θ = C ⋅ cosh(m ⋅ x ) + D ⋅ sinh(m ⋅ x ) Differentiate this expression. dθ = C ⋅ m ⋅ sinh(m ⋅ x ) + D ⋅ m ⋅ cosh(m ⋅ x ) dx (29) (30) Apply the first boundary condition at the base: θ (0) = θ 0 = C sinh(m ⋅ 0) + D cosh(m ⋅ 0) 0 1 (31) So that D = θ0. Now apply the second boundary condition at the tip to find the value of C: dθ ( L ) = 0 = Cm sinh( m ⋅ L) + θ 0 m cosh( m ⋅ L) dx (32) which requires that C = −θ 0 cosh(mL) sinh(mL) (33) This leads to the general temperature profile: θ ( x) = θ 0 cosh m( L − x ) cosh(mL) (34) We may find the heat flow at any value of x by differentiating the temperature profile and substituting it into the Fourier Law: q = − k ⋅ Ac ⋅ dT dθ = − k ⋅ Ac ⋅ dx dx (35) 6 So that the energy flowing through the base of the fin is: q = hPkAc θ 0 tanh( mL) = Mθ 0 tanh( mL) (36) If we compare this result with that for the very long fin, we see that the primary difference in form is in the hyperbolic tangent term. That term, which always results in a number equal to or less than one, represents the reduced heat loss due to the shortening of the fin. Other tip conditions: We have already seen two tip conditions, one being the long fin and the other being the insulated tip. Two other possibilities are usually considered for fin analysis: (i) a tip subjected to convective heat transfer, and (ii) a tip with a prescribed temperature. The expressions for temperature distribution and fin heat transfer for all the four cases are summarized in the table below. Table 3.1 Case Tip Condition A Convection heat transfer: hθ(L)=-k(dθ/dx)x=L B C Adiabatic (dθ/dx)x=L=0 Given temperature: θ(L)= θL Infinitely long fin θ(L)=0 Temp. Distribution cosh m( L − x ) + ( h cosh mL + ( h Fin heat transfer mk mk ) sinh m( L − x ) ) sinh mL M θo sinh mL + ( h ) cosh mL mk cosh mL + ( h ) sinh mL mk cosh m( L − x ) cosh mL (θ L Mθ 0 tanh mL (cosh mL − θ b ) sinh m( L − x ) + sinh m( L − x ) sinh mL θL Mθ 0 θb ) sinh mL D e − mx M θ0 3.4 Fin Effectiveness How effective a fin can enhance heat transfer is characterized by the fin effectiveness, ε f , which is as the ratio of fin heat transfer and the heat transfer without the fin. For an adiabatic fin: εf = qf q = qf hAC (Tb − T∞ ) = hPkAC tanh( mL) = hAC kP tanh( mL) hAC (37) If the fin is long enough, mL>2, tanh(mL)→1, and hence it can be considered as infinite fin (case D in Table 3.1). Hence, for long fins, 7 εf → kP ⎛k⎞ P = ⎜ ⎟ hAC ⎝ h ⎠ AC (38) In order to enhance heat transfer, ε f should be greater than 1 (In case ε f <1, the fin would have no purpose as it would serve as an insulator instead). However ε f ≥ 2 is considered unjustifiable because of diminishing returns as fin length increases. To increase ε f , the fin’s material should have higher thermal conductivity, k. It seems to be counterintuitive that the lower convection coefficient, h, the higher ε f . Well, if h is very high, it is not necessary to enhance heat transfer by adding heat fins. Therefore, heat fins are more effective if h is low. Observations: • • If fins are to be used on surfaces separating gas and liquid, fins are usually placed on the gas side. (Why?) P/AC should be as high as possible. Use a square fin with a dimension of W by W as an example: P=4W, AC=W2, P/AC=(4/W). The smaller the W, the higher is the P/AC, and the higher the ε f .Conclusion: It is preferred to use thin and closely spaced (to increase the total number) fins. The effectiveness of a fin can also be characterized by (Tb − T∞ ) / Rt , f (Tb − T∞ ) / Rt ,h εf = qf q = qf hAC (Tb − T∞ ) = = Rt ,h Rt , f (39) It is a ratio of the thermal resistance due to convection to the thermal resistance of a fin. In order to enhance heat transfer, the fin’s resistance should be lower than the resistance due only to convection. 3.5 Fin Efficiency The fin efficiency is defined as the ratio of the energy transferred through a real fin to that transferred through an ideal fin. An ideal fin is thought to be one made of a perfect or infinite conductor material. A perfect conductor has an infinite thermal conductivity so that the entire fin is at the base material temperature. η= qreal = qideal h ⋅ P ⋅ k ⋅ Ac ⋅ θ L ⋅ tanh(m ⋅ L) h ⋅ ( P ⋅ L) ⋅ θ L (40) 8 Tb x Real situation Simplifying equation (40): Tb x Ideal situation η= k ⋅ Ac θ L ⋅ tanh(m ⋅ L) tanh(m ⋅ L) = h⋅ P L ⋅θL m⋅ L (41) The heat transfer through any fin can now be written as: ⎡ 1 q.⎢ ⎢η .h. A f ⎣ ⎤ ⎥ = (T − T∞ ⎥ ⎦ (42) The above equation provides us with the concept of fin thermal resistance (using electrical analogy) as 1 Rt , f = (43) η.h. A f Overall Fin Efficiency: Overall fin efficiency for an array of fins qf qb Define terms: Ab: base area exposed to coolant Af: surface area of a single fin At: total area including base area and total finned surface, At=Ab+NAf N: total number of fins Heat Transfer from a Fin Array: 9 qt = qb + Nq f = hAb (Tb − T∞ ) + Nη f hAf (Tb − T∞ ) = h[( At − NAf ) + Nη f Af ](Tb − T∞ ) = h[ At − NAf (1 − η f )](Tb − T∞ ) = hAt [1 − NAf At (1 − η f )](Tb − T∞ ) = ηO hAt (Tb − T∞ ) NAf At (1 − η f ) Define overall fin efficiency: ηO = 1 − q t = hAtη O ( T b − T ∞ ) = Tb − T∞ 1 w here R t ,O = R t ,O h Atη O C om pare to heat transfer w ith out fin s q = h A ( T b − T ∞ ) = h ( Ab + N Ab , f )( T b − T ∞ ) = w h ere Ab ,f T o en hance heat transfer Atη O > > A T h at is, to in crease th e effective area η O At . 1 hA is the base area (un exposed) for the fin Thermal Resistance Concept: L1 t A=Ab+NAb,f T1 T ∞ Rb= t/(kbA) T2 Tb T1 T2 Tb T∞ R1=L1/(k1A) Rt ,O = 1 /(hAtηO ) q= T1 − T∞ T1 − T∞ = ∑ R R1 + Rb + Rt ,O 10 Module 4: Learning objectives • • • • • The primary objective of this module is to develop an appreciation for the nature of two- or multi-dimensional conduction problems and the methods that are available for its solutions. For a multi-dimensional problem, the student should be able to determine whether an exact solution is known. This may be done by examining one or more of the many excellent references in which exact solutions to the heat equation are obtained. The student should understand what a conduction shape factor is, and link it to the concept of thermal resistances in 2D problems. The student should be able to determine whether the shape factor is known for the system, and if available, use to solve the heat transfer problem. However, if conditions are such that the use of a shape factor or an exact solution is not possible, the student should be able to use a numerical solution, such as the finite difference method.. The student should appreciate the inherent nature of the discretization process, and know how to formulate the finite difference equations for the discrete points of a nodal network. Although one may find it convenient to solve these equations using hand calculations for a coarse mesh, one should be able to treat fine meshes using standard computer algorithms involving direct or iterative techniques. Module 4 Multi-Dimensional Steady State Heat Conduction Multidimensional Heat Transfer H e a t D iffu s io n E q u a tio n ρc p ∂T ∂ 2T ∂ 2T ∂ 2T & & + + ) + q = k∇ 2 T + q = k( 2 2 2 ∂x ∂y ∂z ∂t This equation governs the Cartesian, temperature distribution for a three-dimensional unsteady, heat transfer problem involving heat generation. For steady state ∂/ ∂t = 0 & No generation q = 0 To solve for the full equation, it requires a total of six boundary conditions: two for each direction. Only one initial condition is needed to account for the transient behavior. Pradip Dutta HMT/M4/V1/May 05 2 Two-Dimensional, Steady State Case For a 2 - D, steady state situation, the heat equation is simplified to ∂2T ∂2T + 2 = 0, it needs two boundary conditions in each direction. 2 ∂y ∂x There are three approaches to solve this equation: Numerical Method: Finite difference or finite element schemes, usually will be solved using computers. Graphical Method: Limited use. However, the conduction shape factor concept derived under this concept can be useful for specific configurations. (see Table 4.1 for selected configurations) Analytical Method: The mathematical equation can be solved using techniques like the method of separation of variables. (refer to handout) Pradip Dutta HMT/M4/V1/May 05 3 Conduction Shape Factor This approach applied to 2-D conduction involving two isothermal surfaces, with all other surfaces being adiabatic. The heat transfer from one surface (at a temperature T1) to the other surface (at T2) can be expressed as: q=Sk(T1-T2) where k is the thermal conductivity of the solid and S is the conduction shape factor. The shape factor can be related to the thermal resistance: q=Sk(T1-T2)=(T1-T2)/(1/kS)= (T1-T2)/Rt where Rt = 1/(kS) 1-D heat transfer can use shape factor also. Ex: Heat transfer inside a plane wall of thickness L is q=kA(ΔT/L), S=A/L Pradip Dutta HMT/M4/V1/May 05 4 Example An Alaska oil pipe line is buried in the earth at a depth of 1 m. The horizontal pipe is a thin-walled of outside diameter of 50 cm. The pipe is very long and the averaged temperature of the oil is 100°C and the ground soil temperature is at -20 °C (ksoil=0.5W/m.K), estimate the heat loss per unit length of pipe. T2 z=1 m From Table 8.7, case 1. L>>D, z>3D/2 2πL 2π (1) = = 3.02 S= ln( 4 z / D) ln( 4 / 0.5) q = kS(T1 − T2 ) = (0.5)(3.02)(100 + 20) T1 Pradip Dutta = 181.2(W ) heat loss for every meter of pipe HMT/M4/V1/May 05 5 Example (contd...) If the mass flow rate of the oil is 2 kg/s and the specific heat of the oil is 2 kJ/kg.K, determine the temperature change in 1 m of pipe length. & q = mC P ΔT , ΔT = q 181.2 = = 0.045(° C ) & mC P 2000 * 2 Therefore, the total temperature variation can be significant if the pipe is very long. For example, 45°C for every 1 km of pipe length. Heating might be needed to prevent the oil from freezing up. The heat transfer can not be considered constant for a long pipe Pradip Dutta HMT/M4/V1/May 05 6 Example (contd...) Ground at -20°C Heat transfer to the ground (q) Length dx Pradip Dutta HMT/M4/V1/May 05 & )Td + T( PCm 7 & TPCm Example (contd...) Heat Transfer at section with a temperature T(x) 2πk(dx) q= (T + 20) = 1.51(T + 20)( dx ) ln(4z / D) & & Energy balance: mC P T − q = mCP (T + dT ) dT dT + 1.51(T + 20) = 0, = −0.000378dx, integrate dx T + 20 T ( x ) = −20 + Ce −0.000378 x , at inlet x = 0, T(0) = 100° C, C = 120 & mCP T(x) = -20 + 120e −0.000378 x Pradip Dutta HMT/M4/V1/May 05 8 Example (contd...) 100 50 T( x ) 0 50 0 1000 2000 x 3000 4000 5000 Temperature drops exponentially from the initial temp. of 100°C It reaches 0°C at x=4740 m, therefore, reheating is required every 4.7 km. Pradip Dutta HMT/M4/V1/May 05 9 Numerical Methods Due to the increasing complexities encountered in the development of modern technology, analytical solutions usually are not available. For these problems, numerical solutions obtained using high-speed computer are very useful, especially when the geometry of the object of interest is irregular, or the boundary conditions are nonlinear. In numerical analysis, two different approaches are commonly used: The finite difference and the finite element methods. In heat transfer problems, the finite difference method is used more often and will be discussed here. Pradip Dutta HMT/M4/V1/May 05 10 Numerical Methods (contd…) The finite difference method involves: Establish nodal networks Derive finite difference approximations for the governing equation at both interior and exterior nodal points Develop a system of simultaneous algebraic nodal equations Solve the system of equations using numerical schemes Pradip Dutta HMT/M4/V1/May 05 11 The Nodal Networks The basic idea is to subdivide the area of interest into subvolumes with the distance between adjacent nodes by Δx and Δy as shown. If the distance between points is small enough, the differential equation can be approximated locally by a set of finite difference equations. Each node now represents a small region where the nodal temperature is a measure of the average temperature of the region. Pradip Dutta HMT/M4/V1/May 05 12 The Nodal Networks (contd…) Example Δx m,n+1 m-1,n Δy m,n m+1, n x=mΔx, y=nΔy Pradip Dutta m+½,n m-½,n intermediate points HMT/M4/V1/May 05 13 m,n-1 & q 1 ∂T Heat Diffusion Equation: ∇ T + = , k α ∂t k where α = is the thermal diffusivity ρ C PV 2 Finite Difference Approximation ∂ = 0, ⇒ ∇2T = 0 ∂t First, approximated the first order differentiation at intermediate points (m+1/2,n) & (m-1/2,n) & No generation and steady state: q=0 and ∂T ΔT ≈ ∂x ( m +1/ 2,n ) Δx ∂T ΔT ≈ ∂x ( m −1/ 2,n ) Δx Pradip Dutta ( m +1/ 2,n ) Tm +1,n − Tm ,n = Δx Tm ,n − Tm −1,n = Δx HMT/M4/V1/May 05 14 ( m −1/ 2,n ) Finite Difference Approximation (contd...) N ext, approximate the second order differentiation at m,n ∂ 2T ∂x 2 ∂ 2T ∂x 2 ≈ m ,n ∂T / ∂x m +1 / 2 ,n − ∂T / ∂x Δx m −1 / 2 , n m ,n Tm +1, n + T m −1, n − 2 Tm , n ≈ (Δx )2 Similarly, the approximation can be applied to the other dimension y ∂ 2T ∂y 2 T m , n + 1 + T m , n −1 − 2 T m , n ≈ (Δy )2 HMT/M4/V1/May 05 15 m ,n Pradip Dutta Finite Difference Approximation (contd...) Tm +1,n + Tm −1,n − 2Tm ,n Tm ,n +1 + Tm ,n −1 − 2Tm ,n ⎛ ∂ 2T ∂ 2T ⎞ + ⎜ ∂x 2 + ∂y 2 ⎟ ≈ 2 ( Δx ) ( Δy ) 2 ⎝ ⎠ m ,n To model the steady state, no generation heat equation: ∇2T = 0 This approximation can be simplified by specify Δx=Δy and the nodal equation can be obtained as Tm +1,n + Tm −1,n + Tm ,n +1 + Tm ,n −1 − 4Tm ,n = 0 This equation approximates the nodal temperature distribution based on the heat equation. This approximation is improved when the distance between the adjacent nodal points is decreased: ΔT ∂T ΔT ∂T Since lim( Δx → 0) = ,lim( Δy → 0) = Δx ∂x Δy ∂y Pradip Dutta HMT/M4/V1/May 05 16 A System of Algebraic Equations The nodal equations derived previously are valid for all interior points satisfying the steady state, no generation heat equation. For each node, there is one such equation. For example: for nodal point m=3, n=4, the equation is T2,4 + T4,4 + T3,3 + T3,5 - 4T3,4 =0 T3,4=(1/4)(T2,4 + T4,4 + T3,3 + T3,5) Nodal relation table for exterior nodes (boundary conditions) can be found in standard heat transfer textbooks (Table 4.2 of our textbook). Derive one equation for each nodal point (including both interior and exterior points) in the system of interest. The result is a system of N algebraic equations for a total of N nodal points. Pradip Dutta HMT/M4/V1/May 05 17 Matrix Form The system of equations: a11T1 + a12T2 + L + a1N TN = C1 a21T1 + a22T2 + L + a2 N TN = C2 M M M M M a N 1T1 + a N 2T2 + L + a NN TN = C N A total of N algebraic equations for the N nodal points and the system can be expressed as a matrix formulation: [A][T]=[C] ⎡ a11 a12 L a1N ⎤ ⎡ T1 ⎤ ⎡ C1 ⎤ ⎢a ⎢T ⎥ ⎢C ⎥ a22 L a2 N ⎥ ⎢ 21 ⎥ , T = ⎢ 2 ⎥ ,C = ⎢ 2 ⎥ where A= M M M ⎥ ⎢ M ⎢M ⎥ ⎢ M ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ a N 1 a N 2 L a NN ⎦ TN ⎦ ⎣ ⎣ ⎣C N ⎦ Pradip Dutta HMT/M4/V1/May 05 18 Numerical Solutions Matrix form: [A][T]=[C]. From linear algebra: [A]-1[A][T]=[A]-1[C], [T]=[A]-1[C] where [A]-1 is the inverse of matrix [A]. [T] is the solution vector. Matrix inversion requires cumbersome numerical computations and is not efficient if the order of the matrix is high (>10) Pradip Dutta HMT/M4/V1/May 05 19 Numerical Solutions (contd…) Gauss elimination method and other matrix solvers are usually available in many numerical solution package. For example, “Numerical Recipes” by Cambridge University Press or their web source at www.nr.com. For high order matrix, iterative methods are usually more efficient. The famous Jacobi & Gauss-Seidel iteration methods will be introduced in the following. Pradip Dutta HMT/M4/V1/May 05 20 Iteration General algebraic equation for nodal point: ∑a T j =1 ij i −1 j + aiiTi + j =i +1 ∑aT ij N j = Ci , Replace (k) by (k-1) for the Jacobi iteration (Example : a31T1 + a32T2 + a33T3 + L + a1N TN = C1 , i = 3) Rewrite the equation of the form: Ti (k ) Ci i −1 aij ( k ) = − ∑ Tj − aii j =1 aii j =i +1 ∑a N aij ii T j( k −1) Pradip Dutta HMT/M4/V1/May 05 21 Iteration (contd…) (k) - specify the level of the iteration, (k-1) means the present level and (k) represents the new level. An initial guess (k=0) is needed to start the iteration. By substituting iterated values at (k-1) into the equation, the new values at iteration (k) can be estimated The iteration will be stopped when max⎜Ti(k)-Ti(k-1)⎟≤ε, where ε specifies a predetermined value of acceptable error Pradip Dutta HMT/M4/V1/May 05 22 Example Solve the following system of equations using (a) the Jacobi methods, (b) the Gauss Seidel iteration method. ⎡ 4 2 1 ⎤ ⎡ X ⎤ ⎡11⎤ ⎢ −1 2 0 ⎥ ⎢ Y ⎥ = ⎢ 3 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ 2 1 4 ⎥ ⎢ Z ⎥ ⎢16⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ Reorganize into new form: 11 1 1 X = - Y- Z 4 2 4 3 1 Y = + X+0*Z 2 2 1 1 Z = 4- X- Y 2 4 4 X + 2Y + Z = 11, − X + 2Y + 0 * Z = 3, 2 X + Y + 4 Z = 16 Pradip Dutta HMT/M4/V1/May 05 23 Example (contd…) (a) Jacobi method: use initial guess X0=Y0=Z0=1, stop when max⎜Xk-Xk-1,Yk-Yk-1,Zk-Zk-1⎥ ≤ 0.1 First iteration: X1= (11/4) - (1/2)Y0 - (1/4)Z0 = 2 Y1= (3/2) + (1/2)X0 = 2 Z1= 4 - (1/2) X0 - (1/4)Y0 = 13/4 Pradip Dutta HMT/M4/V1/May 05 24 Example (contd...) Second iteration: use the iterated values X1=2, Y1=2, Z1=13/4 X2 = (11/4) - (1/2)Y1 - (1/4)Z1 = 15/16 Y2 = (3/2) + (1/2)X1 = 5/2 Z2 = 4 - (1/2) X1 - (1/4)Y1 = 5/2 Converging Process: 13 ⎤ ⎡ 15 5 5 ⎤ ⎡ 7 63 93 ⎤ ⎡ 133 31 393 ⎤ ⎡ [1,1,1], ⎢ 2,2, ⎥ , ⎢ , , ⎥ , ⎢ , , ⎥ , ⎢ , , 4 ⎦ ⎣ 16 2 2 ⎦ ⎣ 8 32 32 ⎦ ⎣ 128 16 128 ⎥ ⎣ ⎦ ⎡ 519 517 767 ⎤ ⎢ 512 , 256 , 256 ⎥ . Stop the iteration when ⎣ ⎦ max X 5 − X 4 , Y 5 − Y 4 , Z 5 − Z 4 ≤ 0.1 Final solution [1.014, 2.02, 2.996] Exact solution [1, 2, 3] Pradip Dutta HMT/M4/V1/May 05 25 Example (contd...) Use initial guess X 0 = Y 0 = Z 0 = 1 X= (b) Gauss-Seidel iteration: Substitute the iterated values into the iterative process immediately after they are computed. 11 1 1 3 1 1 1 − Y − Z, Y = + X , Z = 4 − X − Y 4 2 4 2 2 2 4 11 1 1 Immediate substitution First iteration: X1 = − (Y 0 ) − ( Z 0 ) = 2 4 2 4 3 1 3 1 5 Y 1 = + X 1 = + (2) = 2 2 2 2 2 1 1 1 1 1 1 ⎛ 5 ⎞ 19 1 Z = 4 − X − Y = 4 − (2) − ⎜ ⎟ = 2 4 2 4⎝2⎠ 8 ⎡ 5 19 ⎤ ⎡ 29 125 783 ⎤ ⎡ 1033 4095 24541 ⎤ Converging process: [1,1,1], ⎢ 2, , ⎥ , ⎢ , , ⎥ , ⎢ 1024 , 2048 , 8192 ⎥ ⎣ 2 8 ⎦ ⎣ 32 64 256 ⎦ ⎣ ⎦ The iterated solution [1.009, 1.9995, 2.996] and it converges faster Pradip Dutta HMT/M4/V1/May 05 26 Numerical Method (Special Cases) For all the special cases discussed in the following, the derivation will be based on the standard nodal point configuration as shown to the right. A m,n+1 q1 m-1,n q2 q3 m,n q4 m+1, n Symmetric case: symmetrical relative to the A-A axis. In this case, Tm-1,n=Tm+1,n Therefore the standard nodal equation can be written as m,n-1 A axis A-A Tm +1,n + Tm −1,n + Tm ,n +1 + Tm ,n −1 − 4Tm ,n = 2Tm +1,n + Tm ,n +1 + Tm ,n −1 − 4Tm ,n = 0 Pradip Dutta HMT/M4/V1/May 05 27 Special cases (contd...) A Insulated surface case: If the axis A-A is an insulated wall, therefore there is no heat transfer across A-A. Also, the surface area for q1 and q3 is only half of their original value. Write the energy balance equation (q2=0): Insulated surface q1 + q3 + q4 = 0 m,n+1 q1 m-1,n q2 m,n q4 q3 m+1, n m,n-1 A Tm +1,n − Tm ,n ⎛ Δx ⎞ Tm ,n +1 − Tm ,n ⎛ Δx ⎞ Tm ,n −1 − Tm ,n k⎜ ⎟ +k⎜ ⎟ + k Δy =0 Δy Δy Δx ⎝ 2 ⎠ ⎝ 2 ⎠ 2Tm +1,n + Tm ,n +1 + Tm ,n −1 − 4Tm ,n = 0 This equation is identical to the symmetrical case discussed previously. Pradip Dutta HMT/M4/V1/May 05 28 Special cases (contd...) With internal generation G=gV where g is the power generated per unit volume (W/m3). Based on the energy balance concept: q1 + q2 + q3 + q4 + G q1 + q2 + q3 + q4 + g ( Δx )( Δy )(1) = 0 Use 1 to represent the dimension along the z-direction. k (Tm +1,n + Tm −1,n + Tm ,n +1 + Tm ,n −1 − 4Tm ,n ) + g ( Δx ) = 0 2 g ( Δx ) 2 Tm +1,n + Tm −1,n + Tm ,n +1 + Tm ,n −1 − 4Tm ,n + =0 k Pradip Dutta HMT/M4/V1/May 05 29 Special cases (contd...) Radiation heat exchange with respect to the surrounding (assume no convection, no generation to simplify the derivation). Given surface emissivity ε, surrounding temperature Tsurr. qrad m,n m-1,n q2 q3 q4 m,n-1 m+1,n + + = Tsurr From energy balance concept: q2+q3+q4=qrad Pradip Dutta HMT/M4/V1/May 05 30 Special cases (contd...) Tm ,n −1 − Tm ,n ⎛ Δy ⎞ Tm −1,n − Tm ,n ⎛ Δy ⎞ Tm +1,n − Tm ,n 4 4 k⎜ + k ( Δx ) +k⎜ = εσ ( Δx ) Tm ,n − Tsurr ⎟ ⎟ Δx Δy Δx ⎝ 2 ⎠ ⎝ 2 ⎠ ( ) 4 4 k ( Tm −1,n + Tm +1,n + 2Tm ,n −1 − 4Tm ,n ) = 2εσ ( Δx ) (Tm ,n − Tsurr ) 2εσ ( Δx ) 4 εσ ( Δx ) 4 Tm −1,n + Tm +1,n + 2Tm ,n −1 − 4Tm ,n − Tm ,n = −2 Tsurr k k Non-linear term, can solve using the iteration method Pradip Dutta HMT/M4/V1/May 05 31 Module 4: Worked out problems Problem 1: A two dimensional rectangular plate is subjected to the uniform temperature boundary conditions shown. Using the results of the analytical solution for the heat equation, calculate the temperature at the midpoint (1, 0.5) by considering the first five nonzero terms of the infinite series that must be evaluated. Assess the error from using only the first three terms of the infinite series. Known: Two-dimensional rectangular plate to prescribed uniform temperature boundary conditions. Find: temperature at the midpoint using the exact solution considering the first five nonzero terms: Assess the error from using only the first three terms. Schematic: Assumptions: (1) Two-dimensional, steady-state conduction, (2) constant properties. Analysis: From analytical solution, the temperature distribution is θ ( x, y ) = T − T1 2 = T2 − T1 π ∑ n =1 ∞ nπ x sinh( nπ y / L ) ( −1) n +1 + 1 sin . n L sinh( nπ W / L ) Considering now the point (x, y) = (1.0, 0.5) and recognizing x/L =1/2, y/L=1/4 and W/L=1/2, the distribution has the form θ (1,0.5) = T − T1 2 = T2 − T1 π ∑ n =1 ∞ nπ sinh( nπ / L) (−1) n +1 + 1 sin . n L sinh( nπ / L) When n is even (2,4,6…..), the corresponding term is zero; hence we need only consider n=1,3,5,7 and 9 as the first five non-zero terms. π 3π 5π ⎫ ⎧ sinh sinh sinh ⎪ π⎞ 2 ⎛ 3π ⎞ 2 ⎛ 5π ⎞ ⎛ 4 + sin ⎜ ⎟ 4 + sin⎜ ⎟ 4 +⎪ ⎪ ⎪2 sin⎜ ⎟ ⎝ 2 ⎠ sinh π 3 ⎝ 2 ⎠ sinh 3π 5 ⎝ 2 ⎠ sinh 5π ⎪ ⎪ 2⎪ 2 2 2 ⎪ θ (1,0.5) = ⎨ ⎬ 7π 9π ⎪ π⎪ sinh sinh 2 ⎛ 7π ⎞ ⎪ 4 + 2 sin⎛ 9π ⎞ 4 ⎪ sin⎜ ⎟ ⎜ ⎟ ⎪ 7π 9 ⎝ 2 ⎠ 9π ⎪ 7 ⎝ 2 ⎠ sinh sinh ⎪ ⎪ ⎩ 2 2 ⎭ θ (1,0.5) = 2 π [0.755 + 0.063 + 0.008 − 0.001 + 0.000] = 0.445 Since θ = (T-T1)/ (T2-T1), it follows that T (1, 0.5) = θ(1, 0.5) (T2-T1) +T1=0.445(150-50) +50=94.5°C If only the first term of the series, Eq (2) is considered, the result will be θ(1, 0.5) =0.446 that is, there is less than a 0.2% effect. Problem 2: A long power transmission cable is buried at a depth (ground to cable centerline distance) of 2m. The cable is encased in a thin walled pipe of 0.1 m diameter, and to render the cable superconducting (essentially zero power dissipation), the space between the cable and pipe is filled with liquid nitrogen at 77 K. If the pipe is covered a super insulator(ki=0.005W/m.K) of 0.05-m thickness and the surface of the earth (kg=1.2W/m.K) is at 300K, what is the cooling load in W/m which must be maintained by a cryogenic refrigerator per unit pipe length. Known: Operating conditions of a buried superconducting cable. Find: required cooling load. Schematic: Tg=300K Kg=1.2W/m.K Insulation, ki=0.005W/m.K, Do=0.2m,Di=0.1m Liquid nitrogen.Tn=77K Cable Assumptions: (1) steady-state conditions, (2) constant properties, (3) two-dimensional conduction in soil, (4) one-dimensional conduction in insulation. Analysis: The heat rate per unit length is q' = q' = T g − Tn ' R g + R I' Z=2m T g − Tn [k g (2π / ln(4 z / Do ))] −1 + ln( Do / Di ) / 2π k i where table 4.1 have been used to evaluate the ground resistance. Hence, q' = q' = (300 − 77) K [(1.2W / m.K ) (2π / ln(8 / 0.2))] −1 + ln(2) / 2π × 0.005 W / m.K 223K (0.489 + 22.064)m.K / W q ' = 9.9W / m Comments: the heat gain is small and the dominant contribution to the thermal resistance is made by the insulation. Problem 3: Two parallel pipelines spaced 0.5 m apart are buried in soil having a thermal conductivity of 0.5W/m.K. the pipes have outer-diameters of 100 and 75 mm with surface temperatures of 175°C and 5°C, respectively. Estimate the heat transfer rate per unit length between the two pipe lines. Known: Surfaces temperatures of two parallel pipe lines buried in soil. Find: heat transfer per unit length between the pipe lines. Schematic: Assumptions: (1) steady state conditions, (2) two-dimensional conduction, (3) constant properties, (4) pipe lines are buried very deeply approximating burial in an infinite medium, (5) pipe length>> D1 or D2 and w>D1 or D2 Analysis: the heat transfer rate per length from the hot pipe to the cool pipe is q' = q S = k (T1 − T2 ) L L The shape factor S for this configuration is given in table 4.1 as S= 2πL 2 ⎛ 4 w 2 − D12 − D 2 cosh −1 ⎜ ⎜ 2 D1 D 2 ⎝ ⎞ ⎟ ⎟ ⎠ Substituting numerical values 4 × (0.5m) 2 − (0.1m) 2 − (0.075m) 2 S = 2π / cosh −1 = 2π / cosh −1 (65.63) L 2 × 0.1m × 0.075m S = 2π / 4.88 = 1.29. l hence, the heat rate per unit length is q ' = 1.29 × 0.5W / m.K (175 − 5)°C = 110W / m Comments: The heat gain to the cooler pipe line will be larger than 110W/m if the soil temperature is greater than 5°C. How would you estimate the heat gain if the soil were at 25°C? Problem 4: A furnace of cubical shape, with external dimensions of 0.35m, is constructed from a refractory brick (fireclay). If the wall thickness is 50mm, the inner surface temperature is 600°C, and the outer surface temperature is 75°C, calculate the heat loss from the furnace. Known: Cubical furnace, 350mm external dimensions, with 50mm thick walls. Find: The heat loss, q(W). Schematic: Assumptions: (1) steady-state conditions, (2) two-dimensional conduction, (3) constant properties. Properties: From table of properties, fireclay brick ( T = (T1 + T2 ) / 2 = 610 K ) : k ≈ 1.1W / m.K Analysis: using relations for the shape factor from table 4.1, Plane walls (6) Edges (12) Corners (8) SW = A 0.25 × 0.25m 2 = = 1.25m L 0.05m − S E = 0.54 D = 0.52 × 0.25m = 0.14m S C = 0.15 L = 0.15 × 0.05m = 0.008m The heat rate in terms of the shape factor is q = kS (T1 − T2 ) = k (6 S W + 12 S E + 8S C )(T1 − T2 ) q = 1.1 W (6 × 1.25m + 12 × 0.14m + 0.15 × 0.008m)(600 − 75)°C m.K q = 5.30kW Comments: Be sure to note that the restriction for SE and SC has been met. Problem 5: Consider nodal configuration 4 of table 4.2. Derive the finite-difference equation under steady-state conditions for the following situations. (a) The upper boundary of the external corner is perfectly insulated and the side boundary is subjected to the convection process (T∞, h) (b) Both boundaries of external corner are perfectly insulated. Known: External corner of a two-dimensional system whose boundaries are subjected to prescribed conditions. Find: finite-difference equations for these situations: (a) upper boundary is perfectly insulated and side boundary is subjected to a convection process, (b) both boundaries are perfectly insulated. Schematic: Assumptions: (1) steady-state conditions, (2) two-dimensional conduction, (3) constant properties, (4) no internal generation. Analysis: Consider the nodal point configuration shown in schematic and also as case 4, table 4.2. the control volume about the nodes shaded area above of unit thickness normal to the page has dimensions, (Δx/2)( Δy/2).1. The heat transfer processes at the surface of the CV are identified as q1, q2….perform an energy balance wherein the processes are expressed using the appropriate rate equations. With the upper boundary insulated and the side boundary and the side boundary subjected to a convection process, the energy balance has the form E in − E out = 0 . q1 + q 2 + q 3 + q 4 = 0 ⎛ Δy ⎞ Tm −1, n − Tm, n ⎛ Δy ⎞ ⎛ Δx ⎞ Tm, n −1 − Tm, n + k⎜ k⎜ .1⎟ . 1⎟ + h⎜ .1⎟ (T∞ − Tm, n ) + 0 = 0 Δx Δy ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ Letting Δx=Δy, and regrouping, find Tm, n −1 + Tm −1, n + hΔx ⎛ 1 hΔx ⎞ + 1⎟Tm, n = 0 T∞ − 2⎜ k ⎝2 k ⎠ with both boundaries insulated, the energy balance of Eq(2) would have q3=q4=0. the same result would be obtained by letting h=0 in the finite difference equation, Eq(3). The result is Tm, n −1 + Tm −1, n − 2Tm, n = 0 Comments: Note the convenience resulting formulating the energy balance by assuming that all the heat flow is into the node. Module 4: Short questions 1. For what kind of problems in multidimensional heat transfer are analytical solutions possible? Name some common analytical methods in steady state multidimensional heat transfer? 2. What are the limitations of analytical methods? In spite of their limitations, why are analytical solutions useful? 3. What is meant by “shape factor” in two-dimensional heat transfer analysis? What is the advantage of using such a method? Is there a shape factor in 1D heat transfer? 4. How do numerical solution methods differ from analytical methods? What are the advantages and disadvantages of numerical and analytical methods? 5. What is the basis of energy balance method in numerical analysis? How does it differ from the formal finite difference method using Taylor series approximation? For a specified nodal network, will these two methods result in the same or a different set of equations? 6. Consider a medium in which the finite difference formulation of a general interior node is given in its simplest form as & Tm −1 − 2Tm + Tm +1 g m + =0 2 k Δx (a) Is heat transfer in this medium steady or transient? (b) Is heat transfer one-, two-, or three-dimensional? (c) Is there heat generation in the medium? (d) Is the nodal spacing constant or variable? (e) Is thermal conductivity of the medium constant or variable? 7. Consider a medium in which the finite difference formulation of a general interior node is given in its simplest form as & g l2 Tleft + Ttop + Tright + Tbottom − 4Tnode + m = 0 k (a) Is heat transfer in this medium steady or transient? (b) Is heat transfer one-, two-, or three-dimensional? (c) Is there heat generation in the medium? (d) Is the nodal spacing constant or variable? (e) Is thermal conductivity of the medium constant or variable? 8. What is an irregular boundary? What is a practical way of handling irregular boundary surfaces with the finite difference method? 9. When performing numerical calculations of heat diffusion on a structured Cartesian grid in two dimensions, a simplified form of the equations states that the temperature at a node is simply the average of its four adjacent neighbours. What assumption is NOT required to allow this simplified form a) must have no heat generation b) must not be at a domain boundary c) must have uniform cell dimensions in both directions d) must be a solid medium MODULE 4 MULTI-DIMENSIONAL STEADY STATE HEAT CONDUCTION 4.1 Introduction We have, to this point, considered only One Dimensional, Steady State problems. The reason for this is that such problems lead to ordinary differential equations and can be solved with relatively ordinary mathematical techniques. In general the properties of any physical system may depend on both location (x, y, z) and time (τ). The inclusion of two or more independent variables results in a partial differential equation. The multidimensional heat diffusion equation in a Cartesian coordinate system can be written as: 1 ∂T ∂ 2 T ∂ 2T ∂ 2 T q ⋅ = + + + a ∂τ ∂x 2 ∂y 2 ∂z 2 k L (1) The above equation governs the Cartesian, temperature distribution for a three-dimensional unsteady, heat transfer problem involving heat generation. To solve for the full equation, it requires a total of six boundary conditions: two for each direction. Only one initial condition is needed to account for the transient behavior. For 2D, steady state (∂/ ∂t = 0) and without heat generation, the above equation reduces to: ∂ 2T ∂ 2T + =0 ∂x 2 ∂y 2 (2) Equation (2) needs 2 boundary conditions in each direction. There are three approaches to solve this equation: • • Analytical Method: The mathematical equation can be solved using techniques like the method of separation of variables. Graphical Method: Limited use. However, the conduction shape factor concept derived under this concept can be useful for specific configurations. (see Table 4.1 for selected configurations) Numerical Method: Finite difference or finite volume schemes, usually will be solved using computers. • Analytical solutions are possible only for a limited number of cases (such as linear problems with simple geometry). Standard analytical techniques such as separation of variables can be found in basic textbooks on engineering mathematics, and will not be reproduced here. The student is encouraged to refer to textbooks on basic mathematics for an overview of the analytical solutions to heat diffusion problems. In the present lecture material, we will cover the graphical and numerical techniques, which are used quite conveniently by engineers for solving multi-dimensional heat conduction problems. 4.2 Graphical Method: Conduction Shape Factor This approach applied to 2-D conduction involving two isothermal surfaces, with all other surfaces being adiabatic. The heat transfer from one surface (at a temperature T1) to the other surface (at T2) can be expressed as: q=Sk(T1-T2) where k is the thermal conductivity of the solid and S is the conduction shape factor. The shape factor can be related to the thermal resistance: q=S.k.(T1-T2)=(T1-T2)/(1/kS)= (T1-T2)/Rt where Rt = 1/(kS) is the thermal resistance in 2D. Note that 1-D heat transfer can also use the concept of shape factor. For example, heat transfer inside a plane wall of thickness L is q=kA(ΔT/L), where the shape factor S=A/L. Common shape factors for selected configurations can be found in Table 4.1 Example: A 10 cm OD uninsulated pipe carries steam from the power plant across campus. Find the heat loss if the pipe is buried 1 m in the ground is the ground surface temperature is 50 ºC. Assume a thermal conductivity of the sandy soil as k = 0.52 w/m K. Solution: T2 Z=1 m T1 The shape factor for long cylinders is found in Table 4.1 as Case 2, with L >> D: S = 2⋅π⋅L/ln(4⋅z/D) Where z = depth at which pipe is buried. S = 2⋅π⋅1⋅m/ln(40) = 1.7 m Then q' = (1.7⋅m)(0.52 W/m⋅K)(100 oC - 50 oC) q' = 44.2 W Table 4.1 Conduction shape factors for selected two-dimensional systems [q = Sk(T1-T2)] System Schematic T2 z Isothermal sphere buried in as finite medium T1 T2 Horizontal isothermal cylinder of length L buried in a semi finite medium z T1 D T2 Vertical cylinder in a semi finite medium T1 D 2πL 2 ⎛ 4 w 2 − D12 − D 2 cosh −1 ⎜ ⎜ 2 D1 D 2 ⎝ Restrictions Shape Factor 2πD 1 − D / 4z z>D/2 D L>>D L L>>D z>3D/2 2πL cosh −1 (2 z / D ) 2πL ln(4 z / D) L L>>D 2πL ln(4 L / D) Conduction between two cylinders of length L in infinite medium D1 T1 w D2 T2 L>>D1,D2 L>>w ⎞ ⎟ ⎟ ⎠ Horizontal circular cylinder of length L midway between parallel planes of equal length and infinite width ∞ z z T2 ∞ z>>D/2 L>>2 2πL ln(8 z / πD) T1 T2 D ∞ ∞ Circular cylinder of length L centered in a square solid of equal length T2 w T1 D W>D L>>w 2πL ln(1.08w / D ) Eccentric circular cylinder of length L in a cylinder of equal length D d T1 T2 z D>d L>>D 2πL ⎛ D 2 + d 2 − 4z 2 cosh −1 ⎜ ⎜ 2 Dd ⎝ ⎞ ⎟ ⎟ ⎠ 4.3 Numerical Methods Due to the increasing complexities encountered in the development of modern technology, analytical solutions usually are not available. For these problems, numerical solutions obtained using high-speed computer are very useful, especially when the geometry of the object of interest is irregular, or the boundary conditions are nonlinear. In numerical analysis, three different approaches are commonly used: the finite difference, the finite volume and the finite element methods. Brief descriptions of the three methods are as follows: The Finite Difference Method (FDM) This is the oldest method for numerical solution of PDEs, introduced by Euler in the 18th century. It's also the easiest method to use for simple geometries. The starting point is the conservation equation in differential form. The solution domain is covered by grid. At each grid point, the differential equation is approximated by replacing the partial derivatives by approximations in terms of the nodal values of the functions. The result is one algebraic equation per grid node, in which the variable value at that and a certain number of neighbor nodes appear as unknowns. In principle, the FD method can be applied to any grid type. However, in all applications of the FD method known, it has been applied to structured grids. Taylor series expansion or polynomial fitting is used to obtain approximations to the first and second derivatives of the variables with respect to the coordinates. When necessary, these methods are also used to obtain variable values at locations other than grid nodes (interpolation). On structured grids, the FD method is very simple and effective. It is especially easy to obtain higher-order schemes on regular grids. The disadvantage of FD methods is that the conservation is not enforced unless special care is taken. Also, the restriction to simple geometries is a significant disadvantage. Finite Volume Method (FVM) In this dissertation finite volume method is used. The FV method uses the integral form of the conservation equations as its starting point. The solution domain is subdivided into a finite number of contiguous control volumes (CVs), and the conservation equations are applied to each CV. At the centroid of each CV lies a computational node at which the variable values are to be calculated. Interpolation is used to express variable values at the CV surface in terms of the nodal (CV-center) values. As a result, one obtains an algebraic equation for each CV, in which a number of neighbor nodal values appear. The FVM method can accommodate any type of grid when compared to FDM, which is applied to only structured grids. The FVM approach is perhaps the simplest to understand and to program. All terms that need be approximated have physical meaning, which is why it is popular. The disadvantage of FV methods compared to FD schemes is that methods of order higher than second are more difficult to develop in 3D. This is due to the fact that the FV approach requires two levels of approximation: interpolation and integration. Finite Element Method (FEM) The FE method is similar to the FV method in many ways. The domain is broken into a set of discrete volumes or finite elements that are generally unstructured; in 2D, they are usually triangles or quadrilaterals, while in 3D tetrahedra or hexahedra are most often used. The distinguishing feature of FE methods is that the equations are multiplied by a weight function before they are integrated over the entire domain. In the simplest FE methods, the solution is approximated by a linear shape function within each element in a way that guarantees continuity of the solution across element boundaries. Such a function can be constructed from its values at the corners of the elements. The weight function is usually of the same form. This approximation is then substituted into the weighted integral of the conservation law and the equations to be solved are derived by requiring the derivative of the integral with respect to each nodal value to be zero; this corresponds to selecting the best solution within the set of allowed functions (the one with minimum residual). The result is a set of non-linear algebraic equations. An important advantage of finite element methods is the ability to deal with arbitrary geometries. Finite element methods are relatively easy to analyze mathematically and can be shown to have optimality properties for certain types of equations. The principal drawback, which is shared by any method that uses unstructured grids, is that the matrices of the linearized equations are not as well structured as those for regular grids making it more difficult to find efficient solution methods. 