Question

Electric heater wires are installed in a solid wall having a thickness of 8 cm and k=2.5 W/m.°C. The right face is exposed to an environment with h=50 W/m2°C and k'=30°C, while the left face is exposed to h=75 W/m2°C and T[infinity]=50°C. What is the maximum allowable heat generation rate such that the maximum temperature in the solid does not exceed 300°C.

Answer 1

Answer:

2.46 * 10⁵ W/m³

Explanation:

See attached pictures for detailed explanation.

Answer 2

Answer:

$q^.=2.46*10^5W/m^3$

Explanation:

$Given\\k=2.5W/m\\h_{1} =75(left)\\h_{2} =50(right)\\T_{1} =50^oC\\T_{2} =30^oC$

so

$T=-\frac{q^.x^2}{2k} +c_{1}x+ c_{2} \\T=T_{1} \\at \\x=-0.04\\T=T_{2} \\at\\x=+0.04$

$dT/dx=-q^.x/k+c_{1} \\T=T_{max} =300\\at\\x=c_{1} \frac{k}{q^.} (1)$

$h_{1}(T_{1infinity} -T_{1} )=-k\frac{dT}{dx} |_{x=0.04} (2)\\-k\frac{dT}{dx} |_{x=0.04} =h_{2} (T_{2}-T_{2infinity} (3)$

$300=-\frac{q^.}{2k} [c_{1} \frac{k}{q} ]^2+c_{1} [c_{1} \frac{k}{q} ]+c_{2} (1)$

$75[50+\frac{q^2}{2k} (0.04)^2+c_{1} (0.04)-c_{2} ]=-k[\frac{+q^2(0.04)}{2k} ](2)$

$-k[\frac{-q^.(0.04)}{2k} ]=50[\frac{-q^.(0.04)}{2k} +c_{1} (0.04)+c_{2} -30](3)$

solving above 3 equations for 3 unknowns c1,c2,q

we get $q^.=2.46*10^5W/m^3$