Chapter 3, Problem 3.1P

Consider the plane wall of Figure 3.1, separating hot and cold 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 temperature distribution within the wall and the heat flux in terms of $T_{\infty }{}_{,1},T_{\infty }{}_{,2},h_1,h_2,k,$ and L.

To determine

The temperature distribution within the wall and heat flux.

Answer to Problem 3.1P

The expression of temperature distribution is $T\left(x\right)=\frac{h_1\cdot h_2}{k\cdot h_1+k\cdot h_2+h_1\cdot h_2\cdot L}x+\frac{k\cdot h_1\cdot T_{\infty ,1}+k\cdot h_2\cdot T_{\infty ,2}+h_2\cdot L\cdot h_1\cdot T_{\infty ,1}}{k\cdot h_1+k\cdot h_2+h_1\cdot h_2\cdot L}$ and heat flux is $q^{″}=\frac{h_1\cdot k\cdot h_2}{\frac{1}{h_2}+\frac{1}{h_1}+\frac{L}{k}}\left(T_{\infty ,1}-T_{\infty ,2}\right)$ .

Explanation of Solution

Given:

The temperature of hot fluid is $T_{\infty ,1}$

The temperature of cold fluid is $T_{\infty ,2}$

Formula Used:

The expression of one dimensional heat equation is given by,

  $\rho c_p\frac{dT}{dt}=\frac{d}{dx}\left(k\frac{dT}{dx}\right)+\dot{q}$

The expression of convection heat rate is given by,

  $q_{\text{conv}}=h\left(T_s-T_{\infty }\right)$

Calculation:

Assume steady state, constant material properties with no heat generation.

The heat equation is calculated as,

  $\frac{d}{dx}\left(k\frac{dT}{dx}\right)=0$

Integrate the above equation once.

  $\begin{array}{c}k\frac{dT}{dx}=C_1\\ C_1=\frac{q^{″}}{k}\end{array}$

Integrate above equation again.

  $T\left(x\right)=C_{1}x+C_2$

Apply boundary condition at $x=0$ and $x=L$ .

At $x=0$ , $\begin{array}{c}{q}^{″}=k{\left(\frac{dT}{dx}\right)}_{x=0}\\ =h_1\left(T_{\infty ,1}-T_{s,1}\right)\end{array}$

At $x=L$ , $\begin{array}{c}{q}^{″}=k{\left(\frac{dT}{dx}\right)}_{x=L}\\ =h_2\left(T_{s,2}-T_{\infty ,2}\right)\end{array}$

The surface temperature at $x=0$ is given by,

  $T_{s,1}=C_1\cdot 0+C_2$

The surface temperature at $x=L$ is given by,

  $T_{s,2}=C_1\cdot L+C_2$ .

Solve all the equation to find unknown.

  $C_1=\frac{h_1\cdot h_2}{k\cdot h_1+k\cdot h_2+h_1\cdot h_2\cdot L}$

  $C_2=\frac{k\cdot h_1\cdot T_{\infty ,1}+k\cdot h_2\cdot T_{\infty ,2}+h_2\cdot L\cdot h_1\cdot T_{\infty ,1}}{k\cdot h_1+k\cdot h_2+h_1\cdot h_2\cdot L}$

  $q^{″}=\frac{h_1\cdot k\cdot h_2}{k\cdot h_1+k\cdot h_2+h_1\cdot h_2\cdot L}\left(T_{\infty ,1}-T_{\infty ,2}\right)$

The temperature distribution is calculated as,

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

The heat flux is calculated as,

  $q^{″}=\frac{h_1\cdot k\cdot h_2}{\frac{1}{h_2}+\frac{1}{h_1}+\frac{L}{k}}\left(T_{\infty ,1}-T_{\infty ,2}\right)$

Conclusion:

Therefore, the expression of temperature distribution is $T\left(x\right)=\frac{h_1\cdot h_2}{k\cdot h_1+k\cdot h_2+h_1\cdot h_2\cdot L}x+\frac{k\cdot h_1\cdot T_{\infty ,1}+k\cdot h_2\cdot T_{\infty ,2}+h_2\cdot L\cdot h_1\cdot T_{\infty ,1}}{k\cdot h_1+k\cdot h_2+h_1\cdot h_2\cdot L}$ and heat flux is $q^{″}=\frac{h_1\cdot k\cdot h_2}{\frac{1}{h_2}+\frac{1}{h_1}+\frac{L}{k}}\left(T_{\infty ,1}-T_{\infty ,2}\right)$ .