Question 11.25: A steel pipe, k = 26 Btu/(h·ft.·°F), having an outside diame...
A steel pipe, k = 26 Btu/(h·ft.·°F), having an outside diameter of 3.5 in. and an inside diameter of 3.00 in., and 5 ft. long, is covered with 1 in. of mineral wool, k = 0.026 Btu/ (h·ft.·°F). If the film coefficient on the inside of the pipe is 45 Btu/(h·ft .^{2} ·°F) and on the outside is 0.9 Btu/(h·ft .^{2}·°F), determine the overall heat transfer coefficient. (See Figure 11.29 and Illustrative Problems 11.10 and 11.12.)
Because the same amount of heat traverses each of the paths, we can write
\begin{aligned}\dot{Q} &=\dot{Q}_{i}=A_{i} h_{i}\left(t_{i}-t_{1}\right)=2 \pi r_{1} L h_{i}\left(t_{i}-t_{1}\right) \\&=\frac{2 \pi k_{1} L}{\ln \left(r_{2} / r_{1}\right)}\left(t_{1}-t_{2}\right) \\&=\frac{2 \pi k_{2} L}{\ln \left(r_{3} / r_{2}\right)}\left(t_{2}-t_{3}\right) \\&=A_{o} h_{o}\left(t_{3}-t_{o}\right)=2 \pi r_{3} L h_{o}\left(t_{3}-t_{o}\right)\end{aligned}Solving for the temperature differences and adding them yields
\dot{Q}=\frac{t_{i}-t_{o}}{\frac{1}{2 \pi r_{1} L h_{i}}+\frac{\ln \left(r_{2} / r_{1}\right)}{2 \pi k_{1} L}+\frac{\ln\left(r_{3} / r_{2}\right)}{2 \pi k_{2} L}+\frac{1}{2 \pi r_{3} L h_{o}}} (a)
Although this equation looks formidable, we can simplify it and interpret it in terms of j items discussed previously in this chapter. Thus,
\dot{Q}=\frac{\left(t_{i}-t_{0}\right) 2 \pi L}{\frac{1}{h_{i} r_{1}}+\frac{\ln \left(r_{2} / r_{1}\right)}{k_{1}}+\frac{\ln \left(r_{3} / r_{2}\right)}{k_{2}}+\frac{1}{h_{0} r_{3}}} (b)
Let us now define Uo as the overall heat-transfer coefficient based on the outside pipe surface A_{o} as
\dot{Q}=U_{0} A_{o}\left(t_{i}-t_{0}\right) (c)
If we multiply the numerator and the denominator of Equation b by r_{3}, we obtain
\dot{Q}=\frac{\left(t_{i}-t_{o}\right) 2 \pi L r_{3}}{\frac{r_{3}}{h_{i} r_{1}}+\frac{r_{3}}{k_{1}} \ln \frac{r_{2}}{r_{1}}+\frac{r_{3}}{k_{2}} \ln \frac{r_{3}}{r_{2}}+\frac{r_{3}}{h_{0} r_{3}}} (d)
Comparison of Equations c and d yields
U_{0}=\frac{1}{\frac{1}{h_{i}\left(r_{1} / r_{3}\right)}+\frac{r_{3}}{k_{1}} \ln \frac{r_{2}}{r_{1}}+\frac{r_{3}}{k_{2}} \ln \frac{r_{3}}{r_{2}}+\frac{1}{h_{0}}} (e)
Note that U_{o} is the overall heat-transfer coefficient based on the outside tube surface. If we had multiplied Equation b by r_{1} / r_{1} we would obtain the overall heat-transfer coefficient U_{i}based on the inside surface as
U_{i}=\frac{1}{\frac{1}{h_{i}}+\frac{r_{1}}{k_{1}} \ln \frac{r_{2}}{r_{1}}+\frac{r_{1}}{k_{2}} \ln \frac{r_{3}}{r_{2}}+\frac{1}{h_{0}\left(r_{3} / r_{1}\right)}} (f)
In effect, we have required that U_{0} A_{o}=U_{i} A_{i}. When discussing an overall heat-transfer coefficient, the reference area must be given. Proceeding with the numerical problems yields
\begin{gathered}\frac{1}{h_{i}}=\frac{1}{45}=0.02222 \\\frac{r_{1}}{k_{1}} \ln \frac{r_{2}}{r_{1}}=\frac{3.00}{2} /(26 \times 12) \ln \frac{3.50}{3.00}=0.00074 \\\frac{r_{1}}{k_{2}} \ln \frac{r_{3}}{r_{2}}=\frac{3.00}{2} /(0.026 \times 12) \ln \frac{5.50}{3.50}=2.1730\end{gathered}\begin{aligned}\frac{1}{h_{0}\left(r_{3} / r_{1}\right)} &=\frac{1}{0.9(5.50 / 3.00)}=0.6061 \\\sum &=2.8021\end{aligned}
Therefore,
U_{i}=\frac{1}{2.8021}=0.357 \frac{\text { Btu }}{ h \cdot ft .^{2} \cdot{ }^{\circ} F } \text { (of inside area) }Because U_{o} A_{o}=U_{i} A_{i}, U_{o}=0.357 \times A_{i} / A_{o}=0.357 D_{1} / D_{3}=(0.357)(3.00 / 5.50)=0.195 Btu / \left( h \cdot ft.^{2 \cdot \circ} F \right) (of the outside area).