ﺑﺴﻢ ﷲ اﻟﺮﺣﻤﻦ اﻟﺮﺣﯿﻢWeek number (5), (II) Lecture (10) Convection heat transfer Convection is the mode of heat transfer between a solid surface and the adjacent fluid (liquid or gas) that is in motion. The faster the fluid motion, the greater the convection heat transfer. In the absence of any bulk fluid motion, heat transfer between a solid surface and the adjacent fluid is by pure conduction. The presence of bulk motion of the fluid enhances the heat transfer between the solid surface and the fluid, Convection is classified as “natural (or free)” and “forced convection”, depending on how the fluid motion is initiated. In forced convection, the fluid is forced to flow over a surface or in a pipe by external means such as a pump or a fan, while in natural convection, any fluid motion is caused by natural means such as the buoyancy effect, which manifests itself as the rise of warmer fluid and the fall of the cooler fluid. Convection heat transfer strongly depends on the fluid properties dynamic viscosity “μ”, thermal conductivity “k”, density “ρ”, and specific heat “C” as well as the fluid velocity “u”. It also depends on the geometry and the roughness of the solid surface, in addition to the type of fluid flow. There are a wide variety of fluid flow problems encountered in practice such as, Laminar and turbulent flows Some flows are smooth and orderly while others are rather chaotic. The highly ordered fluid motion characterized by smooth streamlines is called laminar. The flow of high-viscosity fluids such as oils at low velocities is typically laminar. The highly disordered fluid motion that typically occurs at high velocities characterized by velocity fluctuations is called turbulent. The flow of low-viscosity fluids such as air at high velocities is typically turbulent. The flow regime greatly influences the heat transfer rates and the required power for pumping. Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Heat &Mass Transfer, ME209, Sep. 2014 Page 51 Internal versus External Flow A fluid flow is classified as being internal and external, depending on whether the fluid is forced to flow in a confined channel or over a surface. The flow of an unbounded fluid over a surface such as a plate, a wire, or a pipe is external flow. The flow in a pipe or duct is internal flow if the fluid is completely bounded by solid surfaces. Internal flow of water in a pipe and the external flow of air over the same pipe. Velocity boundary layer: The region of the flow above the plate bounded by “δ” in which the effects of the viscous shearing forces caused by fluid viscosity are felt. The development of the boundary layer for flow over a flat plate, and the different flow regimes. Thermal boundary layer: The flow region over the surface in which the temperature variation in the direction normal to the surface is significant Thermal boundary layer on a flat plate (the fluid is hotter than the plate surface). Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Heat &Mass Transfer, ME209, Sep. 2014 Page 52 The rate of convection heat transfer is expressed by Newton’s law of cooling as: Forced Convection ) For flow over flat plate; (ReL = Laminar flow 5×105 > ReL Transient flow 5×105≤ ReL ≤ 2×106 Turbulent flow ReL > 2×106 For flow inside tube (Red = ) Laminar flow 2200 > Red Transient flow 4200 ≥ Red ≥ 2200 Turbulent flow Red > 4200 Governing equations of convection heat transfer Conservation of Mass Equation or Mass balance (Continuity equation) Conservation of Momentum Equation (Momentum Equation) Conservation of Energy Equation (Energy Equation) Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Heat &Mass Transfer, ME209, Sep. 2014 Page 53 Dimensionless Analysis To obtain Convection heat transfer coefficient, the governing equations of convection heat transfer can be solved simultaneously. Also common practice to nondimensionalize the heat transfer coefficient “h” with the Nusselt number “Nu” as follow: h α (u, Cp, ρ, k, L, μ) h = constant [ua Cpb ρc kd Le μf] Where; h: heat transfer coefficient, W/m2 C M/T3θ u: Velocity, m/s L/T Cp: Specific heat, J/Kg C L2/T2 θ ρ: Density, kg/m3 M/L3 ML/T3 θ k: Thermal Conductivity, W/mC Lc: Characteristic Length, m L [Lc = L, for flat plates (Length of plate parallel to flow), Lc = D, for cylinders] μ: Viscosity, Kg/m.s M/LT MT-3θ-1 = Constant [(LcT-1) a (Lc2T-2 θ-1) b (MLc-3) c (MLcT-3 θ-1) d (Lc) e (MLc -1T-1) f] M 1= c + d + f 1 L 0= a + 2b - 3c+ d + e - f 2 T -3 = -a - 2b - 3d - f 3 θ -1=-b - d 4 From 3 and 4 -3=-a -2b -3(1-b)-f f=b-a Substitute in 1 and 2 c=a e = a-1 d = 1-b Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Heat &Mass Transfer, ME209, Sep. 2014 Page 54 Sep. ME209. Mech. Hesham Mostafa. Dr.] Pr “Prandtl number” = = Prof. HTI..h = constant [ua Cpb ρa k1-b Lca-1 μb-a] h = Constant Nu “Nusselt Number” = hL = CRe Pr K [dimensionless convection heat transfer coefficient. Eng. Dept. 2014 Page 55 . Heat &Mass Transfer. Find the amount of