Chapter 22, Problem 23P

To determine

The overall heat transfer coefficient based on the inner and the outer surface of copper tube.

Explanation of Solution

Given:

The heat capacity for hot fluid $\left(C\right)$ is $4440\text{W}/\text{K}$.

The inlet temperature for hot fluid $\left(T_{hi}\right)$ is $150°\text{C}$.

The length of tube $\left(L\right)$ is $1.5\text{m}$.

The thermal conductivity of copper $\left(k\right)$ is $250\text{W}/\text{m}\cdot \text{K}$.

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

The outer diameter of inner tube $\left(D_o\right)$ is $2.25\text{cm}$.

The mass flow rate of hot fluid $\left({\dot{m}}_h\right)$ is $2\text{kg}/\text{s}$.

The exit temperature of hot fluid $\left(T_{\hslash e}\right)$ is $50°\text{C}$.

The inlet temperature for cold fluid $\left(T_{ci}\right)$ is $20°\text{C}$.

The outlet temperature for cold fluid $\left(T_{ce}\right)$ is $70°\text{C}$.

The outside fouling factor ${\left(R_f\right)}_o$ is $0.00015{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$.

The outside fouling factor ${\left(R_f\right)}_i$ is $0.0001{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$.

Calculation:

Refer Table A-19 “Properties of liquid”.

Obtain the following properties of liquid corresponding to the temperature of $100°\text{C}$ as follows:

$\rho =840\text{kg}/{\text{m}}^{\text{3}}$

$c_p=2220\text{kJ}/\text{kg}\cdot \text{K}$

$k=0.1367\text{W}/\text{m}\cdot \text{K}$

$\mu =0.01718\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$

$\mathrm{Pr}=279.1$

Refer Table A-15 “Properties of water”.

Obtain the following properties of liquid corresponding to the average temperature of $45°\text{C}$ as follows:

$\rho =990.1\text{kg}/{\text{m}}^{\text{3}}$

$c_p=4180\text{kJ}/\text{kg}\cdot \text{K}$

$k=0.637\text{W}/\text{m}\cdot \text{K}$

$\mu =0.596\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$

$\mathrm{Pr}=3.91$

Calculate the inner surface areas of the heart exchanger.

  $\begin{array}{c}{A}_i=\pi D_{i}L\\ =\pi \left(2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\left(1.5\text{m}\right)\\ =0.0942{\text{m}}^{\text{2}}\end{array}$

Calculate the outer surface areas of the heart exchanger.

  $\begin{array}{c}{A}_o=\pi D_{o}L\\ =\pi \left(2.25\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\left(1.5\text{m}\right)\\ =0.106{\text{m}}^{\text{2}}\end{array}$

Calculate the volume flow rate of the oil.

  $\begin{array}{c}V=\frac{4{\dot{m}}_h}{\rho \pi D_i^2}\\ =\frac{4\left(2\text{kg}/\text{s}\right)}{\pi \left(840\text{kg}/{\text{m}}^3\right)\left(2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}\\ =7.58\text{m}/\text{s}\end{array}$

Calculate the Reynolds number for the oil.

  $\begin{array}{c}\mathrm{Re}=\frac{\rho V{D}_i}{\mu }\\ =\frac{\left(840\text{kg}/{\text{m}}^{\text{3}}\right)\left(7.58\text{m}/\text{s}\right)\left(2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}{0.01718\text{kg}/\text{m}\cdot \text{s}}\\ =7412\end{array}$

Calculate the mass flow rate of cooling water.

  $\begin{array}{c}C\left(T_{hi}-T_{he}\right)={\dot{m}}_c{c}_{pc}\left(T_{ce}-T_{ci}\right)\\ \left(4440\text{W}/\text{K}\right)\left(150°\text{C}-50°\text{C}\right)={\dot{m}}_c\left(70°\text{C}-20°\text{C}\right)\\ {\dot{m}}_c=2.124\text{kg}/\text{s}\end{array}$

Calculate the hydraulic diameter of the annular space on the shell side.

  $\begin{array}{c}{D}_h=D_o-D_i\\ =6\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)-2.25\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.0375\text{m}\end{array}$

Calculate the average velocity of cooling water.

