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

2.46 * 10⁵ W/m³

Step-by-step explanation:

See attached pictures for detailed explanation.

Answer 2

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

Step-by-step explanation:

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

so

`T=-(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) (k)/(q^.) (1)`

`h_(1)(T_(1infinity) -T_(1) )=-k(dT)/(dx) |_(x=0.04) (2)\\-k(dT)/(dx) |_(x=0.04) =h_(2) (T_(2)-T_(2infinity) (3)`

`300=-(q^.)/(2k) [c_(1) (k)/(q) ]^2+c_(1) [c_(1) (k)/(q) ]+c_(2) (1)`

`75[50+(q^2)/(2k) (0.04)^2+c_(1) (0.04)-c_(2) ]=-k[(+q^2(0.04))/(2k) ](2)`

`-k[(-q^.(0.04))/(2k) ]=50[(-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`