CSU MECH 344 - HW4 - Solutions (11 pages)

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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