heat transfer from 0. Eng. Local Nux (NuL=Nux×2= h L/k) Nux = 0. Tw = 60 C = Constant. Heat &Mass Transfer. L = 0.4637Re .The plate is heated over its entire length to constant temperature 60 C. Mech.4 m From tables.6 < Pr < 50 Nux = 0. Dr.04656 / 1+ Pr Re=(ρuL)/μ 0.0207 / 1+ Pr Note : For laminar flow only.453Rex0.5 Pr1/3 5 (Re < 5×10 ) Nu = Constant 0.332Rex0. average Nu = 2Nux=L All Properties are calculated at Tf = (Tw +T∞ )/2 ----------------------------------------------------------------------------------------Example 12 Flow Pr (From tables) 0. Pr / Wall Pr out of range “Local” / Temperature and RePr > 100 0.6 < Pr < 50 Air at 27C flows over a flat plate with a velocity 2 m/s .5 Pr1/3 Nu = Constant Laminar. Air properties @ Tf = (Tw+T∞)/2 Tf = (Tw+T∞)/2 = (60+27)/2 = 43. Sep.5 K Prof. HTI. 2014 Page 56 .3778Re . Solution T∞= 27 C.ﺑﺴﻢ ﷲ اﻟﺮﺣﻤﻦ اﻟﺮﺣﯿﻢ Week number (6). u = 2 m/s. 0. Dept. Hesham Mostafa. ME209.5 C = 316..4 m of the plate assume unit width. (I) Lecture (11) Convection heat transfer (External Flow) Laminar Flow over flat plates Laminar. Pr / Pr out of range Heat Flux / and RePr > 100 0. Sep. Dr.332 (ReL) 0. Mech.0275 W/m C < 5× 105 Re = (ρ u L)/μ = (uL)/ν = Then the flow is laminar with.6 < Pr < 50 Nux=L = 0.332 ReL0.7)1/3 = Nu = 2 Nux=L = Nu= h L/k .7 μ= kg/ms k = 0. h= W/m2C Q = hA (Tw .4 × 1) (60-27) = W ----------------------------------------------------------------------------------------Example 13 Engine oil at 20 C is forced over 5 m square plate at velocity 2 m/s. HTI. Hesham Mostafa. Calculate the heat lost by the plate. ME209. Tw = Constant. 2014 Page 57 .T∞) = h (0. The plate is heated to uniform temperature equal to 60 C.ρ= Kg/m3 Pr = 0.5 Pr1/3 = 0. Dept.5 (0. Heat &Mass Transfer.. Eng. Solution Oil properties at Tf = (Tw+T∞)/2 Tf = (Tw+T∞)/2 = (60+20)/2 = 40 C Then the flow is laminar Tw = Constant Re < 5 ×105 RePr > 100 Prof. 0. Re < 5×105. .. 0.7 m.144 W/m2 oC h= A= 5×5= 25 m2 Q = hA (Tw . Mech. Dept. Air properties at Tf = (Tw+T∞)/2 Tf = (Tw+T∞)/2 = (60+20)/2 = 40 C = 313 K ρ=1. ME209. Hesham Mostafa.906×10-5 Kg/m3 Pr = 0.533 × 106 Prof. Dr. 2014 Page 58 . Solution T∞= 20 C Tw = 60 C = Constant u = 35m/s L = 0. The plate is maintained at 60 C. NuL = Pr1/3 [0.7 k = 0. Heat &Mass Transfer.T∞) = W ----------------------------------------------------------------------------------------- Laminar-Turbulent Flow (5 × 105 < ReL < 107) Average. Assume unit depth in Z-direction and length equal 0.7 m From tables.3778Re Nu = Pr 0.0275 W/m C ReL = (ρ u L)/μ = 1. Eng.037 ReL0. Sep.13 kg/m3 μ=1. Calculate heat transfer from the plate.04656 1+ Pr Nux=L = / / / Average.8 – 850] Note: All Properties are calculated at Tf = (Tw +T∞)/2 ----------------------------------------------------------------------------------------Example 14 Air at 20C flows over a flat plate with a velocity 35 m/s. HTI. Nu = 2 Nux=L = Nu= h L/k =h×5/0. Heat &Mass Transfer. then the flow is Laminar-Turbulent NuL = Pr1/3 [0.T∞) = h (0. Mech. Eng.8 – 850] = (0.037 ReL0.Since 5 × 105 < ReL < 107. 2014 Page 59 . Sep.8 – 850] = Nu= h L/k .533 × 106)0. Dr.. h= W/m2C Q = hA (Tw .7)1/3 [0. Hesham Mostafa.7 × 1) (60-20) = W ----------------------------------------------------------------------------------------- Prof.037 (1. HTI. Dept. ME209. Determine the rate of heat transfer from the plate if the air flows parallel to the (a) 6-m-long side and (b) the 1. (Air Properties at this pressure and temperature: k = 0.. Heat &Mass Transfer.5-m side. ME209. Solution 1-a) Prof. Dr.ﺑﺴﻢ ﷲ اﻟﺮﺣﻤﻦ اﻟﺮﺣﯿﻢ Solved Examples "Part Four" Ex.02953 W/m ºC. ν = 2. Mech. 4-1) Air at 83. Pr = 0.4 kPa and 20°C flows with a velocity of 8 m/s over a 1. Dept.5 m × 6 m flat plate whose temperature is 140°C. Sep. Eng.548 × 10-5 m2/s). HTI.7154. 2014 Page 60 . Hesham Mostafa. Dept.1-b) Note that the direction of fluid flow can have a significant effect on convection heat transfer to or from a surface .From the above example. HTI.. 2014 Page 61 . Eng. Sep. Mech. Dr. ME209. Prof. we can increase the heat transfer rate by 65 percent by simply blowing the air along the long side of the rectangular plate instead of the short side. Heat &Mass Transfer. Hesham Mostafa. 081 10 7 5 2 1. Dr. wind at 55 km/h is blowing parallel to a 4-m-high and 10-m-long wall of a house.5)C 9081 W 9.C)(40 m 2 )(12 .081 10 7 ) 0.C)(40 m 2 )(12 .7340)1 / 3 2.037 Re L 850) Pr 1 / 3 [0.08 kW s s If the wind velocity is doubled: Re L V L [(110 1000 / 3600) m/s](10 m) 2. Dept.5C are k 0. Sep.037(1.384 10 4 k k 0.7340 Air flows parallel to the 10 m side: The Reynolds number in this case is Re L V L [(55 1000 / 3600)m/s](10 m) 1.7340)1 / 3 1.88 W/m 2 .43 W/m 2 .C L 10 m As wL (4 m)(10 m) = 40 m 2 Q hA (T T ) (32.C h Nu (1.02428 W/m.336 10 4 k k 0.5)C 16.8 871](0. 