4.4 The Finite Difference Method Applied to Heat Transfer Problems: In heat transfer problems, the finite difference method is used more often and will be discussed here in more detail. The finite difference method involves: Establish nodal networks Derive finite difference approximations for the governing equation at both interior and exterior nodal points Develop a system of simultaneous algebraic nodal equations Solve the system of equations using numerical schemes The Nodal Networks: The basic idea is to subdivide the area of interest into sub-volumes with the distance between adjacent nodes by Δx and Δy as shown. If the distance between points is small enough, the differential equation can be approximated locally by a set of finite difference equations. Each node now represents a small region where the nodal temperature is a measure of the average temperature of the region. Example: Δx m,n+1 m-1,n Δy m, m+1, n x=mΔx, y=nΔy m,n-1 m-½,n intermediate points m+½,n Finite Difference Approximation: Heat Diffusion Equation: ∇2T + where α = & q 1 ∂T = , k α ∂t k is the thermal diffusivity ρ C PV ∂ = 0, ⇒ ∇2T = 0 ∂t First, approximated the first order differentiation & No generation and steady state: q=0 and at intermediate points (m+1/2,n) & (m-1/2,n) ∂T ΔT ≈ ∂x ( m +1/ 2,n ) Δx ∂T ΔT ≈ ∂x ( m −1/ 2,n ) Δx = ( m +1/ 2,n ) Tm +1,n − Tm ,n Δx Tm ,n − Tm −1,n Δx = ( m −1/ 2,n ) Next, approximate the second order differentiation at m,n ∂ 2T ∂x 2 ∂ 2T ∂x 2 ≈ m ,n ∂T / ∂x m +1/ 2,n − ∂T / ∂x m −1/ 2,n Δx Tm +1,n + Tm −1,n − 2Tm ,n ( Δx ) 2 ≈ m ,n Similarly, the approximation can be applied to the other dimension y ∂ 2T ∂y 2 ≈ m ,n Tm ,n +1 + Tm ,n −1 − 2Tm ,n ( Δy ) 2 Tm +1,n + Tm −1,n − 2Tm ,n Tm ,n +1 + Tm ,n −1 − 2Tm ,n ⎛ ∂ 2T ∂ 2T ⎞ + ⎜ 2 + 2⎟ ≈ ∂y ⎠ m ,n ( Δx ) 2 ( Δy ) 2 ⎝ ∂x To model the steady state, no generation heat equation: ∇2T = 0 This approximation can be simplified by specify Δx=Δy and the nodal equation can be obtained as Tm +1,n + Tm −1,n + Tm ,n +1 + Tm ,n −1 − 4Tm ,n = 0 This equation approximates the nodal temperature distribution based on the heat equation. This approximation is improved when the distance between the adjacent nodal points is decreased: ΔT ∂T ΔT ∂T Since lim( Δx → 0) = ,lim( Δy → 0) = Δx ∂x Δy ∂y Table 4.2 provides a list of nodal finite difference equation for various configurations. A System of Algebraic Equations • The nodal equations derived previously are valid for all interior points satisfying the steady state, no generation heat equation. For each node, there is one such equation. For example: for nodal point m=3, n=4, the equation is T2,4 + T4,4 + T3,3 + T3,5 - 4T3,4 =0 T3,4=(1/4)(T2,4 + T4,4 + T3,3 + T3,5) • Nodal relation table for exterior nodes (boundary conditions) can be found in standard heat transfer textbooks. • Derive one equation for each nodal point (including both interior and exterior points) in the system of interest. The result is a system of N algebraic equations for a total of N nodal points. Matrix Form The system of equations: a11T1 + a12T2 + L + a1N TN = C1 a21T1 + a22T2 + L + a2 N TN = C2 M a N 1T1 + a N 2T2 + L + a NN TN = CN M M M M A total of N algebraic equations for the N nodal points and the system can be expressed as a matrix formulation: [A][T]=[C] . ⎡ a11 a12 L a1N ⎤ ⎡ T1 ⎤ ⎡ C1 ⎤ ⎢a ⎢T ⎥ ⎢C ⎥ a22 L a2 N ⎥ ⎥ , T = ⎢ 2 ⎥ ,C = ⎢ 2 ⎥ where A= ⎢ 21 M M M ⎥ ⎢ M ⎢M ⎥ ⎢ M ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ a N 1 a N 2 L a NN ⎦ ⎣TN ⎦ ⎣C N ⎦ Tm,n +1 + Tm,n −1 + Tm +1,n + Tm −1, n − 4Tm, n = 0 2(Tm −1,n + Tm,n +1 ) + (Tm +1, n + Tm, n −1 ) + 2 hΔx ⎞ hΔx ⎛ T∞ − 2⎜ 3 + ⎟Tm, n = 0 k ⎠ k ⎝ 2(Tm −1, n + Tm, n +1 + Tm, n −1 ) + 2 hΔx ⎛ hΔx ⎞ T∞ − 2⎜ + 2 ⎟Tm, n = 0 k ⎝ k ⎠ Table 4.2 Summary of nodal finite-difference methods m-1,n y x m,n m,n-1 Case 4. Node at an external corner with convection h,∞ 2(Tm −1,n + Tm,n +1 + Tm,n −1 ) + 2 hΔx ⎛ hΔx ⎞ T∞ − 2⎜ + 1⎟Tm, n = 0 k ⎝ k ⎠ m,n+1 y T1 m-1,n a x x m,n-1 T2 m,n b y m+1,n 2 2 2 2 ⎛ 2 2⎞ Tm +1, n + Tm,n −1 + T1 + T2 − ⎜ + ⎟Tm,n = 0 a +1 b +1 a (a + 1) b)b + 1) ⎝a b⎠ Case 5. Node near a curved surface maintained at a non uniform temperature Numerical Solutions Matrix form: [A][T]=[C]. From linear algebra: [A]-1[A][T]=[A]-1[C], [T]=[A]-1[C] where [A]-1 is the inverse of matrix [A]. [T] is the solution vector. • Matrix inversion requires cumbersome numerical computations and is not efficient if the order of the matrix is high (>10) • Gauss elimination method and other matrix solvers are usually available in many numerical solution package. For example, “Numerical Recipes” by Cambridge University Press or their web source at www.nr.com. • For high order matrix, iterative methods are usually more efficient. The famous Jacobi & Gauss-Seidel iteration methods will be introduced in the following. Iteration General algebraic equation for nodal point: ∑ aijT j + aiiTi + j =1 i −1 j =i +1 ∑aT ij N j = Ci , Replace (k) by (k-1) for the Jacobi iteration (Example : a31T1 + a32T2 + a33T3 + L + a1N TN = C1 , i = 3) Rewrite the equation of the form: Ti ( k ) = Ci i −1 aij ( k ) − ∑ Tj − aii j =1 aii j =i +1 ∑a N aij ii T j( k −1) • • • (k) - specify the level of the iteration, (k-1) means the present level and (k) represents the new level. An initial guess (k=0) is needed to start the iteration. • By substituting iterated values at (k-1) into the equation, the new values at iteration (k) can be estimated The iteration will be stopped when max⎜Ti(k)-Ti(k-1) ⎟ε≤ , where ε specifies a predetermined value of acceptable error Module 5: Learning objectives • The primary objective of this chapter is to learn various methods of treating transient conduction which occurs in numerous engineering applications. The student should learn to recognize several classes of problems in unsteady conduction, and should be able to apply appropriate simplicity conditions before attempting to solve the problems. The simplest case is the lumped capacity condition, and the student should understand under what conditions one can apply this assumption. The first thing a student should do is calculate the Biot number. If this number is much less than unity, he/she may use the lumped capacitance method to obtain accurate results with minimum computational requirements. However, if the Biot number is not much less than unity, spatial effects must be considered and some other method must be used. Analytical results are available in convenient graphical and equation form for the plane wall, the finite cylinder, the sphere, and the semi-infinite solid. The student should know when and how to use these results. If geometrical complexities and /or the form of the boundary conditions preclude the use of analytical solutions, recourse must be made to an approximate numerical technique, such as the finite difference method or the finite volume method. • • • Module 5: Worked out problems Problem 1: A microwave oven operates on the principle that application of a high frequency field causes electrically polarized molecules in food to oscillate. The net effect is a uniform generation of thermal energy within the food, which enables it to be heated from refrigeration temperatures to 90º in as short a time as 30 s. Known: Microwave and radiant heating conditions for a slab of beef. Find: Sketch temperature distributions at specific times during heating and cooling. Schematic: Assumptions: (1) one-dimensional conduction in x, (2) uniform internal heat generation for microwave, (3) uniform surface heating for radiant oven, (4) heat loss from surface of meat to surroundings is negligible during the heat process, (5) symmetry about mid plane. Analysis: Comments: (1) With uniform generation and negligible surface heat loss, the temperature distribution remains nearly uniform during microwave heating. During the subsequent surface cooling, the maximum temperature is at the mid plane. (2) The interior of the meat is heated by conduction from the hotter surfaces during radiant heating, and the lowest temperature is at the mid plane. The situation is reversed shortly after cooling begins, and the maximum temperature is at the mid plane. Problem 2: The heat transfer coefficient for air flowing over a sphere is to be determined by observing the temperature- time history of a sphere fabricated from pure copper. The sphere which is 12.7 mm in diameter is at 66º C before it is inserted into an air stream having a temperature of 27ºC. A thermocouple on the outer surface of the sphere indicates 55ºC, 69 s after the sphere is inserted into an air stream. Assume, and then justify, that the sphere behaves as a space-wise isothermal object and calculate the heat transfer coefficient. Known: The temperature-time history of a pure copper sphere in air stream. Find: The heat transfer coefficient between and the air stream Schematic: Assumptions: (1) temperature of sphere is spatially uniform, (2) negligible radiation exchange, (3) constant properties. Properties: From table of properties, pure copper (333K): =8933 kg/m3, cp=389 J/kg.K, k=389W/m.K Analysis: the time temperature history is given by Rt = ⎛ θ (t ) t = exp⎜ − ⎜ RC θi t t ⎝ ⎞ ⎟ ⎟ ⎠ 1 hAs Where C t = ρVc p As = πD 2 V= θ = T − T∞ πD 3 6 Recognize that when t = 69 s ⎛ t θ (t ) (55 − 27 )° C = = 0.718 = exp⎜ − ⎜ τ ° θi (66 − 27 ) C ⎝ t ⎞ ⎛ 69 s ⎞ ⎟ = exp⎜ − ⎟ ⎟ ⎜ τ ⎟ t ⎠ ⎠ ⎝ And noting that τ t = Rt C t find τ t = 208s Hence, h= ρVc p As τ t = 8933kg / m 3 (π 0.0127 3 m 3 / 6)389 J / kg.K π 0.0127 2 m 2 × 208s h = 35.3 W / m 2 .K Comments: Note that with Lc =D0 / 6 Bi = hLc 0.0127 = 35.3W / m 2 .K × m / 398 W / m.K = 1.88 × 10 − 4 k 6 Hence Bi<0.1 and the spatially isothermal assumption is reasonable. Problem 3: A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and enclosed alternating layers of the storage material and the flow passage. Each layer of the storage material is aluminium slab of width=0.05m which is at an initial temperatures of 25ºC. consider the conditions for which the storage unit is charged by passing a hot gas through the passages, with the gas temperature and convection coefficient assumed to have constant values of T=600ºC and h=100W/m2.K throughout the channel how long will it take to achieve 75% of the maximum possible energy storage? What is the temperature of the aluminium at this time? Known: Configuration, initial temperature and charging conditions of a thermal energy storage unit. Find: Time required achieving 75% of maximum possible energy storage. Temperature of storage medium at this time. Schematic: Assumptions: (1) one-dimensional conduction, (2) constant properties, (3) negligible heat exchange with surroundings. Properties: From any table of properties: Aluminum, pure (T 600K=327C): k=231W/m.K, c= 1033 J/kg.K, =2702kg/m3. Analysis: recognizing the characteristic length is the half thickness, find hL 100 W / m 2 .K × 0.025m = = 0.011 k 231 W / m.K Bi = Hence, the lumped capacitance method may be used. Q = ( ρVc )θ i [1 − exp( −t / τ i )] = − ΔE st − ΔE st , max = ( ρVc )θ i Dividing eq. (1) and (2), the condition sought is for ΔE st / ΔE st , max = 1 − exp( −t / τ th ) = 0.75 Solving for τ th and substituting numerical values, find τ th = ρVc hAs = ρLc h = 2702kg / m 3 × 0.025m ×1033J / kg.K 100W / m 2 .K = 698s Hence, the time required is -exp (-t/698s) =-0.25 T − T∞ = exp(−t / τ th ) Ti − T∞ T = T∞ + (Ti − T∞ ) exp(−t / τ th ) = 600°C − (575°C ) exp(−968 / 698) or t=968s. T=456°C Comments: for the prescribed temperatures, the property temperatures dependence is significant and some error is incurred by assuming constant properties. However, selecting at 600K was reasonable for this estimate. Problem 4: A one-dimensional plane wall with a thickness of 0.1 m initially at a uniform temperature of 250C is suddenly immersed in an oil bath at 30°C. assuming the convection heat transfer coefficient for the wall in the bath is 500 W/m2.K. Calculate the surface temperature of the wall 9 min after immersion. The properties of the wall are k=50 W/m.K, ρ=7835 kg/m3, and c=465 J/kg.K. Known: plane wall, initially at a uniform temperature, is suddenly immersed in an oil bath and subjected to a convection cooling process. Find: Surface temperature of the wall nine minutes after immersion, T (L, 9 min). Schematic: K=50W/m.k =7835kg/m3 c=465J/kg.K T =30 C h=500W/m2.K T ,h T(L,qmin) 2L=0.1m x Ti=T(x,0)=250 C Assumptions: The Biot number for the plane wall is Bi = hLc 500 W / m 2 .K × 0.05 m = = 0.50 k 50 W / m.K Since Bi>0.1, lumped capacitance analysis is not appropriate. Fo = αt L2 = (k / ρ c)t L2 = 50W / m.K / 7835 kg / m 3 × 465 J / kg .K × (9 × 60 ) s = 2 .96 ( 0 .05 m ) 2 And Bi-1=1/0.50 = 2, find θ 0 T (0, t ) − T∞ = ≈ 0.3 θi Ti − T∞ We know that Bi-1=1/0.50 = 2 and for X/L=1, find θ (1, t ) ≈ 0.8 θ0 By combining equation, θ (1, t ) =0.8( θ 0 ) = 0.8(0.3 θ i ) =0.24 θ i Recalling that θ = T ( L, t ) − T∞ and θ i = Ti − T∞ , it follows that T ( L, t ) = T∞ + 0.24 (Ti − T∞ ) = 30°C + 0.24 (250 − 30)°C = 83°C Comments: (1) note that figure provides a relationship between the temperature at any x/L and the centerline temperature as a function of only the Biot number. Fig applies to the centerline temperature which is a function of the Biot number and the Fourier number. The centerline temperature at t=9min follows from equation with T (0, t ) − T∞ = 0.3(Ti − T∞ ) = 0.3(250 − 30)°C = 66°C (2) Since F0>=0.2, the approximate analytical solution for θ* is valid. From table with Bi=0.50, and ζ1 =0.6533 rad and C1=1.0701. Substituting numerical values into equations θ*=0.303 and θ*(1, FO) =0.240 From this value, find T (L, 9 min) =83°C which is identical to graphical result. Problem 5: A long cylinder of 30mm diameter, initially at a uniform temperature of 1000K, is suddenly quenched in a large, constant-temperature oil bath at 350K. The cylinder properties are k=1.7W/m.K, c=1600 J/kg.K, and ρ=400 kg/m3, while the convection coefficient is 50W/m2.K. Calculate the time required for the surface cylinder to reach 500K. Known: A long cylinder, initially at a uniform temperature, is suddenly quenched in large oil bath. Find: time required for the surface to reach 500K. Schematic: Assumptions: (1) one dimensional radial conduction, (2) constant properties Analysis: check whether lumped capacitance methods are applicable. BI c = hLc h(r0 / 2) 50W / m 2 .K (0.015m / 2) = = = 0.221 k k 1.7W / m.K Since BI c >0.1, method is not suited. Using the approximate series solutions for the infinite cylinder, θ * (r * , Fo) = C1 exp(−ς 12 Fo) × J o (ς 12 r * ) Solving for Fo and letting =1, find Fo = − 1 ⎡ θ* ⎤ ln ⎢ ⎥ 2 ⎢ C1 J o (ς 1 ) ⎥ ⎣ ⎦ T (ro , t o ) − T∞ Ti − T∞ = (500 − 350) K = 0.231 (1000 − 350) K ς 12 where θ * (1, F0 ) = From table, Bi=0.441, find ς 1 =0.8882 rad and C1=1.1019. From table find Jo ( ς 12 ) =0.8121. Substituting numerical values into equation, Fo = − 1 (0.8882) 2 ln[0.231 / 1.1019 × 0.8121] = 1.72 From the definition of the Fourier number, Fo = αt ro 2 = Fo .ro 2 ρc k t = 1.72 (0.015m) 2 × 400 kg / m 3 ×1600 J / kg.K / 1.7W / m.K = 145 s Comments: (1) Note that Fo>=0.2, so approximate series solution is appropriate. (2) Using the Heisler chart, find Fo as follows. With Bi-1=2.27, find from fir r/ro=1 that θ (ro , t ) T (ro , t ) − T∞ = ≈ 0.8 or θo T (0, t )0 − T∞ hence T (0, t ) = T∞ + 1 [T (ro , t ) − T∞ ] = 537 K 0.8 θo (537 − 350) K = = 0.29 θ i (1000 − 350) K From fig, with θo =0.29 and Bi-1 =2.27, find Fo ≈ 1.7 and eventually obtain t ≈ 144s. θi Problem 6: In heat treating to harden steel ball bearings (c=500 J/kg.K, ρ=7800 kg/m3, k=50 W/m.K) it is desirable to increase the surface temperature for a short time without significantly warming the interior of the ball. This type of heating can be accomplished by sudden immersion of the ball in a molten salt bath with T∞=1300 K and h= 5000 W/m2.K. Assume that any location within the ball whose temperature exceeds 1000 K will be hardened. Estimate the time required to harden the outer millimeter of a ball of diameter 20 mm if its initial temperature is 300 K. Known: A ball bearing is suddenly immersed in a molten salt bath; heat treatment to harden occurs at locations with T>1000K. Find: time required to harden outer layer of 1mm. Schematic: Assumptions: (1) one-dimensional radial conduction, (2) constant properties, (3) Fo≥0.2. Analysis: since any location within the ball whose temperature exceeds 1000K will be hardened, the problem is to find the time when the location r=9mm reaches 1000K. Then a 1mm outer layer is hardened. Using the approximate series solution, begin by finding the Biot number. Bi = hro 5000 W / m 2 .K (0.020m / 2) = = 1.00 k 50 W / m.K Using the appropriate solution form for a sphere solved for Fo , find Fo = − 1 ς 12 ⎡ ⎤ 1 ln ⎢θ * / C1 sin(ς 1 r * )⎥ * ς 1r ⎢ ⎥ ⎣ ⎦ From table, with Bi=1.00, for the sphere find ς 1 =1.5708 rad and C1 =1.2732. with r* =r/ro= (9mm/10mm)=0.9, substitute numerical values. Fo = − ⎡ (1000 − 1300) K ⎤ 1 ln ⎢ / 1.2732 sin(1.5708 × 0.9rad )⎥ = 0.441 1.5708 × 0.9 (1.5708) ⎣ (300 − 1300) K ⎦ 1 2 From the definition of the Fourier number with α=k/ρc, t = Fo r 2o = Fo .r 2 ρc α kg J ⎛ 0.020 ⎞ = 0.441× ⎜ / 50W / m.K = 3.4 s ⎟ 7800 3 × 500 k 2 ⎠ kg .K m ⎝ 2 Comments: (1) note the very short time required to harden the ball. At this time it can be easily shown the center temperature is T(0,3.4s)=871K. (2) The Heisler charts can also be used. From fig, with Bi-1=1.0 and r/r0=0.9, read θ/θo =0.69(±0.03). since θ = T − T∞ = 1000 − 1300 = −300 K θ i = Ti − T∞ = −1000 K It follows that θ = 0.30 θi since θ θ θo = . θi θo θi then θ θ = 0.69 o , θi θi And then θ o 0.30 = = 0.43(±0.02) θ i 0.69 θo =0.43, Bi-1=1.0, read FO =0.45( ± 0.3) and t=3.5 ( ± 0.2) s. θi From fig at Note the use of tolerances assigned as acceptable numbers dependent upon reading the charts to ±5%. Problem 7: The convection coefficient for flow over a solid sphere may be determined by submerging the sphere, which is initially at 25°C, into the flow, which is at 75°C and measuring its surface temperature at some time during the transient heating process. The sphere has a diameter of 0.1m, and its thermal conductivity and thermal diffusivity are 15 W/m.K and 105 2 m /s, respectively. If the convection coefficient is 300W/m2.K, at what time will a surface temperature of 60°C be recorded? Known: Initial temperatures and properties of solid sphere. Surface temperatures after immersion in a fluid of prescribed temperatures and convection coefficient. Find: The process time Schematic: D=0.1m T(ro,t)=60°C T8 =75 C h=300W/m2.K K=15W/m.K a=10-5m2/s Ti=25 C Assumptions: (1) one-dimensional, radial conduction, (2) constant properties. Analysis: the Biot number is Bi = h(r0 / 3) 300W / m 2 .K (0.05m / 3) = = 0.333 k 15W / m.K Hence the lumped capacitance methods should be used. From equation T − T∞ Ti − T∞ = C1 exp(−ς 12 Fo ) sin(ς 1 r * ) ς 1r * At the surface, r * =1. from table , for Bi=1.0, ς 1 =1.5708 rad and C1=1.2732. hence, 60 − 75 sin 90° +Exp = 0.30 = 1.2732 exp (−1.5708 2 Fo ) 25 − 75 1.5708 (-2.467F0) =0.370 Fo = αt r02 = 0.403 (0.05m) 2 10 −5 m 2 / s t=100s Comments: Use of this technique to determine h from measurement of T (ro) at a prescribed t requires an iterative solution of the governing equations. Module 5: Short questions 1. What is lumped capacity analysis? When is it applicable? What is the physical significance of the Biot number? 2. The Biot number is used when considering a solid body subject to convection in a surrounding fluid. It is a comparison of a) Convection to conduction in the surrounding fluid b) Conduction in the surrounding fluid to conduction in the solid c) Convection at the solid surface to conduction within the solid d) The thermal diffusivity in the solid to the kinematic viscosity in the fluid e) None of the above 3. Consider heat transfer between two identical hot solid bodies and the surrounding air. The first solid is cooled by a fan, while the second one is cooled by natural convection in air. For which case is the lumped capacity assumption more applicable? 4. Consider a hot boiled potato kept on a plate and cooled by natural convection in air. During the first minute, the temperature drops by 10ºC. During the second minute, will the temperature drop be more than, less than or same as that during the first minute? 5. In what medium will the lumped capacity assumption more likely to be valid: in air or in water? 6. Consider a sphere and a cylinder of equal volume and made of copper. Both are heated to the same temperature and then kept in air for cooling. Which one is likely to cool faster? 7. A block of metal is cooled in a water bath. Its unsteady temperature is considered uniform and is thus modelled using a lumped capacitance method. The product of the block’s resistance to convection and its lumped thermal capacitance is a) Bi b) Nu c) Fo d) τ e) None of the above 8. In transient heat transfer analysis, when is it proper to treat an actual cylinder as an infinitely long one, and when is it not? 9. Why are the transient temperature charts prepared using non-dimensionalised quantities such as the Biot and the Fourier numbers and not the actual variables such as thermal conductivity and time? 10. What is the physical significance of the Fourier number? Will the Fourier number of a specific transient heat transfer problem double if the time is doubled? 11. What is a semi-infinite medium? Give examples of solid bodies that can be treated as semi-infinite mediums for the purpose of transient heat transfer studies? Under what conditions can a plane wall be treated as a semi-infinite medium? Multiple choice questions: 1) When modelling the unsteady conduction in a semi-infinite slab with convection at the surface, there is no geometric length scale with which to construct a Biot number. The appropriate length scale for this is therefore 0 .5 (a) (αt ) (b) αt 0 .5 (c) (ht ) (d) ρct (e) none of the above 2) The Biot number is used when considering a solid body subject to convection in a surrounding fluid. It is a comparison of a) Convection to conduction in the surrounding fluid b) Conduction in the surrounding fluid to conduction in the solid c) Convection at the solid surface to conduction within the solid d) The thermal diffusivity in the solid to the kinematic viscosity in the fluid e) None of the above 3) A block of metal is cooled in a water bath. Its unsteady temperature is considered uniform and is thus modelled using a lumped capacitance method. The product of the block’s resistance to convection and its lumped thermal capacitance is a) Bi b) Nu c) Fo d) τ e) None of the above UNSTEADY HEAT TRANSFER Many heat transfer problems require the understanding of the complete time history of the temperature variation. For example, in metallurgy, the heat treating process can be controlled to directly affect the characteristics of the processed materials. Annealing (slow cool) can soften metals and improve ductility. On the other hand, quenching (rapid cool) can harden the strain boundary and increase strength. In order to characterize this transient behavior, the full unsteady equation is needed: ∂T 1 ∂T 2 ρc = k ∇ T , or = ∇ 2T α ∂t ∂t k where α = is the thermal diffusivity ρc “A heated/cooled body at Ti is suddenly exposed to fluid at T∞ with a known heat transfer coefficient . Either evaluate the temperature at a given time, or find time for a given temperature.” Q: “How good an approximation would it be to say the bar is more or less isothermal?” A: “Depends on the relative importance of the thermal conductivity in the thermal circuit compared to the convective heat transfer coefficient”. Biot No. Bi •Defined to describe the relative resistance in a thermal circuit of the convection compared hLc Lc / kA Internal conduction resistance within solid Bi = = = k 1 / hA External convection resistance at body surface Lc is a characteristic length of the body Bi→0: No conduction resistance at all. The body is isothermal. Small Bi: Conduction resistance is less important. The body may still be approximated as isothermal (purple temp. plot in figure) Lumped capacitance analysis can be performed. Large Bi: Conduction resistance is significant. The body cannot be treated as isothermal (blue temp. plot in figure). Transient heat transfer with no internal resistance: Lumped Parameter Analysis Valid for Bi<0.1 Solid Total Resistance= Rexternal + Rinternal GE: dT hA =− (T − T∞ ) mc p dt Solution: BC: T (t = 0 ) = Ti let Θ = T − T∞ , therefore hA dΘ =− Θ mc p dt Lumped Parameter Analysis Θi = Ti − T∞ Θ hA ln =− t Θi mc p Θ =e Θi − hA t mc p −t T − T∞ =e Ti − T∞ mc p hA - To determine the temperature at a given time, or - To determine the time required for the temperature to reach a specified value. Note: Temperature function only of time and not of space! Lumped Parameter Analysis T − T∞ hA T= = exp( − t) ρcV T0 − T∞ α hA ⎛ hLc ⎞⎛ k ⎞ 1 1 t =⎜ ⎟⎜ ⎟ ⎜ ρc ⎟ L L t = Bi L 2 t ρcV ⎝ k ⎠⎝ ⎠ c c c Thermal diffusivity: ⎛ k ⎞ α ≡⎜ ⎟ ⎜ ρc ⎟ ⎝ ⎠ (m² s-1) Lumped Parameter Analysis Define Fo as the Fourier number (dimensionless time) Fo ≡ α Lc t and Biot number Bi ≡ hLC 2 k The temperature variation can be expressed as T = exp(-Bi*Fo) where Lc is a characteristic length scale : realte to the size of the solid invloved in the problem r for example , Lc = o (half - radius) when the solid is a cylinder. 2 r Lc = o (one - third radius) when the solid is sphere 3 Lc = L (half thickness) when the solid is aplane wall with a 2L thickness Spatial Effects and the Role of Analytical Solutions The Plane Wall: Solution to the Heat Equation for a Plane Wall with Symmetrical Convection Conditions 1 ∂T ∂ 2T ⋅ = 2 a ∂τ ∂x T ( x, 0) = Ti ∂T ∂x =0 x =0 ∂T −k ∂x = h[T ( L, t ) − T∞ ] x =l The Plane Wall: Note: Once spatial variability of temperature is included, there is existence of seven different independent variables. How may the functional dependence be simplified? •The answer is Non-dimensionalisation. We first need to understand the physics behind the phenomenon, identify parameters governing the process, and group them into meaningful nondimensional numbers. Dimensionless temperature difference: Dimensionless coordinate: Dimensionless time: The Biot Number: θ T − T∞ θ = = θ i Ti − T∞ * x* = t* = αt L 2 x L = Fo hL Bi = k solid The solution for temperature will now be a function of the other non-dimensional quantities θ * = f ( x * , Fo, Bi ) * ∞ n =1 Exact Solution: θ = ∑ C n exp(− ζ n2 Fo )cos(ζ n x * ) ζ n tan ζ n = Bi 4 sin ζ n Cn = 2ζ n + sin(2ζ n ) The roots (eigenvalues) of the equation can be obtained from tables given in standard textbooks. The One-Term Approximation Fo > 0.2 Variation of mid-plane temperature with time Fo ( x * = 0) θ 0* = T − T∞ ≈ C1 exp(− ζ 12 Fo ) Ti − T∞ From tables given in standard textbooks, one can obtain as a function of Bi. Variation of temperature with location ( x * ) C1 and ζ1 and time ( Fo ): θ * = θ 0* = cos(ζ 1 x * ) Change in thermal energy storage with time: ΔE st = −Q ⎛ sin ζ 1 ⎞ * ⎟θ 0 Q = Q0 ⎜1 − ⎜ ζ1 ⎟ ⎝ ⎠ Q0 = ρcV (Ti − T∞ ) Graphical Representation of the One-Term Approximation: The Heisler Charts Midplane Temperature: Temperature Distribution Change in Thermal Energy Storage Assumptions in using Heisler charts: •Constant Ti and thermal properties over the body •Constant boundary fluid T∞ by step change •Simple geometry: slab, cylinder or sphere Radial Systems Long Rods or Spheres Heated or Cooled by Convection Similar Heisler charts are available for radial systems in standard text books. Important tips: Pay attention to the length scale used in those charts, and calculate your Biot number accordingly. Unsteady Heat Transfer in Semi-infinite Solids Solidification process of the coating layer during a thermal spray operation is an unsteady heat transfer problem. As we discuss earlier, thermal spray process deposits thin layer of coating materials on surface for protection and thermal resistant purposes, as shown. The heated, molten materials will attach to the substrate and cool down rapidly. The cooling process is important to prevent the accumulation of residual thermal stresses in the coating layer. Unsteady Heat Transfer in Semi-infinite Solids(contd…) liquid Coating with density ρ, latent heat of fusion: hsf S(t) solid δ Substrate, k, α Example As described in the previous slide, the cooling process can now be modeled as heat loss through a semi-infinite solid. (Since the substrate is significantly thicker than the coating layer) The molten material is at the fusion temperature Tf and the substrate is maintained at a constant temperature Ti. Derive an expression for the total time that is required to solidify the coating layer of thickness δ. Example Assume the molten layer stays at a constant temperature Tf throughout the process. The heat loss to the substrate is solely supplied by the release of the latent heat of fusion. From energy balance: hsf Δm (solidified mass during Δt) = ΔQ = q" AΔt (energy input) dm = q" A, where m = ρV = ρAS, dt where S is solidified thickness dS ρ = q" Heat transfer from dt h sf the molten material to the substrate (q=q”A) Example (contd...) Identify that the previous situation corresponds to the case of a semiinfinite transient heat transfer problem with a constant surface temperature boundary condition. This boundary condition can be modeled as a special case of convection boundary condition case by setting h=∞, therefore, Ts=T∞). Example (contd...) If the surface temperature is Ts and the initial temperature of the bolck is Ti , the analytical solution of the problem can be found: The temperature distribution and the heat transfer into the block are: T(x,t)-Ts ⎛ x ⎞ = erf ⎜ ⎟ , where erf( ) is the Gaussian error function. Ti − Ts ⎝ 2 αt ⎠ 2 w − v2 It is defined as erf(w)= ∫ e dv π 0 qs"(t)= k(Ts − Ti ) παt Example (contd...) From the previous equation dS k(Tf − Ti ) k(Tf − Ti ) dt ρ hsf =q"= , and ∫ dS = dt πα t ρ hsf πα ∫ t 0 0 t δ 2k(Tf − Ti ) πα ⎛ δρ hsf δ (t ) = t , therefore, δ ∝ t. Cooling time t = 2 ⎜ 4k ⎜ T f − Ti ρ hsf πα ⎝ Use the following values to calculate: k=120 W/m.K, α=4×10-5 m2/s, ρ=3970 kg/m3, and hsf=3.577 ×106 J/kg, Tf=2318 K, Ti=300K, and δ=2 mm ⎞ ⎟ ⎟ ⎠ 2 Example (contd…) 2k(Tf − Ti ) δ (t ) = t = 0.00304 t ρhsf πα 0.004 0.003 δ( t )0.002 0.001 0 0 0.2 0.4 t 0.6 0.8 1 δ(t) ∝ t1/2 Therefore, the layer solidifies very fast initially and then slows down as shown in the figure Note: we neglect contact resistance between the coating and the substrate and assume temperature of the coating material stays the same even after it solidifies. To solidify 2 mm thickness, it takes 0.43 seconds. Example (contd…) What will be the substrate temperature as it varies in time? The temperature distribution is: T ( x, t ) − TS ⎛ x = erf ⎜ Ti − TS ⎝ 2 αt ⎞ ⎟, ⎠ x ⎞ ⎛ ⎜ 79.06 ⎟ t⎠ ⎝ ⎛ x ⎞ T ( x, t ) = 2318 + (300 − 2318)erf ⎜ ⎟ = 2318 − 2018erf ⎝ 2 αt ⎠ Example (contd…) For a fixed distance away from the surface, we can examine the variation of the temperature as a function of time. Example, 1 cm deep into the substrate the temperature should behave as: T ( x = 0.01, t ) = 2318 − 2018erf 79.06 x 0.79 = 2318 − 2018erf t t Example (contd...) 2000 T1( t ) T2( t ) T3( t ) 1600 1200 800 400 0 0 2 4 t Time 6 8 10 At x=1 cm, the temperature rises almost instantaneously at a very fast rate. A short time later, the rate of temp. increase slows down significantly since the energy has to distribute to a very large mass. At deeper depth (x=2 & 3 cm), the temperature will not respond to the surface condition until much later. Temperature x=1 cm x=2 cm x=3 cm Example (contd...) We can also examine the spatial temperature distribution at any given time, say at t=1 second. T ( x, t = 1) = 2318 − 2018erf 79.06 x = 2318 − 2018erf 79.06 x t 3000 Heat penetrates into the substrate as shown for different T1( x ) 2000 time instants. T2( x ) It takes more than 5 seconds for the energy to transfer to a T3( x ) 1000 depth of 5 cm into the substrate The slopes of the temperature 0 0 0.01 0.02 0.03 0.04 0.05 profiles indicate the amount of x conduction heat transfer at that distance (m) t=1 s. instant. t=5 s. t=10 s. Temperature (K) Numerical Methods for Unsteady Heat Transfer Unsteady heat transfer equation, no generation, constant k, twodimensional in Cartesian coordinate: 1 ∂T ∂ 2T ∂ 2T = 2 + 2 α ∂t ∂x ∂y We have learned how to discretize the Laplacian operator into system of finite difference equations using nodal network. For the unsteady problem, the temperature variation with time needs to be discretized too. To be consistent with the notation from the book, we choose to analyze the time variation in small time increment Δt, such that the real time t=pΔt. The time differentiation can be approximated as: ∂T ∂t m ,n + TmP,n1 − TmP,n ≈ , while m & n correspond to nodal location Δt such that x=mΔx, and y=nΔy as introduced earlier. Finite Difference Equations m,n+1 m-1,n m,n m+1, n m,n-1 From the nodal network to the left, the heat equation can be written in finite difference form: Finite Difference Equations (contd…) P +1 P P P P P P P 1 Tm ,n − Tm ,n Tm +1,n + Tm −1,n − 2Tm ,n Tm ,n +1 + Tm ,n −1 − 2Tm ,n = + 2 Δt α ( Δx ) ( Δy ) 2 Assume Δx=Δy and the discretized Fourier number Fo= + TmP,n1 = Fo (TmP+1,n + TmP−1,n + TmP,n +1 + TmP,n −1 ) + (1 − 4 Fo)TmP,n αΔt ( Δx ) 2 This is the explicit, finite difference equation for a 2-D, unsteady heat transfer equation. The temperature at time p+1 is explicitly expressed as a function of neighboring temperatures at an earlier time p Nodal Equations Some common nodal configurations are listed in table for your reference. On the third column of the table, there is a stability criterion for each nodal configuration. This criterion has to be satisfied for the finite difference solution to be stable. Otherwise, the solution may be diverging and never reach the final solution. Nodal Equations (contd…) For example, Fo≤1/4. That is, αΔt/(Δx)2 ≤1/4 and Δt≤(1/4α)(Δx)2. Therefore, the time increment has to be small enough in order to maintain stability of the solution. This criterion can also be interpreted as that we should require the coefficient for TPm,n in the finite difference equation be greater than or equal to zero. Question: Why this can be a problem? Can we just make time increment as small as possible to avoid it? Finite Difference Solution Question: How do we solve the finite difference equation derived? First, by specifying initial conditions for all points inside the nodal network. That is to specify values for all temperature at time level p=0. Important: check stability criterion for each points. From the explicit equation, we can determine all temperature at the next time level p+1=0+1=1. The following transient response can then be determined by marching out in time p+2, p+3, and so on. Example Example: A flat plate at an initial temperature of 100 deg. is suddenly immersed into a cold temperature bath of 0 deg. Use the unsteady finite difference equation to determine the transient response of the temperature of the plate. L(thickness)=0.02 m, k=10 W/m.K, α=10×10-6 m2/s, h=1000 W/m2.K, Ti=100°C, T∞=0°C, Δx=0.01 m x Bi=(hΔx)/k=1, Fo=(αΔt)/(Δx)2=0.1 1 There are three nodal points: 1 interior and two 3 2 exterior points: For node 2, it satisfies the case 1 configuration in table. T2P +1 = Fo(T1P + T3P + T2P + T2P ) + (1 − 4 Fo)T2P = Fo(T1P + T3P ) + (1 − 2 Fo)T2P = 0.1(T1P + T3P ) + 0.8T2P Stability criterion: 1-2Fo ≥ 0 or Fo=0.1 ≤ 1 ,it is satisfied 2 Example For nodes 1 & 3, they are consistent with the case 3 in table. Node 1: T1P +1 = Fo(2T2P + T1P + T1P + 2 BiT∞ ) + (1 − 4 Fo − 2 BiFo)T1P = Fo(2T2P + 2 BiT∞ ) + (1 − 2 Fo − 2 BiFo)T1P = 0.2T2P + 0.6T1P Node 3: T3P +1 = 0.2T2P + 0.6T3P Stability criterion: (1-2Fo-2BiFo) ≥ 0, System of equations T1P +1 = 0.2T2P + 0.6T1P T2P +1 = 0.1(T1P + T3P ) + 0.8T2P T3P +1 = 0.2T2P + 0.6T3P 1 ≥ Fo(1 + Bi ) = 0.2 and it is satisfied 2 Use initial condition, T10 = T20 = T30 = 100, T11 = 0.2T20 + 0.6T10 = 80 T21 = 0.1(T10 + T30 ) + 0.8T20 = 100 T31 = 0.2T20 + 0.6T30 = 80 1 1 Marching in time, T11 = T3 = 80, T2 = 100 T12 = 0.2T21 + 0.6T11 = 0.2(100) + 0.6(80) = 68 T22 = 0.1(T11 + T31 ) + 0.8T21 = 0.1(80 + 80) + 0.8(100) = 96 T32 = 0.2T21 + 0.6T31 = 0.2(100) + 0.6(80) = 68, and so on MODULE 5 UNSTEADY STATE HEAT CONDUCTION 5.1 Introduction To this point, we have considered conductive heat transfer problems in which the temperatures are independent of time. In many applications, however, the temperatures are varying with time, and we require the understanding of the complete time history of the temperature variation. For example, in metallurgy, the heat treating process can be controlled to directly affect the characteristics of the processed materials. Annealing (slow cool) can soften metals and improve ductility. On the other hand, quenching (rapid cool) can harden the strain boundary and increase strength. In order to characterize this transient behavior, the full unsteady equation is needed: 1 ∂T ∂ 2 T ∂ 2T ∂ 2 T q ⋅ = + + + a ∂τ ∂x 2 ∂y 2 ∂z 2 k L (5.1) k is the thermal diffusivity. Without any heat generation and considering spatial ρc variation of temperature only in x-direction, the above equation reduces to: where α = 1 ∂T ∂ 2 T ⋅ = a ∂τ ∂x 2 (5.2) For the solution of equation (5.2), we need two boundary conditions in x-direction and one initial condition. Boundary conditions, as the name implies, are frequently specified along the physical boundary of an object; they can, however, also be internal – e.g. a known temperature gradient at an internal line of symmetry. 