Chapter 2 (1)

► CHAPTER2 The First Law of Thermodynamics Learning Objectives To demonstrate mastery of the material in Chapter 2, you should be able to: ► State and illustrate by example the first law of thermodynamics—that is, the conservation of mass and energy—and its basic concepts, including conversion of energy from one form to another and the transfer of energy from the surroundings to the system by heat, work, and flow of mass. ► Write the integral and differential forms of the first law for (1) closed systems and (2) open systems under steady-state and transient (uniform-state) conditions. Convert these equations between intensive and extensive forms and between mass-based and molar forms. Given a physical problem, evaluate which terms in the equation are important and which terms are negligible or zero and determine whether the ideal gas model or property tables should be used to solve the problem. ► Apply the first law of thermodynamics to identify, formulate, and solve engineering problems for adiabatic and isothermal processes in the following types of systems: rigid tank, expansion/compression in a piston– cylinder assembly, nozzle, diffuser, turbine, pump, heat exchanger, throttling device, filling or emptying of a tank, and Carnot power and refrigeration cycles. ► Create an appropriate hypothetical path to solve these problems with available data. ► Describe the molecular basis for internal energy, heat transfer, work, and heat capacity. ► Describe the difference between a reversible process and an irreversible process, and, given a process, evaluate whether it is reversible or irreversible. ► Explain why it is convenient to use the thermodynamic property enthalpy for (1) streams flowing into and out of open systems and (2) closed systems at constant pressure. Describe the role of flow work and shaft work in open- system energy balances. ► Describe the energy changes associated with sensible heat, latent heat, and chemical reaction on both a macroscopic and a molecular level. Calculate their enthalpy changes using available data such as heat capacity, enthalpies of vaporization, fusion and sublimation, and enthalpies of formation. 36 c02.indd 36 05/11/12 1:35 PM 2.1 The First Law of Thermodynamics ◄ 37 ► 2.1 THE FIRST LAW OF THERMODYNAMICS The first law of thermodynamics states that while energy can be changed from one form to another, the total quantity of energy, E, in the universe is constant.1 This statement can be quantitatively expressed as follows: DEuniv 5 0 (2.1) However, it is very inconvenient to consider the entire universe every time we need to do a calculation. As we have seen, we can break down the universe into the region in which we are interested (the system) and the rest of the universe (the surroundings). The system is separated from the surroundings by its boundary. We can now restate the first law by saying that the energy change of the system must be equal to the energy transferred across its boundaries from the surroundings. Energy can be transferred by heat, Q, by work, W, and, in the case of open systems, by the energy associated with the mass that flows into and out of the system. In essence, the first law then lets us be accountants of the energy in the system, by tracking the “deposits” to and “withdrawals” from the surroundings in much the same way as you would account for the balance of money in your bank account. We will consider explicit forms of the first law for closed and open systems shortly. Forms of Energy The energy within a system can be transformed from one form to another. ►EXERCISE Name the three common forms that energy is divided into. See if you can define each form: Energy is classified according to three specific forms: (1) The macroscopic kinetic energy, EK is the energy associated with the bulk (macroscopic) motion of the system S as a whole. For example, an object of mass m moving at velocity V has a kinetic energy given by: 1 S2 EK 5 mV (2.2) 2 (2) The macroscopic potential energy, EP, is the energy associated with the bulk (macro- scopic) position of the system in a potential field. For example, an object in the Earth’s gravitational field has a potential energy given by: EP 5 mgz (2.3) where z is the height above the surface of the Earth and g is the gravitational con- stant.2 (3) The internal energy, U, is the energy associated with the motion, position, and chemical-bonding configuration of the individual molecules of the substances within the system. Energy is not an absolute quantity but rather is only defined relative to a reference state, so we must be careful to identify the particular reference state that we are using. As you read this text, what is your kinetic energy (assuming you are not riding the bus)? 1 Nuclear reaction presents an interesting case where energy and mass are coupled. However, we will not address this case in this text. 2 Potentials due to surface tension or electric or magnetic fields can also be included. c02.indd 37 05/11/12 1:35 PM 38 ► Chapter 2. The First Law of Thermodynamics If you answer zero, you are correct in the context of a well-defined reference state, the Earth. However, if instead we had considered the sun to be the reference state, the answer would be quite different. The Earth is in motion with a velocity of 30,000 m/s around the sun, and your kinetic energy is on the order of 106 J! In the case of kinetic and S potential energy, we usually define EK (i.e., V ) as zero when there is no motion relative to the Earth and EP (i.e., z) as zero at the surface of the Earth. In fact, these reference states are so obvious they are sometimes implicitly assumed. In this text, we will be care- ful to identify references states explicitly. This effort will become useful in the case of U, where there is more than one convenient reference state. What is the reference state used for U in the steam tables? In your introductory physics course, you focused primarily on changes associated with the first two forms of energy. Since solving first-law problems involves proficiency in relating different forms of energy, it is instructive to review a typical example from mechanics that you may have seen in introductory physics. It is presented in the context in which we will approach problems in this text. You should realize that, as chemical engineers, the form of energy that we will primarily focus on is internal energy, which is not covered in the following example. EXAMPLE 2.1 If a large stone is dropped from a cliff 10 m high, how fast will it be going when it hits the Typical Energy ground? Problem in Mechanics SOLUTION From Equation (2.1), we have: DE 5 DEK 1 DEP 5 0 (E2.1A) We can define the process as throwing the stone off the cliff. Typically, we set up our problem by labeling the thermodynamic states between which our process is going. We can define state 1 as the initial state when the stone is at the top of the cliff and state 2 as when the stone hits the ground. Using Equations (2.2) and (2.3), Equation (E2.1A) becomes: 1 S 1 S a mV 22 2 mV12b 1 1 mgz2 2 mgz1 2 5 0 2 2 By convention, we define the change in a property, D, as “final − initial.” Now, using the reference states described above, we get: 0 0 1 S2 1 S2 a mV 2 2 mV1 b 1 1 mgz2 2 mgz1 2 5 0 2 2 or, 1 S2 mV2 2 mgz1 5 0 2 Finally, solving for the final velocity yields: S V2 5 "2gz1 5 "2 1 9.8 3 m/s2 4 2 1 10 3 m 4 2 5 14 3 m/s 4 This value is equivalent to 31 miles/hr. Ouch! Note that our reference state of energy is arbitrary and if we had chosen different reference states, we would still get the same answer. c02.indd 38 05/11/12 1:35 PM you get used to it and become proficient at solving the type of problems for which it is useful. Changes in internal energy can result in the following: 1. you may have an initial period of discomfort as you gain experience. by molecular kinetic energy we mean motion. G. This motion is referred to as translational motion because the individ- ual molecules are going somewhere—that is. Originally internal energy was viewed simply as all forms of energy that are not associated with bulk motion or bulk position. at equilibrium. U. internal energy. as fast as a jet plane. chemical reaction 1 N2 1 3H2 h 2NH3 2 A change in internal energy that leads to a change in temperature is often termed sen- sible heat. which provide additional modes of molecular kinetic energy—vibrational and rotational motion. the molecules are flying around at significant speeds.indd 39 05/11/12 1:35 PM . on average. the motion of the individual molecules in a system. There are two general components of inter- nal energy—molecular potential energy and molecular kinetic energy. Like macroscopic kinetic energy. is an important form of energy for chemical engi- neering applications.] Additionally. includ- ing the kinetic and potential energies of the molecules themselves. This experience typically translates into relative comfort in including internal energy in the energy balance to solve first-law problems. for example. Do you know how fast the average oxy- gen molecule that you are now breathing is moving? [ANSWER: Oxygen molecules at room temperature are moving. Tlow h Thigh 2. that is. their speeds vary according to a Maxwell–Boltzmann distribution. and Gibbs energy. Molecular poten- tial energy can be either intermolecular (between different molecules) and intramolecular (within the same molecule) in character. Keep in mind that soon we will introduce thermodynamic properties with which you have less experience. but. in this case. These properties are fundamentally no more challenging than energy to learn to work with. changes in temperature. Let’s now examine how we can relate changes in molecular. for example. that is. The trick is that when you work with any thermodynamic property enough. it is very hard to say exactly what energy is. you (hopefully!) are comfortable using it in the context of the problem above. for example. the measured macroscopic property c02. However. however. diatomic and polyatomic molecules (as opposed to atoms) can vibrate and rotate. 2. As we saw in Chapter 1. molecular energy is noticed in the real world in different ways. Likewise. S. A change in inter- nal energy can present itself in several macroscopic manifestations. solid h gas 3. changes in chemical structure. changes in phase. translating—even though the bulk of the gas may not be. that is approximately 450 m/s. Remember that we use the term molecular when we are describing what the species are doing on the atomic scale while we use the term macroscopic to describe behavior in the bulk (or molar) scale of the world in which we live. The ability to apply the abstract property energy to solve engineering problems lies in your experience with it. In the gas phase. we often refer to a change in internal energy that results in phase transformations as latent heat. Internal energy encompasses all forms of molecular energy. it is instructive to take advantage of our knowledge of chemistry and consider a molecular perspective on internal energy. chemical energy to the three macroscopic features described above.1 The First Law of Thermodynamics ◄ 39 One philosophical comment: Energy is inherently a very abstract quantity. such as entropy. However. The type of motion depends on the phase the species are in. Ways We Observe Changes in U As we mentioned. Not all molecules have the same speed. In contrast. Therefore. Similar argu- ments hold for the melting of solids and evaporation of liquids (see point 2 above). their main mode of molecular kinetic energy is in the form of vibrations. the internal energy for an ideal gas is independent of the position of the molecules. or the temperature. Hence. depends only on motion of the molecules. phonons represent part of the internal (or molecular) energy. energy must be added to the system. in turn. resulting in higher internal energy. In this case. the molecular kinetic energy. resulting in a change in the thermodynamic property. is one part of the internal energy. Because an ideal gas exhib- its no intermolecular forces (see Section 1. solids do not have translational motion. which. U.3 Internal Energy of an Ideal Gas We next explore the property dependence for the internal energy for an ideal gas. and. Thus the energetics change as the atoms are rearranged. the internal energy. molecular potential energy. that is. u. we will learn that thermodynamic entropy also plays a role in determining how far a reaction will proceed. The attraction between molecules adds stability and decreases the molecular energy of the system. on the one hand. To sublimate. The solid is held together by bonds between the molecules. internal energy consists of two com- ponents. the mol- ecules’ speed is directly related to the molecular kinetic energy. molecular kinetic energy and molecular potential energy. so it has greater internal energy (see point 1 above). the molecular energy of the attraction of the bonds must be overcome. For example. barring chemical reaction. In Chapter 4. uideal gas 5 f 1 T only 2 (2. So the faster the solid is vibrating. On the other hand. The First Law of Thermodynamics temperature is representative of how fast the gas molecules are moving in the system (formally temperature is proportional to the mean-square velocity). the vapor phase is representative of higher internal energy relative to the solid at the same temperature. Often the bonding in a solid results from electrostatic attraction of the molecules. As we learned with the discussion of point 1 earlier. its molecular potential energy is constant (assuming no chemical reactions). the greater the temperature and the greater the internal energy. ammonia is produced by the reaction: N2 1 3H2 h 2NH3 A triple bond between N atoms and three single bonds between H atoms are replaced by six N i H bonds (three each for two molecules produced). Finally. 40 ► Chapter 2. consider chemical reaction.3). such as the sublimation of a solid into a vapor. the molecules in the vapor are much farther away from one another and have little or no attraction. Again. so U is reduced (see point 3 above). we will consider the thermodynamic properties of real 3 In Chapter 3. the chemical bonds between the atoms in the molecules of the reactants are broken and replaced by the bonds of the products.4) Said another way.indd 40 05/11/12 1:35 PM . that is. However. the product is lower in energy (a lot more sta- ble). The strength of a chemi- cal bond is determined by the overlap of the valence electrons of the constituent atoms. Consider next a phase change. Hence. and temperature is directly related to one of them. Thus. as a gas increases in temperature. An example with which you may have experience is the case with CO2 (dry ice) at atmospheric temperature and pressure. The vibrations of the atoms in a solid are called phonons. the average velocity of its molecules increases (the molecules move faster). they are directly related to the tempera- ture of the solid. on the other hand. In this case. c02. 1 The First Law of Thermodynamics ◄ 41 gases. P 2 EXAMPLE 2.098 3 kJ/kg 4 2 We have used specific energy and converted the units to be consistent with the steam tables.1 5 104.2) on a per-mass basis. we do not expect the properties of a liquid to be significantly affected by pressure.96 2 104.96 3 J/kg 4 We can now go to the steam tables and determine at which temperature saturated water has this energy. Therefore. however.03 atm). Interpolating.005 u^ l 1 at 30 3 °C 4 2 2 u^ l 1 at 25 3 °C 4 2 125.77 2 104.2 In Example 2. solving for the final temperature yields: T2 5 25 1 1 0. Again we will use state 1 to denote the initial state and state 2 to denote the final state. we have: 1S De^ K 5 V22 5 98 3 J/kg 4 5 0. it provides us a large resource to be harvested. In this case.4 From Appendix B. why we are so interested in internal energy. we have: u^ l.86 3 kJ/kg 4 The problem statement says the internal energy of the water increases by the same amount as the energy of the stone. where the molecules are close enough to experience the effect of intermolecular forces. we get: u^ l. resulting in a velocity around 31 miles/hr.2 1 at T2 2 2 u^ 1 1 at 25 3 °C 4 2 104. This example illustrates that a large amount of energy is stored in u relative to the other forms of energy. two independent intensive properties are needed to specify the state of a system with constant composition.1. consequently.indd 41 05/11/12 1:35 PM . ureal gas 5 u 1 T.098 3 kJ/kg 4 So for the final state of water.02°C The temperature of the water barely changes! Thus the energy stored in a stone 10 m up a cliff corresponds to a negligible amount of internal energy. that is. Again we neglect the effect of the pressure difference between the subcooled state and the saturated state.86 5 5 0. Liquid water is subcooled at 25ºC and 1 atm. we considered the potential energy of a stone at the top of a 10-m cliff.1 5 104.1 5 De^ k 5 0.2 5 Du^ 1 u^ l. c02.2 2 u^ l. It will serve you well to catalog your experiences of such tricks for reference for solving problems in the future. such as: ureal gas 5 u 1 T. When Equivalent Energy it fell. it gained kinetic energy.005 2 5 5 25. we can use the temperature tables for saturated liquid water at 25ºC (which is technically at a pressure of 0. 2. How hot would the water end up if its internal energy increased by the same amount? SOLUTION If we write Equation (2. As engineers. and. Du^ 5 u^ l.1: u^ l. 4 We often use this trick to find the properties of subcooled water (or other substances) when the pressure is not appreciably different from the saturation pressure.86 Finally. v 2 or. Consider now an Stored in u equivalent mass of water initially at 25ºC. 5) by A as follows: FE W 5 3 # d 1 Ax 2 5 3 PE # dV 5 3 PEdV cos u 5 23 PEdV (2. In the case of expansion. in this process. and the final state. dx: W 5 3 FE # dx (2. However. Thus. To complete the definition. in the context of powering the steam engine. Whenever we calculate the work. we must account for the real path that the system takes. we can divide and multiply the terms on the right hand side of Equation (2.5)]. electrical. c02. The negative sign in Equation (2. for example.indd 42 05/11/12 1:35 PM . 2. be careful to note which sign convention is chosen for work or you may get tripped up. 1.1a can be obtained from the area under the curve [which is equivalent to graphically integrating the expression in Equation (2. In this text. The most common case of work in engineering ther- modynamics is when a force causes a displacement in the boundary of a system. Heat is the transfer of energy by a temperature gradient. Since work refers to the transfer of energy between the system and the surround- ings. When you go to other sources. mechanical (expansion/compression. no work has been done. whereas all other forms of energy transfer in a closed system occur via work. the sys- tem exchanges energy with the surroundings in the form of work. x for a general process is shown in Figure 2. but also on the specific path that it takes. If the external force is acting on a surface of cross-sectional area A. since the external force and displacement vectors are in opposite directions. If the boundary of the system does not move.6) A where PE is the external pressure to the surface. and magnetic. BTU.5) In contrast to thermodynamic properties. the work on a system depends not only on the initial state. for example. ergs. when the first and second laws of thermodynamics were originally formulated. the system needs to push the surroundings out of the way to increase the boundary. The definition given by Equation (2. The work associated with the process in Figure 2. the opposite sign convention was used: Work from the system to the surroundings was defined as positive (since the engineering objective was to get work out of the system to power a train!). We will examine these terms in more detail below. and so on. The work. We generally associate work with something useful being done by (or to) the system. W. You should be aware that this sign convention is arbitrary. such as joules. Both terms refer to the transfer of energy between the sur- roundings and system.5) is consistent with this sign convention. A plot of FE vs. the transfer of energy between the surround- ings and the system can only be accomplished by heat or by work. rotat- ing shaft). with respect to the direction of displacement. we will say work is positive when energy is transferred from the surroundings to the system and work is negative when energy is transferred from the system to the surroundings. no matter how large the force is. the system expends energy. In a closed system. can be described mathematically by the line integral of the external force. The First Law of Thermodynamics Work and Heat: Transfer of Energy Between the System and the Surroundings In the sciences we need to be very careful about how we use language and define terms such as work and heat. 42 ► Chapter 2. FE. Work There are many forms of work.1a. we need to choose a sign convention for work. of the system.6) results. We choose this convention to be consistent with today’s convention. it has the same units as energy. 3.g.3 illustrates how work is calculated for such a system. Calculate work done by the system Cylinder Assembly during this process. The gas in the cylinder expands until the pressure of the gas matches the pressure of the surroundings. Example 2. SOLUTION The amount of work done can be calculated by applying Equation (2. 2. If Equation (2. V.7) is often encountered in thermodynamics. Again. (b) PE vs. as shown in Figure 2.6) is written on a molar basis (J/mol).1b.7) Equation (2. the work described by this equation will be referred to as “Pv work.1 Graphical determination of the value of work for a system that undergoes a process between states 1 and 2 by integrating: (a) FE vs.2 L. (Continued) c02. A piston– cylinder assembly is a common system that is used to obtain work (e. the work can be obtained from the area under the appropriate curve. The units of pressure and volume have been converted to their SI equivalents in this calculation 3 1 Pa m3 5 1 J 4 . the value for work is negative since the system loses energy to the surroundings as a result of this process. in your automo- bile). The expansion process is initiated by Work in a Piston– releasing the latch.6): V2 W 5 2 3 PEdV (2. we get: w 5 23 PEdν (2.indd 43 05/11/12 1:35 PM .3. as discussed in Section 1.3 Consider the constant pressure expansion that is illustrated in Figure E2.6) V1 Since the external pressure is constant. EXAMPLE 2.” On a molecular scale. Initially the system Calculation of Pv contains 1 mole of gas A at 2 bar within a volume of 10 L. the energy transfer by Pv work can be understood in terms of momentum transfer of the molecules in the system when they bounce off the moving boundary. it can be pulled out of the integral: V2 105 Pa 1023 m3 W 5 2PE 3 dV 5 2PE 1 V2 2 V1 2 5 21 bar B R 1 15.1 The First Law of Thermodynamics ◄ 43 FE PE 1 2 1 2 Work Work x V (a) (b) Figure 2. The final volume is 15.2 2 10 2 LB R 5 2520 J bar L V1 In this case. x.. refers to the transfer of energy between the surroundings and the system where the driving force is provided by a temperature gradient.indd 44 05/11/12 1:35 PM . A positive value indicates that energy is transferred from the surroundings to the system or. Those changes that manifested as changes in temperature were termed sensi- ble heat while changes in phase were termed latent heat. when we generically use the term shaft work. the system is “heating up. A magnet placed on the end of the turbine also rotates. in all cases we are converting the internal energy of the fluid into useful work. Often a rotating shaft is used to deliver energy between the system and the surroundings.3 Example of a process in which energy is transferred from the system to the sur- roundings by Pv work: expansion of a gas in a piston–cylinder assembly. heat provides a path for the unwanted dissipation of energy (such as when your coffee gets cold or your soda gets warm). 44 ► Chapter 2. it expands and cools.” Alternatively.1 we referred to the effects of changes in U on the system. However. Unlike for work. you may think of a positive value for Q to correspond with an increase in internal energy 1 DU 2 of the system (ignoring work). magnetic. since we are referring to changes in the internal energy of the system and not c02. In Section 2. as the fluid passes through the turbine. the sign convention for heat has been historically invariant. is another important type of work encountered in engineering practice.2. leading to the rotation of the shaft at the end of the turbine. In this case. Often. it may indeed be any of these forms of work. Shaft work. This nomenclature is somewhat misleading. The surroundings are maintained at 1 bar. In fact. Energy will transfer spontane- ously from the high-temperature region to the low-temperature region. Heat Heat. For example. Ws. We term such a process adiabatic 1 Q 5 0 2 . It is designed to convert the internal energy of the working fluid into useful work by means of a shaft. which is used to generate a current that can charge a battery and store energy. In the latter case. Q. it is useful to try to insulate the system as well as possible to eliminate unwanted transfer of energy. Note that depending on how we draw the boundary to our system. the transfer of energy from the system to the surroundings (the shaft work) can be mechanical. or electric. consider the turbine shown in Figure 2. the transfer of energy by heat would be reduced to zero.2 L V1 = 10 L 1 mol gas A 1 mol gas A Initial state (1) Final state (2) Figure E2. The First Law of Thermodynamics PE = 1 bar PE = 1 bar Process P2 = 1 bar P1 = 2 bar V2 = 15. The changing magnetic field induces an electrical potential. however. In the ideal case. colloquially. Sometimes this form of energy transfer is part of an engineering design (such as “heating” up your house on a cold day). 