4-2) During a cold winter day. heat transfer coefficient and then heat transfer rate are determined to be hL 0 .88 W/m 2 .336 10 4 ) 32.C h Nu (2. Eng. determine the rate of heat loss from that wall by convection.384 10 4 ) 57.21 kW s s Prof. Hesham Mostafa. Using the proper relation for Nusselt number. Heat &Mass Transfer.8 Nu (0.413 10 5 m 2 /s Thus we have combined laminar and turbulent flow.02428 W/m.Ex.C L 10 m As wL (10 m)(4 m) = 40 m 2 Q hA (T T ) (57. 2014 Page 62 .206 W 16. Mech.02428 W/m.037 Re L 850) Pr 1 / 3 [0. If the air outside is at 5°C and the surface temperature of the wall is 12ºC.413 10 m /s Thus we have combined laminar and turbulent flow. Using the proper relation for Nusselt number.43 W/m 2 .163 10 7 1.C 1.163 10 7 ) 0.413 10 -5 m 2 /s Pr 0.. What would your answer be if the wind velocity was doubled? Solution The properties of air at 1 atm and the film temperature of (Ts + T)/2 = (12+5)/2 = 8.8 871](0. HTI. the average heat transfer coefficient and the heat transfer rate are determined to be hL 0 . ME209.8 Nu (0.037(2. 20)C = 1363 W The radiation heat transfer from the same surface is Q rad As (Ts 4 Tsurr 4 ) (0. Using the proper relations. But the flow is assumed to be turbulent over the entire surface because of the constant agitation of the engine block.4 m) = 0.K 4 )[(80 + 273 K) 4 . ME209.5-m-high. Solution The properties of air the film temperature of (Ts + T)/2 = (80+20)/2 =50C are k 0..C)(0.037 Re L 0. The Reynolds number in this case is Re L V L [(80 1000 / 3600) m/s](0.02735 W/m.C h Nu ( 2076 ) 70.8 m Nu As wL (0.888 10 5 1. Heat &Mass Transfer.8-m-long rectangular block. 0. Hesham Mostafa.C 1. Determine the rate of heat transfer from the bottom surface of the engine block by convection and radiation as the car travels at a velocity of 80 km/h. and the road surface is at 25ºC. 4-3) Consider a hot automotive engine. which can be approximated as a 0.7228)1 / 3 2076 k k 0.95)(0.02735 W/m. HTI. Dept. Assume the flow to be turbulent over the entire surface because of the constant agitation of the engine block.8 Pr1 / 3 0. the heat transfer coefficient. the Nusselt number. The ambient air is at 20ºC. Dr.8 (0. Mech.037(9.7228 Air flows parallel to the 0. Sep.Ex.798 10 -5 m 2 /s Pr 0.(25 + 273 K) 4 ] 132 W Then the total rate of heat transfer from that surface becomes Q total Q conv Q rad (1363 132 ) W 1495 W Prof.95.67 10 -8 W/m 2 . Eng.40-m-wide. The bottom surface of the block is at a temperature of 80ºC and has an emissivity of 0.8 m) 9.798 10 5 m 2 /s .C L 0 . and 0.98 W/m 2 . which is less than the critical Reynolds number.888 105 )0.32 m 2 Q conv hAs (T Ts ) (70.32 m 2 )(80 .8 m)(0.98 W/m 2 .4 m side. and the heat transfer rate are determined to be hL 0. 2014 Page 63 .32 m 2 )(5. C 1.664 Re L 0.2 m 2 )(90 .C h Nu ( 259. 2014 Page 64 .896 10 -5 m 2 /s Pr 0.7 k k 0. Sep.2 m)(0. HTI. Dr.896 10 5 m 2 /s . and the sheet is subjected to air flow at 30°C at a velocity of 3 m/s on both sides along its surfaces normal to the direction of motion of the sheet.899 10 5 1.7202 The width of the cooling section is first determined from W Vt [(15 / 60) m/s](2 s) = 0.30)C = 437 W Prof.Ex.02808 W/m.07 W/m 2 .0282 W/m. ME209.C)(1.5 (0. Dept. Using the proper relation in laminar flow for Nusselt number.2 m 2 Q conv hAs (T Ts ) (6. Heat &Mass Transfer. The temperature of the plastic sheet is 90°C when it is exposed to the surrounding air.2 m Nu As 2 LW 2(1.7) 6. Eng.5 m) = 1.899 10 5 ) 0.2 m) 1.7202)1 / 3 259. The width of the air cooling section is such that a fixed point on the plastic sheet passes through that section in 2 s. Solution The properties of air at the film temperature of (Ts + T)/2 = (90+30)/2 =60C are 1. the average heat transfer coefficient and the heat transfer rate are determined to be hL 0.5 m The Reynolds number is Re L V L (3 m/s)(1. 4-4) The forming section of a plastics plant puts out a continuous sheet of plastic that is 1. Thus the flow is laminar.664(1. Determine the rate of heat transfer from the plastic sheet to the air. Mech..2 m wide and 2 mm thick at a rate of 15 m/min.059 kg/m 3 k 0.C L 1 . Hesham Mostafa.07 W/m 2 .5 Pr 1 / 3 0. which is less than the critical Reynolds number. 2014 Page 65 . The forming section of a plastics plant puts out a continuous sheet of plastic that is 1. The average temperature of the plate is not to exceed 65°C. Heat &Mass Transfer. Mech. An array of power transistors. Prof. The width of the air cooling section is such that a fixed point on the plastic sheet passes through that section in 2 s. ME209. The temperature of the plastic sheet is 90°C when it is exposed to the surrounding air. HTI.ﺑﺴﻢ ﷲ اﻟﺮﺣﻤﻦ اﻟﺮﺣﯿﻢ Exercises "Sheet Four" 4-1. Determine the rate of heat transfer from the plastic sheet to the air. determine the number of transistors that can be placed on this plate. are to be cooled by mounting them on a 25-cm × 25-cm square aluminum plate and blowing air at 35°C over the plate with a fan at a velocity of 4 m/s.. Dr. 4-2. and the sheet is subjected to air flow at 30°C at a velocity of 3 m/s on both sides along its surfaces normal to the direction of motion of the sheet.2 m wide and 2 mm thick at a rate of 15 m/min. Sep. dissipating 6 W of power each. Assuming the heat transfer from the back side of the plate to be negligible and disregarding radiation. Eng. Dept. Hesham Mostafa. . Dept. Heat &Mass Transfer. Dr. ME209. 