Chapter 2, Problem 2.1P

A plane wall, 7.5 cm thick, generates heat internally at the rate of 10⁵ W/m³. One side of the wall is insulated, and the other side is exposed to an environment at 90°C. The convection heat transfer coefficient between the wall and the environment is 500 W/m² K. If the thermal conductivity of the wall is 12 W/m K, calculate the maximum temperature in the wall.

To determine

The maximum temperature in the wall

Answer to Problem 2.1P

Maximum temperature in the wall is 128.43⁰C at it occurs at x = 0.

Explanation of Solution

Given Information:

Thickness of the wall L = 7.5 cm = 0.075 m

Rate of internal heat generation q_g = 10⁵ W/m³

Temperature of environment = 90⁰C

Convective heat transfer coefficient between the wall and the environment h = 500 W/m²K

Thermal conductivity of the wall k = 12W/mK

Assumptions:

1. One dimensional steady state heat transfer.

2. Properties of the wall and convective heat transfer coefficient remains constant.

Explanation: For one dimensional heat transfer with internal heat generation we have:

$\begin{array}{l}\text{k}\frac{{\partial }^2\text{T}}{\partial {\text{x}}^2}+\stackrel{\text{.}}{{\text{q}}_{\text{g}}}=\text{ρc}\frac{\partial \text{T}}{\partial \text{t}}\\ \text{Under steady state conditions:}\frac{\partial \text{T}}{\partial \text{t}}\text{= 0}\\ \text{k}\frac{{\text{d}}^{\text{2}}\text{T}}{{\text{dx}}^{\text{2}}}\text{+}\stackrel{\text{.}}{{\text{q}}_{\text{g}}}\text{=0}\\ \frac{{\text{d}}^{\text{2}}\text{T}}{{\text{dx}}^{\text{2}}}\text{=}\frac{\text{-}\stackrel{\text{.}}{{\text{q}}_{\text{g}}}}{\text{k}}\\ \frac{\text{dT}}{\text{dx}}\text{=}\frac{\text{-}\stackrel{\text{.}}{{\text{q}}_{\text{g}}}}{\text{k}}{\text{x+C}}_{\text{1}}\text{------Equation 1}\end{array}$

$\begin{array}{l}\text{T =}\frac{\text{-}\stackrel{\text{.}}{{\text{q}}_{\text{g}}}}{\text{k}}\frac{{\text{x}}^{\text{2}}}{\text{2}}{\text{+ C}}_{\text{1}}{\text{x + C}}_{\text{2}}\text{----------Equation 2}\\ \text{At x = 0 ; since it is insulated}\frac{\text{dT}}{\text{dx}}=0\\ \text{From Equation 1}\\ \text{0=}\frac{\text{-}\stackrel{\text{.}}{{\text{q}}_{\text{g}}}}{\text{k}}{\text{(0)+C}}_{\text{1}}\\ {\text{C}}_{\text{1}}=0\\ {\text{Let T}}_{\text{2}}\text{be the temperature of the wall at x = L}\\ \text{From Equation 2,}\\ {\text{T}}_{\text{2}}\text{=}\frac{\text{-}\stackrel{\text{.}}{{\text{q}}_{\text{g}}}}{\text{k}}\frac{{\text{L}}^{\text{2}}}{\text{2}}{\text{+ 0*L + C}}_{\text{2}}\\ {\text{T}}_{\text{2}}\text{=}\frac{\text{-}\stackrel{\text{.}}{{\text{q}}_{\text{g}}}}{\text{k}}\frac{{\text{L}}^{\text{2}}}{\text{2}}{\text{+ C}}_{\text{2}}\end{array}$

$\begin{array}{l}{\text{C}}_{\text{2}}={\text{T}}_{\text{2}}+\frac{\stackrel{.}{{\text{q}}_{\text{g}}}}{\text{k}}\frac{{\text{L}}^{\text{2}}}{\text{2}}\text{-----------Equation 3}\\ \text{All the heat that is generated in the wall gets convected to environment at x = L}\\ \text{Total rate of heat generation in the wall}\stackrel{.}{{\text{Q}}_{\text{g}}}=\stackrel{.}{{\text{(q}}_{\text{g}}}*\text{A*L) Watt}\\ \text{At x = L}\\ \stackrel{.}{{\text{(q}}_{\text{g}}}*\text{A*L)}={\text{hA(T}}_2-{\text{T}}_{\infty })\\ {\text{T}}_2=\frac{\stackrel{.}{{\text{q}}_{\text{g}}}\text{L}}{\text{h}}+{\text{T}}_{\infty }\\ {\text{Substitute T}}_2=\frac{\stackrel{.}{{\text{q}}_{\text{g}}}\text{L}}{\text{h}}+{\text{T}}_{\infty }\text{in Equation 3}\\ {\text{C}}_{\text{2}}=\frac{\stackrel{.}{{\text{q}}_{\text{g}}}\text{L}}{\text{h}}+{\text{T}}_{\infty }+\frac{\stackrel{.}{{\text{q}}_{\text{g}}}}{\text{k}}\frac{{\text{L}}^{\text{2}}}{\text{2}}\\ {\text{Substitute C}}_{\text{2}}{\text{and C}}_{\text{1}}\text{values in Equation 2}\end{array}$

$\begin{array}{l}\text{T =}\frac{\text{-}\stackrel{\text{.}}{{\text{q}}_{\text{g}}}}{\text{k}}\frac{{\text{x}}^{\text{2}}}{\text{2}}\text{+}\frac{\stackrel{.}{{\text{q}}_{\text{g}}}\text{L}}{\text{h}}+{\text{T}}_{\infty }+\frac{\stackrel{.}{{\text{q}}_{\text{g}}}}{\text{k}}\frac{{\text{L}}^{\text{2}}}{\text{2}}\text{---------- Equation 4}\\ \\ \text{Equation-4 shows temperature distribution in the wall}\text{.}\\ \text{From Equation 4 it is observed that , temperature decreases as the value of x increases}\text{.}\\ \text{Maximum temperature in the wall occurs at x = 0}\text{.}\\ \text{Substituting x = 0 in Equation - 4 we get}\end{array}$

$\begin{array}{l}\text{T =}\frac{\stackrel{.}{{\text{q}}_{\text{g}}}\text{L}}{\text{h}}+{\text{T}}_{\infty }+\frac{\stackrel{.}{{\text{q}}_{\text{g}}}}{\text{k}}\frac{{\text{L}}^{\text{2}}}{\text{2}}\\ {\text{T}}_{\text{max}}\text{=}\frac{{\text{10}}^{\text{5}}\times 0.075}{500}+90+\frac{{\text{10}}^{\text{5}}\times {0.075}^2}{\text{2}\times \text{12}}\\ {\text{T}}_{\text{max}}\text{=}128.43{}^{\text{0}}\text{C}\end{array}$

Conclusion:

Maximum temperature in the wall occurs at x = 0 and it is 128.43 ⁰C .