1 The First Law of Thermodynamics ◄ 45 Fluid out Fluid in e in Ws N S i Fluid in + – battery Fluid out e out Figure 2. convection. k. If you expose the front side of a chunk of quartz glass. however. these atoms are connected to those next to them on the crystal lattice and the region of high-energy lattice vibration will “spread out. Convection refers to the case of enhanced heat transfer through coupling with fluid flow. to temperature Thigh and the back side to a temperature Tlow. metals tend to conduct well.1 W m21 ºC21. The thermal conductivity of liquids tends to be lower than that of solids. there is an additional mechanism for conduction of energy—drift of free electrons in the valence band. The rotating shaft is coupled to a battery by means of a magnet affixed to its end. On the other hand. phonons refer to the vibrations of the atoms in a solid) will be vibrating at a higher velocity. energy will transfer through the glass. that is. is an effective thermal insulator. It is easiest to think of conduction in terms of a solid body. it has a thermal conductiv- ity of around 42 W m21 ºC21. The rate at which the energy transfers via conduction—that is. One way to enhance the transfer of energy (so that you can eat quickly and your tongue does not get burned) is by blowing on the soup in your spoon. on the other hand. the rate at which the lattice vibration spreads out—is proportional to a property of the material called the thermal conductivity.01–0.6 W m21 ºC21 for most liquids and 0. and radiation. the pho- nons on the hot side (remember. “heat. You will learn how to quantify the rates of these processes in your heat-transfer (or transport processes) class.2 Schematic of a turbine converting the energy from a flowing fluid into shaft work. The end result is a transfer of energy from Thigh to Tlow. consider the example of want- ing to cool a bowl of hot soup. Wood.” which more precisely refers to transfer of energy across the system boundary in a specified way. A thermally conducting material like copper may con- duct energy an order of magnitude faster than glass. the flow of gas carries away hot molecules (moving with high velocity) and replaces them with colder c02. having a thermal conductivity of 385 W m21 ºC21. Typical values range from 0. 2. In the case of metals. This is an example of convection.06–0. for example. On a molecular scale. When you blow on the soup. with greater energy. the underlying mechanisms of each process will be described briefly here.” With time. Can you provide a molecular explanation as to why gases have much lower thermal conductivity than solids? Convection is another mechanism by which energy can be transferred between the system and the surroundings via heat. However. For example. and gases have even lower thermal conductivities than liquids.indd 45 05/11/12 1:35 PM . There are three modes through which energy can be transferred by a tempera- ture gradient: conduction. Liquids and gases may also transfer energy through conduction. having a thermal conductivity around 0. Glass is not very conductive. this nomenclature is deeply rooted in the literature. the phonons on the front side will vibrate less vigorously (thus reducing their temperature) while the phonons on the back side increase in energy. However.07 W m21 ºC21 for most gases. we will often lump all three modes of heat transfer together and just define the heat transfer for the system. We use both amounts of energy transferred. ► 2. It allows us to use data like those available in the appendices of this textbook to solve many different problems. the rate of heat transfer. Each photon that leaves carries some energy with it. due to vibration. not just the visible portion. possible at all!). By light. Hypothetical paths are constructed to make the calculation easier (or. and for these quantities we must follow the actual path the system takes. The change in any property (e. Have you ever seen a piece of metal that is “red hot”? By being red hot. they are also called state functions. the driving force for energy transfer—is greater. and boundary. Much of the methodology of this textbook is based on using judicious choices of hypothetical paths to develop theory or obtain c02. we call it a hypothetical path. and it is unique to thermodynamics.) To a large degree. radiation becomes the dominant mode of heat transfer. it is actually cooling by emitting photons in the red portion of the electromagnetic spectrum. is proportional to temperature: # Q~T 1 conduction and convection 2 but for radiation: # Q ~ T4 1 radiation 2 Thus. We are free to choose any convenient path to calculate the change in a given property. we are referring to all wavelengths of electromagnetic radiation. You should be aware. (Conversely. Radiation refers to the transfer of energy by light. surroundings. Because prop- erties depend only on the initial and final states of the system. 46 ► Chapter 2. In # conduction and convection. they do not depend on path the system takes. at high temperature. Identify all the heat-transfer mechanisms associated with this system. Often a path is constructed so that we can use available physical data.2 CONSTRUCTION OF HYPOTHETICAL PATHS Internal energy and volume are examples of thermodynamic properties. In fact. radiation is associated with accelera- tion of charged particles (electrons and nuclei) near the surface of the object. Clearly describing convection mathematically is more difficult than describing conduction. in some cases. If the path we use for calculation is different from the path the system actually undergoes. that is. what important modes may be present. # Q [J]. the ability to construct hypothetical paths between states allows for efficient collection and organization of experimental data. however. ►EXERCISE Consider a pot of boiling water. Q. On a molecular scale. heat and work are path dependent.g. Thus. Convection depends not only on the conductive properties of the soup and the air but also on the type of flow patterns that are set up. Sketch the system.Du) for a hypothetical path is the same as for the actual process as long as the system starts and ends in the same states as the actual process. The First Law of Thermodynamics fluid. The utility of this construct is enormous. We can take advantage of this feature by constructing our own paths. thermodynamics is built on the path independence of properties.indd 46 05/11/12 1:35 PM . and rates of energy transferred Q 3 J/s or W 4 . In this text. the temperature difference between the soup and the neighboring gas—that is. All objects above absolute zero radiate light. They are independent of the process. The rate of heat transfer by radia- tion is a much stronger function of temperature than either conduction or convection. and cooling occurs more quickly than by conduction alone. Figure 2. Three alternative paths are shown: the real path as T1 T2 well as two convenient Temperature (K) hypothetical paths. we may need to add another step to the hypothetical path.3 Plot of a v2 State 2 (T2. 2.6). as shown in path b.3 can enable you to identify useful hypothetical paths. consider a problem where we need to calculate the change in internal energy of a gas going from state 1 to state 2. we may need to have redrawn our hypothetical path from the choice that we first conceived. Often the data that is available for the isothermal heating is for ideal gases (such as the ideal gas heat capacity that we will cover in Section 2. We will not go through every possible path that may ever be useful. Δu hypothetical path b Step 2b vlarge Step 3b Step 1b Molar volume (m3/mol) State 1 Step 1a v1 (T1. This path consists of first heating the gas at con- stant volume from T1 to T2 (labeled Step 1a) and then isothermally compressing the gas from v1 to v2 (labeled Step 2a). As an example of how we might construct such a path. we consider the calculation of Du for two hypothetical paths shown by the dashed lines in the figure. The next two steps are similar to path a. we first expand the system to a very large molar volume. The goal is to get you to recognize when you can appropriately construct a hypothetical path and have you be able to develop the path and execute such a calculation on your own. which we call path a. The process the system actually undergoes is depicted by the solid line.indd 47 05/11/12 1:35 PM . to solve this problem. However. c02. Consequently. this path may not be good enough to allow us to calculate Du.2 Construction of Hypothetical Paths ◄ 47 solutions to engineering problems of interest. v1) Step 2a Δuhypothetical path a Δuactual Actual path Figure 2. we may use any path we want as long as the path starts at the initial state and ends at the final state. However. vlarge.where the gas behaves as an ideal gas (labeled Step 1b).3 shows the initial and final states on a temperature–volume plot. For example. In this case. undoubtedly the most useful way to learn about constructing hypothetical paths will be when you need to develop your own hypothetical paths in homework to solve the end-of-chapter problems. In that case. we will use hypothetical paths as we develop theory and illustrate their use with examples. However. The calculations are simplified because only one property changes at a time. Drawing a diagram like that shown in Figure 2. where we heat at constant volume (Step 2b) and then compress isothermally (Step 3b). First. In such a case. we consider a two-step hypothetical path. Coming up with clever hypothetical paths can turn into quite a creative endeavor and is a key to problem solving and to understanding thermodynamics. v2) process that takes a sys- tem from state 1 to state 2 in Tv space. In an irreversible process. that is. there can be no friction. Such processes are irreversible processes. In this case we consider force because it is the driving force for mechanical work. we must be able to turn the process around at any point and cool it merely by changing the temperature by an infinitesimal amount. if we are heating a gas. Reversible processes are never completely realized in real life since they can be accomplished only by changing the driving force to a process by an infinitesimal amount. and F) consist of the opposite process. momentum transfer. The First Law of Thermodynamics ►2. that is. the system can be returned to its original state without any net effect on the surroundings. we label the process depicted in Figure E2. They have friction and are carried out with finite driving forces. For example. if we examine the process depicted in Figure E2. isothermal compression between state 2 and state 1.3. after the process occurs. C. it would take more than an infinitesimal force to turn it around and begin compressing the gas inside. the surroundings must be altered. and E) entail isothermal expansion of a piston–cylinder assembly between the same states: state 1 and state 2. Irreversible Processes Real processes are not reversible. in engineering these types of idealizations are often useful.indd 48 05/11/12 1:35 PM . hence. This value tells us the best that we could possibly do. it is useful to know how much work we could get out of a system if a process could be executed reversibly. If the piston is halfway up. However. We could perform a similar analysis on adiabatic processes where there is no energy transfer via heat between the system and the surroundings. This result occurs only when the driving force is infinitesimally small. 48 ► Chapter 2. The other three (B. Likewise. Temperature difference is the driving force for energy transfer. we realize that this condition cannot be accomplished. A defi- nition for reversible processes is as follows: A process is reversible if. The work obtained in an irreversible process is always less than that obtained in the idealization of a reversible process. If a gas undergoes an expansion process.3 REVERSIBLE AND IRREVERSIBLE PROCESSES Reversible Processes The concept of a reversible process is crucial in understanding thermodynamics. (Perhaps a pendulum comes close. For example. We can then compare how well we really do. D. if the system is returned to its original state. An isothermal process results in the limit of fast heat transfer with the surroundings. They represent a limiting case. a process that is perfectly executed. Why? To help solidify these abstract ideas. c02.3 irreversible. We will see that to accomplish a reversible process we must be able to reverse the direc- tion of the process at any point and go the other way by changing the driving force by an infinitesimal amount. Three processes (A. a concrete example is illustrative. Good engineering is as much determining where to focus your resources as it is actual problem solving. and see if it is worth focusing our efforts on improving the pro- cess. we must be able to turn the process around at any point and compress it merely by changing the force on the piston by an infinitesimal amount.) These processes represent an idealization. We use driving force as a generic term that represents some type of influence on a system to change. We will label these cases process A through process F. For a process to be reversible. We will compare the value of work for six processes. 4 m): Az m3 v1 5 ¢ ≤ 5 0.7): v2 v2 w 5 23 PEdv 5 2Patm3 dv 5 24000 3 J/mol 4 (2. In reality. The expansion process continues until once again the pressures equilibrate. the pressure inside the piston can then be found by a force balance: mg P1 5 1 Patm 5 2 3 bar 4 A where Patm is taken to be 1 bar. we apply Equation (2. Note. To find the work. The negative sign indicates we get 4 kJ of work out of the system from this expansion process.4.04 m3. the piston may undergo damped oscillations in its path toward its c02. The surroundings are at atmospheric pressure. the external pres- sure is only that from the atmosphere.7) and discussion]. Hence.4. we have idealized the process to stop precisely where the forces balance. The piston then comes to rest in state 2 where the pressure is given by: P2 5 Patm 5 1 3 bar 4 The ideal gas law can be applied to this isothermal process to give: Pv 5 RT 5 const The volume of state 2 is then given by: P1v1 m3 v2 5 5 0. In our analysis. To calculate the work. The molar volume in state 1 can be found from the area 1 0.3 Reversible and Irreversible Processes ◄ 49 Process A is illustrated in Figure 2. The system contains 1 mole of pure ideal gas. intensive properties and is also labeled in Figure 2.1 m2 2 and the height (0. State 1 is labeled on the Pv diagram in Figure 2. 2. Process A is initiated by removing the 1020-kg mass. and the gas within the piston expands. The pressure of the piston is now greater than that exerted by the surroundings.indd 49 05/11/12 1:35 PM . A 1020-kg mass sits on the piston. at any volume larger than 0. The piston only starts to move once the 1020-kg block is removed.4.4.08 B R P2 mol State 2 is now constrained by two independent. we are not saying anything about the pressure in the system but rather we are graphically illustrating what external pressure the gas expands against. volume is illus- trated in Figure 2.8) v1 v1 The same result can be found by graphically counting the area under the curve in the Pv diagram in Figure 2. The path of external pressure vs. These two properties completely constrain the initial state.04 B R n 1 mol Since the piston is originally at rest.4. we must consider the “external” pressure upon which the gas must expand [see Equation (2. 6 On the other hand. 50 ► Chapter 2. While it is only an approximation of the real behavior. this simplification proves useful in allowing us to compare irreversible processes with reversible processes. Its kinetic energy could be accounted for by the difference in force between the system pressure and the external pressure.8). the piston would oscillate forever. if there were no dissipative mechanism.04 . back toward the equilibrium position. Once past this volume.1 m2 ideal gas 1 mol of pure. the pressure in the system will be less than the pressure of the surroundings.08 3 m3 /mol 4 .08 Molar volume (m3/mol) Figure 2. The First Law of Thermodynamics Weightless.1 m2 m = 1020 Isothermal expansion Patm kg 1 mol of pure. frictionless piston Patm Process A: A = 0. 0. final resting position in state 2.08 3 m3 /mol 4 . the contribution of the oscillating expansions and contractions to the work will exactly cancel. In that case. leading to a molar volume greater than 0. It may again overshoot the equilibrium position. leading to a molar volume less than 0.indd 50 05/11/12 1:35 PM . the kinetic energy of the piston causes it to overshoot its final equilibrium position. A = 0.4 Schematic of an isothermal expansion process (process A).6 In this text.5.4 m ideal gas State 1 State 2 1 2 pressure (bar) "external" Process A 1 2 Fexternal / area or Work is given by the area under the curve . The corresponding plot of the process on a PEv diagram is shown at the bottom. leading to the same value as calculated by Equation (2. 5 We assume all the energy dissipation occurs within the system. This process will inevitably contain a fric- tional dissipative mechanism that causes the piston to come to rest at state 2. c02. we will ignore the oscillatory behavior of these types of processes and approximate them in the simpler context where the system does not overshoot its final equilibrium state. At some point. the motion stops and the piston reverses direction. and so on and so on. where it is then again turned around and expands. Since the external pressure is the same during these oscillations. The corresponding plot of the process on a PEv diagram is shown at the bottom.3 Reversible and Irreversible Processes ◄ 51 m = 1020 kg Patm A = 0. we see it costs us more work to compress the piston back to state 1 than we got from expanding it to state 2. 0. In this case.7): v1 v1 mg J w 5 2 3 PEdv 5 2aPatm 1 b 3 dv 5 8000 c d A mol v2 v2 This value can also be found from the area under the curve. molar volume is plotted in Figure 2.5 and is labeled process B.08 Molar volume (m3/mol) Figure 2. We next want to calculate the work needed to compress the gas from state 2 back to state 1. The work is found by Equation (2. The piston goes down until the pressures equilibrate.1 m2 Process B: Isothermal compression m = 1020 Patm kg 0. The external pressure now exceeds the pressure of the gas initiating the compression process.5 Schematic of an isothermal compression process (process B). This process is illustrated in Figure 2. originally in state 2. 2. The net difference in work 1 8000−4000 5 4000 3 J 4 2 in going from state 1 to state 2 and back to state 1 results in a “net effect on the surrounding.8 m A = 0. at state 1.” c02.indd 51 05/11/12 1:35 PM . Comparing process A and process B. we drop the 1020-kg mass back on the piston. 1 mol of pure. Again. we are not representing the system pressure in this graph.4 m ideal gas ideal gas State 2 State 1 1 Process B 2 pressure (bar) "external" 1 2 Fexternal / area or Work is given by the area under the curve .04 . but rather the force per area that acts on the piston.1 m2 1 mol of pure.5. In process B. the external pressure consists of contributions from both the block and the atmosphere. The external pressure vs. the work can be found graphically from the area under the curve. However.9) v v1 mol v1 v1 v1 c02. At each differential step. The compression process is the opposite of the expansion. the system pressure is no more than 'mg/A different than the external pressure. we want to get the most work out of a system as possible. The expansion process is labeled process C and follows arrows from state 1 to the intermediate state to state 2. Next we again consider expansion from state 1 to 2 (process C) and compression from state 2 to 1 (process D). so these processes are still irreversible. To find the work. we integrate over the external pressure. The work is found by: v1 vint v1 m 2g mg J w 5 2 3 PEdv 5 2B aPatm 1 b 3 dv 1 aPatm 1 b 3 dvR 5 6667 B R A A mol v2 v2 vint In analogy to the expansion process. it still costs us more work to compress from state 2 to state 1 than we get out of the expansion. With the system in state 2. we want it to be as small as possible. we get: v2 v2 v2 P1v1 v2 J w 5 2 3 PEdv 5 2 3 Pdv 5 2 3 dv 5 2P1v1 ln ¢ ≤ 5 25545 c d (2. We did “better” in both expansion and compression processes when we divided the 1020-kg mass into two parts. completing the expansion to state 2.indd 52 05/11/12 1:35 PM . and even better by dividing it into eight parts. The First Law of Thermodynamics Examining our definition of a reversible process. followed by placement of the second block.6. If we want to do the best possible. 52 ► Chapter 2. The expansion is carried out as follows: The system is originally in state 1 when the first 510-kg mass is removed.053 m3 /mol. Thus. a 510-kg block is placed on the piston until it compresses to the intermediate state. The second 510-kg mass is then removed. but now we use two 510-kg masses instead of one larger 1020-kg mass. respectively. process D is “better” than process B in that it costs us less work to compress the system back to state 1.7. we can divide the 1020-kg mass into many “differential” units and take them off one at a time for expansion or place them on one at a time for compression. However.7. to a close approximation: P 5 PE The process paths are illustrated in the Pv diagram in Figure 2.5 bar and a molar volume of 0. since the external pressure is equal to the system pressure. These three states are shown in the Pv diagram in Figure 2. Process C is “better” than process A in that it allows us to extract more work from the system. This causes the gas to expand to an intermediate state at a pressure of 1. The work the system delivers to the surroundings is given by: v2 vint v2 m 2g J w 5 2 3 PEdv 5 2B aPatm 1 b 3 dv 1 Patm 3 dvR 5 24667 B R A mol v1 v1 vint Again. These pro- cesses are labeled process E and process F. Of course. and are illustrated in Figure 2. we see that processes A and B are irreversible. and so on. Presumably we would do better by dividing it into four parts. When we have to put work into a system. Similarly. Moreover. we are never more than slightly out of equilibrium. we get more work out of the reversible expansion than the irreversible expansions. we could turn the process around the other way and compress the piston by adding differential masses instead of removing them. From our definition. The corresponding plots of the process on a PEv diagram is shown at the bottom.1 m2 Process C: m = 510 Isothermal expansion Patm kg 0. the reversible compression costs us less work than the irreversible processes. Hence. At any point during the expansion. we see that these processes are reversible.1 m2 Isothermal compression 1 mol of pure.3 Reversible and Irreversible Processes ◄ 53 m = 510 kg m = 510 kg Patm m = 510 kg A = 0.8 m Process D: A = 0. In a reversible process.08 Molar volume (m3/mol) Figure 2. we can return to state 1 by supplying the same amount of work that we got from the system in the expansion process.4 m 1 mol of pure.6 Schematic of two-step isothermal expansion (process C) and compression (process D) processes. 0.indd 53 05/11/12 1:35 PM .04 . Similarly the work of compression is: v1 v1 v1 P1v1 v1 J w 5 2 3 PEdv 5 2 3 Pdv 5 2 3 dv 5 2 P1v1 ln ¢ ≤ 5 5545 c d v v2 mol v2 v2 v2 In processes E and F. The reversible case represents the limit of what is possible in the real world—it c02. ideal gas ideal gas 1 2 pressure (bar) "external" Process D 1 2 Fexternal / area or Process C Work is given by the area under the curve . we can go from state 1 to 2 and back to state 1 without a net effect on the surroundings. 2. process A. that is. never has a mass on the piston. gives us the most work we can get out or the least work we have to put in! Moreover. the more the molecule’s speed will increase and the higher its kinetic energy. Why do we get less work out of the irreversible expansion (process A) than the reversible expansion (process E)? Work is the transfer of energy between the system and the surroundings. The heavier the bat.1 m2 Process E: Patm Isothermal expansion 0. Those of you who are baseball fans may consider an analogy to the size of a hitter’s bat. ideal gas ideal gas 1 2 pressure (bar) Process F "external" PE = P Process E 1 2 Fexternal / area or .4 m 1 mol of pure. only in a reversible process can we substitute the system pressure for the external pressure. On the other hand.7 Schematic of infinitesimal-step. The irreversible expansion. thus the molecules of the gas are running into something “smaller” and will not transfer as much energy. The corresponding plots of the processes on a PEv diagram is shown at the bottom. 0.08 Molar volume (m3/mol) Figure 2.8 m Process F: A = 0.indd 54 05/11/12 1:35 PM . 