Question

A plane wall of length L, constant thermal conductivity k and no thermal energy generation undergoes one-dimensional, steady-state conduction. The left hand side surface of the wall is subjected to a constant heat flux q The right hand side surface has a uniform temperature of T,. Using the standard approach, find the temperature distribution, the heat flux and rate of heat transfer.

Answer 1

Answer:

Temperature distribution is $T(x)=\dfrac{q}{k}(L-x)+T$

Heat flux=q

Heat rate=q A  

Explanation:

We know that for no heat generation and at steady state

$\dfrac{\partial^2 T}{\partial x^2}=0$

$\dfrac{\partial T}{\partial x}=a$

$T(x)=ax+b$

a and are the constant.

Given that heat flux=q

We know that heat flux given as

$q=-K\dfrac{dT}{dx}$

From above we can say that

$a= -\dfrac{q}{K}$

Alos given that when x= L temperature is T(L)=T

$T= -\dfrac{q}{K}L+b$

$b=T+\dfrac{q}{K}L$

So temperature T(x)

$T(x)=-\dfrac{q}{K}x+T+\dfrac{q}{K}L$

$T(x)=\dfrac{q}{k}(L-x)+T$

So temperature distribution is $T(x)=\dfrac{q}{k}(L-x)+T$

Heat flux=q

Heat rate=q A         (where A is the cross sectional area of wall)