2014 Page 66 . (II) “Mid-Term Exam” 30 Marks Prof. Sep. Mech. Eng. Hesham Mostafa. HTI.ﺑﺴﻢ ﷲ اﻟﺮﺣﻤﻦ اﻟﺮﺣﯿﻢ Week number (6). Mech. and conduit are usually used interchangeably for flow sections. In general. tube. Sep. which explains the overwhelming popularity of circular tubes in heat transfer equipment.ﺑﺴﻢ ﷲ اﻟﺮﺣﻤﻦ اﻟﺮﺣﯿﻢ Week number (7). (I) Lecture (12) Internal Flow It can be noticed that most fluids. flow sections of circular cross section are referred to as pipes (especially when the fluid is a liquid). the circular tube gives the most heat transfer for the least pressure drop. For a fixed surface area. ME209. The terms pipe. and the flow sections of noncircular cross section as ducts (especially when the fluid is a gas). are transported in circular pipes. Turbulent Flow inside Tube Red > 4200 Prof. Dr. HTI. Hesham Mostafa. Noncircular pipes are usually used in applications such as the heating and cooling systems of buildings where the pressure difference is relatively small and the manufacturing and installation costs are lower. Dept. 2014 Page 67 . duct. Eng. Small diameter pipes are usually referred to as tubes. This is because pipes with a circular cross section can withstand large pressure differences between the inside and the outside without undergoing any distortion. Heat &Mass Transfer.. especially liquids. Dr. 2014 Page 68 .Red = (ρ u d)/μ Tb2 Tb1 TW Tw > Tb1 Where: Tw > Tb2 Q = m Cp (Tb2.Tb1) =h A Z Tb: Bulb Temperature.Tb) av. Dept.8 Pr1/3 For fully developed turbulent flow. Hesham Mostafa. Heat &Mass Transfer..4 for heating for fluid = 0.3 for cooling for fluid Note: All Properties are calculated at Tb1 If the effect of viscosity variation was considered as: Nud = 0.) For constant wall temperature. m2 Z = (Tw . Sep. J/Kg C h: Heat transfer Coefficient W/m2C A: Heat transfer Area.Tbav. C Cp: Specific heat at constant pressure. = (Tw . ME209.8 Prn Where n = 0. C Tw: Wall Temperature. μ μ . Mech. Empirical correlations for Nud For turbulent flow inside tube: For fully developed turbulent flow Red > 4200 0.6 < Pr < 100 Nud = 0. For constant heat flux. Eng.027 Red0. HTI.023 Red0. Red > 4200 Note: All Properties are calculated at Tb1 expect at TW Prof. Tube inner surface temperature was maintained at 100 C. HTI.14].027 Red0.83 × 10 Kg/m2 Re = (ρ u D)/μ = 9555 > 4200 Then the flow is Turbulent Nud = 0.8 Pr1/3 μ μ .8 Pr1/3 (/w) 0. Calculate tube length.052 u = 0.62W/mC . Water properties at Tb1=30 C ρ= 995 Kg/m3 μ =8 × 10-4 Kg/m2 Pr = 5.. ME209.If the effect of entrance region was considered as: . Solution d = 0. Mech. Eng.av expect at TW ----------------------------------------------------------------------------------------Example (15) Water at 30 C enters a tube 5 cm in diameter and leaves at 50 C with mass flow rate of 0.036 Red0.3 kg/s. Red > 4200 Note: All Properties are calculated at Tb1 Laminar Flow inside Tube Red < 2200 Nud = 1. 2014 Page 69 . Note: All Properties are calculated at Tb.3 Kg/s = ρ u a = ρ u (π/4) d2 = 995 u (π/4) 0. Dr.05 m m = 0. Dept. 10 < L/d < 400. Heat &Mass Transfer. Prof.8 Pr1/3 For Turbulent Flow. Sep. Hesham Mostafa.153 m/s From tables.4 Cp = 4177 J/kgC K = 0. at Tw = 100 C From tables μ = 2. [Nud=0. Nud = 0.027 Red0.86 (Red Pr) 1/3 (d/L) μ 1/3 μ . Hesham Mostafa.027 (9555)0. Mech. h= m = ρ u a =ρ a (π/4) d2 = W/m2C kg/s Q = h A (Tw . Dept. × = 83.Tb) av.4)1/3 Nu= h d/k .Tb1) Prof.58 m ----------------------------------------------------------------------------------------Example 16 Air at 100 C is heated as it flow through a tube with 25 mm in diameter at a velocity 10 m/s.) = m Cp (Tb2. 2014 Page 70 .Nud = 0.8 (5. .6 Then the flow is Turbulent Nud = 0. ME209.Tb av.3(4177)(50-30) =1032 [π (0. Eng. Air properties at Tb1 Re = (ρ u D)/μ = Pr = > 4200 > 0.8 Prn Since the flow is heated then n = 0. Solution Heat flux is Constant then (Tw . . HTI.Tb) av. Also find exist temperature.025 m From tables. Calculate heat transfer Per unit length of the tube if a constant heat flux condition is maintained at the tube wall and the wall temperature is 20 C above air temperature all along the tube length..023 Red0.05) L] [100 – ( L = 2.4 Nud = Nu= h d/k . Dr. = m.Tb1) )]= 0. Cp (Tb2. . Sep.25 h= 1032 W/m2C Q = h A (Tw . Heat &Mass Transfer. = 20 C u = 10 m/s d = 25 mm = 0. 025 m u = 0. av.654 W/mC d= 0. Prof. Heat &Mass Transfer.Tb1) Tb2 = C From tables..025/2) >10 Nud = 1.01 Cp = 4179 J/kgC K = 0. Cp (Tb2. Eng.Tb.100) = W Tb2 = C ----------------------------------------------------------------------------------------Example 17 Water at 60C enters a tube 25 mm in a diameter at amen velocity 2 cm/s. Water properties at Tb1=60 C ρ= 983.86 [Rex0. Hesham Mostafa.2 Kg/m3 μ =4. Dept. HTI. Nud = Nu= h d/k .8 Pr (d/L)] 1/3 μ μ .02 m/s . Solution From tables. Dr. ME209. Find exit water temperature.) = m. h= m = ρ(@Tb1 u a =ρ a (π/4) d2 = W/m2C kg/s Q = h A (Tw . 