5.2 Biot and Fourier numbers In some transient problems, the internal temperature gradients in the body may be quite small and insignificant. Yet the temperature at a given location, or the average temperature of the object, may be changing quite rapidly with time. From eq. (5.1) we can note that such could be the case for large thermal diffusivity α . A more meaningful approach is to consider the general problem of transient cooling of an object, such as the hollow cylinder shown in figure 5.1. Ts > T∞ Fig. 5.1 For very large ri, the heat transfer rate by conduction through the cylinder wall is approximately ⎛ T − Ti ⎞ ⎛ Ti − Ts ⎞ (5.3) q ≈ − k ( 2πro l )⎜ s ⎟ ⎜ r − r ⎟ = k ( 2πro l )⎜ L ⎟ ⎝ ⎠ ⎝ o i ⎠ where l is the length of the cylinder and L is the material thickness. The rate of heat transfer away from the outer surface by convection is q = h (2πro l )(Ts − T∞ ) (5.4) where h is the average heat transfer coefficient for convection from the entire surface. Equating (5.3) and (5.4) gives Ti − Ts h L = = Biot number Ts − T∞ k The Biot number is dimensionless, and it can be thought of as the ratio Bi = resistance to internal heat flow resistance to external heat flow (5.5) Whenever the Biot number is small, the internal temperature gradients are also small and a transient problem can be treated by the “lumped thermal capacity” approach. The lumped capacity assumption implies that the object for analysis is considered to have a single massaveraged temperature. In the derivation shown above, the significant object dimension was the conduction path length, L = ro − ri . In general, a characteristic length scale may be obtained by dividing the volume of the solid by its surface area: V (5.6) L= As Using this method to determine the characteristic length scale, the corresponding Biot number may be evaluated for objects of any shape, for example a plate, a cylinder, or a sphere. As a thumb rule, if the Biot number turns out to be less than 0.1, lumped capacity assumption is applied. In this context, a dimensionless time, known as the Fourier number, can be obtained by multiplying the dimensional time by the thermal diffusivity and dividing by the square of the characteristic length: dimensionless time = αt L2 = Fo (5.7) 5.3 Lumped thermal capacity analysis The simplest situation in an unsteady heat transfer process is to use the lumped capacity assumption, wherein we neglect the temperature distribution inside the solid and only deal with the heat transfer between the solid and the ambient fluids. In other words, we are assuming that the temperature inside the solid is constant and is equal to the surface temperature. q = h As (T − T∞ ) Solid T(t) ρ, c, V T∞ h Fig. 5.2 The solid object shown in figure 5.2 is a metal piece which is being cooled in air after hot forming. Thermal energy is leaving the object from all elements of the surface, and this is shown for simplicity by a single arrow. The first law of thermodynamics applied to this problem is ⎛ heat out of object ⎞ ⎛ decrease of internal thermal ⎞ ⎟=⎜ ⎜ ⎟ ⎟ ⎜ energy of object during time dt ⎟ ⎜ during time dt ⎠ ⎠ ⎝ ⎝ Now, if Biot number is small and temperature of the object can be considered to be uniform, this equation can be written as h As [T (t ) − T∞ ]dt = − ρcVdT h As dT =− dt (T − T∞ ) ρcV (5.8) or, (5.9) Integrating and applying the initial condition T (0) = Ti , hA T (t ) − T∞ =− s t Ti − T∞ ρcV Taking the exponents of both sides and rearranging, T (t ) − T∞ = e −bt Ti − T∞ where h As (1/s) b= ρcV ln (5.10) (5.11) (5.12) Note: In eq. 5.12, b is a positive quantity having dimension (time)-1. The reciprocal of b is usually called time constant, which has the dimension of time. Question: What is the significance of b? Answer: According to eq. 5.11, the temperature of a body approaches the ambient temperature T∞ exponentially. In other words, the temperature changes rapidly in the beginning, and then slowly. A larger value of b indicates that the body will approach the surrounding temperature in a shorter time. You can visualize this if you note the variables in the numerator and denominator of the expression for b. As an exercise, plot T vs. t for various values of b and note the behaviour. Rate of convection heat transfer at any given time t: & Q (t ) = hAs [T (t ) − T∞ ] Total amount of heat transfer between the body and the surrounding from t=0 to t: Q = mc [T (t ) − Ti ] Maximum heat transfer (limit reached when body temperature equals that of the surrounding): Q = mc [T∞ − Ti ] 5.4 Spatial Effects and the Role of Analytical Solutions If the lumped capacitance approximation can not be made, consideration must be given to spatial, as well as temporal, variations in temperature during the transient process. The Plane Wall: Solution to the Heat Equation for a Plane Wall with Symmetrical Convection Conditions • For a plane wall with symmetrical convection conditions and constant properties, the heat equation and initial/boundary conditions are: 1 ∂T ∂ 2 T ⋅ = a ∂τ ∂x 2 T ( x, 0) = Ti ∂T ∂x −k =0 x =0 ∂T ∂x = h[T ( L, t ) − T∞ ] x =l Note: Once spatial variability of temperature is included, there is existence of seven different independent variables. T = T ( x, t , Ti , T∞ , h, k , α ) How may the functional dependence be simplified? • The answer is Non-dimensionalisation. We first need to understand the physics behind the phenomenon, identify parameters governing the process, and group them into meaningful non-dimensional numbers. Non-dimensionalisation of Heat Equation and Initial/Boundary Conditions: The following dimensionless quantities are defined. θ T − T∞ Dimensionless temperature difference: θ * = = θ i Ti − T∞ x Dimensionless coordinate: x* = L αt Dimensionless time: t * = 2 = Fo L hL The Biot Number: Bi = k solid The solution for temperature will now be a function of the other non-dimensional quantities θ * = f ( x * , Fo, Bi ) Exact Solution: θ * = ∑ C n exp(− ζ n2 Fo )cos(ζ n x * ) ∞ n =1 4 sin ζ n ζ n tan ζ n = Bi 2ζ n + sin(2ζ n ) The roots (eigenvalues) of the equation can be obtained from tables given in standard textbooks. Cn = Fo > 0.2 The One-Term Approximation * Variation of mid-plane ( x = 0) temperature with time ( Fo ) T − T∞ θ 0* = ≈ C1 exp(− ζ 12 Fo ) Ti − T∞ From tables given in standard textbooks, one can obtain C1 and ζ 1 as a function of Bi. Variation of temperature with location ( x * ) and time ( Fo ): θ * = θ 0* = cos(ζ 1 x * ) Change in thermal energy storage with time: ΔE st = −Q ⎛ sin ζ 1 ⎞ * ⎟θ 0 Q = Q0 ⎜1 − ⎜ ζ1 ⎟ ⎝ ⎠ Q0 = ρcV (Ti − T∞ ) Can the foregoing results be used for a plane wall that is well insulated on one side and convectively heated or cooled on the other? Can the foregoing results be used if an isothermal condition (Ts ≠ Ti ) is instantaneously imposed on both surfaces of a plane wall or on one surface of a wall whose other surface is well insulated? Graphical Representation of the One-Term Approximation: The Heisler Charts Midplane Temperature: Temperature Distribution Change in Thermal Energy Storage § Assumptions in using Heisler charts: l Constant Ti and thermal properties over the body l Constant boundary fluid T∞ by step change l Simple geometry: slab, cylinder or sphere § Limitations: l Far from edges l No heat generation (Q=0) l Relatively long after initial times (Fo >0.2) Radial Systems Long Rods or Spheres Heated or Cooled by Convection Bi = hr0 / k Fo = αt / r02 Similar Heisler charts are available for radial systems in standard text books. Important tips: Pay attention to the length scale used in those charts, and calculate your Biot number accordingly. 5.5 Numerical methods in transient heat transfer: The Finite Volume Method Considering the steady convection-diffusion equation: ∂ ( ρφ ) + div ( ρφ u) = div (Γ grad φ ) + Sφ ∂t The time and control volume integrations give: t + Δt t + Δt t + Δt ⎛ t + Δt ∂ ⎞ ⎛ ⎛ ⎞ ⎞ ⎛ ⎞ ⎜ ∫ ⎟dV + ∫ ⎜ ∫ n ⋅ (ρφ u )dA ⎟dt = ∫ ⎜ ∫ n ⋅ (Γ grad φ )dA ⎟dt + ∫ ⎜ ∫ Sφ dV ⎟dt (ρφ )dt ⎟ ∫ ⎜ ∂t ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ CV ⎝ t t ⎝A t ⎝A t ⎝ CV ⎠ ⎠ ⎠ ⎠ • Unsteady one-dimensional heat conduction: ∂T ∂ ⎛ ∂T ⎞ ρc = ⎜k ⎟+S ∂t ∂x ⎝ ∂x ⎠ t W (δx)WP w P Δx (δx)PE e E x Consider the one-dimensional control volume. Integration over the control volume and over a time interval gives: t + Δt t + Δt t + Δt ⎛ ⎛ ∂T ⎞ ⎞ ⎛ ∂ ⎛ ∂T ⎞ ⎞ ⎛ ⎞ ⎜ ∫ ⎜ ρc dV ⎟dt = ∫ ⎜ ∫ ⎜ k dV ⎟dt + ∫ ⎜ ∫ SdV ⎟dt ⎟ ⎟ ∫ ⎜ CV ⎝ ∂t ⎟ ⎟ ⎜ ∂x ⎝ ∂x ⎠ ⎜ ⎟ ⎠ ⎠ t ⎝ t ⎝ cv t ⎝ CV ⎠ ⎠ Re-written e t + Δt t + Δt t + Δt ⎛ ⎛ ∂T ⎞ ⎛ ∂T ⎞ ⎞ ∂T ⎞ ⎟dV = ∫ ⎜ ⎛ kA ⎜ ∫ ρc ⎟dt + ∫ (S ΔV )dt ∫ ⎜ t ∂t dt ⎟ ⎜ ⎜ ∂x ⎟ − ⎜ kA ∂x ⎟ ⎟ ⎠e ⎝ ⎠w ⎠ w⎝ t ⎝⎝ t ⎠ If the temperature at a node is assumed to prevail over the whole control volume, applying the central differencing scheme, we have: t + Δt t + Δt ⎛⎛ T − TW ⎞ ⎞ T − TP ⎞ ⎛ ⎟ ⎟dt + ∫ (S ΔV )dt ⎟ − ⎜ kw A P ρc (TP − TP0 )ΔV = ∫ ⎜ ⎜ k e A E ⎜⎜ δx PE ⎟ ⎜ δxWP ⎟ ⎟ ⎠ ⎝ ⎠⎠ t ⎝⎝ t An assumption about the variation of TP, TE and Tw with time. By generalizing the approach by means of a weighting parameter θ between 0 and 1: t + Δt IT = ∫ T dt = [θT P t P + (1 − θ )TP0 Δt ⎞ ⎛ TP − TW ⎟ − ⎜ kw ⎟ ⎜ δxWP ⎠ ⎝ ⎞⎤ ⎟⎥ ⎟ ⎠⎦ ⎞⎤ ⎟⎥ + S Δx ⎟ ⎠⎦ ] Therefore, ⎡⎛ T − T ⎛T −T 0 ⎞ ρc⎜ P P ⎟Δx = θ ⎢⎜ k e E P ⎜ ⎜ Δt ⎟ δx PE ⎝ ⎠ ⎣⎝ 0 ⎡⎛ T 0 − TP0 ⎞ ⎛ TP0 − TW ⎟ − ⎜ kw + (1 − θ ) ⎢⎜ k e E ⎜ δx PE ⎟ ⎜ δxWP ⎠ ⎝ ⎣⎝ Re-arranging: ⎡ Δx ⎛ k k +θ⎜ e + w ⎢ ρc ⎜ δx ⎝ PE δxWP ⎣ Δt ⎞⎤ k k 0 ⎟⎥TP = e θTE + (1 − θ )TE0 + w θTW + (1 − θ )TW ⎟ δx PE δxWP ⎠⎦ ⎡ Δx k k ⎤ + ⎢ ρc − (1 − θ ) e − (1 − θ ) w ⎥TP0 + S Δx δx PE δxWP ⎦ ⎣ Δt Compared with standard form: 0 0 a PTP = aW θTW + (1 − θ )TW + a E θTE + (1 − θ )TE0 + a P − (1 − θ )aW − (1 − θ )a E TP0 + b where [ ] [ ] 0 a P = θ (aW + a E ) + a P Δx 0 a P = ρc Δt kw aW = δxWP k aE = e δx PE [ ] [ ] [ ] b = S Δx When θ = 0, the resulting scheme is “explicit”. When 0 < θ ≤ 1, the resulting scheme is “implicit”. When θ = 1, the resulting scheme is “fully implicit”. When θ = 1/2, the resulting scheme is “the Crank-Nicolson”. Explicit scheme 0 0 a PTP = aW θTW + (1 − θ )TW + a E θTE + (1 − θ )TE0 + a P − (1 − θ )aW − (1 − θ )a E TP0 + b [ ] [ ] [ ] The source term is linearised as b = S u + S pT p0 and set θ = 0 The explicit discretisation: 0 0 a PTP = aW TW + a E TE0 + a P − (aW + a E ) TP0 + S u where 0 aP = aP Δx 0 a P = ρc Δt kw aW = δxWP k aE = e δx PE The scheme is based on backward differencing and its Taylor series trunction error accuracy is first-order with respect to times. All coefficient must be positive in the discretised equation: 0 a P − (aW + a E − S P ) > 0 or k k Δx ρc − ( w + e ) > 0 Δt δxWP δx PE or Δx 2 k ρc > Δt Δx or ( Δx ) 2 Δt < ρc 2k It becomes very expensive to improve spatial accuracy. This method is not recommended for general transient problems. Nevertheless, provided that the time step size is chosen with care, the explicit scheme described above is efficient for simple conduction calculations. [ ] Crank-Nicolson scheme 0 0 a PTP = aW θTW + (1 − θ )TW + a E θTE + (1 − θ )TE0 + a P − (1 − θ )aW − (1 − θ )a E TP0 + b Set θ = 1/2 0 ⎛ T + TW ⎞ ⎡ 0 a E aW ⎤ 0 ⎛ T + TE0 ⎞ ⎟ + aW ⎜ W a PTP = a E ⎜ E ⎜ 2 ⎟ ⎟ ⎜ 2 ⎟ + ⎢a P − 2 − 2 ⎥TP + b ⎦ ⎝ ⎠ ⎠ ⎣ ⎝ where 1 1 0 a P = (a E + aW ) + a P − S P 2 2 Δx 0 a P = ρc Δt kw aW = δxWP k aE = e δx PE 1 b = S u + S pT p0 2 The method is implicit and simultaneous equations for all node points need to be solved at each time step. All coefficient must be positive in the discretised equation: a + aW 0 aP > E 2 or (Δx) 2 Δt < ρc k This is only slightly less restrictive than the explicit method. The Crank-Nicolson method is based on central differencing and hence it is second-order accurate in time. So, it is normally used in conjunction with spatial central differencing. [ ] [ ] [ ] The fully implicit scheme 0 0 a PTP = aW θTW + (1 − θ )TW + a E θTE + (1 − θ )TE0 + a P − (1 − θ )aW − (1 − θ )a E TP0 + b Set θ = 1 0 a PTP = a E TE + aW TW + a PTP0 where 0 a P = a P + a E + aW − S P Δx 0 a P = ρc Δt k aW = w δxWP k aE = e δx PE A system of algebraic equations must be solved at each time level. The accuracy of the scheme is first-order in time. The time marching procedure starts with a given initial field of temperature T0. The system is solved after selecting time step Δt. All coefficients are positive, which makes the implicit scheme unconditionally stable for any size of time step. The implicit method is recommended for general purpose transient calculations because of its robustness and unconditional stability. [ ] [ ] [ ] Module 6: Learning objectives • In this chapter attempts have been made to develop, in a logical fashion, the mathematical and physical bases of convection transport. To test your comprehension of the material, you should challenge yourself with appropriate question. What are the velocity, thermal, and concentration boundary layers? Under what conditions do they develop, and why are they of interest to the engineer? How do laminar and turbulent boundary layer is laminar or turbulent? There are numerous processes that affect momentum, energy, and species transfer in a boundary layer. What are they? How are they represented mathematically? What are the boundary layer approximations, and in what way do they alter the conservation equations? What are the relevant dimensionless groups for the various boundary layers? How may they be physically interpreted? How will the use of these groups facilitate convection calculations? How are velocity, thermal, and concentration boundary layer behaviors analogous? How the effects of turbulence may be treated in a boundary layer analysis? Finally, what is the central problem of convection? In this chapter convection correlations that may be used to estimate convection transfer rates for a variety of external flow conditions have been developed. For simple surfaces geometries these results may be derived from a boundary layer analysis, but in most cases they are obtained from generalization based on experiment. The student should know when and how to use various expressions, and he/she should be familiar with the general methodology of a convection calculation. Internal flow is encountered in numerous applications, and it is important to appreciate its unique features. What is the nature of fully developed flow, and how does it differ from flow in the entry region? How does the Prandtl number influence boundary layer development in the entry region? How do thermal conditions in the fluid depend on the surface condition? For example, how do the surface and mean temperatures vary with x for the case of uniform surface heat flux? Or how do mean temperatures and the surface heat flux vary for the case of uniform surface temperature? The student must be able to perform engineering calculations that involve an energy balance and appropriate convection correlations. The methodology involves determining whether the flow is laminar or turbulent and establishing the length of the entry region. After deciding whether one is interested in local conditions (at a particular axial location) or in average conditions (for the entire tube), the convection correlation may be selected and used with the appropriate form of the energy balance to solve the problem. Several features that complicate internal flows have not been considered in this chapter. For example situation may exists for which there is a prescribed axial variation in Ts or qsn, rather than uniform surface conditions. Among other things, such a variation would preclude the existence of a fully developed region. There may also exist surface roughness effects, circumferential heat flux or temperature variations, widely varying fluid properties, or transition flow conditions. • • • • • • We have considered convective flows that originate in part or exclusively form buoyancy forces, and we have introduced the dimensionless parameters needed to characterize such flows. The student should be able to discern when free convection effects are important and to quantify the associated heat transfer rates. An assortment of empirical correlations has been provided for this purpose. In short, by the end of the module, the student should have a fundamental understanding of the convection process and its mathematical description. Convection • Heat transfer in the presence of a fluid motion on a solid surface •Various mechanisms at play in the fluid: - advection → physical transport of the fluid - diffusion → conduction in the fluid - generation → due to fluid friction •But fluid directly in contact with the wall does not move relative to it; hence direct heat transport to the fluid is by conduction in the fluid only. U∞ y U∞ u(y) q” Ts y y T∞ T(y) T∞ T(y) Ts ′′ q conv But ∂T ∂y ∂T = −kf ∂y ⎞ ⎟ = h (Ts − T∞ ) ⎟ ⎠ y =0 U∞ ⎞ ⎟ ⎟ ⎠ y = 0 depends on the whole fluid motion, and both fluid flow and heat transfer equations are needed Convection Free or natural convection (induced by buoyancy forces) Convection forced convection (driven externally) May occur with phase change (boiling, condensation) Heat transfer rate q = h( Ts-T ∞ )W h=heat transfer coefficient (W /m2K) (h is not a property. It depends on geometry ,nature of flow, thermodynamics properties etc.) Typical values of h (W/m2K) Free convection: gases: 2 - 25 liquid: 50 - 100 Forced convection: gases: 25 - 250 liquid: 50 - 20,000 Boiling/Condensation: 2500 -100,000 Convection rate equation U∞ y U∞ u(y) q” Ts Main purpose of convective heat transfer analysis is to determine: • flow field • temperature field in fluid • heat transfer coefficient, h q’’=heat flux = h(Ts - T∞) q’’ = -k(∂T/ ∂y)y=0 Hence, h = [-k(∂T/ ∂y)y=0] / (Ts - T∞) y T∞ T(y) The expression shows that in order to determine h, we must first determine the temperature distribution in the thin fluid layer that coats the wall. Classes of convective flows: • extremely diverse • several parameters involved (fluid properties, geometry, nature of flow, phases etc) • systematic approach required • classify flows into certain types, based on certain parameters • identify parameters governing the flow, and group them into meaningful non-dimensional numbers • need to understand the physics behind each phenomenon Common classifications: A. Based on geometry: External flow / Internal flow B. Based on driving mechanism Natural convection / forced convection / mixed convection C. Based on number of phases Single phase / multiple phase D. Based on nature of flow Laminar / turbulent How to solve a convection problem ? • Solve governing equations along with boundary conditions • Governing equations include 1. conservation of mass 2. conservation of momentum 3. conservation of energy • In Conduction problems, only (3) is needed to be solved. Hence, only few parameters are involved • In Convection, all the governing equations need to be solved. ⇒ large number of parameters can be involved Forced convection: Non-dimensional groupings • Nusselt No. Nu = hx / k = (convection heat transfer strength)/ (conduction heat transfer strength) • Prandtl No. Pr = ν/α = (momentum diffusivity)/ (thermal diffusivity) • Reynolds No. Re = U x / ν = (inertia force)/(viscous force) Viscous force provides the dampening effect for disturbances in the fluid. If dampening is strong enough ⇒ laminar flow Otherwise, instability ⇒ turbulent flow ⇒ critical Reynolds number δ δ Laminar Turbulent FORCED CONVECTION: external flow (over flat plate) An internal flow is surrounded by solid boundaries that can restrict the development of its boundary layer, for example, a pipe flow. An external flow, on the other hand, are flows over bodies immersed in an unbounded fluid so that the flow boundary layer can grow freely in one direction. Examples include the flows over airfoils, ship hulls, turbine blades, etc. •Fluid particle adjacent to the solid surface is at rest T Ts •These particles act to retard the motion of adjoining layers •⇒ boundary layer effect Momentum balance: inertia forces, pressure gradient, viscous forces, body forces q x Energy balance: convective flux, diffusive flux, heat generation, energy storage h=f(Fluid, Vel ,Distance,Temp) Hydrodynamic boundary layer One of the most important concepts in understanding the external flows is the boundary layer development. For simplicity, we are going to analyze a boundary layer flow over a flat plate with no curvature and no external pressure variation. U∞ Dye streak U∞ U∞ U∞ laminar transition turbulent Boundary layer definition Boundary layer thickness (δ): defined as the distance away from the surface where the local velocity reaches to 99% of the free-stream velocity, that is u(y=δ)=0.99U∞. Somewhat an easy to understand but arbitrary definition. Boundary layer is usually very thin: δ/x usually << 1. Hydrodynamic and Thermal boundary layers As we have seen earlier,the hydrodynamic boundary layer is a region of a fluid flow, near a solid surface, where the flow patterns are directly influenced by viscous drag from the surface wall. 0<u<U, 0<y<δ The Thermal Boundary Layer is a region of a fluid flow, near a solid surface, where the fluid temperatures are directly influenced by heating or cooling from the surface wall. 0<t<T, 0<y<δt The two boundary layers may be expected to have similar characteristics but do not normally coincide. Liquid metals tend to conduct heat from the wall easily and temperature changes are observed well outside the dynamic boundary layer. Other materials tend to show velocity changes well outside the thermal layer. Effects of Prandtl number, Pr δ δT δT δ, δT δ Pr >>1 ν >> α e.g., oils Pr = 1 ν=α e.g., air and gases have Pr ~ 1 (0.7 - 0.9) T − TW u similar to U∞ T∞ − TW (Reynold’s analogy) Pr <<1 ν << α e.g., liquid metals Boundary layer equations (laminar flow) • Simpler than general equations because boundary layer is thin T∞ U∞ y x U∞ δ δT TW ∂u ∂v + =0 ∂x ∂y • Equations for 2D, laminar, steady boundary layer flow Conservation of mass : ∂u ∂u dU ∞ ∂ ⎛ ∂u ⎞ = U∞ + ⎜ν Conservation of x - momentum : u + v ⎜ ∂y ⎟ ⎟ ∂x ∂y dx ∂y ⎝ ⎠ ∂T ∂T ∂ ⎛ ∂T ⎞ +v = ⎜α Conservation of energy : u ⎜ ∂y ⎟ ⎟ ∂x ∂y ∂y ⎝ ⎠ • Note: for a flat plate, U ∞ is constant , hence dU ∞ =0 dx Boundary layer thic kness Exact solutions: Blasius δ 4.99 x = Re x τw 0.664 = Skin friction coefficien t C f = 1 2 Re x 2 ρU ∞ ⎛ ⎜ Re = U ∞ x , τ = μ ∂u w ⎜ x ∂y ν ⎝ 1 1.328 Average drag coefficien t C D = ∫ C f dx = L0 Re L Local Nusselt number Average Nusselt number Nu x = 0.339 Re x Pr 1 2 1 2 1 3 1 ⎞ ⎟ ⎟ y =0 ⎠ U∞L ⎞ ⎛ ⎜ Re L = ⎟ ν ⎠ ⎝ L N u = 0.678 Re L Pr 3 Heat transfer coefficient • Local heat transfer coefficient: Nu x k 0 .339 k Re x Pr = hx = x x 2 1 1 3 • Average heat transfer coefficient: h = Nu k L 0 .678 k Re L Pr = L 2 1 1 3 • Recall: q w = h A(Tw − T∞ ), heat flow rate from wall • Film temperature, Tfilm For heated or cooled surfaces, the thermophysical properties within the boundary layer should be selected based on the average T film = 1 (Tw + T∞ ) 2 temperature of the wall and the free stream; Heat transfer coefficient Convection Coefficient, h. U∞ Thermal Boundary Layer, δt x Hydrodynamic Boundary Layer, δ Laminar and turbulent b.l. Laminar Region Turbulent Region Turbulent boundary layer * Re x increases with x. Beyond a critical value of Reynolds number ( Re x = Re xc ), the flow becomes transitio nal and eventually turbulent . Re xc = U ∞ xc ( For flow over flat plate, xc ≈ 5 × 10 5 ) ν * Turbulent b.l. equations are similar to laminar ones, but infinitely more difficult to solve. * We will mainly use correlatio ns based on experiment al data : C f = 0.059 Re − 0.2 x C D = 0.072 Re L − (Re x > 5 × 10 5 ) 1 0.072 Re 0.8 − 1.328 Re 0.5 xc xc Re L 1 3 1 1 3 ( ) ) Nu k etc. x Nu x = 0.029 Re 0.8 Pr x N u = 0.036 Re 0.8 Pr 3 − Pr L (0.036 Re 0 .8 xc − 0.664 Re 0.5 xc * Calculate heat trans fer coefficien t in usual way : h = Laminar Boundary Layer Development 1 δ( x ) 0.5 • Boundary layer growth: δ ∝ √x • Initial growth is fast • Growth rate dδ/dx ∝ 1/√x, decreasing downstream. 0 0.5 x 1 0 10 τ w( x ) 5 • Wall shear stress: τw ∝ 1/√x • As the boundary layer grows, the wall shear stress decreases as the velocity gradient at the wall becomes less steep. 0 0.5 x 0 Example Determine the boundary layer thickness, the wall shear stress of a laminar water flow over a flat plate. The freestream velocity is 1 m/s, the kinematic viscosity of the water is 10-6 m2/s. The density of the water is 1,000 kg/m3. The transition Reynolds number Re=Ux/ν=5×105. Determine the distance downstream of the leading edge when the boundary transitions to turbulent. Determine the total frictional drag produced by the laminar and turbulent portions of the plate which is 1 m long. If the free stream and plate temperatures are 100 °C and 25 °C, respectively, determine the heat transfer rate from the plate. δ ( x) = 5 νx U∞ = 5 × 10 −3 x ( m ). Therefore, for a 1m long plate, the boundary layer grows by 0.005(m), or 5 mm, a very thin layer. The wall shear stress, τ w 2 0.332 ρU ∞ = = 0.332U ∞ Re x ρμU ∞ x = 0.0105 ( Pa ) x The transition Reyn olds number: Re = U ∞ xtr ν = 5 × 10 5 , xtr = 0.5( m ) Example (cont..) The total frictional drag is equal to the integration of the wall shear stress: FD = xtr ∫τ 0 w (1)dx = xtr ∫ 0.332U 0 ∞ ρμU ∞ x 2 0.664 ρU ∞ dx = = 0.939( N ) Re xtr Define skin friction coefficient: C f Cf = τw 0.664 for a laminar boundary layer. = 2 1 ρU Re x ∞ 2 Forced convection over exterior bodies • Much more complicated. •Some boundary layer may exist, but it is likely to be curved and U∞ will not be constant. • Boundary layer may also separate from the wall. • Correlations based on experimental data can be used for flow and heat transfer calculations • Reynolds number should now be based on a ρ U∞D Re D = characteristic diameter. μ • If body is not circular, the equivalent diameter Dh is used 4 × Area Dh = Perimeter CD = Drag force 2 1 ρ U ∞ Anormal 2 ; Nu = hD k ; h = Nuk D Flow over circular cylinders Pr .62 N u = C Re m .25 D Prs Re D 1 − 40 40 - 10 3 C m 0.75 0.4 0.51 0.5 10 3 - 2 × 10 5 0.26 0.6 2 × 10 5 - 10 6 0.08 0.7 All properties at free stream temperature, Prs at cylinder surface temperature Flow over circular cylinders Flow patterns for cross flow over a cylinder at various Reynolds numbers FORCED CONVECTION: Internal flow • Thermal conditions Laminar or turbulent entrance flow and fully developed thermal condition e.g. pipe flow Thermal entrance region, xfd,t For laminar flows the thermal entrance length is a function of the Reynolds number and the Prandtl number: xfd,t/D ≈ 0.05ReDPr, where the Prandtl number is defined as Pr = ν/α and α is the thermal diffusitivity. For turbulent flow, xfd,t ≈ 10D. Thermal Conditions • For a fully developed pipe flow, the convection coefficient is a constant and is not varied along the pipe length. (as long as all thermal and flow properties are constant also.) h(x) constant x xfd,t • Newton’s law of cooling: q”S = hA(TS-Tm) Question: since the temperature inside a pipe flow is not constant, what temperature we should use. A mean temperature Tm is defined. Energy Transfer Consider the total thermal energy carried by the fluid as ∫ ρVC TdA = (mass flux) (internal energy) v A Now image this same amount of energy is carried by a body of fluid with the same mass flow rate but at a uniform mean temperature Tm. Therefore Tm can be defined as Tm = ∫ ρVC TdA v A & mC v Consider Tm as the reference temperature of the fluid so that the total heat transfer between the pipe and the fluid is governed by the Newton’s cooling law as: qs”=h(Ts-Tm), where h is the local convection coefficient, and Ts is the local surface temperature. Note: usually Tm is not a constant and it varies along the pipe depending on the condition of the heat transfer. Energy Balance Example: We would like to design a solar water heater that can heat up the water temperature from 20° C to 50° C at a water flow rate of 0.15 kg/s. The water is flowing through a 5 cm diameter pipe and is receiving a net solar radiation flux of 200 W per unit length (meter). Determine the total pipe length required to achieve the goal. Example (cont.) Questions: (1) How do we determine the heat transfer coefficient, h? There are a total of six parameters involving in this problem: h, V, D, ν, kf, cp. The last two variables are thermal conductivity and the specific heat of the water. The temperature dependence is implicit and is only through the variation of thermal properties. Density ρ is included in the kinematic viscosity, ν=μ/ρ. According to the Buckingham theorem, it is possible for us to reduce the number of parameters by three. Therefore, the convection coefficient relationship can be reduced to a function of only three variables: Nu=hD/kf, Nusselt number, Re=VD/ν, Reynolds number, and Pr=ν/α, Prandtl number. This conclusion is consistent with empirical observation, that is Nu=f(Re, Pr). If we can determine the Reynolds and the Prandtl numbers, we can find the Nusselt number, hence, the heat transfer coefficient, h. Convection Correlations ⇒ Laminar, fully developed circular pipe flow: hD Nu D = = 4.36, when q s " = constant, (page 543, ch. 10-6, ITHT) kf Nu D = 3.66, when Ts = constant, (page 543, ch. 10-6, ITHT) Note: the therma conductivity should be calculated at Tm . ⇒ Fully developed, turbulent pipe flow: Nu = f(Re, Pr), Nu can be related to Re & Pr experimentally, as shown. ln(Nu) Fixed Pr ln(Nu) Fixed Re slope m slope n ln(Re) ln(Pr) Empirical Correlations Dittus-Boelter equation: Nu D = 0.023 Re 4 / 5 Pr n , (eq 10-76, p 546, ITHT) where n = 0.4 for heating (T s > Tm ), n = 0.3 for cooling (Ts < Tm ). The range of validity: 0.7 ≤ Pr ≤ 160, Re D ≥ 10, 000, L / D ≥ 10. Note: This equation can be used only for moderate temperature difference with all the properties evaluated at Tm. Other more accurate correlation equations can be found in other references. Caution: The ranges of application for these correlations can be quite different. For example, the Gnielinski correlation is the most accurate among all these equations: ( f / 8)(Re D − 1000) Pr Nu D = (from other reference) 1/ 2 2/3 1 + 12.7( f / 8) (Pr − 1) It is valid for 0.5 < Pr < 2000 and 3000 < Re D < 5 × 10 6 . All properties are calculated at Tm . Example (cont.) In our example, we need to first calculate the Reynolds number: water at 35°C, Cp=4.18(kJ/kg.K), μ=7x10-4 (N.s/m2), kf=0.626 (W/m.K), Pr=4.8. & 4m 4(0.15) = = = 5460 −4 μ π D μ π (0.05)(7 × 10 ) Re > 4000, it is turbulent pipe flow. Use the Gnielinski correlation, from the Moody chart, f = 0.036, Pr = 4.8 ( f / 8)(Re D − 1000) Pr (0.036 / 8)(5460 − 1000)(4.8) = Nu D = = 37.4 1/ 2 2/3 1/ 2 2/3 1 + 12.7( f / 8) (Pr − 1) 1 + 12.7(0.036 / 8) (4.8 − 1) kf 0.626 h= Nu D = (37.4) = 469(W / m 2 . K ) D 0.05 A ρVD Re = = μ & m D Energy Balance Question (2): How can we determine the required pipe length? Use energy balance concept: (energy storage) = (energy in) minus (energy out). energy in = energy received during a steady state operation (assume no loss) & q '( L ) = mC P (Tout − Tin ), & mC P (Tin − Tout ) (0.15)(4180)(50 − 20) L= = = 94( m ) 200 q' q’=q/L Tin Tout Temperature Distribution Question (3): Can we determine the water temperature variation along the pipe? Recognize the fact that the energy balance equation is valid for any pipe length x: & q '( x ) = mC P (T ( x ) − Tin ) T ( x ) = Tin + 200 q' x = 20 + x = 20 + 0.319 x & (0.15)(4180) mC P It is a linear distribution along the pipe Question (4): How about the surface temperature distribution? From local Newton's cooling law: q = hA(Ts − Tm ) ⇒ q ' Δ x = h (π D Δ x )(Ts ( x ) − Tm ( x )) q' 200 Ts ( x ) = + Tm ( x ) = + 20 + 0.319 x = 22.7 + 0.319 x (°C ) π Dh π (0.05)(469) At the end of the pipe, Ts ( x = 94) = 52.7( °C ) Temperature variation for constant heat flux 60 50 T m( x ) T s( x ) 40 Constant temperature difference due to the constant heat flux. 30 20 0 20 40 x 60 80 100 Note: These distributions are valid only in the fully developed region. In the entrance region, the convection condition should be different. In general, the entrance length x/D≈10 for a turbulent pipe flow and is usually negligible as compared to the total pipe length. Internal Flow Convection -constant surface temperature case Another commonly encountered internal convection condition is when the surface temperature of the pipe is a constant. The temperature distribution in this case is drastically different from that of a constant heat flux case. Consider the following pipe flow configuration: Constant Ts dx Tm,o Tm,i & Energy change = mC p [(Tm + dTm ) − Tm ] & = mC p dTm Tm Tm+dTm Energy in = hA(Ts − Tm ) qs=hA(Ts-Tm) Energy change = energy in & mC p dTm = hA(Ts − Tm ) Temperature distribution & mC p dTm = hA(Ts − Tm ), Note: q = hA(Ts − Tm ) is valid locally only, since Tm is not a constant dTm hA , where A = Pdx, and P is the perimeter of the pipe =− & mC P (Tm − Ts ) Integrate from the inlet to a diatance x downstream: Tm ( x ) x hP x dTm P ∫Tm ,i (Tm − Ts ) = − ∫0 mC P dx = − mC P ∫0 hdx & & ln(Tm − Ts ) | Tm ( x ) Tm ,i Ph x , where L is the total pipe length =− & mC P and h is the averaged convection coefficient of the pipe between 0 & x. x 1 x h = ∫ hdx , or ∫ hdx = hx 0 0 x Temperature distribution Tm ( x ) − Ts Ph = exp( − x ), for constant surface temperature & Tm ,i − Ts mC P Constant surface temperature Ts T( x) Tm(x) x The difference between the averaged fluid temperature and the surface temperature decreases exponentially further downstream along the pipe. Log-Mean Temperature Difference For the entire pipe: Tm ,o − Ts Δ To h ( PL ) = = exp( − ) & Δ Ti Tm ,i − Ts mC P & or mC P = − hAs Δ To ln( ) Δ Ti & & q = mC P (Tm ,o − Tm ,i ) = mC P ((Ts − Tm ,i ) − (Ts − Tm ,o )) & = mC P ( Δ Ti − Δ To ) = hAs Δ To − Δ Ti = hAs Δ Tlm Δ To ln( ) Δ Ti where Δ Tlm = Δ To − Δ Ti is called the log mean temperature difference. Δ To ln( ) Δ Ti This relation is valid for the entire pipe. External Heat Transfer Can we extend the previous analysis to include the situation that some external heat transfer conditions are given, rather than that the surface temperature is given. Example: Pipe flow buried underground with insulation. In that case, the heat transfer is first from the fluid to the pipe wall through convection; then followed by the conduction through the insulation layer; finally, heat is transferred to the soil surface by conduction. See the following figure: Soil (ks) temperature Ts Soil resistance Resistance of insulator Diameter D, insulation thickness t Convection resistance Overall Heat Transfer Coefficient From the previous example, the total thermal resistance can be written as Rtotal=Rsoil+Rinsulator+Rconvection. The heat transfer can be expressed as: q=ΔTlm/Rtot=UAs ΔTlm by defining the overall heat transfer coefficient UAs=1/Rtot. (Consider U as an equivalent heat transfer coefficient taking into consideration of all heat transfer modes between two constant temperature sources.) We can replace the convection coefficient h by U in the temperature distribution equation derived earlier: Tm ,o − Tsoil Δ To UAs 1 ) = exp( − ) = = exp( − & & Tm ,i − Tsoil Δ Ti mC P mC P Rtot Free Convection A free convection flow field is a self-sustained flow driven by the presence of a temperature gradient. (As opposed to a forced convection flow where external means are used to provide the flow.) As a result of the temperature difference, the density field is not uniform also. Buoyancy will induce a flow current due to the gravitational field and the variation in the density field. In general, a free convection heat transfer is usually much smaller compared to a forced convection heat transfer. It is therefore important only when there is no external flow exists. cold Flow is unstable and a circulatory pattern will be induced. T↓ ⇒ ρ↑ Τ↑ ⇒ ρ↓ hot Basic Definitions Buoyancy effect: Surrounding fluid, cold, ρ∞ Warm, ρ Hot plate Net force=(ρ∞- ρ)gV The density difference is due to the temperature difference and it can be characterized by ther volumetric thermal expansion coefficient, β: β =− 1 ∂ρ 1 ρ∞ − ρ 1 Δρ ( )P ≈ − =− ρ ∂T ρ T∞ − T ρ ΔT Δρ ≈ βΔT Grashof Number and Rayleigh Number Define Grashof number, Gr, as the ratio between the buoyancy force and the viscous force: 3 3 Gr = g β Δ TL ν2 = g β (TS − T∞ ) L ν2 • Grashof number replaces the Reynolds number in the convection correlation equation. In free convection, buoyancy driven flow sometimes dominates the flow inertia, therefore, the Nusselt number is a function of the Grashof number and the Prandtle number alone. Nu=f(Gr, Pr). Reynolds number will be important if there is an external flow. (combined forced and free convection. • In many instances, it is better to combine the Grashof number and the Prandtle number to define a new parameter, the Rayleigh number, Ra=GrPr. The most important use of the Rayleigh number is to characterize the laminar to turbulence transition of a free convection boundary layer flow. For example, when Ra>109, the vertical free convection boundary layer flow over a flat plate becomes turbulent. Example Determine the rate of heat loss from a heated pipe as a result of natural (free) convection. T∞=0°C D=0.1 m Ts=100°C Film temperature( Tf): averaged boundary layer temperature Tf=1/2(Ts+T ∞)=50 °C. kf=0.03 W/m.K, Pr=0.7, ν=2×10-5 m2/s, β=1/Tf=1/(273+50)=0.0031(1/K) Ra = g β (TS − T∞ ) L3 ν2 (9.8)(0.0031)(100 − 0)(0.1) 3 Pr = (0.7) = 7.6 × 10 6. (2 × 10 −5 ) 2 0.387 Ra 1 / 6 }2 = 26.0 (equation 11.15 in Table 11.1) Nu D = {0.6 + [1 + (0.559 / Pr) 9 /16 ]8 / 27 kf 0.03 (26) = 7.8(W / m 2 K ) h= Nu D = 0.1 D q = hA(TS − T∞ ) = (7.8)(π )( 0.1)(1)(100 − 0) = 244.9(W ) Can be significant if the pipe are long. MODULE 6: Worked-out Problems Problem 1: For laminar free convection from a heated vertical surface, the local convection coefficient may be expressed as hx=Cx -1/4, where hx is the coefficient at a distance x from the leading edge of the surface and the quantity C, which depends on the fluid properties, is independent of x. − − Obtain an expression for the ratio h x / h x , where h x is the average coefficient between the leading edge (x=0) and the x location. Sketch the − variation of hx and h x with x. Known: Variation of local convection coefficient with x for free convection from a heated vertical plate. Find: ratio of average to local convection coefficient. Schematic: Ts B oundary layer, h x =C x -1/4 w here C is a constant x Analysis: It follows that average coefficient from 0 to x is given by hx = hx = − − 1 C −1/4 ∫ h x dx = x ∫ x dx x0 0 4C 4 4 x3/4 = Cx −1/4 = h x 3x 3 3 − x x h 4 Hence x = hx 3 The variation with distance of the local and average convection coefficient is shown in the sketch. − hx h x − hx 4 = hx 3 hx =Cx Comments: note that hx / h x − − −1/4 =4/3, independent of x. hence the average coefficients for an entire plate of length L is h L =4/3L, where hL is the local coefficient at x=L. note also that the average exceeds the local. Why? Problem 2: Experiments to determine the local convection heat transfer coefficient for uniform flow normal to heated circular disk have yielded a radial Nusselt number distribution of the form n h(r)D ⎡ ⎛ r ⎞ ⎤ Nu D = = ⎢1 + ⎜ a ⎟ ⎥ ⎜ ⎟ k ⎢ ⎝ ro ⎠ ⎥ ⎣ ⎦ Where n and a are positive. The Nusselt number at the stagnation point is correlated in terms of the Reynolds number (ReD=VD/ν) and Prandtl Nu o = h(r = 0)D = 0.814Re 1/2 Pr 0.36 D k − − Obtain an expression for the average Nusselt number, Nu D = h D / k , corresponding to heat transfer from an isothermal disk. Typically boundary layer development from a stagnation point yields a decaying convection coefficient with increasing distance from the stagnation point. Provide a plausible for why the opposite trend is observed for the disk. Known: Radial distribution of local convection coefficient for flow normal to a circular disk. Find: Expression for average Nusselt number. Schematic: Assumptions: Constant properties. Analysis: The average convection coefficient is h= h= − − − 1 As 1 πro2 As ∫ hdA ro s ∫ DNu 0 k 0 [1 + a(r/ro ) n ]2π2πr ro kNuo ⎡ r 2 ar n + 2 ⎤ h= ⎥ ⎢ + ro3 ⎣ 2 (n + 2)ron ⎦ 0 Where Nuo is the Nusselt number at the stagnation point (r=0).hence, n+2 ⎡ (r/ro )2 hD a ⎛ r ⎞ ⎤ NuD = = 2Nu o ⎢ + ⎜ ⎟ ⎥ k (n + 2) ⎝ ro ⎠ ⎥ ⎢ 2 ⎣ ⎦ − − ro o NuD = Nu o [1 + 2a/(n + 2)] NuD = [1 + 2a/(n + 2)]0.841Re 1/2 Pr 0.36 D − − Comments: The increase in h(r) with r may be explained in terms of the sharp turn, which the boundary layer flow must take around the edge of the disk. The boundary layer accelerates and its thickness decreases as it makes the turn, causing the local convection coefficient to increase. Problem 3: In a flow over a surface, velocity and temperature profiles are of the forms u(y)=Ay+By2-Cy3 and T(y)=D+Ey+Fy2-Gy3 Where the coefficients A through G are constants. Obtain expressions for friction coefficients Cf and the convection coefficient h in terms of u∞, T∞ and appropriate profile coefficients and fluid properties. Known: form of the velocity and temperature profiles for flow over a surface. Find: expressions for the friction and convection coefficients. Schematic: y y u∞,T∞ T(y)=D+Ey+Fy2-Gy3 Ts=T(0)=D 2 3 U(y)=Ay-By -Cy Analysis: The shear stress at the wall is τs = µ ∂u ∂y = µ[ A + 2 By − 3Cy 2 ] y =0 = Aµ . y =0 Hence, the friction coefficient has the form, Cf = Cf = τs 2Aµ = 2 2 ρu ∞ /2 ρu ∞ 2Aν 2 u∞ The convection coefficient is h= h= − k f (∂T/∂y) y =0 Ts − T∞ − kfE D − T∞ = − k f [E + 2Fy − 3Gy 2 ]y =0 D − T∞ Comments: It is a simple matter to obtain the important surface parameters from knowledge of the corresponding boundary layers profile. However is rarely simple matter to determine the form of the profile. Problem 4: In a particular application involving airflow over a heated surface, the boundary layer temperature distribution may be approximated as T − Ts u y⎞ ⎛ = 1 − exp⎜ − Pr ∞ ⎟ T∞ − Ts v ⎠ ⎝ Where y is the distance normal to the surface and the Prandtl number, Pr=cpµ/k=0.7, is a dimensionless fluid property. If T∞ =400K, Ts=300K, and u∞/v=5000m-1, what is the surface heat flux? Known: Boundary layer temperature distribution Find: Surface heat flux. Schematic: T( y) −Ts u y = 1− exp(− Pr ∞ ) T∞ −Ts ν Properties: Air ( TS = 300k ): k = 0.0263 W/m.k Analysis: Applying the Fourier’s law at y=0, the heat flux is q " = −k s u y⎤ ∂T ⎡ u ⎤ ⎡ = −k(T∞ − Ts )⎢Pr ∞ ⎥exp⎢− Pr ∞ ⎥ ∂y y =0 ν ⎦ ν ⎦ y =0 ⎣ ⎣ q " = −k(T∞ − Ts )Pr s u∞ ν q " = −0.02063w/m .K(100K) 0.7x5000 1/m s Comments: (1) Negligible flux implies convection heat transited surface (2) Note use of k at Ts to evaluate from q s" Fourier’s law. Problem 5: Consider a lightly loaded journal bearing using oil having the constant properties µ=10-2 kg/s-m and k=0.15W/m. K. if the journal and the bearing are each mentioned at a temperature of 400C, what is the maximum temperature in the oil when the journal is rotating at 10m/s? Known: Oil properties, journal and bearing temperature, and journal speed for lightly loaded journal bearing. Find: Maximum oil temperature. Schematic: Assumptions: (1) steady-state conditions, (2) Incompressible fluid with constant properties, (3) Clearances is much less than journal radius and flow is Couette. Analysis: The temperature distribution corresponds to the result obtained in the text example on Couette flow. 2 µ 2⎡y ⎛ y ⎞ ⎤ T(y) = To + U ⎢ −⎜ ⎟ ⎥ 2k ⎢L ⎝ L ⎠ ⎥ ⎣ ⎦ The position of maximum temperature is obtained from dT µ 2 ⎡ 1 ⎛ 2y ⎞ =0= U ⎢ −⎜ 2 ⎟ dy 2k ⎢L ⎝ L ⎠ ⎣ ⎤ ⎥ ⎥ ⎦ y=L/2. Or, The temperature is a maximum at this point since d 2 T/dy 2 < 0. 