Chapter 11, Problem 11.1P

(Case 1)

Interpretation Introduction

Interpretation: The overall heat transfer coefficient based on the inside as well as the outside area of the given condenser tube is to be calculated.

Concept Introduction:

The rate of heat transfer is written as:

$q=\frac{1}{R}\left(T_o-T_i\right)$ ....... (1)

Here, $R$ is the thermal resistance, $T_i$ is the inside temperature of the surface, and $T_o$ is the outside temperature of the surface.

Further the thermal resistance can be written as:

$R=\frac{1}{h_i{A}_i}+\frac{\ln \left(\frac{r_o}{r_i}\right)}{2\pi k_m}+\frac{1}{h_o{A}_o}$ ....... (2)

Here, $h_i$ is the inner heat transfer coefficient, $h_o$ is the outer heat transfer coefficient, $k_m$ is the thermal conductivity of the material, $A_i$ is the inner heat transfer area, $A_o$ is the outer heat transfer area, $r_o$ is the outer radius of the pipe, and $r_i$ is the inner radius of the pipe.

Overall heat transfer coefficient for the inner surface is calculated as:

$U_i=\frac{1}{R{A}_i}$ ....... (3)

Overall heat transfer coefficient for the outer surface is calculated as:

$U_o=\frac{1}{R{A}_o}$ ....... (4)

Answer to Problem 11.1P

Overall heat transfer coefficient for the inner surface is, $U_i=6156\frac{\text{W}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}$

Overall heat transfer coefficient for the outer surface is, $U_o=5651\frac{\text{W}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}$

Explanation of Solution

Given information:

Temperature of the water flowing inside the tube, $T_W={10}^{\circ }\text{C}$

Condenser tube is $\frac{3}{4}\text{ in}\text{.}$ 16 BWG

Temperature of the saturated steam flowing outside of the tube, $T_S={105}^{\circ }\text{C}$

Inside heat transfer coefficient, $h_i=12\frac{\text{kW}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}$

Outside heat transfer coefficient, $h_o=14\frac{\text{kW}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}$

Thermal conductivity of the tube material, $k_m=120\frac{\text{W}}{\text{m}{\cdot }^{\circ }\text{C}}$

For a $\frac{3}{4}\text{ in}\text{.}$ 16 BWG tube, its inner and outer radius is taken as:

$\begin{array}{c}{r}_i=\frac{3}{4}\text{ in}\text{.}\\ =19.05\text{ mm}\\ =0.01905\text{ m}\\ r_o=r_i+t\\ =\text{19}\text{.05 mm}+1.7\text{ mm}\\ =20.75\text{ mm}\\ =0.02075\text{ m}\end{array}$

Calculate the inner and outer area for the unit length of the pipe as:

$\begin{array}{c}{A}_i=2\pi r_{i}L\\ =2\pi \left(0.01905\text{ m}\right)\left(1\text{ m}\right)\\ =0.1197{\text{ m}}^2\\ A_o=2\pi r_{o}L\\ =2\pi \left(0.02075\text{ m}\right)\left(1\text{ m}\right)\\ =0.1304{\text{ m}}^2\end{array}$

Now, calculate the thermal resistance for the given tube using equation (2) as:

$\begin{array}{c}R=\frac{1}{h_i{A}_i}+\frac{\ln \left(\frac{r_o}{r_i}\right)}{2\pi k_m}+\frac{1}{h_o{A}_o}\\ =\frac{1}{\left(12\frac{\text{kW}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}\right)\left(0.1197{\text{ m}}^2\right)}+\frac{\ln \left(\frac{0.02075\text{ m}}{0.01905\text{ m}}\right)}{2\pi \left(120\frac{\text{W}}{\text{m}{\cdot }^{\circ }\text{C}}\times \frac{\text{kW}}{1000\text{ W}}\right)}+\frac{1}{\left(14\frac{\text{kW}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}\right)\left(0.1304{\text{ m}}^2\right)}\\ =1.357\frac{{}^{\circ }\text{C}}{\text{kW}}\end{array}$

Use equation (3) to calculate the overall heat transfer coefficient for the inner surface as:

$\begin{array}{c}{U}_i=\frac{1}{R{A}_i}\\ =\frac{1}{\left(1.357\frac{{}^{\circ }\text{C}}{\text{kW}}\right)\left(0.1197{\text{ m}}^2\right)}\\ =6.156\frac{\text{kW}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}\\ =6156\frac{\text{W}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}\end{array}$

Use equation (4) to calculate the overall heat transfer coefficient for the outer surface as:

$\begin{array}{c}{U}_o=\frac{1}{R{A}_o}\\ =\frac{1}{\left(1.357\frac{{}^{\circ }\text{C}}{\text{kW}}\right)\left(0.1304{\text{ m}}^2\right)}\\ =5.651\frac{\text{kW}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}\\ =5651\frac{\text{W}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}\end{array}$

(case 2)

Interpretation Introduction

Interpretation: The overall heat transfer coefficient based on the inside as well as the outside area of the given steel pipe is to be calculated.