2014 Page 71 .014× (0.7 × 10-4 Kg/m2 Pr = 3.58 × 10− 4 Then the flow is Laminar Re Pr (d/L) = 1040 ×3. at Tw = 80 C From tables μ Re = (ρ u D)/μ = < 2200 = 3. Mech. Tube length 3 m and wall temperature is constant and equal to 80 C. Water properties at Tbav. Sep.Q = h (πdL) (20) = m Cp (Tb2. ME209. Heat &Mass Transfer. Dept. h= m = ρ@Tb1 u a =ρ a (π/4) d2 = W/m2C kg/s Q = h A (Tw . Mech. Eng.8 Pr (d/L)] 1/3 μ μ . Dr. HTI.86 Kg/m2 μ= >10 [Rex0. 2014 Page 72 . Hesham Mostafa.. Sep.Tb1) Tb2 = Q= C W ----------------------------------------------------------------------------------------- Prof.) = m Cp (Tb2. Nud = Nu= h d/k . = (Tb1 +Tb2)/2 ρ= Kg/m3 K= W/mC Re = (ρ u D)/μ = Pr = Cp = J/kgC < 2200 Re Pr (d/L) = Nud = 1.Tb av.Tbav. The cylinder surface is maintained at 150 C.ﺑﺴﻢ ﷲ اﻟﺮﺣﻤﻦ اﻟﺮﺣﯿﻢ Week number (7).966 Kg/m3. Heat &Mass Transfer. ME209.05 m Re = (ρ u d)/μ = Pr = > 0. Mech. Sep. Calculate heat loss per unit length of the cylinder.6 From table 3 using “Re” c= n= Nud = c Redn Pr1/3 Nud = Nu= h d/k . Pr = 0. Dr. (II) Lecture (13) External Flow Flow across singe tube Red = (ρ u d)/μ Nud = c Redn Pr1/3 All properties are calculated at Tf = (Tw+T∞)/2 c.7 × 10-4 Kg/m2. Solution From tables. u = 50 m/s μ =4..0312 W/mC d = 50 mm = 0. Hesham Mostafa. Air properties at Tf = (Tw+T∞)/2 ρ= 0. HTI. Eng. 2014 Page 73 . n from table 3 using “Re” Example 18 Air at 35 C flows across 50 mm diameter cylinder at a velocity 50 m/s. Dept.T∞) = h (πdL) (150-35) = W Prof.695. h= W/m2C Q = h A (Tw . k = 0. Hesham Mostafa. ME209.) = = m.005823 Kg/s 35.08W/m2C Q = h A (Tw . HTI. Cp (Tb2.16956 m2 m = ρ u a =ρ u (π/4) d2 = 1.4 Nud =40.023 Red0.1774(7)(π/4) (0. 2014 Page 74 . n=0. d=30 mm. Nud=0. Heat &Mass Transfer. Mech.08 ×0. L=1.8462 ×10-5 kg/ms k = 0. Dept.005823 (94.03 m Re = (ρ u d)/μ = (1. Sep.5 watt.708 Cp = 1005 J/kgC d = 30 mm = 0.618) Solution a) For Flow inside the tube From tables. if it is exposed to air at 27 C in the following cases: a) Air flow inside the tube with a velocity 7 m/s..005823 (1005) (Tb2. Prof.8 Prxn Since the flow is heated then n = 0.8 = 0.03)2 = 0.6 < Pr < 100 Then the flow is fully developed Turbulent flow Nud = 0.1774 Kg/m3 μ=1. Air properties at Tb1 = 27 C =300 K ρ = 1. Eng.11 Nu= h d/k .1696× (127 - ) = = 0.03) 1.Tb av.Tb1) Tbav.1774 ×7×0.8 Pr0. = (Tb1 + Tb2 ) /2 A= πdL= π (0.Example 19 Find the amount of heat transfer from a hot cylinder maintained at 127 C.4-27) = 394. h= 35. Dr.27) Tb2 = 94.8 m.4 C Q =1005× 0.8462 ×10-5 =13394 > 4200 0. ( c=0.023 Rex0.03)/ 1.4 b) Air flow across the tube with a velocity 7 m/s.02624 W/m C u = 7 m/s Pr = 0.193. HTI..T∞) = h (πdL)(127-27) = 978 W ----------------------------------------------------------------------------------------- Prof. Eng. ME209.998 ×7×0.07 × 10-5 = 1.03 m Re = (ρ u d)/μ = (0. Mech.998Kg/m3 u = 7 m/s μ =2. Heat &Mass Transfer. Dept.b) For Air flow across the tube From tables.03003 W/mC d = 30 mm = 0. Hesham Mostafa. Sep. h = 57.618 Nud = c Redn Pr1/3 Nud = 57.07 × 10-5 Kg/ms Pr = 0.03)/ 2. Dr.012× 104 From table 3 using “Re” c =0.695 k = 0.6 Nu= h d/k . 2014 Page 75 .7 W/m2C Q = h A (Tw . Air properties at Tf = (Tw+T∞)/2 = 77 C = 350 K ρ= 0.193 n = 0. In such equipment. one fluid moves through the tubes while the other moves over the tubes in a perpendicular direction. Hesham Mostafa. (I) Lecture (14) Flow across tube banks Cross-flow over tube banks is commonly encountered in practice in heat transfer equipment such as the condensers and evaporators of power plants. In a heat exchanger that involves a tube bank. HTI. Eng. Dept. and air conditioners. Mech. refrigerators. SP u∞ Sn M (High) SP u∞ SD Sn N (Deep) Prof. Heat &Mass Transfer.. and the fluid flows through the space between the tubes and the shell. ME209.ﺑﺴﻢ ﷲ اﻟﺮﺣﻤﻦ اﻟﺮﺣﯿﻢ Week number (8). Sep. especially when the fluid is a liquid. the tubes are usually placed in a shell (and thus the name shell-and-tube heat exchanger). 2014 Page 76 . Dr. .2. av.maxn Pr1/3 c.T.2+T∞. HTI.) = Cp (T∞. = (T∞. Heat &Mass Transfer. Dept. 2014 Page 77 Sn-d .1) T.T∞.03 m Prof. ME209.Sn: Normal distance between two tube centers (> d) SP: Parallel distance between two tube centers (> d) d: Diameter of a the tube (Apply for all tubes) Tw: Tube wall temperature (Apply for all tubes) For In-line arrangement m = ρ∞ u∞ (Sn L) M = ρ∞ umax (Sn-d) L M umax = u∞ (Sn / Sn-d) u∞ Sn Red. Sp/d Nud = h d/k All properties at Tf = (Tw+T∞)/2. av. Hesham Mostafa.1)/2 --------------------------------------------------------------------Example 20 Air at 24oC flows across a bank of tubes (10 rows high & 10 rows deep) with a speed of 10 m/s. Tube diameter was 30 mm. Mech. Eng. Sep. expect ρ∞ @ T∞1 A=πdL (NM) ε: Correction factor from table 5 for “N” Q = ε h A (Tw . Dr. n from table 4 for Sn/d . Solution T∞1 =24 C N = 10 M = 10 Sn= Sp= 45 mm (Inline arrangement) u∞ = 10 m/s Tw = 130 C d = 30 mm = 0. 