2 µ 2 ⎡1 1 ⎤ µ 2 Tm, ax = T(l/2) = To + U ⎢ − ⎥ = To + U 2k 8k ⎣2 4 ⎦ hence Tmax = 40 ° C + Tmax 10 −2 kg/s.m(10m/s) 2 8 × 0.15W/m.K ° = 40.83 C Comments: Note that Tmax increases with increasing µ and U, decreases with increasing k, and is independent of L. Problem 6: Consider two large (infinite) parallel plates, 5mm apart. One plate is stationary, while the other plate is moving at a speed of 200m/s. both plates are maintained at 27°C. Consider two cases, one for which the plates are separated by water and the other for which the plates are separated by air. For each of the two fluids, which is the force per unit surface area required to maintain the above condition? What is the corresponding requirement? What is the viscous dissipation associated with each of the two fluids? What is the maximum temperature in each of the two fluids? Known: conditions associated with the Couette flow of air or water. Find: (a) Force and power requirements per unit surface area, (2) viscous dissipation,(3) maximum fluid temperature. Schematic: Assumptions: (1) Fully developed Couette flow, (2) Incompressible fluid with constant properties. Properties: Air (300K); µ=184.6*10-7 N.s/m2, k=26.3*10-3W/m.K; water (300K): µ=855*10—6N.s/m2,k=0.613W/m.K Analysis: (a) the force per unit area is associated with the shear stress. Hence, with the linear velocity profile for Couette flow τ=µ (du/dy) = µ (U/L). Air : Water : τ air = 184.6 × 10 -7 N.s/m 2 × τ water 200m/s = 0.738N/m 2 0.005m 200m/s = 855 × 10 -6 N.s/m 2 × = 34.2N/m 2 0.005m With the required power given by P/A= τ .U Air : Water : (P/A) air = (0.738N/m 2 ) × 200m/s = 147.6W/m 2 (P/A) water = (34.2N/m 2 ) × 200m/s = 6840W/m 2 (b) The viscous dissipation is µϕ = µ(du/dy) 2 = µ(U/L) 2 .hence 2 Air : (µµφ air ⎡ 200m/s ⎤ = 2.95 × 10 4 W/m 3 = 184.6 × 10 N.s/m × ⎢ 0.005m ⎥ ⎣ ⎦ -7 2 2 ⎡ 200m/s ⎤ 6 3 Water : (µµφ water = 855 × 10 -6 N.s/m 2 × ⎢ ⎥ = 1.37 × 10 W/m ⎣ 0.005m ⎦ The location of the maximum temperature corresponds to ymax=L/2. Hence Tmax=To+µU2/8k and Air : Water : (Tmax ) air = 27°C + (Tmax) water 184.6 × 10 -7 N.s/m 2 × (200m/s)2_ = 30.5°C 8 × 0.0263W/m.K 855 × 10 -6 N.s/m 2 × (200m/s)2 = 27°C + = 34.0°C 80.613W/m.K Comments: (1) the viscous dissipation associated with the entire fluid layer, µφ (LA), must equal the power, P. (2) Although ( µφ ) water >> ( µφ ) air , kwater>>kair. Hence, Tmax,water ≈ Tmax,air . Problem 7: A flat plate that is 0.2m by 0.2 m on a side is orientated parallel to an atmospheric air stream having a velocity of 40m/s. the air is at a temperature of T∞=20°C, while the plate is maintained at Ts=120°C. The sir flows over the top and bottom surfaces of the plate, and measurement of the drag force reveals a value of 0.075N. What is the rate of heat transfer from both sides of the plate to the air? Known: Variation of hx with x for flow over a flat plate. Find: Ratio of average Nusselt number for the entire plate to the local Nusselt number at x=L. Schematic: Analysis: The expressions for the local and average Nusselt number are h L L (CL−1/2 )L CL1/2 = = Nu L = k k k h L Nu L = L k where 2C 1/2 C 1 L = 2CL−1/2 h L = ∫ h x dx = ∫ x −1/2 dx = L L0 L0 − L L − Hence 2CL−1/2 (L) 2CL−1/2 NuL = = k k and - NuL NuL - = 2. Comments: note the manner in which note that 1 Nu L ≠ ∫ Nu x dx L0 − L NuL - is defined in terms of h L . Also − Problem 8: For flow over a flat plate of length L, the local heat transfer coefficient hx is known to vary as x-1/2, where x is the distance from the leading edge of the plate. What is the ratio of the average Nusslet number for the entire plate to the local Nusslet number at x=L (NuL)? Known: Drag force and air flow conditions associated with a flat plate. Find: Rate of heat transfer from the plate. Schematic: Assumptions: (1) Chilton-Colburn analogy is applicable. Properties: Air(70°C,1atm): ν=20.22*10-6m2/s. ρ=1.018kg/m3, cp=1009J/kg.K, pr=0.70, Analysis: the rate of heat transfer from the plate is q = 2 h(L)2 (Ts − T∞ ) − Where − − − h may be obtained from the Chilton-Colburn analogy, − − C h Pr 2/3 j H = f = S tPr 2/3 = ρu ∞ c p 2 Cf 1 τ s 1 (0.075N/2)(0.2m) 2 = = = 5.76 × 10 − 4 2 3 2 2 2 ρu ∞ /2 2 1.018kg/m (40m/s) /2 hence, h= - − − Cf ρu ∞ c p Pr − 2/3 2 h = 5.76 × 10 − 4 (1.018kg/m 3 )40m/s(1009J/kg.K)(0.70) − 2/3 h = 30W/m 2 .K The heat rate is q = 2(30W/m 2 .K)(0.3m) 2 (120 − 20)°C q = 240W Comments: Although the flow is laminar over the entire surface ( Re L = u ∞ L / ν = 40m / s × 0.2m / 20.22 ×10 −6 m 2 / s = 4.0 ×10 5 ) , the pressure gradient is zero and the Chilton-Colburn analogy is applicable to average, as well as local, surface conditions. Note that the only contribution to the drag force is made by the surface shear stress. Problem 9: Consider atmospheric air at 25°C in parallel flow at 5m/s over both surfaces of 1-mlong flat plate maintained at 75°C. Determine the boundary layer thickness, the surface shear stress, and the heat flux at the trailing edge. Determine the drag force on the plate and the total heat transfer from the plate, each per unit width of the plate. Known: Temperature, pressure, and velocity of atmospheric air in parallel flow over a Plate of prescribed length and temperature. Find: (a) Boundary layer thickness, surface shear stress and heat flux at trailing edges, (b) drag force and total heat transfer flux per unit width of plate. Schematic: Fluid x) Ts=75°C ∞=5m/s T∞=25C P∞=1 atm L=1m Assumptions: (1) Critical Reynolds number is 5*105, (2) flow over top and bottom surfaces Properties: (Tf=323K, 6m2/s,k=0.028W/m.K,pr=0.707 1atm) Air: ρ=1.085kg/m3,ν=18.2*10- Analysis: (a) calculate the Reynolds number to know the nature of flow Re L = u∞L 5 m/s × 1 m = = 2.75 × 10 5 −6 2 ν 18.2 × 10 m /s Hence the flow is laminar, and at x=L δ = 5LRe L τ s,L −1/2 = 5 × 1m/(2.75 × 10 5 )1/2 = 9.5mm −1/2 1.085 kg (5m/s) 2 0.664/(2.75 × 10 5 ) 1/2 3 2 m 2 τ s,L = 0.0172kg/m.s = 0.0172N/m 2 = (ρρ 2 /2)0.664Re L = Using the correct correlation, Nu L = hence, h L = 155.1(0.028W/m.K)/1m = 4.34W/m 2 .K q" s(L) = hL(Ts − T∞ ) = 4.34W/m 2 .K(75°C − 25°C)217W/m 2 hL L 1/2 = 0.332Re L Pr 1/3 = 0.332(2.75105)1/2 (0.707)1/3 = 155.1 k (b) The drag force per unit area plate width is D ' = 2 L τ s L where the factor of two is included to account for both sides of the plate. Hence with τ s L = (ρρ 2 ∞ /2)1.328Re L The drag is τ s L = 0.0343N/m 2 D ' = 2(1m)0.0343N/m 2 = 0.0686N/m also with h L = 2hL = 8.68W/m 2 .K q " = 2Lh L (Ts − T∞ ) = 2(1m)8.68W/m 2 .K(75 − 25)°C = 868W/m − − −1/2 − = (1.085kg/m 3 /2)(5m/s) 2 1.328(2.75 × 10 5 ) −1/2 Problem 10: Engine oil at 100°C and a velocity of 0.1m/s flows over both surfaces of a 1-m-long flat plate maintained at 20°C. Determine a. The velocity and thermal boundary thickness at the trailing edge. b. The local heat flux and surface shear stress at the trailing edge. c. The total drag force and heat transfer per unit area width of the plate. Known: Temperature and velocity of engine oil Temperature and length of flat plate. Find: (a) velocity and thermal boundary thickness at the trailing edge, (b) Heat flux and surface shear stress at the trailing edge, (c) total drag force and heat transfer per unit plate width. Schematic: Assumptions: engine oil (Tf=33K): ρ=864kg/m3,ν=86.1*106m2/s, k=0.140W/m /K, Pr=1081. Analysis: (a) calculate the Reynolds number to know the nature of flow ReL = u∞L 0.1m/s × 1m = = 1161 ν 86.1 * 10 -6 m 2 /s Hence the flow is laminar at x=L, and δ = 5LRe L −1/2 = 5(1m)(1161) −1/2 = 0.147m δ t = δPr −1/3 = 0.147(1081) −1/3 = 0.0143m (b) The local convection coefficient and heat flux at x=L are hL = k 0.140W/m.K −1/2 0.3325LRe L Pr 1/3 = 0.332(1161)(1081)1/3 = 16.25W/m 2 .K L 1m q" = hL(Ts − T∞ ) = 16.25W/m 2 .K(20 − 100)°C = −1300W/m 2 Also the local shear stress is τ s L = (ρρ 2 ∞ /2)0.664Re L − − −1/2 = 864kg/m 3 (0.1m/s) 2 0.664(1161) −1/2 2 τ s L = 0.0842kg/m.s 2 = 0.0842N/m 2 − (c) With the drag force per unit width given by D ' = 2 L τ s L where the factor of 2 is included to account for both sides of the plate, is follows that τ s L = (ρρ 2 ∞ /2)1.328Re L _ − −1/2 = 2(1m)864kg/m 3 (0.1m/s) 2 1.328(1161) −1/2 = 0.673N/m with h L = 2h L = 32.5W/m 2 .K, it also follows that q = 2L h L (Ts − T∞ ) = 2(1m)32.5W/m 2 .K(20 − 100) ° C = −5200W/m _ Comments: Note effect of Pr on (δ/δt). Problem 11: Consider water at 27°C in parallel flow over an isothermal, 1-m-long, flat plate with a velocity of 2m/s. Plot the variation of the local heat transfer coefficient with distance along the plate. What is the value of the average coefficient? Known: velocity and temperature of air in parallel flow over a flat plate of prescribed length. Find: (a) variation of local convection coefficient with distance along the plate, (b) Average convection coefficient. Schematic: Assumptions: (1) Critical Reynolds number is 5*105. Properties: Water (300K):ρ =997kg/m3,µ=855*10-6N.s/m2, ν=µ/ρ=0.858*10-6m2/s, k=0.613W/m. K, Pr=5.83 Analysis: (a) With Re L = u∞L 2m/s × 1m = = 2.33 × 10 5 ν 0.858 × 10 −6 m 2 /s Boundary layer conditions are mixed and x c = L(Re x,c /Re L ) = 1m(5 × 10 5 /2.33 × 10 6 ) = 0.215m k ⎛u ⎞ for x ≤ 0.215m, hx = 0.332Re 1/2 Pr 1/3 = 0.332k ⎜ ∞ ⎟ Pr 1/3 x −1/2 x x ⎝ ν ⎠ x(m) hx(W/m2.K) 0.1 1768 0.215 1206 4/5 1/2 k ⎛u ⎞ for x > 0.215m, hx = 0.0296Re 4/5 Pr 1/3 = 0.0296k ⎜ ∞ ⎟ Pr 1/3 x −0.2 x x ⎝ ν ⎠ x(m) 0.0215 0.4 4871 0.6 0.8 1.0 h x (W/m 2 .K) 5514 4491 4240 4055 The Spatial variation of the local convection coefficient is shown above (b) The average coefficient is hL = _ _ k 0.613W/m.K (0.037Re 4/5 − 871)Pr 1/3 = [0.0379(2.33 × 10 6 ) 4/5 − 871(5.83)1/3 L L 1m h L = 4106W/m 2 .K Problem 12: A circular cylinder of 25-mm diameter is initially at 150C and is quenched by immersion in a 80C oil bath, which moves at a velocity of 2m.s in cross flow over the cylinder. What is the initial rate of heat loss unit length of the cylinder? Known: Diameter and initial temperature of a circular cylinder submerged in an oil bath f prescribed temperature and velocity. Find: initial rate of heat loss unit per length. Schematic: Assumptions: (1) Steady-state conditions, (2) uniform surface temperature. (Ts=423K): Prs=98. Properties: Engine oil (T∞=353K): ν=38.1*10-6m2/s, k=0.138W/m. K, Pr∞=501; Analysis: The initial heat loss per unit length is q = h πD(Ts − T∞ ) where h may be computed from the Zhukauskas relation. when Re D = 2m/s(0.025m) VD = = 1312 ν 38.1 × 10 =6 m 2 /s _ _ Find C = 0.26 and m = 0.6 from table 7.4 . hence k 0.138W/m.K ⎛ 501 ⎞ h = CRe m Pr n (Pr∞ /Prs )1/4 = 0.26(1312) 0.6 (501) 0.37 ⎜ ⎟ D D 0.025m ⎝ 98 ⎠ _ 1/4 h = 1600W.m 2 .K q' = 1600W/m 2 .K (π )0.025m(150 − 80)°C = 8.8W/m _ Comments: Evaluating properties at the film temperature, Tf=388K(ν=14.0*10-6m2/s, k=0.135W/m. K, Pr=196), find ReD=3517. Problem 13: An uninsulated steam pipe is used to transport high-temperature steam from one building to one another. The pipe is 0.5-m diameter, has a surface temperature of 150C, and is exposed to ambient air at -10C.the air moves in cross flow over the pipe with a velocity of 5m.s.What is the heat loss per unit length of pipe? Known: Diameter and surface temperature of uninsulated steam pipe. Velocity and temperature of air in cross flow. Find: Heat loss per unit length. Schematic: Air D=0.5mm Ts=150 C V=5m/s T∞=263K Assumptions: (1) steady-state conditions, (2) uniform surface temperature Properties: Air (T=263K, 1atm):ν=12.6*10-6m2/s, k=0.0233W/m. K, Pr=0.72;(Ts=423K, 1atm); Prs=0.649. Analysis: the heat loss per unit length is q = h πD(Ts − T∞ ) _ where h may be computed from the Zhukauskas relation. when Re D = 2m/s(0.025m) VD = = 1312 ν 38.1 × 10 =6 m 2 /s _ Find C = 0.26 and m = 0.6 from table 7.4 . hence k 0.138W/m.K ⎛ 501 ⎞ h = CRe m Pr n (Pr∞ /Prs )1/4 = 0.26(1312) 0.6 (501) 0.37 ⎜ ⎟ D D 0.025m ⎝ 98 ⎠ _ 1/4 h = 1600W.m 2 .K _ Hence the heat rate is q' = 16.3W/m 2 .K (π )0.5m(150 - ( −10)°C = 4100W/m Comments: Note that q’αDm, in which case the heat loss increases significantly with increasing D. Problem 14: Atmospheric air at 25°C and velocity of 0.5m/s flows over a 50-W incandescent bulb whose surface temperature is at 140°C. The bulb may be approximated as a sphere of 50-mm diameter. What is the rate of heat loss by convection to the air? Known: Conditions associated with airflow over a spherical light bulb of prescribed diameter and surface temperature. Find: Heat loss by convection. Schematic: Assumptions: (1) steady-state conditions, (2) uniform surface temperature. 0.71,µ=183.6*10-7N.s/m2; Air (Ts=140°C, 1atm): µ=235.5*10-7N.s/m2 Properties: Air (Tf=25°C, 1atm): ν=15.71*10-6m2/s, k=0.0261W/m. /K.Pr=. Analysis: q = h πD(Ts − T∞ ) _ where h may be computed from the Whitaker relation. when h= k [2 + (0.4Re 1/2 + 0.06Re 2/3 )Pr 0.4 (µµ/ s ) 1/4 ] D D D Where VD 0.5m/s × 0.05m = = 1591 Re D = ν 15.71 × 10 −6 m 2 /s _ _ hence 1/4 ⎫ 0.0261W/m.K ⎧ ⎪ ⎪ 1/2 2/3 0.4 ⎛ 183.6 ⎞ h= 2 + [0.4(1591) + 0.06(1591) ](0.71) ⎜ ⎟ ⎬ ⎨ 0.05m ⎪ ⎝ 235.5 ⎠ ⎪ ⎩ ⎭ _ h = 11.4W/m 2 .K and the heat rate is q = 11.4 W π(0.05m) 2 (140 − 25)°C = 10.3W _ m 2 .K Comments: (1) The low value of h suggests that heat transfer by free convection may be significant and hence that the total loss by convection exceeds 10.3W (2) The surface of the bulb also dissipates heat to the surroundings by radiation. Further, in an actual light bulb, there is also heat loss by conduction through the socket. (3) The Correlation has been used its range of application (µ/µs )<1. _ Problem 15: Water at 27° C flows with a mean velocity of 1m/s through a 1-km-long cast iron pipe at 0.25 m inside diameter. (a) Determine the pressure drop over the pipe length and the corresponding pump power requirement, if the pipe surface is clean. (b) If the pipe surface roughness is increased by 25% because of contamination, what is the new pressure drop and pump power requirement. Known: Temperature and velocity of water in a cast iron pipe of prescribed dimensions. Find: pressure drops and power requirement for (a) a clean surface and (b) a surface with a 25% larger roughness. Schematic: Assumptions: (1) Steady, fully developed flow. Properties: Water (300K):ρ=1000 kg/m3, µ=855*10-6 N.s/m2. Analysis: (a) from eq.8.22, the pressure drop is 2 ρu m ∆p = f L 2D e=2.6*10-4 m for clean cast iron; hence e/D=1.04*10-3. With Re D = um D 1m/s × 0.25m = 2.92 × 10 5 = ν 855 × 10 −6 N.s/m 2 /1000kg/m 3 find from fig 8.3 that f ≈ 0.021. hence , (1000kg/m 3 )(1m/s) 2 ∆P = 0.021 1000m = 4.2 × 10 4 kg/s 2 .m 2(0.25m) ∆P = 4.2 × 10 4 N/m 2 = 0.42bar The pump power requirement is P = ∆p. V = ∆p(πD 2 /4)u m P = 4.2 × 10 4 N/m 2 (π × 0.25 2 /4)m 2 × 1m/s = 2.06kW . (b) Increasing ε by 25% it follows that ε=3.25*10-4m and e/D=0.0013. With ReD unchanged, from fig 8.3, it follows that f ≈ 0.0225. Hence ∆P 2 = (f 2 /f 1 )∆p1 = 0.45 bar P2 = (f 2 /f 1 )P1 = 2.21kW Comments: (1) Note that L/D=4000>>(xfd,h/D) ≈10 for turbulent flow and the assumption of fully developed conditions is justified. (2) Surface fouling results in increased surface and increases operating costs through increasing pump power requirements. Problem 16: Consider flow in a circular tube. Within the test section length (between 1 and 2) a constant heat flux q”s is maintained. (a) For the two cases identified, sketch, qualitatively, the surface temperature Ts(x) and the fluid mean temperature Tm(x) as a function of distance along the test section x. in case A flow is hydro dynamically and thermally fully developed. In case B flow is not developed. (b) Assuming that the surface flux q”s and the inlet mean temperature Tm,1 are identical for both cases, will the exit mean temperature Tm,2 for case A be greater than, equal to , or less than Tm,2 for case B? Briefly explain why? Known: internal flow with constant surface heat flux, q s" . Find: (a) Qualitative temperature distributions (x), under developing and fully developed flow, (b) exit mean temperature for both situations. Schematic: q”s=constant Flow 1 2 Assumptions: (a) Steady-state conditions, (b) constant properties, (c) incompressible flow. Analysis: Based upon the analysis, the constant surface heat flux conditions, dTm = constant dx Hence, regardless of whether the hydrodynamic or thermal boundary layer is fully developed, is follows that Tm(x) T m,2 is linear will be the same for all flow conditions. The surface heat flux can be written as q " = h[Ts − Tm (x)] s Under fully developed flow and thermal conditions, h=hfd is a constant. When flow is developing h> hfd . Hence, the temperature distributions appear as below. Problem 17: A thick-walled, stainless steel (AISI 316) pipe of inside and outside diameter Di=20mm and Do=40m is heated electrically to provide a uniform heat generation rate of q = 10 6 W / m 3 . The outer surface of the pipe is insulated while water flows through the pipe at a rate of m = 0.1kg / s . . (a) If the water inlet temperature is Tm,1=20°C and the desired outlet temperature is Tm,o=40°C, what is the required pipe length (b) What are the location and value of the maximum pipe temperature? Known: Inner and outer diameter of a steel pipe insulated on the outside and experiencing uniform heat generation. Flow rate and inlet temperature of water flowing through the pipe. Find: (a) pipe length required to achieve desired outlet temperature, (b) location and value of maximum pipe temperature. Schematic: Assumptions: (1) steady-state conditions, (2) constant properties, (3) negligible kinetic energy, potential energy and flow work changes, (4) onedimensional radial conduction in pipe wall, (5) outer surface is adiabatic. Properties: Stainless steel 316 (T≈ 400K): k=15W/m.k; water ( Tm = 303K ); cp=4178J/kg. K, k=0.617W/m. K,ρ=803*10-6N.s/m2, Pr=5.45 Analysis: (a) performing an energy balance for a control volume about the inner tube, it follows that . 2 m c p (Tm,o − Tm,i ) = q = q(ππ/4)( 0 − D i2 )L . . L= mc p (Tm,o − Tm,i ) 2 q(ππ/4)( 0 − D i2 )L . . = (0.1kg/s)4178(J/kg.K)20°C 10 W/m 3 (ππ/4)[(0.4m 2 − (0.02) 2 ] 6 L = 8.87m (b) The maximum wall temperature exists at the pipe exit (x=L) and the insulated surface (r=ro). The radial temperature distribution in the wall is of the form q 2 T(r) = − r + C 1 λnr + C 2 4k considering the boundary conditions ; . . . r = ro ; C −q dT ⎞ =0=− ro + 1 ⎟ dr ⎠ r =ro 2k ro . . 2 o qro2 C1 = 2k . . - q 2 qr λnri + C 2 r = r1 : T(ri ) = Ts = ri + 2k 4k q 2 qro2 λnri + Ts C2 = ri 2k 4k The temperature distribution and the maximum wall temperature (r=ro) are q 2 2 qro2 r T(r) = (r - ri ) + λn + Ts 4k 2k ri r q 2 2 qro2 = T(ro ) = (ro - ri ) + λn o + Ts 4k 2k ri . . . . Tw,max where Ts, the inner suraface temperture of the wall at exit, it follows from . . 2 2 q(π(π/4) o − D i2 )L q(D o − D i2 ) q = = = h(Ts − Tm,o ) πD i L 4D i " s Where h is the local convection coefficient at the exit. With 4m 4 × 0.1kg/s Re D = = = 7928 πD i µ π(0.02m)803 × 10 −6 N.s/m 2 . The flow is turbulent and, with (L/Di)=(8.87m/0.02m)=444>>(xfd/D)≈10,it is also fully developed. Hence, from the Dittus-Boelter correlation, h= k 0.617W/m.K (0.023Re 4/5 pr 0.4 ) = 0.023(7928) 4/5 5.45 0.4 = 1840W/m 2 .K D Di 0.02m Hence the inner surface temperature of the wall at the exit is 2 q(D o − D i2 ) 10 6 W/m 3 [(0.04m) 2 − (0.02m) 2 ] + Tm,o = + 40°C = 48.2°C Ts = 4D i 4 × 180W/m 2 .K(0.02m) . and 10 6 W/m 3 10 6 W/m 3 (0.02) 2 0.02 [(0.02) 2 − (0.01) 2 ] + λn + 48.2°C = 52.4°C 415W/m.K 215W/m.K 0.01 Tw,max = − Comments: The physical situation corresponds to a uniform surface heat flux, and Tm increases linearly with x. in the fully developed region, Ts also increases linearly with x. T Ts(x) Tm(x) x xfd Problem 18: The surface of a 50-mm diameter, thin walled tube is maintained thin walled tube is maintained at 100°C. In one case air is cross flow very the tube with a temperature of 25°C and a velocity of 30m/s. In another case air is in fully developed flow through the tube with a temperature of 25°C and a mean velocity of 30m/s. compare the heat flux from the tube to the air for the two cases. Known: surface temperature and diameter of a tube. Velocity and temperature of air in cross flow. Velocity and temperature of air in fully developed internal flow. Find: convection heat flux associated with the external and internal flows. Schematic: Air V=30m/s T =25 C Ts=100 C Air D=0.05m Um=30m/s Tm=25°C Assumptions: (1) steady-state conditions, (2) uniform cylinder surface temperature, (3) fully developed internal flow Properties: Air (298K): ν=15.71*10-6m2/s, k=0.0261W/m.K, Pr=0.71 Analysis: for the external and internal flow Re D = VD ν = um D ν = 30m / s × .05m = 9.55 × 10 4 2 −6 15.71 × 10 m / s From the Zhukauskas relation for the external flow, with C=0.26 and m=0.6 N u D = C Re m Pr(Pr/ Prs )1 / 4 = 0.26(9.55 × 10 4 ) 0.6 (0.71) 0.37 (1)1 / 4 = 223 D _ Hence, the convection coefficient and heat rate are h= _ _ k 0.0261W / m.K NuD = × 223 = 116.4W / m 2 .K D 0.05m q " = h(Ts − T∞ ) = 116.4W / m 2 .k (100 − 25) ° C = 8.73 × 10 3 W / m 2 Using the Dittus-Boelter correlation, for the internal flow, which is Turbulent, NU D = 0.023 Re 4 / 5 Pr 0.4 = 0.023(9.55 × 10 4 ) 4 / 5 (0.71) 0.4 = 193 D _ _ 2 h= k _ 0.0261W / m.K Nu D = × 193 = 101W / m2.K D 0.05 and the heat flux is q " = h(Ts − Tm ) = 101W / m 2 .K (100 − 25)°C = 7.58 × 10 3 W / m 2 Comments: Convection effects associated with the two flow conditions are comparable. Problem 19: Cooling water flows through he 25.4 mm diameter thin walled tubes of a stream condenser at 1m/s, and a surface temperature of 350K is maintained by the condensing steam. If the water inlet temperature is 290 K and the tubes are 5 m long, what is the water outlet temperature? Water properties may be evaluated at an assumed average temperature of 300K Known: Diameter, length and surface temperature of condenser tubes. Water velocity and inlet temperature. Find: Water outlet temperature. Schematic: L=5m Tm,i=290K Um=1m/s D=0.0254m Ts=350K Assumptions: (1) Negligible tube wall conduction resistance, (2) Negligible kinetic energy, potential energy and flow work changes. Properties: Water (300K):ρ=997kg/m3, cp=4179J/kg.K, µ=855*106kg/s.m,k=0.613W/m.K, Pr=5.83 Analysis: Tm,o = Ts − (Ts − Tm,i )exp[−(ππDLm c p ) h] ρu m D 997kg/m 3 (1m/s)0.0254m = = 29,618 µ 855 × 10 −6 kg/s.m . _ Re D = The flow is turbulent. Since L/D=197, it is reasonable to assume fully developed flow throughout the tube. Hence h = NuD(k/D) = 176(0.613W/m.K/0.0254m) = 4248W/m 2 .K With m = ρu m (ππ 2 /4) = (ππ/4)997k/m 3 (1 m/s)(0.0254m) 2 = 0.505kg/s ⎡ π(0.0254m)5m(4248W/m 2 .K) ⎤ Tm,o = 350K − (60K)exp⎢ ⎥ ≈ 323K ⎣ (0.505kg/s(4179J/kg.K) ⎦ . _ Comments: The accuracy of the calculations may be improved slightly by reevaluating properties at Tm = 306.5K . _ Problem 20: The air passage for cooling a gas turbine vane can be approximated as a tube of 3-mm diameter and 75-mm length. If the operating temperature of the vane is 650°C, calculate the outlet temperature of the air if it enters the tube at 427°C and 0.18kg/h. Known: gas turbine vane approximation as a tube of prescribed diameter and length maintained at an known surface temperature. Air inlet temperature and flow rate. Find: outlet temperature of the air coolant. Schematic: m = 0.18kg / h . Assumptions: (1) Steady-state conditions, (2) negligible Kinetic and potential energy changes. Properties: Air (assume Tm = 780 K ,1 atm) : cp=1094J/kg. K, k=0.0563 W/m. K, µ=363.7*10-7 N.s/m2, Pr=0.706; Pr=0.706;(Ts=650°C=923K, 1atm): µ=404.2*10-7N.s/m2. Analysis: For constant wall temperatures heating, Ts − Tm ,o Ts − Tm ,i _ ⎞ ⎛ ⎜ PL h ⎟ = exp⎜ − . ⎟ ⎜ mc ⎟ p ⎠ ⎝ _ Where P=πD. for flow in circular passage, 4m 4 × 0.18kg/h(1/3600s/h) Re D = = πDµ π(0.03m)363.7 × 10 −7 N.s/m . The flow is laminar, and since L/D=75mm/3mm=25, the Sieder-Tate correlation including combined entry length fields. hD ⎛ Re Pr ⎞ Nu D = = 1.86⎜ D ⎟ k ⎝ L/D ⎠ _ _ _ 1/3 ⎛µ⎞ ⎜ ⎟ ⎜µ ⎟ ⎝ s⎠ 0.14 0.0563W/m.K ⎛ 584 × 0.706 ⎞ h= 1.86⎜ ⎟ 0.003m 25 ⎝ ⎠ 1/3 ⎛ 363.7 × 10 −7 ⎞ ⎜ ⎟ ⎜ 404.2 × 10 −7 ⎟ ⎠ ⎝ 0.14 = 87.5W/m 2 .K Hence, the air outlet temperature is ⎛ π(0.003m) × 0.075m × 87.5W/m 2 .K ⎞ ⎟ = exp⎜ − ⎜ (650 − 427)°C (0.18/3600)kg/s × 1094J/kg.K ⎟ ⎠ ⎝ 650 − Tm,o Tm,o=578°C Comments: (1) based upon the calculations for Tm,o=578°C, Tm =775K which is in good agreement with our assumption to evaluate the thermo physical properties. _ Problem 21 A household oven door of 0.5-m height and 0.7-m width reaches an average surface temperature of 32°C during operation. Estimate the heat loss to the room with ambient air at 22°C. If the door has an emissivity of 1.0 and the surroundings are also at 22°C, comment on the heat loss by free convection relative to that by radiation. Known: Oven door with average surface temperature of 32°C in a room with ambient temperature at 22°C. Find: Heat loss to the room. Also, find effect on heat loss if emissivity of door is unity and the surroundings are at 22°C. Schematic: Assumptions: (1) Ambient air in quiescent, (2) surface radiation effects are negligible. Properties: Air (Tf=300K, 1atm): ν=15.89*10-6 m2/s, k=0.0263W/m. K, α=22.5*10-6 m2/s, Pr=0.707, β=1/Tf=3.33*10-3K-1 Analysis: the heat rate from the oven door surface by convection to the ambient air is _ q = h A s (Ts − T∞ ) L=0.5m 0.7 m Where h can be estimated from the free convection correlation for a vertical plate, 0.387Ra 1/6 hL ⎧ ⎪ L Nu L = = ⎨0.825 + k ⎪ 1 + (0.492/Pr) 9/16 ⎩ _` _ _ [ ] ⎫ ⎪ 8/27 ⎬ ⎪ ⎭ 2 The Rayleigh number, Ra L = gβ (Ts − T∞ )L2 9.8m/s 2 (1/300k))32 − 22)K0.5 3 m3 = = 1.142 × 10 4 2 2 −6 −6 να 15.89 × 10 m /s × 22.5 × 10 m /s Substituting numerical values into equation, find 0.387(1.142 × 10 8 ) 1/8 ⎫ hL ⎧ ⎪ ⎪ Nu L = = ⎨0.825 + ⎬ = 63.5 9/16 8/27 k ⎪ ⎪ 1 + (0.492/Pr) ⎩ ⎭ _ 0.0263W/m.K k _` h L = Nu L = × 63.5 = 3.34W/m 2 .K L 0.5m _` _ 2 [ ] The heat rate using equation is q = 3.34W/m 2 .k(0.5 × 0.7)m 2 (32 − 22)K = 11.7W Heat loss by radiation, assuming ε=1 is 4 q rad = εA s σ(Ts4 − Tsur ) q rad = 1(0.5 × 0.7)m 2 × 5.6710 −8 W/m 2 .K 4 [(273 + 32) 4 − (273 + 22) 2 ]k 4 = 21.4W Note that heat loss by radiation is nearly double that by free convection. Comments: (1) Note the characteristics length in the Rayleigh number is the height of the vertical plate (door). Problem 22 An Aluminum alloy (2024) plate, heated to a uniform temperature of 227°C, is allowed to cool while vertically suspended in a room where the ambient air and surroundings are at 27°C. The late is 0.3 m square with a thickness of 15 mm and an emissivity of 0.25. a. Develop an expression for the time rate of change of the plate temperature assuming the temperature to be uniform at any time. b. Determine the initial rate of cooling of the plate temperature is 227°C. c. Justify the uniform plate temperature assumption. Known: Aluminum plate alloy (2024) at uniform temperature of 227°C suspended in a room where the ambient air and the surroundings are at 27°C Find: (1) expression for the time rate of change of the plate, (2) Initial rate of cooling (K/s) when the plate temperature is 227°C. (3) justify the uniform plate temperature assumption. Schematic: Properties: Aluminium alloy 2024 (T=500K):ρ=2270 kg/m3, k=186W/m.K, c=983 J/kg.K; Air (Tf=400K, 1atm):ν=26.41*106m2/s,k=0.0338 W/m.K,α=38.3*10-6 m2/s, Pr=0.690. Analysis :( a) from an energy balance on the plate considering free convection and radiation exchange − E out = E st . 4 − h L 2A s (Ts − T∞ ) − ε2A s σ(Ts4 − Tsur ) = ρA s λc _ . . Where Ts is the plate temperature assumed to be uniform at any time. (b) To evaluate (dt/dx), estimate h L . Find first the Rayleigh number Ra L = gβ (Ts − T∞ )L2 9.8m/s 2 (1/400k)(227 − 27)K × 0.3m 3 = = 1.308 × 10 8 −6 −6 2 2 να 26.41 × 10 m /s × 38.3 × 10 m /s _ dT dT − 2 . 4 = or [h L (Ts − T∞ ) + εσ(Ts4 − Tsur ) dt dt ∫ λc Substituting numerical values, find ⎧ 0.670Ra1/4 ⎪ L Nu L = ⎨0.68 + ⎪ 1 + (0.492/Pr)9/16 ⎩ _` [ ] ⎫ ⎧ 0.670(1.308 × 108 ) ⎫ ⎪ ⎪ ⎪ ⎬ = ⎨0.68 + ⎬ = 55.5 4/9 9/16 4/9 ⎪ ⎪ ⎪ 1 + (0.492/0.690) ⎭ ⎩ ⎭ 2 [ ] 2 h L = Nu k/L = 55.5× 0.0338W/m. K/0.3m= 6.25W/m .K _ _` dT −2 2 4 = × 6.25W/m .K(227− 27)K+ 0.25(5.67 10−8 W/m2 .K)(500 − 3004 )K × 3 dt 2770Kg/m × 0.015m× 983J/kg.K = 0.099K/s (c) The uniform temperature assumption is justified if the Biot number criterion is satisfied. With L c ≡ (V/A s ) = (A s .λ/A s ) = λ and htot = hconv + hrad , Bi = htot λ / k ≤ 0.1 . Using the linearized radiation coefficient relation find, _ _ _ _ 2 h rad = εσ(Ts + Tsur )(Ts2 + Tsur ) = 0.25(5.67 × 10 − 8W/m 2 .K 4 )(500 + 300)(5002 + 3002)K 3 = 3.86W/m 2 .K _ Hence Bi=(6.25+3.86) W/m2.k (0.015m)/186W/m. K=8.15*10-4. Since Bi<<0.1, the assumption is appropriate. Problem 23 The ABC Evening News Report in news segment on hypothermia research studies at the University of Minnesota claimed that heat loss from the body is 30 times faster in 1 0°C water than in air at the same temperature. Is that a realistic statement? Known: Person, approximated as a cylinder, experiencing heat loss in water or air at 10°C. Find: Whether heat loss from body in water is 30 times that in air. Assumptions: (1) Person can be approximated as a vertical cylinder of diameter D=0.3 m and length L=1.8m, at 25°C, (2) Loss is only from the lateral surface. Properties: Air ( T =(25+10)° C/2=290K, 1atm); K=0.0293 W/m. K, ν==19.91*10-6m2/s, α=28.4*10-6 m2/s; Water (290K); k=0.598 W/m.K; ν=µvf=1.081*10-6m2/s, α=k vf /cp =1.431*10-7m2/s, βf=174*10-6K-1. Analysis: in both water (wa) an air(a), the heat loss from the lateral surface of the cylinder approximating the body is _ _ q = h πDL(Ts − T∞ ) Where Ts and T∞ are the same for both situations. Hence, q wa h wa = _ qa ha _ Vertical cylinder in air: Ra L = gββ∆T 3 9.8m/s 2 (1/290K)(25 − 10)k(1.8m) 3 = 5.228 ×10 9 = −6 −6 2 2 να 19.91*10 m /s × 28.410 m /s with C = 0.1 and n = 1/3, _ _ h L Nu L = L = CRa n = 0.1(5.228 × 10 9 )1/3 = 173.4 L k h L = 2.82W/m 2 .K _ Vertical cylinder in water. Ra L = 9.8m/s 2 × 174 × 10 −6 K −1 (25 − 10)K(1.8m) 3 = 9.643 × 10 11 −6 −7 2 2 1.081 × 10 m /s × 1.431 × 10 m /s with C = 0.1 and n = 1/3, _ _ hL Nu L = = CRa n = 0.1(9.643 × 1011 ) 1/3 = 978.9 L k h L = 328W/m 2 .K _ Hence, from this analysis we find q wa 328W/m 2 .K = = 117 qa 2.8W/m 2 .K which compares poorly with the claim of 30 Problem 24 In a study of heat losses from buildings, free convection heat transfer from room air at 305 K to the inner surface of a 2.5-m-high wall at 295 K is simulated by performing laboratory experiments using water in a smaller test cell. In the experiments the water and the inner surface of the test cell are maintained at 300 and 290K, respectively. To achieve similarity between conditions in the room and the test cell, what is the required test cell height? If the average Nusselt number for the wall may be correlated exclusively in terms of the Rayleigh number, what is the ratio f the average convection coefficient for the room wall to the average coefficient for the test cell wall? Known: Air temperature and wall temperature and height for a room. Water temperature and wall temperature for a simulation experiment. Find: Required test cell height for similarity. Ratio and height convection coefficient for the two cases. Schematic: La=2.5m Assumptions: (1) Air and water are quiescent; (2) Flow conditions correspond to free convection boundary layer development on an isothermal vertical plate, (3) constant properties. Properties: Air (Tf =300K, 1 atm); K=0.0293 W/m.K, ν==15.9*10-6m2/s, α=22.5*10-6 m2/s; β=1/Tf=3.33*10-3K-1, k=0.0263W/m.K; water (Tf=295K):ρ=998kg/m3, µ=959*10-6N.s/m2, cp=4181 J/kg.K, β=227.5*106 -1 K , k=0.606 W/m.K; Hence ν=µ/ ρ =9.61/s, α=k / ρ cp =1.45*10-7m2/s. Lw Analysis: Similarity requires that RaL,a=RaL,w where Ra L = hence, gβ (Ts -T ∞ )L3 να ⎡ 9.61 * 1.45 * 10 −14 3.33 * 10 −3 ⎤ Lw ⎡ (αν ) w β air ⎤ =⎢ =⎢ ⎥ −12 −3 ⎥ La ⎣ (αν ) a β H 2 o ⎦ ⎣15.9 * 22.5 * 10 0.228 * 10 ⎦ Lw = 2.5m(0.179) = 0.45m 1/ 3 if Ra L,a = Ra L,w , it follows that Nu L,a = Nu L,w .hence ha hw _ _ _ _ = L w k a 0.45 0.0263 = = 7.81 * 10 −3 La k w 2.5 0.606 Comments: Similitude allows us to obtain valuable information for one system by performing experiments for a smaller system and a different fluid. Problem 25 A square plate of pure aluminum, 0.5 m on a side and 16 mm hick, is initially at 300C and is suspended in a large chamber. The walls of the chamber are maintained at 27C, as is the enclosed air. If the surface emissivity of the plate temperature during the cooling process? Is it reasonable to assume a uniform plate temperature during the cooling process? Known: Initial temperature and dimensions of an aluminum plate. Conditions of the plate surroundings. Find: (a) initial cooling rate, (2) validity of assuming negligible temperature gradients in the plate during the cooling process. Schematic: Assumptions: (1) plate temperature is uniform; (2) chamber air is quiescent, (3) Plate surface is diffuse-gray, (4) Chamber surface is much larger than that of plate, (5) Negligible heat transfer from edges. Properties: Aluminum (573k); k=232W/m.k, cp=1022J/kg.K, ρ=2702 kg/m3: Air (Tf=436K, 1 atm):υ=30.72*10-6m2/s,α=44.7*10-6m2/s,k=0.0363W/m.K, Pr=0.687, β=0.00229K-1. Analysis: (a) performing an energy balance on the plate, L 4 − q = −2 As [h(Ts − T∞ ) + εσ (T 4 − Tsur )] = E st = ρVc p [dT / dt ] _ . 4 dT / dt = −2 As [h(Ts − T∞ ) + εσ (T 4 − Tsur )] / ρwc p _ Ra L = _ gβ (Ts -T ∞ )L3 να = 9.8m/s 2 × 0.00229K - 1(300 − 27)K(0.5m)3 = 5.58 × 10 8 2 2 −6 −6 30.72 *10 m /s × 44.7 × 10 m /s h= _ ⎫ 0.0363 ⎧ 0.670 Ra 1 / 4 k⎧ 0.670(5.58 × 10 8 )1 / 4 ⎫ L = 0.68 + 0.68 + ⎨ ⎬ ⎨ ⎬ L⎩ 0 .5 ⎩ [1 + (0.492 / Pr) 9 / 16 ] 4 / 9 ⎭ [1 + (0.492 / 0.687) 9 / 16 ] 4 / 9 ⎭ h = 5.8W / m 2 .K Hence the initial cooling rate is dT 2 5.8W / m 2 .K (300 − 27)C ° + 0.25 × 5.67 × 10 −8 W / m 2 .K 4 [(573K ) 4 − (300 K ) 4 ] =− dt 2702kg / m 3 (0.016m)1022 J / kg.k dT = −0.136 K / s dt { } (b) To check the Validity of neglecting temperature gradients across the plate thickness, calculate " Bi = heff ( w / 2) / k where heff = qtot /(Ti − T∞ ) = (1583 + 1413)W / m 2 / 273K = 11.0W / m 2 .k hence Bi = (11W/m2.K)(0.008m)/232W/m.K = 3.8 × 10 -4 And the assumption is excellent. Comments: (1) Longitudinal (x) temperature gradients are likely to me more severe than those associated with the plate thickness due to the variation of h with x. (2) Initially q”conv ≈q”rad. Problem 26 Beer in cans 150 mm long and 60 mm in diameter is initially at 27°C and is to be cooled by placement in a refrigerator compartment at 4°C. In the interest of maximizing the cooling rate, should the cans be laid horizontally or vertically in the compartment? As a first approximation, neglect heat transfer from the ends. Known: Dimensions and temperature of beer can in refrigerator compartment. Find: orientation which maximum cooling rate. Air, T∞=4°C P=1 atm L=150mm D=0.06m Ts=27 C Schematic: Horizontal, (h) Vertical, (V) Assumptions: (1) End effects are negligible, (2) Compartment air is quiescent, (3) constant properties. Properties: Air (Tf=288.5K, 1 atm): υ=14.87*10-6 m2/s, k=0.0254W/m.K, α=21.0*10-6m2/s, Pr=0.71, β=1/Tf=3.47*10-3K-1. Analysis: The ratio of cooling rates may be expressed as _ _ qv qh = hv _ hh πDL (Ts − T∞ ) hv = = πDL (Ts − T∞ ) _ hh For the vertical surface, find Ra L = gβ (Ts -T ∞ )L3 να = 9.8m/s 2 × 3.47 × 10 -3 K -1 (23°C) 3 L = 2.5 × 10 9 L3 (14.87 × 10 -6 m 2 /s)(21 × 10 -6 m 2 /s) Ra L = 2.5 × 10 9 (0.15) 3 = 8.44 × 10 6 Using the correct correlation ⎧ 0.387(8.44 × 10 6 )1 / 6 ⎫ NuL = ⎨0.825 + ⎬ = 29.7 [1 + (0.492 / 0.71) 9 / 16 ]8 / 27 ⎭ ⎩ _ _ _ 2 hence, hL = h v = Nu L k 0.0254W / m.K = 29.7 = 5.03 W / m 2 .K L 0.15m Using the correct correlation, ⎧ 0.387(5.4 × 10 5 )1 / 6 ⎫ ⎪ ⎪ = 12.24 Nu D = ⎨0.60 + 6 8 / 27 ⎬ ⎪ ⎪ [1 + (0.559 / 0.71) 9 / 1 ] ⎩ ⎭ _ _ _ _ 2 hD = h h = Nu D 0.0254W / m.K k = 12.24 = 5.18W / m 2 .K 0.06m D hence, qv 5.03 = = 0.97 q h 5.18 Comments: in view of the uncertainties associated with equations and the neglect of end effects, the above result is inconclusive. The cooling rates are are approximately the same. Module 6: Short questions 1. How does forced convection differ from natural convection? Is convection caused by wind forced or natural convection? 2. How does external forced convection differ from internal forced convection? Can a heat transfer system involve both internal and external convection at the same time? Give an example. 3. In which mode of heat transfer is the heat transfer coefficient usually higher: natural convection or forced convection? 4. What is the physical significance of the Nusselt number? 5. Distinguish between incompressible flow and incompressible fluid. 6. In forced convection, distinguish between upstream velocity and free stream velocity. For which type of flow are the two same? 7. What is the difference between skin friction drag and pressure drag? Which is usually significant for slender bodies such as airfoils? 8. For laminar flow over a flat plate, how do the local heat transfer coefficient and the friction coefficient vary with distance from the leading edge? 9. Define hydrodynamic entry length for flow in a tube. Is the entry length longer in laminar or turbulent flow? 10. How is thermal entry length defined for flow in a tube? 11. Consider the flow of oil in a circular tube. How will the hydrodynamic and thermal entry lengths compare if the flow is laminar? How would they compare if the flow is turbulent? 12. Consider the flow of mercury in a circular tube. How will the hydrodynamic and thermal entry lengths compare if the flow is laminar? How would they compare if the flow is turbulent? 