54 ► Chapter 2. This change of momentum with time represents the net energy transferred between the system and surroundings.04 . If we sum up all the molecules. c02. the more energy can be transferred to the baseball. it is the work. the mass on the piston is larger than the corresponding reversible process. we see that the net energy transfer (work) is greater. in the irreversible compression process.1 m2 Isothermal compression 1 mol of pure. as the gas molecules bounce off the piston. their change in net z momentum between before and after a collision is determined by the movement of the piston. The greater the force the piston exerts on a given molecule as it rebounds off the piston. It therefore imparts more energy to those molecules and costs more work. In this case. reversible isothermal expansion (process E) and com- pression (process F) processes. The First Law of Thermodynamics mT = 1020 kg mT = 1020 kg Patm ∂m ∂m A = 0. We have done better. In the next section.10) wrev For example.11) wirrev For example.4 THE FIRST LAW OF THERMODYNAMICS FOR CLOSED SYSTEMS Integral Balances In this section. is: wirrev hexp 5 (2. In this figure. When we address systems undergoing chemical reaction.72 wE 25545 where wi represents the work of process i. irreversible processes is to solve the problem for the idealized. open systems will be treated. Figure 2. For a pure species or a mixture of constant composition. hexp. Thus. we compare how much work we actually get out to the idealized. the efficiency of expansion.69 wB 8000 or 69% efficient. the number of moles remains constant: n1 5 n2 Since mass cannot enter or leave a closed system. the two-block process C is 84% efficient. For an expansion process. ► 2. the changes in the energy within the system 1 D 5 final 2 initial 2 are equal to the energy transferred from the c02. the efficiency of process B would be: wF 5545 hcomp 5 5 5 0.indd 55 05/11/12 1:35 PM . efficiencies are defined so that if we can operate a process reversibly. 2. reversible process.4 The First Law of Thermodynamics for Closed Systems ◄ 55 Efficiency We can compare the amount of work required in an irreversible process to that of a reversible process by defining the efficiency factor.8 shows a schematic of a closed system that undergoes a process from initial state 1 to final state 2. we consider energy balances for closed systems. On the other hand. the number of moles or the mass of a particular component may change. There are two ways in which to catalog the amount of material in the system— by mass or by moles. In both cases. we say process A is 72% efficient. to determine the efficiency of a compression process. h. hcomp we compare the reversible work to the work we must actually put in: wrev hcomp 5 (2. Each way can be convenient. surroundings. the efficiency of Process A would be: wA 24000 hexp 5 5 5 0. we would have 100% efficiency. While the total mass must be conserved. and boundary are delineated. One strategy for actual. care must be taken. in comparison. In a closed system. Thus. mass cannot transfer across the system boundary. reversible process and then correct for the irreversibilities using an assigned efficiency factor. the two forms are equivalent and can be interconverted using the molecular weight. In the absence of chemical reaction. while the real processes are less efficient. the system. positive for energy transfer from the sur- roundings to the system. u^ 3 J/kg 4 . Writing down the first law in quantitative terms. For this case. surroundings by either heat or work.13b) respectively. it would be would be wrong to write them DQ or DW.8 also illustrates the sign convention that we have defined for heat and work. for example. You should be able to convert any equation to a mass basis that uses the corresponding specific property.13a) and. we get:7 change in energy transferred c energy in s 5 c from surroundings s system to system DU 1 DEK 1 DEP 5 Q 1 W ('''')''''* (')'* property: depends on (2. Du 5 q 1 w (2.12b) We often neglect macroscopic kinetic and potential energy. utilizing the appropriate intensive thermodynamic properties. we can rewrite Equation (2. The First Law of Thermodynamics +Q +W System Surroundings property ΔEK = EK. All three forms of energy are Boundary considered. The terms on the right-hand side are process dependent and the real path of the system must be used. Since the composition of a closed system remains constant (barring chemical reac- tion). They can be calculated using the real path or any hypothetical path we cre- ate. internal energy will be u [J/mol]. namely.12a) using intensive properties by dividing through by the total number of moles:8 Du 1 DeK 1 DeP 5 q 1 w (2.indd 56 05/11/12 1:35 PM . the extensive and intensive forms of the closed system energy balances become: DU 5 Q 1 W (2. 56 ► Chapter 2.2 − EP. 8 We will sometimes write balance equations on a molar basis.8 Illustration of closed system and sign conventions for heat and work.2 − EK.1 ΔEP = EP. 7 Heat and work already refer to the amount of energy transferred. c02. Figure 2. hence.1 ΔU = U2 − U1 Figure 2. We reserve the D for state function that depends just on the initial and final state of the system.12a) depends only path on state 1 and 2 The properties on the left-hand side of Equation (2.12a) depend only on the initial and final states. For example. we use the inexact differential d with heat and work to remind us that we must keep track of the path when we integrate to get these quantities.4 through 2. the gas has a pres- Closed System sure of 20.5 5 const (Continued) c02.4 The First Law of Thermodynamics for Closed Systems ◄ 57 Differential Balances The first law can also be written in differential form for each differential step in time during the process. What is the heat transferred in each case. 2.0 m3. The pressure volume relationship during this process is given by: Pv1.4 Consider a piston–cylinder assembly containing 10. The energy balances above are often differentiated with respect to time. Initially.0 kg of water. dU 5 dQ 1 dW extensive 3 J 4 (2. dU 1 dEK 1 dEP 5 dQ 1 dW extensive 3 J 4 du 1 deK 1 deP 5 dq 1 dw intensive 3 J/mol 4 or. yielding: dU dEK dEP # # 1 1 5Q1W extensive 3 W 4 dt dt dt du deK deP 1 1 5 q# 1 w# intensive 3 W/mol 4 dt dt dt where the rate of heat transfer and the rate of work [J/s or W] are denoted with a dot over the corresponding variable. Can you explain the difference in relation to the efficiency factor? Example 2. Numerical solutions are obtained by integration of the resulting dif- ferential energy balance. The system undergoes a reversible process Energy Balance in which it is compressed to 100 bar.indd 57 05/11/12 1:35 PM .0 bar and occupies a volume of 1. in contrast.14) du 5 dq 1 dw intensive 3 J/mol 4 We use the exact differential d with the energy terms to indicate that they depend only on the final and initial states. Again. neglecting kinetic and potential energy.15) dt du 5 q# 1 w# intensive 3 W/mol 4 dt ►EXERCISE Consider the six processes depicted in Figures 2. we often neglect kinetic and potential energy to give: dU # # 5Q1W extensive 3 W 4 (2. Common forms of energy balance over a differential element can be written by analogy to the equations just presented.7. We define the initial state as state 1 and the final state as state 2.4 (Appendix B).4. v^ 2.5 5 const 5 P1v^ 1. we see the kg temperature is 212.0 m3 v^ 5 5 5 0.0 kg The values in the steam tables are in units of Pa. Therefore. (c) Calculate the heat transferred during this process. (b) To calculate the work it is useful to draw a schematic of the process. we can use the steam tables to find other properties. Because this is a reversible process. as shown in Figure E2.4 Initial and final states of the expansion process. H2O Process P1 = 20 [bar] v1 = 0.10 B R m 10. Once values for these properties are determined.100 B R.indd 58 05/11/12 1:35 PM .4°C (or slightly above). we see that at P 5 2 MPa to three significant figures the volume m3 at the saturation temperature is 0. We can calculate v^ 2 from the equation in the problem statement: Pv^ 1. (d) What is the final temperature? SOLUTION (a) We need two independent intensive properties to constrain the state of the system. we need to know the specific volume of the final state. so we convert 20 bar to 2 MPa. The First Law of Thermodynamics (a) What is the initial temperature? (b) Calculate the work done during this process. 58 ► Chapter 2.1 [m3/ kg] T1 = 212.4 [° C] P2 = 100 [bar] State 1 State 2 Figure E2. looking at Table B.5 1 5 P2v ^ 1. we can write: w^ 5 23 PEdv^ 5 23 Pdv^ To find the upper limit on the integrand. The specific volume of the initial state can be determined as follows: V 1.2. If we look at Table B.5 2 c02. 03279 5 525ºC c02.5 R 5 284 B R 0.0342 P1v^ 1.6 B R kg Thus.03279 5 3045.03564 2 0.we get: v2 2 v 1 T 5 500ºC 2 T2 5 500ºC 1 3 550ºC 2 500ºC 4 B R v 1 T 5 550ºC 2 2 v 1 T 5 500ºC 2 0.03279 5 500ºC 1 3 50ºC 4 B R 0. we need to interpolate: u2 2 u 1 T 5 500ºC 2 v2 2 v 1 T 5 500ºC 2 5 u 1 T 5 550ºC 2 2 u 1 T 5 500ºC 2 v 1 T 5 550ºC 2 2 v 1 T 5 500ºC 2 Solving for u2.1 kg The sign of work is positive because we are adding energy to the system during the compression.5 1 m3 v^ 2 5 ¢ ≤ 5 1 0. u2 5 u 1 T 5 500ºC 2 1 3 u 1 T 5 550ºC 2 2 u 1 T 5 500ºC 2 4 v2 2 v 1 T 5 500ºC 2 B R v 1 T 5 550ºC 2 2 v 1 T 5 500ºC 2 0.2 3 0. heat transfers from the system to the surroundings.0342 2 0. In part (a).03279 kJ 5 3094. (c) To find heat.indd 59 05/11/12 1:35 PM .1 0.4 The First Law of Thermodynamics for Closed Systems ◄ 59 Solving for v^ 2 gives: 1/1.8 1 3 3144. we find: kJ u1 5 2600.1 v^ 1.11. From Table B. we can apply the first law: q 5 Du 2 w Because we solved for work in part (b).5 v^ 2 v^ 1 0.0342 B R P2 kg Now we can solve for work: 0. Looking at Table B. we look at Table B.5 2 1/1. (d) To find T2.03564 2 0.5 P1v^ 1.0342 0.0342 2 0.5 1 B 0. In this case.4.5 1 1 1 kJ w^ 5 23 Pdv^ 5 23 dv^ 5 2P1v^ 1. we found that state 1 is approximately saturated vapor at 20 bar.3 B R kg For u2. we must also interpolate.8 4 B R 0.5 2 0.5 5 0.0342 0. we need only to determine the internal energy in states 2 and 1 from the steam tables. 2.2.4 at a pressure of 10 MPa (100 bar).5 2 3045. kJ q 5 Du 2 w 5 u2 2 u1 2 w 5 210 B R kg Since the value of q is positive. for a nonreacting system. A generic open system with two streams in and two streams out is shown in Figure 2.9. but we must also account for the energy change in the system associated with the streams flowing into and out of the system. Thus the mass balance can be written: dm # # a b 5 a min 2 a mout (2. mass can flow into and out of the system. Mass can be converted to moles using the molecular weight. the mass flow rate can be written as: S m# 5 AV (2.9. 60 ► Chapter 2. The piston shown in the plot is hypothetical. there are δ(Eflow)in = − PinAinδx PE = Pin in Imaginary piston Stream 2 in Ws In out min System mout ûin Stream 1 ûout (êK)in (êK)out (êP)in (êP)out ŵflow = (PVˆ )in ŵflow = (PVˆ )out Qsys out Figure 2. In the general case of many streams in and many streams out. at velocity V. we consider the case of the generic open system illustrated in Figure 2. The accumulation of mass in the system is equal to the difference of the total rate of mass in minus the total rate of mass out. c02. it is convenient to express the amount of chemical species in terms of moles (molar basis) rather than mass. it illustrates the point that flow work is always associated with fluid flowing into or out of the system. For a stream flowing through a cross-sectional S area A. Flow Work The energy balance for an open system contains all the terms associated with an energy balance for a closed system. however.16) dt sys in out where m# is the mass flow rate in [kg/s]. Compared to the closed-system analysis we performed in Section 2. The First Law of Thermodynamics ►2. the balances developed here will be true for any number of inlet or outlet streams.9 Schematic of an open system with two streams in and two streams out. To accomplish this task. we must sum over all the in and out streams. Let’s look at the contribution to the energy balance from the inlet stream labeled stream 1. This open system happens to have two streams in and two streams out.4.17) v^ In many situations. We must keep track of the mass in the system since it can change with time. In setting up the balance equations.5 THE FIRST LAW OF THERMODYNAMICS FOR OPEN SYSTEMS Material Balance In open systems.indd 60 05/11/12 1:35 PM . it is convenient to discuss rates: mass flow rates [kg/s] and rates of energy transfer [J/s or W]. We first consider conservation of mass. as depicted in Figure 2. By a similar argument. Flow work is the work the inlet fluid must do on the sys- tem to displace fluid within the system so that it can enter. The energy balance is written as [on a rate basis. It may be visualized by placing an imaginary piston in front of the material that is about to enter the system. therefore. we must sum over all the in and out streams. we can show the flow work of any outlet stream is given by: # 1 Wflow 2 out 5 2m# outPoutv^ out We can write the total energy transfer due to work in the system in terms of shaft work. The imaginary piston acts like the real piston shown in Figure 2.15)]: 1 1 0 5 a m# in(u^ 1 V2 1 gz)in 2 a m# out( u^ 1 V2 1 gz)out S S in 2 out 2 energy flowing energy flowing into the system out of the system steady state with the inlet with the outlet steams steams c02.indd 61 05/11/12 1:35 PM . and potential energy. we consider a system at steady-state.17) was used. We have set PE 5 Pin and eliminated the negative sign. but the flowing streams can also have (macroscopic) kinetic energy. that is. therefore. we can write the flow work in intensive form: # 1 Wflow 2 in 1 w^ flow 2 in 5 5 Pinv^ in m# in We conclude that the flow work of any inlet stream is given by the term 1 Pv^ 2 in. there is no accumulation of energy or mass in the system with time.5. and flow work. If we divide by mass flow rate. Ws.9. 1 e^ P 2 in. 1 e^ k 2 in. Second. as discussed earlier. as follows: # # # # W 5 Wshaft 1 Wflow 5 BWs 1 a m# in 1 Pv^ 2 in 1 a m# out 1 2Pv^ 2 out R in out The shaft work is representative of the useful work that is obtained from the system. the inlet stream adds energy into the system by supplying work. 2. The rate of work exerted by the fluid to enter the system is. # Ws generically represents any possible way to achieve useful work. by analogy to Equation (2. given by: # dx S # v 1 Wflow 2 in 5 2PE Ain 5 Pin AinVin 5 Pin m in ^ in dt where Equation (2. u^ in.2) and (2. since the direction of velocity is the negative of the normal vector from the piston. it does not include flow work! The flow work does not provide a source of power. typically the most important form of energy is the specific internal energy. In the general case of many streams in and many streams out.3) shows these last two terms can be written as 12 V 2 and gz respectively. First. the molecules flowing into the system carry their own energy. the so-called flow work. First.5 The First Law of Thermodynamics for Open Systems ◄ 61 two additional ways in which energy can be transferred from the surroundings. While shafts are commonly used to get work out of an open system. S Inspection of Equations (2. We can now include the new ways in which energy can exchange between the sys- tem and surroundings in our energy balance for open systems. associated with them. it is merely the “cost” of pushing fluid into or out of the open system. We give it the name enthalpy. Steady-State Energy Balances In summary.19) Take a look at Equation (2. As we will learn in the next section. it is convenient to group these terms together (so that we don’t forget one).19). The term h^ in includes both the internal energy of the stream and the flow 2 work it adds to enter the system.20) in out c02. In cases where we can neglect (macroscopic) kinetic and potential energy. the additional energy associated with the flowing inlet stream is given by h^ in [as well as 1 S2 V and gz]. Recall the internal energy of an ideal gas depends only on T. enthalpy. Thus. enthalpy also is convenient to use with closed systems at constant pressure. like the internal energy. and v are all properties. Application of the defi- nition of enthalpy and the definition of an ideal gas gives: hideal gas 5 u 1 Pv 5 u 1 RT 5 f 1 T only 2 since Pv 5 RT for an ideal gas. we get: 1S 1S # # 0 5 a m# in C 1 u^ 1 Pv^ 2 1 V2 1 gz D in 2 a m# out C 1 u^ 1 Pv^ 2 1 V2 1 gz D out 1 Q 1 Ws in 2 out 2 Enthalpy The inlet streams that flow into open systems always have terms associated with both internal energy and flow work. u 1 Pv (2.18) Since u.indd 62 05/11/12 1:35 PM . The First Law of Thermodynamics # # 1Q 1 BWs 1 a m# in 1 Pv^ 2 in 1 a m# out 1 2Pv^ 2 out R in out flow work flow work from the inlet from the outlet streams streams Rearranging. is also a property. integral energy balance becomes: # # # 0 5 a minh^ in 2 a m# outh^ out 1 Q 1 Ws (2. Similarly. Thus. We next develop the relationship between temperature and enthalpy for an ideal gas. this new group. P. h [J/mol]: h . therefore. the combined internal energy and flow work leaving the system as a result of the exiting streams are given by h^ out. the steady-state. Enthalpy provides us a property that is a convenient way to account for these two contributions of flowing streams to the energy in open systems. 62 ► Chapter 2. the enthalpy of an ideal gas depends only on T. See if you can identify the physical meaning of each term. the steady-state energy balances can be written: 1S 1S 0 5 a m# in (h^ 1 V2 1 gz)in 2 a m# out (h^ 1 V2 1 gz)out in 2 out 2 # # 1 Q 1 Ws (2. in start-up as the equipment “warms up. Ws. the left-hand side represents accumulation of energy within the system.21) and (2.21). (b) What are ways to find h of a system? (c) For the energy balance depicted in Equation (2.22) dt sys in out In Equations (2. SOLUTION The steady-state energy balance is given by Equation (2. There is no flow work associated with this term. the unsteady- state energy balance becomes: dU dEK dEP 1S 1S ¢ 1 1 ≤ 5 a m# in ah^ 1 V2 1 gzb 2 a m# out ah^ 1 V2 1 gzb dt dt dt sys in 2 in out 2 out # # 1 Q 1 Ws (2.5 Steam enters a turbine with a mass flow rate of 10 kg/s. integral energy balance becomes: dU # # # # a b 5 a minh^ in 2 a mouth^ out 1 Q 1 Ws (2. h^ is appropri- ate. heat dissipation is negligible. It will save you many mistakes down the road! ►EXERCISE (a) Simplify Equations (2. On the other hand. It is worthwhile for you to take a moment and reconcile the use of U and h^ above. the first two terms on the right-hand side account for energy flowing in and out of the system. hence. These terms must account for both the internal energy and flow work of the flowing streams. The inlet pressure is 100 bar and the Calculation of Work inlet temperature is 500ºC. for example. If we label the inlet stream “1” and the outlet stream “2. At steady-state. which terms correspond to accumula- tion? To energy in? To energy out? EXAMPLE 2. the appropri- ate property is U. The outlet contains saturated steam at 1 bar.19) and (2.21) for the case of one stream in and one stream out. there is one stream in and one stream out.21) In cases where we can neglect (macroscopic) kinetic and potential energy.5A) becomes: # 0 5 m# 1h^ 1 2 m# 2h^ 2 1 Ws (E2.. Unsteady-state is important.indd 63 05/11/12 1:35 PM .20): # # 0 5 a m# inh^ in 2 a m# outh^ out 1 Q 1 Ws (E2. In this case.5A) in out For a turbine.22). 2.” In the case when the inlet and outlet streams stay constant with time. the unsteady-state. Moreover.” Equation (E2. respectively.5 The First Law of Thermodynamics for Open Systems ◄ 63 Transient Energy Balance Another useful form of the energy balance for open systems is for unsteady-state con- ditions.5B) (Continued) c02. calcu- from the First Law late the power (in kW) generated by the turbine. # # since Q . Unsteady-State Analysis For the choice of system shown in Figure E2.5 3 kJ/kg 4 while state 1 is superheated steam at 500ºC and 100 bar 1 5 10 Mpa 2 : h^ 1 5 3373.5C) Plugging in Equation (E2.6 Steam at 10 MPa. This is the equivalent power to that delivered by approximately 70 automobiles running simultaneously.6A).5 2 3373. and then the valve is closed.6 3 kJ/kg 4 Plugging these numerical values into Equation (E2. Let us define the initial state of the tank (empty). SOLUTION (a) This problem can be solved several ways. For state 2. the mass balance can be written as: m# 1 5 m# 2 5 m# (E2.6 2 3 kJ/kg 4 5 26981 3 kW 4 Thus.5D) We can look up values for specific enthalpy from the steam tables. state 1. Since we have one stream in and no streams out. EXAMPLE 2.5B) gives: # Ws 5 m# 1 h^ 2 2 h^ 1 2 (E2. Transient Process (b) Explain why the final temperature in the tank is not the same as that of the steam flowing in the pipe.indd 64 05/11/12 1:35 PM .6A. The inlet flow stream will be designated by “in. this turbine generates approximately 7 MW of power.5C) into (E2. We first solve it with the fixed bound- ary illustrated in Figure (E2. The First Law of Thermodynamics At steady-state. The valve is opened and the tank fills with steam until the Temperature for a pressure is 10 MPa. Note the negative sign. 64 ► Chapter 2.5D) gives: # Ws 5 10 3 kg/s 4 1 2675. The process takes place adiabatically. and the final state (filled to 10 MPa). (a) Determine the final temperature of the steam in the tank.16). state 2. we must use an unsteady-state energy balance since the mass and energy inside the tank (system) increase with time. which indicates that we are getting useful work from the system.” The conservation of mass is given by Equation (2. we use saturated steam at 1 bar 1 5 100 kPa 2 : h^ 2 5 2675. 450ºC is flowing in a pipe. Connected to this pipe Calculation of Final through a valve is an evacuated tank.6A. we get: dm # a b 5 min dt sys Separating variables and integrating gives: m2 t # 3 dm 5 3 mindt m1 5 0 0 c02. as shown in Figure E2. An alternative solution with a moving system boundary is then presented. Separating variables and integrating both sides with respect to time from the initial empty state to the final state when the tank is at a pressure of 10 MPa gives: U2 t t # ^ ^ # 3 dU 5 3 minhindt 5 hin 3 mindt (E2.22): 0 0 0 dU # ^ # ^ # # a b 5 minhin 2 mouthout 1 Q 1 Ws dt sys where the terms associated with flow out. Equation (E2. is given by Equation (2. Hence. in mass (specific) units. 450°C System Figure E2. the inlet specific enthalpy must be equal to the final specific internal energy of the system. 2. We have moved the specific enthalpy out of the integral.6C) since state 1 identically has a value for U of 0.indd 65 05/11/12 1:35 PM . The system boundary is indicated with dashed lines. since the properties of the inlet stream remain constant throughout the process. Equation (E2. steam at P 5 10 MPa and T 5 450°C has a specific enthalpy: h^ in 5 3241 3 kJ/kg 4 According to Equation (E2. the final state has two independent intensive properties (Continued) c02.6C) simplifies to: u^ 2 5 h^ in (E2.6A) 0 The unsteady energy balance.6B) U1 5 0 0 0 Steam 10 MPa. we get: t m2 5 3 m# indt (E2. heat.6B) becomes: m2u^ 2 5 m2h^ in (E2. Applying Equation (E2.5 The First Law of Thermodynamics for Open Systems ◄ 65 Since state 1 identically has a value for m of 0.6D). and work are zero.6A) and integrating.6D) From the steam tables.6A Flow of super- Empty boundary heated steam from a supply tank to be filled line to fill an empty tank. An external pressure of 10 MPa.6G) into (E2. for steam at P 5 10 MPa and u^ 5 3241 3 kJ/kg 4 .6E) and (E2. The First Law of Thermodynamics specified: P 5 10 MPa and u^ 5 3241 3 kJ/kg 4 . the final temperature of the system is: T2 5 600°C Closed-System Analysis Alternatively. no mass flow crosses the boundary.6E) PE = 10 MPa Steam 10 MPa 450°C System boundary Pin = 10 MPa Tin = 450°C Equivalent Diaphragm Empty Vacuum tank to be filled Q= 0 Figure E2. 66 ► Chapter 2. is shown on the right of Figure E2. the cylinder contains steam at 10 MPa and 450ºC.6D). given by Equation (E2. We consider the mass that starts in the pipe and eventually ends up in the tank as part of the system.6B. in terms of our familiar piston–cylinder assembly.6B are equivalent? A mass balance on either closed system depicted in Figure (E2.. From the steam tables.6. Work is given by: V W52 3 PEdv 5 minPinv^ in (E2.6G) V1min v^ in Substituting Equations (E2. Q 5 0. thus. acts against the piston until the pressures equilibrate. With this choice of system. Above the diaphragm.6F) Since the process is adiabatic. and the system is taken from state “in” to state 2. the surroundings transfer energy to the system via work.6A.6B. As mass flows into the tank. Likewise. An equivalent closed system. During the compression process. Below the diaphragm is a vacuum. The adiabatic chamber is separated by a diaphragm that plays the same role as the valve in the original system. The system boundary in this case is illustrated on the left side of Figure E2. Can you see that the two processes depicted in Figure E2. we have a closed system.6F) and rearranging gives: m2u^ 2 5 min 1 u^ in 1 Pinv^ in 2 5 minh^ in 5 m2h^ in which is identical to the result we obtained for the unsteady-state open system of Figure E2. identical to the inlet steam. The process is initiated by removing the diaphragm.indd 66 05/11/12 1:35 PM .6B A closed system approach to solving the problem in Example 2.6B) gives: min 5 m2 (E2. the energy balance is: DU 5 m2u^ 2 2 minu^ in 5 Q 1 W (E2. c02. the boundary contracts. we can use a different choice of system to solve this problem. representative of the steam in the pipe outside the system boundary. The container is connected to a heat source (in this case.3. In this case. heat capacity data are crucial in this problem-solving meth- odology. cv.10a. it is often convenient to choose measured properties (T. either P or v can be chosen according to convenience. increasing T from 450ºC to 600ºC. It does not have to be the path of the actual process. In fact. and the enthalpy. h. Thus. As heat is supplied. While any thermodynamic properties can be used. This setup is known as a constant- volume calorimeter. since u is a thermodynamic property. This information is often obtained in the form of heat capacity (or specific heat). we must know the tempera- ture dependence of u to calculate Du. the molecules of A move faster. 