1 m long and surface temperature 130 oC. Tube bank was arranged in an in-line arrangement with Sp =Sn =45 mm.max = (ρ umax d)/μ umax Nud = c Red. Calculate heat lost from tube bank and exit air temperature. ) = m.T av. Sep.1915 kg/m3. Calculate the total heat transfer per unit length of the tube and exit air temperature. Prof. Sp/d=1. Dr. Hesham Mostafa.84 For N= 10 from table 5. ME209. The surfaces of the tubes are maintained at 65 C .4 C Q =159004W --------------------------------------------------------------------Example 21 Air flow at 10 C flows across a bank of tubes 15 rows high and 5 rows deep at velocity of 7 m/s measured at a point in the flow before enters the tube bank. Heat &Mass Transfer.5 mm.From tables.T∞1) T..65 Nu= h d/k . = (T∞.03003 W/m C.max Pr1/3 = 184. HTI. 2014 Page 78 . W/m2C h=184. Mech.36 Kg/s Q = ε h A (Tw . Cp = 1009 J/kgC umax = u∞ (Sn /( Sn-d)) = 10 (45/(45-30)) = 30 m/s Re = (umax d)/ ν = 43290 Sn/d= 1.697 k = 0.2+T∞. Sp/d c = 0. Cp (T∞2. av.425 m2 Air density at 297 K.5 From table 4 for for Sn/d. n = 0. Eng. ε =1 A=πdL (NM) = π×0.03×1× (10×10) = 9. Dept.79×10-6 m2 /s Pr = 0.62 Nud = c Red.1)/2 T∞2= 53. m = ρ∞ u∞ (Sn L) M = 5.5 .Tube diameter is 25 mm arranged in an in-line manner so that the spacing in both normal and parallel direction to flow is 37.278 . Air properties at Tf = (Tw+T∞)/2 Tf = (Tw+T∞)/2 = (130+ 24)/2 = 77 C = 350 K ν=20. ρ∞=1. Mech.Solution T∞1 =10 C N=5 u∞ = 7 m/s Tw = 65 C M = 15 Sn= Sp= 37. HTI.) = m Cp (T∞.5 mm d = 25 mm = 0..5/(37. Dr.T∞.5 . n = 0.706 k = 0. Eng. ε =0. Dept.5 C = 310.5-25)) = 21 m/s Red.025 m From tables.137 kg/m3 μ=1.Tb. maxn Pr1/3 = Nu= h d/k . Heat &Mass Transfer.5 K ρ=1.027 W/m C umax = u∞ (Sn / Sn-d) = 7 (37. av.92 A=πdL(NM) = π(25×10-3)×1× (5×15) = m = ρ∞ u∞ (Sn L) M = T∞2= Q= m2 Kg/s C W --------------------------------------------------------------------- Prof.max = (ρ umax d)/μ = 31500 Sn/d= 1.1) For N= 5 from table 5. Sep. Hesham Mostafa.62 Nud = c Red. Sp/d c = 0.278 .2. Sp/d=1. W/m2C h= Q = ε h A (Tw .5 From table 4 for Sn/d .89×10-5 Kg/m3 Pr = 0. ME209. Air properties at Tf = (Tw+T∞)/2 Tf = (Tw+T∞)/2 = (65 + 10)/2 = 37. 2014 Page 79 . Heat &Mass Transfer. Mech.54 cm in diameter at a velocity 10 m/s.. Dept. 5-1) Air at 2 atm 200C is heated as it flow through a tube with 2. ME209. Dr. Eng. 2014 Page 80 . Sep.ﺑﺴﻢ ﷲ اﻟﺮﺣﻤﻦ اﻟﺮﺣﯿﻢ Solved Examples "Part Five" Ex. Calculate heat transfer per unit length of the tube if a constant heat flux condition is maintained at the tube wall and the wall temperature is 20 C above air temperature all along the tube length. How much would the bulk temperature increase over a 3 m length of the tube? Solution Prof. Hesham Mostafa. HTI. 5-2) A long 10-cm-diameter steam pipe whose external surface temperature is 110°C passes through some open area that is not protected against the winds. HTI.. Dept. ME209. Determine the rate of heat loss from the pipe per unit of its length when the air is at 1 atm pressure and 10°C and the wind is blowing across the pipe at a velocity of 8 m/s.193 n = 0. Dr. Mech. Eng. Sep.Ex. 2014 Page 81 . Hesham Mostafa. Heat &Mass Transfer.618 Nud = c Rexn Prx1/3 = 124 Prof. Solution From table 3 for “Re” c = 0. Sep.C Nu (82. Heat &Mass Transfer. so that in steady operation all the heat generated in the heater is transferred to the water in the tube and the wall temperature remains constant.4244 m / s Ac ( 0.637 W/m.02 m)(7 m)](Ts (80 10) )C 2 Ts .02 m) 2 / 4 Vm Re Vm D h (0. e 113.101) 0. ME209.C / 0.1 kg/m 3 k 0.8 Pr n . The outer surface of the heater is well insulated.132 kg/s)(4180 J/kg.4 82.1 kg/m 3 )(0.602 10 6 m 2 /s .4244 m/s)(0. determine the power rating of the resistance heater. 5-3) Water is to be heated from 10°C to 80°C as it flows through a 2cm-internaldiameter 7-m-long tube.023 Re 0.Ex.023(14. Solution The properties of water at the inlet bulk temperature of 10C are 990.02 m Then the inner surface temperature of the pipe at the exit becomes Q hA (T T ) s s bav 38. The tube is equipped with an electric resistance heater that provides uniform heating throughout the surface of the tube.C)[ (0.101 0.3C Prof.02 m) 14.C Pr 3. Dept.132 kg/s Q m C p (Te Ti ) (0. If the system is to provide hot water at a rate of 8 L/min. Eng. Dr.8 (3.4 k Nu 0.921 kg/min 0. Mech. Hesham Mostafa.637 W/m.C)(80 10)C 38. 2014 Page 82 .79) 2637 W/m 2 .602 10 -6 m 2 /s C p 4180 J/kg.79 Nu Heat transfer coefficient is h k 0. the flow is fully developed turbulent flow hDh 0.91) 0..008 m 3 /min ) 7. HTI.91 The power rating of the resistance heater is m V (990 .C Dh 0. which is greater than > 4200. n 0. Therefore.627 W (2637 W/m2 .627 W The velocity of water and the Reynolds number are V (8 10 3 / 60) m 3 / s 0. 000 4/5 0. it was noticed that a 12-m-long section of a 10-cm-diameter steam pipe is completely exposed to the ambient air. HTI. is estimated to be 0ºC. and the average temperature of the surfaces surrounding the pipe. Determine the amount of heat lost from the steam during a 10-h-long work day.3 1/ 4 k 1 (0. n from table 3 using Re] The heat transfer coefficient is h k 0.632 10 4 0.5 Pr 1 / 3 0.1 m The rate of heat loss by convection is As DL (0. 