Concept Introduction:

The rate of heat transfer is written as:

$q=\frac{1}{R}\left(T_o-T_i\right)$ ....... (1)

Here, $R$ is the thermal resistance, $T_i$ is the inside temperature of the surface, and $T_o$ is the outside temperature of the surface.

Further the thermal resistance can be written as:

$R=\frac{1}{h_i{A}_i}+\frac{\ln \left(\frac{r_o}{r_i}\right)}{2\pi k_m}+\frac{1}{h_o{A}_o}$ ....... (2)

Here, $h_i$ is the inner heat transfer coefficient, $h_o$ is the outer heat transfer coefficient, $k_m$ is the thermal conductivity of the material, $A_i$ is the inner heat transfer area, $A_o$ is the outer heat transfer area, $r_o$ is the outer radius of the pipe, and $r_i$ is the inner radius of the pipe.

Overall heat transfer coefficient for the inner surface is calculated as:

$U_i=\frac{1}{R{A}_i}$ ....... (3)

Overall heat transfer coefficient for the outer surface is calculated as:

$U_o=\frac{1}{R{A}_o}$ ....... (4)

Answer to Problem 11.1P

Overall heat transfer coefficient for the inner surface is, $U_i=19.51\frac{\text{W}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}$

Overall heat transfer coefficient for the outer surface is, $U_o=14.04\frac{\text{W}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}$

Explanation of Solution

Given information:

Temperature of the air flowing inside the pipe, $T_{air}={15}^{\circ }\text{C}$

Velocity of air flowing inside the pipe, $v=6\text{ m/s}$ .

Outside diameter of the steel pipe, $d_o=25\text{ mm}$

Thickness of the pipe wall, $t=3.5\text{ mm}$

Inside heat transfer coefficient, $h_i=20\frac{\text{W}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}$

Outside heat transfer coefficient, $h_o=1200\frac{\text{W}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}$

Thermal conductivity of the pipe material, $k_m=45\frac{\text{W}}{\text{m}{\cdot }^{\circ }\text{C}}$

For the given steel pipe, its inner and outer radius is taken as:

$\begin{array}{c}{r}_o=\frac{d_o}{2}\\ =\frac{25\text{ mm}}{2}\\ =12.5\text{ mm}\\ =0.0125\text{ m}\\ r_i=r_o-t\\ =\text{12}\text{.5 mm}-3.5\text{ mm}\\ =9\text{ mm}\\ =0.009\text{ m}\end{array}$

Calculate the inner and outer area for the unit length of the pipe as:

$\begin{array}{c}{A}_i=2\pi r_{i}L\\ =2\pi \left(0.009\text{ m}\right)\left(1\text{ m}\right)\\ =0.0565{\text{ m}}^2\\ A_o=2\pi r_{o}L\\ =2\pi \left(0.0125\text{ m}\right)\left(1\text{ m}\right)\\ =0.0785{\text{ m}}^2\end{array}$

Now, calculate the thermal resistance for the given tube using equation (2) as:

$\begin{array}{c}R=\frac{1}{h_i{A}_i}+\frac{\ln \left(\frac{r_o}{r_i}\right)}{2\pi k_m}+\frac{1}{h_o{A}_o}\\ =\frac{1}{\left(20\frac{\text{W}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}\right)\left(0.0565{\text{ m}}^2\right)}+\frac{\ln \left(\frac{0.0125\text{ m}}{0.009\text{ m}}\right)}{2\pi \left(45\frac{\text{W}}{\text{m}{\cdot }^{\circ }\text{C}}\right)}+\frac{1}{\left(1200\frac{\text{W}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}\right)\left(0.0785{\text{ m}}^2\right)}\\ =0.9073\frac{{}^{\circ }\text{C}}{\text{W}}\end{array}$

Use equation (3) to calculate the overall heat transfer coefficient for the inner surface as:

$\begin{array}{c}{U}_i=\frac{1}{R{A}_i}\\ =\frac{1}{\left(0.9073\frac{{}^{\circ }\text{C}}{\text{W}}\right)\left(0.0565{\text{ m}}^2\right)}\\ =19.51\frac{\text{W}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}\end{array}$

Use equation (4) to calculate the overall heat transfer coefficient for the outer surface as:

$\begin{array}{c}{U}_o=\frac{1}{R{A}_o}\\ =\frac{1}{\left(0.9073\frac{{}^{\circ }\text{C}}{\text{W}}\right)\left(0.0785{\text{ m}}^2\right)}\\ =14.04\frac{\text{W}}{{\text{m}}^2{\cdot }^{\circ }\text{C}}\end{array}$

(case 3)

Interpretation Introduction

Interpretation:The overall heat transfer coefficient based on the inside as well as the outside area of the given schedule 40 steel pipe is to be calculated.

Concept Introduction:

The rate of heat transfer is written as:

$q=\frac{1}{R}\left(T_o-T_i\right)$ ....... (1)

Here, $R$ is the thermal resistance, $T_i$ is the inside temperature of the surface, and $T_o$ is the outside temperature of the surface.