2. The other independent property is typically also a measurable property. In the next section.13b)] to this system and get: Du 5 q closed system. Figure 2. cv To measure the heat capacity at constant volume. for an ideal gas. Du. is provided through the resistive heater. an experimental setup as conceptu- ally shown in Figure 2. A typical data set for pure species A is also shown in Figure 2. T.23) c02. and there is a direct relationship between T and u.1). it is only this component that contributes to u. Moreover. Likewise for Dh. as we go from T1 to T2 at constant volume.10a can be used. for a pure species are constrained by specifying two independent intensive properties. const V (2.6B is equivalent to work done by an equivalent constant external pressure acting on the piston shown on the right. As we learned in Section 1. q. we will explore how heat capacities are experimentally determined and how they are reported. In step 1. the closed-system analysis helps us see the importance of accounting for flow work in energy balances on open systems. this type of energy change is labeled sensible heat. it is necessary to be able to determine how the energy (or enthalpy) of the species in the system changes during a process.indd 67 05/11/12 1:35 PM . Since we can “sense” the result of the heat input with the thermocouple. Similarly. we can specify a hypothetical path to calculate the change in internal energy. The flow work of steam flowing into the system shown on the left of Figure E2. the internal energy. we can apply the first law [Equation (2. which is one component of u (see Section 2. This closed system consists of pure species A within a rigid container.3 illustrates a common hypothetical path used to calculate Du. Therefore. that is. the temperature dependence of h used to find Dh can be found through reported heat capacity values. The experiment is conducted as follows: As a known amount of heat. T and v are chosen as independent properties. u. Since there is no work done in this process. ►2.6 Thermochemical Data for U and H ◄ 67 (b) The fluid in the system receives flow work from the fluid behind it to get into the system.6 THERMOCHEMICAL DATA FOR U AND H Heat Capacity: cv and cP In order to perform energy balances on both closed and open systems. temperature is a measure of the molecular kinetic energy. or v) as the independent properties. of the system is measured. P. and the temperature increases. since it can be measured in the lab (or field). the tempera- ture. a resist- ance heater) and is otherwise well insulated. This work component adds energy from the incoming stream to the system. Temperature is almost always chosen as one of the independent properties. Heat Capacity at Constant Volume. Equation (2. (a) constant-volume calorimeter to obtain cv. We define the heat capacity at constant volume.25) Experiment: Data: T cv ≡ ∂u ∂T v V = const A A q = Δu A A A A A Slope = cv at T2 A A A T1 T2 q Slope = cv at T1 T (a) Constant-Volume Calorimeter: Determination of cv Experiment: Data: cP ≡ ∂h ∂T P P = const T A A q = Δh A A A A A Slope = cP at T2 A A A T1 T2 q Slope = cP at T1 T Pure substance.10 Schematics of the experimental determination of heat capacity. the slope of the curve gives us the heat capacity at any temperature. A (b) Constant-Pressure Calorimeter: Determination of cP Figure 2. By taking the slope of this curve as a function of temperature.indd 68 05/11/12 1:35 PM . In these data. we can get: cv 5 cv 1 T 2 We can then fit the data to a polynomial expression of the form: cv 5 a 1 BT 1 CT2 1 DT22 1 ET3 (2. (b) constant-pressure calorimeter to determine cP. heat capacity changes with T. the amount of heat supplied is plotted on the y-axis and the temperature measured is on the x-axis. The First Law of Thermodynamics Note that in Figure 2. cv.23) shows heat input is identical to Du.24) 'T v Hence.10a. However. as: 'u cv . c02. 68 ► Chapter 2. the heat capacity at T1 is less than that at T2. Typically. a b (2. we increase u.26) T1 T1 Heat capacity should only be used for temperature changes between the same phase. Consider an ideal gas. Heat capacities for gases. On a molecular level. however. this depiction is in terms of the piston–cylinder assembly that we have previously examined. D. the contribution is R per mole. The amount the internal energy increases with temperature is quantified by the heat capacity according to Equation (2. respectively. We will not formally address that prob- lem here. C. In summary. c02.24) shows us that if we increase T. The first mode is related to the center-of-mass motion of the molecules through space. as discussed shortly.indd 69 05/11/12 1:35 PM . in u with temperature to three possible modes in which the molecules can obtain kinetic energy. 9 Except at very high temperature when electrons occupy excited states. monatomic gases have heat capacities given by this value. The kinetic energy due to vibration is much more interesting.25) manifests itself in the vibrational mode. and solids can be intepreted in a similar way. In Chapter 1. It is related to the specific quantized energy levels of the molecule. is 3R/2. We can associate the increase in molecular kinetic energy and. but in a system that can expand as it is heated so as to keep the pressure constant.26). Equation (2. The more energy it gains. the additional kinetic energy due to rotational motion for linear and nonlinear molecules is RT and 3RT/2 per mole. In fact. Heat Capacity at Constand Pressure. cP. rota- tional.6 Thermochemical Data for U and H ◄ 69 Parameters a. A conceptual representation of the experimental setup to measure cP is shown in Figure 2. The distribution of these levels depends on temperature. and E are then tabulated and can be used any time we want to know how the internal energy of species A changes with temperature at constant vol- ume. it is useful to realize that the temperature dependence of the heat capacity indicated by Equation (2. This translational energy contributes kT/2 per molecule (or RT/2 per mole) to the kinetic energy in each direction that the molecule moves. To account for the vibrational modes of kinetic energy.10b. At low temperature. cv can be attributed to molecular structure and the ways in which each species exhibits translational. 2. Since molecules translate through space in three directions. We can then find Du by integration: T2 T2 Du 5 3 cv dT 5 3 3 a 1 BT 1 CT2 1 DT22 4 dT (2. The contribution of translational motion to cv. B. we may want to know how this increase in energy manifests itself. therefore. cp The heat capacity at constant pressure. and vibrational kinetic energy. At high temperature. liquids. is measured in a similar manner. the translational motion contributes 3RT/2 per mole to the internal energy of the molecules. the vibrational contribution goes to zero and the heat capacity is given by the translational and rotational modes only. When a phase change occurs. we saw that the Maxwell–Boltzmann distribution characterizes the velocities of the mol- ecules at a given temperature.9 Diatomic and polyatomic molecules can manifest kinetic energy in rotational and vibrational modes as well. the latent heat must also be considered. where the vibrational motion is fully active. While the actual apparatus may look dif- ferent. we would need to resort to quantum mechanics. only gas A is no longer held within a rigid container. Except at extremely low temperature. the larger cv. given by the derivative of the internal energy with respect to temperature. Hence. we get: Dh 5 q closed system. an energy balance tells us the heat supplied at constant pressure is just equal to the change in the thermodynamic property.10b and can be fit to the polynomial form: cP 5 A 1 BT 1 CT2 1 DT22 1 ET3 (2. D.10b.30) The parameters A. const P (2. is reported in terms of the property Dhrxn. for closed systems at constant P. If species A is in the liquid or solid phase. the piston depicted c02.28) Hence. inspec- tion of Equation (2. and E are reported for some ideal gases in Appendix A.27) So Equation (2. the energetics of a chemical reac- tion. the energy balance contains a term for work: Du 5 q 1 w 5 q 2 PDv (2. The First Law of Thermodynamics Since the system is now doing Pv work as it expands. Relations between cP and cV By comparing Figure 2.2. Therefore. 70 ► Chapter 2. B. the property h couples internal energy and work. that is. we define the heat capacity at constant pressure as: 'h cP . the experimentally measured heat can be related directly to a thermodynamic property. For example. However.27) can be rewritten as: Du 1 D 1 Pv 2 5 q since at constant pressure. Therefore. Heat capacity parameters at constant pressure of some liquids and solids are reported in this appendix. In this way. the so-called enthalpy of reaction.indd 70 05/11/12 1:35 PM . This equation holds.28) suggests a second common use of enthalpy. Enthalpy—A Second Common Use Recall that we “constructed” the property enthalpy to account for both the internal energy and flow work for streams flowing into and out of open systems. its vol- ume expansion upon heating should be relatively small. and: D 1 Pv 2 5 PDv 1 vDP 5 PDv Applying the definition for enthalpy [Equation (2.18)]. a b (2. it accounts for both the change in internal energy and the Pv work as the system boundary changes in order to keep pres- sure constant. enthalpy. C. In both cases.10a to Figure 2. typical data for species A are presented in Figure 2. in general. the molar volumes of liquids and solids do not change much with temperature. experiments that are conveniently done in closed systems at constant P are reported using the thermodynamic property enthalpy. we can estimate the difference in cv and cP for the different phases of matter. In this case.29) 'T P Again. in this case. DP 5 0. As its name suggests. It is usually reported between 298 K and a given temperature. the only change in molecular energy is in molecular kinetic energy. the mean heat capacity is the average of cP between two temperatures. the enthalpy change becomes: Dh 5 cP 1 T 2 298 2 Solving for cP gives: T e cPdT 298 cP 5 (2. a b 5 c d 5a b 1a b 5a b 1R (2. T.6 Thermochemical Data for U and H ◄ 71 in Figure 2.indd 71 05/11/12 1:35 PM . Problems 5. For the case of an ideal gas. we can figure out the relationship between cP and cv by applying the ideal gas law to the definition for cP as follows: 'h ' 1 u 1 Pv 2 'u 'RT 'u cP .21.10b will not move significantly and the value of work in Equation (2.32) 'T P dT 'T v Plugging Equation (2. Thus Equation (2. for an ideal gas. c02. Use of cP eliminates the need for integration and can make the mechanics of problem solving easier. heat capacity data are often reported in terms of the mean heat capacity.20 and 5.32) into (2. 10 We will revisit the relation between cP and cv of liquids in Chapter 5. the internal energy depends on temperature only. 'u du 'u a b 5 5a b (2. Therefore.29) and Equation (2. when doing calculations using these data.31) 'T P 'T P 'T P 'T P 'T P since Pv 5 RT for an ideal gas. Thus. the volume expansion of a gas will be significant. 2. the mathematical average of the continu- ous function cP over temperature between 298 K and T. Mean Heat Capacity For many gases. Hence.31) gives: cP 5 cv 1 R for ideal gases Values of heat capacity for gases are almost always reported for the ideal gas state. you must choose a hypothetical path where the change in temperature occurs when the gas behaves ideally. cP.24) are approximately equivalent and. consequently:10 cP < cv liquids and solids On the other hand. by definition. However.33) T 2 298 Note that Equation (2. that is.33) is also.27) will be small compared to q. B 2 D 773 Dh 5 R c AT 1 T 2 d 2 T 473 Substituting in values: Dh 5 8.2: h^ 1 1 at 1 MPa. we must multiply by mass.121 3 105 5 A 1 BT 1 DT22 5 3.121 3 105 a 2 b R 5 10.314 3 3. c02.470 1 1. Calculate the Heat Input heat input required using the following sources for data: Calculations Using (a) Heat capacity (b) Steam tables Different Data Sources SOLUTION (a) Since this process occurs at constant pressure. we get the following integral expression: T2 773 Dh 5 3 cPdT 5 R 3 3 A 1 BT 1 DT22 4 dT T1 473 Integrating. water is not an ideal gas but rather has attractive intermolecular interactions.28) can be written as: Q 5 nDh If we assume water is an ideal gas.991 c d 773 473 mol and.018 3 kg/mol 4 2 1 650.5 3 kJ/kg 4 Since the steam tables give us the specific enthalpy. 72 ► Chapter 2. In accordance with the discussion above. Appendix B. The extra energy needed in part (b) results from that needed to pull the water molecules apart.725 3 1023 1 7732 2 4732 2 1 1 J 20.981 3 J 4 (b) From the steam tables. MWH2O 5 0.4 3 kJ/kg 4 So.2 to calculate Dh: ideal gas cP 0.470 1 300 2 1 0. the system will expand as T increases.450 3 1023T 1 R T2 Using the definition of heat capacity. Thus. 200°C 2 5 2827. we must use the molecular weight for water.418 3 J 4 The answer for part (b) is approximately 6% higher than for part (a). The extensive version of Equation (2. We will learn more of these things in Chapter 4.indd 72 05/11/12 1:35 PM . 500°C 2 5 3478. Dh^ 5 h^ 2 2 h^ 1 5 650.5 3 kJ/kg 4 2 1 103 3 J/kJ 4 2 5 23.9 3 kJ/kg 4 h^ 2 1 at 1 MPa. Q 5 nDh 5 21. The First Law of Thermodynamics EXAMPLE 2.7 Consider heating 2 moles of steam from 200ºC and 1 MPa to 500ºC and 1 MPa. At 1 MPa. we use the values of heat capacity given in Appendix A. enthalpy is the appropriate property to calculate the heat input.018 3 kg/mol 4 : Q 5 mDh^ 5 1 2 3 mol 4 2 1 0. 2. for air between Determination of T1 5 298 K and T2 5 300 to 1000 K.8 Use the data available in Appendix A. Using Mean Heat SOLUTION This process occurs at steady-state with one stream in and one stream out.95 EXAMPLE 2. Mean Heat Capacity for Air SOLUTION Using the definition of cP from Equation (2.17 400 3003.8B) 2 T 298 T1 298 where for air: A 5 3.575 3 1023.8C). cP.8C) The solution to Equation (E2.8B) and (E2.93 102 29.97 700 12146.75 202 29.2: T2 T T B 2 D 3 cPdT 5 R 3 3 A 1 BT 1 DT 4 dT 5 RBAT 1 T 2 R 22 (E2.19) can be written as follows: 0 0 0 0 0 S S # V 2 # V 2 # # 0 5 n1 ah 1 MW 1 MWgzb 2 n2 ah 1 MW 1 MWgzb 1 Q 1 Ws 2 1 2 2 (Continued) c02. and D 5 20.71 1000 21727.02 502 30.8A).51 402 30. B 5 0.33): T 3 cPdT Dh 298 cP 5 5 (E2. using (E2.2 to calculate the mean heat capacity.6 Thermochemical Data for U and H ◄ 73 EXAMPLE 2. is presented at intervals of 100 K in Table E2.87 602 30.22 800 15292.35 2 29.016 3 105 (E2.8.8A) 1 T 2 298 2 T 2 298 The heat capacity can be integrated with respect to temperature using the parameters in Appendix A. Heat Calculation Determine the heat rate required using the mean heat capacity data from Example 2. T Dh cP [K] 3 J/mol 4 T 2 298 3 J/mol K 4 300 58.355. in intervals of 100 K.9 You need to preheat a stream of air flowing steadily at 10 mol/min from 600 K to 900 K.8.indd 73 05/11/12 1:35 PM .45 500 6001.89 702 30. TABLE E2.59 302 29.46 900 18485. hence Capacity for Air Equation (2.8 Calculated Values for Mean Heat Capacity of Air at Different Temperatures.71 600 9049. 97 1 600 2 298 2 4 5 94. Equation (E2. as shown in Figure E2. A mole balance yields: n# 1 5 n# 2 5 n# air so the first law balance simplifies to: # Q 5 n# air 1 h900 2 h600 2 5 n# air 3 1 h900 2 h298 2 2 1 h600 2 h298 2 4 (E2.363 3 J/min 4 EXAMPLE 2. 10 cm above Piston–Cylinder the base of the cylinder.8. (b) Determine the heat transferred.10 Air is contained within a piston–cylinder assembly. The system is then reversibly heated to 100ºC. h600 is the enthalpy at 600 K. we get: # Q 5 n# Dh 5 10 3 30.33): 1 h900 2 h298 2 5 cP. 74 ► Chapter 2.01 m2. In this state.9B) 1 h600 2 h298 2 5 cP. The cross- Spring-Assembled sectional area of the piston is 0. Initially the piston is at 1 bar and 25ºC.9A) was rewritten to use the definition for mean heat capacity given by Equation (2.10A. the system pressure is always balanced by the external pressure and the work done is given by: v2 W 5 2 3 PdV (E2.000 3 N/m 4 and x is the displacement length from its uncompressed position. it exerts a force on the piston according to: F 5 2kx where k 5 50.900 1 900 2 298 2 (E2.indd 74 05/11/12 1:35 PM . As the spring is compressed. (a) Determine the work done.9A) and using the values from Table E2.9C) Substituting Equations (E2.10A) v1 We can draw a free-body diagram to determine how all the forces acting on the piston balance.9C) into (E2. The First Law of Thermodynamics where we have set the bulk kinetic and potential energies and shaft work to zero. can be written in terms of the change in volume: V 2 V1 DV x5 5 A A c02. and h298 is the enthalpy at 298 K. SOLUTION (a) Since the process is reversible. the spring exerts no force on the piston.9A) where h900 is the enthalpy of air at 900 K. x.71 1 900 2 298 2 2 29. The displacement of the spring.9B) and (E2.10A. as shown in Figure E2.600 1 600 2 298 2 (E2. The initial state of the system for Example 2.10.01 m2 Air 10 cm P1 = 1 bar Figure E2.10B) A A Plugging Equation (E2. where DV 5 V 2 V1.10C).00116 3 m3 4 which can be plugged back into Equation (E2.10A Piston–cylinder assembly with a T1 = 25°C spring attached to the piston.10 C) to get: W 5 2166 3 J 4 (b) To find the heat transferred during this process.10B Schematic of the forces acting on the piston in Figure E2.C) A2 2A2 V1 0 To solve Equation (E2.6 Thermochemical Data for U and H ◄ 75 F = −kx Pext = 1 bar A = 0.10 is shown. We can then equate the force per area acting on each side of the piston to get: kx kDV P 5 Pext 1 5 Pext 1 2 (E2. 2. Applying the ideal gas law gives: P1V1 P2V2 V2 k 1 V2 2 V1 2 5 5 ¢Pext 1 ≤ T1 T2 T2 A2 Solving this quadratic equation for V2 gives: V2 5 0.10A as FAir = PAirA the gas expands. we must find V2. we can apply the first law to this closed system: DU 5 Q 1 W (E2.indd 75 05/11/12 1:35 PM .10B) into (E2.10D) (Continued) c02.10A) and integrating gives: V2 DV5 1V2 2V1 2 kDV k 1 V2 2 V1 2 2 W 5 2 3 PextdV 2 3 d 1 DV 2 5 2Pext 1 V2 2 V1 2 2 (E2. |Fspring| = kx Fext = PextA Piston Figure E2. there is a substantial change in internal energy associated with it (see Section 2.1). the change in enthalpy from the liquid phase to the solid phase is reported as the enthalpy of fusion. so that the temperature will stay constant during the phase change. Liquids are held together by attractive forces between the molecules. Dhfus: Dhfus 5 hs 2 hl c02. Du 5 1. Like heat capacities.10D): Q 5 DU 2 W 5 803 3 J 4 Latent Heats When a substance undergoes a phase change. we must supply enough energy to overcome the forces of attraction.355. it is equal to the enthalpy of the vapor minus the enthalpy of the liquid. We need to be able to determine a value for this energy change if we want to apply the first law to a process involving a phase change. A given amount of liquid substance A is placed in a well-insulated closed system at constant pressure. A measured amount of heat is supplied until A becomes all vapor.2: A 5 3. The First Law of Thermodynamics The change in internal energy is given by: T2 T2 T2 Du 5 3 cvdT 5 3 1 cP 2 R 2 dT 5 R 3 3 1 A 2 1 2 1 BT 1 DT22 4 dT T1 T1 T1 T2 B 2 D 5 RB 1 A 2 1 2 T 1 T 2 R 2 T T1 The heat capacity parameters for air can be found in Appendix A. as depicted in Figure 2.11. A typical experimental setup is schematically shown in Figure 2. B 5 0. 76 ► Chapter 2.016 3 105 Thus.28) shows that if we measure the heat absorbed as A changes phase.580 3 J/mol 4 P1V1 and. To vaporize a liquid. For example. and D 5 20. The temperature of the subcooled liquid and the superheated vapor increases as energy is supplied via heat. We term this dif- ference the enthalpy of vaporization: Dhvap 5 hv 2 hl If we need to calculate the energetics of a vapor condensing to a liquid.11. the energetics characteristic of a given phase change are reported based on accessible measured data. DU 5 nDu 5 ¢ ≤Du 5 638 3 J 4 RT1 We can now solve for the heat transfer from Equation (E2. A schematic for the data acquired in this experiment is presented on the right of Figure 2. Similarly. It is only the heat input at constant temperature during the phase transition that is reported as the enthalpy of vaporization. Examination of Equation (2.575 3 1023.11.indd 76 05/11/12 1:35 PM . We chose a system at constant pressure. consider the vaporization of a liquid. we simply use the negative of Dhvap. we calculate the change in enthalpy of the liquid from T to Tb. the enthalpy of vaporization is typically reported at 1 bar. Latent means “hidden. The latent heat changes with temperature. described previously. Tb. given Dhvap? The term to describe the change of enthalpy during a phase transition at constant pressure is latent heat. For example. since this is the value we have for Dhvap. T 5 3 clP dT 1 Dhvap. In step 1. we calculate the change in enthalpy of the vapor from the normal boiling point to T.” and it is called “latent” because we cannot “sense” the heat input by detecting the temperature change. It consists of three steps. Figure 2. we usually only know its value at one state.11 Schematic of the experimental determination of enthalpy of vaporization. 2. However. using heat capacity data. In step 3. Tb 1 3 DcvlP dT T Tb Tb Step 2 Tb Δhvap. we get: Tb T T Dhvap. And the change from the solid phase to the vapor phase is the enthalpy of sublimation. we vaporize the liquid at the normal boiling point. as is the case with “sensible” heat. c02.Tb Figure 2. Adding together the three steps.2).6 Thermochemical Data for U and H ◄ 77 Experiment: Data: State 1 State 2 P = const A Vapor q = Δh A A A Δhvap T A A A T A A AAA Liquid A A AA A A AAA Tvap T q q Figure 2.Tb Temperature Step 3 Tb T ∫ c |pdT ∫ cvpdT Step 1 Tb T T Δhvap. the so-called normal boiling point. Tb 1 3 cvP dT 5 D hvap.T at any T based on the measured value to which we have access at Tb.indd 77 05/11/12 1:35 PM . we need to construct an appropriate hypothetical path (see Section 2.12 Hypothetical path to calculate Dhvap at temperature T from Liquid Vapor data available at Tb and heat capacity Phase data. To find Dhvap at another pressure (and therefore another temperature). In step 2. Duvap. Dhsub: Dhsub 5 hv 2 hs How would you find the internal energy of vaporization.12 illustrates a path for the calculation of Dhvap. 11C) The enthalpy change can be divided into three parts: (1) the sensible heat required to raise liquid hexane to its boiling point.68.314B3.8°C 298. Equation (2.11A) and solving for the heat-transfer rate. Thus. and (3) the sensible heat required to raise hexane in the vapor state to 100ºC. 78 ► Chapter 2.11B) Plugging (E2.3dT 5 9485 3 J/mol 4 5 9.2 and. Dh 5 3 R 1 A 1 BT 1 CT 2 2 dT v. It exits as a vapor at 100ºC. EXAMPLE 2. we get: # Q 5 n# 1 h2 2 h1 2 (E2. Determination of What is the required heat input to the heater? Take the enthalpy of vaporization at 68.indd 78 05/11/12 1:35 PM . 68.8°C 5 28. that is.8°C v.8°C 1 Dh (E2. the enthalpy difference becomes: h2 2 h1 5 Dh 1 Dhvap.11B) into (E2.8°C S 100°C Values for the first and third terms can be found from the appropriate heat capacity data: 342 Dh 5 3 216.11D) l. latent heat. The First Law of Thermodynamics where we used the following definition: DcvlP 5 cvP 2 c lP This procedure can likewise be applied to determine the values of Dhfus and Dhsub at different temperatures than that at which they are reported.22 2 3422 2 2 c02.8°C S 100°C 342 53.025 1 373.49 3 kJ/mol 4 l.68.88 3 kJ/mol 4 SOLUTION This process occurs in an open system with one stream in and one stream out.2 373. 68. A mole balance gives: n# 1 5 n# 2 5 n# (E2.” In this case.11A) where we have set the macroscopic kinetic and potential energy and the shaft work equal to zero.20) can be written on a molar basis as: 0 # # # # 0 5 n 1h 1 2 n 2h 2 1 Q 1 Ws (E2.8ºC to be: Heat Required to Evaporate Hexane Dhvap.11 10 mol/sec of liquid hexane flows into a steady-state boiler at 25ºC. Can you draw a schematic? We label the inlet state as “state 1” and the outlet as “state 2.722 3 1023 5 8.25°C S 68.25°C S 68.2 2 342 2 1 1 373. (2) the enthalpy of vaporization. Hence.12 Schematic of evaporation of saturated liquid H2O in a rigid.49 1 28. 2. A mass balance gives: mv2 5 m1l 1 mv1 The internal energy in state 1 must account for that of both the saturated liquid and the saturated vapor.11C) gives the rate at which heat must be supplied: # Q 5 a10 3 mol/s 4 b a 1 9. We label the initial state as “state 1” and the final state as “state 2.20 2 3 kJ/mol 4 b 5 435 3 kJ/s 4 EXAMPLE 2.85 1 5. the pressure in the vapor phase will increase.12.indd 79 05/11/12 1:35 PM . energy is required for both the evaporation of water (latent heat) and the increase in temperature (sensible heat). (Remember.11. we can write the first law according to Equation (2. As water boils.12A) as: P v v Liquid P1 = 10 kPa Vapor process P2 = ? P2 2 Liquid-vapor l 10 kPa 1 Q v1 = v2 v State 1 State 2 Figure E2. The increase in pressure will require an increase in the temperature of the system for boiling to proceed. The Determination of system pressure is at 10 kPa. a given pressure constrains the temperature of the two-phase region. This result contrasts with Figure 2. we can write Equation (E2. closed system. where enthalpy is used.12A) Since our system is at constant volume. What is the minimum amount of heat needed to evaporate all Heat Required the liquid? to Evaporate H2O SOLUTION A schematic of the process is shown in Figure E2.6 Thermochemical Data for U and H ◄ 79 16.3 kg of saturated vapor. This heating process occurs in a closed system at constant volume.11D) and plugging the result into Equation (E2.791 3 1026 2 1 373.” The left-hand side of the figure shows the physical process while the right-hand side represents it on a Pv diagram. in that case.) Thus.20 3 kJ/mol 4 Summing together values for enthalpy in Equation (E2.12 A rigid vessel contains 50. (Continued) c02. there is Pv work.23 2 3423 2 R 3 5 5. there is no work done and the heat needed equals the change in internal energy. since. Since bulk kinetic and potential energy are negligible.0 kg of saturated liquid water and 4.13a): DU 5 Q 1 W (E2. 