# CSU MECH 344 - HW4 - Solutions (11 pages)

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## HW4 - Solutions

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- Pages:
- 11
- School:
- Colorado State University- Fort Collins
- Course:
- Mech 344 - Heat and Mass Transfer

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PROBLEM 3 81 KNOWN Plane wall with internal heat generation which is insulated at the inner surface and subjected to a convection process at the outer surface FIND Maximum temperature in the wall SCHEMATIC ASSUMPTIONS 1 Steady state conditions 2 One dimensional conduction with uniform volumetric heat generation 3 Inner surface is adiabatic ANALYSIS The temperature at the inner surface is given by Eq 3 48 and is the maximum temperature within the wall 2 2k Ts To qL The outer surface temperature follows from Eq 3 51 Ts T qL h Ts 92 C 0 3 106 W m3 0 1m 500W m 2 K 92 C 60 C 152 C It follows that To 0 3 106 W m3 0 1m 2 25W m K 152 C 2 To 60 C 152 C 212 C COMMENTS The heat flux leaving the wall can be determined from knowledge of h Ts and T using Newton s law of cooling q conv h Ts T 500W m 2 K 152 92 C 30kW m 2 This same result can be determined from an energy balance on the entire wall which has the form E g E out 0 where E g qAL and E out q conv A Hence q conv qL 0 3 106 W m3 0 1m 30kW m 2 PROBLEM 3 98 KNOWN Radii and thermal conductivities of reactor fuel element and cladding Fuel heat generation rate Temperature and convection coefficient of coolant FIND a Expressions for temperature distributions in fuel and cladding b Maximum fuel element temperature for prescribed conditions c Effect of h on temperature distribution SCHEMATIC ASSUMPTIONS 1 Steady state conditions 2 One dimensional conduction 3 Negligible contact resistance 4 Constant properties ANALYSIS a From Eqs 3 54 and 3 28 the heat equations for the fuel f and cladding c are 1 d dTf q r r dr dr kf 1 d dTc r r dr dr 0 r r1 0 r1 r r2 Hence integrating both equations twice dTf dr dTc dr qr 2k f C1 kf r C3 kcr Tf 2 qr C 1 ln r C2 4k f k f C Tc 3 ln r C4 kc 1 2 3 4 The corresponding boundary conditions are dTf dr r 0 0 k f dT dTf k c c dr r r dr r r 1 1 Tf r1 Tc r1 kc dTc h Tc r2 T dr r r 5 6 7 8 2 Note that Eqs 7 and 8 are obtained from surface energy balances at r1 and r2 respectively Applying Eq 5 to Eq 1 it follows that C1 0 Hence Tf 2 qr C2 4k f From Eq 6 it follows that 2 C ln r qr 1 C2 3 1 C4 4k f kc 9 10 Continued PROBLEM 3 98 Cont Also from Eq 7 1 qr C 3 2 r1 2 qr C3 1 2 or 11 C C Finally from Eq 8 3 h 3 ln r2 C 4 T or substituting for C3 and solving for C4 r2 kc C4 12 qr 12 qr ln r2 T 2r2 h 2k c Substituting Eqs 11 and 12 into 10 it follows that C2 12 12 qr 2 ln r1 qr 2 qr 2 qr 1 1 1 ln r2 T 4k f 2k c 2r2 h 2k c 2 r qr ln 2 1 T 4k f 2k c r1 2r2 h Substituting Eq 13 into 9 C2 12 qr 12 qr 13 2 r qr ln 2 1 T 4k f 2k c r1 2r2 h Substituting Eqs 11 and 12 into 4 Tf Tc q r12 r 2 12 qr 14 12 qr 2 r qr ln 2 1 T 2k c r 2r2 h 15 b Applying Eq 14 at r 0 the maximum fuel temperature for h 2000 W m2 K is Tf 0 2 108 W m3 0 006 m 4 2 W m K 2 2 108 W m3 0 006 m 2 108 W m3 0 006 m 2 25 W m K 2 ln 0 009 m 0 006 m 2 2 0 009 m 2000 W m K 2 300 K Tf 0 900 58 4 200 300 K 1458 K c Temperature distributions for the prescribed values of h are as follows 600 1300 Temperature Tc K Temperature Tf K 1500 1100 900 700 500 300 0 0 001 0 002 0 004 0 005 Radius in fuel element r m h 2000 W m 2 K h 5000 W m 2 K h 10000 W m 2 K 0 006 500 400 300 0 006 0 007 0 008 0 009 Radius in cladding r m h 2000 W m 2 K h 5000 W m 2 K h 10000 W m 2 K Continued PROBLEM 3 98 Cont Clearly the ability to control the maximum fuel temperature by increasing h is limited and even for h Tf 0 exceeds 1000 K The overall temperature drop Tf 0 T is influenced principally by the low thermal conductivity of the fuel material 12 4k f 2 108 COMMENTS For the prescribed conditions Eq 14 yields Tf 0 Tf r1 qr W m3 0 006 m 3 8 W m K 900 K in which case with no cladding and h Tf 0 1200 K To reduce Tf 0 below 1000 K for the prescribed material it is necessary to reduce q PROBLEM 3 105 KNOWN Carton of apples modeled as 80 mm diameter spheres ventilated with air at 5 C and experiencing internal volumetric heat generation at a rate of 4000 J kg day FIND a The apple center and surface temperatures when the convection coefficient is 7 5 W m2 K and b Compute and plot the apple temperatures as a function of air velocity V for the range 0 1 V 1 m s when the convection coefficient has the form h C1V0 425 where C1 10 1 W m2 K m s 0 425 SCHEMATIC ASSUMPTIONS 1 Apples can be modeled as spheres 2 Each apple experiences flow of ventilation air at T 5 C 3 One dimensional radial conduction 4 Constant properties and 5 Uniform heat generation ANALYSIS a From Eq C 24 the temperature distribution in a solid sphere apple with uniform generation is T r o2 qr r2 1 Ts 6k r 2 o 1 To determine Ts perform an energy balance on the apple as shown in the sketch above with volume V 4 3 ro3 E in E out E g 0 q cv q 0 Ts T q 4 ro3 3 0 7 5 W m 2 K 4 0 0402 m 2 Ts 5o C 38 9 W m3 4 0 0403 m3 3 0 h 4 ro2 2 where the volumetric generation rate is q 4000 J kg day q 4000 J kg day 840 kg m3 1day 24 hr 1hr 3600 s q 38 9 W m3 and solving for Ts find Ts 5 14o C From Eq 1 at r 0 with Ts find T 0 38 9 W m3 0 …

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