Chapter 3, Problem 3.1P

Consider the plane wall of Figure 3.1, separating hot andcold fluids at temperatures $T_{\infty ,1}$ and $T_{\infty ,2},$ respectively.Using surface energy balances as boundary conditions at $x=0$ and $x=L$ (see Equation 2.34), obtain the temperaturedistribution within the wall and the heat flux interms of $T_{\infty ,1},T_{\infty ,2},h_1,h_2,k,$ and L.

To determine

The temperature distribution within the wall and the heat flux.

Answer to Problem 3.1P

The temperature distribution: $T\left(x\right)=-\frac{\left(T_{\infty ,1}-T_{\infty ,2}\right)}{\left[\frac{1}{h_1}+\frac{1}{h_2}+\frac{L}{k}\right]}\left[\frac{x}{k}+\frac{1}{h_1}\right]+T_{\infty ,1}$

The heat flux: $q_x^{"}=\frac{\left(T_{\infty ,1}-T_{\infty ,2}\right)}{\left[\frac{1}{h_1}+\frac{1}{h_2}+\frac{L}{k}\right]}$

Explanation of Solution

Given information:

Temperature of hot fluid is $T_{\infty ,1}$.

Temperature of cold fluid is $T_{\infty ,2}$.

Figure of the plane wall:

  

Calculations:

  

From the general solution of the heat diffusion equation:

  $T\left(x\right)=C_{1}x+C_2\mathrm{................}\left(1\right)$

  $\text{where }{C}_1\text{ and }{C}_2\text{ are constants of integration}\text{.}$

  $\text{Now applying the surface energy balance conditions:}$

  $\text{At }x=0:{\left[\text{ -k}\frac{dT}{dt}\right]}_{x=0}=h_1\left[T_{\infty ,1}-T\left(0\right)\right]\mathrm{.................}\left(2\right)$

  $\text{At }x=L:{\left[\text{ -k}\frac{dT}{dx}\right]}_{x=L}=h_2\left[T\left(L\right)-T_{\infty ,2}\right]\mathrm{.................}\left(3\right)$

  $Fromequation\left(1\right)and\left(2\right)withx=0:$

  $-k\left(C_1+0\right)=h_1\left[T_{\infty ,1}-\left(C_1\cdot 0+C_2\right)\right]\mathrm{.................}\left(4\right)$

  $And,fromequation\left(1\right)and\left(3\right)withx=L:$

  $-k\left(C_1+0\right)=h_2\left[\left(C_1\cdot L+C_2\right)-T_{\infty ,2}\right]\mathrm{.................}\left(5\right)$

  $\text{Solving equation }\left(\text{4}\right)\text{ and }\left(\text{5}\right){\text{ for C}}_{\text{1}}{\text{ and C}}_{\text{2}}:$

  $C_1=-\frac{\left(T_{\infty ,1}-T_{\infty ,2}\right)}{k\left[\frac{1}{h_1}+\frac{1}{h_2}+\frac{L}{k}\right]}$

  $and,C_1=-\frac{\left(T_{\infty ,1}-T_{\infty ,2}\right)}{h_1\left[\frac{1}{h_1}+\frac{1}{h_2}+\frac{L}{k}\right]}+T_{\infty ,1}$

Substituting in equation (1), the temperature distribution is:

  $\Rightarrow T\left(x\right)=-\frac{\left(T_{\infty ,1}-T_{\infty ,2}\right)}{\left[\frac{1}{h_1}+\frac{1}{h_2}+\frac{L}{k}\right]}\left[\frac{x}{k}+\frac{1}{h_1}\right]+T_{\infty ,1}$

Now find the heat flux using the fourier’s law:

  $\begin{array}{l}{q}_x^{"}=-k\frac{dT}{dx}\\ or,\\ q_x^{"}=-k{C}_1\\ q_x^{"}=-k\left(-\frac{\left(T_{\infty ,1}-T_{\infty ,2}\right)}{k\left[\frac{1}{h_1}+\frac{1}{h_2}+\frac{L}{k}\right]}\right)\\ \Rightarrow q_x^{"}=\frac{\left(T_{\infty ,1}-T_{\infty ,2}\right)}{\left[\frac{1}{h_1}+\frac{1}{h_2}+\frac{L}{k}\right]}\end{array}$

Conclusion: The temperature distribution within the wall is $T\left(x\right)=-\frac{\left(T_{\infty ,1}-T_{\infty ,2}\right)}{\left[\frac{1}{h_1}+\frac{1}{h_2}+\frac{L}{k}\right]}\left[\frac{x}{k}+\frac{1}{h_1}\right]+T_{\infty ,1}$ and the heat flux is $q_x^{"}=\frac{\left(T_{\infty ,1}-T_{\infty ,2}\right)}{\left[\frac{1}{h_1}+\frac{1}{h_2}+\frac{L}{k}\right]}$ .