13. How does Rayleigh number differ from Grashof number? Multiple choice questions: 1) In a laminar boundary layer near a heated flat plate of zero emissivity, the heat flux away from the plate is due to a) conduction b) macroscopic eddies c) thermal radiation d) subspace waves e) None of the above 2) A rectangular plate with the dimensions L × 2L is at a fixed temperature different than the fluid that surrounds it. Two orientations are considered with the plate parallel to an oncoming irrotational uniform flow 1) With the longer side normal to the oncoming flow 2)With the shorter side normal to the flow. Choose the true statement below a) The heat transfer is always higher for configuration 1) b) The heat transfer is always higher for configuration 2) c) The Nusselt number increases with x until Rex=2300 d) The heat transfer on either is higher if the boundary layer is tripped at x = 0 e) None of the above 3) In a staggered array of 1 cm tubes, the oncoming flow of 2 m/s gas (ν = 3 × 10-5m2/s) enters the bundle to cool it. If SL = 2 cm, ST = 3 cm, the Reynolds number used to calculate Nusselt number is: a) 1000 b) 1111 c) 3333 d) 10000 e) None of the above 4) Choose the false statement for steady laminar flow in a circular pipe with uniform wall temperature unequal to the inlet fluid temperature a) hD/k = 3.66 once the flow is fully developed b) The Nusselt number in the development region is greater than the Nusselt number in the fully developed flow c) When the flow is fully developed, the fluid temperature equals the wall temperature d) The Reynolds number is less than 500000 e) None of the above 5) Two identical heated plates are exposed to still air, thus generating natural convection. One plate is vertical with a Grashof number of Gr1, while the other faces down at an angle of 45 degrees and has a Grashof number of Gr2. The ratio Gr2/Gr1 is a) 0.5 b) 0.7071 c) 1 d) 1.414 e) None of the above 6) A 1 cm cylinder at 400 K is surrounded by still air at 300 K is inside the orbiting space shuttle ( ε = 1.0). If ν2 = 2 × 10-10 m4/s2, the Grashof number is a) 109 b) 10-9 c) 14014 d) 7007 e) None of the above MODULE 6 CONVECTION 6.1 Objectives of convection analysis: Main purpose of convective heat transfer analysis is to determine: - flow field - temperature field in fluid - heat transfer coefficient, h How do we determine h ? Consider the process of convective cooling, as we pass a cool fluid past a heated wall. This process is described by Newton’s law of Cooling: q=h·A·(TS-T∞) U∞ y U∞ u(y) q” Ts y T∞ T(y) Near any wall a fluid is subject to the no slip condition; that is, there is a stagnant sub layer. Since there is no fluid motion in this layer, heat transfer is by conduction in this region. Above the sub layer is a region where viscous forces retard fluid motion; in this region some convection may occur, but conduction may well predominate. A careful analysis of this region allows us to use our conductive analysis in analyzing heat transfer. This is the basis of our convective theory. At the wall, the convective heat transfer rate can be expressed as the heat flux. ′′ qconv = − k f ∂T ⎞ ⎟ = h (Ts − T∞ ) ∂y ⎟ y =0 ⎠ U∞ y T∞ T(y) Ts −kf Hence, h = ∂T ∂y (Ts − T∞ ) ∂T ⎞ ⎟ ∂y ⎟ y =0 ⎠ But ⎞ depends on the whole fluid motion, and both fluid flow and heat transfer ⎟ ⎟ ⎠ y=0 equations are needed The expression shows that in order to determine h, we must first determine the temperature distribution in the thin fluid layer that coats the wall. 2.2 Classes of Convective Flows Free or natural convection (induced by buoyancy forces) Convection Forced convection (induced by external means) • • • • • • extremely diverse several parameters involved (fluid properties, geometry, nature of flow, phases etc) systematic approach required classify flows into certain types, based on certain parameters identify parameters governing the flow, and group them into meaningful nondimensional numbers need to understand the physics behind each phenomenon Common classifications: A. Based on geometry: External flow / Internal flow B. Based on driving mechanism Natural convection / forced convection / mixed convection C. Based on number of phases Single phase / multiple phase D. Based on nature of flow Laminar / turbulent May occur with phase change (boiling, condensation) Table 6.1. Typical values of h (W/m2K) Free convection gases: 2 - 25 liquid: 50 – 100 gases: 25 - 250 liquid: 50 - 20,000 2500 -100,000 Forced convection Boiling/Condensation 2.3 How to solve a convection problem ? • • Solve governing equations along with boundary conditions Governing equations include 1. conservation of mass 2. conservation of momentum 3. conservation of energy In Conduction problems, only (3) is needed to be solved. Hence, only few parameters are involved In Convection, all the governing equations need to be solved. ⇒ large number of parameters can be involved • • 2.4 FORCED CONVECTION: external flow (over flat plate) An internal flow is surrounded by solid boundaries that can restrict the development of its boundary layer, for example, a pipe flow. An external flow, on the other hand, are flows over bodies immersed in an unbounded fluid so that the flow boundary layer can grow freely in one direction. Examples include the flows over airfoils, ship hulls, turbine blades, etc U U U < U∞ • • • Fluid particle adjacent to the solid surface is at rest These particles act to retard the motion of adjoining layers ⇒ boundary layer effect Inside the boundary layer, we can apply the following conservation principles: Momentum balance: inertia forces, pressure gradient, viscous forces, body forces Energy balance: convective flux, diffusive flux, heat generation, energy storage 2.5 Forced Convection Correlations Since the heat transfer coefficient is a direct function of the temperature gradient next to the wall, the physical variables on which it depends can be expressed as follows: h=f(fluid properties, velocity field ,geometry,temperature etc.) As the function is dependent on several parameters, the heat transfer coefficient is usually expressed in terms of correlations involving pertinent non-dimensional numbers. Forced convection: Non-dimensional groupings Nusselt No. Nu = hx / k = (convection heat transfer strength)/ (conduction heat transfer strength) • Prandtl No. Pr = ν/α = (momentum diffusivity)/ (thermal diffusivity) • Reynolds No. Re = U x / ν = (inertia force)/(viscous force) Viscous force provides the dampening effect for disturbances in the fluid. If dampening is strong enough ⇒ laminar flow Otherwise, instability ⇒ turbulent flow ⇒ critical Reynolds number For forced convection, the heat transfer correlation can be expressed as Nu=f (Re, Pr) • The convective correlation for laminar flow across a flat plate heated to a constant wall temperature is: U∞ x Nux = 0.323·Rex½ · Pr1/3 where Nux ≡ h⋅x/k Rex ≡ (U∞⋅x⋅ρ)/μ Pr ≡ cP⋅μ/k Physical Interpretation of Convective Correlation The Reynolds number is a familiar term to all of us, but we may benefit by considering what the ratio tells us. Recall that the thickness of the dynamic boundary layer, δ, is proportional to the distance along the plate, x. Rex ≡ (U∞⋅x⋅ρ)/μ ∝ (U∞⋅δ⋅ρ)/μ = (ρ⋅U∞2)/( μ⋅U∞/δ) The numerator is a mass flow per unit area times a velocity; i.e. a momentum flow per unit area. The denominator is a viscous stress, i.e. a viscous force per unit area. The ratio represents the ratio of momentum to viscous forces. If viscous forces dominate, the flow will be laminar; if momentum dominates, the flow will be turbulent. Physical Meaning of Prandtl Number The Prandtl number was introduced earlier. If we multiply and divide the equation by the fluid density, ρ, we obtain: Pr ≡ (μ/ρ)/(k/ρ⋅cP) = υ/α The Prandtl number may be seen to be a ratio reflecting the ratio of the rate that viscous forces penetrate the material to the rate that thermal energy penetrates the material. As a consequence the Prandtl number is proportional to the rate of growth of the two boundary layers: δ/δt = Pr1/3 Physical Meaning of Nusselt Number The Nusselt number may be physically described as well. Nux ≡ h⋅x/k If we recall that the thickness of the boundary layer at any point along the surface, δ, is also a function of x then Nux ∝ h⋅δ/k ∝ (δ/k⋅A)/(1/h⋅A) We see that the Nusselt may be viewed as the ratio of the conduction resistance of a material to the convection resistance of the same material. Students, recalling the Biot number, may wish to compare the two so that they may distinguish the two. Nux ≡ h⋅x/kfluid Bix ≡ h⋅x/ksolid The denominator of the Nusselt number involves the thermal conductivity of the fluid at the solid-fluid convective interface; The denominator of the Biot number involves the thermal conductivity of the solid at the solid-fluid convective interface. Local Nature of Convective Correlation Consider again the correlation that we have developed for laminar flow over a flat plate at constant wall temperature Nux = 0.323·Rex½ · Pr1/3 To put this back into dimensional form, we replace the Nusselt number by its equivalent, hx/k and take the x/k to the other side: h = 0.323·(k/x)⋅Rex½ · Pr1/3 Now expand the Reynolds number h = 0.323·(k/x)⋅[(U∞⋅x⋅ρ)/μ]½ · Pr1/3 We proceed to combine the x terms: h = 0.323·k⋅[(U∞⋅ρ)/( x⋅μ)]½ · Pr1/3 And see that the convective coefficient decreases with x½. Convection Coefficient, h. U∞ Thermal Boundary Layer, δt x Hydrodynamic Boundary Layer, δ We see that as the boundary layer thickens, the convection coefficient decreases. Some designers will introduce a series of “trip wires”, i.e. devices to disrupt the boundary layer, so that the buildup of the insulating layer must begin anew. This will result in regular “thinning” of the boundary layer so that the convection coefficient will remain high. Use of the “Local Correlation” A local correlation may be used whenever it is necessary to find the convection coefficient at a particular location along a surface. For example, consider the effect of chip placement upon a printed circuit board: U∞ Chip 1 Chip 2 Chip 3 X1 X2 X3 Here are the design conditions. We know that as the higher the operating temperature of a chip, the lower the life expectancy. With this in mind, we might choose to operate all chips at the same design temperature. Where should the chip generating the largest power per unit surface area be placed? The lowest power? Life expectancy of Chip Operating Temperature of Chip Averaged Correlations If one were interested in the total heat loss from a surface, rather than the temperature at a point, then they may well want to know something about average convective coefficients. For example, if we were trying to select a heater to go inside an aquarium, we would not be interested in the heat loss at 5 cm, 7 cm and 10 cm from the edge of the aquarium; instead we want some sort of an average heat loss. Average Convection Coefficient, hL U∞ Local Convection x Coefficient, hx. The desire is to find a correlation that provides an overall heat transfer rate: Q = hL⋅A⋅[Twall-T∞] = ∫ hx ⋅ [Twall − T∞ ] ⋅ dA = ∫0 hx ⋅ [Twall − T∞ ] ⋅ dx L where hx and hL, refer to local and average convective coefficients, respectively. Compare the second and fourth equations where the area is assumed to be equal to A = (1⋅L): hL⋅L⋅[Twall-T∞] = ∫0 hx ⋅ [Twall − T∞ ] ⋅ dx L Since the temperature difference is constant, it may be taken outside of the integral and cancelled: hL⋅L= ∫0 hx ⋅ dx L This is a general definition of an integrated average. Proceed to substitute the correlation for the local coefficient. k ⎡ U∞ ⋅ x ⋅ ρ ⎤ hL⋅L= ∫0 0.323 ⋅ ⋅ ⎢ ⎥ x ⎣ μ ⎦ L 0.5 ⋅ Pr 1/3 ⋅ dx Take the constant terms from outside the integral, and divide both sides by k. ⎡ U∞ ⋅ ρ ⎤ L ⎡ 1⎤ 1/ 3 hL⋅L/k = 0.323 ⋅ ⎢ ⎥ ⋅ Pr ⋅ ∫0 ⎢ x ⎥ ⋅ dx ⎣ ⎦ ⎣ μ ⎦ 0.5 0.5 Integrate the right side. ⎡ U∞ ⋅ ρ ⎤ hL⋅L/k = 0.323 ⋅ ⎢ ⎥ ⎣ μ ⎦ 0.5 x 0.5 ⋅ Pr ⋅ 0.5 1/3 L 0 The left side is defined as the average Nusselt number, NuL. Algebraically rearrange the right side. 0.323 ⎡U ∞ ⋅ ρ ⎤ ⋅ NuL = 0 .5 ⎢ μ ⎥ ⎣ ⎦ 0.5 ⋅ Pr ⋅ L 1 3 0.5 ⎡U ⋅ L ⋅ ρ ⎤ = 0.646 ⋅ ⎢ ∞ ⎥ μ ⎣ ⎦ 0 .5 ⋅ Pr 1 3 The term in the brackets may be recognized as the Reynolds number, evaluated at the end of the convective section. Finally, NuL = 0.646 ⋅ Re 0.5 ⋅ Pr L 1 3 This is our average correlation for laminar flow over a flat plate with constant wall temperature. Reynolds Analogy In the development of the boundary layer theory, one may notice the strong relationship between the dynamic boundary layer and the thermal boundary layer. Reynold’s noted the strong correlation and found that fluid friction and convection coefficient could be related. This refers to the Reynolds Analogy. Conclusion from Reynold’s analogy: Knowing the frictional drag, we know the Nusselt Number. If the drag coefficient is increased, say through increased wall roughness, then the convective coefficient will increase. coefficient is decreased. If the wall friction is decreased, the convective Turbulent Flow We could develop a turbulent heat transfer correlation in a manner similar to the von Karman analysis. It is probably easier, having developed the Reynolds analogy, to follow that course. The local fluid friction factor, Cf, associated with turbulent flow over a flat plate is given as: Cf = 0.0592/Rex0.2 Substitute into the Reynolds analogy: (0.0592/Rex0.2)/2 = Nux/RexPr1/3 Rearrange to find Nux = 0.0296⋅Rex0.8⋅Pr1/3 Local Correlation Turbulent Flow Flat Plate. In order to develop an average correlation, one would evaluate an integral along the plate similar to that used in a laminar flow: Laminar Region Turbulent region crit hL⋅L = ∫0 hx dx = ∫0 hx ,la min ar ⋅ dx + ∫ Lcrit hx ,turbulent ⋅ dx L L L Note: The critical Reynolds number for flow over a flat plate is 5⋅105; the critical Reynolds number for flow through a round tube is 2000. The result of the above integration is: Nux = 0.037⋅(Rex0.8 – 871)⋅Pr1/3 Note: Fluid properties should be evaluated at the average temperature in the boundary layer, i.e. at an average between the wall and free stream temperature. Tprop = 0.5⋅(Twall+ T∞) 2.6 Free convection Free convection is sometimes defined as a convective process in which fluid motion is caused by buoyancy effects. Tw Heated boundary layer T∞ < Tboundry. layer < Tw ρ∞ < ρboundry. layer Velocity Profiles Compare the velocity profiles for forced and natural convection shown below: U∞ > 0 U∞ = 0 Forced Convection Free Convection Coefficient of Volumetric Expansion The thermodynamic property which describes the change in density leading to buoyancy in the Coefficient of Volumetric Expansion, β. β≡− Evaluation of β • • 1 ∂ρ ⋅ ρ ∂T P = Const . Liquids and Solids: β is a thermodynamic property and should be found from Property Tables. Values of β are found for a number of engineering fluids in Tables given in Handbooks and Text Books. Ideal Gases: We may develop a general expression for β for an ideal gas from the ideal gas law: P = ρ⋅R⋅T Then, ρ = P/R⋅T Differentiating while holding P constant: dρ dT =− P = Const . P ρ ⋅ R⋅T ρ =− 2 = − 2 R⋅T R⋅T T Substitute into the definition of β β = 1 Tabs Ideal Gas Grashof Number Because U∞ is always zero, the Reynolds number, [ρ⋅U∞⋅D]/μ, is also zero and is no longer suitable to describe the flow in the system. Instead, we introduce a new parameter for natural convection, the Grashof Number. Here we will be most concerned with flow across a vertical surface, so that we use the vertical distance, z or L, as the characteristic length. g ⋅ β ⋅ Δ T ⋅ L3 Gr ≡ ν2 Just as we have looked at the Reynolds number for a physical meaning, we may consider the Grashof number: ⎛ Buoyant Force ⎞ ⎛ Momentum ⎞ ρ ⋅ g ⋅ β ⋅ Δ T ⋅ L3 2 ⎟ ⋅⎜ ⎟ 2 3 ( ) ⋅ ( ρ ⋅ U max ) ⎜ 2 ⎝ ⎠ ⎝ ρ ⋅ g⋅ β ⋅ ΔT ⋅ L Area Area ⎠ L Gr ≡ = = 2 2 U max μ2 ⎛ ViscousForce ⎞ 2 μ ⋅ 2 ⎜ ⎟ L ⎝ ⎠ Area Free Convection Heat Transfer Correlations The standard form for free, or natural, convection correlations will appear much like those for forced convection except that (1) the Reynolds number is replaced with a Grashof number and (2) the exponent on Prandtl number is not generally 1/3 (The von Karman boundary layer analysis from which we developed the 1/3 exponent was for forced convection flows): Nux = C⋅Grxm⋅Prn NuL = C⋅GrLm⋅Prn Local Correlation Average Correlation Quite often experimentalists find that the exponent on the Grashof and Prandtl numbers are equal so that the general correlations may be written in the form: Nux = C⋅[Grx⋅Pr]m NuL = C⋅[GrL⋅Pr]m Local Correlation Average Correlation This leads to the introduction of the new, dimensionless parameter, the Rayleigh number, Ra: Rax = Grx⋅Pr RaL = GrL⋅Pr So that the general correlation for free convection becomes: Nux = C⋅Raxm NuL = C⋅RaLm Local Correlation Average Correlation Laminar to Turbulent Transition Just as for forced convection, a boundary layer will form for free convection. The insulating film will be relatively thin toward the leading edge of the surface resulting in a relatively high convection coefficient. At a Rayleigh number of about 109 the flow over a flat plate will transition to a turbulent pattern. The increased turbulence inside the boundary layer will enhance heat transfer leading to relative high convection coefficients, much like forced convection. Turbulent Flow Ra < 109 Ra > 109 Laminar flow. [Vertical Flat Plate] Turbulent flow. [Vertical Flat Plate] Laminar Flow Generally the characteristic length used in the correlation relates to the distance over which the boundary layer is allowed to grow. In the case of a vertical flat plate this will be x or L, in the case of a vertical cylinder this will also be x or L; in the case of a horizontal cylinder, the length will be d. Critical Rayleigh Number Consider the flow between two surfaces, each at different temperatures. Under developed flow conditions, the interstitial fluid will reach a temperature between the temperatures of the two surfaces and will develop free convection flow patterns. The fluid will be heated by one surface, resulting in an upward buoyant flow, and will be cooled by the other, resulting in a downward flow. Note that for enclosures it is customary to develop correlations which describe the overall (both heated and cooled surfaces) within a single correlation. Q T1 L Free Convection Inside an Enclosure If the surfaces are placed closer together, the flow patterns will begin to interfere: T2 Q Q Q Q T2 T1 T1 L Free Convection Inside an Enclosure With Partial Flow Interference L Free Convection Inside an Enclosure With Complete Flow Interference T2 Q In the case of complete flow interference, the upward and downward forces will cancel, canceling circulation forces. This case would be treated as a pure convection problem since no bulk transport occurs. The transition in enclosures from convection heat transfer to conduction heat transfer occurs at what is termed the “Critical Rayleigh Number”. Note that this terminology is in clear contrast to forced convection where the critical Reynolds number refers to the transition from laminar to turbulent flow. Racrit = 1000 Racrit = 1728 (Enclosures With Horizontal Heat Flow) (Enclosures With Vertical Heat Flow) The existence of a Critical Rayleigh number suggests that there are now three flow regimes: (1) No flow, (2) Laminar Flow and (3) Turbulent Flow. In all enclosure problems the Rayleigh number will be calculated to determine the proper flow regime before a correlation is chosen. Module 7: Learning objectives • One of the most common applications of heat transfer is to design equipment for exchanging heat from one fluid to another. Such devices are generally called Heat Exchangers. Because there are many important applications, heat exchanger research and development has had a long history. Such activity is by no means complete, however as many talented workers continue to seek ways of improving deign and performance. From the view point of energy conservation and space constraint, there has been a steady and substantial increase in research activity. a focal point for this work has been heat transfer enhancement, which includes the search for special heat exchanger surfaces through which enhancement may be achieved. In this chapter we have attempted to develop tools that will allow you to perform approximate heat exchanger calculations. Although we have restricted ourselves to heat exchangers involving separation of hot and cold fluids by stationary walls, there are other important options. for example, evaporative heat exchangers enable direct contact between liquid and gas and because of latent energy effects, large heat transfer rates per unit volume are possible. Also for gas-to-gas heat exchange, use is often made of regenerators in which the same space is alternately occupied by hot and cold gases. In a fixed regenerator such as a packed bed, the hot and cold gases alternately enter a stationary, porous solid in a rotary regenerator, the porous solid is a rotation wheel, which alternately exposes its surfaces to the continuously flowing hot and cold gases. • • • What are heat exchangers for? Heat exchangers are practical devices used to transfer energy from one fluid to another To get fluid streams to the right temperature for the next process – reactions often require feeds at high temp. To condense vapours To evaporate liquids To recover heat to use elsewhere To reject low-grade heat To drive a power cycle Application: Power cycle Steam Turbine Boiler Feed water Heater Condenser Main Categories Of Exchanger Heat exchangers Recuperators Regenerators Wall separating streams Wall separating streams Direct contact Most heat exchangers have two streams, hot and cold, but some have more than two Recuperators/Regenerators Recuperative: Has separate flow paths for each fluid which flow simultaneously through the exchanger transferring heat between the streams Regenerative Has a single flow path which the hot and cold fluids alternately pass through. Compactness Can be measured by the heat-transfer area per unit volume or by channel size Conventional exchangers (shell and tube) have channel size of 10 to 30 mm giving about 100m2/m3 Plate-type exchangers have typically 5mm channel size with more than 200m2/m3 More compact types available Double Pipe Simplest type has one tube inside another - inner tube may have longitudinal fins on the outside However, most have a number of tubes in the outer tube - can have very many tubes thus becoming a shell-and-tube Shell and Tube Typical shell and tube exchanger as used in the process industry Shell-Side Flow Plate-Fin Exchanger Made up of flat plates (parting sheets) and corrugated sheets which form fins Brazed by heating in vacuum furnace Configurations Heat Transfer Considerations: Overall heat transfer coefficient Internal and external thermal resistances in series 1 1 1 = = UA (UA )c (UA )h R ′′,c R ′′,h 1 1 1 f f = + + Rw + + (hηo A )h (ηo A )h UA (hηo A )c (ηo A )c A is wall total surface area on hot or cold side R”f is fouling factor (m2K/W) Rw wall Fin ηo is overall surface efficiency (if finned) Heat Transfer Considerations (contd…): Fouling factor Material deposits on the surfaces of the heat exchanger tube may add further resistance to heat transfer in addition to those listed above. Such deposits are termed fouling and may significantly affect heat exchanger performance. Scaling is the most common form of fouling and is associated with inverse solubility salts. Examples of such salts are CaCO3, CaSO4, Ca3(PO4)2, CaSiO3, Ca(OH)2, Mg(OH)2, MgSiO3, Na2SO4, LiSO4, and Li2CO3. Corrosion fouling is classified as a chemical reaction which involves the heat exchanger tubes. Many metals, copper and aluminum being specific examples, form adherent oxide coatings which serve to passivity the surface and prevent further corrosion. Heat Transfer Considerations (contd…): Chemical reaction fouling involves chemical reactions in the process stream which results in deposition of material on the heat exchanger tubes. When food products are involved this may be termed scorching but a wide range of organic materials are subject to similar problems. Freezing fouling is said to occur when a portion of the hot stream is cooled to near the freezing point for one of its components. This is most notable in refineries where paraffin frequently solidifies from petroleum products at various stages in the refining process, obstructing both flow and heat transfer. Biological fouling is common where untreated water is used as a coolant stream. Problems range from algae or other microbes to barnacles. Heat Transfer Considerations (contd…): Fluid Seawater and treated boiler feedwater (below 50oC) Seawater and treated boiler feedwater (above 50oC) River water (below 50oC) Fuel Oil Regrigerating liquids Steam (non-oil bearing) R”, m K/Watt 0.0001 0.0002 0.0002-0.001 0.0009 0.0002 0.0001 2 Basic flow arrangement in tube in tube flow t1 t2 t2 t1 T1 T1 Temperature Parallel Flow T2 T 1 T 1 Temperature C ounter Flow T 2 T2 t1 t2 T 2 t2 t1 Position Position Heat Exchanger Analysis Log mean temperature difference (LMTD) method . Want a relation Q = UA ΔTm Where ΔTmis some mean ΔT between hot and cold fluid Heat Exchanger Analysis(contd…) Counterflow Note Th ,out can be < Tc ,out Parallel flow T ' s' cannot cross Heat Exchanger Analysis (contd…) Energy balance (counterflow) on element shown & & & (1) dQ = −mh ch dTh = −mc cc dTc & m = mass flow rate of fluid c = specific heat Rate Equation & dQ = UdA(T − T ) h c ( 2) & − dQ dTc = & mc c c Now from (1) & − dQ dTh = & mh ch &⎛ 1 − 1 ⎞ ⎟ ∴ d (Th − Tc ) = dQ ⎜ ⎜m c ⎟ & ⎝ & c c mh ch ⎠ Heat Exchanger Analysis (contd…) & Subtract dQ from (2), ⎛ 1 d (Th − Tc ) 1 ⎞ = U⎜ ⎜ m c − m c ⎟dA &h h ⎟ Th − Tc ⎝ &c c ⎠ Integrate 1 → 2 ⎛ Th 2 − Tc 2 ⎞ ⎛ 1 1 ⎞ ln⎜ ⎜ T − T ⎟ = UA⎜ m c − m c ⎟ ⎟ ⎜ & &h h ⎟ c1 ⎠ ⎝ h1 ⎝ c c ⎠ Total heat transfer rate & & & & Q = mh ch (Th1 − Th 2 ) and Q = mc cc (Tc1 − Tc 2 ) Heat Exchanger Analysis (contd…) & Substitute for mc and put ΔT1 = Th1 − Tc1 END 1 ΔT2 = Th 2 − Tc 2 & = UA⎡ ΔT2 − ΔT1 ⎤ Q ⎢ ln (ΔT / ΔT )⎥ 2 1 ⎦ ⎣ & Q = UA(LMTD ) END 2 LMTD is Log Mean Temperature Difference • Remember – 1 and 2 are ends, not fluids • Same formula for parallel flow (but ΔT’s are different) •Counterflow has highest LMTD, for given T’s therefore smallest area for Q. Heat Exchanger Analysis (contd…) Condenser Evaporator Multipass HX Flow Arrangements In order to increase the surface area for convection relative to the fluid volume, it is common to design for multiple tubes within a single heat exchanger. With multiple tubes it is possible to arrange to flow so that one region will be in parallel and another portion in counter flow. 1-2 pass heat exchanger, indicating that the shell side fluid passes through the unit once, the tube side twice. By convention the number of shell side passes is always listed first. Multipass HX Flow Arrangements (contd…) The LMTD formulas developed earlier are no longer adequate for multipass heat exchangers. Normal practice is to calculate the LMTD for counter flow, LMTDcf, and to apply a correction factor, FT, such that Δθ eff = FT ⋅ LMTDCF The correction factors, FT, can be found theoretically and presented in analytical form. The equation given below has been shown to be accurate for any arrangement having 2, 4, 6, .....,2n tube passes per shell pass to within 2%. Multipass HX Flow Arrangements (contd…) ⎡ 1− P ⎤ R + 1 ln ⎢ ⎦ ⎣1 − R ⋅ P ⎥ FT = ⎡ 2 − P R +1− R 2 +1 (R − 1) ln ⎢ 2 ⎢ 2 − P R +1+ R +1 ⎣ 2 ( ( )⎤⎥ )⎥⎦ 1− X1/ Nshell Effectiven : P = ess , forR ≠1 1/ Nshell R −X Po P= , for R = 1 N shell − Po ⋅ (N shell − 1) T1 − T2 Capacity ratio R = t 2 − t1 t 2 − t1 Po = T1 − t1 X = Po ⋅ R − 1 Po − 1 T,t = Shell / tube side; 1, 2 = inlet / outlet Multipass HX Flow Arrangements (contd…) 1.0 FT R=10.0 R=0.1 0.5 0.0 P 1.0 Effectiveness-NTU Method How willexisting H. Ex. perform for given inlet conditions ? & Qactual Defineeffectiven : ε = ess & Q max & whereQmaxis for an infinitelylong H.Ex. Onefluid ΔT → ΔTmax = Th,in − Tc,in & & & and sinceQ = (mc )ΔT = (mc )ΔT A A B B = CA ΔTA = CB ΔTB then only thefluid withlesserof CA , CB heat capacityrate can haveΔTmax Effectiveness-NTU Method(contd…) & i.e. Q max = C min ΔTmax and ε = & or, Q = εC min (Th .in − Tc.in ) C min (Th .in − Tc.in ) & Q Want expression for ε which does not contain outlet T' s & Substitute back into Q = UA(LMTD) ......... ⎡ - UA ⎛ C min ⎞⎤ ⎜1 − 1 - exp⎢ ⎜ C ⎟⎥ ⎟ C min ⎝ max ⎠ ⎦ ⎣ ε= ⎡ - UA ⎛ C min ⎞⎤ C min ⎜1 − 1− exp⎢ ⎜ C ⎟⎥ ⎟ C max C min ⎝ max ⎠ ⎦ ⎣ ⎛ C ⎜ NTU , min ∴ε = ε ⎜ C max ⎝ ⎞ ⎟ ⎟ ⎠ and No. of transfer units (size of HEx.) NTU = UA C min Charts for each Configuration Procedure: Determine Cmax, Cmin/Cmax Get UA/Cmin, → ε from chart & Q = ε C min (Th.in − Tc.in ) Charts for each Configuration Procedure: Determine Cmax, Cmin/Cmax Get UA/Cmin, → ε from chart & Q = ε C min (Th.in − Tc.in ) Effectiveness-NTU Method(contd…) NTU max = UA C min ⇒ A= NTU max C min U • NTUmax can be obtained from figures in textbooks/handbooks First, however, we must determine which fluid has Cmin • For the type of HEX used in this problem & & m g c pg (T1 − T2 ) = mw cw (t1 − t 2 ) ⇒ & m g c pg & mw cw (t1 − t 2 ) = (T1 − T2 ) Examination of the last equation, subject to values given, indicated that gas will have Cmin. Effectiveness-NTU Method(contd…) t −t . . ⎛ kg ⎞⎛ J ⎞ 85 − 35 = ⎛ W⎞ ⎟ C = m g c pg = m g cw 2 1 = ⎜ 2.5 ⎟⎜ 4179 ⎟ ⎜ 4,882 ⎜ min °C ⎠ T −T s ⎟⎜ kg .°C ⎟ 200 − 93 ⎝ ⎠⎝ ⎠ 1 2 ⎝ . ⎞ ⎛ kg ⎞⎛ Cmax = m g cw = ⎜ 2.5 ⎟⎜ 4179 J ⎟ = ⎛10,448 W ⎞ ⎟ ⎜ ⎜ °C ⎠ s ⎟⎜ kg.°C ⎟ ⎝ ⎝ ⎠⎝ ⎠ = Effectiven ess can be calculated using = − − = − − = = Effectiveness-NTU Method (contd…) ⎫ C min =0.467⎪ ⎪ Cmax ⎪ ε =0.649 ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ → → NTU =1.4 max W⎞ 1.4 4,882 ⎟ NTU C max min = °C⎟ =38.0m2 ⎠ A= U 180 W m2°C ⎛ ⎜ ⎜ ⎝ Module 7: Short questions 1. What are the heat transfer mechanisms involved during heat transfer from the hot fluid to the cold fluid? 2. In heat exchange between air and water across a tube wall, it is proposed to use fins to enhance the overall heat transfer coefficient. Would you put the fins on the air side or on the water side? 3. When is a heat exchanger classified as compact? 4. How does a cross flow heat exchanger differ from a counter flow one? 5. What is the role of baffles in a shell-and-tube heat exchanger? What is the implication about pressure drop? 6. Under what conditions is the effectiveness NTU method preferred over LMTD method as a method of analysis of a heat exchanger? 7. Can temperature of the hot fluid drop below the inlet temperature of the cold fluid at any location in a heat exchanger? 8. Can temperature of the cold fluid rise above the inlet temperature of the hot fluid at any location in a heat exchanger? 9. Consider two double pipe counterflow heat exchangers that are identical except that one is twice as long as the other one. Which of the exchangers is more likely to have a higher effectiveness? 10. Can effectiveness be greater than one? 11. Under what conditions can a counter flow heat exchanger have an effectiveness of one? What would be your answer for a parallel flow heat exchanger? Multiple choice questions: 1) In a thin walled heat exchanger with no fouling, the overall heat transfer coefficient is a) A(hi−1 + ho−1 ) b) (hi−1 + ho−1 ) −1 −1 c) A(hi−1 + ho−1 ) d) (hi−1 + ho−1 ) e) None of the above 2) In a liquid to gas heat exchanger, it is best to put extended surfaces on the gas side because a) This reduces fouling b) The gas side heat transfer coefficient is highest c) It reduces drag in high speed flows d) All of the first three e) None of the above 3) When applying the ε - NTU method for heat exchangers, when one fluid is condensing steam, the heat capacity ratio Cr is effectively a) 0 b) 1 c) π d) ∞ e) None of the above 4) On one side of a heat exchanger, air enters at 72.82ºC and leaves at 90ºC. On the other side of the heat exchanger is condensing steam at one atmosphere. The value for ΔTlmtd is a) 10 K b) 17.18 K c) 27ºC d) 100ºC e) None of the above 5) Select the FALSE statement concerning the ε -NTU method for heat exchangers a) qmax = Cmin (Th,i - Tc,i) b) ε = q/qmax c) NTU = UA/Cmin d) q = ε Cmin (Th,i - Tc,i) e) None of the above Module 7: Solved Problems 1. A thin-walled concentric tube heat exchanger of 0.19-m length is to be used to heat deionized water from 40 to 60°C at a flow rate of 5 kg/s. the deionized water flows through the inner tube of 30-mm diameter while hot process water at 95°C flows in the annulus formed with the outer tube of 60-mm diameter. The thermo physical properties of the fluids are: Considering a parallel-flow configuration of the exchanger, determine the minimum flow rate required for the hot process water. Determine the overall heat transfer coefficient required for the conditions of part a. Considering a counter flow configuration, determine the minimum flow rate required for the hot process water. What is the effectiveness of the exchanger for this situation? Known: Thin-walled concentric tube, Parallel flow heat exchanger of prescribed diameter and length with process and deionized water. Inlet and outlet temperatures and flow rate of desired water. Inlet temperature and outlet temperature and flow rate of deionized water. Inlet temperature of process water. Find: (1) minimum flow rate required for the hot process water, (b) required overall heat transfer coefficient and whether it is possible to accomplish this heating, and (c) for CF arrangements minimum process water flow required and the effectiveness? Schematic: Assumptions: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes. Analysis: (a) from overall energy balances, q = ( m c ) h (Th,i − Th,o ) = ( m c ) h (Tc ,o − Tc ,i ) . . For a fixed term Th,i , (m )h will be a minimum when Th,o is a minimum. With the parallel flow configuration, this requires that Th,o=Tc,o=60°C. Hence, m h, min = . . ( mc ) c (Tc ,o − Tc ,i ) c h (Th,i − Th,o ) . 5kg / s × 4181J / kg . K (60 − 40) ° C = = 2.85kg / s 4197 J / kg . K (95 − 60)°C (b)From the rate equation and the log mean temperature relation, q = UAΔTlm , PF ΔTlm , PF = ΔT1 - ΔT2 ΔT ⎞ ln⎛ 1 ⎜ ΔT2 ⎟ ⎝ ⎠ And since ΔT2=0, ΔTlm=0 so that UA=∞. Since A=πDL is finite, U must be extremely large. Hence, the heating cannot be accomplished with this arrangement. . (c) With the CF arrangements m h will be a minimum when Tho is a minimum. This requires that Th,o is a minimum. This requires that Th,o is a minimum. This requires that Th,o=Tc,i=40°C. Hence, from the overall energy balance, m= . 5kg / s × 4181J / kg . K (60 − 40) K = 1.81kg / s 4197 J / kg . K (95 − 40) K For this condition, Cmin=Ch which is cooled from Th,i to Tc,i, hence ε=1 Comments: For the counter flow arrangement, the heat exchanger must be infinitely long. 2. An automobile radiator may be viewed as a cross-flow heat exchanger with both fluids unmixed. Water, which has flow rate of 0.05kg/s, enters the radiator at 400K and is to leave at 330 K. The water is cooled by air which enters at 0.75kg/s and 300K. If the overall heat transfer coefficient is 200W/m2.K, what is the required heat transfer surface area? Known: flow rate and inlet temperature for automobile radiator. Overall heat transfer coefficient. Find: Area required to achieve a prescribed outlet temperature. Schematic: Water Th,i=400K mh=0.05kg/s Air Tc,i=300K mc=0.75kg/s Tc,o Th,o=330K Assumptions: (1) Negligible heat loss to surroundings and kinetic and potential energy changes, (2) Constant properties. Analysis: The required heat transfer rate is q = ( m c ) h (Th,i − Th,o ) = 0.05kg / s(4209J / kg . K )70 K = 14,732W . Using the ε-NTU method, C min = C h = 210.45W / K C max = C c = 755.25W / K , hence , C min / C max (Th ,i − Tc ,i ) = 210.45W / K (100 K ) = 21,045W and ε = q / q max = 14,732W / 21,045W = 0.700 From figure, NTU≈1.5, hence A = NTU (C min / U ) = 1.5 × 210.45W / K ( 200W / m 2 . K ) = 1.58m 2 Comments: (1) the air outlet temperature is Tc ,o = Tc , i + q / C c = 300 K + (14,732W / 755.25W / K ) = 319.5 K (2) Using the LMTD approach, ΔTlm=51.2 K, R=0.279 and P=0.7. Hence from fig F≈0.95 and A = q / FUΔTlm = (14,732W ) /[0.95( 200W / m 2 . K )51.2 K ] = 1.51m 2 . 3. Saturated water vapor leaves a steam turbine at a flow rate of 1.5kg/s and a pressure of 0.51 bars. The vapor is to be completely condensed to saturated liquid in a shell-and –tube heat exchanger which uses city water as the cold fluid. The water enters the thin-walled tubes at 17°C and is to leave at 57°C. assuming an overall heat transfer coefficient of 200W/m2.K, determine the required heat exchanger surface area and the water flow rate. After extended operation, fouling causes the overall heat transfer coefficient to decrease to 100W/m2.K, and to completely condense the vapor, there must be an attendant reduction in the vapor flow rate. For the same water inlet temperature and flow rate, what is the new vapor flow rate required for complete condensation? Known: Pressure and initial flow rate of water vapor. Water inlet and outlet temperatures. Initial and final overall heat transfer coefficients. Find: (a) Surface area for initial U and water flow rate, (b) Vapour flow rate for final U. Schematic: Assumptions: (1) Negligible heat loss to surroundings, (2) Negligible wall conduction resistance. Properties: Table for sat.Water: (T c = 310 K ) : c p,c = 4178J / kg . K ; (p = 0.51 bars) : Tsa t = 355K, h fg = 2304kJ/kg. − Analysis: (a) The required heat transfer rate is q = m h h fg = 1.5kg / s( 2.304 × 10 6 J / kg ) = 3.46 × 10 6 W . And the corresponding heat capacity rate for the water is C c = C min = q /(Tc ,o − Tc ,i ) = 3.48 × 10 6 W / 40 K = 86,400W / K hence , ε = q /(C min [Th,i − Tc ,i ]) = 3.46 × 10 6 W / 86,400W / K (65 K ) = 0.62 since C min /C max = 0, NTU = -ln(1 - ε ) = − ln(1 − 0.62) = 0.97 And A = NTU (C min / U ) = 0.97(86,400W / K / 2000W / m 2. K ) = 41.9m 2 m c = C c / c p ,c = 86,400W / K / 4178J / kg . K = 20.7 kg / s . (b) using the final overall heat transfer coefficient, find Since C min /C max = 0, ε = 1 − exp(− NTU ) = 1 − exp(−0.485) = 0.384 hence, q = εC min (Th,i − Tc ,i ) = 0.384(886,400W / K )65 K = 2.16106W m h = q / h fg = 2.16 × 10 6 W / 2.304 × 10 6 J / kg = 0.936kg / s . Comments: The significant reduction (38%) in m h represents a significant loss in turbine power. Periodic cleaning of condenser surfaces should be employed to minimize the adverse effects of fouling. . 4. Water at 225 kg/h is to be heated from 35 to 95°C by means of a concentric tube heat exchanger. Oil at 225kg/h and 210°C, with a specific heat of 2095 J/kg.K, is to be used as the hot fluid, If the overall heat transfer coefficient based on the outer diameter of the inner tube if 550W/m2.K,determine the length of the exchanger if the outer diameters is 100mm. Known: Concentric tube heat exchanger. Find: Length of the exchanger Schematic: Assumptions: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes, (3) Constant properties. Properties: Table for Water: (Tc = ( 35 + 95) ° C / 2 = 338 K ) : c p ,c = 4188J / kg . K _ Analysis: From rate equation with Ao=πDoL, L=q/UoDoΔT l m The heat rate, q, can be evaluated from an energy balance on the cold fluid, q = m c c c (Tc , 0 − Tc ,i ) = . 225kg / h × 4188J / kg . K (95 − 35) K = 15,705W 3600 s / h In order to evaluate ΔT l m, we need to know whether the exchanger is operating in CF or PF. From an energy balance on the hot fluid, find Th,o = Th ,i − q / m h c h = 210 ° C − 15,705W / . 225kg / h J × 2095 = 90.1°C 3600 s / h kg . K Since Th,o<Tc,o it follows that HXer operation must be CF. From eq. for log mean temperature difference, ΛTlm ,CF = ΔT1 − ΔT2 ( 210 − 95) − (90.1 − 35) = °C = 81.5°C ln( ΔT1 / ΔT2 ) ln(115 / 55.1) Substituting numerical values, the HXer length is L = 15,705W / 550W / m 2 . Kπ (0.10m ) × 81.4 K = 1.12m Comments: The ε–NTU method could also be used. It would be necessary to perform the hot fluid energy balance to determining CF operation existed. The capacity rate is Cmin/Cmax=0.50. From eq. for effectiveness, and from with q evaluated from an energy balance on the hot fluid, ε= Th, i − Th,o Th ,i − Tc , i = 210 − 90.1 = 0.69 210 − 35 From fig, find NTU≈1.5 giving L = NTU .C min / U oπDo ≈ 1.5 × 130.94 W W 550 2 .π (0.10m ) ≈ 1.14m K m .K Note the good agreement by both methods. 5. Consider a very long, concentric tube heat exchanger having hot and cold water inlet temperatures of 85 and 15°C. The flow rate of the hot water is twice that of the cold water. Assuming equivalent hot and cold water specifies heats; determine the hot water outlet temperature for the following modes of operation (a) Counter flow, (b) Parallel flow. Known: A very long, concentric tube heat exchanger having hot and cold water inlet temperatures of 85 and 15°C, respectively: flow rate of the hot water is twice that of the cold water. Find: outlet temperatures for counter flow and parallel flow operations. Schematic: Assumptions: (1) equivalent hot and cold water specific heats, (2) Negligible Kinetic and potential energy changes, (3) No eat loss to surroundings. Analysis: the heat rate for a concentric tube Heat exchanger with very large surface area Operating in the counter flow mode is q = q max = C min (Th,i − Tc ,i ) Combining the above relation and rearranging, find Th,o = − C C min (Th,i − Tc ,i ) + Th,i = − c (Th,i − Tc ,i ) + Th,i Ch Ch Substituting numerical values 1 Th,o = − (85 − 15)°C + 85°C = 50°C 2 For parallel flow operation, the hot and cold outlet temperatures will be equal; that is Tc,o=Th,o. Hence C c (Tc ,o − Tc ,i ) = C h (Th ,i − Th,o ) Setting Tc,o=Th,o and rearranging ⎡ ⎤ ⎡ C ⎤ C Th,o = ⎢Th,i + c Tc ,i ⎥ / ⎢1 + c ⎥ Ch ⎣ ⎦ ⎣ Ch ⎦ 1 ⎡ ⎤ ⎡ Th,o = ⎢85 + × 15⎥ °C / ⎢1 + 2 ⎣ ⎣ ⎦ 1⎤ ° ⎥ = 61.7 C 2⎦ Comments: Note that while ε =1 for CF operation, for PF operation find ε= q/qmax=0.67. 