Further, the thermal resistance can be written as:

$R=\frac{1}{h_i{A}_i}+\frac{\ln \left(\frac{r_o}{r_i}\right)}{2\pi k_m}+\frac{1}{h_o{A}_o}$ ....... (2)

Here, $h_i$ is the inner heat transfer coefficient, $h_o$ is the outer heat transfer coefficient, $k_m$ is the thermal conductivity of the material, $A_i$ is the inner heat transfer area, $A_o$ is the outer heat transfer area, $r_o$ is the outer radius of the pipe, and $r_i$ is the inner radius of the pipe.

Overall heat transfer coefficient for the inner surface is calculated as:

$U_i=\frac{1}{R{A}_i}$ ....... (3)

Overall heat transfer coefficient for the outer surface is calculated as:

$U_o=\frac{1}{R{A}_o}$ ....... (4)

Answer to Problem 11.1P

Overall heat transfer coefficient for the inner surface is, $U_i=89.26\frac{\text{Btu}}{{\text{ft}}^2\cdot \text{h}{\cdot }^{\circ }\text{F}}$

Overall heat transfer coefficient for the outer surface is, $U_o=71.19\frac{\text{Btu}}{{\text{ft}}^2\cdot \text{h}{\cdot }^{\circ }\text{F}}$

Explanation of Solution

Given information:

Gauge pressure of condensation, $P=50\frac{{\text{lb}}_{\text{f}}}{{\text{in}}^2}$

1-in. Schedule 40 steel pipe is used.

Temperature of the oil flowing inside the pipe, $T_{oil}={100}^{\circ }\text{F}$

Inside heat transfer coefficient, $h_i=130\frac{\text{Btu}}{{\text{ft}}^2\cdot \text{h}{\cdot }^{\circ }\text{F}}$

Outside heat transfer coefficient, $h_o=14000\frac{\text{Btu}}{{\text{ft}}^2\cdot \text{h}{\cdot }^{\circ }\text{F}}$

Thermal conductivity of the pipe material, $k_m=26\frac{\text{Btu}}{\text{ft}\cdot \text{h}{\cdot }^{\circ }\text{F}}$

For the given 1-in. Schedule 40 steel pipe, its inner and outer radius is taken as:

$\begin{array}{c}{r}_o=0.0548\text{ ft}\\ r_i=r_o-t\\ =0.0548\text{ ft}-0.0111\text{ ft}\\ =0.0437\text{ ft}\end{array}$

Calculate the inner and outer area for the unit length of the pipe as:

$\begin{array}{c}{A}_i=2\pi r_{i}L\\ =2\pi \left(0.0437\text{ ft}\right)\left(1\text{ m}\right)\\ =0.2746{\text{ ft}}^2\\ A_o=2\pi r_{o}L\\ =2\pi \left(0.0548\text{ ft}\right)\left(1\text{ m}\right)\\ =0.3443{\text{ ft}}^2\end{array}$

Now, calculate the thermal resistance for the given tube using equation (2) as:

$\begin{array}{c}R=\frac{1}{h_i{A}_i}+\frac{\ln \left(\frac{r_o}{r_i}\right)}{2\pi k_m}+\frac{1}{h_o{A}_o}\\ =\frac{1}{\left(130\frac{\text{Btu}}{{\text{ft}}^2\cdot \text{h}{\cdot }^{\circ }\text{F}}\right)\left(0.2746{\text{ ft}}^2\right)}+\frac{\ln \left(\frac{0.0548\text{ ft}}{0.0437\text{ ft}}\right)}{2\pi \left(26\frac{\text{Btu}}{\text{ft}\cdot \text{h}{\cdot }^{\circ }\text{F}}\right)}+\frac{1}{\left(14000\frac{\text{Btu}}{{\text{ft}}^2\cdot \text{h}{\cdot }^{\circ }\text{F}}\right)\left(0.3443{\text{ ft}}^2\right)}\\ =0.0408\frac{\text{h}{\cdot }^{\circ }\text{F}}{\text{Btu}}\end{array}$

Use equation (3) to calculate the overall heat transfer coefficient for the inner surface as:

$\begin{array}{c}{U}_i=\frac{1}{R{A}_i}\\ =\frac{1}{\left(0.0408\frac{\text{h}{\cdot }^{\circ }\text{F}}{\text{Btu}}\right)\left(0.2746{\text{ ft}}^2\right)}\\ =89.26\frac{\text{Btu}}{{\text{ft}}^2\cdot \text{h}{\cdot }^{\circ }\text{F}}\end{array}$

Use equation (4) to calculate the overall heat transfer coefficient for the outer surface as:

$\begin{array}{c}{U}_o=\frac{1}{R{A}_o}\\ =\frac{1}{\left(0.0408\frac{\text{h}{\cdot }^{\circ }\text{F}}{\text{Btu}}\right)\left(0.3443{\text{ ft}}^2\right)}\\ =71.19\frac{\text{Btu}}{{\text{ft}}^2\cdot \text{h}{\cdot }^{\circ }\text{F}}\end{array}$