Chapter 10, Problem 10.25P

To determine

The expression for the maximum temperature in the stainless tube and in the boron nitride (bn), and to express the results in terms of geometric parameters $\left(r_1,r_2,r_3,r_4\right)$ , thermal conductivities $\left(K_{ss},k_{bn}\right)$ and the boiling parameters $\left(C,T_{\text{sat}},T_s\right)$ .

The value of the maximum temperature of the stainless steel tube and boron nitride tube is $-\frac{\dot{q}}{2k_{ss}}\left(r_1^2-r_2^2\right)+\left(\frac{\dot{q}{r}_1^2}{2k_{ss}}\right)\ln \left(\frac{r}{r_2}\right)+T_s+\left(R_{23}+R_{34}\right)C{\left(T_s-T_{\text{sat}}\right)}^3$ .

Given:

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

The required diagram for the resistance of the concentric cylinder circuit is shown in Figure 2

  

Figure 2

In the above diagram the thermal resistive circuit is represented by $R_{23}$ and resistance of the boron nitride is represented by $R_{34}$ .

The expression for the heat generation for the boiling and the evaporation rate is given by,

  ${\dot{q}}_{\text{boil}}={q^{″}}_x{A}_x$

The expression for the heat generation rate is evaluated as,

  $\begin{array}{l}{\dot{q}}_{\text{boil}}={q^{″}}_x{A}_x\\ \dot{q}\left(r_2^2-r_1^2\right)=\left(2\pi r_4\right)C{\left(T_s-T_{\text{sat}}\right)}^3\\ \dot{q}=\left(\frac{2r_4}{r_2^2-r_1^2}\right)C{\left(T_s-T_{\text{sat}}\right)}^3\end{array}$ ...... (1)

The expression for the maximum temperature of the stainless steel and boron nitride is evaluated as,

  $T\left(r\right)=-\frac{\dot{q}}{2k_{ss}}{r}^2+C_1\ln r+C_2$  ........... (2)

The first boundary condition are given by,

  $\begin{array}{l}r=\eta \\ \frac{dT}{dr}=0\end{array}$

Apply above boundary condition in equation (2).

  $\begin{array}{l}-\frac{\dot{q}}{2k_{ss}}2r_1+\frac{C_1}{\eta }+0=0\\ C_1=\pm \frac{q{r}_1^2}{k_{ss}}\end{array}$

The second boundary conditions are given by,

  $\begin{array}{l}r=r_2\\ T\left(r_2\right)=T_2\end{array}$

Apply above boundary equation in equation (2).

  $\begin{array}{l}{T}_2=-\frac{\dot{q}}{2k_{ss}}2r_2^2+\frac{\dot{q}{r}_1^2}{k_{ss}}\ln r_2+C_2=0\\ C_2=T_2+\frac{\dot{q}}{2k_{ss}}{r}_2^2-\frac{\dot{q}{r}_1^2}{k_{ss}}\ln r_2\end{array}$

Rewrite equation (2) in terms of the above equations.

  $\begin{array}{c}T\left(r\right)=-\frac{\dot{q}}{2k_{ss}}{r}^2+\left(\frac{\dot{q}{r}_1^2}{k_{ss}}\right)\ln r+T_2+\left(\frac{\dot{q}}{2k_{ss}}\right)r_2^2-\frac{\dot{q}{r}_1^2}{k_{ss}}\ln r_2\\ =-\frac{\dot{q}}{2k_{ss}}\left(r^2-r_2^2\right)+\left(\frac{\dot{q}{r}_1^2}{2k_{ss}}\right)\ln \left(\frac{r}{r_2}\right)+T_2\end{array}$

From above and from equation (1) the final temperature equation is calculated as,

  $\frac{T_2-T_s}{R_{23}+R_{34}}=\left(2\pi r_4\right)C{\left(T_s-T_{\text{sat}}\right)}^3$

The maximum temperature at inner stainless steel tube is calculated as,

  $T_{\max }=-\frac{\dot{q}}{2k_{ss}}\left(r_1^2-r_2^2\right)+\left(\frac{\dot{q}{r}_1^2}{2k_{ss}}\right)\ln \left(\frac{r}{r_2}\right)+T_s+\left(R_{23}+R_{34}\right)C{\left(T_s-T_{\text{sat}}\right)}^3$

The thermal resistance for he boron and the stainless steel is given by,

  $\begin{array}{l}{R}_{23}=\frac{\ln \left(\frac{r_3}{r_2}\right)}{2\pi k_{\text{bn}}}\\ R_{34}=\frac{\ln \left(\frac{r_4}{r_3}\right)}{2\pi k_{\text{ss}}}\end{array}$

Conclusion:

Therefore, the value of the maximum temperature of the stainless steel tube and boron nitride tube is $-\frac{\dot{q}}{2k_{ss}}\left(r_1^2-r_2^2\right)+\left(\frac{\dot{q}{r}_1^2}{2k_{ss}}\right)\ln \left(\frac{r}{r_2}\right)+T_s+\left(R_{23}+R_{34}\right)C{\left(T_s-T_{\text{sat}}\right)}^3$ .