Solution Manual for Fundamentals of Heat and Mass Transfer, 7th Edition

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PROBLEM 1.1KNOWN:Thermal conductivity, thickness and temperature difference across a sheet of rigidextruded insulation.FIND:(a) The heat flux through a 2 m×2 m sheet of the insulation, and (b) The heat ratethrough the sheet.SCHEMATIC:ASSUMPTIONS:(1) One-dimensional conduction in the x-direction, (2) Steady-stateconditions, (3) Constant properties.ANALYSIS:From Equation 1.2 the heat flux is12xT - TdTq= -k= kdxL′′Solving,"xW10 Kq= 0.029×m K0.02 m⋅x2Wq= 14.5 m′′<The heat rate is2xx2Wq= qA = 14.5× 4 m= 58 Wm′′ ⋅<COMMENTS:(1)Be sure to keep in mind the important distinction between the heat flux(W/m2) and the heat rate (W). (2) The direction of heat flow is from hot to cold. (3) Note thata temperaturedifferencemay be expressed in kelvins or degrees Celsius.qcondA = 4 m2T2T1k = 0.029⋅Wm KxL = 20 mmT1– T2= 10˚CqcondA = 4 m2T2T1k = 0.029⋅Wm KxL = 20 mmT1– T2= 10˚C

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PROBLEM 1.2KNOWN:Thickness and thermal conductivity of a wall. Heat flux applied to one face andtemperatures of both surfaces.FIND:Whether steady-state conditions exist.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internal energygeneration.ANALYSIS:Under steady-state conditions an energy balance on the control volume shown is2inoutcond12() /12 W/m K(50 C30 C) / 0.01 m24,000 W/mqqqk TTL′′′′′′===−=⋅°−°=Since the heat flux in at the left face is only 20 W/m2, the conditions are not steady state.<COMMENTS:If the same heat flux is maintained until steady-state conditions are reached, thesteady-state temperature difference across the wall will beΔT=2/20 W/m0.01 m /12 W/m K0.0167 Kq L k′′=×⋅=which is much smaller than the specified temperature difference of 20°C.q” = 20 W/m2L= 10 mmk= 12 W/m·KT1= 50°CT2= 30°Cq″cond

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PROBLEM 1.3KNOWN:Inner surface temperature and thermal conductivity of a concrete wall.FIND:Heat loss by conduction through the wall as a function of outer surface temperatures ranging from-15 to 38°C.SCHEMATIC:ASSUMPTIONS:(1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3)Constant properties.ANALYSIS:From Fourier’s law, ifxq′′and k are each constant it is evident that the gradient,xdT dxqk′′= −, is a constant, and hence the temperature distribution is linear. The heat flux must beconstant under one-dimensional, steady-state conditions; and k is approximately constant if it dependsonly weakly on temperature. The heat flux and heat rate when the outside wall temperature is T2= -15°Care()212x25 C15 CdTTTqkk1W m K133.3 W mdxL0.30 m− −−′′ = −==⋅=oo.(1)22xxqqA133.3 W m20 m2667 W′′=×=×=.(2)<Combining Eqs. (1) and (2), the heat rate qxcan be determined for the range of outer surface temperature,-15≤T2≤38°C, with different wall thermal conductivities, k.For the concrete wall, k = 1 W/m⋅K, the heat loss varies linearly from +2667 W to -867 W and is zerowhen the inside and outer surface temperatures are the same. The magnitude of the heat rate increaseswith increasing thermal conductivity.COMMENTS:Without steady-state conditions and constant k, the temperature distribution in a planewall would not be linear.-20-10010203040Ambient air temperature, T2 (C)-1500-500500150025003500Heat loss, qx (W)Wall thermal conductivity, k = 1.25 W/m.Kk = 1 W/m.K, concrete wallk = 0.75 W/m.KOutside surface-20-10010203040Ambient air temperature, T2 (C)-1500-500500150025003500Heat loss, qx (W)Wall thermal conductivity, k = 1.25 W/m.Kk = 1 W/m.K, concrete wallk = 0.75 W/m.KOutside surface

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PROBLEM 1.4KNOWN:Dimensions, thermal conductivity and surface temperatures of a concrete slab. Efficiencyof gas furnace and cost of natural gas.FIND:Daily cost of heat loss.SCHEMATIC:ASSUMPTIONS:(1) Steady state, (2) One-dimensional conduction, (3) Constant properties.ANALYSIS:The rate of heat loss by conduction through the slab is()()12TT7 CqkLW1.4 W / m K 11m8 m4312 Wt0.20 m−°==⋅×=<The daily cost of natural gas that must be combusted to compensate for the heat loss is()()gd6fq C4312 W$0.02 / MJCt24 h / d3600s / h$8.28 / d0.910 J / MJη×=Δ=×=×<COMMENTS:The loss could be reduced by installing a floor covering with a layer of insulationbetween it and the concrete.