3 2 1 14. When the atoms in molecules rearrange by undergoing a chemical reaction.15 MPa If this value did not match a table entry.16 B R m2 54. The energy change upon reaction is an important component in applying the first law to reacting systems.1 m3 m3 v^ v2 5 v 5 5 1. but we need to find a value for a thermodynamic property. 80 ► Chapter 2. it is more common to report the change in enthalpy of reaction. Dhrxn.2) gives: u^ l1 5 191.67 2 5 63. Looking up the internal energy of state 2 gives: u^ v2 5 2519.9 3 kJ/kg 4 We now need to constrain state 2. we would have to interpolate. as illustrated on the Pv diagram in Figure E2. since experiments are more conveniently executed at constant pressure.1 m3 We can now solve for the specific volume of state 2: V2 63.3 2 1 2519.001 2 1 1 4.8 2 2 1 4.12. Looking up values for internal energy in the steam tables for saturated water: pressure (Appendix B. It can be characterized by a change in internal energy. Durxn. The First Law of Thermodynamics U2 2 U1 5 1 mv2u^ v2 2 2 1 ml1u^ l1 1 mv1u^ v1 2 5 Q (E2.12B) State 1 is completely constrained. this process occurs at constant volume. c02. We know we have saturated vapor.9 2 4 5 117 3 106 J Enthalpy of Reactions A large amount of energy is “stored” in the chemical bonds within molecules.6 3 kJ/kg 4 Solving Equation (E2. the energy stored within the bonds of the products is typically different from that of the reactants. significant amounts of energy can be absorbed or liberated during chemical reactions.12B) for heat gives: Q 5 1 mv2u^ v2 2 2 1 ml1u^ l1 1 mv1u^ v1 2 5 1 54. Thus.6 2 2 3 1 50 2 1 191. however.3 2 1 2437. Since the container is rigid.8 3 kJ/kg 4 u^ v1 5 2437. We can find the volume of the vessel as follows: V1 5 V2 5 ml1v^ l1 1 mv1v^ v1 5 1 50 2 1 0. we find the specific volume of saturated vapor matches at: P2 5 0.indd 80 05/11/12 1:35 PM .3 kg kg Looking up the value. defines a reference state as the atomic form of each element in the system. Reactions that release energy are said to be exothermic.34) The reactants contain three bonds per molecule of oxygen reacting: single bonds between hydrogen atoms in each of the two H2 molecules and a double bond between the oxygen atoms in O2. we used atoms for our reference state. These bonds are covalent in nature and vary in energy based on how the interacting valence electrons overlap.41 Source: Derived from average bond enthalpies reported in G. The two product H2O molecules have four oxygen–hydrogen single bonds. and νH2O 5 2 In the previous discussion. 2.1 Bond Dissociation Energies of H. 0. νi. C. negative for reactants.13.6 Thermochemical Data for U and H ◄ 81 For example. when these atoms re- form into two water molecules. 0. νreactants . respectively. By convention. energy is released.5 eV for every two mol- ecules of water produced by this reaction.1 eV of energy 1 1 eV 5 1. it is unitless and positive for products. and zero for inerts. The energy differences between states depicted by the arrows are based on the bond energies. Let’s consider the energetics of this reaction by the following path: One way to do this calculation is to first pull apart the reactant molecules into their constituent atoms and then have the atoms recombine to form the product molecules. To generalize to any reaction.1. For example. in Reaction (2. νinerts 5 0. consider the reaction of two molecules of hydrogen gas with one mol- ecule of oxygen to form gaseous water at 298 K and 1 bar. However. D.6 3 10219 J 2 to dissociate two molecules of H2 and one molecule of O2 into four H atoms and two O atoms. and. Bond energies for the three different types of bonds in this system are reported in Table 2. 1969). Pimentel and R.O Bonds Bond Energy [eV/molecule] HiH 4. νO2 5 21. in essence. The reaction stoichiometry can be expressed as follows: 2H2 1 g 2 1 O2 1 g 2 h 2H2O 1 g 2 (2.13 O iH 4.1 eV of energy. The negative sign indicates that the products are more stable than the reactants.34): νH2 5 22. This path. It can be found as the number before the corresponding species in a balanced chemical reaction. A schematic of the energetics in this reaction path is shown in Figure 2. c02. consequently. since this choice made it straightforward to see how the energy of molecules changes with atomic TABLE 2. they release 17. while reactions that absorb energy are termed endothermic. Understanding Chemical Thermodynamics (San Francisco: Holden-Day.indd 81 05/11/12 1:35 PM . Spratley. The net energy change represents the internal energy of reaction and is found to be 23. we introduce the stoichiometric coefficient.50 O wO 5. νproducts . It takes 14. The stoichiometric coefficient equals the proportion of a given species consumed or pro- duced in a reaction relative to the other species. Such a calculation path for the enthalpy of reaction at 298 K is illustrated in Figure 2. The enthalpy of formation can be represented as: Dhf elements S species i The enthalpy of formation of a species containing only one element. since.34) into their atoms. we may be interested in a system in which water is reacting at higher temperatures.3 also reports the enthalpy of reaction for gaseous water at 298 K and 1 atm. DUrxn. the most stable form of oxygen is O2 gas while the most stable form of carbon is solid graphite.H2O The resulting energy dif- ference characterizes the internal energy change of reaction. making the c02. it is straightforward to calculate the enthalpy of reaction. in nature. Appendix A.H.14. 82 ► Chapter 2. the enthalpy of formation is representative of a hypothetical (but impor- tant!) change of state that is often useful. Enthalpies in the form of Dhf are the most common thermochemical data available to calculate the enthalpy of reaction. For example. species seldom exist as atoms at 298 K and 1 bar. as it is found in nature. hydrogen and oxygen are found as diatomic gases at 25ºC and 1 atm.83 3 kJ/mol 4 . In the dashed (calculation) path. where it is a vapor. The first step in obtaining the enthalpy of reaction at the system T would be finding it at 298 K. Dhf.5 eV H2O. However.6 eV sociate the reactants and products of Reaction H2. the reactants are first decom- posed into their constituent elements. With the enthalpies of formation available. is identically zero.13 Schematic O.1 eV 17. rearrangement. as given by Dh2. The constituent elements are then allowed to react to form products.O2 (2.indd 82 05/11/12 1:35 PM . Example 2. Appendix A. We are free to pick any refer- ence state that we desire as long as we stick to it. The value for the enthalpy of formation for this reaction is found in Appendix A. the enthalpy of formation of liquid water is defined by the reaction: 1 H2 1 g 2 1 O2 1 g 2 S H2O 1 l 2 2 since the elements found in water.H2. Note that the stoichiometric coefficients of the reactants are negative.14 illustrates such a calculation.3 to be Dh°f. The enthalpy difference between a given molecule and this reference state is defined as the enthalpy of forma- tion.H. Although water cannot physically exist in this state. A more convenient reference state can be defind as the pure elements in their most stable form at 298 K for the temperature of interest and 1 bar. this reference state is inconvenient in practice. The First Law of Thermodynamics H.298 5 2285.14. For example.H Figure 2. This part of the path is labeled Dh1. as found in nature. ΔUrxn = −3.O representation of the energy needed to dis- Energy . For example. at 298 K and 1 bar.3 shows some representative values for 25ºC and 1 atm. that is.298 b 1 a νi aDh°f. In Chapter 9 we will learn how to quantify the extent to which a chemical reaction proceeds at equilibrium. 298 2 CO2 1 3 1 Dh°f. 298 2 H2 2 1 Dh°f. 1 Dh°f 2 i.51 2 1 3 1 0 2 2 1 2241.3: Dh°298 5 1 2393. 298 2 H2O 2 1 Dh°f. Equating the two paths yields: Dh°rxn. Taking values for the others from Appendix A. In this case.35) can be written as follows: Dh°rxn 5 a νi 1 Dh°f. Dh°rxn 5 a νih°i 5 a νi aDh°f b (2. signs for Dh1 consistent with the definition of enthalpy of formation above.298)i products Enthalpy reactants Reactants 298.15 K.66 2 5 49.14 Calculation path of Dh°rxn from the standard enthalpies of formation.82 2 2 1 2200.298 Products 298. the enthalpy of reaction can be determined by scaling each species’Dhf by its stoichiometric coefficient. 298 2 CH3OH The enthalpy of formation of H2 is zero by definition.0 3 kJ/mol 4 The sign of the enthalpy of reaction is positive.indd 83 05/11/12 1:35 PM . c02. 1 bar Δh1 = Σνi (Δh°f.298 b reactants i products i 5 a νi aDh°f.15 K 1 bar Figure 2.6 Thermochemical Data for U and H ◄ 83 Elemental form found in nature 298. 298 2 i 5 1 Dh°f. Example 2.298 5 Dh1 1 Dh2 5 a νi aDh°f. For incomplete reactions. Equation (2. if enthalpies of formation are available for all the species in the chemical reaction of interest. indicating that this reaction is endothermic. there remain some reactants in the out- let stream.16 illustrates such a case.35) i Often a reaction does not go to completion. In summary.298 b i Thus. 2.13 Calculate the enthalpy of reaction at 298 K for the following reaction: Determination of Enthalpy of Reaction H2O 1 g 2 1 CH3OH 1 g 2 S CO2 1 g 2 1 3H2 1 g 2 SOLUTION The enthalpy of reaction can be found from the enthalpy of formation data presented in Appendix A. since that is the form hydrogen takes at 298 K and 1 bar. we must account only for the enthalpy of reaction for the species that did react in our energy balance.3.15 K 1 bar Δh°rxn. EXAMPLE 2.298)i Δh2 = Σνi (Δh°f. As you will recall. Take the average power density for solar radiation over a 24-hour period to be 200 W/m2.1 and 1% for crop plants (it goes up to around 3% for microalgae grown in bioreactors). defined as the energy content of the biomass that can be harvested divided by the average energy of solar radiation impinging on the same area. enthalpy is a “constructed” property that is useful to characterize the energy for processes in closed systems at constant pressure because it accounts for both the change in internal energy and the Pv work needed to move the system boundary. It is a biofuel that can be generated biologically through fermentation of corn.7B R 3 106 B R Vr cm3 m3 n5 5 5 1. How does the crop area needed for corn compare to the area calculated in part (a)? SOLUTION (a) First we need to calculate the amount of energy produced by the consumption of 3 million m3 of petroleum. The energy efficiency of photosynthesis. The U.3 3 1016 3 J 4 A5 5 5 5.7 3 106 B 2 R m 2s day h m We can now find the area: H 9.indd 84 05/11/12 1:35 PM . Assuming complete combustion.4 3 1010 3 m2 4 Er J 1.298 2 H2O 2 1 Dh°f. we can write the balanced equation as: 25 C8H18 1 l 2 1 O2 1 g 2 S 8CO2 1 g 2 1 9H2O 1 g 2 2 Using Equation (2. In this example. 84 ► Chapter 2. petroleum consumption is approximately 3 million m3 per day.7 3 10 B 2 R 6 m c02.70 g/cm3. The density of octane is 0. (a) Assuming a solar cell efficiency of 10%. This value can be found with the enthalpy of reaction.298 2 i 5 8 1 Dh°f. you wish to explore the feasibility of two Alternative Energy alternative energy sources to petroleum.3 3 1013 3 kJ 4 5 29.2B R mol Thus.600B R 5 1.10 3 24B R 3 3. You may assume that the fuel value of petroleum can be represented by Sources octane.35) and finding the appropriate values in Appendix A.8 3 1010 3 mol 4 MW g 114.S.298 2 C8H18 5 25. we need to deter- mine the number of moles of octane consumed in a day: g cm3 3 3 106 3 m3 4 3 0. how much area would be needed to provide the equivalent energy to that used by petroleum in the United States? (b) Bioethanol provides another possible alternative energy source.073 B R mol Why is the sign negative? To find the extensive value for energy. The First Law of Thermodynamics Example 2.298 2 CO2 1 9 1 Dh°f.14 Alternative energy sources that are renewable and that can reduce greenhouse gas emissions Land Areas for are actively being developed.3 gives: kJ Dh°rxn 5 a yi 1 Dh°f. as approximated by octane. is typically between 0.3 3 1016 3 J 4 The energy density of the sun can be found from information given in the problem statement: J h s J Er 5 200B R 3 0. H 5 nh 5 29. 2.6 Thermochemical Data for U and H ◄ 85 This represents solar cells filling a square parcel of around 150 miles on a side—approximately one-fifth the size of the state of Arizona. (b) The range of area for the biofuel can be found from the ratio of the efficiency to the solar cells. For 1% efficiency, this calculation gives A 5 5.4 3 1011 3 m2 4 and for 0.1% efficiency, A 5 5.4 3 1012 3 m2 4 . These represent square parcels with sides approximately 500 and 1,500 miles, respectively. The latter size is over half the land area of the United States. Comment: the preceding analysis examined the area requirements for two alternative energy sources. However, comparison of the two alternatives is more complex; we must assess the total cost of the systems, including water, capital, operations, and maintenance costs. We should also consider the risks from manufacturing and possible interactions with the food supply and climate change. Example 2.15 Propane is placed in an adiabatic, constant pressure combustion chamber at 25°C and allowed Adiabatic Flame to react as follows. What is the final temperature in each case? Assume complete combustion. Temperature (a) It is mixed with a stoichiometric amount of oxygen and reacts to form H2O and CO2. (b) It is mixed with a stoichiometric amount of air and reacts to form H2O and CO2. Calculations (c) It is mixed with a stoichiometric amount of air, and the carbon distribution in the product stream contains 90% CO2 and 10% CO. SOLUTION (a) For part (a), a balanced equation of the chemical reaction is written as follows: C3H8 1 5O2 S 3CO2 1 4H2O (E15.1) A schematic of this process, assuming Reaction E15.1 goes to completion, is shown in Figure E2.15A. An energy balance on the closed system at constant pressure gives [see Equation (2.28)]: DH 5 a 1 nihi 2 2 2 a 1 nihi 2 1 5 Q 5 0 (E15.2) We now need a path to calculate DH. A convenient choice is illustrated by the solid lines in Figure E2.15B. That figure also shows the overall energy balance constraint of Equation (E15.2) as a dashed line. Because enthalpy of reaction data are available at 298 K (Appendix A.3), we choose a hypothetical path where we first completely combust propane at 298 K, then P = const. Combustion process CO2 H2O C3H8, O2 T= ? Well T = 298 [K] insulated Initial state (1) Final state (2) Figure E2.15A Schematic of complete combustion of propane in a stoichiometric mixture of oxygen at constant pressure. The closed system is adiabatic. (Continued) c02.indd 85 05/11/12 1:35 PM 86 ► Chapter 2. The First Law of Thermodynamics 2 T2 2 Σi ni ∫ cP,i dT =0 T ΔH 298 T 298 1 ΔHrxn,298 Figure E2.15B Hypothetical path for C3H8, CO2, calculation of enthalpy in complete O2 H2O combustion of propane. heat the products to a temperature T2, which makes the enthalpy change between states 1 and 2 zero. Using this hypothetical path, Equation (E15.2) becomes: T2 DHrxn,298 1 3 a 1 ni 2 2 1 cp 2 idT 5 0 298 The enthalpy of reaction can be found based on the reaction stoichiometry given by Reaction E15.1: Dhrxn,298 5 a νi 1 Dhof 2 i 5 νCO2 1 Dhof 2 CO2 1 νH2O 1 Dhof 2 H2O 1 νC3Hs 1 Dhof 2 C3Hs 1 νO2 1 Dhof 2 O2 Using values from Appendix A.3, we get: J Dhrxn,298 5 3 1 2393.51 2 1 4 1 2241.82 2 2 1 1 2103.85 2 2 0 5 22.044 3 106 B R mol and using a basis of 1 mol C3H8, the extensive enthalpy of reaction is: DHrxn,298 5 nC3H8Dhrxn,298 5 22.044 3 106 3 J 4 The enthalpy of reaction must be counteracted by an equal but opposite increase in the sensible heat. This effect will allow us to calculate T2. The sensible heat can be found through a summation of heat capacities, as follows: T2 T2 T2 3 a 1 ni 2 2 1 cp 2 i dT 5 3 nCO2 1 cp 2 CO2 dT 1 3 nH2O 1 cp 2 H2O dT (E15.3) 298 298 298 Values for heat capacity parameters can be found in Appendix A.3. They are summarized in Table E15.1: Table E2.15 Heat Capacity Parameters for Species in State 2 Species A B D CO2 5.457 1.045 3 10 3 21.157 3 105 H 2O 3.470 1.45 3 1023 1.21 3 104 N2 3.280 5.93 3 1024 4.00 3 103 CO 3.376 5.57 3 1024 23.10 3 103 c02.indd 86 05/11/12 1:35 PM 2.6 Thermochemical Data for U and H ◄ 87 Integrating each term on the right of Equation E15.3 gives: T2 BCO2 3 nCO2 1 cp 2 CO2 dT 5 1 3 2 RBACO2 1 T2 2 298 2 1 2 1 T22 2 1 298 2 2 2 298 1 1 2 DCO2 ¢ 2 ≤R (E15.4A) T2 298 T2 BH2O 2 3 nH2O 1 cp 2 H2O dT 5 1 4 2 RBAH2O 1 T2 2 298 2 1 2 1 T2 2 1 298 2 2 2 298 1 1 2 DH 2O ¢ 2 ≤R (E15.4B) T2 298 The only unknown, T2, can be solved for implicitly by the value at which the sum of the two preceding equations equal2DHrxn,298, 2.044 3 106 3 J 4 . T2 is found to be: T2 5 4,910 3 K 4 This value is known as the adiabatic flame temperature, and in this case is quite large. The adiabatic flame temperature indicates the maximum temperature a reactor can reach for a given fuel. If there is heat transfer out of the system, the temperature will be lower. If you used stoichiometric amount of air, do you think the adiabatic flame temperature would be higher, lower, or the same? How about excess air? How about if some CO formed as well? We will see the results in the cases that follow. (b) If we use air instead of pure oxygen, the reaction described in part (a) remains the same; thus, the enthalpy of reaction remains 22.044 3 106 3 J 4 . However, we must now account for the inert N2 which does not participate in the reaction, in the sensible heat term. This species will provide more “thermal mass,” and consequently, we expect the adiabatic flame temperature to be lower. We have 5 moles of oxygen, so we calculate: 0.79 nN2 5 ¢ ≤5 5 18.8 3 mol 4 0.21 The value of nN2 is large compared to the two reaction products so we expect T2 will reduce significantly. To calculate the sensible heat, we must add a term for N2 as follows: T2 T2 T2 T2 3 a 1 ni 2 2 1 cp 2 i dT 5 3 nCO2 1 cp 2 CO2 dT 1 3 nH2O 1 cp 2 H2O dT 1 3 nN2 1 cp 2 N2 dT 298 298 298 298 Using the following equation in addition to Equations (E15.4A ) and (E15.4B): T2 BN2 2 3 nN2 1 cp 2 N2 dT 5 1 18.8 2 RBAN2 1 T2 2 298 2 1 2 1 T2 2 1 298 2 2 2 298 1 1 2 DN2 ¢ 2 ≤R (E15.4C) T2 298 (Continued) c02.indd 87 05/11/12 1:35 PM The only unknown. can similiarly be solved for implicitly by the value at which the sum of the three preceding equations equal2DHrxn. 2. we need to consider the following reaction: 7 C3H8 1 O2 S 3CO 1 4H2O (E15.96 3 10 3 J 4 II DHrxn. the total enthalpy of reaction is: C3H8Dhrxn.370 [K] found in part (b).1 3 ≤ 5 18.5) 2 with. The First Law of Thermodynamics Again. T2 is found to be: T2 5 2.indd 88 05/11/12 1:35 PM . DhIIrxn.298 5 νCO 1 Dhof 2 CO 1 νH2O 1 Dhof 2 H2O 2 νC3H8 1 Dhof 2 C3H8 2 νO2 1 Dhof 2 O2 J 5 21.21 2 sensible heat. for a 1 mol basis of C3H8. 0. In addition to Reaction E15.298.298 5 21.9 and the second reaction (E15. we multiple the first reaction (E15.1) by nIC3H8 5 0.1.7 3 mol 4 .298 1 nII 6 and the following species concentrations are obtained: nCO2 5 2. 88 ► Chapter 2. (c) We must now account for CO production as well. we must add a term for CO to the expression we used in part (b): T2 T2 T2 T2 T2 3 a 1 ni 2 2 1 cp 2 idT 5 3 nCO2 1 cp 2 CO2dT 1 3 nCO 1 cp 2 COdT 1 3 nH2O 1 cp 2 H2OdT 1 3 nN2 1 cp 2 N2dT 298 298 298 298 298 T2 is found to be: T2 5 2. c02.2 3 mol 4 .370 3 K 4 Note T2 has dropped significantly from the value of 4.1.1.195 3 106 B R mol Because the carbon distribution in the product stream contains 90% CO2 and 10% CO. To calculate the 0.350 3 K 4 Note T2 is similar to the value of 2.298 5 nIC3H8DhIrxn.5) by nII C3H8 5 0. and nN2 5 ¢ ≤ ¢0.910 [K] found in part (a). Thus. T2.3 and are reported in Table E15.044 3 106 3 J 4 .3 3 mol 4 .79 7 nH2O 5 4 3 mol 4 .9 3 5 1 0. values for heat capacity parameters for N2 can be found in Appendix A. nCO 5 0. 2.6 Thermochemical Data for U and H ◄ 89 Example 2.16 Consider an isobaric chemical reactor where the following two simultaneous chemical reac- Energy Balance for tions occur: Multiple Incomplete 1 Reactions CO 1 O2 S CO2 (1) 2 1 C 1 O2 S CO (2) 2 Initially, the reactor contains 4 mol of CO, 4 mol of O2, and 2 mol of C. At the end of the reaction process, the reactor contains 2 mol of CO and 2 mol of O2. If the initial temperature is 25°C and the final temperature is 225°C, determine the amount of heat transferred during the process. SOLUTION An energy balance gives: 498 Q 5 DH 5 DHrxn 1 3 a 1 ni 2 2 1 cp 2 i dT 298 where the extensive enthalpy change of the system has been decomposed into a component for the reaction and a component for the sensible heat in a way similar to that shown in Figure E2.15B. To determine the enthalpy of reaction, we need thermochemical data for each of the two reactions listed above. For Reaction 1 we have: 1 Dh1rxn,298 5 a νi 1 Dhf 2 i 5 1 Dhf 2 CO2 2 1 Dhof 2 CO 2 1 Dhf 2 O2 o o o 2 J Dh1rxn,298 5 1 2393.51 2 2 1 1 2110.53 2 2 0 5 22.83 3 105 B R (E2.16A) mol Similarly, for Reaction 2: o 1 Dh2rxn,298 5 a νi 1 Dhof 2 i 5 1 Dhf 2 CO 2 1 Dhf 2 C 2 1 Dhof 2 O2 o 2 J Dh1rxn,298 5 1 1 2110.53 2 2 0 2 0 5 21.11 3 105 B R (E2.16B) mol The number of moles in the denominator of Equations (E2.16A) and (E2.16B) refer to the moles of CO or C, respectively, which have reacted. We can define this amount as the extent of reaction, j. We will learn more about the extent of reaction when we cover Chemical Reaction Equilibrium in Chapter 9. Thus, the number of moles of any species i can be related to the extent of k reactions through the stoichiometry of the reactions and their extent: ni 5 n0i 1 a νijk (E2.16C) k c02.indd 89 05/11/12 1:35 PM 90 ► Chapter 2. The First Law of Thermodynamics where n0i is the initial number of moles of i and ni is the final number of moles after reaction. Using Equation (E2.16C), the number of moles of CO can be written: nCO 5 n°CO 2 j1 1 j2 Substituting in values we get: 2 mol 5 4 mol 2 j1 1 j2 Similarly for O2, we have: nO2 5 n0O2 2 12 j1 2 12 j2 and, 2 mol 5 4 mol 2 12 j1 2 12 j2 Solving the simultaneous equations gives: j1 5 3 mol j2 5 1 mol To calculate the extensive enthalpy of reaction we can multiply the molar enthalpy of reaction for each reaction by its extent: DHrxn 5 j1Dhrxn,1 1 j2Dhrxn,2 5 29.59 3 105 3 J 4 (E2.16D) Finally, to calculate the sensible heat, we need to determine the final species concentrations. Using Equation (E2.16C), we have: nCO,2 5 4 2 j1 1 j2 5 2 mol nO2,2 5 4 2 12 j1 2 12 j2 5 2 mol nCO2,2 5 0 1 j1 5 3 mol nC,2 5 2 2 j2 5 1 mol so, 498 3 a 1 ni 2 2 1 cp 2 i dT 5 52,000 3 J 4 (E2.16E) 298 where the data from Appendix A.3 were used, similarly to Example 2.15. Calculating the amount of energy transferred by heat: 498 Q 5 DH 5 DHrxn 1 3 a 1 ni 2 2 1 cp 2 i dT 5 29.1 3 105 3 J 4 298 where the values given by Equations (E2.16D) and (E.2.16E) were used. The negative sign indicates that heat must be removed from the reactor to the surroundings. c02.indd 90 05/11/12 1:35 PM 2.6 Thermochemical Data for U and H ◄ 91 EXAMPLE 2.17 How would you calculate the enthalpy of reaction, Dhrxn, at any temperature T, given data for Enthalpy of Reaction enthalpy of formation at 298 K and heat capacity parameters, available in Appendix A? at Different T SOLUTION Since enthalpy is a thermodynamic property, we can construct a hypothetical path that utilizes the available data. The enthalpy of reaction at any temperature T can then be found from the path illustrated in Figure E2.17. The reactants are first brought to 298 K. They are then allowed to react under standard conditions to make the desired products. The prod- ucts are then brought back up to the system temperature, T. Adding these three steps gives the following integral: T Dhrxn,T 5 Dhrxn,298 1 3 ¢ a nicP,i ≤ dT i 298 Substituting in Equations (2.35) and (2.30) gives: T o Di Dhrxn,T 5 a νi 1 Dh f , 298 2 i 1 3 ¢R a νi ¢Ai 1 BiT 1 CiT 2 1 2 1 EiT 3≤ ≤dT T (E2.17) i 298 When standard enthalpies of formation and heat capacity parameters are available, Equation (E2.17) can be solved explicitly for Dhrxn, T at any given T. Note the similarity between the paths in Figure 2.12 and Figure E2.17. In thermodynamics we can often apply concepts developed to solve one type of problem to many other cases. T Reactants Products Δhrxn,T Δh1 Δh3 Δh°rxn,298 Reactants Products 298.15 K Δhrxn,T = Δh1 + Δh°rxn.298 + Δh3 298 Δh1 = − ∫R Σνi (Ai + BiT + CiT 2 + DiT −2 + EiT 3)dT T reactants T Δh3 = − ∫R Σνi (Ai + BiT + CiT 2 + DiT −2 + EiT 3)dT 298 products Figure E2.17 Calculation path of DhoT at temperature T from heat capacity data and the enthalpy of reaction at 298 K. c02.indd 91 05/11/12 1:35 PM Can you predict the signs of DU. and W ? Since the internal energy of an ideal gas is only a function of temperature.36) Applying the ideal gas relationship: nRT V5 P Positive (+) Negative (−) Zero (0) ΔU Q W Process Ideal gas Ideal gas Constant T Constant T reservoir reservoir Initial state (1) Final state (2) Figure 2.9). we can integrate over the system pressure (see Section 2. See if you can predict the signs of DU. c02. saving costs and resources.15 An ideal gas in a piston–cylinder assembly undergoing a reversible.7 REVERSIBLE PROCESSES IN CLOSED SYSTEMS One useful application of thermodynamics is in the calculation of work and heat effects for many different processes by applying the first law. A thermal reservoir contains enough mass so that its temperature does not noticeably change dur- ing the process. Q. We will look at nonideal gases in Chapter 5. The First Law of Thermodynamics ►2. and W for this process in the table. we go through two such examples of these types of calculations using an ideal gas undergoing reversible processes.15. isothermal expan- sion. The gas is kept at constant temperature by keeping it in contact with a thermal reservoir. Isothermal Expansion (Compression) Consider a reversible. the specific process must be defined in order to perform the necessary cal- culations. DU 5 0 For a reversible process.3): W 5 23 PdV (2. Q. Reversible. In this section.indd 92 05/11/12 1:35 PM . 