Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 50 W/m -K and a thickness L =0.25 m, with no internal heat generation. Determine the heat flux and the unknown quantity for each case and sketch the temperature distribution, indicating the direction of the heat flux.
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Answer:solution:dT/dx =T2-T1/L & q_x = -k*(dT/dx)Case (1) dT/dx= (-20-50)/0.35==> -280 K/m q_x =-50*(-280)*10^3==>14 kWCase (2) dT/dx= (-10+30)/0.35==> 80 K/m q_x =-50*(80)*10^3==>-4 kWCase (2) dT/dx= (-10+30)/0.35==> 80 K/m q_x =-50*(80)*10^3==>-4 kWCase (3) q_x =-50*(160)*10^3==>-8 kWT2=T1+dT/dx*L=70+160*0.25==> 110° CCase (4) q_x =-50*(-80)*10^3==>4 kWT1=T2-dT/dx*L=40+80*0.25==> 60° CCase (5) q_x =-50*(200)*10^3==>-10 kWT1=T2-dT/dx*L=30-200*0.25==> -20° Cnote:all graph are attached
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