Chapter 17, Problem 111P

(a)

To determine

The average surface temperature of the pipe disregarding the flanges.

Explanation of Solution

Given:

Length of the pipe is 3 m.

Thickness of the pipe is 0.4 cm.

Thermal conductivity of the pipe is $52\text{W}/\text{m}\cdot \text{K}$.

Outer diameter of the pipes is 10 cm.

Thickness of the flanges is 1 cm.

Outer diameter of the flange is 20 cm.

Temperature of steam is $200°\text{C}$.

The heat transfer coefficient of the inside is $180\text{W}/{\text{m}}^2\cdot \text{K}$.

Temperature of ambient air is $12°\text{C}$.

The heat transfer coefficient of the outside is $25\text{W}/{\text{m}}^2\cdot \text{K}$.

Calculation:

The total thermal resistance is,

  $\begin{array}{c}{R}_{total}=R_i+R_{cond}+R_o\\ =\frac{1}{h_i{A}_i}+\frac{\ln \left(r_2/r_1\right)}{2\pi kL}+\frac{1}{h_o{A}_o}\\ =\frac{1}{\left(180\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(\pi \times 0.092\text{ m}\times \text{6 m}\right)}+\frac{\ln \left(5/4.6\right)}{2\pi \left(52\text{W}/\text{m}\cdot \text{K}\right)\left(6\text{ m}\right)}\\ +\frac{1}{\left(25\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(\pi \times 0.1\text{ m}\times \text{6 m}\right)}\\ =0.02447°\text{C}/\text{W}\end{array}$

The rate of heat transfer is,

  $\begin{array}{c}\dot{Q}=\frac{T_{\infty 1}-T_{\infty 2}}{R_{total}}\\ =\frac{\left(200-12\right)°\text{C}}{0.02447°\text{C}/\text{W}}\\ =7684\text{ W}\end{array}$

Calculate the surface temperature of the pipe.

  $\begin{array}{l}\dot{Q}=\frac{T_2-T_{\infty 2}}{R_o}\\ 7684\text{ W}=\frac{\left(T_2-12\right)°\text{C}}{\frac{1}{\left(25\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(\pi \times 0.1\text{ m}\times \text{6 m}\right)}}\\ T_2=175°\text{C}\end{array}$

Thus, the average surface temperature of the pipe disregarding the flanges is $175°\text{C}$.

(b)

To determine

The efficiency of the fin and the rate of heat transfer from the flanges.

Explanation of Solution

Calculation:

Calculate the value of $\frac{r_{2c}}{r_1}$.

  $\begin{array}{c}\frac{r_{2c}}{r_1}=\frac{r_2+\frac{t}{2}}{r_1}\\ =\frac{0.1+\frac{0.02}{2}}{0.05}\\ =2.2\end{array}$

Calculate the value of $\epsilon$.

  $\begin{array}{c}\epsilon =L_c{}^{3/2}\sqrt{\frac{h}{k{A}_p}}\\ ={\left(L+\frac{t}{2}\right)}^{3/2}\sqrt{\frac{h}{kt}}\\ ={\left(0.05\text{ m}+\frac{0.02}{2}\text{ m}\right)}^{3/2}\sqrt{\frac{25\text{W}/{\text{m}}^2\cdot \text{K}}{\left(52\text{W}/\text{m}\cdot \text{K}\right)\left(0.02\text{ m}\right)}}\\ =0.29\end{array}$

Obtain the efficiency of the fin from Figure 17 – 44 corresponding to the above values as ${\eta }_{fin}=0.93$

Thus, the efficiency of the fin is $0.93$.

Calculate the heat transfer rate from the flanges.

  $\begin{array}{c}{\dot{Q}}_{finned}={\eta }_{fin}{\dot{Q}}_{\max }\\ ={\eta }_{fin}h{A}_{fin}\left(T_b-T_{\infty }\right)\\ ={\eta }_{fin}h\left[2\pi \left(r_2{}^2-r_1{}^2\right)+2\pi r_{2}t\right]\left(T_b-T_{\infty }\right)\\ =\left(0.93\right)\left(25\text{W}/{\text{m}}^2\cdot \text{K}\right)2\pi \left[\begin{array}{l}{\left(0.1\text{m}\right)}^2-{\left(0.05\text{m}\right)}^2\\ +\left(0.1\text{ m}\right)\left(0.02\text{ m}\right)\end{array}\right]\left(175-12\right)\text{K}\\ =226\text{}\text{W}\end{array}$

Thus, the heat transfer rate from the flanges is $226\text{}\text{W}$.

(c)

To determine

The length of the pipe equivalent to the flanges for heat transfer purpose.

Explanation of Solution

Calculation:

The heat transfer rate per unit length of the pipe is,

  $\frac{7684\text{ W}}{6\text{ m}}=1271\text{ W/m}$

The equivalent length of the pipe is,

  $\begin{array}{c}{L}_{eqv}=\frac{226\text{ W}}{1281\text{}\text{W/m}}\\ =0.176\text{ m}\\ =17.6\text{ cm}\end{array}$

Thus, the length of the pipe equivalent to the flanges for heat transfer purpose is $17.6\text{ cm}$.