92 ► Chapter 2. This information allows engineers to use energy more efficiently. The intent is to gain some experience with applying the first law to get values for work and heat as well as to develop expressions that are useful in understanding the Carnot cycle (Section 2. A schematic of a piston– cylinder assembly undergoing such a process is shown in Figure 2. Since heat and work are path dependent. isothermal expansion of an ideal gas. can you predict the signs of DU. 2. reversible expansion (as opposed to isothermal).38) and (2. for this case.44) where we applied the product rule.39) P1 Since P2 .15? How do Equations (2. From Equation (2. that T is constant): nRT dV 5 2 dP (2. This process is illustrated in Figure 2. Solving Equation (2.41) and for a reversible process: dW 5 2PdV (2.42) into Equation (2. and P: d 1 nRT 2 5 d 1 PV 2 5 PdV 1 VdP (2. we get: P2 Q 5 DU 2 W 5 2nRT ln (2. We will assume that the heat capacity of this gas does not change with temperature. since this process is adiabatic.43) and rearranging gives: cvVdP 5 2 1 cv 1 R 2 PdV 5 2cPPdV (2. the sign for W is negative and for Q is positive.41) and (2.14): 0 dU 5 dQ 1 dW (2.39) change if the gas under- goes a compression process instead of an expansion? Adiabatic Expansion (Compression) with Constant Heat Capacity Consider when the same ideal gas undergoes an adiabatic.7 Reversible Processes in Closed Systems ◄ 93 the differential in volume can be transformed into a differential in pressure (remember- ing. Q.43) We can use the ideal gas law to relate the measured properties T.40) yields: ncvdT 5 2PdV (2. that is. Did you get the sign right in the table in Figure 2.37) into Equation (2. P1.42) Substituting Equations (2. Again.40) where the heat transfer was set to zero.indd 93 05/11/12 1:35 PM .45) c02.24) we get: dU 5 ncvdT (2.36) and integrating gives: 2 nRT P2 W53 dP 5 nRT ln (2.44) for dT and then plugging back into Equation (2.37) P2 Substituting Equation (2.16. V.38) 1 P P1 Now applying the first law. constant heat capacity. the first law for a closed sys- tem in differential form is obtained from Equation (2. and W ? Neglecting macroscopic kinetic and potential energy. PVk 5 const (2.48) Now integrating for work: const 1 1 W 5 23 PdV 5 23 const V2kdV 5 B k21 2 k21 R k 2 1 V2 V1 1 nR 5 3 P2V2 2 P1V1 4 5 3 T2 2 T1 4 k21 k21 From the first law: 1 nR DU 5 W 5 3 P2V2 2 P1V1 4 5 3 T2 2 T1 4 k21 k21 c02.45): cP dV dP 2 5 (2. In this example. V2 P2 2k ln¢ ≤ 5 ln ¢ ≤ (2.46) cv V P Now we integrate Equation (2. Separating variables in Equation (2. cv is constant.indd 94 05/11/12 1:35 PM . See if you can predict the signs of DU.47) as: 2k V2 V2 V1 k 2k ln¢ ≤ 5 ln ¢ ≤ 5 ln ¢ ≤ V1 V1 V2 so.47) V1 P1 where k 5 cP/cv. 94 ► Chapter 2. ln 1 P1V1k 2 5 ln 1 P2V2k 2 or. and W for this process in the table. Well- insulated insulated Initial state (1) Final state (2) Figure 2.16 An ideal gas in a piston–cylinder assembly undergoing a reversible. The First Law of Thermodynamics Positive (+) Negative (−) Zero (0) ΔU Q W Process Ideal gas Ideal gas Cv ≠ Cv (T ) Well. Applying mathematical relationships of the natural logarithm. Q. adiabatic expan- sion. we can rewrite the left-hand side of Equation (2.46) from the initial state 1 to the final state 2. c02. Heat. However. all the energy delivered as work is provided by the surroundings in the form of heat. The isothermal expan- sion of an ideal gas follows Equation (2. adiabatic expansion of an ideal gas with constant heat capacity has g 5 k 5 cP/cv. the energy for work is provided by the internal energy of the gas in the system. Most cases will consist of one stream in and one stream out. each case represents a limit. respectively.50) can be rewritten in molar terms as: S S 2 2 0 5 n# 1 ¢ h 1 MW 1 MWgz ≤ 1 1 n# 2 ¢ h 1 MW V V 1 MWgz ≤ 2 (2. In both cases. These systems will be analyzed at steady-state. when the proper- ties at any place in the system do not change with time. for the adiabatic case. and Work for an Ideal Gas Undergoing a Reversible Process Isothermal Adiabatic.19) becomes: 0 5 m# 1 1 h^ 1 12V2 1 gz 2 1 1 m# 2 1 h^ 1 12 V2 1 gz 2 2 S S (2. which will be labeled streams 1 and 2. expansion of a piston provides useful energy to the surroundings in the form of work. Can you think of another example of a polytropic process? TABLE 2. 2.2.8 OPEN-SYSTEM ENERGY BALANCES ON PROCESS EQUIPMENT In this section.49) Both the processes in this section can be considered polytropic. In the isothermal process. For these cases the mass balance becomes: m# 1 5 m# 2 And the energy balance. we will examine examples of how to apply the first law to common types of process equipment.50) Equation (2. A process is defined as polytropic if it follows the relation: PVg 5 const (2. On the other hand. An intermediate case exists where there is some heat adsorbed from the sur- roundings as well as some “cooling” of the gas in the system.8 Open-System Energy Balances on Process Equipment ◄ 95 ►Summary A summary of the two cases presented in this section is shown in Table 2.50 molar) 2 2 where the molecular weight is used in the macroscopic kinetic and potential energy terms to convert from a mass basis to a molar basis.indd 95 05/11/12 1:35 PM . cv 2 cv 1 T 2 DU 0 nR 3 T2 2 T1 4 k21 Q P2 2nRT ln P1 0 W P2 nR nRT ln 3 T2 2 T1 4 P1 k21 ► 2.2 Summary of Expressions for Change in Internal Energy. Equation (2.49) with g 5 1 while the reversible. The First Law of Thermodynamics It is important to remember that the examples in this section are restricted to cases when steady-state can be applied.50): 0 0 0 0 0 S S # V 2 # V 2 # # 0 5 n1 (h 1 MW 1 MWgz)1 2 n2(h 1 MW 1 MWgz)2 1 Q 1 Ws (E2. This steady-state process occurs in an open system with one stream in and one stream out. B 5 0. What is the temperature of the air upon exiting the diffuser and entering the compressor? Calcualtion Calculation SOLUTION A schematic diagram of the system.15.000 m. we must use the unsteady form of the energy balance.18 The intake to the engine of a jet airliner consists of a diffuser that must reduce the air velocity Diffuser Final to zero so that it can enter the compressor.indd 96 05/11/12 1:35 PM . In a nozzle the flow is constricted.18A) 2 2 where the negligible terms have been set to zero. An example of a process calculation through a diffuser follows. we can write the first law using Equation (2.355.3. and D 5 20. T1 = 10° C T2 = ? → → V1 = 350 m/s V2 ≈ 0 m/s Figure E2. A mole balance gives: n# 1 5 n# 2 so that Equation (E2. Nozzles and Diffusers These process devices convert between internal energy and kinetic energy by changing the cross-sectional area through which a fluid flows. increasing eK. Note that the reference state for potential energy is set at 10.000 m where the temperature is 10ºC. If we are interested in start-up or shutdown of these processes.18B) 2 T1 Looking up the value for heat capacity for air in Appendix A.016 3 105 c02.575 3 1023.18 Schematic of the dif- Diffuser fuser in Example 2.18A) can be simplified to: T2 1 S2 eK 5 1 MW 2 V1 5 1 h2 2 h1 2 5 3 cP. we get: A 5 3. 96 ► Chapter 2.air dT (E2. A diffuser increases the cross-sectional area to decrease the bulk flow velocity. including the information that we know. is shown in Figure E2. or the case where there are fluctuations in feed or operating conditions. In this case.18. EXAMPLE 2. Consider a jet flying at a cruising speed of 350 m/s Temperature at an altitude of 10. 19. we get the following integral expression: T2 T2 B 2 D T2 3 cPdT 5 R 3 3 A 1 BT 1 DT 4 dT 5 R c AT 1 2 T 2 T d 22 (E2. They are commonly found in power plants and used to produce energy locally as part of chemical plants. 2. since they are compressible. This system is at steady-state. The term compressor is reserved for gases. 250 m above.4. rearranging. However.18C). Turbines and Pumps (or Compressors) These processes involve the transfer of energy via shaft work. # # # V2 Ws 5 m2 1 gz 2 2 5 gz2 v^ 2 (Continued) c02. their enthalpy is equal. SOLUTION Can you draw a schematic of this process? We need to write the energy balance. Pump Power Calculate the minimum power needed by the pump. When working with macroscopic potential energy. the exit temperature is the same as the inlet. T2. with one stream in and one stream out. 1 S 2 # # 0 5 m1 (h 1 V 2 1 gz)1 2 m2 (h 1 V 1 gz)2 1 Q 1 Ws 2 2 or.8 Open-System Energy Balances on Process Equipment ◄ 97 Using the definition of heat capacity. Typically they are used to raise the pressure of a fluid. Thus. therefore.001 m3 /s of water from a well to your house on a mountain. EXAMPLE 2. Equation (E2.18C) 283 T1 283 Using Equation (E2.indd 97 05/11/12 1:35 PM . they can also be used to increase its potential energy. Pumps and compressors use shaft work to achieve a desired outcome. which can be solved implicitly to give: T2 5 344 3 K 4 The temperature of the air increases because the kinetic energy of the inlet stream is being converted to internal energy. 1 S .19 You wish to pump 0. Since there are no frictional losses. the first law simplifies to: 0 0 0 0 0 0 . A turbine serves to generate power as a result of a fluid passing through a set of rotating blades. neglecting the friction between the flow- Calculation ing water and the pipe.1 and illustrated in Example 2. This process was described in Section 2.18B) becomes: 1 S B 1 1 1 MW 2 V12 5 RBA 1 T2 2 283 2 1 1 T22 2 2832 2 2 D ¢ 2 ≤R 2 2 T2 283 We now have one equation with one unknown. as illustrated in Example 2. it is often convenient to write the balance on a mass (rather than mole) basis. We will neglect the bulk kinetic energy of the water at the inlet and outlet and the heat loss through the pipe. The air must leave no cooler than 20ºC. In this case. liq.20 You plan to use a heat exchanger to bring a stream of saturated liquid CO2 at 0ºC to a Heat Exchanger superheated vapor state at 10ºC. energy is removed from the engine block to keep it from overheating during combustion.21.20A.5 3 kW 4 1 0. The radiator in your automobile is an example of a heat exchanger. Solving for shaft work gives: # 1 0. The flow rate of CO2 is 10 mol/min. The enthalpy of vaporization for CO2 at 0ºC is given by: Calculation Dh^ vap.20. The hot stream available Flow Rate to the heat exchanger is air at 50ºC. CO2 5 236 3 kJ/kg 4 at 0° C What is the required flow rate of air? SOLUTION First. An alternative design allows the fluids to be mixed directly.20A Schematic of heat exchanger with boundary 1 depicted. mol Boundary 1 nCO2 = 10 min Sat. Heat Exchangers These processes are designed to “heat up” or “cool down” fluids through thermal con- tact with another fluid at a different temperature.indd 98 05/11/12 1:36 PM . We next need to c02.8 3 m/s2 4 2 1 250 3 m 4 2 5 2. the heat transferred # from the air stream to evaporate and warm the CO2 stream is labeled Q. Why? The actual work needed would be greater due to frictional losses.20A. EXAMPLE 2. In this application. but not mass. A calculation on a system employing this design is given in Example 2. There are several possible choices for our system boundary. Vapor CO2 CO2 T1 = 0°C T2 = 10°C Q Air Air T4 = 20°C T3 = 50°C Heat exchanger Figure E2.001 3 m3 /s 4 2 Ws 5 B R 1 9. 98 ► Chapter 2. labeled “boundary 1” in Figure E2. An example of such an open feedwater heater is given in Example 2. let’s draw a diagram of the system including the information that we know.001 3 m3 /kg 4 2 Note that the sign for work is positive. can pass. The First Law of Thermodynamics # where V is the volumetric flow rate. We will choose a boundary around the CO2 stream. The most common design is when the two streams are separated from each other by a wall through which energy. shown in Figure E2. Thus.CO2dT 5 R c A 1 T2 2 T1 2 1 1 T 2 T 122 2 D ¢ 2 b R 5 353 c d T1 2 2 T2 T1 mol where the numerical values for the heat capacity parameters. the energy transferred via heat to boundary 1 is: # Q 5 100. The latent heat is given by: kJ kg J Dhvap. mol nCO2 = 10 min Sat. 2.400 c d kg kmol mol and the sensible heat is given by: T2 B 2 1 1 J 3 cP. A mole balance yields: n# 1 5 n# 2 5 n# CO2 so the first-law balance on boundary 1 simplifies to: # Q 5 n# CO2 1 h2 2 h1 2 To determine the change in enthalpy.20B. CO2dT T1 These can be found.20B Schematic of heat exchanger with boundary 2 depicted. which is labeled boundary 2 in Figure E2.3. that is: T2 1 h2 2 h1 2 5 Dhvap. as follows. liq. The appropriate energy balance is for an open system at steady-state with one stream in and one stream out is: 0 0 0 0 0 S S # V2 # V2 # # 0 5 n1 (h 1 MW 1 MWgz)1 2 n2(h 1 MW 1 MWgz)2 1 Q 1 Ws 2 2 where we have set the bulk kinetic and potential energies and shaft work to zero. B. We do this by choosing a different system boundary in the heat exchanger. and D.CO2 1 3 cP.753 3 J/min 4 Now that we know the rate at which energy must be supplied to the CO2 stream. A. are given in Appendix A. in [J/mol]. we can find the flow required for the air.8 Open-System Energy Balances on Process Equipment ◄ 99 perform a first-law balance around boundary 1.indd 99 05/11/12 1:36 PM .CO2 5 ¢236 c d ≤ ¢44 c d ≤ 5 10. (Continued) c02. Vapor CO2 CO2 T1 = 0°C T2 = 10°C Q Air Air T4 = 20°C T3 = 50°C Boundary 2 Heat exchanger Figure E2. we must account for the latent heat (vaporization) and the sensible heat of the CO2 stream. a mass balance at steady-state gives: 0 5 m# 1 1 m# 2 2 m# 3 (E2.3. Rearranging Equation (E2. so Equation (2. EXAMPLE 2.21C) We can look up values for the enthalpies from the steam tables (Appendix B). so: h^ 1 5 3238.21 Superheated water vapor at a pressure of 200 bar.indd 100 05/11/12 1:36 PM . an energy balance reduces to: # # # 0 5 m1h^ 1 1 m2h^ 2 2 m3h^ 3 (E2. 100 ► Chapter 2.21B) and substituting into (E2.21.21B) Rearranging Equation (E2. as shown in Figure E2. In that case.20A) # Note that we must be careful about signs! We have included a negative sign on Q since the heat that enters boundary 1 must leave boundary 2.21A) gives: 0 5 m# 1h^ 1 1 m# 2h^ 2 2 1 m# 1 1 m# 2 2 h^ 3 (E2. the superheated steam is at 500ºC and 200 bar 1 5 20 MPa 2 . The First Law of Thermodynamics A balance similar to that above yields: # 2Q 5 n# air 1 h4 2 h3 2 (E2. this problem could have been solved with a system boundary around the entire heat exchanger. a temperature of 500ºC. For state 1.50) does not apply. If we assume that the rate of heat transfer and the bulk kinetic energy of the streams are negligible and the bulk potential energy and shaft work are set to zero. a first-law balance would give: 0 5 n# CO2 1 h2 2 h1 2 1 n# air 1 h4 2 h3 2 which could then be solved for n# air. What flow rate is needed for the liquid stream? SOLUTION The first step is to draw a diagram of the system with the known information. Heater Calculation This process is accomplished by mixing this stream with a stream of liquid water at 20ºC and 100 bar. we get: 1 100.air dT 2 T4 T3 T3 Looking up values for the heat capacity parameters in Appendix A. This example has two inlet streams in. and a flow rate Open Feedwater of 10 kg/s is to be brought to a saturated vapor state at 100 bar in an open feedwater heater.2 3 kJ/kg 4 c02.21A) Similarly.20A) gives: # # # Q Q Q n# air 5 2 5 2 T4 5 1 h4 2 h3 2 B 2 1 1 RBA 1 T4 2 T3 2 1 1 T4 2 T 23 2 2 D ¢ 2 ≤ R 3 cP.753 3 J/min 4 2 n# air 5 5 123 3 mol/min 4 877 3 J/mol 4 Alternatively. 8 Open-System Energy Balances on Process Equipment ◄ 101 Subcooled liquid T2 = 20°C P2 = 100 bar Superheated Open Saturated vapor vapor feedwater P3 = 100 bar heater T1 = 500°C P1 = 200 bar kg m1 = 10 s Figure E2. Consequently. For state 2.50) on a mass basis. A steady-state energy balance with one stream in and one stream out is appropriate for this system. What is the exit Throttling Device temperature? Calculation SOLUTION First. the residence time of the passing fluid is small. We will assume that the bulk kinetic energy of the stream is negligible and that the porous plug is sufficiently small as not to allow a significant rate of heat transfer. Hence. let’s draw a diagram of the system.21C) and plugging in values gives: m# 1 1 h^ 1 2 h^ 3 2 kg m# 2 5 5 1. EXAMPLE 2. such as a partially opened valve or a porous plug. Since there is also no shaft work. there is little energy loss by the transfer of heat.7 3 kJ/kg 4 Finally. the energy balance reduces to a very simple equation.95 c d 1 h^ 3 2 h^ 2 2 s Throttling Devices These components are used to reduce the pressure of flowing streams. The pressure reduction can be accomplished by simply placing a restriction in the flow line. rearranging Equation (E2. 2. we get: S 0 0 S 0 0 0 0 # ^ V2 # ^ V2 # # 0 5 m1 1 h 1 1 gz 2 1 2 m2 1 h 1 1 gz 2 2 1 Q 1 Ws 2 2 (Continued) c02. we use subcooled liquid at 20ºC and 100 bar: h^ 2 5 93. as the next example illustrates. Since these devices occupy a relatively small volume.indd 101 05/11/12 1:36 PM .22.3 3 kJ/kg 4 and the saturated vapor at 100 bar (10 MPa) for state 3 is: h^ 3 5 2724. It exits at 1 bar. Rewriting Equation (2. as shown in see Figure E2.22 Water at 350ºC flows into a porous plug from a 10-MPa line.21 Schematic of the open feedwater heater in Example 2.21. we can neglect heat transfer. Hence. to explore the maximum possible efficiency that could be obtained by the steam engines of his time. c02.9 THERMODYNAMIC CYCLES AND THE CARNOT CYCLE A thermodynamic cycle describes a set of processes through which a system returns to the same state that it was in initially.3 11 In general. There are many different examples of ther- modynamic cycles. we will learn that a Carnot cycle represents the most efficient type of cycle we can possibly have. the energy balance reduces this system to an isenthalpic process:11 h^ 1 5 h^ 2 Looking up the value for the inlet stream from Appendix B.22 Schematic of the throttling device in Example 2. we have two intensive properties to constrain the exit state: h^ 2 and P2.3 250 2974. all the properties have the same values they had originally. we can repeat the cycle continuously. The advantage of executing a thermodynamic cycle is that by having the system return to its initial state.3 2 2875. a French engineer.3 Interpolation gives: h^ 2 2 h^ at 200 2923. The First Law of Thermodynamics T1 = 350°C P1 = 10 MPa P2 = 1 bar Porous plug Figure E2. in this section. we find: h^ 1 5 2923.indd 102 05/11/12 1:36 PM . 102 ► Chapter 2. Typically cycles are used to produce power or provide refrigeration.12 In Chapter 3. Inspection of the steam tables shows that T2 is somewhere between 200 and 250ºC.4 3 kJ/kg 4 Since the enthalpy of stream 2 equals that of stream 1.4. Since the system returns to its initial state after the cycle has been completed.3 T2 5 200 1 3 DT 4 B ^ R 5 200 1 3 50 4 c d 5 224 3 °C 4 hat 250 2 h^ at 200 2974. P 5 100 kPa T[ºC] h^ 3 kJ/kg 4 200 2875. 12 This cycle was conceived in 1824 by Sadi Carnot. The following are taken from the superheated steam table at 100 kPa. we examine one such cycle—the Carnot cycle.22.4 2 2875. To find the temperature of stream 2. we must use linear interpolation. ► 2. the energy balance of throttling processes reduces to this simple form. 9 Thermodynamic Cycles and the Carnot Cycle ◄ 103 QH Constant TH Isothermal expansion Adiabatic State 1 State 2 Adiabatic compression T1 T2 = T1 Well insulated expansion P1 P2 State 4 State 3 T 4 = T3 T3 State 1 State 4 State 2 State 3 P4 P3 T1 T4 T2 T3 P1 P4 P2 P3 Isothermal compression QC Constant TC Figure 2. as indicated on the diagram. alternating with two adiabatic processes.17.indd 103 05/11/12 1:36 PM . work is done by the system on the surroundings. In this step. The first step of a Carnot cycle is a reversible isothermal expansion. TH. Consider a gas that is initially in state 1 at a pressure P1 and a temperature T1 as shown at the top of Figure 2. The system returns to its initial state (state 1) through an adiabatic compression. in which the gas is exposed to a hot reservoir at temperature. These processes were analyzed. QC to the cold reservoir. it gains energy via heat. the sign of Wnet is negative.17 An ideal gas undergoing a Carnot cycle. T4 2 . both T and P decrease. the gas goes through four reversible processes through which it returns to its initial state. that is. in Section 2. Two processes occur isothermally. The Carnot cycle consists of four reversible processes by which the gas is returned to its original state. In both expansion processes. which takes the system from state 1 to state 2 (at P2 and T2 ). we get useful work out. During this process. 2.17 shows an ideal gas in a piston–cylinder assembly undergoing a Carnot cycle. The net work obtained in a Carnot cycle is given by the sum of the work obtained in all four processes: 2Wnet 5 0 W12 0 1 0 W23 0 2 0 W34 0 2 0 W41 0 (2. it is isothermally compressed by being placed in contact with a cold thermal reservoir at temperature TC. This process takes the system to state 4 1 P4. The system then undergoes two revers- ible compression processes. In this cycle. QH. individually.51) Since the overall effect of the power cycle is to deliver work from the system to the surroundings.7. The piston–cylinder assembly is then transferred into an adiabatic (well-insulated) environment and expanded further to state 3. First. Figure 2. The gas loses an amount of energy via heat. the pres- sure decreases while the temperature stays the same. The subscript “ij” on the terms for work in c02. 5 (2.indd 104 05/11/12 1:36 PM .18 Alternative representation for the Carnot cycle. (a) Carnot “engine”.17.52). (b) Carnot refrigerator. Thus. c02.51) and (2. The First Law of Thermodynamics Equation (2. TC (a ) (b ) Figure 2. TH QH QH Carnot Carnot Wnet Wnet engine refrigeration QC QC Cold reservoir. and expelled to the cold reservoir. Since the cycle returns the system to its original state. h. its internal energy must have the same value as at the start of the cycle. of the cycle is defined as the net work obtained divided by the heat absorbed from the hot reservoir: net work Wnet h. TH Hot reservoir. This schematic gives an overview of the energy transferred between the Carnot engine and the surroundings. DUcycle 5 0 5 Wnet 1 Qnet (2. we see that: QH 0 QC 0 2Wnet 5 Qnet 5 Q12 1 Q23 1 Q34 1 Q41 5 0 QH 0 2 0 QC 0 We see that the net work obtained is the difference in heat absorbed from the hot reservoir.52) Comparing Equations (2. The net work obtained from a Carnot cycle can also be calculated by applying the first law to the entire cycle. Inside the circle labeled “Carnot engine” are the four processes depicted in Figure 2. QH. Efficiency The efficiency. Absolute val- ues are used to explicitly distinguish the steps where we get work out from those where we must put work in. An alternative way of schematically representing a Carnot cycle is shown in Figure 2.51) refers to the work obtained in going from state i to state j.18a. 104 ► Chapter 2. QC.53) heat absorbed from the hot reservoir QH Hot reservoir. TC Cold reservoir. Can you draw the analogous cycle to Figure 2. isothermal compression at 300 K. which is defined as follows: QC COP 5 (2. your freezer at home needs electricity to keep the ice cream cold.13 A refrigeration cycle allows us to cool a system so that we can store some ice cream and so on.1 bar and 1000 K to 300 K. The effectiveness of a refrigeration cycle is measured by its coefficient of performance. Efficiency (i) A reversible. This gas undergoes a Carnot Carnot Cycle cycle.9 Thermodynamic Cycles and the Carnot Cycle ◄ 105 For a given amount of energy available from the hot reservoir via QH.54) that the higher the COP. and DU for each of the steps in the Carnot cycle. isothermal expansion at 1000 K from state 1 at 10 bar to state 2 at 0. For example. which is described below. in general. It takes work from the surroundings to accomplish this task. SOLUTION (a) We will analyze each of the steps separately. the direction of heat transfer is opposite that of the power cycle depicted in Figure 2. We label each state in a manner consistent with Figure 2.17. with a little help from the results of Section 2. we can see there was much engineering that remained to be done! c02. suggest two ways to make the above process more efficient. (iii) A reversible. We then expel the energy via heat QH to the hot reservoir. adiabatic compression from 300 K to 1000 K and 10 bar. We supply work to the cycle in order to absorb energy via heat QC from the cold reservoir. (b) Draw the cycle on a Pv diagram. COP. isothermal expansion from 10 bar to 0. In this case we want to expel heat from a cold reservoir. W. Figure 2. These first steam engines had efficiencies of only about 1%. (ii) A reversible. (e) Calculate the efficiency of the cycle. the greater the effi- ciency. (d) Compare h to 1 2 1 Tc /TH 2 .18b shows a schematic way of represent- ing the energy transferred in a refrigeration cycle. say the high temperature in the hot reser- voir is obtained from the combustion of coal. the more work we obtain. (e) If what is found in part (d) is true.7: nRT P2 W53 dP 5 nRT ln 5 238.54) Wnet We can see from Equation (2.indd 105 05/11/12 1:36 PM . Thus. the internal energy change for an ideal gas at constant temperature is: DU 5 0 We can calculate the work using a result from Section 2.23 Consider 1 mole of an ideal gas in a piston–cylinder assembly. Indeed.1 bar. cv 5 1 3/2 2 R.7. For example.17 that goes in the circle labeled “Carnot refrigerator”? EXAMPLE 2. The heat capacity is constant. 