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PROBLEM 1.5KNOWN:Thermal conductivity and thickness of a wall. Heat flux through wall. Steady-stateconditions.FIND:Value of temperature gradient in K/m and in°C/m.SCHEMATIC:L =20 mmk= 2.3 W/m·Kq”x= 10 W/m2xASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties.ANALYSIS:Under steady-state conditions,"210 W/m4.35 K/m4.35C/m2.3 W/m KxdTqdxk= −= −= −= −°⋅<Since the K units here represent a temperaturedifference, and since the temperature difference is thesame in K and°C units, the temperature gradient value is the same in either units.COMMENTS:A negative value of temperature gradient means that temperature is decreasing withincreasingx, corresponding to a positive heat flux in thex-direction.

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PROBLEM 1.6KNOWN:Heat flux and surface temperatures associated with a wood slab of prescribedthickness.FIND:Thermal conductivity, k, of the wood.SCHEMATIC:ASSUMPTIONS:(1) One-dimensional conduction in the x-direction, (2) Steady-stateconditions, (3) Constant properties.ANALYSIS:Subject to the foregoing assumptions, the thermal conductivity may bedetermined from Fourier’s law, Eq. 1.2. Rearranging,()LW0.05mk=q40TTm40-20Cx212′′=−ok = 0.10 W / m K.⋅<COMMENTS:Note that the°C or K temperature units may be used interchangeably whenevaluating a temperature difference.

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PROBLEM 1.7KNOWN:Inner and outer surface temperatures of a glass window of prescribed dimensions.FIND:Heat loss through window.SCHEMATIC:ASSUMPTIONS:(1) One-dimensional conduction in the x-direction, (2) Steady-stateconditions, (3) Constant properties.ANALYSIS:Subject to the foregoing conditions the heat flux may be computed fromFourier’s law, Eq. 1.2.()TTqkL15-5CWq1.4 m K0.005mq2800 W/m.12xx2x−′′ =′′ =⋅′′ =oSince the heat flux is uniform over the surface, the heat loss (rate) isq = qxAq = 2800 W / m23m2′′ ××q = 8400 W.<COMMENTS:A linear temperature distribution exists in the glass for the prescribedconditions.

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PROBLEM 1.8KNOWN:Net power output, average compressor and turbine temperatures, shaft dimensions andthermal conductivity.FIND:(a) Comparison of the conduction rate through the shaft to the predicted net power output ofthe device, (b) Plot of the ratio of the shaft conduction heat rate to the anticipated net power output ofthe device over the range0.005 m≤L≤1 mand feasibility of aL= 0.005 m device.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Net power output isproportional to the volume of the gas turbine.PROPERTIES:Shaft (given):k= 40 W/m⋅K.ANALYSIS:(a) The conduction through the shaft may be evaluated using Fourier’s law, yielding()()232()40W/m K(1000400) C"/ 4(7010m)/ 492.4W1mhcck TTqqAdLππ−−⋅−°===×=The ratio of the conduction heat rate to the net power output is6692.4W18.510510 WqrP−===××<(b) The volume of the turbine is proportional toL3. DesignatingLa= 1 m,da= 70 mm andPaas theshaft length, shaft diameter, and net power output, respectively, in part (a),d=da×(L/La);P=Pa×(L/La)3and the ratio of the conduction heat rate to the net power output is()()()222323266222()()()/ 4// 4/"4(/)40W/m K(1000400) C(7010m)1m / 510 W18.510m4=hchchcaaaaacaak TTk TTk TTdd L Ld LPqALLrPPPL LLLLππππ−−−−−====⋅−°××××=Continued…L =1md =70 mmP =5 MWTurbineCompressorShaftTh=1000°CTc=400°CCombustionchamberL =1md =70 mmP =5 MWTurbineCompressorShaftTh=1000°CTc=400°CCombustionchamberk= 40 W/m·K

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PROBLEM 1.8 (Cont.)The ratio of the shaft conduction to net power is shown below. AtL= 0.005 m = 5 mm, the shaftconduction to net power output ratio is 0.74. The concept of the very small turbine is not feasible sinceit will be unlikely that the large temperature difference between the compressor and turbine can bemaintained.<COMMENTS:(1) The thermodynamics analysis does not account for heat transfer effects and istherefore meaningful only when heat transfer can be safely ignored, as is the case for the shaft in part(a). (2) Successful miniaturization of thermal devices is often hindered by heat transfer effects thatmust be overcome with innovative design.Ratio of shaft conduction to net power00.20.40.60.81L (m)0.00010.0010.010.11r

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PROBLEM 1.9KNOWN:Width, height, thickness and thermal conductivity of a single pane window andthe air space of a double pane window. Representative winter surface temperatures of singlepane and air space.FIND:Heat loss through single and double pane windows.SCHEMATIC:ASSUMPTIONS:(1) One-dimensional conduction through glass or air, (2) Steady-stateconditions, (3) Enclosed air of double pane window is stagnant (negligible buoyancy inducedmotion).ANALYSIS:From Fourier’s law, the heat losses areSingle Pane:()TT35C212qkA1.4 W/m K 2m19, 600 WggL0.005m−==⋅=o<Double Pane:()TT25C212qk A0.024 2m120 WaaL0.010 m−===o<COMMENTS:Losses associated with a single pane are unacceptable and would remainexcessive, even if the thickness of the glass were doubled to match that of the air space. Theprincipal advantage of the double pane construction resides with the low thermal conductivityof air (~ 60 times smaller than that of glass). For a fixed ambient outside air temperature, useof the double pane construction would also increase the surface temperature of the glassexposed to the room (inside) air.

