Solution Manual For Fundamentals of Heat and Mass Transfer, 8th Edition

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PROBLEM 1.1KNOWN:Temperature distribution in wall of Example 1.1.FIND:Heat fluxes and heat rates atx= 0 andx = L.SCHEMATIC:ASSUMPTIONS:(1) One-dimensional conduction through the wall, (2) constant thermal conductivity,(3) no internal thermal energy generation within the wall.PROPERTIES:Thermal conductivity of wall (given):k= 1.7 W/m·K.ANALYSIS:The heat flux in the wall is by conduction and is described by Fourier’s law,xdTqk dx′′ = −(1)Since the temperature distribution isT(x) =a+bx, the temperature gradient isdTbdx=(2)Hence, the heat flux is constant throughout the wall, and is()21.7 W/m K1000 K/m1700 W/mxdTqkkbdx′′ = −= −= −⋅× −=<Since the cross-sectional area through which heat is conducted is constant, the heat rate is constant and is()2()1700 W/m1.2 m × 0.5 m1020 WxxqqWH′′=××=×=<Because the heat rate into the wall is equal to the heat rate out of the wall, steady-state conditions exist.<COMMENTS:(1)If the heat rates were not equal, the internal energy of the wall would be changingwith time. (2) The temperatures of the wall surfaces areT1= 1400 K andT2= 1250 K.

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PROBLEM 1.2KNOWN:Thermal conductivity, thickness and temperature difference across a sheet of rigidextruded insulation.FIND:(a) The heat flux through a 3 m×3 m sheet of the insulation, (b) the heat rate throughthe sheet, and (c) the thermal conduction resistance of the sheet.SCHEMATIC:ASSUMPTIONS:(1) One-dimensional conduction in the x-direction, (2) Steady-stateconditions, (3) Constant properties.ANALYSIS:(a) From Equation 1.2 the heat flux is12x2T - TdTW12 KWq= -k= k= 0.029×= 13.9dxLm K0.025 mm′′⋅<(b) The heat rate is2xx2Wq= qA = 13.9× 9 m= 125 Wm′′ ⋅<(c) From Eq. 1.11, the thermal resistance ist,condxRT / q= 12 K /125 W0.096 K/W= ∆=<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. (4) The conductionthermal resistance for a plane wall could equivalently be calculated from Rt,cond= L/kA.qcondA = 4 m2T2T1k = 0.029xL = 20 mmT1– T2= 10˚CqcondA = 4 m2T2T1k = 0.029xL = 20 mmT1– T2= 10˚C9 m212°C25 mmqcondA = 4 m2T2T1k = 0.029xL = 20 mmT1– T2= 10˚CqcondA = 4 m2T2T1k = 0.029xL = 20 mmT1– T2= 10˚C9 m212°C25 mm

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PROBLEM 1.3KNOWN: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 be∆T=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 1.4KNOWN: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 rangingfrom -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− −−′′ = −==⋅=.(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 1.5KNOWN: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:Furnace,= 0.90ηfNatural gas,C= $0.01/MJgConcrete, k = 1.4 W/m-Kt = 0.2 mL = 11 mW = 8 mT= 17 C1oT= 10 C2oqWarm airASSUMPTIONS:(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×=∆=×=×<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 1.6KNOWN: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′′=-k = 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 and thermal resistance of a glass window ofprescribed dimensions.FIND:Heat loss through window. Thermal conductivity of glass.SCHEMATIC:A =1m x 3m = 3m2,Rt,cond=1.19 x 10-3K/WASSUMPTIONS:(1) One-dimensional conduction in the x-direction, (2) Steady-stateconditions, (3) Constant properties.ANALYSIS:From Eq. 1.11,()3t,cond15-5CTTq8400 WR1.1910K/W12x--===×<The thermal resistance due to conduction for a plane wall is related to the thermalconductivity and dimensions according tot,condR= L/kATherefore32t,condkL/(RA)0.005 m / (1.1910K/W3 m )1.40 W/m K-==××=⋅<COMMENTS:The thermal conductivity value agrees with the value for glass in Table A.3.

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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: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 1.10KNOWN: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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PROBLEM 1.10 (Cont.)1x1x1T(x0)Tqln bcTaqcTln ba==′′−+=′′=+Therefore, the temperature distribution is given by()xx1qqTln axbTln baa′′′′= −+++<x1qbTlnaaxb′′=++<COMMENTS:Temperature distributions are not linear in many situations, such as when thethermal conductivity varies spatially or is a function of temperature. Non-linear temperaturedistributions may also evolve if internal energy generation occurs or non-steady conditions exist.