2. the less work it takes to produce a desired level of cooling. Perform the following analysis: (a) Calculate Q. By definition. (iv) A reversible.287 3 J 4 P P1 (Continued) 13 The steam engine was patented by James Watt in 1765. adiabatic expansion from 0.1 bar.18a. A high efficiency means we can reduce the amount of coal we need to combust to produce a given amount of work. (i) The first process is a reversible. 23A) is: 0. we know PVk 5 const for the polytropic. adiabatic processes (ii) and (iv). 106 ► Chapter 2. the work given by Equation (E2.49 bar W 5 nRT ln 5 11. We first find k: cP cv 1 R k5 5 5 1.67 1 nRT4 2 1. we apply the first law: QH 5 DU 2 W 5 38.486 3 J 4 0.indd 106 05/11/12 1:36 PM .67 1 nRT3 2 1.287 3 J 4 (ii) The second process is a reversible.1 bar and 1000 K to state 3 at 300 K.67 1.67 2 5 5 7347 5 P20.49 bar 341 Thus. To find the heat. we get: 1.67 cv cv Setting PVk equal for states 2 and 3 gives: 1 nRT2 2 1. Again: DU 5 0 and. the change in internal energy becomes: DU 5 ncv 1 T3 2 T2 2 5 28730 3 J 4 Applying the first law gives: W 5 DU 5 28730 3 J 4 (iii) The third process is a reversible.67 3 Solving for P3. adiabatic expansion from 0.67 PVk 5 P1V1. The First Law of Thermodynamics The negative sign indicates that the system is performing work on the surroundings (we are getting useful work out).67 1 5 5 341 5 P10. The pressure decreases during this process. nRT P4 W 5 23 PdV 5 3 dP 5 nRT ln (E2.5 P4 5 B R 5 0.7.0049 bar c02.5 1 nRT3 2 1. isothermal compression at 300 K.67 P3 5 B R 5 0. 1 nRT4 2 1.67 P40. we now need to find P3 and P4.67 P0. By the definition of an adiabatic process: Q50 At constant heat capacity.23A) P P3 However. From Section 2.0049 bar 7347 Similarly for P4: 1 nRT1 2 1.67 PVk 5 P2V 1.67 and. (c) The efficiency is given by Equation (2.730 0 (iii) State 3 to 4 0 11. Isotherms. (ii) .486 3 J 4 (iv) The fourth process is a reversible. the change in internal energy becomes: DU 5 ncv 1 T1 2 T4 2 5 8730 3 J 4 Applying the first law gives: W 5 DU 5 8730 3 J 4 TABLE E2. A sketch (not to scale) of the Pv diagram is presented in Figure E2. and v for Carnot Cycle in Example 2.730 28.8 kJ after one cycle. adiabatic compression from state 4 at 300 K and 0. 2.9 Thermodynamic Cycles and the Carnot Cycle ◄ 107 The work is positive for this compression process. The work for a reversible process is given by the area under the Pv curve.23B) heat absorbed from the hot reservoir 38.70 (E2.8314 3 300 0. Again.730 8.800 TABLE E2. After this process.800 26.287 In practice. electrical power plants have efficiencies around 40%.23. The results are presented in Table E2.53) net work 26.23A Results of Calculations for Carnot Cycle in Example 2.287 (ii) State 2 to 3 28. 5 5 0.23 Process DU 3 J 4 W [J] Q [J] (i) State 1 to 2 0 238.23 State T [K] P [bar] v 3 m3 /mol 4 1 1000 10 0. for this adiabatic compression: Q50 At constant heat capacity. . hence.52 bar back to state 1 at 1000 K and 10 bar (process 4 S 1).287 38. and Q for the four processes and the totals for the cycle are presented in Table E2.486 211.486 (iv) State 4 to 1 8.10 4 300 0. the gas can repeat steps (i).730 0 Total 0 226.1 0.800 h.indd 107 05/11/12 1:36 PM . the net work is given by the shaded area in the box in Figure E2. for processes (i) and (iii) are also labeled. QC 5 DU 2 W 5 211.49 0. TH and TC. W . We get a net work of 26.23A. P.23B.051 A summary of DU.0049 5. (b) To sketch this process on a Pv diagram. (Continued) c02. From the first law.23B T.23. we first calculate the molar volume at each state using the ideal gas law.0083 2 1000 0. . The shaded Isotherm area represents the net work v obtained from one cycle.indd 108 05/11/12 1:36 PM .23B) and (E2. larger. we can also reach this conclusion by realizing that the isotherms in Figure E2.23 are fixed on the Pv plane. Note that these options will push the isotherms depicted in Figure E2. in extensive form. written in extensive form. indeed. we get: TC h512 (E2. Energy balances have been developed for closed systems and for open systems.21) c02. ►2.23 up and down.12a) For open systems. Equation (E2.23D) is true in general.23D) holds. respectively. We will learn in Chapter 3 that.23 Pv diagram of TC 3 a Carnot cycle. the process can be made more efficient by raising TH or lowering TC. However.10 SUMMARY The first law of thermodynamics states that the total energy in the universe is a constant.7 (E2. (d) Applying the relation in the problem statement. is: DU 1 DEK 1 DEP 5 Q 1 W (2.23D) TH (e) If Equation (E2. we get: TC 300 12 5 12 5 0. is: dU dEK dEP 1 S # 1 S a 1 1 b 5 a m# in ah^ 1 V 2 1gzb 2 a m out ah^ 1 V 2 1gzb 1Q# 1 W# s dt dt dt sys in 2 in out 2 out (2. which represents net work. The First Law of Thermodynamics P 1 TH Isotherm Q H = −W Q=0 2 wnet Q=0 4 −Q C =W Figure E2. 108 ► Chapter 2. Thus either raising TH or lowering TC will serve to make the shaded box. the integral equation of the first law for a closed system. it is convenient to write the first law on a rate bases.23C) TH 1000 Comparing the values of Equations (E2. For exam- ple. The integral equation.23C). Changes in internal energy present themselves in several macro- scopic manifestations. u 1 Pv (2. Real processes are irreversible. pumps. h. you should understand the concepts well enough so that you are not restricted to the systems discussed above but rather are able to apply the first law to any system of interest. experiments conveniently done in closed systems at constant pressure are reported using enthalpy. In such cases. We can compare the amount of work required in an irreversible process to that of a reversible process by defining the efficiency fac- tor. and changes in chemical structure. is defined as: h .indd 109 05/11/12 1:36 PM . that is. The thermodynamic property enthalpy. chemical reaction. The reversible case represents the limit of what is possible in the real world—it gives us the most work we can get out or the least work we have to put in. Moreover. They have friction and are carried out with finite driving forces. the surroundings must be altered. only in a reversible process can we substitute the system pressure for the external pressure in calculating Pv work. the system can be returned to its original state without any net effect on the surroundings. Our strategy for actual.10 Summary ◄ 109 We have also developed these equations in intensive forms. A change in internal energy that is manifested as a change in temperature is often termed sensible heat. therefore. We applied the first law to many engineering systems. and throttling devices. For some processes.30) c02. reversible process and then correcting for the irreversibilities using an assigned efficiency factor. We must also identify whether the ideal gas model or property tables are needed to solve the problem. A process is reversible if. Steady-state open systems were illustrated by nozzles. we must account for both the internal energy of the stream and the flow work associated with it entering or leaving the system. and rotational. This result occurs only when the driving force is infinitesimally small. changes in phase. Enthalpy also describes the combined effects of internal energy and Pv work for closed systems at constant P. if the system is returned to its original state. in u with temperature to three possible modes in which the molecules can obtain kinetic energy: translational. For example. On a molecular level. it is convenient to define a hypotheti- cal path so that we can use available data to solve the problem. Given a physical problem. Enthalpy accounts for both these effects. In an irreversible process. internal energy encompasses the kinetic and potential ener- gies of the molecules. Transient open-system problems included filling or emptying of a tank. Examples of closed systems included rigid tanks and adiabatic or isothermal expansion/compression in a piston–cylinder assembly. 2. We can associate the increase in molecular kinetic energy and. heat exchangers. after the process occurs. experi- mental data for the energetics associated with phase changes and chemical reaction are typically reported using enthalpy. turbines. h. Data for the change in energy due to sensible heat can be obtained using heat capacity or using property tables. diffusers. However. including changes in temperature. The heat capacity for a species is often fit to a polynomial in temperature of the form: cP 5 A 1 BT 1 CT 2 1 DT22 1 ET 3 (2. we must determine which form to use and which terms in these equations are important and which terms are neg- ligible or zero. irreversible processes is often solving a problem for the idealized. Therefore. while the Carnot power and refrigeration cycles provided examples of thermodynamic cycles. on a mass and a molar basis. and for differential increments.18) We first recognized the usefulness of h in describing streams flowing into and out of open systems. vibra- tional. Process 1 Process 2 T1 P = const V = const T2 A A A A A A A A A A A A A A A A A A A A A A q q c02. the same amount of heat. q. using these values. The First Law of Thermodynamics The parameters A. The heat capacity of gas A is greater than the heat capacity of gas B. An ideal gas presents a special case: All its internal energy is realized as molecular kinetic energy—it is a function of T only. Process 1 occurs at constant volume. (a) Which gas has the greater final temperature? Explain. however. is supplied to equal amounts (in moles) of different gases.2 In the two processes shown in the following figure. Both gases are initially at room temperature. the same amount of heat. These processes take place at constant volume.11 PROBLEMS Conceptual Problems 2. process 2 occurs at constant pressure. is it convenient to use the property enthalpy? Explain. D.indd 110 05/11/12 1:36 PM . Heat capacity parameters at constant pressure of some liquids and solids are also reported in this appendix. The internal energy changes associated with chemical reactions can be attributed to the energetic differences between the chemical bonds of the reactants and the products. is supplied to the same gas. Which gas has the greater final temperature? Explain. fusion. (b) For which process. Data for latent heats are reported as enthalpies of vaporization. gas A. This energy is associated with the different degrees of attraction between the molecules in the different phases.2. In both processes. q. the initial temperature and initial number of moles are equal.1 In the two processes shown in the following figure. if any. gas A and gas B. We refer to a change in energy that results in phase transformations as latent heat. and E are reported for some ideal gases in Appendix A. B. Process 1 Process 2 T1 T2 V = const V = const A B A B A B A A B B A A B B A B A A A B A B q q 2. ►2. we can construct hypothetical paths to find the latent heats at any pressure. The enthalpy of a given chemical reaction can be determined from reported enthalpies of formation. These values are typically reported at 1 bar. 110 ► Chapter 2. C. and sublimation. 6 An ideal gas flows into a well-insulated tank that is initially at vacuum. Air in P1 Turbine T1 P2 Air out T2 2. Water T1 T2 P1 = 10 MPa P2 = 1 bar Porous plug 2. an ideal gas flows through a porous plug. stay the same. 2. Does the temperature increase. Does the temperature increase.3 In the steady-state process shown in the following figure. stay the same. Ideal Gas T1 T2 P1 = 10 MPa P2 = 1 bar Porous plug 2. Does the temperature increase. or decrease? Explain. steam flows through a porous plug. or decrease? Explain. or lower)? Explain.indd 111 05/11/12 1:36 PM . air flows through a turbine.11 Problems ◄ 111 2. as shown in the follow- ing figure. Ideal Gas Ideal Gas Tin Tin Pin Pin Ideal Gas Initially P2 = Pin Vacuum T2 State 1 State 2 c02. or decrease? Explain.4 In the steady-state process shown in the following figure. How does T2 compare to Tin (higher.5 In the steady-state process shown in the following figure. stay the same. the same. and its pressure drops from 10 MPa to 1 bar. and its pressure drops from 10 MPa to 1 bar. c02. M. In Process B. M. # # (a) 0 5 m# 1h1 1 m# 2h2 2 m# 3h3 1 Q 2 Ws (b) h2 2 h1 5 cP 1 T2 2 T1 2 (c) u2 2 u1 5 cP 1 T2 2 T1 2 (d) cP 5 cv 1 R (e) h 5 u 1 Pv (f) w 5 23 Pdv (g) Du 5 q 1 w (h) q 5 0 2. the value calculated in Example 2.indd 112 05/11/12 1:36 PM . is removed in incremental amounts. Put the final temperatures in order from highest to lowest. (c) Consider an irreversible process that occurs from the same initial state (State 1) to the same final state (State 2).0 m3 to a pressure of 100 bar. we solved a problem where 10.4? Explain. an equivalent mass.4? Explain. Weightless. 112 ► Chapter 2. or equal to. less than.0 kg of water was reversibly compressed in a piston–cylinder assembly from a pressure of 20 bar and a volume of 1. M. the heat was q 5 210 B R. The First Law of Thermodynamics 2. In Process C. Explain. Will the magnitude of work required for the compression be greater than. Qualitatively answer the following questions. frictionless piston frictionless piston Patm mT =M Well mT=M Well Insulated ∂m Insulated ∂m Patm Patm Process B M Process D Patm T1 TB M T1 P1 P1 TD State 1 State B State 1 State D 2. and kg kg the final temperature was T2 5 525°C.4. You do not have to do any calculations. state 1. is added in incremental amounts. less than. In Process A.8 State the conditions under which the following equations apply (try to be as specific as you can with the limitations). the value calcu- lated in Example 2. (a) Consider an adiabatic process that occurs from the same initial state (State 1) to the same final state (State 2). but you must choose the right answer and explain your reasoning.7 Four processes are shown in the following figure. an equivalent mass.4? Explain. Each process occurs in a well-insulated pis- ton cylinder assembly and starts from the same initial state. is removed from the piston. less than. or equal to. Weightless. In Process D. is placed on the piston. (b) Consider an isothermal process that occurs from the same initial state (State 1) to the same final pressure 1 P2 2 . Weightless. a large block of mass. or equal to) the value calculated in Example 2. Will the magnitude of work required for the compression be (greater than. Will the heat transfer be greater than. M. a large block of mass. the work was calculated to be w 5 285 B R.9 In Example 2. kJ kJ In this example. frictionless piston frictionless piston Well Patm Well Insulated M Insulated Patm Patm M M M Process A Process C Patm TA M T1 T1 P1 P1 TC State 1 State A State 1 State C Weightless. Now consider the dryer in your dwelling (or at the laundromat). we have seen that the temperature of a gas in a piston–cylinder assembly cools upon expansion. If you hold it to your lips. you have been told that it is more efficient to leave your house warm all day rather than turn off the heat during the day and reheat the house when you get home at night. 2. the surroundings.14 On a hot summer day. Illustrate the alternative processes. you are typically away from your house all day. However. 2. What is the efficiency of the process? Can you suggest any ways to make it more efficient? Numerical Problems 2. Determine the final pressure of the container. Draw a schematic of the system. less than.indd 113 05/11/12 1:36 PM . estimate how much energy is used to actually dry the towel. and the energy transferred by work and by heat during this process.12 Take a thick rubber band and expand it by stretching it. The degree to which the spring compresses is related to its spring constant. you will sense that it is hotter. it is important to maintain the body at this temperature.16 Explain why ice often forms on the valve of a tank of compressed gas (high pressure) when it is opened to the atmosphere and the gas escapes. The water undergoes a process in which it is heated to a final temperature of 540°C. Will the heat transfer be greater than. Estimate how much energy it takes to dry the towel after it has been washed. Come up with and sketch as many ways as you can think of to raise the temperature of the water. For normal metabolic pro- cesses. 2.20 The normal human body temperature is approximately 37°C. Explain these opposite results in the context of an energy balance. Hypothermia is a condition when the core body temperature drops below 35°C.18 Consider that towel you used to dry yourself with after your last shower. Choosing the entire apartment as the system. the value calcu- lated in Example 2. Explain this result. you get the paradox- ical result that at higher temperatures the water drops take longer to evaporate than at lower temperatures.11 Consider the compression of a spring by placing a large mass on it. State all assumptions that you make. perform a first-law analysis to decide whether this idea has merit. You think that thermodynamics may be able to resolve this issue. 2. k. The force exerted by the spring on the mass is given by: F 5 2kx where x represents the distance the spring compresses from its relaxed position. What is your choice to save power? Justify your answer.4? Explain. 2. 2. 2.17 (a) What requires more heat input: to raise the temperature of a gas in a constant-pressure cylinder or in a constant-volume bomb? Explain. However.10 Consider a cup of cold water. 2. 2.11 Problems ◄ 113 (d) Consider an irreversible process that occurs from the same initial state (State 1) to the same final state (State 2). You may assume the body contains 60% water. Where does the H2O come from? 2. (b) Explain why you feel less comfortable on a hot summer day when the (relative) humidity is higher even though the temperature is the same. However.19 A rigid container contains saturated water at a pressure of 2 bar and a quality of 0. and the boundary.42. it will evaporate. After being washed it must be dried so it can be used again. Does the com- pression of a spring represent potential or internal energy? 2. your roommate suggests that you open the refrigerator to cool off your apartment.15 You are making plans to stay warm in the winter.13 If you sprinkle water on a very hot skillet. Due to your busy schedule. c02. You know it costs a lot to operate the electric heaters to keep your house warm. If the towel is the only item placed in the dryer. or equal to. label it path B. until the forces again balance.26 A 5-kg aluminum block sits in your lab. 2. in J/kg. 2. A block of mass m 5 20. will the final temperature be greater than. and its initial height is 0.408 kg. (b) Suppose that the process is done isothermally.23 Consider a process that takes steam from an initial state where P 5 1 bar and T 5 400°C to a state where P 5 0. and estimate the amount of energy lost by heat for the onset of hypo- thermia. W. (i) Draw the path on the same Pv diagram. F = −kx m = 20. is transferred during the process? 2. During the compression process. for the process? What is the work done. Draw the area that represents the work.7172 m m1 = 0. Du.17 3 105 3 J/kg 4 of work is done on the system.indd 114 05/11/12 1:36 PM . 114 ► Chapter 2. The cross-sectional area of the piston is 0.2 kg State 1 (a) Precisely draw the process on a Pext-v diagram. A spring with a linear spring constant. Du.5 L of an ideal gas at 30ºC and 8 bar. Answer the following questions for each of the following reversible processes: (a) The first process is isothermal expansion. (ii) Calculate the following: w. (ii) Calculate the following: w.0 bar. Initially the spring exerts no force on the very thin m piston. (b) How much heat. How much heat in [J] must be supplied? c02.24 Consider a piston–cylinder assembly that contains 2. (b) The second process is heating at constant pressure followed by cooling at constant volume. equal to. Q? What is the rate of heat input. (b) ideal gas heat capacity. which is at 21ºC. N k 5 1.408 kg Patm = 1 bar H2O A = 0.7172 m above the base of the cylinder. q. (a) Write an energy balance for this process (you may neglect changes in potential and kinetic energy).50 m2. Dh. You may consider N2 to behave as an ideal gas. Dh.25 Five moles of nitrogen are expanded from an initial state of 3 bar and 88ºC to a final state of 1 bar and 88ºC.20 kg of pure water. The First Law of Thermodynamics (a) Consider a 70 kg person. q. or less than 30ºC? Explain. What is the total heat input. is attached to the piston. 6. What is the change in internal energy. Say it takes roughly 10 min to bring 1 L of H2#O taken from the tap at 25ºC to boil. as shown in the following figure.21 Consider the piston–cylinder assembly containing 0. The initial pressure is 1. Calculate the change in internal energy for this process using the following sources for data: (a) the steam tables. State any assumptions that you make. Q? 2. The gas reversibly expands to 5 bar. (b) What mechanisms does your body have to allow more heat to be lost than that estimated in part a before the onset of hypothermia? 2. is then placed on the piston. during the process? What is the heat transferred. You wish to increase the tempera- ture of the block to 50ºC. (i) Draw the path on a Pv diagram and label it path A. 2.62 3 105 B R.5 bar and T 5 200°C.22 Consider boiling water to make a pot of tea.5 m2 h1 = 0. Q? (c) If the process is done adiabatically (instead of isothermally). causing the gas to compress. DU. you come across the following experiment that was performed on a closed system containing pure gas A within a rigid container. As a known amount of heat. The left side con- tains an ideal gas 3 cP 5 30 J/ 1 mol K 2 4 at 10 bar and 300 K. The right side contains nothing. filling the entire volume.740 [J] of work. The system is well insulated.33 You have 0.4 mol of gas A within a piston–cylinder assembly.11 Problems ◄ 115 2. Determine the final temperature.28. It undergoes a process whereby it expands against a constant pressure of 20 bar. (a) What is the final temperature of the system? (b) How much work is done during the process? You may assume gas A behaves as an ideal gas during this process. The initial pressure is 20 bar. After the pin is removed. Its initial volume is 10 L and initial pressure.indd 115 05/11/12 1:36 PM .) 2. well-insulated 2 -m3 tank into two equal parts. The compartments are separated by a well-insulated piston that is held in place by a restraining pin. A. but no heat is transferred through the piston. and eventually the temperature in the tank equalizes. the temperature. Calculate the value for the work. and the initial temperature is 675 K. One compartment is initially at a pressure of 1 bar and the other is at 5 bar. pressure and volumes of each compartment. During the process. The container was connected to a heat source (in this case a resistance heater) and was otherwise well insulated. 2 bar. describe the process by which you can obtain the maximum work from the system. the piston moves. T.3 350 735. A small hole forms in the membrane.30 The insulated vessel shown below has two compartments separated by a membrane.31 A membrane divides a rigid.27 A piston–cylinder assembly contains 3 kg of steam at a pressure of 100 bar and a temperature of 400ºC.28? 2. The system then undergoes an adiabatic process where the piston expands against constant external pressure of 1 bar until the forces balance. Q.32 A well-insulated rigid container contains two 10 L compartments initially at 300 K. On one side is 1 kg of steam at 400ºC and 200 bar. (Thanks to Prof. was provided through the resistive heater. H2O T1 = 400°C Vacuum P1 = 200 bar Insulation 2. Each compartment contains argon gas. The gas is allowed to expand against a constant external pressure of 1 bar until it reaches mechanical equilibrium. In checking the lab notes for data for gas A. The membrane ruptures. of the system was measured. 2. Is this a reversible process? What is the final temperature of the system? How much work was obtained? For gas A: cV 5 1 5/2 2 R. The final pressure is 100 bar. 2.28 Consider a piston–cylinder assembly that contains 1 mole of ideal gas. Water is not an ideal gas under these conditions.8 c02. What is the final temperature? Why is T lower than that calcu- lated in Problem 2. The other side is evacuated. The fol- lowing data were collected for one mole of species A: T [K] Q [J] 293 0 300 30. the piston generates 748. it is a vacuum.29 For the well-insulated piston–cylinder assembly containing 1 mole of ideal gas described in Problem 2. until the forces balance. Determine the final temperature of the steam and the volume of the vessel. Determine the final temperature in K and the heat transferred (in [J]) during the process. 2. gas slowly leaks out from the left side. Frank Foulkes for providing the idea for this problem. What is the final temperature? What is the final pressure? 2. 35 Repeat Problem 2. Conceptually explain your answer.” The heat capacity of a protein in solution can be measured using a system similar to the one schematically shown in Figure 2. the protein unfolds and becomes “denatured. which include van der Waals interactions and hydrogen bonding (which we will discuss further in Chapter 4).9 500 2505. All other conditions remain the same as in Problem 2. the protein folds back on itself and is held together by a large number of intramolecular interactions. A plot of heat capacity versus temperature for a protein is shown in the following figure. 