Solution Manual for Heat Transfer , 1st Edition

Preview (16 of 2666 Pages)

100%

Solution Manual for Heat Transfer , 1st Edition - Page 1 preview image

Loading page ...

P1.1-2 (1-1 in text): Conductivity of a dilute gasSection 1.1.2 provides an approximation for the thermal conductivity of a monatomic gas at idealgas conditions. Test the validity of this approximation by comparing the conductivity estimatedusing Eq. (1-18) to the value of thermal conductivity for a monotonic ideal gas (e.g., lowpressure argon) provided by the internal function in EES.Note that the molecular radius,σ, isprovided in EES by the Lennard-Jones potential using the functionsigma_LJ.a.)What is the value and units of the proportionality constant required to make Eq. (1-18) anequality?Equation (1-18) is repeated below:2vcTkMWσ∝(1)Equation (1) is written as an equality by including a constant of proportionality (Ck):2vkcTkCMWσ=(2)Solving forCkleads to:2kvkMWCcTσ=(3)which indicates thatCkhas units m-kg1.5/s-kgmol05-K0.5.The inputs are entered in EES for Argon at relatively low pressure (0.1 MPa) and 300 K."Problem 1.1-2"$UnitSystem SI MASS RAD PA K J$TABSTOPS0.2 0.4 0.6 0.8 3.5 inT=300 [K]"temperature"F$='Argon'"fluid"P_MPa=0.1 [MPa]"pressure, in MPa"P=P_MPa*convert(MPa, Pa)"pressure"The conductivity, specific heat capacity, Lennard-Jones potential, and molecular weight ofArgon (k,cv,σ, andMW) are evaluated using EES' built-in funcions.Equation (3) is used toevaluate the proportionality constant.k=conductivity(F$,T=T,P=P)"conductivity"cv=cv(F$,T=T,P=P)"specific heat capacity at constant volume"MW=molarMass(F$)"molecular weight"sigma=sigma_LJ(F$)"Lennard-Jones potential"C_k=k*sigma^2*sqrt(MW/T)/cv"constant of proportionality"

Solution Manual for Heat Transfer , 1st Edition - Page 2 preview image

Loading page ...

Solution Manual for Heat Transfer , 1st Edition - Page 3 preview image

Loading page ...

which leads toCk= 2.619x10-24m-kg1.5/s-kgmol0.5-K0.5.b.) Plot the value of the proportionality constant for 300 K argon at pressures between 0.01 and100 MPa on a semi-log plot with pressure on the log scale.At what pressure does theapproximation given in Eq. (1-18) begin to fail at 300 K for argon?Figure 1 illustrates the constant of proportionality as a function of pressure for argon at 300 K.The approximation provided by Eq. (1-18) breaks down at approximately 1 MPa.0.0010.010.11101000x10010-242x10-243x10-244x10-245x10-246x10-247x10-248x10-24Pressure (MPa)Ck(m-kg1.5/s-kgmol0.5-K0.5)Figure 1: Constant of proportionality in Eq. (3) as a function of pressure for argon at 300 K.

Solution Manual for Heat Transfer , 1st Edition - Page 4 preview image

Loading page ...