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PROBLEM 1.11KNOWN:Thickness, diameter and inner surface temperature of bottom of pan used to boilwater. Rate of heat transfer to the pan.FIND:Outer surface temperature of pan for an aluminum and a copper bottom.SCHEMATIC:ASSUMPTIONS:(1) One-dimensional, steady-state conduction through bottom of pan.ANALYSIS:From Fourier’s law, the rate of heat transfer by conduction through the bottomof the pan isTT12qkAL−=Hence,qLTT12kA=+where()222AD/ 40.22m/ 40.038 m .ππ===Aluminum:()()600W 0.008 mT110C110.5C12240 W/m K 0.038 m=+=⋅<Copper:()()600W 0.008 mT110C110.3C12390 W/m K 0.038 m=+=⋅<COMMENTS:Although the temperature drop across the bottom is slightly larger foraluminum (due to its smaller thermal conductivity), it is sufficiently small to be negligible forboth materials. To a good approximation, the bottom may be consideredisothermalat T≈110°C, which is a desirable feature of pots and pans.Aluminum(k=240 W/m-K)orCopper(k=390 W/m-K)L = 5 mmD = 200 mmq = 600 WT1T= 110 C2oL = 8 mmD = 220 mm

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PROBLEM 1.12KNOWN:Hand experiencing convection heat transfer with moving air and water.FIND:Determine which condition feels colder. Contrast these results with a heat loss of 30 W/m2undernormal room conditions.SCHEMATIC:ASSUMPTIONS:(1) Temperature is uniform over the hand’s surface, (2) Convection coefficient isuniform over the hand, and (3) Negligible radiation exchange between hand and surroundings in the caseof air flow.ANALYSIS:The hand will feel colder for the condition which results in the larger heat loss. The heatloss can be determined from Newton’s law of cooling, Eq. 1.3a, written as()sqh TT∞′′ =−For the air stream:()22airq40 W mK 308K1,520 W m′′=⋅− −=<For the water stream:()22waterq900 W mK 3010 K18, 000 W m′′=⋅−=<COMMENTS:The heat loss for the hand in the water stream is an order of magnitude larger than whenin the air stream for the given temperature and convection coefficient conditions. In contrast, the heatloss in a normal room environment is only 30 W/m2which is a factor of 50 times less than the loss in theair stream. In the room environment, the hand would feel comfortable; in the air and water streams, asyou probably know from experience, the hand would feel uncomfortably cold since the heat loss isexcessively high.

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PROBLEM 1.13KNOWN:Power required to maintain the surface temperature of a long, 25-mm diameter cylinderwith an imbedded electrical heater for different air velocities.FIND:(a) Determine the convection coefficient for each of the air velocity conditions and displaythe results graphically, and (b) Assuming that the convection coefficient depends upon air velocity ash = CVn, determine the parameters C and n.SCHEMATIC:V(m/s)124812¢Pe(W/m)45065898315071963h (W/m2⋅K)22.032.248.173.896.1ASSUMPTIONS:(1) Temperature is uniform over the cylinder surface, (2) Negligible radiationexchange between the cylinder surface and the surroundings, (3) Steady-state conditions.ANALYSIS:(a) From an overall energy balance on the cylinder, the power dissipated by theelectrical heater is transferred by convection to the air stream. Using Newton’s law of cooling on aper unit length basis,()()esPhDTTπ∞′ =−whereeP′is the electrical power dissipated per unit length of the cylinder. For the V = 1 m/scondition, using the data from the table above, find()2h450 W m0.025 m 30040C22.0 W mKπ=×−=⋅<Repeating the calculations, find the convection coefficients for the remaining conditions which aretabulated above and plotted below. Note that h is not linear with respect to the air velocity.(b) To determine the (C,n) parameters, we plotted h vs. V on log-log coordinates. Choosing C =22.12 W/m2⋅K(s/m)n, assuring a match at V = 1, we can readily find the exponent n from the slope ofthe h vs. V curve. From the trials with n = 0.8, 0.6 and 0.5, we recognize that n = 0.6 is a reasonablechoice. Hence, C = 22.12 and n = 0.6.<024681012Air velocity, V (m/s)20406080100Coefficient, h (W/m^2.K)Data, smooth curve, 5-points1246810Air velocity, V (m/s)1020406080100Coefficient, h (W/m^2.K)Data , smooth curve, 5 pointsh = C * V^n, C = 22.1, n = 0.5n = 0.6n = 0.8COMMENTS:Radiation may not be negligible, depending on surface emissivity.

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