Consider a 1.5m high and .6m wide plate whose thickness is.15m. One side of the

Question

Consider a 1.5m high and .6m wide plate whose thickness is.15m. One side of the plate is maintained at a constant temperatureof 500K while the other side is maintained at 350K. The thermalconductivity of the plate can be assumed to vary linearly in thattemperature range as k(T)=k0(1+T) wherek0=25W/m K and =8.7x10-4K-1.Disregarding the edge effects and assuming steady one-dimensionalheat transfer, determine the rate of heat conduction through theplate. The answer is 30.8kW. Please Help!! Consider a 1.5m high and .6m wide plate whose thickness is.15m. One side of the plate is maintained at a constant temperatureof 500K while the other side is maintained at 350K. The thermalconductivity of the plate can be assumed to vary linearly in thattemperature range as k(T)=k0(1+T) wherek0=25W/m K and =8.7x10-4K-1.Disregarding the edge effects and assuming steady one-dimensionalheat transfer, determine the rate of heat conduction through theplate. The answer is 30.8kW. Please Help!!

Explanation / Answer

-d(k*dT/dx)/dx =0               steady state heat equation k*dT/dx = C1 k0*(1 + *T)*dT = C1*dx k0*T + (k0*/2)T2 = C1*x + C2 T + (/2)T2 = (C1/k0)*x + (C2/k0) = C3 *x + C4 T + (/2)T2 = C3 *x + C4 Apply x = 0, T = 500 K 500 + (/2)*5002 = C3*0 + C4 C4 = 500 + (8.7e-4K-1 /2)*(5002) = 608.75 Apply x = 0.15, T = 350 K 350 + (/2)*3502 = C3*0.15 + 608.75 C3 = -1369.75 Therefore, T + (/2)T2 = -1369.75 *x + 608.75 T + (4.65e-4)T2 = -1369.75 *x + 608.75 (4.65e-4)T2 + T +1369.75 *x - 608.75 = 0 Now solve for T by quadratic formula T = -1/(2*4.65e-4) ± (1/2*4.65e-4)*[1 -4*4.65e-4*(1369.75*x - 608.75) ] T = -1075.26882 ± 1075.26882*[1 - 0.00186*(1369.75*x- 608.75)] T = -1075.26882 ± 1075.26882*[2.132275 -2.547735*x] T = -1075.26882 + 1075.26882*[2.132275 - 2.547735*x] positive because T > 0 Heat flux = -k*dT/dx dT/dx = 1075.26882*0.5*-2.547735 / [2.132275 -2.547735*x]1/2 dT/dx = -1369.75/[2.132275 - 2.547735*x]1/2 Heat Flux = -k0*(1+T)*-1369.75/[2.132275 -2.547735*x]1/2 Evaluate heat flux at x = 0, or T = 500 K Heat Flux = -k0*(1+*500)*-1369.75/[2.132275 -2.547735*0]1/2 Heat Flux = (25*(1+8.7e-4*500)*1369.75/[2.132275]1/2) Heat Flux = 33652.0566 W/m2 Heat Flow = Heat Flux*Area = 33652.0566 W/m2 *1.5m*0.6m = 30.3 kW

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