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PROBLEM 1.10KNOWN:Dimensions of freezer compartment. Inner and outer surface temperatures.FIND:Thickness of styrofoam insulation needed to maintain heat load below prescribedvalue.SCHEMATIC:ASSUMPTIONS:(1) Perfectly insulated bottom, (2) One-dimensional conduction through 5walls of area A = 4m2, (3) Steady-state conditions, (4) Constant properties.ANALYSIS:Using Fourier’s law, Eq. 1.2, the heat rate isq = qA = kTLAtotal′′ ⋅ΔSolving for L and recognizing that Atotal= 5×W2, findL =5 kT Wq2Δ()()50.03 W/m K35 --10C4mL =500 W2⎡⎤×⋅⎣⎦oL = 0.054m = 54mm.<COMMENTS:The corners will cause local departures from one-dimensional conductionand a slightly larger heat loss.

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PROBLEM 1.11KNOWN:Heat flux at one face and air temperature and convection coefficient at other face of planewall. Temperature of surface exposed to convection.FIND:If steady-state conditions exist. If not, whether the temperature is increasing or decreasing.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) No internal energy generation.ANALYSIS:Conservation of energy for a control volume around the wall givesstinoutgdEEEEdt=−+&&&[]stinin222()()20 W/m20 W/mK(50 C30 C)380 W/mssdEq AhA TTqh TTAdtAA∞∞′′′′=−−=−−=−⋅°−°= −⎡⎤⎣⎦SincedEst/dt≠0, the system is not at steady-state.<SincedEst/dt< 0, the stored energy is decreasing, therefore the wall temperature is decreasing.<COMMENTS:When the surface temperature of the face exposed to convection cools to 31°C,qin=qoutanddEst/dt= 0 and the wall will have reached steady-state conditions.q” = 20 W/m2Ts= 50°Ch =20 W/m2·KT∞= 30°CAirq”conv

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PROBLEM 1.12KNOWN:Dimensions and thermal conductivity of food/beverage container. Inner and outersurface temperatures.FIND:Heat flux through container wall and total heat load.SCHEMATIC:ASSUMPTIONS:(1) Steady-state conditions, (2) Negligible heat transfer through bottomwall, (3) Uniform surface temperatures and one-dimensional conduction through remainingwalls.ANALYSIS:From Fourier’s law, Eq. 1.2, the heat flux is()0.023 W/m K202CTT221qk16.6 W/mL0.025 m⋅−−′′ ===o<Since the flux is uniform over each of the five walls through which heat is transferred, theheat load is()qqAqH 2W2WWWtotal1212′′′′ ⎡⎤=×=++×⎣⎦()()2q16.6 W/m0.6m 1.6m1.2m0.8m0.6m35.9 W⎡⎤=++×=⎣⎦<COMMENTS:The corners and edges of the container create local departures from one-dimensional conduction, which increase the heat load. However, for H, W1, W2>> L, theeffect is negligible.

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PROBLEM 1.13KNOWN:Masonry wall of known thermal conductivity has a heat rate which is 80% of thatthrough a composite wall of prescribed thermal conductivity and thickness.FIND:Thickness of masonry wall.SCHEMATIC:ASSUMPTIONS:(1) Both walls subjected to same surface temperatures, (2) One-dimensional conduction, (3) Steady-state conditions, (4) Constant properties.ANALYSIS:For steady-state conditions, the conduction heat flux through a one-dimensional wall follows from Fourier’s law, Eq. 1.2,′′q= kTLΔwhereΔT represents the difference in surface temperatures. SinceΔT is the same for bothwalls, it follows thatL= Lkkqq121221⋅′′′′.With the heat fluxes related as′′ =′′q0.8 q12L= 100mm 0.75 W / m K0.25 W / m K10.8= 375mm.1⋅⋅×<COMMENTS:Not knowing the temperature difference across the walls, we cannot find thevalue of the heat rate.

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PROBLEM 1.14KNOWN:Expression for variable thermal conductivity of a wall.Constant heat flux.Temperature at x = 0.FIND:Expression for temperature gradient and temperature distribution.SCHEMATIC:ASSUMPTIONS:(1) One-dimensional conduction.ANALYSIS:The heat flux is given by Fourier’s law, and is known to be constant, thereforexdTqkconstantdx′′ = −=Solving for the temperature gradient and substituting the expression for k yieldsxxqqdTdxkaxb′′′′= −= −+<This expression can be integrated to find the temperature distribution, as follows:xqdT dxdxdxaxb′′= −+∫∫Sincexqconstant′′ =, we can integrate the right hand side to find()xqTln axbca′′= −++where c is a constant of integration. Applying the known condition that T = T1at x = 0,we can solve for c.Continued…q”xk= ax+ bT1

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