5-4) During a plant visit.19) 18. including the sky.02662 W/m.02662 W/m.632 10 4 ) 0.000 1 (0.C Steam pipe 1. ME209. There are also light winds in the area at 10 km/h.5 (0.5)C = 5001 W The rate of heat loss by radiation is Q rad As (Ts 4 Tsurr 4 ) (0.702 10 5 m 2 /s The Nusselt number corresponding this Reynolds number is determined to be Nu hD 0.77 m 2 )(5. Dr.19 (You can solve the problem using the empirical equation given in the notes) [Nud = c Rexn Prx1/3>>>>>>>>>>>>>> c.702 10 -5 m 2 /s Pr 0.62(1. Hesham Mostafa.7255)1 / 3 1.361 10 5 kJ/day Prof.3 1 1/ 4 282.559 kJ/s )(10 h/day 3600 s/h ) 2.7255) 2 / 3 5/8 4/5 71.8.4 / 0.4 / Pr) 2 / 3 Re 5 / 8 1 282. The temperature measurements indicate that the average temperature of the outer surface of the steam pipe is 75ºC when the ambient temperature is 5ºC. Eng.632 10 4 1.95 W/m 2 .8)(3.95 W/m 2 . Heat &Mass Transfer.1 m) 1. Dept. Solution Properties The properties of air at 1 atm and the film temperature of (Ts + T)/2 = (75+5)/2 = 40C Wind V = 10 km/h k 0.62 Re 0.C D 0.7255 Ts = 75C The Reynolds number is Re V D (10 1000/3600) m/s(0. Mech.Ex.K 4 ) (75 273 K ) 4 (0 273 K ) 4 1558 W The total rate of heat loss then becomes Q total Q conv Q rad 5001 1558 6559 W The amount of heat loss from the steam during a 10-hour work day is Q Q total t (6.67 10 -8 W/m 2 . 2014 Page 83 . Sep.C)(3.77 m 2 )(75 . The emissivity of the outer surface of the pipe is 0.77 m 2 Q hAs (T s T ) (18.C Nu (71.1 m )(12 m) 3.. 675 Pr 1 / 3 0.5 m) 1.0 W Prof.Ex. Solution The properties of air at 1 atm and the film temperature of (Ts + T)/2 = (65+30)/2 = 47.2 k (You can solve the problem using the empirical equation given in the notes) [Nud = c Rexn Prx1/3>>>>>>>>>>>>>> c.758 10 4 1.2 m 2 Q hAs (T s T ) (15.C)(1. and thus are cooled by air at 30°C flowing over the duct with a velocity of 200 m/min. HTI.C D 0 .2 m) 3. determine the total power rating of the electronic devices that can be mounted into the duct.758 10 4 ) 0. 5-5) The components of an electronic system are located in a 1.774 10 5 m 2 /s Using the relation for a square duct.7235 The Reynolds number is Re V D (200/60) m/s(0.102 Re 0. Dept. 2014 Page 84 .02717 W/m. If the surface temperature of the duct is not to exceed 65°C.24 W/m 2 . ME209.30) C = 640. Sep.C Nu (112. The components in the duct are not allowed to come into direct contact with cooling air.102(3.5 m 2 /s Pr 0.5C are k 0.2 m )(1.2 m Then the rate of heat transfer from the duct becomes As ( 4 0..675 (0. Eng.2) 15. the Nusselt number is determined to be Nu hD 0. Mech.C 1. n from table 3 using Re] The heat transfer coefficient is h k 0.5-mlong horizontal duct whose cross section is 20 cm ×20 cm.7235)1 / 3 112. Hesham Mostafa.774 10.2 m 2 )(65 . Dr.24 W/m 2 . Heat &Mass Transfer.02717 W/m. Mech. Determine the heat transfer coefficient.. Dept.Ex. It is estimated 15% of the power dissipation is lost through the cumulative effect of surface radiation and conduction through the cumulative effect of surface radiation and conduction through the endpieces. HTI. the heater power dissipation was measured to be P = 46 W.3ºC. The cylinder is heated internally by an electrical heater and is subjected to a cross flow of air in a low-speed wind tunnel. Eng. while the average cylinder surface temperature was determined to be Ts = 128. 2014 Page 85 . 5-6) Experiments have been conducted on a metallic cylinder 12. Heat &Mass Transfer. respectively. Hesham Mostafa. Solution Prof.4 ºC.7 mm in diameter and 94 mm long. Sep. Dr. ME209. Under a specific set of operating conditions for which the upstream air velocity and temperature maintained at u = 10 m/s and 26. ME209. Sep. The outer diameter of the tubes is 1. Dept. In an industrial facility. Dr. Eng. and has a resistance of 0. Mech. Air enters the duct at 20ºC and 1 atm with a mean velocity of 4. Hesham Mostafa. Prof. The tube length is 10 cm and is maintained at a constant temperature of 60. Water enters a 3 mm diameter tube at 21 and leaves at 32. There are 6 rows (N) in the flow direction with 10 tubes (M) in each row.ﺑﺴﻢ ﷲ اﻟﺮﺣﻤﻦ اﻟﺮﺣﯿﻢ Exercises "Sheet Five" 5-1. Determine the rate of heat transfer per unit length of the tubes (L=1 m). 5-3. air is to be preheated before entering a furnace by geothermal water at 120ºC flowing through the tubes of a tube bank located in a duct.5 cm. The flow rate is such that the Reynolds number is 600. HTI. Calculate the water flow rate. 5-2. and the tubes are arranged in-line with longitudinal and transverse pitches of Sn = Sp = 3 cm (SL =ST).. Determine the surface temperature of the wire during a windy day when the air temperature is 10°C and the wind is blowing across the transmission line at 40 km/h.002 ohm per meter length. and flows over the tubes in normal direction. Heat &Mass Transfer.5 m/s. 2014 Page 86 . A 6-mm-diameter electrical transmission line carries an electric current of 50 A. heat transfer from electric baseboard heaters or steam radiators.8 m/s2 Lc Characteristic length m = L “for vertical wall or cylinder” = d “for horizontal Cylinder” c . and VCRs. TVs.