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kati45 [8]
3 years ago
14

A plane wall of length L, constant thermal conductivity k and no thermal energy generation undergoes one-dimensional, steady-sta

te 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.
Engineering
1 answer:
KIM [24]3 years ago
6 0

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)

   

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