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).

Question

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.)

Accepted Answer

Because the same amount of heat traverses each of the paths, we can write

[latex]\begin{aligned}\dot{Q} &=\dot{Q}_{i}=A_{i} h_{i}\left(t_{i}-t_{1} ight)=2 \pi r_{1} L h_{i}\left(t_{i}-t_{1} ight) \&=\frac{2 \pi k_{1} L}{\ln \left(r_{2} / r_{1} ight)}\left(t_{1}-t_{2} ight) \&=\frac{2 \pi k_{2} L}{\ln \left(r_{3} / r_{2} ight)}\left(t_{2}-t_{3} ight) \&=A_{o} h_{o}\left(t_{3}-t_{o} ight)=2 \pi r_{3} L h_{o}\left(t_{3}-t_{o} ight)\end{aligned}[/latex]

Solving for the temperature differences and adding them yields

[latex]\dot{Q}=\frac{t_{i}-t_{o}}{\frac{1}{2 \pi r_{1} L h_{i}}+\frac{\ln \left(r_{2} / r_{1} ight)}{2 \pi k_{1} L}+\frac{\ln\left(r_{3} / r_{2} ight)}{2 \pi k_{2} L}+\frac{1}{2 \pi r_{3} L h_{o}}}[/latex] (a)

Although this equation looks formidable, we can simplify it and interpret it in terms of j items discussed previously in this chapter. Thus,

[latex]\dot{Q}=\frac{\left(t_{i}-t_{0} ight) 2 \pi L}{\frac{1}{h_{i} r_{1}}+\frac{\ln \left(r_{2} / r_{1} ight)}{k_{1}}+\frac{\ln \left(r_{3} / r_{2} ight)}{k_{2}}+\frac{1}{h_{0} r_{3}}}[/latex] (b)

Let us now define Uo as the overall heat-transfer coefficient based on the outside pipe surface [latex]A_{o}[/latex] as

[latex]\dot{Q}=U_{0} A_{o}\left(t_{i}-t_{0} ight)[/latex] (c)

If we multiply the numerator and the denominator of Equation b by [latex]r_{3}[/latex], we obtain

[latex]\dot{Q}=\frac{\left(t_{i}-t_{o} ight) 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}}}[/latex] (d)

Comparison of Equations c and d yields

[latex]U_{0}=\frac{1}{\frac{1}{h_{i}\left(r_{1} / r_{3} ight)}+\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}}}[/latex] (e)

Note that [latex]U_{o}[/latex] is the overall heat-transfer coefficient based on the outside tube surface. If we had multiplied Equation b by [latex]r_{1} / r_{1}[/latex] we would obtain the overall heat-transfer coefficient [latex]U_{i}[/latex]based on the inside surface as

[latex]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} ight)}}[/latex] (f)

In effect, we have required that [latex]U_{0} A_{o}=U_{i} A_{i}[/latex]. When discussing an overall heat-transfer coefficient, the reference area must be given. Proceeding with the numerical problems yields

[latex]\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 imes 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 imes 12) \ln \frac{5.50}{3.50}=2.1730\end{gathered}[/latex]

[latex]\begin{aligned}\frac{1}{h_{0}\left(r_{3} / r_{1} ight)} &=\frac{1}{0.9(5.50 / 3.00)}=0.6061 \\sum &=2.8021\end{aligned}[/latex]

Therefore,

[latex]U_{i}=\frac{1}{2.8021}=0.357 \frac{ ext { Btu }}{ h \cdot ft .^{2} \cdot{ }^{\circ} F } ext { (of inside area) }[/latex]

Because [latex]U_{o} A_{o}=U_{i} A_{i}, U_{o}=0.357 imes 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 ight)[/latex] (of the outside area).

Question 11.25: A steel pipe, k = 26 Btu/(h·ft.·°F), having an outside diame...

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