6. A concentric tube heat exchanger uses water, which is available at 15C, to cool ethylene glycol from 100 to 60C. The water and glycol flow rates are each 0.5 kg/s. What are the maximum possible heat transfer rate and effectiveness of the exchanger? Which is preferred, a parallel –flow or counter flow mode of operation? Known: Inlet temperatures and flow rate for a concentric tube heat exchanger. Find: (a) Maximum possible heat transfer rate and effectiveness, (b) Proffered mode of operation. Schematic: Assumptions: (1) Steady-state operation, (2) Negligible KE and PE changes, (3) Negligible heat loss to surroundings, (4) Fixed overall heat transfer and coefficient. Properties: Table: Ethylene glycol ( T in = 80°C ); cp=2650J/kg.K; Water (Tm ≈ 30°C ) : c p = 4178J / kg . K _ _ Analysis: (a) Using the ε-NTU method, find C min = C h = m h c p , h = ( 0.5kg / s )( 2650 J / kg . K ) = 1325W / K . q ma x = C min (Th. i − Tc ,i ) = (1325W / K )(100 − 15)°C = 1.13 × 10 5 W q = m h c p ,h (Th. i − Tc ,i ) = 0.5kg / s( 2650J / kg . K )(100 − 60)°C = 0.53 × 10 5 W . ε = q / q max = 0.53 × 10 5 / 1.13 × 10 5 = 0.47 (b) Tc ,o = Tc ,i + q . = 15°C + m c c p ,c 0.53 × 10 5 = 40.4°C 0.5kg / s × 4178 J / kg . K Since Tc,o<Th,o, a parallel flow mode of operation is possible. However, with (Cmin/Cmax) = ( m c p ,h / m c c p , c ) =0.63, h . . From fig (NTU)PF≈0.95, (NTU)CF≈0.75 Hence (ACF/APF)= (NTU) CF/ (NTU) PF ≈(0.75/0.95)=0.79 Because of the reduced size requirement, hence capital investment, the counter flow mode of operation is proffered. MODULE 7 HEAT EXCHANGERS 7.1 What are heat exchangers? Heat exchangers are practical devices used to transfer energy from one fluid to another. Around the household, we are accustomed to seeing the condensers and evaporators used in air conditioning units. In automobiles we see radiators and oil coolers. In the power industry we see boilers, condensers, economizers, pre-heaters and numerous other heat exchangers. Within the process industry, we find heat exchangers used extensively for a variety of purposes. Because of the wide variety of uses for heat exchangers, their construction may vary widely. We will consider only the more common types here, but the considerations included are common to all types. In industrial drawings it is common to use the abbreviation HX to indicate heat exchangers. We will use this terminology here to shorten the discussions. 7.2 Heat Transfer Considerations The energy flow between hot and cold streams, when viewed from one end of the heat exchanger, will appear as shown in Figure to the right. Heat transfer will occur by convection to the outside of the inner tube, by conduction across the tube and by convection to the cooler fluid from the inside tube surface. Since the heat transfer occurs across the smaller tube, it is this internal surface which controls the heat transfer process. By convention, it is the outer surface, termed Ao, of this central tube t T di do which is referred to in describing heat exchanger area. Applying the electrical analogy, an equivalent thermal resistance may be defined for this tube. ⎜r ⎟ ln⎛ o r ⎞ ⎝ i⎠ 1 1 + R= + 2π ⋅ kl hi Ai ho Ao If we define the heat exchanger coefficient, Uc, as: Uc ≡ 1 RAo Substituting the value of R above this yields: 1 1 = + U c ho ⎜r ⎟ ro ln⎛ o r ⎞ ⎝ i⎠ A + o k hi Ai Both convective coefficients, ho and hi, can be evaluated from experimentally developed convective correlations. Areas and radii are determined from the geometry of the internal tube. The thermal conductivity, k, corresponds to that for the material of the internal tube. In this fashion each of the terms are generally available for determining Uc and the term is well defined for most heat exchangers. 7.3 Fouling Material deposits on the surfaces of the heat exchanger tube may add further resistances to heat transfer in addition to those listed above. Such deposits are termed fouling and may significantly affect heat exchanger performance. The heat exchanger coefficient, Uc, determined above may be modified to include the fouling factor Rf. 1 1 = + R" Ud Uc • Scaling is the most common form of fouling and is associated with inverse solubility salts. Examples of such salts are CaCO3, CaSO4, Ca3(PO4)2, CaSiO3, Ca(OH)2, Mg(OH)2, MgSiO3, Na2SO4, LiSO4, and Li2CO3. The characteristic which is termed inverse solubility is that, unlike most inorganic materials, the solubility decreases with temperature. The most important of these compounds is calcium carbonate, CaCO3. Calcium carbonate exists in several forms, but one of the more important is limestone. The material frequently crystallizes in a form closely resembling marble, another • • • • • form of calcium carbonate. Such materials are extremely difficult to remove mechanically and may require acid cleaning. Corrosion fouling is classified as a chemical reaction which involves the heat exchanger tubes. Many metals, copper and aluminum being specific examples, form adherent oxide coatings which serve to passivate the surface and prevent further corrosion. Metal oxides are a type of ceramic and typically exhibit quite low thermal conductivities. Even relative thin coatings of oxides may significantly affect heat exchanger performance and should be included in evaluating overall heat transfer resistance. Chemical reaction fouling involves chemical reactions in the process stream which results in deposition of material on the heat exchanger tubes. When food products are involved this may be termed scorching but a wide range of organic materials are subject to similar problems. This is commonly encountered when chemically sensitive process fluids are heated to temperatures near that for chemical decomposition. Because of the no flow condition at the wall surface and the temperature gradient which exists across this laminar sub–layer, these regions will operate at somewhat higher temperatures than the bulk and are ideally suited to promote favorable conditions for such reactions. Freezing fouling is said to occur when a portion of the hot stream is cooled to near the freezing point for one of its components. This is most notable in refineries where paraffin frequently solidifies from petroleum products at various stages in the refining process, obstructing both flow and heat transfer. Biological fouling is common where untreated water is used as a coolant stream. Problems range from algae or other microbes to barnacles. During the season where such microbes are said to bloom, colonies several millimeters deep may grow across a tube surface virtually overnight, impeding circulation near the tube wall and retarding heat transport. Viewed under a microscope, many of these organisms appear as loosely intertwined fibers–much like the form of fiberglass insulation Traditionally these organisms have been treated which chlorine, but present day concerns on possible contamination to open water bodies has severely restricted the use of oxidizers in open discharge systems. Particulate fouling results from the presence of Brownian sized particles in solution. Under certain conditions such materials display a phenomenon known as thermophoresis in which motion is induced as a result of a temperature gradient. Thermodynamically this is referred to as a cross-coupled phenomenon and may be viewed as being analogous to the Seabeck effect. When such particles accumulate on a heat exchanger surface they sometimes fuse, resulting in a buildup having the texture of a sandstone. Like scale these deposits are difficult to remove mechanically. Most of the actual data on fouling factors is tightly held be a few specialty consulting companies. The data which is commonly available is sparse. An example is given below: Fluid Seawater and treated boiler feedwater (below 50oC) Seawater and treated boiler feedwater (above 50oC) River water (below 50oC) Fuel Oil Regrigerating liquids Steam (non-oil bearing) R”, m K/Watt 0.0001 0.0002 0.0002-0.001 0.0009 0.0002 0.0001 2 Table: Representative Fouling Factors r ln( o r ) 1 1 i + + + R" k ho ⎛ ri ⎞ ⎜ ⎟ hi ⋅ ⎝ r ⎠ o 1 = Ud 7.4 Basic Heat Exchanger Flow Arrangements Basic flow arrangements are as shown in the Figure below. Parallel and counterflow provide alternative arrangements for certain specialized applications. In parallel flow both the hot and cold streams enter the heat exchanger at the same end and travel to the opposite end in parallel streams. Energy is transferred along the length from the hot to the cold fluid so the outlet temperatures asymptotically approach one another. In counter flow the two streams enter at opposite ends of the heat exchanger and flow in opposite directions. Temperatures within the two streams tend to approach one another in a nearly linearly fashion resulting in a much more uniform heating pattern. Shown below the heat exchangers are representations of the axial temperature profiles for each. Parallel flow results in rapid initial rates of heat exchange but rates rapidly decrease as the temperatures of the two streams approach one another. Counter flow provides for relatively uniform temperature differences and, consequently, lead toward relatively uniform heat rates throughout the length of the unit. t1 t2 t2 t1 T1 T1 Temperature Parallel Flow T2 T1 T1 Counter Flow T2 T2 t1 t2 Temperature T2 t2 t1 Position Position Basic Flow Arrangements for Tube in Tube Heat Exchangers. 7.5 Log Mean Temperature Differences Heat flows between the hot and cold streams due to the temperature difference across the tube acting as a driving force. As seen in the Figure below, the difference will vary with axial position within the HX so that one must speak in terms of the effective or integrated average temperature differences. T1 1 Counter Flow T1 T2 Parallel Flow T2 t1 t2 2 t2 t1 2 1 Position Position Temperature Differences Between Hot and Cold Process Streams Working from the three heat exchanger equations shown above, after some development it if found that the integrated average temperature difference for either parallel or counter flow may be written as: Δθ = LMTD = θ1 − θ 2 ⎛θ ⎞ ln⎜ 1 ⎟ ⎝θ2 ⎠ The effective temperature difference calculated from this equation is known as the log mean temperature difference, frequently abbreviated as LMTD, based on the type of mathematical average that it describes. While the equation applies to either parallel or counter flow, it can be shown that Δθeff will always be greater in the counter flow arrangement. This can be shown theoretically from Second Law considerations but, for the undergraduate student, it is generally more satisfying to arbitrarily choose a set of temperatures and check the results from the two equations. The only restrictions that we place on the case is that it be physically possible for parallel flow, i.e. θ1 and θ2 must both be positive. Another interesting observation from the above Figure is that counter flow is more appropriate for maximum energy recovery. In a number of industrial applications there will be considerable energy available within a hot waste stream which may be recovered before the stream is discharged. This is done by recovering energy into a fresh cold stream. Note in the Figures shown above that the hot stream may be cooled to t1 for counter flow, but may only be cooled to t2 for parallel flow. Counter flow allows for a greater degree of energy recovery. Similar arguments may be made to show the advantage of counter flow for energy recovery from refrigerated cold streams. 7.6 Applications for Counter and Parallel Flows We have seen two advantages for counter flow, (a) larger effective LMTD and (b) greater potential energy recovery. The advantage of the larger LMTD, as seen from the heat exchanger equation, is that a larger LMTD permits a smaller heat exchanger area, Ao, for a given thermal duty, Q. This would normally be expected to result in smaller, less expensive equipment for a given application. This should not lead to the assumption that counter flow is always a superior. Parallel flows are advantageous (a) where the high initial heating rate may be used to advantage and (b) where the more moderate temperatures developed at the tube walls are required. In heating very viscous fluids, parallel flow provides for rapid initial heating. The quick decrease in viscosity which results may significantly reduce pumping requirements through the heat exchanger. The decrease in viscosity also serves to shorten the distance required for flow to transition from laminar to turbulent, enhancing heat transfer rates. Where the improvements in heat transfer rates compensate for the lower LMTD parallel flow may be used to advantage. A second feature of parallel flow may occur due to the moderation of tube wall temperatures. As an example, consider a case where convective coefficients are approximately equal on both sides of the heat exchanger tube. This will result in the tube wall temperatures being about the average of the two stream temperatures. In the case of counter flow the two extreme hot temperatures are at one end, the two extreme cold temperatures at the other. This produces relatively hot tube wall temperatures at one end and relatively cold temperatures at the other. Temperature sensitive fluids, notably food products, pharmaceuticals and biological products, are less likely to be “scorched” or “thermally damaged” in a parallel flow heat exchanger. Chemical reaction fouling may be considered as leading to a thermally damaged process stream. In such cases, counter flow may result in greater fouling rates and, ultimately, lower thermal performance. Other types of fouling are also thermally sensitive. Most notable are scaling, corrosion fouling and freezing fouling. Where control of temperature sensitive fouling is a major concern, parallel flow may be used to advantage. 7.7 Multipass Flow Arrangements In order to increase the surface area for convection relative to the fluid volume, it is common to design for multiple tubes within a single heat exchanger. With multiple tubes it is possible to arrange to flow so that one region will be in parallel and another portion in counter flow. An arrangement where the tube side fluid passes through once in parallel and once in counter flow is shown in the Figure below. Normal terminology would refer to this arrangement as a 1-2 pass heat exchanger, indicating that the shell side fluid passes through the unit once, the tube side twice. By convention the number of shell side passes is always listed first. The primary reason for using multipass designs is to increase the average tube side fluid velocity in a given arrangement. In a two pass arrangement the fluid flows through only half the tubes and any one point, so that the Reynold’s number is effectively doubled. Increasing the Reynolds’s number results in increased turbulence, increased Nusselt numbers and, finally, in increased convection coefficients. Even though the parallel portion of the flow results in a lower effective ΔT, the increase in overall heat transfer coefficient will frequently compensate so that the overall heat exchanger size will be smaller for a specific service. The improvements achievable with multipass heat exchangers is sufficiently large that they have become much more common in industry than the true parallel or counter flow designs. The LMTD formulas developed earlier are no longer adequate for multipass heat exchangers. Normal practice is to calculate the LMTD for counter flow, LMTDcf, and to apply a correction factor, FT, such that Δθ eff = FT ⋅ LMTDCF The correction factors, FT, can be found theoretically and presented in analytical form. The equation given below has been shown to be accurate for any arrangement having 2, 4, 6, .....,2n tube passes per shell pass to within 2%. ⎡ 1− P ⎤ R 2 + 1 ln ⎢ ⎣1 − R ⋅ P ⎥ ⎦ FT = ⎡2 − P R + 1− R2 + 1 ⎤ ⎥ ( R − 1) ln ⎢ ⎢ ⎥ 2 ⎢2 − P R + 1 + R + 1 ⎦ ⎥ ⎣ ( ( ) ) where the capacity ratio, R, is defined as: R= T1 − T2 t 2 − t1 The effectiveness may be given by the equation: P= 1 − X 1/ N shell R − X 1/ N shell Po − Po ⋅ ( N shell − 1) provided that R≠1. In the case that R=1, the effectiveness is given by: P= N shell where Po = t2 − t1 T1 − t1 and X = Po ⋅ R − 1 Po − 1 As an alternative to using the formulas for the correction factors, which can become tedious for non-computerized calculations, charts are available. Several are included in standard texts. Experience has shown that, due to variability in reading charts, considerable error can be introduced into the calculations and the equations are recommended. When charts are used, they should be reproduced at a sufficiently large scale, and considerable care should be used in making interpolations. 7.8 Limitations of Multipass Arrangements Since the 1-2 heat exchanger uses one parallel pass and one counter current, it follows that the maximum heat recovery for these units should be between that of parallel and counter flow. As a practical limit it is important that nowhere Figure . Temperature Profiles for a 1-2 HX with a in the unit should the cold fluid Temperature Cross. temperature exceed that of the hot fluid. If so, then heat transfer is obviously in the wrong direction. Such a situation can arise in a multipass heat exchanger as seen in Figure 6. This unit represents a cold fluid, located on the tube side of the heat exchanger, making two passes through the unit, the hot fluid, on the shell side, traveling across the unit only once. Here the cold fluid is heated to a temperature slightly above that of the hot fluid near the exit for the two streams. At this axial location, near the left end of the unit, the temperature of the cold fluid in the first pass remains well below that of the hot fluid so that considerable heat transfer occurs. The cold fluid in the second pass is slightly above that of the hot fluid at the same location. The small temperature difference between the second pass cold fluid and the hot stream, indicates that only a small amount of heat will be transferred between these streams. Overall heat will flow from hot to cold fluid, but a portion of the heat transfer surface is being used in a counter productive way. This condition is termed as a temperature cross. In the limit the hot fluid exit temperature could be cooled to the average of the cold fluid inlet and exit temperature. This would, however, be highly inefficient and would require an excessively large surface area. Some engineers advocate that good design should not permit a temperature cross, indicating that the 1-2 should operate with the same heat recovery limit as a true parallel flow. The preferred method of attaining additional heat recovery is to stage heat exchangers in series so that no temperature cross occurs in any unit. An equivalent solution is to put multiple 1–n arrangements within a single shell. A 2–4 unit is the equivalent of 2 1–2 units provided that the total heat transfer area is equal. Similarly a 3–6 unit is the equivalent of 3 1–2 units with equal overall area. Other engineers suggest that a small temperature cross may be acceptable and may provide a less expensive design than the more complex alternatives. If one were to plot the locus of points where the temperature cross occurs for the 1-2 heat exchanger on the temperature correction chart, it would be found to correspond to a relatively narrow range of FT values ranging from about 0.78 to 0.82. Lower values of FT may be taken as an indication that a temperature cross will occur. A second consideration is that at lower FT values the slope, dFT/dP, becomes extremely steep. This is an indication that the temperature efficiency, P, is asymptotically approaching its upper limit and the design has no margin to accommodate uncertainties. A good rule of thumb is that the minimum slope of dFT/dP, which is negative, should not fall below -1.5. Instead a 2-4 or even a 36 should be selected to provide the needed operational design margin.. Similar restrictions exist for these designs as well. In the limit a counter flow design may be the only suitable selection for high heat recovery applications. 7.9 Effectiveness-NTU Method: In our previous discussions, we have been looking at practical HX designs using the LMTDCF with a Ft correction factor to account for the mixed flow conditions. Now we wish to consider an alternate, more recent approach that is in common use today. This is the effectiveness-NTU method. Effectiveness, ε Consider two counter-flow heat exchangers, one in which the cold fluid has the larger ΔT (smaller m⋅cp) and a second in which the cold fluid has the smaller ΔT (larger m⋅cp): Δt > ΔT M⋅Cp > m⋅cp T2 T1 ΔT > Δt m⋅cp > M⋅Cp T1 t2 T2 t2 t1 t1 We may see in the first case that, because the cold fluid heat capacity is small, its temperature changes rapidly. If we seek efficient energy recovery, we see that in the limit a HX could be designed in which the cold fluid exit temperature would reach that of the hot fluid inlet. In the second case, the hot fluid temperature changes more rapidly, so that in the limit the hot fluid exit temperature would reach that of the cold fluid inlet. The effectiveness is the ratio of the energy recovered in a HX to that recoverable in an ideal HX. ε= & m ⋅ c p ⋅ (T1 − t1 ) & m ⋅ c p ⋅ ( t 2 − t1 ) Δt > ΔT & M ⋅ C p ⋅ (T1 − T2 ) ε= & M ⋅ C p ⋅ (T1 − t1 ) ΔT > Δt Canceling identical terms from the numerator and denominator of both terms: ε= (t − t ) (T − t ) 2 1 1 1 Δt > ΔT ε= (T − T ) (T − t ) 1 2 1 1 ΔT > Δt We see that the numerator, in the two cases, is the temperature change for the stream having the larger temperature change. The denominator is the same in either case: ε= (T − t ) 1 1 Δ Tmax In the LMTD-Ft method an effectiveness was defined: P= t 2 − t1 T1 − t1 Note that the use of the upper case T in the numerator, in contrast to our normal terminology, does not indicate that the hot fluid temperature change is used here. The max subscript over-rides the normal terminology and indicates that this refers to the side having the larger temperature change. Number of Transfer Units (NTU) Recall that the energy flow in any HX is described by three equations: Q = U⋅A⋅Δθeff Q = -M⋅Cp⋅ΔT Q = m⋅cp⋅Δt HX equation 1st Law Equation 1st Law Equation We may generalize the latter two expressions, using ε-NTU terminology as follows: Q = (M⋅Cp)min⋅ΔTmax Q = (M⋅Cp)max⋅ΔTmin Again the use of the upper case letters is over-ridden by the use of the subscripts. If we eliminate Q between the HX equation and one of the 1st Law equations, U⋅A⋅Δθeff = (M⋅Cp)min⋅ΔTmax This expression may be made non-dimensional by taking the temperatures to one side and the other terms to the other side: NTU ≡ ( U⋅A & M ⋅ Cp ) = min Δ Tmax Δ θ eff Physically we see that a HX with a large product U⋅A and a small (M⋅Cp)min should result in a high degree of energy recovery, i.e. should result in a large effectiveness, ε. Capacity Ratio, CR The final non-dimensional ratio needed here is the capacity ratio, defined as follows: CR ≡ ( M ⋅ Cp ) min ( M ⋅ Cp ) max = Δ Tmin Δ Tmax In the LMTD-Ft method a capacity ratio was defined: R= T1 − T2 t 2 − t1 ε-NTU Relationships In the LMTD-Ft method, we found a general equation which described Ft for all 1-2N, 2-4N, 3-6N, etc. heat exchangers. Another relationship, not given here, is required for cross flow arrangements. In a similar fashion, we may develop a number of functional relationships showing ε = ε (NTU, CR) or, alternatively: NTU = NTU(ε,CR) These relationships are shown in tables in standard text books. For example, we find that the relationship for a parallel flow exchanger is: 1− e ε= 1 − CR ⋅ e − NTU ⋅(1− CR ) − NTU ⋅ ( 1− C R ) Note: These correlations are not general. Specific correlations will be given for different kind of HX. The ε-NTU method offers a number of advantages to the designer over the traditional LMTD-Ft method. One type of calculation where the ε-NTU method may be used to clear advantage would be cases in which neither fluid outlet temperature is known. Module 8: Learning objectives • • • • In this chapter, the focus is on convection processes associated with the change in phase of a fluid, particularly those processes that can occur at a solid-liquid interface, namely, boiling and condensation. For these processes, the latent heat associated with the phase change are significant. Because there is phase change, heat transfer to and from the fluid can occur without influencing the fluid temperature. Hence, in boiling and condensation, large heat transfer rates may be achieved with small temperature differences. The heat transfer coefficient associated with boiling and condensation depends on several parameters, such as surface tension between liquid-vapour interface, latent heat, density difference between liquid and vapour, length scale, specific heat and viscosity. It is apparent that boiling and condensation are complicated processes for which the existence of generalized relations in somewhat limited. This chapter identifies the essential physical features of the processes and presents correlations suitable for the approximate engineering calculations. • Condensation and Boiling Until now, we have been considering convection heat transfer in homogeneous single-phase (HSP) systems Boiling and condensation, however, provide much higher heat transfer rates than those possible with the HSP systems Condensation Condensation occurs when the temperature of a vapor is reduced below its saturation temperature Condensation heat transfer Film condensation Drop wise condensation Heat transfer rates in drop wise condensation may be as much as 10 times higher than in film condensation Laminar Film condensation on a vertical wall (VW) y x Tsat (y ⎞ μ ∂u ⎟ l ∂y ⎟y ⎠ x y Δy A g A A T( ( (x) Condensate Film ∂u ⎞ μ ⎟ l ∂y ⎟y+Δy ⎠ A (ρ − ρv)gAΔy l Laminar Film condensation on a vertical wall (cont..) ⎡ 4 xk l (Tsat − Tw )ν l ⎤ δ( x ) = ⎢ ⎥ ⎢ h fg g (ρ l − ρ v ) ⎥ ⎣ ⎦ 1/ 4 ⎡ h fg g (ρ l − ρ v )k ⎤ h(x) = ⎢ ⎥ ⎢ 4 x (Tsat − Tw )ν l ⎥ ⎦ ⎣ 3 l 1/ 4 Average coeff. where L is the plate length. Total heat transfer rate : Condensation rate : ⎡ h fg g (ρl − ρ v )k h L = 0.943⎢ ⎢ L(Tsat − Tw )ν l ⎣ q = h L A(Tsat − Tw ) h L A(Tsat − Tw ) q & = m= h fg h fg 3 l ⎤ ⎥ ⎥ ⎦ 1/ 4 Example Laminar film condensation of steam Saturated steam condenses on the outside of a 5 cm-diameter vertical tube, 50 cm high. If the saturation temperature of the steam is 302 K, and cooling water maintains the wall temperature at 299 K, determine: (i) the average heat transfer coefficient, (ii) the total condensation rate, and (iii) the film thickness at the bottom of the tube. Given: Film condensation of saturated steam Required: (i) Average heat transfer coefficient, (ii) total condensation rate, (iii) and film thickness 1. Effect of tube curvature negligible 2. Effect of liquid sub cooling negligible 3. Laminar Example (contd...) y x Tsat (y The Average heat trasn sfer coefficent is given by : _ h= ⎡ ' g (ρ − ρ v ) k 3 ⎤ ⎢h l l⎥ ⎢ fg ⎥ 0 . 943 ⎢ L (Tsat − T w ) v ⎥ ⎢ l⎥ ⎢ ⎥ ⎣ ⎦ 1/ 4 A g T( ( (x) Condensate Film Evaluate hfg at the saturation temperature of 302 K From Table of water properties : h fg = 2 .432 ×10 6 J / kg ρ v = 0 .03 kg / m 3 Example (contd...) Also, for water k l = 0.611 W/mK ρ l = 996 kg/m3 ν l = 0.87 × 10 -6 m 2 /s ⎡ h fh g (ρ l − ρ v )k ⎤ h = 0.943⎢ ⎥ L(Tsat − Tw )ν l ⎦ ⎣ 3 l 1/ 4 ⎡ ( 2.432 × 10 −6 )(9.81)(996 − 0.03)(0.611) 3 ⎤ = 0.943⎢ ⎥ (0.5)(3)(0.87 × 10 −6 ) ⎢ ⎥ ⎣ ⎦ 1/ 4 = 7570 W/m 2 K (ii) The total condensation rate is : & Q h AΔT (7570)(3)π (0.05)(0.5) & = = = 7.33 × 10 −4 kg/s m= ( 2.432 × 10 6 ) h fg h fg Example (contd...) (iii) The film thickness is ⎛ 3ν l Γ ⎞ δ =⎜ ρ v << ρ l ⎜ρg⎟ ⎟ ⎝ l ⎠ The mass flow rate per unit width of film Γ is : & m (7.33 × 10 −4 ) Γ= = = 4.67 × 10 −3 kg/ms πD (π )(0.05) ⎛ 3(0.87 × 10 )( 4.67 × 10 ) ⎞ ⎟ Hence, δ = ⎜ ⎜ ⎟ (996)(9.81) ⎝ ⎠ -6 −3 1/ 3 1/ 3 = 1.08 × 10 −4 m Boiling Boiling occurs when the surface temperature Tw exceeds the saturation temperature Tsat corresponding to the liquid pressure Heat transfer rate : q′′ = h (Tw − Tsat ) = hΔTe s where ΔTe = Tw − Tsat (excess temperature) Boiling process is characterized by formation of vapor bubbles, which grow and subsequently detach from the surface Bubble growth and dynamics depend on several factors such as excess temp., nature of surface, thermo physical properties of fluid (e.g. surface tension, liquid density, vapor density, etc.). Hence, heat transfer coefficient also depends on those factors. Pool Boiling Curve q′′ s (ΔTe ) = Modes of Pool Boiling Free convection boiling ΔTe ≈ 5 C Nucleate boiling Transition boiling Film boiling o 5o C ≤ ΔTe ≤ 30 o C 30 o C ≤ ΔTe ≤ 120 o C ΔTe ≥ 120 o C Module 8: Solved Problems 1. A long, 1-mm-diameter wire passes an electrical current dissipating 3150W/m and reaches a surface temperature of 126°C when submerged in water at 1atm. What is the boiling heat transfer coefficient? Estimate the value of the correlation coefficient Cs,f. Known: long wire, 1–mm-diameter, reaches a surface temperature of 126°C in water at 1atm while dissipating 3150W/m. Find: (1) Boiling heat transfer coefficient and (2) correlation coefficient Cs,f, if nucleate boiling occurs Schematic: Assumptions: (1) Steady-state conditions, (2) Nucleate boiling. Table: Water (saturated, 1atm) Ts=100C, ρl=1/vf=957.9kg/m3, ρf=1/vg=0.5955kg/m3, cp,l=4217 J/kg.K, μl=279*10-6 N.s/m2, prl=1.76, hfg=2257 KJ/kg, σ=58.9*10-3N/m. Analysis: (a) For the boiling process, the rate equation can be rewritten as h= _ " qs q' = s /(Ts − Tsat ) (Ts − Tsat ) πD h= _ 3150W / m W /(126 − 100)°C = 1.00 × 10 6 2 / 26 ° C = 38,600W / m 2 .K π × 0.001m m Note that heat flux is very close to q”max, and nucleate boiling exists. (b) For nucleate boiling, the Rohsenow correlation may be solved for Cs,f, to give 1 3 C s,f ⎧ μ h ⎫ ⎡ g ( ρ l − ρν ) ⎤ = ⎨ l 'f ,g ⎬ ⎢ ⎥ σ ⎦ ⎩ qs ⎭ ⎣ 1/ 6 ⎡ c p ,l Δ Te ⎤ ⎢ n ⎥ ⎢ h f , g Prl ⎥ ⎦ ⎣ Assuming the liquid surface combination is such that n=1 and substituting numerical values with ΔTe=Ts-Tsat, find 1/ 3 1/ 6 ⎧ 279 × 10 −6 N .s / m × 22257 × 10 3 J / kg ⎫ C s, f = ⎨ ⎬ 1.00 × 10 6 W / m 2 ⎩ ⎭ ⎛ 4217 J / kg.K × 26 K ⎞ ×⎜ ⎜ 2257 × 10 3 J / kg × 1.76 ⎟ ⎟ ⎝ ⎠ C s , f = 0.017 ⎡ 9.8m / s 2 (957.9 − 0.5955kg / m 3 ⎤ ⎢ ⎥ 58.9 × 10 −3 N / m ⎣ ⎦ Comments: By comparison with the values Cs , f for other watersurface combinations (given in standard tables), the Cs , f value for the wire is quite large suggesting that its surface must be highly polished. Note that the value of the boiling heat transfer coefficient is much larger that for other convection processes previously encountered. 2. The bottom of a copper pan, 150 mm in diameter, is maintained at 115°C by the heating element of an electric range. Estimate the power required to boil the water in this pan. Determine the evaporation rate. What is the ratio of the surface heat flux to the critical heat flux? What pan temperature is required to achieve the critical heat flux? Known: copper pan 150 mm in diameter and filled with water at 1atm, maintained at 115°C. Find: the power required to boil the water and the evaporation rate; ratio of the heat flux to the critical heat flux; pan temperature is required to achieve the critical heat flux. Schematic: . m Assumptions: (1) Nucleate pool boiling, (2) Copper pan is polished surface. Properties: Table: Water (1atm) Tsat=100°C, ρl=957.9kg/m3, ρv=0.5955kg/m3, cp,l=4217 J/kg.K, μl=279*10-6N.s/m2, prl=1.76, hfg=2257 KJ/kg, σ=58.9*10-3N/m. Analysis: the power requirement for boiling and the evaporation rate can be expressed as follows, " q boil = q s . As m = q boil / h f , g . The heat flux for nucleate pool boiling can be estimated using the Ronsenow Correlation. 3 ⎡ g ( ρ l − ρν ) ⎤ q s = μl h f ,g ⎢ ⎥ σ ⎣ ⎦ " 1/ 2 ⎡ c p ,l ΔTe ⎤ ⎢ n ⎥ ⎢ C s , f h f , g Prl ⎥ ⎣ ⎦ Selecting Cs , f =0.013 and n=1 from standard table for the polished copper finish, find ⎧ 279 × 10 −6 N .s / m × 2257 × 10 3 J / kg ⎫⎡ 9.8m / s 2 (957.9 − 0.5955kg / m 3 ⎤ C s, f = ⎨ ⎬⎢ ⎥ 58.9 × 10 −3 N / m ⎩ ⎦ ⎭⎣ ⎛ 4217 J / kg.K × 26 K ⎞ ×⎜ ⎜ 2257 × 10 3 J / kg × 1.76 ⎟ ⎟ ⎝ ⎠ C s , f = 4.619 × 10 5 W / m 2 1/ 6 The power and evaporation rate are q boil = 4.619 × 10 5 W / m 2 × . π 4 (0.150m) 2 = 8.16kW m boil = 8.16kW / 2257 × 10 3 J / kg = 3.62 × 10 −3 kg / s = 13kg / h The maximum or critical heat flux was found as Q”max=1.26MW/m2. Hence, the ratio of the operating to maximum heat flux is " qs = 4.619 × 10 5 W / m 2 / 1.26MW / m 2 = 0.367 " q max From the boiling curve, ΔTe ≈30°C will provide the maximum heat flux 3. A silicon chip of thickness L=25 mm and thermal conductivity ks=135W/m.K is cooled by boiling a saturated fluorocarbon liquid (Tsat=57°C) on its surface. The electronic circuits on the bottom of the chip are perfectly insulated. Properties of the saturated fluorocarbons are cp,l=110J?kg.K, hfg=84,400J/kg, ρl=1619.2kg/m3, ρν=13.4kg/m3,σ=8.1*103 kg/s2,μl=440*10-6kg/m.s and prl=9.01. In addition the nucleate boiling constants are Cs,f=0.005 and n=1.7. What is the steady-state temperature To at the bottom of the chip? If, during testing of the chip, q”o is increases to 90% of the critical heat flux, what is the new steady-state value of To? Known: Thickness and thermal conductivity of a silicon chip. Properties of saturated fluorocarbon liquid on top side. Find: (a) Temperature at bottom surface of chip for a prescribed heat flux, (b) Temperature of bottom surface at 90% of CHF. Schematic: Assumptions: (1) steady-state conditions, (2) uniform heat flux and adiabatic sides, hence one-dimensional conduction in chip, (3) Constant properties, (4) Nucleate boiling in liquid. Properties: Saturated fluorocarbon (given): c p ,l = 1100 J / kg.K , hf,g=84,400 J/kg, ρ l =1619.2kg/m3, ρν=13.4kg/m3,σ=8.1*10-3kg/s2, μ l =440*10-6kg/m-s, Pr℘ =9.01. " " " Analysis: (a) Energy balances yield qo = qcond = k s (To − Ts ) / L = qb . Obtain Ts from the Rohsenow correlation. Ts − Tsat 1/ 3 Ts − Tsat = 0.005(84,400 J / kg )9.011.7 1100 J / kg.K ⎛ ⎞ 5 × 10 4 W / m 2 ⎜ ⎜ 440 × 10 −6 kg / m.s × 84,400 J / kg ⎟ × ⎟ ⎝ ⎠ = 15.9°C C s , f h f , g Prln ⎛ q s' ⎜ = ⎜μ h c p ,l ⎝ l f ,g ⎞⎡ ⎤ σ ⎟⎢ ⎟ g (ρ − ρ ) ⎥ l ν ⎦ ⎠⎣ 1/ 6 ⎡ ⎤ 8.1 × 10 −3 kg / s 2 ⎢ 2 3⎥ ⎣ 9.807 m / s (1619.2 − 13.4)kg / m ⎦ Ts = (15.9 + 57) ° C = 72.9°C 1/ 6 From the rate equation, To = Ts + " qo 5 × 10 4 W / m 2 × 0.0025m = 72.9 ° C + = 73.8°C ks 135W / m.K (b) With the heat rate 90% of the critical heat flux (CHF) q " max ⎡σg ( ρ l − ρ v ) ⎤ = 0.149hfgρv ⎢ ⎥ ρv 2 ⎣ ⎦ 1/ 4 = 0.149 × 84,400 J / kg × 13.4kg / m 3 1/ 4 ⎡ 8.1 × 10 −3 × 9.807m / s 2 (1619.2 − 13.4)kg / m 3 ⎤ ×⎢ ⎥ 13.4kg / m 3 ⎣ ⎦ " q max = 15.5 × 10 4 W / m 2 " q " = 0.9q max = 13.9 × 10 4 W/m 2 o ΔTe = ΔTe ) a (q " /q " a )1/3 = 15.9°C × 1.41 = 22.4 ° C o o, = 79.4C + 13.9 × 10 4 W/m 2 × .0025 = 82.0 ° C 135W/m.K Ts = 22.4°C + 57°C = 79.4 ° C To Comments: Pool boiling is not adequate for many VLSI chip design 4. As strip steel leaves the last set of rollers in a hot rolling mill, it is quenched by planar water jets before being coiled. Due to the large plate temperatures, film boiling is achieved shortly downstream of the jet impingement region. Consider conditions for which the strip steel beneath the vapor blanket is at a temperature of 907K and has an emissivity of 0.35. Neglecting the effects of the strip and jet motions and assuming convection within the film to be approximated by that associated with a large horizontal cylinder of 1-m-diameter, estimate the rate of heat transfer per unit surface area from the strip to the wall. Known: Surface temperature and emissivity of strip steel. Find: heat flux across vapor blanket. Schematic: Assumptions: (1) Steady-state conditions, (2) Vapor/jet interface is at Tsat for P=1atm, (3) Negligible effect of jet and strip motion. Properties: Table: saturated Water (100°C 1atm) ρl=1/vf=957.9kg/m3, hfg=2257 KJ/kg: saturated water vapor (Tf=640K): ρv=175.4kg/m3, cp,v=42 J/kg.K, μv=32*10-6N.s/m2,k=0.155W/m.K,νv=0.182*10-6m2/s. Analysis: The heat flux is q "x = h ΔTe Where ΔTe = 907K − 373K = 534K and - 4/3 _ h = h - 4/3 conv + h rad h _ −1/3 with h 'fg = h fg + 0.80c p, v (Ts − Tsat ) = 2.02 × 10 7 J/kg ⎡ 9.8m / s 2 (957.9 − 175.4)kg / m 3 (2.02 × 10 −7 J / kg )(1m) 3 ⎤ N u D = 0.62⎢ ⎥ 0.182 × 10 −6 m 2 / s (0.155W / m.K )(907 − 373) K ⎣ ⎦ _ 1/ 4 = 6243 hence, h conv = N u D k v / D = 6243W / m 2 .K (0.155W / m.K / 1m) = 968W / m 2 .K h rad = _ 4 εσ (Ts4 − Tsat ) _ _ Ts − Tsat = 0.35 × 5.67 × 10 −8 W / m 2 .k (907 4 − 373 4 ) K 4 (907 − 373) K h rad = 24W / m2.K _ hence, and h = 968W / m2.K + (3 / 4)(24W / m2.K ) = 986W / m2.K q " = 986W / m 2 .k (907 − 373) K = 5.265 × 10 5 W / m 2 s _ 5. Saturated steam at 0.1 bars condenses with a convection coefficient of 6800 W/m2.K on the outside of a brass tube having inner and outer diameters of 16.5 and 19mm, respectively. The convection coefficient for water flowing inside the tube is 5200W/m2.K. Estimate the steam condensation rate per unit length of the tube when the mean water temperature is 30°C. Known: saturated steam condensing on the outside of a brass tube and water flowing on the inside of the tube; convection coefficients are prescribed. Find: Steam condensation rate per unit length of the tube. Schematic: Assumptions: (1) Steady-state conditions. Properties: Table: Water, vapor (0.1 bar): Tsat≈320K, hfg=2390*103J/kg; Table: Brass ( T = (Tm + Tsat ) / 2 ≈ 300 K ); k=110W/m.K. Analysis: The condensation rate per unit length is written as m = q ' / h fg . ' ' _ (1) Where the heat rate follows from equation using overall heat transfer coefficient q ' = U o .π D o ( T sat − T m ) (2) −1 ⎡ 1 D /2 D D 1⎤ U o = ⎢ + o ln o + o ⎥ k Di Di hi ⎦ ⎣ ho (3) 1 0.0095m 19 19 1 ⎡ ⎤ ln Uo = ⎢ + + 2 2 ⎣ 6800W / m .K 110W / m.K 16.5 16.5 5200W / m .K ⎥ ⎦ −1 U o = 147.1 × 10 −6 + 12.18 × 10 −6 + 192.3 × 10 −6 W / m 2 .K = 2627W / m 2 .K Combining equations, (1) and (2) and substituting numerical values, find m = U oπDo (Tsat − Tm ) / h ' . ' . ' fg m = 2627W / m 2 .Kπ (0.019m)(320 − 303) K / 2410 × 10 3 J / kg = 1.11 × 10 3 kg / s Comments: (1) Note from evaluation of equation. (3) That the thermal resistance of the brass tube is not negligible. (2) With Ja = c p ,l (Tsat − Ts ) / h f , g , h ' fg = h fg [1 + 0.68 Ja ]. note from expression for Uo, that the internal resistance is the largest. Hence, estimate Ts ,o ≈ To − ( Ro / ΣR)(To − Tm ) ≈ 313K . Hence h 'fg ≈ 2390 × 10 3 J / kg[1 + 0.68 × 4179 J / kg.K (320 − 313) K / 2390 × 10 3 J / kg ] h 'fg = 2410kJ / kg Where c p,l for water(liquid) is evaluated at Tf = (Ts ,o + To ) / 2 ≈ 317 K Module 8: Short questions 1. What mechanisms are responsible for the very high heat transfer coefficients in nucleate boiling? 2. Does the amount of heat absorbed as 1 kg of saturated liquid water boils at 100ºC have to be equal to the amount of heat released as 1 kg of saturated water vapour condenses at 100ºC? 3. What is the difference between evaporation and boiling? 4. What is the difference between pool boiling and flow boiling? 5. What is the difference between subcooled and saturated boiling? 6. Suggest some methods of enhancing pool boiling heat transfer coefficient. 7. What is the meaning of burnout point in a boiling curve? How is burnout avoided in the design of steam boilers? 8. Using concepts of thermodynamics, explain how condensation occurs. 9. Why is higher heat transfer coefficient generally associated with dropwise condensation than with film condensation? MODULE 8 BOILING AND CONDENSATION 8.1 Boiling: General considerations • Boiling is associated with transformation of liquid to vapor at a solid/liquid interface due to convection heat transfer from the solid. • Agitation of fluid by vapor bubbles provides for large convection coefficients and hence large heat fluxes at low-tomoderate surface-to-fluid temperature differences • Special form of Newton’s law of cooling: ΔTe = Ts − Tsat ) ′ q s′ = h (Ts − Tsat ) = hΔTe where Tsat is the saturation temperature of the liquid, and is the excess temperature. 