Problem 1.2-3 (1-2 in text): Conduction through a WallFigure P1.2-3 illustrates a plane wall made of a very thin (thw= 0.001 m) and conductive (k=100 W/m-K) material that separates two fluids, A and fluid B. Fluid A is atTA= 100°C and theheat transfer coefficient between the fluid and the wall isAh= 10 W/m2-K while fluid B is atTB= 0°C withBh= 100 W/m2-K.thw= 0.001 mk= 100 W/m-K2100 C10 W/m -KAATh=°=20 C100 W/m -KBBTh=°=Figure P1.2-3: Plane wall separating two fluidsa.) Draw a resistance network that represents this situation and calculate the value of eachresistor (assuming a unit area for the wall,A= 1 m2).Heat flowing from fluid A to fluid B must pass through a fluid A-to-wall convective resistance(Rconv,A), a resistance to conduction through the wall (Rcond), and a wall-to-fluid B convectiveresistance (Rconv,B).These resistors are in series.The network and values of the resistors areshown in Figure 2.TA= 100°C,1conv AARhA=wcondtRk A=K0.1 WK0.0001 W,1cond BBRhA=K0.01 WTB= 0°CFigure 2: Thermal resistance network representing the wall.b.) If you wanted to predict the heat transfer rate from fluidAtoBvery accurately, then whichparameter (e.g.,thw,k, etc.) would you try to understand/measure very carefully and whichparameters are not very important? Justify your answer.The largest resistance in a series network will control the heat transfer.For the wall above, thelargest resistance isRconv,A.Therefore, I would focus on predicting this resistance accurately.This would suggest thatAhis the most important parameter and the others do not matter much.

Solution Manual for Heat Transfer , 1st Edition - Page 5 preview image

Loading page ...

Problem 1.2-8 (1-3 in text): Frozen GuttersYou have a problem with your house.Every spring at some point the snow immediatelyadjacent to your roof melts and runs along the roof line until it reaches the gutter. The water inthe gutter is exposed to air at temperature less than 0°C and therefore freezes, blocking the gutterand causing water to run into your attic. The situation is shown in Figure P1.2-8.snow melts at this surface2,15 W/m -KoutoutTh=snow,ks= 0.08 W/m-K222 C,10 W/m -KininTh=°=Lins= 3 inchinsulation,kins= 0.05 W/m-Kplywood,0.5 inch,0.2 W/m-KppLk==Ls= 2.5 inchFigure P1.2-8: Roof of your house.The air in the attic is atTin= 22°C and the heat transfer coefficient between the inside air and theinner surface of the roof isinh= 10 W/m2-K.The roof is composed of aLins= 3.0 inch thickpiece of insulation with conductivitykins= 0.05 W/m-K that is sandwiched between twoLp= 0.5inch thick pieces of plywood with conductivitykp= 0.2 W/m-K. There is anLs= 2.5 inch thicklayer of snow on the roof with conductivityks= 0.08 W/m-K.The heat transfer coefficientbetween the outside air at temperatureToutand the surface of the snow isouth= 15 W/m2-K.Neglect radiation and contact resistances for part (a) of this problem.a.) What is the range of outdoor air temperatures where you should be concerned that yourgutters will become blocked by ice?The input parameters are entered in EES and converted to base SI units (N, m, J, K) in order toeliminate any unit conversion errors; note that units should still be checked as you work theproblem but that this is actually a check on the unit consistency of the equations."P1.2-8: Frozen Gutters"$UnitSystem SI MASS RAD PA K J$TABSTOPS0.2 0.4 0.6 0.8 3.5 inT_in=converttemp(C,K,22)"temperature in your attic"L_ins=3 [inch]*convert(inch,m)"insulation thickness"L_p=0.5 [inch]*convert(inch,m)"plywood thickness"k_ins=0.05 [W/m-K]"insulation conductivity"k_p=0.2 [W/m-K]"plywood conductivity"k_s=0.08 [W/m-K]"snow conductivity"L_s=2.5 [inch]*convert(inch,m)"snow thickness"h_in=10 [W/m^2-K]"heat transfer coefficient between attic air and inner surface of roof"h_out=15 [W/m^2-K]"heat transfer coefficient between outside air and snow"A=1 [m^2]"per unit area"

Solution Manual for Heat Transfer , 1st Edition - Page 6 preview image

Loading page ...

