Chapter 22, Problem 47P

To determine

The length of tube that must be used in the heat exchanger.

Explanation of Solution

Given:

The density of ethylene glycol $\left(\rho \right)$ is $1109\text{kg}/{\text{m}}^{\text{3}}$.

The specific heat of ethylene glycol $\left(c_p\right)$ is $2428\text{J}/\text{kg}\cdot \text{K}$.

The thermal conductivity $\left(k\right)$ is $0.253\text{W}/\text{m}\cdot \text{K}$.

The Prandlt number $\left(\mathrm{Pr}\right)$ is $148.5$.

The dynamic viscosity $\left(\mu \right)$ is $0.01545\text{kg}/\text{m}\cdot \text{s}$.

The inside diameter of tube $\left(D_i\right)$ is $2\text{cm}$.

The outside diameter of tube $\left(D_o\right)$ is $2.5\text{cm}$.

The mass flow rate of ethylene glycol $\left(\dot{m}\right)$ is $2.5\text{kg}/\text{s}$.

The inlet temperature of ethylene glycol $T_i$ is $25°\text{C}$.

The outlet temperature of ethylene glycol $T_e$ is $40°\text{C}$.

The thermal conductivity of copper $\left(k\right)$ is $386\text{W}/\text{m}\cdot \text{K}$.

The temperature of saturated vapor $\left(T_g\right)$ is $110°\text{C}$.

Calculation:

The figure below shows the schematic diagram of the heat exchanger.

Figure-(1)

Calculate the heat transfer in the heat exchanger.

    $\begin{array}{c}\dot{Q}=\dot{m}{c}_p\left(T_e-T_i\right)\\ =\left(2.5\text{kg}/\text{s}\right)\left(2428\text{J}/\text{kg}\cdot \text{K}\right)\left(\begin{array}{l}\left(40°\text{C}+273\right)\text{K}-\\ \left(25°\text{C}+273\right)\text{K}\end{array}\right)\\ =91050\text{W}\end{array}$

Calculate the velocity of fluid.

    $\begin{array}{c}V=\frac{\dot{m}}{\rho A_c}\\ =\frac{\dot{m}}{\rho \left(\frac{\pi }{4}{D}_i^2\right)}\\ =\frac{2.5\text{kg}/\text{s}}{\left(1109\text{kg}/{\text{m}}^{\text{3}}\right)\frac{\pi }{4}{\left(2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}\\ =7.176\text{m}/\text{s}\end{array}$

Calculate the Reynolds number.

    $\begin{array}{c}\mathrm{Re}=\frac{\rho VD}{\mu }\\ =\frac{\left(1109\text{kg}/{\text{m}}^{\text{3}}\right)\left(7.176\text{m}/\text{s}\right)\left(2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}{0.01545\text{kg}/\text{m}\cdot \text{s}}\\ =10302\end{array}$

Calculate the Nusselt number for water.

    $\begin{array}{c}Nu=0.023{\mathrm{Re}}^{0.8}{\mathrm{Pr}}^{0.4}\\ =0.023{\left(10302\right)}^{0.8}{\left(148.5\right)}^{0.4}\\ =275.9\end{array}$

Calculate the heat transfer coefficient on the inner side.

    $\begin{array}{c}{h}_i=\frac{k}{D}Nu\\ =\frac{0.253\text{W}/\text{m}\cdot \text{K}}{2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}\left(275.9\right)\\ =3490\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Assume the wall temperature of $100°\text{C}$.

Calculate the heat transfer coefficient on the outer side.

    $\begin{array}{c}{h}_o=9200{\left(T_g-T_w\right)}^{-0.25}\\ =9200{\left(\left(110°\text{C}+273\right)\text{K}-\left(80°\text{C}+273\right)\text{K}\right)}^{-0.25}\\ =3931\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the average temperature of ethylene glycol.

    $\begin{array}{c}{T}_{avg}=\frac{T_i+T_e}{2}\\ =\frac{40°\text{C}+25°\text{C}}{2}\\ =32.5°\text{C}\end{array}$

Now check if the assumed value is correct or not.

    $\begin{array}{c}{h}_i{A}_i{\left(T_w-T_{avg}\right)}^{-0.25}=h_o{A}_o{\left(T_g-T_w\right)}^{-0.25}\\ h_i\left(\pi D_{i}L\right){\left(T_w-T_{avg}\right)}^{-0.25}=h_o\pi D_{o}L{\left(T_g-T_w\right)}^{-0.25}\\ \left[\begin{array}{l}\left(3490\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\\ {\left(T_w-\left(32.5°\text{C}+273\right)\text{K}\right)}^{-0.25}\end{array}\right]=\left[\begin{array}{l}\left(5174\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(2.5\text{cm}\frac{1\text{m}}{100\text{cm}}\right)\\ {\left(\left(110°\text{C}+273\right)\text{K}-T_w\right)}^{-0.25}\end{array}\right]\\ T_w=\left(355.84\text{K}-273\right)°\text{C}\\ =82.84°\text{C}\end{array}$

The assumed value is near to the obtained value. Thus it correct.

Calculate the overall heat coefficient based on the outer surface.

    $\begin{array}{c}{U}_o=\frac{1}{\frac{D_o}{h_i{D}_i}+\frac{D_o\ln \left(D_o/D_i\right)}{2k}+\frac{1}{h_o}}\\ U_o=\left[\frac{1}{\begin{array}{l}\frac{2.5\text{cm}}{\left(3490\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)2\text{cm}}+\frac{2.5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\ln \left(2.5\text{cm}/2\text{cm}\right)}{2\left(386\text{W}/\text{m}\cdot \text{K}\right)}\\ +\frac{1}{3931\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}}\end{array}}\right]\\ U_o=1613\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the log mean temperature difference.

    $\begin{array}{c}\Delta T_{lm}=\frac{\left(T_g-T_e\right)-\left(T_g-T_i\right)}{\ln \left(\frac{\left(T_g-T_e\right)}{\left(T_g-T_i\right)}\right)}\\ =\frac{\left(\begin{array}{l}\left(110°\text{C}+273\right)\text{K}-\\ \left(40°\text{C}+273\right)\text{K}\end{array}\right)-\left(\begin{array}{l}\left(110°\text{C}+273\right)\text{K}-\\ \left(25°\text{C}+273\right)\text{K}\end{array}\right)}{\ln \left(\frac{\left(110°\text{C}-40°\text{C}\right)}{\left(110°\text{C}-25°\text{C}\right)}\right)}\\ =77.26\text{K}\end{array}$

Calculate the length of tube.

    $\begin{array}{c}L=\frac{\dot{Q}}{U_o{A}_o\Delta T_{lm}}\\ =\frac{\dot{Q}}{U_o\left(\frac{\pi }{4}{D}_o^2\right)\Delta T_{lm}}\\ =\frac{91050\text{W}}{1613\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\left(\frac{\pi }{4}\left(2.5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\right)\left(77.26\text{K}\right)}\\ =9.30\text{m}\end{array}$

Thus, the length of the tube is $9.30\text{m}$.