8.2 Special cases Pool Boiling:Liquid motion is due to natural convection and bubble-induced mixing. Forced Convection Boiling:Fluid motion is induced by external means, as well as by bubble-induced mixing. Saturated Boiling:Liquid temperature is slightly larger than saturation temperature Subcooled Boiling:Liquid temperature is less than saturation temperature 8.3 The boiling curve The boiling curve reveals range of conditions associated with saturated pool boiling on a q ′s′ vs. ΔTe plot. Water at Atmospheric Pressure Free Convection Boiling ( ΔTe <5ºC) Little vapor formation. Liquid motion is due principally to single-phase natural convection. Onset of Nucleate Boiling – ONB ( ΔTe ≈ 5ºC) Nucleate boiling (5ºC < ΔTe <30ºC) Isolated Vapor Bubbles (5ºC < ΔTe <10ºC) Liquid motion is strongly influenced by nucleation of bubbles at the surface. h and q s′′ rise sharply with increasing ΔTe Heat transfer is principally due to contact of liquid with the surface (single-phase convection) and not to vaporization Jets and Columns (10ºC < ΔTe <30ºC) Increasing number of nucleation sites causes bubble interactions and coalescence into jets and slugs. Liquid/surface contact is impaired. q ′′ continues to increase with ΔTe while h begins to decrease s Critical Heat Flux - CHF, ( ΔTe ≈ 30ºC) Maximum attainable heat flux in nucleate boiling. ′′ qmax ≈ 1 MW/m 2 for water at atmospheric pressure. Potential Burnout for Power-Controlled Heating ′′ An increase in q s′′ beyond qmax causes the surface to be blanketed by vapor and its temperature to spontaneously achieve a value that can exceed its melting point If the surface survives the temperature shock, conditions are characterized by film boiling Film Boiling Heat transfer is by conduction and radiation across the vapor blanket A reduction in q s′′ follows the cooling the cooling curve continuously to the Leidenfrost point corresponding to the ′′ minimum heat flux q min for film boiling. ′′ A reduction in q ′s′ below q min causes an abrupt reduction in surface temperature to the nucleate boiling regime Transition Boiling for Temperature-Controlled Heating Characterised by continuous decay of q ′s′ (from with increasing ΔTe ′′ q max to ′′ q min ) Surface conditions oscillate between nucleate and film boiling, but portion of surface experiencing film boiling increases with ΔTe Also termed unstable or partial film boiling. 8.4 Pool boiling correlations Nucleate Boiling Rohsenow Correlation, clean surfaces only, ±100% errors ⎡ g ( ρl − ρ v ) ⎤ q′′ = μl h fg ⎢ s ⎥ σ ⎣ ⎦ 1/ 2 ⎛ c p ,l Δ Te ⎞ ⎜ ⎜ C h Pr n ⎟ ⎟ ⎝ s , f fg l ⎠ 3 Cs , f , n → Surface/Fluid Combination ( Critical heat flux: ′′ qmax ⎡σ g ( ρl − ρ v ) ⎤ = 0.149h fg ρv ⎢ ⎥ ρv2 ⎣ ⎦ 1/ 4 Film Boiling 3 hconv D = C ⎡ g ( ρl − ρv ) h′fg D ⎤ Nu D = ⎢ ⎥ ν v kv (Ts − Tsat ) ⎦ kv ⎣ 1/ 4 Geometry Cylinder(Hor.) Sphere C 0.62 0.67 8.5 Condensation: General considerations • Condensation occurs when the temperature of a vapour is reduced below its saturation temperature • Condensation heat transfer Film condensation Dropwise condensation • Heat transfer rates in dropwise condensation may be as much as 10 times higher than in film condensation 8.6 Laminar film condensation on a vertical wall y y Δy x Tsat (y x A g μ ∂u ⎟ l ∂y ⎟ y ⎠ ⎞ A A ⎞ μ ∂u ⎟ l ∂y ⎟ y+Δy ⎠ A T( ( (x) Condensate Film (ρ − ρv )gAΔy l ⎡ 4 xk l (Tsat − Tw )ν l ⎤ δ ( x) = ⎢ ⎥ ⎢ h fg g ( ρ l − ρ v ) ⎥ ⎣ ⎦ ⎡ h fg g ( ρ l − ρ v )k l3 ⎤ h( x ) = ⎢ ⎥ 4 x (Tsat − Tw )ν l ⎦ ⎣ 1/ 4 1/ 4 Average coeff. where L is the plate length. Total heat transfer rate : Condensation rate : ⎡ h g ( ρ l − ρ v )k l3 ⎤ hL = 0.943⎢ fg ⎥ ⎣ L(Tsat − Tw )ν l ⎦ q = hL A(Tsat − Tw ) & m= h A(Tsat − Tw ) q = L h fg h fg 1/ 4 Module 9: Learning objectives • In this chapter of radiation heat transfer, many new important ideas will be introduced. The treatment and terminology is quite different from those of conduction and convection heat transfer. In order to avoid confusion and to link radiation heat transfer with other modes of heat transfer, the subject matter has been developed in a systematic fashion, and careful rereading of the material should make you more comfortable with its application. You should be able to answer the following questions, which have mostly to do with terminology and definitions. What is radiation, and what position does thermal radiation occupy in the electromagnetic spectrum? What are the physical origins of thermal radiation? What are meant by the term irradiation, emissive power, and radiosity? What material property characterizes the ability of a surface to emit thermal radiation? What processes and associated materials properties characterize the manner in which a surface responds to irradiation? What is an opaque surface? A semitransparent surface? You should recognize the difference between directional and spectral radiation properties on the one hand and hemispherical and total properties on the other. Moreover, you should be able to proceed from knowledge of the former to determination of the latter. You should also appreciate the unique role of the black body in the description of thermal radiation. In what sense is the black body ideal? Why is it an idealization, and how it may be approximated in practice? What is the Planck distribution? How is it altered by increasing surface temperature? What are the Wein-Stefan –Boltzman laws? Relations between emissivity and absorptivity are often extremely important in radiative exchange calculations. What is Kirchhoff’s law, and what restrictive conditions are inherent in its deviation? What is a gray surface and under what conditions might the assumption of grey surface be particularly suspect? Finally what are the characteristics of solar radiation? In what region of the spectrum is this radiation concentrated, and how is it altered due to passage through the earth’s atmosphere? What is the nature of its directional distribution at the earth’s surface? In this chapter, we also focused on the analysis of radiation exchange between the surfaces of an enclosure, and in treating this exchange we introduced the view factor concept. Because knowing this geometrical quantity is essential to determining radiation exchange between any two diffuse surfaces, and you should be familiar with the means by which it may be determined. You should also be adept at performing radiations calculations for an enclosure of isothermal, opaque, diffuse, and gray surfaces of uniform radiosity and irradiation. Moreover, you should be familiar with the results that apply to simple cases such as the two surfaces enclosure or the three-surface enclosure with reradiating surface. • • • • Thermal energy emitted by matter as a result of vibrational and rotational movements of molecules, atoms and electrons. The energy is transported by electromagnetic waves (or photons). Radiation requires no medium for its propagation, therefore, can take place also in vacuum. All matters emit radiation as long as they have a finite (greater than absolute zero) temperature. The rate at which radiation energy is emitted is usually quantified by the modified StefanBolzmann law: where the emissivity, ε , is a property of the surface characterizing how effectively the surface radiates compared to a "blackbody" (0<ε<1 ). E=q/A (W/m2) is the surface emissive power. σ is the Stefan-Boltzmann constant (s=5.67x10-8 W/(m2K4)). Tb is the absolute surface temp. (in K) Gα=αG Absorbed irradiation Electromagnetic radiation spectrum Thermal radiation spectrum range: 0.1 to 100 mm It includes some ultraviolet (UV) radiation and all visible (0.4-0.76 mm) and infrared radiation (IR). The Planck Distribution The Planck law describes theoretical spectral distribution for the emissive power of a black body. It can be written as where C1=3.742x108 (W.μm4/m2) and C2=1.439x104 (μm.K) are two constants. The Planck distribution is shown in the following figure as a function of wavelength for different body temperatures. Spectral blackbody emissive power Planck Distribution • At given wavelength, the emissive power increases with increasing temperature • As the temperature increases,more emissive energy appear at shorter wavelengths • For low temperature (>800 K), all radiant energy falls in the infrared region and is not visible to the human eyes. That is why only very high temperature objects, such as molten iron, can glow. • Sun can be approximated as a blackbody at 5800 K Solar Irradiation Angles and Arc Length We are well accustomed to thinking of an angle as a two dimensional object. It may be used to find an arc length: α L L = r·α Solid Angle We generalize the idea of an angle and an arc length to three dimensions and define a solid angle, Ω, which like the standard angle has no dimensions. The solid angle, when multiplied by the radius squared will have dimensions of length squared, or area, and will have the magnitude of the encompassed area. A = r2·dΩ r Projected Area The area, dA1, as seen from the prospective of a viewer, situated at an angle θ from the normal to the surface, will appear somewhat smaller, as cos θ·dA1. This smaller area is termed the projected area. Aprojected = cos θ·Anormal θ dA1 dA1·cos θ Intensity The ideal intensity, Ib, may now be defined as the energy emitted from an ideal body, per unit projected area, per unit time, per unit solid angle. dq I= cos θ ⋅ dA1 ⋅ dΩ Spherical Geometry Since any surface will emit radiation outward in all directions above the surface, the spherical coordinate system provides a convenient tool for analysis. The three basic coordinates shown are R, φ, and θ, representing the radial, azimuthal and zenith directions. R·sin θ dA2 dA1 θ R Δφ In general dA1 will correspond to the emitting φ surface or the source. The surface dA2 will correspond to the receiving surface or the target. Note that the area proscribed on the hemisphere, dA2, may be written as: dA = [( R ⋅ sin θ ) ⋅ dϕ ] ⋅ [ R ⋅ dθ ] 2 dA2 = R 2 ⋅ sin θ ⋅ dϕ ⋅ dθ ] Recalling the definition of the solid angle, dA = R2·dΩ we find that: dΩ = R2·sin θ·dθ·dφ Real Surfaces Thus far we have spoken of ideal surfaces, i.e. those that emit energy according to the Stefan-Boltzman law: Eb = σ·Tabs4 Real surfaces have emissive powers, E, which are somewhat less than that obtained theoretically by Boltzman. To account for this reduction, we introduce the E emissivity, ε. ε≡ Eb Emissive power from any real surface is given by: E = ε·σ·Tabs4 Incident Radiation, G Receiving Properties Targets receive radiation in one of three Reflected ways; they absorption, reflection or Radiation transmission. •Absorptivity, α, the fraction of incident Absorbed radiation absorbed. Radiation •Reflectivity, ρ, the fraction of incident radiation reflected. • Transmissivity, τ, the fraction of incident radiation transmitted. Transmitted Radiation We see, from Conservation of Energy, that: α+ρ+τ =1 In this course, we will deal with only opaque surfaces, τ = 0, so that: α+ρ=1 Relationship Between Absorptivity,α, and Emissivity,ε Consider two flat, infinite planes, surface A and surface B, both emitting radiation toward one another. Surface B is assumed to be an ideal emitter, i.e. εB = 1.0. Surface A will emit radiation according to the Stefan-Boltzman law as: EA = εA·σ·TA4 and will receive radiation as: GA = αA·σ·TB4 The net heat flow from surface A will be: q΄΄ = εA·σ·TA4 - αA·σ·TB4 Surface A, TA Surface B, TB Now suppose that the two surfaces are at exactly the same temperature. The heat flow must be zero according to the 2nd law. If follows then that: αA = εA Thermodynamic properties of the material, α and ε may depend on temperature. In general, this will be the case as radiative properties will depend on wavelength, λ. The wave length of radiation will, in turn, depend on the temperature of the source of radiation. The emissivity, ε, of surface A will depend on the material of which surface A is composed, i.e. aluminum, brass, steel, etc. and on the temperature of surface A. The absorptivity, α, of surface A will depend on the material of which surface A is composed, i.e. aluminum, brass, steel, etc. and on the temperature of surface B. Black Surfaces Within the visual band of radiation, any material, which absorbs all visible light, appears as black. Extending this concept to the much broader thermal band, we speak of surfaces with α = 1 as also being “black” or “thermally black”. It follows that for such a surface, ε = 1 and the surface will behave as an ideal emitter. The terms ideal surface and black surface are used interchangeably. Diffuse Surface: Refers to directional independence of the intensity associated with emitted,reflected ,or incident radiation. Grey Surface: A surface for which the spectral absorptivity and the emissivity are independent of wavelength over the spectral regions of surface irradiation and emission. Relationship Between Emissive Power and Intensity By definition of the two terms, emissive power for an ideal surface, Eb, and intensity for an ideal surface, Ib. Eb = hemisphere ∫I b ⋅ cos θ ⋅ dΩ Replacing the solid angle by its equivalent in spherical angles: Eb = ∫ ∫ 0 2⋅π π 2 0 I b ⋅ cos θ ⋅ sin θ ⋅ dθ ⋅ dϕ Integrate once, holding Ib constant: Integrate a second time. (Note that the derivative of sin θ is cos θ·dθ.) Eb = 2 ⋅ π ⋅ I b ⋅ ∫ π 2 0 2 cos θ ⋅ sin θ ⋅ dθ π 2 sin θ Eb = 2 ⋅ π ⋅ I b ⋅ 2 = π ⋅ Ib 0 Eb = π·Ib dq I= cos θ ⋅ dA1 ⋅ dΩ Next we will project the receiving surface onto the hemisphere surrounding the source. First find the projected area of surface dA2, dA2·cos θ2. (θ2 is the angle between the normal to surface 2 and the position vector, R.) Then find the solid angle, Ω, which encompasses this area. dq = I ⋅ cos θ ⋅ dA1 ⋅ dΩ dA2·cos θ2 dA2 R To obtain the entire heat transferred from a finite area, dA1, to a finite area, dA2, we integrate over both surfaces: q1→ 2 = ∫ ∫ I ⋅ cos θ1 ⋅ dA1 ⋅ cos θ 2 dA2 R2 A2 A1 Total energy emitted from surface 1: qemitted = E1·A1 = π·I1·A1 View Factors-Integral Method Define the view factor, F1-2, as the fraction of energy emitted from surface 1, which directly strikes surface 2. I ⋅ cos θ1 ⋅ dA1 ⋅ cos θ 2 dA2 q1→ 2 A2 A1 R2 = F1→ 2 = π ⋅ I ⋅ A1 qemitted ∫ ∫ dr F1→ 2 1 = ⋅ A1 ∫ ∫ cos θ1 ⋅ cosθ 2 ⋅ dA1 ⋅ dA2 D A2 A1 π ⋅ R2 dAj Aj Example Consider a diffuse circular disk of diameter D and area Aj and a plane diffuse surface of area Ai << Aj. The surfaces are parallel, and Ai is located at a distance L from the center of Aj. Obtain an expression for the view factor Fij. θj R θi dAi dAi R L F1→2 1 = ⋅ A1 ∫ ∫ cos θ1 ⋅ cosθ 2 ⋅ dA1 ⋅ dA2 A2 A1 π ⋅ R2 Since dA1 is a differential area F1→ 2 = ∫ ( R ) ⋅ 2π ⋅ r ⋅ dr L 2 F1→ 2 = ∫ cos θ1 ⋅ cos θ 2 ⋅ dA2 A2 A2 π ⋅R 2 F1→ 2 = ∫ L2 ⋅ 2 ⋅ r ⋅ dr R4 F1→ 2 = π ⋅ R2 A2 Let ρ2 ≡ L2 + r2 = R2. Then 2·ρ·dρ = 2·r·dr. 2 ∫ L2 ⋅ 2 ⋅ ρ ⋅ dρ A2 ρ4 D F1→ 2 = − 2 ⋅ L ⋅ ρ −2 A2 2 ⎡ 1 ⎤ 2 = −L ⋅ ⎢ 2 2⎥ ⎢ L + ρ ⎥0 ⎦ ⎣ 2 D F1→2 4 1⎤ 2 D2 ⎡ = − L2 ⋅ ⎢ − 2⎥ = 2 2 L ⎦0 4 ⋅ L2 + D 2 ⎣4 ⋅ L + D Enclosures In order that we might apply conservation of energy to the radiation process, we must account for all energy leaving a surface. We imagine that the surrounding surfaces act as an enclosure about the heat source which receive all emitted energy. For an N surfaced enclosure, we can then see that: ∑F j =1 N i, j =1 This relationship is known as the Conservation Rule”. This relationship is known as “Reciprocity”. Reciprocity Ai ⋅ Fi → j = A j ⋅ F j →i Example: Consider two concentric spheres shown to the right. All radiation leaving the outside of surface 1 will strike surface 2. Part of the radiant energy leaving the inside surface of object 2 will strike surface 1, part will return to surface 2. Find F2,1 . Apply reciprocity. 1 2 A2 ⋅ F2,1 = A1 ⋅ F1, 2 ⇒ F2,1 A1 A1 ⎡ D1 ⎤ = ⋅ F1, 2 = =⎢ ⎥ A2 A2 ⎣ D2 ⎦ 2 Associative Rule Consider the set of surfaces shown to the right: Clearly, from conservation of energy, the fraction of energy leaving surface i and striking the combined surface j+k will equal the fraction of energy emitted from i and striking j plus the fraction leaving surface i and striking k. i j k Fi ⇒ ( j + k ) = Fi ⇒ j + Fi ⇒ k Radiosity This relationship is known as the “Associative Rule”. ε·Eb ρ·G G Radiosity, J, is defined as the total energy leaving a surface per unit area and per unit time. J ≡ ε·Eb + ρ·G Net Exchange Between Surfaces Consider the two surfaces shown. Radiation will travel from surface i to surface j and will also travel from j to i. qi→j = Ji·Ai· Fi→j likewise, qj→i = Jj·Aj· Fj→j The net heat transfer is then: qj→i (net) = Ji·Ai· Fi→j - Jj·Aj· Fj→j From reciprocity we note that F1→2·A1 = F2→1·A2 so that qj→i (net) = Ji·Ai· Fi→j - Jj· Ai· Fi→j = Ai· Fi→j·(Ji – Jj) Jj Ji Net Energy Leaving a Surface The net energy leaving a surface will be the difference between the energy leaving a surface and the energy received by a surface: q1→ = [ε·Eb – α·G]·A1 Combine this relationship with the definition of Radiosity to eliminate G. J ≡ ε·Eb + ρ·G G = [J - ε·Eb]/ρ q1→ = {ε·Eb – α·[J - ε·Eb]/ρ}·A1 Assume opaque surfaces so that α + ρ = 1 ρ = 1 – α, and substitute for ρ. q1→ = {ε·Eb – α·[J - ε·Eb]/(1 – α)}·A1 Put the equation over a common denominator: q1→ ⎡ (1 − α ) ⋅ ε ⋅ Eb − α ⋅ J + α ⋅ ε ⋅ Eb ⎤ ⎡ ε ⋅ Eb − α ⋅ J ⎤ =⎢ ⎥ ⋅ A1 = ⎢ ⎥ ⋅ A1 1−α 1−α ⎣ ⎦ ⎣ ⎦ assume that α = ε q1→ ⎡ ε ⋅ Eb − ε ⋅ J ⎤ ⎡ ε ⋅ A1 ⎤ =⎢ ⎥ ⋅ A1 = ⎢ 1 − ε ⎥ ⋅ (Eb − J ) 1− ε ⎣ ⎦ ⎣ ⎦ Electrical Analogy for Radiation We may develop an electrical analogy for radiation, similar to that produced for conduction. The two analogies should not be mixed: they have different dimensions on the potential differences, resistance and current flows. Equivalent Current Equivalent Resistance Potential Difference Ohms Law I R ΔV Net Energy Leaving Surface q1? ⎡1 − ε ⎤ ⎢ε ⋅ A⎥ ⎣ ⎦ 1 A1 ⋅ F1→2 Eb - J Net Exchange Between Surfaces qi? j J1 – J2 Example: Consider a grate fed boiler. Coal is fed at the bottom, moves across the grate as it burns and radiates to the walls and top of the furnace. The walls are cooled by flowing water through tubes placed inside of the walls. Saturated water is introduced at the Coal Chute bottom of the walls and leaves at the top at a quality of about 70%. After the vapour is separated from the water, it is circulated through the superheat tubes at the top of the boiler. Since the steam is undergoing a sensible heat addition, its temperature will rise. It is common practice to subdivide the super-heater tubes into sections, each having nearly uniform temperature. In our case we will use only one superheat section using an average temperature for the entire region. Surface 3 Superheat Tubes Surface 2 Water Walls SH section Surface 1 Burning Coal Energy will leave the coal bed, Surface 1, as described by the equation for the net energy leaving a surface. We draw the equivalent electrical network as seen to the right: The heat leaving from the surface of the coal may proceed to either the water walls or to the super-heater section. That part of the circuit is represented by a potential difference between Radiosity: 1 A2 ⋅ F2→3 J2 J3 R12 J1 R1 Eb1 J1 R1 Eb1 ⎡ 1− ε ⎤ ⎢ ⎥ ⎣ ε ⋅ A1 ⎦ σ·T14 J2 R12 1 A1 ⋅ F1→ 2 J3 R13 1 J1 A1 ⋅ F1→3 R1 Eb1 1 A1 ⋅ F1→ 2 R13 1 A1 ⋅ F1→3 It should be noted that surfaces 2 and 3 will also radiate to one another. It remains to evaluate the net heat flow leaving (entering) nodes 2 and 3. ⎡ 1− ε 2 ⎤ ⎢ ⎥ ⎣ ε 2 ⋅ A2 ⎦ R2 J2 R12 1 A2 ⋅ F2→3 1 A1 ⋅ F1→ 2 J3 Eb2 J1 1 R13 Eb3 A1 ⋅ F1→3 ⎡ ε ⋅ A1 ⎤ ⎢ 1− ε ⎥ ⎣ ⎦ ⎡ ε ⋅ A1 ⎤ ⎢ 1− ε ⎥ ⎣ ⎦ R1 Eb1 Insulated surfaces. In steady state heat transfer, a surface cannot receive net energy if it is insulated. Because the energy cannot be stored by a surface in steady state, all energy must be re-radiated back into the enclosure. Insulated surfaces are often termed as re-radiating surfaces. • Black surfaces: A black, or ideal surface, will have no surface resistance: ⎡1 − ε ⎤ ⎡1 − 1 ⎤ ⎢ ε ⋅ A ⎥ = ⎢1⋅ A ⎥ = 0 ⎣ ⎦ ⎣ ⎦ • In this case the nodal Radiosity and emissive power will be equal. Large surfaces: Surfaces having a large surface area will behave as black surfaces, irrespective of the actual surface properties: Consider the case of an object, 1, placed inside a large enclosure, 2. The system will consist of two objects, so we proceed to construct a circuit with two radiosity nodes. Now we ground both Radiosity nodes through a surface resistance. (1-ε1)/(ε1⋅A1) Eb1 σ⋅T14 R1 J1 1/(A1⋅F1→2) • ⎡1 − ε ⎤ ⎡1 − ε ⎤ ⎢ ε ⋅ A ⎥ = ⎢ε ⋅ ∞ ⎥ = 0 ⎦ ⎣ ⎦ ⎣ J1 1/(A1⋅F1→2) J2 J2 (1-ε2)/(ε2⋅A2) R2 Eb2 σ⋅T24 R12 Since A2 is large, R2 = 0. The view factor, F1→2 = 1 (1-ε1)/(ε1⋅A1) Eb1 σ⋅T14 R1 J1 1/(A1⋅F1→2) R12 J2 Eb2 σ⋅T24 Sum the series resistances: RSeries = (1-ε1)/(ε1⋅A1) + 1/A1 = 1/(ε1⋅A1) Ohm’s law: i = ΔV/R or by analogy: q = ΔEb/RSeries = ε1⋅A1⋅σ⋅(T14 – T24) Returning for a moment to the coal grate furnace, let us assume that we know (a) the total heat being produced by the coal bed, (b) the temperatures of the water walls and (c) the temperature of the super heater sections. Apply Kirchoff’s law about node 1, for the coal bed: q1 + q 2→1 + q3→1 = q1 + Similarly, for node 2: J 2 − J1 J 3 − J1 + =0 R12 R13 q 2 + q1→ 2 + q3→ 2 = And for node 3: Eb 2 − J 2 J 1 − J 2 J 3 − J 2 + + =0 R2 R12 R23 q3 + q1→3 + q2→3 ⎡ 1 1 − ⎢− ⎢ R12 R13 1 ⎢ ⎢ R12 ⎢ 1 ⎢ ⎢ R13 ⎣ Eb 3 − J 3 J 1 − J 3 J 2 − J 3 = + + =0 R3 R13 R23 ⎡ ⎤ ⎤ 1 ⎢ ⎥ ⎥ − q1 ⎥ R13 ⎥ ⎡ J1 ⎤ ⎢ 1 ⎥ ⋅ ⎢ J ⎥ = ⎢ − Eb 2 ⎥ ⎥ ⎢ 2⎥ ⎢ R ⎥ R23 2 ⎥ ⎥ ⎢ ⎥ ⎢ 1 1 1 ⎥ ⎣ J 3 ⎦ ⎢ Eb 3 ⎥ − − − − ⎢ R3 ⎥ R3 R13 R23 ⎥ ⎦ ⎣ ⎦ The three equations must be solved simultaneously. Since they are each linear in J, matrix methods may be used: 1 R12 1 1 1 − − − R2 R12 R13 1 R23 Surface 1: Find the coal bed temperature, given the heat flow: q1 = Eb1 − J 1 σ ⋅ T − J 1 ⎡ q ⋅ R + J1 ⎤ = ⇒ T1 = ⎢ 1 1 ⎥ R1 R1 σ ⎣ ⎦ 4 1 0.25 Surface 2: Find the water wall heat input, given the water wall temperature: Eb 2 − J 2 σ ⋅ T24 − J 2 q2 = = R2 R2 Surface 3: (Similar to surface 2) Find the water wall heat input, given the water wall temperature: Eb3 − J 3 σ ⋅ T34 − J 3 q3 = = R3 R3 Module 9: Worked out problems 1. A spherical aluminum shell of inside diameter D=2m is evacuated and is used as a radiation test chamber. If the inner surface is coated with carbon black and maintained at 600K, what is the irradiation on a small test surface placed in the chamber? If the inner surface were not coated and maintained at 600K, what would the irradiation test? Known: Evacuated, aluminum shell of inside diameter D=2m, serving as a radiation test chamber. Find: Irradiation on a small test object when the inner surface is lined with carbon black and maintained at 600K.what effect will surface coating have? Schematic: Assumptions: (1) Sphere walls are isothermal, (2) Test surface area is small compared to the enclosure surface. Analysis: It follows from the discussion that this isothermal sphere is an enclosure behaving as a black body. For such a condition, the irradiation on a small surface within the enclosure is equal to the black body emissive power at the temperature of the enclosure. That is G 1 = E b (Ts ) = σTs4 G 1 = 5.67 × 10 − 8 W / m 2 . K (600 K ) 4 = 7348W / m 2 The irradiation is independent of the nature of the enclosure surface coating properties. Comments: (1) The irradiation depends only upon the enclosure surface temperature and is independent of the enclosure surface properties. (2) Note that the test surface area must be small compared to the enclosure surface area. This allows for inter-reflections to occur such that the radiation field, within the enclosure will be uniform (diffuse) or isotropic. (3) The irradiation level would be the same if the enclosure were not evacuated since; in general, air would be a non-participating medium. 2 Assuming the earth’s surface is black, estimate its temperature if the sun has an equivalently blackbody temperature of 5800K.The diameters of the sun and earth are 1.39*109 and 1.29*107m, respetively, and the distance between the sun and earth is 1.5*1011m. Known: sun has an equivalently blackbody temperature of 5800K. Diameters of the sun and earth as well as separation distances are prescribed. Find: Temperature of the earth assuming the earth is black. Schematic: Assumptions: (1) Sun and earth emit black bodies, (2) No attenuation of solar irradiation enroute to earth, and (3) Earth atmosphere has no effect on earth energy balance. Analysis: performing an energy balance on the earth E in − E out = 0 Ae , p .G S = Ae , s . E b (Te ) (πDe2 / 4)G S = πDe2σTe4 Te = (G S / 4σ ) 1/ 4 . . Where As,p and Ae,s are the projected area and total surface area of the earth, respectively. To determine the irradiation GS at the earth’s surface, perform an energy bounded by the spherical surface shown in sketch E in − E out = 0 2 πD S .σTS4 = 4π [ RS ,e − De / 2]2 G S . . π (1.39 × 10 9 m ) 2 × 5.67 × 10 − 8 W / m 2 . K ( 5800 K ) 4 = 4π [1.5 × 1011 − 1.29 × 10 7 / 2]2 m 2 × G S G S = 1377 .5W / m 2 Substituting numerical values, find Te = (1377.5W / m 2 / 4 × 5.67 × 10 −8 W / m 2 . K 4 )1 / 4 = 279 K Comments: (1) The average earth’s temperature is greater than 279 K since the effect of the atmosphere is to reduce the heat loss by radiation. (2) Note carefully the different areas used in the earth energy balance. Emission occurs from the total spherical area, while solar irradiation is absorbed by the projected spherical area. 3 The spectral, directional emissivity of a diffuse material at 2000K has the following distribution. Determine the total, hemispherical emissivity at 2000K.Determine the emissive power over the spherical range 0.8 t0 2.5 μm and for the directions 0≤θ≤30°. Known: Spectral, directional emissivity of a diffuse material at 2000K. Find: (1) The total, hemispherical emissivity, (b) emissive power over the spherical range 0.8 t0 2.5 μm and for the directions 0≤θ≤30°. Schematic: Assumptions: (1) Surface is diffuse emitter. Analysis: (a) Since the surface is diffuse,ελ,θ is independent of direction; from Eq. ελ,θ = ελ ε (T ) = ∫ ε λ (λ ) E λ ,b (λ , T )dλ / E b (T ) 0 ∞ ε (T ) = ∫ ε 1 ) E λ ,b (λ ,2000)dλ / E b ) + ∫ ε 2 E λ ,b (λ ,2000)dλ / E b 0 0 1.5 1.5 Written now in terms of F (0→λ), with F (0→1.5) =0.2732 at λT=1.5*2000=3000μm.K, find ε ( 2000 K ) = ε 1 F( 0→1.5 ) + ε 2 [1 − F( 0→1.5 ) ] = 0.2 × 0.2732 + 0.8[1 − 0.2732] = 0.636 (b) For the prescribed spectral and geometric limits, 2.5 2π π / 6 ΔE = 0.8 0 ∫ ∫ ∫ ε λ θ Iλ , 0 ,b (λ , T ) cosθ sinθ dθ dφ dλ where I λ ,e (λ ,θ , φ ) = ε λ ,θ I λ ,b (λ , T ). Since the surface is diffuse, ε λ ,θ = ε λ , and nothing I λ ,b is inde of direction and equal to E λ ,b/ π , we can write ⎧ 2π π / 6 ⎫ E (T ) 0∫8 . ΔE = ⎨ ∫ ∫ cos θ sin θ dθ dφ ⎬ b π ⎩0 0 ⎭ 1.5 ε 1 E λ ,b (λ , T )dλ E b (T ) + 2.5 1.5 ∫ε 2 E λ ,b (λ , T )dλ E b (T ) Or in terms F (0→λ) values, ⎧ 2π sin2 θ π / 6⎫ σT 4 ΔE = ⎨φ × ε 1[F(0→1.5) − F(0→0.8) ] + ε 2 [F(0→2.5) − F(0→1.5) ]} ⎬ 2 0⎭ π ⎩ 0 { } From table F( 0→ 0.8 ) λ T = 0.8 × 2000 = 1600 μm . K λ T = 2.5 × 2000 = 5000 μm . K = 0.0197 F( 0→ 2.5 ) = 0.6337 sin 2 π / 6 5.67 × 10 −8 2000 4 W {0.2[0.2732 − 0.0197] + [0.80.6337 − 0.2732]} ΔE = 2π × 2 π m2 ΔE = 0.25 × (5.67 × 10 − 8 × 2000 4 W / m 2 × 0.339 = 76.89W / m 2 4. A diffusely emitting surface is exposed to a radiant source causing the irradiation on the surface to be 1000W/m2.The intensity for emission is 143W/m2.sr and the reflectivity of the surface is 0.8.Determine the emissive power ,E(W/m2),and radiosity ,J(W/m2),for the surface. What is the net heat flux to the surface by the radiation mode? Known: A diffusely emitting surface with an intensity due to emission of Is=143W/m2.sr and a reflectance ρ=0.8 is subjected to irradiation=1000W/m2. Find: (a) emissive power of the surface, E (W/m2), (b) radiosity, J (W/m2), for the surface, (c) net heat flux to the surface. Schematic: Ie( G=1000W/m2 W/m2.sr Surface, Assumptions: (1) surface emits in a diffuse manner. Analysis: (a) For a diffusely emitting surface, Is(θ) =Ie is a constant independent of direction. The emissive power is E = πI e = πsr × 143W / m 2 . sr = 449W / m 2 Note that π has units of steradians (sr). (b) The radiosity is defined as the radiant flux leaving the surface by emission and reflection, J = E + ρG = 449W / m 2 + 0.8 × 1000W / m 2 = 1249W / m 2 (c) The net radiative heat flux to the surface is determined from a radiation balance on the surface. " " " q net = q rad ,in − q rad ,out " q net = G − J = 1000W / m 2 − 1249W / m 2 = −249W / m 2 Comments: No matter how the surface is irradiated, the intensity of the reflected flux will be independent of direction, if the surface reflects diffusely. 5. Radiation leaves the furnace of inside surface temperature 1500K through an aperture 20mm in diameter. A portion of the radiation is intercepted by a detector that is 1m from the aperture, as a surface area of 10-5m2, and is oriented as shown. If the aperture is open, what is the rate at which radiation leaving the furnace is intercepted by the detector? If the aperture is covered with a diffuse, semitransparent material of spectral transmissivity τλ=0.8 for λ≤2μm and τλ=0 for λ>2μm, what is the rate at which radiation leaving the furnace is intercepted by the detector? Known: Furnace wall temperature and aperture diameter. Distance of detector from aperture and orientation of detector relative to aperture. Find: Rate at which radiation leaving the furnace is intercepted by the detector, (b) effect of aperture window of prescribed spectral transmissivity on the radiation interception rate. Schematic: Assumptions: (1) Radiation emerging from aperture has characteristics of emission from a black body, (2) Cover material is diffuse, (3) Aperture and detector surface may be approximated as infinitesimally small. Analysis: (a) the heat rate leaving the furnace aperture and intercepted by the detector is q = I e As cos θ w a − a Ie = E b (T f ) π σT f4 5.67 × 10 −8 (1500) 4 = = = 9.14 × 10 4 W / m 2 . sr π π w s−a = A n As . cos θ 2 10 − 5 m 2 cos 45° = = = 0.70710 − 5 . sr r2 r2 (1m ) 2 hence q = 9.14 × 10 4 W / m 2 . sr[π (0.02)m 2 / 4] cos 30° × 0.707 × 10 −5 sr = 1.76 × 10 −4 W (b) With the window, the heat rate is q = τ ( I e Aa cos θ 1 w a − a ) where τ is the transmissivity of the window to radiation emitted by the furnace wall. τ = ∫ τ λ Gλ d λ ∫ τ λ E λ 0 ∞ ∞ .b (T f )dλ ,b ∫ Gλ d λ 0 ∞ = 0 ∫ Eλ 0 ∞ dλ = 0.8 ∫ ( E λ ,b / E b )dλ = 0.8 F( 0→ 2 μm ) 0 2 with λ T = 2μm × 1500K = 3000 μm.K, from table F(0 → 2 μm) = 0.273. hence with 0.273 × 0.8 = 0.218, find q = 0.218 × 1.76 × 10 -4 W = 0.384 × 10 -4 W 6.A horizontal semitransparent plate is uniformly irradiated from above and below, while air at T=300K flows over the top and bottom surfaces. providing a uniform convection heat transfer coefficient of h=40W/m2.K.the total, hemispherical absorptivity of the plate to the irradiation is 0.40.Under steady-state conditions measurements made with radiation detector above the top surface indicate a radiosity(which includes transmission, as well as reflection and emission) of J=5000W/m2,while the plate is at uniform temperature of T=350K Determine the irradiation G and the total hemispherical emissivity of the plate. Is the plate gray for the prescribed conditions? Known: Temperature, absorptivity, transmissivity, radiosity and convection conditions for a semi-transparent plate. Find: Plate irradiation and total hemispherical emissivity. Schematic: Assumptions: From an energy balance on the plate E in − E out " 2G = 2q conv + 2 J . Solving for the irradiation and substituting numerical values, G=40W/m2.K (350-300) K+5000W/m2=7000W/m2 From the definition of J J = E + ρG + τG = E + (1 − α )G Solving for the emissivity and substituting numerical values, ε = J − (1 − α )G = ( 5000 W / m 2 ) − 0 .6( 7000 W / m 2 ) 5 .67 × 10 − 8 σT 4 W / m 2 .K 4 ( 350 K ) 4 = 0 .94 Hence α ≠ε And the surface is not gray for the prescribed conditions. Comments: The emissivity may also be determined by expressing the plate energy balance as " 2αG = 2q conv 2 E hence εσT 4 = αG − h(T − T∞ ) ε= 0.4(7000W / m 2 ) − 40W / m 2 . K (50 K ) = 0.94 5.67 × 10 − 8 W / m 2 . K 4( 350 K ) 4 7 An opaque, gray surface at 27°C is exposed to irradiation of 1000W/m2, and 800W/m2 is reflected. Air at 17°C flows over the surface and the heat transfer convection coefficient is 15W/m2.K.Determine the net heat flux from the surface. Known: Opaque, gray surface at 27°C with prescribed irradiation, reflected flux and convection process. Find: Net heat flux from the surface. Schematic: Assumptions: (1) Surface is opaque and gray, (2) Surface is diffuse, (3) Effects of surroundings are included in specified irradiation. Analysis: From an energy balance on the surface, the net heat flux from the surface is " " " q net = E out − E in . " " q net = q conv + E + G ref − G = h(Ts − T∞ ) + εσTs4 + G ref − G ε = α = 1 − ρ = 1 − (G ref / G ) = 1 − (800 / 1000) = 1 − 0.8 = 0.2 where ρ = G ref /G. the net heat flux from the surface q " = 15W / m 2 . K ( 27 − 17 ) K + 0.2 × 5.67 × 10 − 8 W / m 2 . K 4 ( 27 + 273) 4 K 4 + 800W / m 2 − 10 net " q net = (150 + 91.9 + 800 − 1000)W / m 2 = 42W / m 2 Comments: (1) For this situation, the radiosity is J = G ref + E = (800 + 91.9)W / m 2 = 892W / m 2 The energy balance can be written involving the radiosity (radiation leaving the surface) and the irradiation (radiation to the surface). " " q net ,out = J − G + q conv = ( 892 − 1000 + 150 )W / m 2 = 42W / m 2 Note the need to assume the surface is diffuse, gray and opaque in order that Eq (2) is applicable. 8. A small disk 5 mm in diameter is positioned at the center of an isothermal, hemispherical enclosure. The disk is diffuse and gray with an emissivity of 0.7 and is maintained at 900 K. The hemispherical enclosure, maintained at 300 K, has a radius of 100 mm and an emissivity of 0.85. Calculate the radiant power leaving an aperture of diameter 2 mm located on the enclosure as shown. Known: Small disk positioned at center of an isothermal, hemispherical enclosure with a small aperture. Find: radiant power [μW] leaving the aperture. Schematic: Assumptions: (1)Disk is diffuse-gray,(2) Enclosure is isothermal and has area much larger than disk,(3) Aperture area is very small compared to enclosure area, (4) Areas of disk and aperture are small compared to radius squared of the enclosure. Analysis: the radiant power leaving the aperture is due to radiation leaving the disk and to irradiation on the aperture from the enclosure. That is q ap = q1→ 2 + G 2 .A2 The radiation leaving the disk can be written in terms of the radiosity of the disk. For the diffuse disk q1→ 2 = 1 π J 1 . A1 cos θ 1 .w 2−1 and with ε = α for the gray behavior, the radiosity is J 1 = ε 1 E b (T1 ) + ρG1 = ε 1σT14 + (1 − ε 1 )σT34 Where the irradiatin G 1 is the emissive power of the black enclosure, E b (T3 ); G 1 = G 2 = E b (T3 ).The solid angle ω 2-1follows ω 2−1 = A2 / R 2 Combining equations. (2), (3) and (4) into eq.(1) with G2=σT43, the radiant power is q ap = 1 π 1 σ [ε 1T14 + (1 − ε 1 )T34 ] A1 cos θ 1 . A2 + A2σT34 R2 q ap = π 5.67 × 10 − 8 W / m 2 . K 4 [0.7(900 K ) 4 + (1 − 0.7 )( 300 K ) 4 π 4 (0.005m ) 2 cos 45° × π / 4(0.002m ) 2 (0.100m ) 2 + π 4 (0.002m )25.67 × 10 − 8 W / m 2 . K 4 ( 300 K ) 4 q ap = ( 36.2 + 0.19 + 1443)μW = 1479 μW Comments: Note the relative magnitudes of the three radiation components. Also, recognize that the emissivity of the enclosure’s ε3 doesn’t enter into the analysis. Why? Module 9: Short questions 1. What is thermal radiation? How does it differ from other forms of electromagnetic radiation? 2. What is a black body? Does a black body actually exist? 3. Define the total and spectral black body emissive powers. How are they related to each other? 4. Consider two identical bodies, one at 1000K and the other at 1500 K. Which body emits more radiation in the shorted wavelength region? 5. What does the solid angle represent, and how does it differ from a plane angle? What is the value of solid angle associated with a sphere? 6. For a diffusely emitting surface, how is the emissive power related to the intensity of the emitted radiation? 7. For diffusely incident radiation, how is irradiation on a surface related to the intensity of the incident radiation? 8. Explain what is the greenhouse effect. 9. What does the view factor represent? When is the view factor form a surface to itself not zero? 10. Consider an enclosure consisting of five surfaces. How many view factors does this geometry involve? How many of these view factors can be determined by the application of reciprocity and the summation rules? 11. How does radiosity for a surface differ from the emitted energy? For what kind of surfaces are these two quantities identical? Multiple choice questions: 1) Radiosity is a) The sum of the emissive power and the irradiance b) The difference between the emissive power and the irradiance c) The sum of the emissive power with the product of the reflectivity and irradiance d) The difference between the emissive power and the product of the reflectivity and irradiance e) None of the above 2) The irradiance to a small non-black body at equilibrium in a large evacuated isothermal cavity is equal to its a) Radiosity b) Emissive power c) Black emissive power d) Intensity e) None of the above 3) Given a surface of known temperature, the Planck spectrum can be integrated to obtain a) Transmissivity b) View Factor c) Mean path length d) TNRR (total normal radiative reflection) e) None of the above 4) If the spectral emissivity of a surface decreases with increasing wavelength, then for increasing temperature its overall emissivity will a) Decrease b) Remain equal c) Increase d) Depend on other factors e) None of the above 5) The sum of the reflectivity, absorptivity and transmissivity is equal to a) Emissivity b) Planck’s constant c) 0 d) 1 e) None of the above 6) For a surface i, surrounded by N surfaces, the sum ∑ F ji 1 N is equal to a) 0 b) 1 c) ∑ Fij 1 N d) G e) None of the above MODULE I RADIATION HEAT TRANSFER Radiation Definition Radiation, energy transfer across a system boundary due to a ΔT, by the mechanism of photon emission or electromagnetic wave emission. Because the mechanism of transmission is photon emission, unlike conduction and convection, there need be no intermediate matter to enable transmission. The significance of this is that radiation will be the only mechanism for heat transfer whenever a vacuum is present. Electromagnetic Phenomena. We are well acquainted with a wide range of electromagnetic phenomena in modern life. These phenomena are sometimes thought of as wave phenomena and are, consequently, often described in terms of electromagnetic wave length, λ. Examples are given in terms of the wave distribution shown below: X Rays UV Visible Light 0.4-0.7μm Light, Thermal Radiation Microwaves ⎪ 10-5 ⎪ 10-4 ⎪ 10-3 ⎪ 10-2 ⎪ 10-1 ⎪ 10-0 ⎪ 101 ⎪ 102 ⎪ 103 ⎪ 104 105 Wavelength, λ, μm One aspect of electromagnetic radiation is that the related topics are more closely associated with optics and electronics than with those normally found in mechanical engineering courses. Nevertheless, these are widely encountered topics and the student is familiar with them through every day life experiences. From a viewpoint of previously studied topics students, particularly those with a background in mechanical or chemical engineering, will find the subject of Radiation Heat Transfer a little unusual. The physics background differs fundamentally from that found in the areas of continuum mechanics. Much of the related material is found in courses more closely identified with quantum physics or electrical engineering, i.e. Fields and Waves. At this point, it is important for us to recognize that since the subject arises from a different area of physics, it will be important that we study these concepts with extra care. Stefan-Boltzman Law Both Stefan and Boltzman were physicists; any student taking a course in quantum physics will become well acquainted with Boltzman’s work as he made a number of important contributions to the field. Both were contemporaries of Einstein so we see that the subject is of fairly recent vintage. (Recall that the basic equation for convection heat transfer is attributed to Newton.) Eb = σ⋅Tabs4 where: Eb = Emissive Power, the gross energy emitted from an ideal surface per unit area, time. σ = Stefan Boltzman constant, 5.67⋅10-8 W/m2⋅K4 Tabs = Absolute temperature of the emitting surface, K. Take particular note of the fact that absolute temperatures are used in Radiation. It is suggested, as a matter of good practice, to convert all temperatures to the absolute scale as an initial step in all radiation problems. You will notice that the equation does not include any heat flux term, q”. Instead we have a term the emissive power. The relationship between these terms is as follows. Consider two infinite plane surfaces, both facing one another. Both surfaces are ideal surfaces. One surface is found to be at temperature, T1, the other at temperature, T2. Since both temperatures are at temperatures above absolute zero, both will radiate energy as described by the Stefan-Boltzman law. The heat flux will be the net radiant flow as given by: q" = Eb1 - Eb2 = σ⋅T14 - σ⋅T24 Plank’s Law While the Stefan-Boltzman law is useful for studying overall energy emissions, it does not allow us to treat those interactions, which deal specifically with wavelength, λ. This problem was overcome by another of the modern physicists, Max Plank, who developed a relationship for wavebased emissions. Ebλ = ƒ(λ) We plot a suitable functional relationship below: Ebλ, W/m2⋅μm Area = Eb Wavelength, λ, μm We haven’t yet defined the Monochromatic Emissive Power, Ebλ. An implicit definition is provided by the following equation: Eb = ∫0 Ebλ ⋅ dλ We may view this equation graphically as follows: ∞ Ebλ, W/m2⋅μm Area = Eb Wavelength, λ, μm A definition of monochromatic Emissive Power would be obtained by differentiating the integral equation: Ebλ ≡ dEb dλ The actual form of Plank’s law is: Ebλ = λ ⋅ e 5 [ C1 C2 λ ⋅T −1 ] where: Where: C1 = 2⋅π⋅h⋅co2 = 3.742⋅108 W⋅μm4/m2 C2 = h⋅co/k = 1.439⋅104 μm⋅K h, co, k are all parameters from quantum physics. We need not worry about their precise definition here. This equation may be solved at any T, λ to give the value of the monochromatic emissivity at that condition. Alternatively, the function ∞ may be substituted into the integral Eb = ∫0 Ebλ ⋅ dλ to find the Emissive power for any temperature. While performing this integral by hand is difficult, students may readily evaluate the integral through one of several computer programs, i.e. MathCad, Maple, Mathmatica, etc. Eb = ∫0 Ebλ ⋅ dλ = σ ⋅ T 4 Emission Over Specific Wave Length Bands Consider the problem of designing a tanning machine. As a part of the machine, we will need to design a very powerful incandescent light source. We may wish to know how much energy is being emitted over the ultraviolet band (10-4 to 0.4 μm), known to be particularly dangerous. ∞ Eb ( 0.0001 → 0.4) = ∫0.001⋅μm Ebλ ⋅ dλ 0. 