The problem may be represented by the resistance network shown in Figure 2.Figure 2: Resistance network representing the roof of your house.The network includes resistances that correspond to convection with the inside and outside air:,1conv outoutRhA=(1),1conv ininRhA=(2)whereAis 1 m2in order to accomplish the problem on a per unit area basis.There are alsoconduction resistances associated with the insulation, plywood and snow:insinsinsLRkA=(3)pppLRkA=(4)sssLRk A=(5)R_conv_out=1/(h_out*A)"outer convection resistance"R_s=L_s/(k_s*A)"snow resistance"R_p=L_p/(k_p*A)"plywood resistance"R_ins=L_ins/(k_ins*A)"insulation resistance"R_conv_in=1/(h_in*A)"inner convection resistance"Which leads toRconv,out= 0.07 K/W,Rs= 0.79 K/W,Rp= 0.06 K/W,Rins= 1.52 K/W andRconv,in= 0.10 K/W.Therefore, the dominant effects for this problem are conduction through theinsulation and the snow; the other effects (convection and the plywood conduction) are notterribly important since the largest resistances will dominate in a series network.If the snow at the surface of the room is melting then the temperature at the connection betweenRsandRpmust beTs= 0°C (see Figure 2). Therefore, the heat transferred through the roof (qinFigure 2) must be:

Solution Manual for Heat Transfer , 1st Edition - Page 7 preview image

Loading page ...

(),2insconv inpinsTTqRRR−=++(6)The temperature of the outside air must therefore be:(),outssconv outTTq RR=−+(7)T_s=converttemp(C,K,0)"roof-to-snow interface temperature must be melting point of water"q_dot=(T_in-T_s)/(R_conv_in+2*R_p+R_ins)"heat transfer from the attic to the snow when melting point is reached"T_out=T_s-q_dot*(R_s+R_conv_out)"outside temperature required to reach melting point at roof surface"T_out_C=converttemp(K,C,T_out)"outside temperature in C"which leads toTout= -10.8°C. If the temperature is below this then the roof temperature will bebelow freezing and the snow will not melt.If the temperature is above 0°C then the water willnot refreeze upon hitting the gutter. Therefore, the range of temperatures of concern are -10.8°C<Tout< 0°C.b.) Would your answer change much if you considered radiation from the outside surface of thesnow to surroundings atTout? Assume that the emissivity of snow isεs= 0.82.The modified resistance network that includes radiation is shown in Figure 3.Figure 3: Resistance network representing the roof of your house and including radiation.The additional resistance for radiation is in parallel with convection from the surface of the snowas heat is transferred from the surface by both mechanisms.The radiation resistance can becalculated approximately according to:314radsRTAεσ=(8)whereTis the average temperature of the surroundings and the snow surface. In order to get aquick idea of the magnitude of this resistance we can approximateTwith its largest possiblevalue (which will result in the largest possible amount of radiation); the maximum temperatureof the snow is 0°C:e_s=0.82 [-]"emissivity of snow"

Solution Manual for Heat Transfer , 1st Edition - Page 8 preview image

Loading page ...

R_rad=1/(4*T_s^3*e_s*sigma#*A)"radiation resistance"which leads toRrad= 0.26 K/W.Notice thatRradis much larger thanRconv,out; the smallestresistance in a parallel combination dominates and therefore the impact of radiation will beminimal.Furthermore,Rconv,outis not even a very important resistance in the original seriescircuit shown in Figure 2.

Solution Manual for Heat Transfer , 1st Edition - Page 9 preview image

Loading page ...