50 45 40 cp [kJ/(mol K)] 35 30 25 20 15 40 50 60 70 80 T °C Answer the following questions: (a) Physically explain.0 600 3915. where is the protein in its native state where is it denatured.6 650 4498.4 450 1933. how do you think the value you calculated in part B will change? (More Q.1 700 5155. The First Law of Thermodynamics 400 1240.10b. Under conditions in which the inter- actions are overcome. except consider now the combustion products contain CO (g) as well as CO2 (g) and H2O (g) with the ratio of CO2 / CO produced being 4/1. State any assumptions that you make in solving this problem. what is the cause of the “hump”? c02. It undergoes an isothermal process in which it reacts and completely combusts to form CO2 (g) and H2O (g).34. The backbone of a protein is a long polypeptide chain. In its natural or “native” state. Take the heat capacity of liquid n-octane to be: J cP 5 250 B R mol K (a) What is the final system pressure? (b) How much heat must be removed to keep the chamber at 100°C? (c) Consider the same amount of C8H18 reacts with a stoichiometric amount of O2 and forms the same products. No calculations! 2.7 2. 2.36 Approximately half the dry mass of the human body consists of proteins.indd 116 05/11/12 1:36 PM . No change in Q).5 550 3248.34 One kg of liquid n-octane 1 C8H18 2 is placed in a closed rigid container with 50% excess air at 100°C and 1 bar. Less Q.34. 116 ► Chapter 2. The resulting structure formed by the protein is important for its function. indd 117 05/11/12 1:36 PM . the solid oxide fuel cell.39 Consider the piston–cylinder assembly containing a pure gas shown below. They are based on producing energy by the following reaction: 1 H 2 1 g 2 1 O2 1 g 2 S H 2O 1 g 2 2 One type of fuel cell. Initially the spring exerts no force on the very thin piston. How much heat was exchanged? (c) Can you think of a process by which the system could go from the initial state to the final state with no net heat exchange with the surroundings? Describe such a process. Initially the gas has a pressure of 20 bar and occupies a volume of 1.32 mol/s H2 and 0.0 m3.) (a) Explain physically what the first and second terms on the right-hand side of the preceding heat rate equation represent.2 3 1028 1 T4 2 T40 2 where T is in K and T0 is the ambient temperature.5 5 constant What is the final temperature and internal energy of the system? Sketch this process on a Pv dia- gram.1 m2 A = 0. the initial pressure is 1 bar. Under these conditions. Dhd.1 1 T 2 T0 2 2 4. which can be taken to be 293 K. and the cross-sectional area of the piston is 0. (a) The system now undergoes a reversible process in which it is compressed to 100 bar.1 m2 Process A Process B c02. and sketch it on a Pv diagram.37 Fuel cells are a promising alternative energy technology. 2. The heat loss rate (in W) is given by: # Q 5 27.38 Consider a piston–cylinder assembly containing 10 kg of water.16 mol/s O2. water does not behave as an ideal gas. (c) What would cP versus T look like if denaturation occurred in a two-step process? 2. How much heat was exchanged? (b) Consider a different process by which the system gets to the same final state as in part (a). and the reaction goes to completion. Jason Keith for providing the information for this problem. The initial vol- ume of the gas is 0.05 m3. operates at high temperatures. forcing it to compress. Explain what physical process this number represents. (b) Calculate the temperature at which the rate of heat loss is equal to the rate of heat generation. putting in numerical values wherever possible. Draw the area that represents the work for this process. a large block is placed on the piston. A solid oxide fuel cell is fed 0.11 Problems ◄ 117 (b) Estimate the enthalpy change associated with the denaturation process.1 m2. Calculate the work done during this process. 2. The pressure–volume relationship is given by: Pv1. In this case. F = − kx F = − kx m = 1020 kg m = 2040 m = 1020 kg Patm Patm kg A = 0. Calculate the work done during this process. 2. Sketch this process on a Pv diagram. Draw the area that represents the work for this process. (Thanks to Prof. The First Law of Thermodynamics (a) A block of mass m 5 2040 kg is then placed on the piston. The cross-sectional area is 0.B = 10 bar T1. contain H2O and are separated by a thin metallic piston.41 Consider the well-insulated.” Draw the area that represents the work for this process. and the heat transferred. 2. DU. and the piston moves until the pressure and temperature in the two compartments become equal. Q.indd 118 05/11/12 1:36 PM . Surroundings Temperature bath TB = 200°C P = 600 kPa Pure H2O V = 0.] (b) Consider instead a process in which a block of mass 1020 kg is placed on the piston in the original initial state. Latch Thin metallic piston Well-insulated wall A B P1.03 m3.1 m2. the final pressure. Consider the tank and the piston–cylinder assembly as the system and the constant- temperature bath as the surroundings. the change in internal energy. 95% of the mass of water is vapor).1 m3.5 m3 Quality = 95% H2O Vinitial = 0.” Draw the area that represents the work for this process.5 m3 is connected to a piston–cylinder assembly by a valve. Two compartments. rigid container shown below. What is the value of the work? [Hint: First find the value for the spring constant. W. by the piston. Draw the process on the same PV diagram. The rigid tank contains saturated water with a quality of 95% (i. the right com- partment is initially at 10 bar and 700ºC. The final volume is 0. The piston is initially held in place by a latch. and the distance that the piston moved.A = 250°C T1. Determine the final temperature. You may assume that the force exerted by the spring on the piston varies linearly with x and that the spring is very “tight. The left compartment is initially at 20 bar and 250ºC. Side B is 50 cm long. Precisely draw the process on a PV diagram. What is the value of the work? 2. A and B. k. What is the value of the work? (c) Describe the process in which it will take the least amount of work to compress the piston. of the system. and the final pressure is 2 3 105 Pa. The piston–cylinder assembly initially has a volume of 0.” so that the volume of the gas is never less than the final volume. and after the gas inside has been compressed another block of mass 1020 kg is placed on the piston. The water flows into the piston–cylinder until equilibrium is obtained. Both vessels contain pure water.A = 20 bar P1.” Draw the area that represents the work for this process. as shown below.e.. labeling it “process A. 118 ► Chapter 2. They are immersed in a constant-temperature bath at 200ºC and 600 kPa. determine the work done.40 A rigid tank of volume 0. Initially the valve is closed and both units are in equilibrium with the surroundings (the bath). Draw the process on the same PV diagram. Side A is 10 cm long. For this process. The latch is removed.1 m3 The valve is then opened. labeling it “process B.B = 700°C H2O H2O 10 cm 50 cm c02. labeling it “process C. The process takes place adiabatically. exposing the tank to a 50 bar line at room temperature until the pressure of the cylinder Argon 50 bar Pinitial = 10 bar c02. You look at your watch.44 You have a rigid container of volume 0.M. The valve is then opened.11 Problems ◄ 119 2. You may assume the outside temperature stays constant at 2ºC from 6:00 P. To accomplish this task. Seeing as you will probably be there all night and you need a diversion. much to the chagrin of a busy class of chemical engineers who have an outrageously long computer pro- ject due the following day. to 6:00 A.01 m3 that you wish to contain water at its critical point.e. The valve is opened and the tank fills with steam until the pressure is 6 MPa. Take CO2 to be an ideal gas.. however. Connected to this pipe through a valve is a tank of volume 0. You may take the heat transfer to be given by the following expression: q 5 h 1 T 2 Tsurr 2 where h is a constant. you start with pure saturated water at 1 bar and heat it.).42 When you open a can of soda (or beer). Well. The circulation fans will stay on.M. (b) Calculate the temperature at 6:00 A.M. it has been determined that the steam in ChE Hall will be shut down at 6:00 P. and the temperature has already fallen halfway from the comfortable 22ºC it was maintained at during the day to the 2ºC of the outside temperature (i. 2. 400ºC is flowing in a pipe.M. This tank initially contains saturated water vapor at 1 MPa. 2. and then the valve is closed. 2. Determine the temperature in the tank right as the valve is closed. you decide to estimate what the temperature will be at 6:00 A.45 In an attempt to save money to compensate for a budget shortfall.M. the cylinder contains 10 bar of argon at room temperature. the temperature is 12ºC at 10:00 P. (a) Plot what you think the temperature will be as a function of time. with a constant heat capacity of cP 5 37 3 J/ 1 mol K 2 4 . leaving the entire building at approximately the same temperature. Explain. compressed CO2 expands irreversibly against the atmosphere as it bubbles up through the drink.M. Does the value you calculate account for all the work that is required? Explain. (a) How much water do you need? (b) What is the quality of water that you need to begin this process? (c) How much heat in [J] will it take? 2. Assume that the process is adiabatic and that the CO2 has an initial pressure of 3 bar.43 Find the Pv work required to blow up a balloon to a diameter of 1 ft.4 m3. What is the final temperature of the CO2 after it has reached atmospheric pressure? 2.. and turned back on at 6:00 A. it is already 10:00 P.indd 119 05/11/12 1:36 PM . things aren’t going as quickly as you might have hoped and it is getting cold in the computer lab. 2.M.46 Steam at 6 MPa.47 Consider filling a cylinder of compressed argon from a high-pressure supply line as shown below.M. Before filling. PE = 1 bar Valve CO2 P = 3 bar. At this point the pressure of the pipe and tank are equal. as shown below. you connect a well-insulated tank of volume 0. Determine the temperature (in [K]) of the steam flowing in the pipe. (a) What is the temperature right after the valve is closed? (b) If the cylinder sits in storage for a long time.1 m3? How much CO2 has entered the tank? Take CO2 to be an ideal gas. For argon take cP 5 1 5/2 2 R and the molecular weight to be 40 kg/kmol. with a constant heat capacity of cP 5 37 3 J/ 1 mol K 2 4 . It initially contains saturated water at a temperature of 200ºC and a quality of 0.4.48 A well-insulated piston–cylinder assembly is connected to a CO2 supply line by a valve. The First Law of Thermodynamics reaches 50 bar. What is the temperature of the CO2 when the volume inside the piston–cylinder assembly reaches 0. You may assume the steam in the pipe stays at the same temperature and pressure throughout this process. The top of the tank contains a pressure-regulating valve that maintains the vapor at constant pressure.4 m3 to this pipe through a valve. 120 ► Chapter 2. The valve is opened. You may use the ideal gas model. How much heat (in [kJ]) is required? You may assume that there is no pressure drop in the exit line. This tank initially is at vacuum. c02.49 A rigid tank has a volume of 0. The process takes place adiabatically. This system undergoes a process whereby it is heated until all the liquid vaporizes. The valve is then closed.50 You wish to measure the temperature and pressure of steam flowing in a pipe. and no more steam flows through the valve. The temperature right after the valve is closed is measured to be 800ºC.01 m3. how much heat is transferred (in kJ/kg)? (c) What is the pressure of the cylinder when it is shipped (after it was stored for a long time)? 2. T = 283 K 2. and the tank fills with steam until the pressure is 9 MPa. The valve is then closed. Initially there is no CO2 in the piston–cylinder assembly. To do this task. and CO2 flows in. Value maintains pressure in system constant v T1 = 200°C x1 = 0.indd 120 05/11/12 1:36 PM .4 V = 0. The valve is then opened.01 m3 Il 2. in °C? Under these conditions. 2. in m3 /s.indd 121 05/11/12 1:36 PM . You may assume the pressure of each stream stays constant as it flows through the heat exchanger (i.52 Methane vapor enters a valve at 3 bar and 25°C and leaves at 1 bar. neglect the pressure drop of the flowing streams). 2. an ideal gas enters a compressor with a mass flow rate of 10.60 1 0. required of the inlet steam to keep it from condensing at the exit? mol nco = 10 2 s Heat Exchanger Tout = 300 [°C] Tin = 150 [°C] CO2 CO2 Pin = 1 [bar] Steam Steam Tin = 400 [°C] Pin = 40 [bar] Insulation 2.53 You wish to heat a stream of CO2 at pressure 1 bar.0°C.51 An ideal gas undergoes an adiabatic. To do this task. The gas exits at 25. from 150°C to 300°C in a countercurrent heat exchanger.00 bar and the inlet temperature is 25. The entire system is well insulated. It is undesir- able for the steam to condense in the heat exchanger tubes. The ideal gas heat capacity is given by: cP 5 3. what is the exit temperature. as shown. reversible expansion process in a closed system: (a) If cp is constant.0 [mol/s].0°C. c02.500 3 1023 T R where T is in [K].54 At steady-state.e. as shown in the following figure. The inlet pressure is 1. you may assume methane is an ideal gas.11 Problems ◄ 121 Steam Steam T=? T=? P = 9 MPa P = 9 MPa H2 O Initially P2 = 9 MPa vacuum T2 = 800°C State 1 State 2 2.0 bar and 65. If the methane under- goes a throttling process.. What is the minimum volumetric flow rate. flowing at 10 mol/s. you have been asked to use a stream of high- pressure steam available at 40 bar and 400°C. show that: k/k21 T2 P2 5¢ ≤ T1 P1 (b) Determine the relationship between temperatures and pressures for an ideal gas if the heat capacity is given by: cP 5 A 1 BT 1 CT2 2. 2. It leaves at 200°C and 1 bar. state with negligible velocity at 4 MPa.55 Answer the following questions: (a) Argon gas enters a heat exchanger at a volumetric flow rate of 4. there is a finite amount of heat transfer. Compute the heat supplied into the unit per mole of ethylene that passes through. The gas mixture exits at 200°C and 1 bar. 6 mol/s of the benzene leaves as vapor. (c) A stream of argon gas flowing at 4. or less than that calculated in Part A? Explain. where 0. c02. calculate the power (in kW) required.0 3 m3 /min 4 .58 An electric generator coupled to a waterfall produces an average electric power output of 5 kW. The power is used to charge a storage battery. (b) Isobutane gas enters a heat exchanger at a flow rate of 3. equal to. (in [kJ]) in 10 hours of operation.0 3 m3 /min 4 . You may assume ideal gas behavior. what is the average velocity (in m/s) of the water? Where does the rest of the energy go? 2. Determine the power developed by the turbine (in kW). will the actual power required be greater than. (b) In the real process. determine the rate of heat input required (in W).6 m.5 bar. and the rest leaves as liquid. Compute the heat supplied.59 Air enters a well-insulated turbine operating at steady. You may assume ideal gas behavior. Compute the heat supplied. You may assume ideal gas behavior. The exit velocity and temperature are 90 m/s and 100ºC. 122 ► Chapter 2.0 3 m3 /min 4 .5 bar. 2. Air in V1 = 0 Turbine P1 = 4 MPa T1 = 300°C V2 = 90 m/s Air out P2 = 100 kPa T2 = 100°C 2. respectively.indd 122 05/11/12 1:36 PM . If the compressor operates between the same initial state and final state as in Part A. in W. the gas strips water from the liquid.57 In a biological system. in W.0 3 m3 /min 4 .56 A stream of pure liquid benzene flowing at a rate of 10 mol/s at 40°C and 10 bar enters a vessel where it is flashed. 100°C and 2 bar is mixed at steady-state with a second stream containing isobutane gas flowing at 3.60 Ethylene 1 C2H4 2 at 100ºC passes through a heater and emerges at 200ºC. (a) Determine the total amount of energy stored in the battery.765 J/mol. The air expands to an exit pressure of 100 kPa. Heat transfer from the battery occurs at a constant rate of 1 kW. The First Law of Thermodynamics (a) Assuming the compressor is adiabatic. At steady-state. N2 is often bubbled through a fermentor to maintain anaerobic con- ditions. The flash vessel operates at a pressure of 5 bar.5 g/min of water are evaporated. As the N2 bubbles through the fermentor. Compute the heat supplied. a temperature of 100°C and a pressure of 1. (a) Does heat need to be added of removed to compensate for the water evaporation? (b) What is the resulting heat load on the fermentor? 2. in W. 100°C and 1.1°C and has an enthalpy of vaporization of 30. (b) If the water flow rate is 200 kg/s and conversion of kinetic energy to electric energy has an efficiency of 50%. It leaves at 200°C and 1 bar. a temperature of 100°C and a pressure of 2 bar. The diameter of the exit is 0. Consider an isothermal continuous fermentation process operated at 30°C. 300ºC. 2. You may assume air behaves like an ideal gas throughout the process. You may assume ideal gas behavior. benzene boils at 80. At 1 atm. What is the steady-state exit velocity? What is the outlet cross-sectional area? 2. as shown. Inlet Sensor tube Main tube Valve Outlet Gas in Gas out Upstream temperature Valve control sensor (T1) Downstream Power Electronics temperature supply sensor (T2) (a) Flow rates are typically reported as standard cubic centimeters per minute (SCCM). The outlet is at 10 bar. which represents the volume the gas would have at a “standard” pressure of 1. 2. (a) At what rate must heat be added (in kW) to raise the stream to 500ºC? (b) Consider the same molar flow rate of n-hexane and the same rate of heat input calculated in part (a). a constant amount of heat is provided to the heating coil. You may assume air behaves as an ideal gas.267 1 5. The exhaust temperature is 308ºC. The process occurs at steady- state. At steady-state.61 Propane at 350ºC and 600 cm3 / mol is expanded in a turbine. to which a constant fraction of the flowing gas is diverted.66 A stream of air is compressed in an adiabatic. It consists of a main tube and a sensor tube. 2. It exits at a velocity of 500 m/s.indd 123 05/11/12 1:36 PM .5 bar. What is the exit temperature? What is the exit velocity? 2. does 1 SCCM correspond to? c02.65 Consider a diffuser operating at steady-state with an outlet twice the area of the inlet.64 Propane enters a nozzle at 5 bar and 200ºC. 2. 2. The outlet is at 1. and the exhaust pressure is atmospheric.68 A mass flow controller (MFC) is used to accurately control the molar flow rate of gases into a system. What is the work obtained? You may assume ideal gas behavior. It exits as saturated vapor at 100 kPa. The inlet is at 300 K and 1 bar.5 bar. Without doing any calculations. explain whether you expect the final temperature of n-hexane to be greater or less than 500ºC. A control valve can then be opened or closed to ensure the desired flow rate. 2.0135 bar and a “standard” temperature of 0ºC.62 Consider 20 mol/s of CO flowing through a heat exchanger at 100ºC and 0. You may assume ideal gas behavior.11 Problems ◄ 123 2. A schematic of an MFC is shown below.324 3 1023T R (a) What is the molar flow rate of the exit stream? (b) What is the temperature of the exit stream? 2.67 Sulfur dioxide 1 SO2 2 with a volumetric flow rate of 5000 cm3 /s at 1 bar and 100ºC is mixed with a second SO2 stream flowing at 2500 cm3 /s at 2 bar and 20ºC. Estimate the minimum power that the compressor uses. Air flows in with a velocity of 300 m/s. The mass flow rate is 1 kg/s. take the heat capacity at constant pressure to be: cP 5 3. and a temperature of 70ºC. a pressure of 1 bar.63 Steam enters a well-insulated nozzle at 10 bar and 200ºC. The temperature difference is measured by upstream and downstream temperature sensors. steady-state flow process at 50 mol/s. what is the exit temperature? Assume ideal gas conditions. For SO2. What molar flow rate (in [mol/s]). In the sensor tube. For ice. 2.72 Compare the change in internal energy for the following two processes: (a) Water is heated from its freezing point to its boiling point at 1 atm.5 kg of saturated vapor at 10 kPa. The ice is originally at 210°C when removed from the freezer and put in the glass. (b) adiabatic. rotational. 2. Assume the glass is adiabatic and that thermal equilibrium is obtained. and the fraction of gas diverted to the sensor tube. where it exits at 100 kPa. initially saturated liquid at 10 kPa.76 Consider the cooling of a glass of tap water by the addition of ice. Consider controlling SiH4 instead of N2. State the assumptions that you make. 2. and vibrational) in which molecules can exhibit kinetic energy.indd 124 05/11/12 1:36 PM .2. each in kJ. Determine the work and the heat transfer for the pro- cess. to which 100 gm of ice is added. The glass contains 400 ml of tap water at room temperature. 2.68 bar.70 Steam at 8 MPa and 500ºC flows through a throttling device.66 3 kJ/mol 4 You may treat propane as an ideal gas. 124 ► Chapter 2.75 A rigid vessel contains 5 kg of saturated liquid water and 0.69 Using data from the steam tables. are heated to saturated vapor while the pressure is maintained constant. Dhvap 5 16. At 0ºC. and NH3 at 300 K. (b) Saturated liquid water is vaporized at 1 atm. Repeat for the change in enthalpy. Account for their relative magnitudes in terms of the three modes (translational.77 One mole of saturated liquid propane and 1 mole of saturated vapor are contained in a rigid container at 0ºC and 4. How much heat must be supplied to evaporate all of the propane. ideal gas expands reversibly from 500 kPa and 300 K to 100 kPa in a piston– cylinder assembly.0 3 kJ/mol 4 . What amount of heat must be transferred to the system to evaporate all the liquid? 2. the heat input to the heating coil. come up with an expression for the ideal gas heat capacity of H2O in the form: cv 5 A 1 BT Compare your answer to the values in Appendix A. (a) What is the final temperature? (b) What % of the cooling is achieved by latent heat (the melting of ice)? 2.73 Calculate the values of the heat capacity of Ar. Dhfus 5 26. What is the exit temperature? 2.78 Calculate the enthalpy of reaction at 298 K for the following reactions: (a) CH4 1 g 2 1 2O2 1 g 2 S CO2 1 g 2 1 2H2O 1 g 2 (b) CH4 1 g 2 1 2O2 1 g 2 S CO2 1 g 2 1 2H2O 1 l 2 (c) CH4 1 g 2 1 H2O 1 g 2 S CO 1 g 2 1 3H2 1 g 2 (d) CO 1 g 2 1 H2O 1 g 2 S CO2 1 g 2 1 H2 1 g 2 (e) 4NH3 1 g 2 1 5O2 1 g 2 S 4NO 1 g 2 1 6H2O 1 g 2 c02. 2. O2. 2. (c) Instead of recalibrating the MFC for any gas that is used.74 Two kilograms of water. The First Law of Thermodynamics (b) Consider controlling the flow of N2. What conver- sion factor must be applied? 2. Calculate the work done if this process is (a) isothermal. conversion factors allow you to cor- rect the MFC readout for different gases. Develop an equation for the molar flow rate of N2 in SCCM in terms of the measured temperature difference.71 A monatomic. The state of each stream is labeled on the plot and defined in the table below. You may assume all of the methane that does react forms H2O and CO2. Q.82 The following data are from a system that undergoes a thermodynamic cycle among three states. and W. The reactants are initially at 298 K. 2.075 100 Quality 90% sat. 2. Assume that the acetylene reacts completely to form CO2 and H2O: (a) It is combusted in a stoichiometric mixture of O2. 2.84 A Rankine cycle is shown below. a condenser. (c) It is combusted with twice the amount of air as needed stoichiometrically. The hot reservoir is at 800ºC and the cold reser- voir is at 25ºC. and a boiler. Determine the net work produced and the efficiency of the cycle. If the reactants are fed into the reactor at 25ºC and 1 bar and the exiting gases leave at 1000ºC and 1 bar. The reactants are initially at 298 K.2 bar and 1 bar. This cycle is used to generate power with water as the working fluid. Fill in the missing values for DU. vapor liquid c02.79 Calculate the adiabatic flame temperature of acetylene gas at a pressure of 1 bar under the following conditions.2 bar and 60 bar.075 0.11 Problems ◄ 125 2. the reaction does not proceed to completion. sat. Compare the answers and comment. The pressure ranges between 0. Because of faulty operation of the adiabatic burner. Is this a power cycle or a refrigera- tion cycle? Process DU 3 kJ 4 W [kJ] Q [kJ] State 1 to 2 350 State 2 to 3 800 800 State 3 to 1 2750 2500 2. Kinetic and potential energy effects are negligible.81 In an experiment. Determine the coef- ficient of performance. The mass flow rate of water is 100 kg/s.indd 125 05/11/12 1:36 PM . methane is burned with the theoretically required amount of oxygen for complete combustion. a compressor. It consists of four unit processes in a thermodynamic cycle: a turbine. (a) propane (b) butane (c) pentane 2.83 One mole of air undergoes a Carnot cycle. 2. The pressure ranges between 0.83 One mole of air undergoes a Carnot refrigeration cycle.80 Calculate the adiabatic flame temperature of the following species in a stoichiometric mix- ture of air at a pressure of 1 bar. 2. determine the percentage of methane that passes through the reactor unburned. (b) It is combusted in a stoichiometric mixture of air. Answer the following questions: State 1 2 3 4 T[ºC] 520 80 P [bar] 100 0. The hot reservoir is at 25ºC and the cold reservoir is at 215+C. Assume that they react com- pletely to form CO2 and H2O. (d) Determine the net power developed in the cycle. Include the vapor–liquid dome. QH. The First Law of Thermodynamics Rankine cycle WS Turbine 1 Fuel 2 QH air Cooling Boiler QC water 4 Condenser 3 WC Compressor (a) Sketch all four processes on a Pv diagram. # # (c) Determine the heat-transfer rates in the boiler. 126 ► Chapter 2. QC. and the condenser. (b) From the plot above. rate of heat tranfer in the boiler c02.indd 126 05/11/12 1:36 PM . (e) What is the thermal efficiency. explain why the net power is negative. h of the cycle? net power delivered by the plant h.