# Practice problem solutions chapters 3-6.pdf - PROBLEM 3.64...

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PROBLEM 3.64KNOWN:Wall of thermal conductivity k and thickness L with uniform generationq; strip heater with uniform heat flux oq ;′′prescribed inside and outside air conditions (hi, T,i, ho, T,o).FIND:(a) Sketch temperature distribution in wall if none of the heat generated within the wall is lost to the outside air, (b) Temperatures at the wall boundaries T(0) and T(L) for the prescribed condition, (c) Value of ′′qorequired to maintain this condition, (d) Temperature of the outer surface, T(L), if oq=0 but q′′corresponds to the value calculated in (c).SCHEMATIC:ASSUMPTIONS:(1) Steady-state conditions, (2) One-dimensional conduction, (3) Uniform volumetric generation, (4) Constant properties. ANALYSIS:(a) If none of the heat generated within the wall is lost to the outsideof the chamber, the gradient at x = 0 must be zero. Since qis uniform, the temperature distribution is parabolic, with T(L) > T,i. (b) To find temperatures at the boundaries of wall, begin with the general solution to the appropriate form of the heat equation (Eq.3.40). ()212qT xxC x+C2k=+(1) From the first boundary condition, x=o1dT0 C0.dx==(2) Two approaches are possible using different forms for the second boundary condition. Approach No. 1:With boundary condition ( )1T 0T=()21qT xxT2k=+(3) To find T1, perform an overall energy balance on the wallinoutgEEE0+=()(),i2,iqLh T LTqL=0 T LTTh+= =+(4) Continued …
PROBLEM 3.64 (Cont.)and from Eq. (3) with x = L and T(L) = T2, ()222112,iqqqLqLT LLT or TTL2k2kh2k=+=+=++(5,6) Substituting numerical values into Eqs. (4) and (6), find 322T50C+1000 W/m0.200 m/20 W/mK=50 C+10 C=60 C=×<()231T60C+1000 W/m0.200 m/ 24 W/mK=65 C.=××<Approach No. 2:Using the boundary condition ()x=L,idTkh T LTdx=yields the following temperature distribution which can be evaluated at x = 0,L for the required temperatures, ()()22,iqqLT xxLT.2kh=++(c) The value of oq′′when T(0) = T1= 65°follows from the circuit 1,oooTTq1/ h′′ =()22oq5 W/mK65-25C=200 W/m.′′ =<(d) With q=0,the situation is represented by the thermal circuit shown. Hence,oabqqq′′′′′′=+1,o1,iooiTTTTq1/ hL/k+1/h′′=+which yields 1T55 C.=<
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PROBLEM 3.66 KNOWN:Diameter, thermal conductivity and microbial energy generation rate in cylindrical hay bales. Ambient conditions. FIND:The maximum hay temperature for q= 1, 10, and 100 W/m3. SCHEMATIC: D= 2 mAirT= 0°C, h= 25 W/m2·K q= 1, 10 or 100 W/m3.rTsk= 0.04 W/mKASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional heat transfer (4) Uniform volumetric generation, (5) Negligible radiation, (6) Negligible conduction to or from the ground. PROPERTIES:k = 0.04 W/mK (given). ANALYSIS: The surface temperature of the dry hay is (Eq. 3.60) 321W/m1m0 C+0.02 C2225W/mKosqrTTh×= +=°×<whereas Ts= 0.2°C and 2.0°C for the moist and wet hay, respectively. <The maximum hay temperature occurs at the centerline, r= 0. From Eq. 3.58, for the dry hay, 232max1W/m(1m)0.02 C6.27 C440.04 W/mKosqrTTk×=+=+°=°×<whereas Tmax= 62.7°C and 627°