Problem P1.2-11 (1-4 in text)Figure P1.2-11(a) illustrates a composite wall. The wall is composed of two materials (A withkA= 1 W/m-K and B withkB= 5 W/m-K), each has thicknessL= 1.0 cm. The surface of the wall atx= 0 is perfectly insulated. A very thin heater is placed between the insulation and material A;the heating element provides25000 W/mq′′ =of heat.The surface of the wall atx= 2Lisexposed to fluid atTf,in= 300 K with heat transfer coefficientinh= 100 W/m2-K.L= 1 cmL= 1 cmx25000 W/mq′′ =insulatedmaterial AkA= 1 W/m-Kmaterial BkB= 5 W/m-K,2300 K100 W/m -Kf ininTh==Figure P1.2-11(a): Composite wall with a heater.You may neglect radiation and contact resistance for parts (a) through (c) of this problem.a.) Draw a resistance network to represent this problem; clearly indicate what each resistancerepresents and calculate the value of each resistance.The input parameters are entered in EES:“P1.2-11: Heater"$UnitSystem SI MASS RAD PA K J$TABSTOPS0.2 0.4 0.6 0.8 3.5 in"Inputs"q_flux=100 [W/m^2]"heat flux provided by the heater"L = 1.0 [cm]*convert(cm,m)"thickness of each layer"k_A=1.0 [W/m-K]"conductivity of material A"k_B=5.0 [W/m-K]"conductivity of material B"T_f_in=300 [K]"fluid temperature at inside surface"h_in=100 [W/m^2-K]"heat transfer on inside surface"A=1 [m^2]"per unit area"The resistance network that represents the problem shown in Figure 2 is:Figure 2: Resistance network.The resistances due to conduction through materials A and B are:

Solution Manual for Heat Transfer , 1st Edition - Page 10 preview image

Loading page ...

AALRkA=(1)BBLRkA=(2)whereAis the area of the wall, taken to be 1 m2in order to carry out the analysis on a per unitarea basis. The resistance due to convection is:,1conv ininRhA=(3)"part (a)"R_A=L/(k_A*A)"resistance to conduction through A"R_B=L/(k_B*A)"resistance to conduction through B"R_conv_in=1/(h_in*A)"resistance to convection on inner surface"which leads toRA= 0.01 K/W,RB= 0.002 K/W, andRconv,in= 0.01 K/W.b.) Use your resistance network from (a) to determine the temperature of the heating element.The resistance network for this problem is simple; the temperature drop across each resistor isequal to the product of the heat transferred through the resistor and its resistance. In this simplecase, all of the heat provided by the heater must pass through materials A, B, and into the fluidby convection so these resistances are in series. The heater temperature (Thtr) is therefore:(),,htrf inABconv inTTRRRq A′′=+++(4)T_htr=T_f_in+(R_A+R_B+R_conv_in)*q_flux*A"heater temperature"which leads toThtr= 410 K.c.) Sketch the temperature distribution on the axes provided below. Make sure that the sketch isconsistent with your solution from (b).The temperatures atx=Landx= 2Lcan be computed according to:(),,xLf inBconv inTTRRq A=′′=++(5)2,,xLf inconv inTTRq A=′′=+(6)T_L=T_f_in+(R_B+R_conv_in)*q_flux*A"temperature at x=L"T_2L=T_f_in+R_conv_in*q_flux*A"temperature at x=2L"

Solution Manual for Heat Transfer , 1st Edition - Page 11 preview image

Loading page ...

which leads toTx=L= 360 K andTx=2L= 350 K. The temperature distribution is sketched on theaxes in Figure 3.Figure 3: Sketch of temperature distribution.Notice that the temperature drop through the two larger resistances (RAandRB) are much largerthan the temperature drop across the small resistance,RB.Figure P1.2-11(b) illustrates the same composite wall shown in Figure P1.2-11(a), but there is anadditional layer added to the wall, material CwithkC= 2.0 W/m-K andL= 1.0 cm.L= 1 cmL= 1 cmx25000 W/mq′′ =insulatedmaterial AkA= 1 W/m-Kmaterial BkB= 5 W/m-K,2300 K100 W/m -Kf ininTh==L= 1 cmmaterial CkC= 2 W/m-KFigure P1.2-11(b): Composite wall with Material C.Neglect radiation and contact resistance for parts (d) through (f) of this problem.d.) Draw a resistance network to represent the problem shown in Figure P1.2-11(b); clearlyindicate what each resistance represents and calculate the value of each resistance.There is an additional resistor corresponding to conduction through material C,RC, as shownbelow:

Solution Manual for Heat Transfer , 1st Edition - Page 12 preview image

Loading page ...