ﺑﺴﻢ ﷲ اﻟﺮﺣﻤﻦ اﻟﺮﺣﯿﻢ Week number (8).. Mech. ME209. Dr. (II) Lecture (15) Free Convection Many familiar heat transfer applications involve natural convection as the primary mechanism of heat transfer. Natural convection in gases is usually accompanied by radiation of comparable magnitude except for low-emissivity surfaces. m from table 6 All properties at Tf = (Tw+T∞)/2 T GrPr = GrPr = GrPr = T∞ T L T T∞ T∞ Prof. 2014 Page 87 . Nu = c (GrPr) m = hLc/K T Gr = Grashof number = Β Thermal expansion T∞ = From tables in case of water = 1/ Tf in case of ideal gas (Air) g: Gravitational acceleration =9. Sep. and heat transfer from the bodies of animals and human beings. Heat &Mass Transfer. HTI. heat transfer from the refrigeration coils and power transmission lines. Dept. Eng. Some examples are cooling of electronic equipment such as power transistors. Hesham Mostafa. 9×106 ×0.Example 22 Find the amount of heat transfer from a hot cylinder of 30 cm in diameter.25 Nu=C (Gr Pr)m = (h Lc /k)=( hd/k) h = 3. Hesham Mostafa.3 m β=1/ Tf = 1/ 290 Gr = T − T∞ Gr = 5. Eng. Solve in case of horizontal and vertical position.0254 w/m oC Lc= d =0.48×10-5 k=0.53. m = 0.7081 = 81.9×106 Gr Pr = 5.. Air properties at Tf =Tw+T∞/2 Tf = (27+7)/2=17 oC=290 K From air tables Pr= 0. ME209.71 υ= 1. Heat &Mass Transfer.Vertical Position Lc =L=2 m Gr = T − T∞ = Prof. Dr.69××106 From table 6 c = 0. HTI. Mech.9 w/m2 oC Q=hA (Tw-T∞) =h (π dL) (Tw-T∞) = Watt 2. Solution a) For Still air (free convection) 1-Horizontal Position From tables. 2014 Page 88 . Dept. 2m long if Tw=27oC exposed to: a) Still air at 7 oC b) Still water at 7 oC. Sep. . Sep.015 × 1010 (From water tables) Lc =L=2 m Gr Pr = ( Gr Pr = )L T − T∞ c. ME209.Gr Pr = c. m (From table 6) Nu=C (Gr Pr) m = (hLc/k) h= w/m2 oC Q = hA (Tw-T∞) =h (π d L) (Tw-T∞) = Watt Prof. Eng. Water properties at Tf = (Tw+T∞)/2 Tf = (27+7)/2=17 oC = 1.For Water (free convection) 1-Horizontal Position From tables.3 m Gr Pr = ( Gr Pr = )L T − T∞ c. Heat &Mass Transfer.Vertical Position = 1. m (From table 6) Nu = C (Gr Pr) m = (hLc /k) w/m2 oC h= Q=hA (Tw-T∞) =h (π dL) (Tw-T∞) = Watt b. HTI. Dr. Mech.015 × 1010 (From water tables) Lc= d =0. m (From table 6) Nu=C (Gr Pr) m = (hLc/k) = (hd/k) W/m2 oC h= Q=hA (Tw-T∞) =h (π d L) (Tw-T∞) 2. Hesham Mostafa. 2014 Page 89 . Dept. Mech. Eng. ME209. 6. Dr.. 2014 Page 90 . Sep. HTI. Calculate the heat transfer if the plate is 10m wide.1) A large vertical plate 4. Dept. Hesham Mostafa.ﺑﺴﻢ ﷲ اﻟﺮﺣﻤﻦ اﻟﺮﺣﯿﻢ Solved Examples "Part Six" Ex. Solution Prof. Heat &Mass Transfer.0 m high is maintained at 60 oC and exposed to atmospheric air at 10 oC. Heat &Mass Transfer. Dr. HTI. Solution Prof. Sep. Eng. Hesham Mostafa.Ex. Dept.Calculate the free convection heat loss per unit length of heater. 6. 2014 Page 91 . ME209. Mech.2) A 2 cm diameter horizontal heater is maintained at a surface temperature of 38ºC and submerged in water at 27ºC.. Sep. Hesham Mostafa. The wire is exposed to air at 0 ºC.Ex. Eng. 2014 Page 92 .3) A fine wire having a diameter of 0. Mech. Calculate the electric power necessary to maintain the wire temperature if the length is 50 cm. HTI.02 mm is maintained at a constant temperature of 54ºC by an electric current. Dept. 6. ME209. Heat &Mass Transfer.. Dr. Solution Prof. Sep. Hesham Mostafa. 2014 Page 93 . Calculate the free convection heat loss per meter length. Heat &Mass Transfer.. Eng. Mech. Dr. Solution Prof. Dept. ME209.Ex.4) A horizontal pipe 0. 6.3048 m in a diameter is maintained at a temperature of 250 ºC in a room where the ambient air 15 ºC. HTI. HTI. 2014 Page 94 . 6-3. Calculate the heat transfer coefficient and the average temperature of the plate and the amount of heat lost by this isothermal surface at this average temperature. Dr. Calculate the value of heat transfer coefficient for the plate. Mech. Hesham Mostafa. The surface temperature of the pipe is 140 ºC. A horizontal pipe 8 cm in diameter is located in where atmospheric air is at ºC 20. A 25 by 25 cm vertical plate is fitted with an electric heater which produces a constant heat flux of 1000 W/m2. Eng.4 mm in diameter is placed horizontally in a container of water at 38 ºC and is electrically heated so that the surface temperature at 93 ºC. Calculate the heat loss by the wire 6-4. The plate is submerged in water at 15 ºC. Prof. The ambient air is at 20 ºC. ME209. 6-2. Heat &Mass Transfer.. Calculate the free convection heat loss per meter of pipe. A 30 cm square vertical plate is hated electrically such that a constant heat flux condition is maintained with a total heat dissipation of 30 W. Sep. Dept. A 10 cm length of platinum wire 0.ﺑﺴﻢ ﷲ اﻟﺮﺣﻤﻦ اﻟﺮﺣﯿﻢ Exercises "Sheet Six" 6-1.