Chapter 17, Problem 122RQ

(a)

To determine

The thickness of the insulation needed to reduce the heat loss by 95 percent.

Explanation of Solution

Given:

Thermal conductivity of the steel pipe $\left(k\right)$ is $61\text{W}/\text{m}\cdot \text{K}$.

Heat transfer coefficient inside the pipe $\left(h_1\right)$ is $105\text{W}/{\text{m}}^2\cdot \text{K}$.

Heat transfer coefficient outside the pipe $\left(h_2\right)$ is $14\text{W}/{\text{m}}^2\cdot \text{K}$.

Inner diameter of the pipe $\left(D_i\right)$ is 10 cm.

Outer diameter of the pipe $\left(D_o\right)$ is 12 cm.

Calculation:

Determine the inner surface area of the pipe.

  $\begin{array}{c}{A}_1=\pi D_{i}L\\ =\pi \left(0.1\text{ m}\right)\left(1\text{ m}\right)\\ =0.3142{\text{ m}}^2\end{array}$

Determine the outer surface area of the pipe.

  $\begin{array}{c}{A}_2=\pi D_{o}L\\ =\pi \left(0.12\text{ m}\right)\left(1\text{ m}\right)\\ =0.377{\text{ m}}^2\end{array}$

Determine the surface area of the insulation.

  $\begin{array}{c}{A}_3=\pi D_{3}L\\ =\pi D_3\left(1\text{ m}\right)\\ =3.1416D_3{\text{ m}}^2\end{array}$

Determine the individual thermal resistances.

  $\begin{array}{c}{R}_i=\frac{1}{h_i{A}_i}\\ =\frac{1}{\left(105\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(0.3142{\text{ m}}^2\right)}\\ =0.03031°\text{C}/\text{W}\end{array}$

  $\begin{array}{c}{R}_1=R_{\text{pipe }}\\ =\frac{\ln \left(r_2/r_1\right)}{2\pi k_{1}L}\\ =\frac{\ln \left(\frac{6\text{ cm}}{5\text{ cm}}\right)}{2\pi \left(61\text{W}/\text{m}\cdot °\text{C}\right)\left(1\text{ m}\right)}\\ =0.00048°\text{C}/\text{W}\end{array}$

  $\begin{array}{c}{R}_2=R_{\text{insulation }}\\ =\frac{\ln \left(r_3/r_2\right)}{2\pi k_{2}L}\\ =\frac{\ln \left(D_3/0.12\right)}{2\pi \left(0.038\text{W}/\text{m}\cdot °\text{C}\right)\left(1\text{ m}\right)}\\ =\frac{\ln \left(D_3/0.12\right)}{0.23876}°\text{C}/\text{W}\end{array}$

  $\begin{array}{c}{R}_{\text{o},\text{steel }}=\frac{1}{h_o{A}_o}\\ =\frac{1}{\left(14\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(0.3770{\text{ m}}^2\right)}\\ =0.18947°\text{C}/\text{W}\end{array}$

  $\begin{array}{c}{R}_{\text{oinsulation}}=\frac{1}{h_0{A}_0}\\ =\frac{1}{\left(14\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(3.1416D_3{\text{ m}}^2\right)}\\ =\frac{0.02274}{D_3}°\text{C}/\text{W}\end{array}$

  $\begin{array}{c}{R}_{\text{total no insulation }}=R_i+R_1+R_{\text{o}\text{.steel }}\\ =0.03031°\text{C}/\text{W}+0.00048°\text{C}/\text{W}+0.18947°\text{C}/\text{W}\\ =0.22026°\text{C}/\text{W}\end{array}$

  $\begin{array}{c}{R}_{\text{total insulation }}=R_i+R_1+R_2+R_{\text{o,insulation }}\\ =0.03031+0.00048+\frac{\ln \left(D_3/0.12\right)}{0.23876}+\frac{0.02274}{D_3}\\ =0.03079+\frac{\ln \left(D_3/0.12\right)}{0.23876}+\frac{0.02274}{D_3}°\text{C}/\text{W}\end{array}$

Determine the steady rate of heat loss from the steam per meter pipe length in case of no insulation.

  $\begin{array}{c}\dot{Q}=\frac{T_{\infty 1}-T_{\infty 2}}{R_{\text{total}}}\\ =\frac{235°\text{C}-20°\text{C}}{0.22026°\text{C}/\text{W}}\\ =976.1\text{ W}\end{array}$

Determine the diameter of insulated material.

  $\begin{array}{l}{\dot{Q}}_{\text{insulation }}=\frac{T_{\infty 1}-T_{\infty 2}}{R_{\text{total insulation }}}\\ -\left(0.05\times 976.1\right)\text{W}=\frac{235°\text{C}-20°\text{C}}{\left(0.03079+\frac{\ln \left(D_3/0.12\right)}{0.23876}+\frac{0.02274}{D_3}\right)°\text{C}/\text{W}}\\ D_3=0.3355\text{ m}\end{array}$

Determine the thickness of insulation.

  $\begin{array}{c}t=\frac{D_3-D_2}{2}\\ =\frac{33.55\text{ cm}-12\text{ cm}}{2}\\ =10.8\text{ cm}\end{array}$

Thus, the thickness of insulation is $\underset{\_}{10.8\text{ cm}}$.

(b)

To determine

The thickness of the insulation needed to reduce the exposed surface temperature of insulated pipe to $40°\text{C}$.

Explanation of Solution

Determine the diameter of insulated material.

  $\begin{array}{l}{\dot{Q}}_{\text{insulation }}=\frac{T_{\infty 1}-T_{\infty 2}}{R_{\text{total insulation }}}\\ {\dot{Q}}_{\text{insulation }}=\frac{T_2-T_{\infty 2}}{R_{\text{o},\text{ insulation }}}\\ \frac{\left(235°\text{C}-20°\text{C}\right)}{\left(0.03079°\text{C}/\text{W}+\frac{\ln \left(D_3/0.12\right)}{0.23876}+\frac{0.02274}{D_3}\right)}=\frac{\left(40°\text{C}-20°\text{C}\right)}{\frac{0.02274}{D_3}°\text{C}/\text{W}}\\ D_3=0.1644\text{ m}\end{array}$

Determine the thickness of the insulation.

  $\begin{array}{c}t=\frac{D_3-D_2}{2}\\ =\frac{16.44\text{ cm}-12\text{ cm}}{2}\\ =2.22\text{ cm}\end{array}$

Thus, the thickness of insulation is $\underset{\_}{2.22\text{ cm}}$.