Notice that the boundary condition at the end ofRCcorresponds to the insulated wall; that is, noheat can be transferred through this resistance. The resistance to conduction through material Cis:CCLRkA=(7)"part (b)"k_C=2.0 [W/m-K]"conductivity of material C"R_C=L/(k_C*A)"resistance to conduction through C"which leads toRC= 0.005 K/W.e.) Use your resistance network from (d) to determine the temperature of the heating element.Because there is no heat transferred throughRC, all of the heat must still go through materials Aand B and be convected from the inner surface of the wall. Therefore, the answer is not changedfrom part (b),Thtr= 410 K.f.) Sketch the temperature distribution on the axes provided below. Make sure that the sketch isconsistent with your solution from (e).The answer is unchanged from part (c) except that there is material to the left of the heater.However, no heat is transferred through material C and therefore there is no temperature gradientin the material.

Solution Manual for Heat Transfer , 1st Edition - Page 13 preview image

Loading page ...

Figure P1.2-11(c) illustrates the same composite wall shown in Figure P1.2-11(b), but there is acontact resistance between materials A and B,20.01 K-m /WcR′′ =, and the surface of the wall atx= -Lis exposed to fluid atTf,out= 400 K with a heat transfer coefficientouth= 10 W/m2-K.L= 1 cmL= 1 cmx25000 W/mq′′=material AkA= 1 W/m-Kmaterial BkB= 5 W/m-K,2300 K100 W/m -Kf ininTh==L= 1 cmmaterial CkC= 2 W/m-K20.01 K-m /WcR′′=,2400 K10 W/m -Kf outoutTh==Figure P1.2-11(c): Composite wall with convection at the outer surface and contact resistance.Neglect radiation for parts (g) through (i) of this problem.g.) Draw a resistance network to represent the problem shown in Figure P1.2-11(c); clearlyindicate what each resistance represents and calculate the value of each resistance.The additional resistances associated with contact resistance and convection to the fluid at theouter surface are indicated.Notice that the boundary condition has changed; heat provided bythe heater has two paths (outqandinq) and so the problem is not as easy to solve.

Solution Manual for Heat Transfer , 1st Edition - Page 14 preview image

Loading page ...

The additional resistances are computed according to:,1conv outoutRhA=(8)ccontactRRA′′=(9)"part (c)"R``_c=0.01 [K-m^2/W]"area specific contact resistance"h_out=10 [W/m^2-K]"heat transfer coefficient"T_f_out=400 [K]"fluid temperature on outside surface"R_contact=R``_c/A"contact resistance"R_conv_out=1/(h_out*A)"convection resistance on outer surface"which leads toRcontact= 0.01 K/W andRconv,out= 0.1 K/W.h.) Use your resistance network from (j) to determine the temperature of the heating element.It is necessary to carry out an energy balance on the heater:inoutq Aqq′′=+(10)The heat transfer rates can be related toThtraccording to:(),,htrf ininAcontactBconv inTTqRRRR−=+++(11)(),,htrf outoutCconv outTTqRR−=+(12)Theseare3equationsin3unknowns,Thtr,outqandinq,andthereforecanbesolvedsimultaneously in EES (note that the previous temperature calculations from part (b) must becommented out):{T_htr=T_f_in+(R_A+R_B+R_conv_in)*q_flux*A"heater temperature"T_L=T_f_in+(R_B+R_conv_in)*q_flux*A"temperature at x=L"

Solution Manual for Heat Transfer , 1st Edition - Page 15 preview image

Loading page ...

