Question

A slab of thickness L with one side (x = 0) is insulted and its other side (x = L) is maintained at a constant temperature $T_o$ as shown in figure below.


A uniformly distributed internal heat source produces heat in the slab at the rate of S $W/m^3$.
Assume the heat conduction to be steady and 1-D along x-direction. Then find the heat flux at x = L

• A. SL
• B. $\frac{SL}{4}$
• C. 0
• D. $\frac{SL}{2}$

Answer 1

The correct option is A SL


1-D steady heat equation with uniform heat generation

$\frac{{\partial }^{2}T}{\partial x^2}+\frac{S}{k}=0$

$\frac{{\partial }^{2}T}{\partial x^2}=-\frac{S}{k}$
Integrating w.r.t. 'x'

$\frac{\partial T}{\partial x}=-\frac{S}{k}x+C_1$

$atx=0,\frac{\partial T}{\partial x}=0$

$0=0+C_1\Rightarrow C_1=0$

$\frac{\partial T}{\partial x}=-\frac{S}{k}x$

Heat flux at $(x=L)=-k{\left(\frac{\partial T}{\partial x}\right)}_{x=L}$

$q_{x=L}=-k{\left(\frac{\partial T}{\partial x}\right)}_{x=L}$

$q=-k(-\frac{S}{k}\times L+0)$

$\therefore q=S.L$
Alternate solution
Since entire heat generated in the slab is crossing the right side face
Heat flux crossing the right face = $\frac{Totalgenerated}{Area}$
q = $\frac{Volumetricheatgenerationrate\times volume}{Area}$
q = $\frac{S\times AL}{A}$
q = SL