4⋅μm With a computer available, evaluation of this integral is rather trivial. Alternatively, the text books provide a table of integrals. The format used is as follows: Ebλ ⋅ dλ Eb ( 0.001 → 0.4) ∫0.001⋅μm Ebλ ⋅ dλ ∫0 Ebλ ⋅ dλ ∫0. = ∞ = ∞ − = F (0 → 0.4) − F (0 → 0.0001) ∞ Eb ∫0 Ebλ ⋅ dλ ∫0 Ebλ ⋅ dλ ∫0 Ebλ ⋅ dλ 0.4⋅ μm 0.4⋅μm 0.0001⋅ μm Referring to such tables, we see the last two functions listed in the second column. In the first column is a parameter, λ⋅T. This is found by taking the product of the absolute temperature of the emitting surface, T, and the upper limit wave length, λ. In our example, suppose that the incandescent bulb is designed to operate at a temperature of 2000K. Reading from the table: T, K λ⋅T, μm⋅K 0.0001 2000 0.2 0.4 2000 600 F(0.4→0.0001μm) = F(0→0.4μm)- F(0→0.0001μm) λ., μm F(0→λ) 0 0.000014 0.000014 This is the fraction of the total energy emitted which falls within the IR band. To find the absolute energy emitted multiply this value times the total energy emitted: EbIR = F(0.4→0.0001μm)⋅σ⋅T4 = 0.000014⋅5.67⋅10-8⋅20004 = 12.7 W/m2 Solar Radiation The magnitude of the energy leaving the Sun varies with time and is closely associated with such factors as solar flares and sunspots. Nevertheless, we often choose to work with an average value. The energy leaving the sun is emitted outward in all directions so that at any particular distance from the Sun we may imagine the energy being dispersed over an imaginary spherical area. Because this area increases with the distance squared, the solar flux also decreases with the distance squared. At the average distance between Earth and Sun this heat flux is 1353 W/m2, so that the average heat flux on any object in Earth orbit is found as: Gs,o = Sc·f·cos θ Where Sc = Solar Constant, 1353 W/m2 f = correction factor for eccentricity in Earth Orbit, (0.97<f<1.03) θ = Angle of surface from normal to Sun. Because of reflection and absorption in the Earth’s atmosphere, this number is significantly reduced at ground level. Nevertheless, this value gives us some opportunity to estimate the potential for using solar energy, such as in photovoltaic cells. Some Definitions In the previous section we introduced the Stefan-Boltzman Equation to describe radiation from an ideal surface. Eb = σ·Tabs4 This equation provides a method of determining the total energy leaving a surface, but gives no indication of the direction in which it travels. As we continue our study, we will want to be able to calculate how heat is distributed among various objects. For this purpose, we will introduce the radiation intensity, I, defined as the energy emitted per unit area, per unit time, per unit solid angle. Before writing an equation for this new property, we will need to define some of the terms we will be using. Angles and Arc Length We are well accustomed to thinking of an angle as a two dimensional object. It may be used to find an arc length: α L L = r·α Solid Angle We generalize the idea of an angle and an arc length to three dimensions and define a solid A = r2·dΩ angle, Ω, which like the standard angle has no dimensions. The solid angle, when multiplied by r the radius squared will have dimensions of length squared, or area, and will have the magnitude of the encompassed area. Projected Area The area, dA1, as seen from the prospective of a viewer, situated at an angle θ from the normal to the dA1 surface, will appear somewhat smaller, as cos θ·dA1. This smaller area is termed the projected area. Aprojected = cos θ·Anormal Intensity The ideal intensity, Ib, may now be defined as the energy emitted from an ideal body, per unit projected area, per unit time, per unit solid angle. θ dA1·cos θ I= dq cos θ ⋅ dA1 ⋅ dΩ Spherical Geometry Since any surface will emit radiation outward in all directions above the surface, the spherical coordinate system provides a convenient tool for analysis. The three basic coordinates shown are R, φ, and θ, R·sin θ representing the radial, azimuthal and zenith directions. In general dA1 will correspond to the emitting surface or the source. The surface dA2 will correspond to the receiving surface or the target. Note that the area proscribed on the hemisphere, dA2, may be written as: dA1 dA2 θ R Δφ φ dA2 = [( R ⋅ sin θ ) ⋅ dϕ ] ⋅ [ R ⋅ dθ ] or, more simply as: dA2 = R2 ⋅ sin θ ⋅ dϕ ⋅ dθ ] Recalling the definition of the solid angle, dA = R2·dΩ we find that: Real Surfaces Thus far we have spoken of ideal surfaces, i.e. those that emit energy according to the Stefan-Boltzman law: Eb = σ·Tabs4 dΩ = R2·sin θ·dθ·dφ Real surfaces have emissive powers, E, which are somewhat less than that obtained theoretically by Boltzman. To account for this reduction, we introduce the emissivity, ε. ε≡ E Eb so that the emissive power from any real surface is given by: E = ε·σ·Tabs4 Receiving Properties Targets receive radiation in one of three ways; they absorption, reflection or transmission. To account for these characteristics, we introduce three additional properties: • Absorptivity, α, the fraction of incident radiation absorbed. Reflected Radiation Incident Radiation, G Absorbed Radiation Transmitted Radiation • Reflectivity, ρ, the fraction of incident radiation reflected. • Transmissivity, τ, the fraction of incident radiation transmitted. We see, from Conservation of Energy, that: α+ρ+τ =1 In this course, we will deal with only opaque surfaces, τ = 0, so that: α+ρ=1 Opaque Surfaces Relationship Between Absorptivity,α, and Emissivity,ε Surface A, TA Surface B, TB Consider two flat, infinite planes, surface A and surface B, both emitting radiation toward one another. Surface B is assumed to be an ideal emitter, i.e. εB = 1.0. Surface A will emit radiation according to the Stefan-Boltzman law as: EA = εA·σ·TA4 and will receive radiation as: GA = αA·σ·TB4 The net heat flow from surface A will be: q΄΄ = εA·σ·TA4 - αA·σ·TB4 Now suppose that the two surfaces are at exactly the same temperature. The heat flow must be zero according to the 2nd law. If follows then that: αA = εA Because of this close relation between emissivity, ε, and absorptivity, α, only one property is normally measured and this value may be used alternatively for either property. Let’s not lose sight of the fact that, as thermodynamic properties of the material, α and ε may depend on temperature. In general, this will be the case as radiative properties will depend on wavelength, λ. The wave length of radiation will, in turn, depend on the temperature of the source of radiation. The emissivity, ε, of surface A will depend on the material of which surface A is composed, i.e. aluminum, brass, steel, etc. and on the temperature of surface A. The absorptivity, α, of surface A will depend on the material of which surface A is composed, i.e. aluminum, brass, steel, etc. and on the temperature of surface B. In the design of solar collectors, engineers have long sought a material which would absorb all solar radiation, (α = 1, Tsun ~ 5600K) but would not re-radiate energy as it came to temperature (ε << 1, Tcollector ~ 400K). NASA developed an anodized chrome, commonly called “black chrome” as a result of this research. Black Surfaces Within the visual band of radiation, any material, which absorbs all visible light, appears as black. Extending this concept to the much broader thermal band, we speak of surfaces with α = 1 as also being “black” or “thermally black”. It follows that for such a surface, ε = 1 and the surface will behave as an ideal emitter. The terms ideal surface and black surface are used interchangeably. Lambert’s Cosine Law: A surface is said to obey Lambert’s cosine law if the intensity, I, is uniform in all directions. This is an idealization of real surfaces as seen by the emissivity at different zenith angles: 0 .2 .4 .6 .8 1.0 0 .2 .4 .6 .8 1.0 Dependence of Emissivity on Zenith Angle, Typical Metal. Dependence of Emissivity on Zenith Angle, Typical Non-Metal. The sketches shown are intended to show is that metals typically have a very low emissivity, ε, which also remain nearly constant, expect at very high zenith angles, θ. Conversely, non-metals will have a relatively high emissivity, ε, except at very high zenith angles. Treating the emissivity as a constant over all angles is generally a good approximation and greatly simplifies engineering calculations. Relationship Between Emissive Power and Intensity By definition of the two terms, emissive power for an ideal surface, Eb, and intensity for an ideal surface, Ib. Eb = b hemisphere ∫I ⋅ cos θ ⋅ dΩ Replacing the solid angle by its equivalent in spherical angles: Eb = ∫ 2 ⋅π 0 ∫ π 2 0 I b ⋅ cos θ ⋅ sin θ ⋅ dθ ⋅ dϕ Integrate once, holding Ib constant: Eb = 2 ⋅ π ⋅ I b ⋅ ∫ 2 cos θ ⋅ sin θ ⋅ dθ 0 π Integrate a second time. (Note that the derivative of sin θ is cos θ·dθ.) sin θ Eb = 2 ⋅ π ⋅ I b ⋅ 2 2 π 2 = π ⋅ Ib 0 Eb = π·Ib Radiation Exchange During the previous lecture we introduced the intensity, I, to describe radiation within a particular solid angle. I= dq cos θ ⋅ dA1 ⋅ dΩ This will now be used to determine the fraction of radiation leaving a given surface and striking a second surface. Rearranging the above equation to express the heat radiated: dq = I ⋅ cos θ ⋅ dA1 ⋅ dΩ Next we will project the receiving surface onto the hemisphere surrounding the source. First find the projected area of surface dA2, dA2·cos θ2. (θ2 is the angle between the normal to surface 2 and the position vector, R.) dA2 Then find the solid angle, Ω, which encompasses this area. Substituting into the heat flow equation above: dA2·cos θ2 dq = I ⋅ cos θ1 ⋅ dA1 ⋅ cos θ 2 dA2 R2 R To obtain the entire heat transferred from a finite area, dA1, to a finite area, dA2, we integrate over both surfaces: I ⋅ cos θ 1 ⋅ dA1 ⋅ cos θ 2 dA2 q1→2 = ∫ ∫ A2 A1 R2 To express the total energy emitted from surface 1, we recall the relation between emissive power, E, and intensity, I. qemitted = E1·A1 = π·I1·A1 View Factors-Integral Method Define the view factor, F1-2, as the fraction of energy emitted from surface 1, which directly strikes surface 2. F1→2 = q1→2 qemitted I ⋅ cos θ1 ⋅ dA1 ⋅ cos θ 2 dA2 ∫A ∫A R2 = 2 1 π ⋅ I ⋅ A1 after algebraic simplification this becomes: F1→2 = 1 cos θ1 ⋅ cosθ 2 ⋅ dA1 ⋅ dA2 ⋅∫ ∫ A1 A2 A1 π ⋅ R2 dr dAj Aj Example Consider a diffuse circular disk of diameter D and area Aj and a plane diffuse surface of area Ai << Aj. The surfaces are parallel, and Ai is located at a distance L from the Obtain an center of Aj. expression for the view factor Fij. D θj R θi dAi R L dAi The view factor may be obtained from: F1→2 = 1 cos θ1 ⋅ cosθ 2 ⋅ dA1 ⋅ dA2 ⋅∫ ∫ A1 A2 A1 π ⋅ R2 Since dAi is a differential area F1→2 = ∫ cosθ1 ⋅ cosθ 2 ⋅ dA1 A1 π ⋅ R2 Substituting for the cosines and the differential area: F1→2 = ∫ After simplifying: (L R ) ⋅ 2π ⋅ r ⋅ dr 2 A1 π ⋅ R2 F1→2 L2 ⋅ 2 ⋅ r ⋅ dr =∫ A1 R4 Let ρ2 ≡ L2 + r2 = R2. Then 2·ρ·dρ = 2·r·dr. F1→2 = ∫ After integrating, L2 ⋅ 2 ⋅ ρ ⋅ dρ A1 ρ4 1 ⎤ 2 2 ⎡ = −L ⋅ ⎢ 2 2⎥ ⎣ L + ρ ⎦0 D F1→2 = − 2 ⋅ L2 ⋅ ρ −2 2 A2 Substituting the upper & lower limits F1→2 4 1⎤ 2 D2 2 ⎡ = −L ⋅ ⎢ − = 4 ⋅ L2 + D 2 L2 ⎥ 0 4 ⋅ L2 + D 2 ⎣ ⎦ D This is but one example of how the view factor may be evaluated using the integral method. The approach used here is conceptually quite straight forward; evaluating the integrals and algebraically simplifying the resulting equations can be quite lengthy. Enclosures In order that we might apply conservation of energy to the radiation process, we must account for all energy leaving a surface. We imagine that the surrounding surfaces act as an enclosure about the heat source which receive all emitted energy. Should there be an opening in this enclosure through which energy might be lost, we place an imaginary surface across this opening to intercept this portion of the emitted energy. For an N surfaced enclosure, we can then see that: ∑F j =1 N i, j =1 This relationship is known as the “Conservation Rule”. Example: Consider the previous problem of a small disk radiating to a larger disk placed directly above at a distance L. 2 3 1 From our conservation rule we have: The view factor was shown to be given by the relationship: F1→2 D2 = 4 ⋅ L2 + D 2 Here, in order to provide an enclosure, we will define an imaginary surface 3, a truncated cone intersecting circles 1 and 2. ∑F j =1 N i, j = F1,1 + F1, 2 + F1,3 Since surface 1 is not convex F1,1 = 0. Then: F1→3 Reciprocity D2 = 1− 4 ⋅ L2 + D 2 We may write the view factor from surface i to surface j as: Ai ⋅ Fi→ j = ∫ A j Ai ∫ cos θ i ⋅ cos θ j ⋅ dAi ⋅ dA j π ⋅ R2 Similarly, between surfaces j and i: A j ⋅ F j →i = ∫ A j Ai ∫ cos θ j ⋅ cos θ i ⋅ dA j ⋅ dAi π ⋅ R2 Comparing the integrals we see that they are identical so that: Ai ⋅ Fi→ j = A j ⋅ F j→i This relationship is known as “Reciprocity”. Example: Consider two concentric spheres shown to right. All radiation leaving the outside of surface 1 will strike surface 2. Part of the radiant energy leaving 1 the inside surface of object 2 will strike surface 1, part will return to surface 2. To find the fraction of energy leaving surface 2 which strikes surface 1, we apply reciprocity: A2 ⋅ F2,1 = A1 ⋅ F1, 2 ⇒ F2,1 = A1 A D ⋅ F1, 2 = 1 = 1 A2 A2 D2 the 2 Associative Rule Consider the set of surfaces shown to the right: Clearly, from conservation of energy, the fraction of energy leaving surface i and striking the combined surface j+k will equal the fraction of energy emitted from i and striking j plus the fraction leaving surface i and striking k. i j k Fi ⇒ ( j + k ) = Fi ⇒ j + Fi ⇒ k This relationship is known as the “Associative Rule”. Radiosity We have developed the concept of intensity, I, which let to the concept of the view factor. We have discussed various methods of finding view factors. There remains one additional concept to introduce before we can consider the solution of radiation problems. Radiosity, J, is defined as the total energy leaving a surface per unit area and per unit time. This may initially sound much like the definition of emissive power, but the sketch below will help to clarify the concept. J ≡ ε·Eb + ρ·G ε·Eb ρ·G G Net Exchange Between Surfaces Consider the two surfaces shown. Radiation will travel from surface i to surface j and will also travel from j to i. qi→j = Ji·Ai· Fi→j likewise, qj→i = Jj·Aj· Fj→j The net heat transfer is then: qj→i (net) = Ji·Ai· Fi→j - Jj·Aj· Fj→j From reciprocity we note that F1→2·A1 = F2→1·A2 so that qj→i (net) = Ji·Ai· Fi→j - Jj· Ai· Fi→j = Ai· Fi→j·(Ji – Jj) Net Energy Leaving a Surface The net energy leaving a surface will be the difference between the energy leaving a surface and the energy received by a surface: ε·Eb ρ·G G Ji Jj q1→ = [ε·Eb – α·G]·A1 Combine this relationship with the definition of Radiosity to eliminate G. J ≡ ε·Eb + ρ·G G = [J - ε·Eb]/ρ q1→ = {ε·Eb – α·[J - ε·Eb]/ρ}·A1 Assume opaque surfaces so that α + ρ = 1 ρ = 1 – α, and substitute for ρ. q1→ = {ε·Eb – α·[J - ε·Eb]/(1 – α)}·A1 Put the equation over a common denominator: ⎡ ε ⋅ Eb − α ⋅ J ⎤ ⎡ (1 − α ) ⋅ ε ⋅ Eb − α ⋅ J + α ⋅ ε ⋅ Eb ⎤ q1→ = ⎢ ⋅ A1 = ⎢ ⎥ ⋅ A1 ⎥ 1−α 1−α ⎦ ⎣ ⎦ ⎣ If we assume that α = ε then the equation reduces to: ⎡ ε ⋅ Eb − ε ⋅ J ⎤ ⎡ ε ⋅ A1 ⎤ q1→ = ⎢ ⋅ A1 = ⎢ ⋅ (Eb − J ) ⎥ ⎣1− ε ⎥ ⎦ ⎦ ⎣ 1− ε Electrical Analogy for Radiation We may develop an electrical analogy for radiation, similar to that produced for conduction. The two analogies should not be mixed: they have different dimensions on the potential differences, resistance and current flows. Equivalent Current Ohms Law Net Energy Leaving Surface Net Exchange Between Surfaces Equivalent Resistance Potential Difference I q1→ qi→j R ⎡1 − ε ⎤ ⎢ε ⋅ A⎥ ⎣ ⎦ 1 A1 ⋅ F1→2 ΔV Eb - J J1 – J2 Example: Consider a grate fed boiler. Coal is fed at the bottom, moves across the grate as it burns and radiates to the walls and top of the furnace. The walls are cooled by flowing water through tubes placed inside of the walls. Saturated water is introduced at the bottom of the walls and leaves at the top at a quality of about 70%. After the vapour is separated from the water, it is circulated through the superheat tubes at the top of the boiler. Since the steam is undergoing a sensible heat addition, its temperature will rise. It is common practice to subdivide the super-heater tubes into sections, each having nearly uniform temperature. In our case we will use only one superheat section using an average temperature for the entire region. Surface 3 SH Coal Chute section Superheat Tubes Surface 2 Water Walls Surface 1 Burning Coal Energy will leave the coal bed, Surface 1, as described by the equation for the net energy leaving R1 a surface. We draw the equivalent electrical Eb1 network as seen to the right: J1 ⎡1− ε ⎤ ⎢ε ⋅ A ⎥ 1⎦ ⎣ 4 σ·T1 The heat leaving from the surface of the coal may proceed to either the water walls or to the super-heater section. That part of the circuit is represented by a potential difference between Radiosity: J2 It should be noted that surfaces 2 and 3 1 A1 ⋅ F1→2 1 A1 ⋅ F1→3 J3 R12 R13 will also radiate to one another. J2 1 A1 ⋅ F1→3 1 A2 ⋅ F2→3 J1 R1 Eb1 1 A1 ⋅ F1→2 J3 R12 R1 Eb1 R13 J2 1 A2 ⋅ F2→3 1 A1 ⋅ F1→2 J3 R3 ⎡1 − ε 3 ⎤ ⎢ ⎥ ⎣ ε ⋅ A3 ⎦ It remains to evaluate the net heat flow leaving (entering) nodes 2 and 3. R2 Eb2 ⎡ 1 − ε2 ⎤ ⎢ε ⋅ A ⎥ ⎣ 2 12 ⎦ A1 ⋅ F1→3 Eb3 R13 R1 R12 Eb1 ⎡ ε ⋅ A1 ⎤ ⎢1− ε ⎥ ⎦ ⎣ Alternate Procedure for Developing Networks • Count the number of surfaces. (A surface must be at a “uniform” temperature and have uniform properties, i.e. ε, α, ρ.) • Draw a radiosity node for each surface. • Connect the Radiosity nodes using view factor resistances, 1/Ai·Fi→j. • Connect each Radiosity node to a grounded battery, through a surface resistance, 1 − ε ε ⋅ A . [ ] This procedure should lead to exactly the same circuit as we obtain previously. Simplifications to the Electrical Network • Insulated surfaces. In steady state heat transfer, a surface cannot receive net energy if it is insulated. Because the energy cannot be ⎡1 − ε ⎤ R3 ⎢ ⎥ stored by a surface in ⎣ε ⋅ A ⎦ steady state, all energy must be re-radiated back into the enclosure. Insulated surfaces are often termed as reradiating surfaces. 3 3 Electrically cannot flow through a battery if it is not grounded. Surface 3 is not grounded so that the battery and surface resistance serve no purpose and are removed from the drawing. • Black surfaces: A black, or ideal surface, will have no surface resistance: ⎡1 − ε ⎤ ⎡1 − 1⎤ ⎢ ε ⋅ A ⎥ = ⎢1 ⋅ A ⎥ = 0 ⎣ ⎦ ⎣ ⎦ In this case the nodal Radiosity and emissive power will be equal. This result gives some insight into the physical meaning of a black surface. Ideal surfaces radiate at the maximum possible level. Nonblack surfaces will have a reduced potential, somewhat like a battery with a corroded terminal. They therefore have a reduced potential to cause heat/current flow. • Large surfaces: Surfaces having a large surface area will behave as black surfaces, irrespective of the actual surface properties: ⎡1 − ε ⎤ ⎡1 − ε ⎤ ⎢ε ⋅ A ⎥ = ⎢ε ⋅ ∞ ⎥ = 0 ⎣ ⎦ ⎣ ⎦ Physically, this corresponds to the characteristic of large surfaces that as they reflect energy, there is very little chance that energy will strike the smaller surfaces; most of the energy is reflected back to another part of the same large surface. After several partial absorptions most of the energy received is absorbed. Solution of Analogous Electrical Circuits. • Large Enclosures Consider the case of an object, 1, placed inside a large enclosure, 2. The system will consist of two objects, so we proceed to construct a circuit with two radiosity nodes. J1 1/(A1⋅F1→2) J2 Now we ground both Radiosity nodes through a surface resistance. (1-ε1)/(ε1⋅A1) Eb1 σ⋅T14 R1 J1 1/(A1⋅F1→2) J2 (1-ε2)/(ε2⋅A2) R2 Eb2 σ⋅T24 R12 Since A2 is large, R2 = 0. The view factor, F1→2 = 1 (1-ε1)/(ε1⋅A1) Eb1 σ⋅T14 R1 J1 1/(A1⋅F1→2) R12 J2 Eb2 σ⋅T24 Sum the series resistances: RSeries = (1-ε1)/(ε1⋅A1) + 1/A1 = 1/(ε1⋅A1) Ohm’s law: i = ΔV/R or by analogy: q = ΔEb/RSeries = ε1⋅A1⋅σ⋅(T14 – T24) You may recall this result from Thermo I, where it was introduced to solve this type of radiation problem. • Networks with Multiple Potentials Systems with 3 or more grounded potentials will require a slightly different solution, but one which students have previously encountered in the Circuits course. The procedure will be to apply Kirchoff’s law to each of the Radiosity junctions. J2 1 A1 ⋅ F1→2 J3 R12 J1 Eb1 R13 Eb3 ∑q i =1 3 i =0 In this example there are three junctions, so we will obtain three equations. This will allow us to solve for three unknowns. Radiation problems will generally be presented on one of two ways: o The surface net heat flow is given and the surface temperature is to be found. o The surface temperature is given and the net heat flow is to be found. Returning for a moment to the coal grate furnace, let us assume that we know (a) the total heat being produced by the coal bed, (b) the temperatures of the water walls and (c) the temperature of the super heater sections. Apply Kirchoff’s law about node 1, for the coal bed: q1 + q 2→1 + q3→1 = q1 + J 2 − J1 J 3 − J1 + =0 R12 R13 Similarly, for node 2: q2 + q1→2 + q3→2 = Eb 2 − J 2 J 1 − J 2 J 3 − J 2 + + =0 R2 R12 R23 (Note how node 1, with a specified heat input, is handled differently than node 2, with a specified temperature. And for node 3: q3 + q1→3 + q2→3 = Eb 3 − J 3 J 1 − J 3 J 2 − J 3 + + =0 R3 R13 R23 The three equations must be solved simultaneously. Since they are each linear in J, matrix methods may be used: ⎡ 1 1 − ⎢− ⎢ R12 R13 1 ⎢ ⎢ R12 ⎢ 1 ⎢ R13 ⎢ ⎣ 1 R12 1 1 1 − − − R2 R12 R13 1 R23 ⎡ ⎤ ⎤ 1 ⎢ ⎥ ⎥ R13 ⎥ ⎡ J 1 ⎤ ⎢ − q1 ⎥ 1 ⎥ ⋅ ⎢ J ⎥ = ⎢− E b 2 ⎥ ⎥ ⎢ 2 ⎥ ⎢ R2 ⎥ R23 ⎢ ⎥ 1 1 1 ⎥ ⎣ J 3 ⎦ ⎢ Eb3 ⎥ ⎢− ⎥ ⎥ − − − R3 R13 R23 ⎥ ⎢ R3 ⎥ ⎦ ⎣ ⎦ The matrix may be solved for the individual Radiosity. Once these are known, we return to the electrical analogy to find the temperature of surface 1, and the heat flows to surfaces 2 and 3. Surface 1: Find the coal bed temperature, given the heat flow: E b1 − J 1 σ ⋅ T14 − J 1 ⎡ q ⋅ R + J1 ⎤ = ⇒ T1 = ⎢ 1 1 q1 = ⎥ σ R1 R1 ⎣ ⎦ 0.25 Surface 2: Find the water wall heat input, given the water wall temperature: Eb 2 − J 2 σ ⋅ T24 − J 2 = q2 = R2 R2 Surface 3: (Similar to surface 2) Find the water wall heat input, given the water wall temperature: Eb 3 − J 3 σ ⋅ T34 − J 3 q3 = = R3 R3 Module 10: Learning objectives • • • The objective of this module is to bring in the concept of mass transfer, which is mass in transit as a result of species concentration difference in a mixture. For this module, it is assumed that the student already has a good concept of conduction and convection heat transfer, so that the mass transfer concepts are taught with the help of drawing analogy from heat transfer. It is important for the student to understand the context in which the term mass transfer is used. The student should understand that the driving potential for mass transfer is concentration gradient, and one can obtain a mass transfer flux due the concentration gradient. The student should understand that there are modes of mass transfer that are similar to conduction and convection modes in heat transfer. Equivalent nondimensional numbers will also be introduced to describe the relative effects of various parameters governing mass transfer. • What is Mass Transfer? “Mass transfer specifically refers to the relative motion of species in a mixture due to concentration gradients.” Analogy between Heat and Mass Transfer Since the principles of mass transfer are very similar to those of heat transfer, the analogy between heat and mass transfer will be used throughout this module. Mass transfer through Diffusion Conduction dT ⎡ J ⎤ q′′ = − k dy ⎢ m 2 s ⎥ ⎦ ⎣ (Fourier’s law) ρ is the density of the gas mixture DAB is the diffusion coefficient Mass Diffusion j ′′ = − ρD AB A dξ A dy ⎡ kg ⎤ ⎢m2s ⎥ ⎣ ⎦ (Fick’s law) ξ A = ρ A / ρ is the mass concentration of component A Mass transfer through Diffusion A B j ′′ A ′ jB′ The sum of all diffusion fluxes must be zero: ∑ ji′′ = 0 ξ A + ξB = 1 d d ξ A = − ξB dy dy DBA = DAB = D Heat and Mass Diffusion: Analogy • Consider unsteady diffusive transfer through a layer Heat conduction, unsteady, semi-infinite plate ∂T ∂ ⎛ ∂T ⎞ = ⎜k ρc ⎟ ∂t ∂x ⎝ ∂x ⎠ ∂T k ∂ 2T ∂ 2T = =α 2 2 ∂t ρc ∂x ∂x Similarity transformation: PDE → ODE d 2θ dθ * + 2η =0 2 dy dη x η= 4αt Heat and Mass Diffusion: Analogy Solution: T − T0 = 1 − erf ⎛ x ⎞ ⎜ ⎟ Tu − T ⎝ 4αt ⎠ Temperature field Heat flux dT & q′′ x =0 = −k dx x =0 kcρ (Tu − T0 ) = (Tu − T ) = πt παt k Heat and Mass Diffusion: Analogy Diffusion of a gas component, which is brought in contact with another gas layer at time t=0 Differential equation: ∂ρ i ∂ ξi = ρD 2 ∂t ∂x ∂ξ i ∂ 2ξ i =D 2 ∂t ∂x 2 Transient diffusion Initial and boundary conditions: ξi (t > 0, x = 0) = ξi ,u ξi (t = 0, x ) = ξi ,o ξi (t > 0, x → ∞ ) = ξi ,o Heat and Mass Diffusion: Analogy Solution: ξ i − ξ i ,o ⎛ x ⎞ = 1 − erf ⎜ ⎟ ξ u − ξ i ,o ⎝ 4 Dt ⎠ Concentration field ρD (ξi ,Ph − ξi ,o ) Diffusive mass flux ji′′ x =0 = πDt Diffusive mass transfer on a surface (Mass convection) Fick's Law, diffusive mass flow rate: ∂ξ j ′′ = − ρD A ∂y y =0 = − ρD ξ ∞ − ξ w ∂ξ * L ∂y * y* =0 mass transfer coefficient j ′′ = hmass (ξ w − ξ ∞ ) A ⎡ kg ⎤ hmass ⎢ 2 ⎥ ⎣m s⎦ hmass L ∂ξ * = Sh = * ∂y ρD = f (Re, Sc) Dimensionless mass transfer number, the Sherwood number Sh y * =0 Sh = C Re m Sc n Note: Compare with energy eqn. and Nusselt No.: The constants C and the exponents m and n of both relationships must be equal for comparable boundary conditions. Diffusive mass transfer on a surface.. Dimensionless number to represent the relative magnitudes of heat and mass diffusion in the thermal and concentration boundary layers Lewis No. Sc α Thermal diffusivity Le = = = Pr D Mass diffusivity Analogy between heat and mass transfer Comparing the correlation for the heat and mass transfer Hence, hmass ⎛ Sc ⎞ =⎜ ⎟ h / c p ⎝ Pr ⎠ n −1 Sh ⎛ Sc ⎞ =⎜ ⎟ Nu ⎝ Pr ⎠ n For gases, Pr ≈ Sc, hence: hmass =1 h / cp Lewis relation 1 MODULE 10: SOLVED PROBLEMS Problem 1: A thin plastic membrane is used to separate Helium from a gas stream. Under state conditions, the concentration of helium in the membrane is known to be 0.02 and 0.005 kmol/m3 at the inner and outer surfaces, respectively. If the membrane is 1 mm thick and the binary diffusion coefficient of helium with respect to the plastic is 10-9 m3/s, what is the diffusion flux? Solution: Known: Molar concentrations of He at the inner and outer surfaces of a plastic membrane; diffusion coefficient and membrane thickness. To calculate: Molar diffusion flux Schematic: Assumptions: Steady state, 1D diffusion in a plane wall, stationary medium, uniform C = CA + CB Analysis: The molar flux may be obtained from N ′′, x = A DAB 10−9 m 2 /s (C A,1 − C A,2 ) = (0.02 − 0.005) kmol/m3 L 0.001 m N ′′, x = 1.5 × 10−8 kmol/s.m2 A 2 Problem 2: Oxygen is maintained at pressures of 2 bars ans 1 bar on opposite sides of a rubber membrane that is 0.5 mm thick, and the entire system is at 25 C. What is the molar diffusion flux of O2 through the membrane? What are the molar concentrations of O2 on both sides of the membrane (outside the rubber)? Solution: Known: Oxygen pressures on opposite sides of a rubber membrane. To find: Molar diffusion flux of oxygen; Molar concentration of oxygen outside the rubber. Schematic: Assumptions: Steady state, 1D diffusion, stationary medium of uniform total molar concentrations, C = CA + CB; perfect gas behaviour. Properties given: Oxygen-rubber (298 K): DAB = 0.21 × 10-9 m2/s; S = 3.12 × 10-3 kmol/m3.bar. Analysis: (a) For the assumed conditions dC A C ( 0) − C A ( L ) = DAB A dx L −3 −3 C A (0) = Sp A,1 = 6.24 × 10 kmol/m N ′′, x = J ′′, x = − DAB A A C A ( L) = Sp A, 2 = 3.12 × 10 −3 kmol/m3 Hence: (6.24 × 10 −3 − 3.12 × 10 −3 ) kmol/m3 ′ N A′, x = 0.21 × 10 m /s 0.0005 m −9 2 ′ N A′, x = 1.31 × 10 −9 kmol/s.m2 (b) From the perfect gas law: C A,1 = PA,1 2 bar = = 0.087 kmol/m3 3 RT (0.08314 m .bar/kmol.K C A, 2 = 0.5C A,1 = 0.0404 kmol/m3 Module 10: Short questions 1. How does mass transfer differ from bulk flow? Can mass transfer occur in a homogeneous medium? 2. Give examples for (a) liquid-to-gas (b) solid-to-liquid, (c) solid-to-gas, and (d) gas-to-liquid mass transfer. 3. What is the driving force for (a) heat transfer (b) electric current (c) fluid flow and (d) mass transfer? 4. What is an impermeable surface in mass transfer? How is it expressed mathematically? To what does it correspond in heat transfer? 5. What is concentration boundary layer? How is it defined for flow over flat plate? 6. What is the physical significance of the Schmidt number? What is the heat transfer equivalent of this number? What does Sc = 1 signify? 7. What is the physical significance of the Lewis number? What is the heat transfer equivalent of this number? What does Sc = 1 signify? 8. What is the physical significance of the Sherwood number? What is the heat transfer equivalent of this number? What does Sh = 1 signify for a plain fluid layer? MODULE 10 MASS TRANSFER Mass transfer specifically refers to the relative motion of species in a mixture due to concentration gradients. In many technical applications, heat transfer processes occur simultaneously with mass transfer processes. The present module discusses these transfer mechanisms. Since the principles of mass transfer are very similar to those of heat transfer, the analogy between heat and mass transfer will be used throughout this module. 10.1 Mass transfer through diffusion In Module 2 "Conduction", the Fourier equation was introduced, which relates the heat transfer to an existent temperature gradient: q′′ = −k dT ⎡ J ⎤ dy ⎢ m 2 s ⎥ ⎣ ⎦ (Fourier’s law) (10.1) For mass transfer, where a component A diffuses in a mixture with a component B an analogous relation for the rate of diffusion, based on the concentration gradient can be used dξ ⎡ kg ⎤ j ′′ = − ρD AB A ⎢ 2 ⎥ (Fick’s law) (10.2) A dy ⎣ m s ⎦ where ρ is the density of the gas mixture, DAB the diffusion coefficient and ξ A = ρ A / ρ the mass concentration of component A. The sum of all diffusion fluxes must be zero, since the diffusion flow is, by definition, superimposed to the net mass transfer: ∑ ji′′ = 0 (10.3) With ξ A + ξ B = 1 we get d d ξ A = − ξB dy dy which yields DBA = DAB = D (10.4) The analogy between diffusive heat transfer (heat conduction) and diffusive mass transfer (diffusion) can be illustrated by considering unsteady diffusive transfer through a layer. In heat conduction we can calculate the temperature field in a semi-infinite plate, whose surface temperature is forced to suddenly change to a constant value at t = 0. The derivation will be repeated here in summarized form. Heat conduction, unsteady, semi-infinite plate: ρc ∂T ∂ ⎛ ∂T ⎞ = ⎜k ⎟ ∂t ∂x ⎝ ∂x ⎠ ∂T k ∂ 2T ∂ 2T = =α 2 ∂t ρc ∂x 2 ∂x Fig. 10.1: Unsteady heat conduction (10.5) This partial differential equation can be transformed into an ordinary differential equation d 2θ dθ * + 2η =0 (10.6) dy 2 dη x with η = 4αt and yields T − T0 ⎛ x ⎞ = 1 − erf ⎜ ⎟ Tu − T ⎝ 4αt ⎠ (10.7) Fig. 10.2 Temperature field The heat flux at the wall can be calculated from the temperature gradient. & q′′ x =0 = − k dT dx = k x =0 παt (T u − T0 ) = kcρ (T − T ) πt u (10.8) The corresponding example in mass transfer is the diffusion of a gas component, which is brought in contact with another gas layer at time t=0. The transient field of concentration with pure diffusion results from a balance of component i Fig. 10.3: Transient diffusion ∂ρ i ∂ 2ξ i = ρD 2 ∂t ∂x 2 ∂ξ i ∂ξ = D 2i ∂x ∂t (10.9) The initial and boundary conditions for the semi-infinite fluid layer with a fixed concentration at the interface are: ξi (t > 0, x = 0) = ξi ,u ξi (t = 0, x ) = ξi ,o (10.10) ξi (t > 0, x → ∞ ) = ξi ,o The solution, i.e. the field of concentration, is analogue to the one above. ξ i − ξ i ,o ⎛ x ⎞ = 1 − erf ⎜ ⎟ ξ u − ξ i ,o ⎝ 4 Dt ⎠ (10.11) Fig. 10.4 Concentration field The diffusive mass flux of component i at the interface can be derived as ji′′ x =0 = ρD (ξi ,Ph − ξi ,o ) πDt (10.12) Fig. 10.5 Gas absorption in a falling film A practical example from chemical engineering is the absorption of a gas component i in a falling film. Considering a large film thickness or short contact times the theoretical results (penetration theory) can be used. The absorbed mass flux is mi = Aji′′ x =0 =A ρD (ξi ,Ph − ξi ,o ) πDt (10.13) Finally, the contact time can be calculated from the film velocity and the film length: L t= f (10.14) uf 10.2 Mass transfer in a flowing medium If we balance the net masses flowing in and out of a control volume of a fluid mixture, i.e. the sum of the convective and diffusive mass flows of the component i & &, &, mi′′ = mi′′conv + mi′′diff = ξ i ρu + ji′′ (10.15) we get for a steady state flow without sources the conservation equation for the component i under investigation: Fig. 10.6 Balance of mass flows on a control volume ∂ (ξi ρu + ji′′,x ) + ∂ (ξi ρv + ji′′, y ) + ∂ (ξi ρw + ji′′,z ) = 0 ∂x ∂y ∂z (10.16) This equation can be differentiated partially and rewritten applying the equation of continuity. This yields the form of the conservation equation for component i (the index i will be disregarded further on): ρu ∂ξ ∂ξ ∂ξ ∂ ⎛ ∂ξ ⎞ ∂ ⎛ ∂ξ ⎞ ∂ ⎛ ∂ξ ⎞ + ρv + ρw = ⎜ ρD ⎟ + ⎜ ρD ⎟ + ⎜ ρD ⎟ ∂x ∂y ∂z ∂x ⎝ ∂x ⎠ ∂y ⎝ ∂y ⎠ ∂z ⎝ ∂z ⎠ (10.17) Assuming constant material properties and introducing the Schmidt number Sc = μ , we get the simplified form: ρD ∂ξ ∂ξ ∂ξ μ ⎛ ∂ 2ξ ∂ 2ξ ∂ 2ξ ⎞ ρu + ρv + ρw = ⎜ 2 + 2 + 2 ⎟ ∂x ∂y ∂z Sc ⎝ ∂x ∂y ∂z ⎠ (10.18) which in analogy to the energy equation of the Module 6 on convection, can be made dimensionless by introducing dimensionless parameters. In a physical μ υ Momentum diffusivity sense, Sc = = = ρD D Mass diffusivity If we define x y z u v w ξ − ξw x * = ; y * = ; z * = ; u * = ; v * = ; w* = ; ξ * = L L L u∞ u∞ u∞ ξ∞ − ξw we get the dimensionless equation of mass conservation * * ∂ξ * 1 ⎛ ∂ 2ξ * ∂ 2ξ * ∂ 2ξ * ⎞ * ∂ξ * ∂ξ +v +w = + + u ⎜ ⎟ ∂x * ∂y * ∂z * Re Sc ⎝ ∂x *2 ∂y *2 ∂z *2 ⎠ * (10.19) from which can be concluded that the scaled concentration field must depend on the dimensionless coordinates and the dimensionless numbers Re and Sc: ξ * = f ( x * , y * , z * , Re, Pr) (10.20) Note the analogy to heat transfer, where in Module 6 "Convection", for the temperature field, the following was valid: T * = f ( x * , y * , z * , Re, Pr) (10.21) 10.3 Diffusive mass transfer on a surface The heat flux was determined in Module 6 "Convection" from the gradient of the temperature at the wall: ∂T & ′′ qw = − k ∂y T∞ − Tw ∂T * =−k y =0 L ∂y * y * =0 (10.22) This heat flux was represented using an empirical equation for the heat transfer coefficient: ′′ qw = h (Tw − T∞ ) (10.23) which written in dimensionless form was introduced as the Nusselt number Nu: hL ∂T * = Nu = * k ∂y y * =0 = f (Re, Pr) (10.24) For many practical cases, the Nusselt laws are written in the form: Nu = C Re m Pr n (10.25) We will proceed accordingly to describe mass transfer. For the diffusive mass flow rate, Fick's Law is rewritten using dimensionless quantities: ξ ∞ − ξ w ∂ξ * ∂ξ (10.26) j ′′ = − ρD A y = 0 = − ρD ∂y L ∂y * y =0 and compared to an empirical equation using the mass transfer coefficient ⎡ kg ⎤ hmass ⎢ 2 ⎥ ⎣m s⎦ * j ′′ = hmass (ξ w − ξ ∞ ) A (10.27) By the comparison, we get the dimensionless mass transfer number, the Sherwood number Sh: hmass L ∂ξ * = Sh = * ∂y ρD y * =0 = f (Re, Sc) (10.28) Sherwood number correlations, in turn, can be written using appropriate dimensionless numbers Sh = C Re m Sc n (10.29) Since the type of the mass conservation equation and energy equation are the same, the constants C and the exponents m and n of both relationships must be equal for comparable boundary conditions. It seems like we need one more dimensionless number to represent the relative magnitudes of heat and mass diffusion in the thermal and concentration boundary layers. That is the Lewis number, defined as Le = Sc α Thermal diffusivity = = Pr D Mass diffusivity 10.4 Analogy between heat and mass transfer Comparing the correlation for the heat and mass transfer, we can find their ratio Sh ⎛ Sc ⎞ =⎜ ⎟ Nu ⎝ Pr ⎠ and hence n 10.30) hmass ⎛ Sc ⎞ =⎜ ⎟ h / c p ⎝ Pr ⎠ n −1 (10.31) For gases, the Prandtl and the Schmidt number are almost equal. In this case a simple approximation for the relationship between the mass and heat transfer coefficient can be derived, which is the so-called Lewis relation hmass =1 h / cp Lewis relation (10.32) 10.5 Evaporation on a liquid surface When vapour from the surface of the liquid A enters the surrounding gas mixture, consisting of the component A and B, then this process is called evaporation. This mass transfer is determined by diffusive processes, i.e. diffusive resistances. The process differs from that of vaporisation, since in the latter the transferred vapour component is transferred to a pure vapour environment. Diffusion resistances are not relevant for the process of vaporisation. The net mass flow of component A from the liquid surface to the gas consists of a convective and diffusive part &A &A &A m′′ = m′′,conv + m′′,diff = ξ A ρv + j ′′ A (10.33) Fig. 10.7 Mass balance on a liquid surface Usually, it is assumed that the gas component B cannot penetrate the liquid surface, hence for the net flow of component B: ′ &′ &B &′ mB′ = m′′,conv + mB′,diff = ξ B ρv + jB′ (10.34) ′ With j ′′ + jB′ = 0 and ξ A + ξ B = 1 we get the net evaporation flow on the surface A & &A m′′ = m′′,w = 1 1 − ξ A ,w 123 F j ′′,w A The net mass flow is obviously increased by the factor F, the Stefan factor, compared to the diffusion flow. This factor takes into account that the wall is only permeable for the evaporating component A ("semi-permeable wall"). If we express the diffusive flow by equation 10.27, by the mass transfer coefficient and the concentration difference, we get: & m′′ = hmass or in dimensionless form ξw − ξ∞ 1 − ξw (10.35) & m′′ Sh ξ w − ξ ∞ = ρu∞ ReSc 1 − ξ w 13 2 B (10.36) with ξ as the mass concentration of the transferred component and B the driving potential for the mass transfer.