T_2L=T_f_in+R_conv_in*q_flux*A"temperature at x=2L"}q_flux*A=q_dot_in+q_dot_out"energy balance on the heater"q_dot_in=(T_htr-T_f_in)/(R_A+R_contact+R_B+R_conv_in)"heat flow to inner fluid"q_dot_out=(T_htr-T_f_out)/(R_C+R_conv_out)"heat flow to outer fluid"which leads toThtr= 446 K.The other intermediate temperatures shown on the resistancediagram can be computed:xLhtrAinTTR q=−=−(13)()xLhtrAcontactinTTRRq=+=−+(14)()2xLhtrAcontactBinTTRRRq==−++(15)xLhtrCoutTTRq=−=−(16)"intermediate temperatures"T_Lm=T_htr-R_A*q_dot_inT_Lp=T_htr-(R_A+R_contact)*q_dot_inT_2L=T_htr-(R_A+R_contact+R_B)*q_dot_inT_mL=T_htr-R_C*q_dot_outwhich leads toTx=L-= 400.4 K,Tx=L+= 354.7 K,Tx=2L= 345.6 K, andTx=-L= 443.8 K.i.)Sketch the temperature distribution on the axes provided below.

Solution Manual for Heat Transfer , 1st Edition - Page 16 preview image

Loading page ...

Problems 1.2-12 (1-5 in text): Floor HeaterYou have decided to install a strip heater under the linoleum in your bathroom in order to keepyour feet warm on cold winter mornings.Figure P1.2-12 illustrates a cross-section of thebathroom floor. The bathroom is located on the first story of your house and isW= 2.5 m widebyL= 2.5 m long. The linoleum thickness isthL= 5 mm and has conductivitykL= 0.05 W/m-K.The strip heater under the linoleum is negligibly thin.Beneath the heater is a piece of plywoodwith thicknessthp= 5 mm and conductivitykp= 0.4 W/m-K. The plywood is supported byths=6 cm thick studs that areWs= 4 cm wide with thermal conductivityks= 0.4 W/m-K. The center-to-center distance between studs isps= 25.0 cm.Between each stud are pockets of air that canbe considered to be stagnant with conductivitykair= 0.025 W/m-K. A sheet of drywall is nailedto the bottom of the studs. The thickness of the drywall isthd= 9.0 mm and the conductivity ofdrywall iskd= 0.1 W/m-K. The air above in the bathroom is atTair,1= 15°C while the air in thebasement is atTair,2= 5°C.The heat transfer coefficient on both sides of the floor ish= 15W/m2-K. You may neglect radiation and contact resistance for this problem.strip heaterlinoleum,kL= 0.05 W/m-KthL= 5 mmthp= 5 mmplywood,kp= 0.4 W/m-Kths= 6 cmWs= 4 cmstuds,ks= 0.4 W/m-Kps= 25 cmthd= 9 mmdrywall,kd= 0.1 W/m-Kair,ka= 0.025 W/m-K2,115 C,15 W/m -KairTh=°=2,25 C,15 W/m -KairTh=°=Figure P1.2-12: Bathroom floor with heater.a.) Draw a thermal resistance network that can be used to represent this situation. Be sure tolabel the temperatures of the air above and below the floor (Tair,1andTair,2), the temperatureat the surface of the linoleum (TL), the temperature of the strip heater (Th), and the heat inputto the strip heater (hq) on your diagram.The resistance diagram corresponding to this problem is shown in Figure 2.

Preview Mode

This document has 2666 pages. Sign in to access the full document!

Report

Study Now!

Document Details

Related Documents

A system is a group of parts that work together to

Solution Manual for Introduction to Robotics: Mechanics and Control, 4th Edition

Solution Manual For Thermal Radiation Heat Transfer, 6th Edition

Solution Manual For Thermodynamics: An Interactive Approach, 1st Edition

Solution Manual for Fluid Mechanics for Engineers, 2017 Edition

Solution Manual For Applied Strength Of Materials, 5th Edition

Solution Manual for Machine Elements in Mechanical Design, 6th Edition

Mechanical Vibrations: Theory and Applications 1st Edition Solution Manual

Solution Manual For System Dynamics, 4th Edition

Solution Manual For Design Of Machine Elements, 8th Edition

Company

Explore

Study Tools