  $\begin{array}{c}V=\frac{4\dot{m}}{\rho \pi D_h^2}\\ =\frac{4\left(2.124\text{kg}/\text{s}\right)}{\left(990.1\text{kg}/{\text{m}}^{\text{3}}\right)\pi {\left(0.0375\text{m}\right)}^2}\\ =1.942\text{m}/\text{s}\end{array}$

Calculate the number for the flow of water.

  $\begin{array}{c}\mathrm{Re}=\frac{\rho V{D}_h}{\mu }\\ =\frac{\left(990.1\text{kg}/{\text{m}}^{\text{3}}\right)\left(1.942\text{m}/\text{s}\right)\left(0.0375\text{m}\right)}{0.596\times {10}^{-2}\text{kg}/\text{m}\cdot \text{s}}\\ =120980\end{array}$

Calculate the friction factor for oil.

  $\begin{array}{c}f={\left(0.79\ln \mathrm{Re}-1.64\right)}^{-2}\\ ={\left(0.79\ln \left(7412\right)-1.64\right)}^{-2}\\ =0.03429\end{array}$

Calculate the friction factor for water.

  $\begin{array}{c}f={\left(0.79\ln \mathrm{Re}-1.64\right)}^{-2}\\ ={\left(0.79\ln \left(120980\right)-1.64\right)}^{-2}\\ =001728\end{array}$

Calculate the Nusselt number for oil.

  $\begin{array}{c}Nu=\frac{\left(f/8\right)\left(\mathrm{Re}-1000\right)\mathrm{Pr}}{1+12.7\sqrt{f/8}\left({\mathrm{Pr}}^{2/3}-1\right)}\\ =\frac{\left(0.03429/8\right)\left(7412-1000\right)279.1}{1+12.7\sqrt{0.03429/8}\left({279.1}^{2/3}-1\right)}\\ =214.94\end{array}$

Calculate the inner convective heat transfer coefficient for oil.

  $\begin{array}{c}{h}_i=\frac{k}{D}Nu\\ =\frac{\left(0.1367\text{W}/\text{m}\cdot \text{K}\right)\left(214.94\right)}{0.02\text{m}}\\ =1469.11\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the Nusselt number for water.

  $\begin{array}{c}Nu=\frac{\left(f/8\right)\left(\mathrm{Re}-1000\right)\mathrm{Pr}}{1+12.7\sqrt{f/8}\left({\mathrm{Pr}}^{2/3}-1\right)}\\ =\frac{\left(0.01728/8\right)\left(120980-1000\right)3.91}{1+12.7\sqrt{0.01728/8}\left({3.91}^{2/3}-1\right)}\\ =540.52\end{array}$

Calculate the inner convective heat transfer coefficient for water.

  $\begin{array}{c}{h}_o=\frac{k}{D_h}Nu\\ =\frac{\left(0.637\text{W}/\text{m}\cdot \text{K}\right)\left(540.52\right)}{0.0375\text{m}}\\ =9188.25\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the overall heat transfer coefficient.

    $\begin{array}{c}\frac{1}{U{A}_s}=\frac{1}{h_i{A}_i}+\frac{{\left(R_f\right)}_i}{A_i}+\frac{\ln \left(\frac{D_o}{D_i}\right)}{2\pi kL}+\frac{{\left(R_f\right)}_o}{A_o}+\frac{1}{h_o{A}_o}\\ =\left[\begin{array}{l}\frac{1}{\left(1469.1\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(0.0942{\text{m}}^{\text{2}}\right)}+\frac{0.0015{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}}{0.0942{\text{m}}^{\text{2}}}+\\ \frac{\ln \left(\frac{2.25\text{cm}}{2\text{cm}}\right)}{2\pi \left(25\text{W}/\text{m}\cdot \text{K}\right)\left(1.5\text{m}\right)}+\frac{0.0001{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}}{0.106{\text{m}}^{\text{2}}}\\ +\frac{1}{\left(9188.25\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(0.106{\text{m}}^{\text{2}}\right)}\end{array}\right]\\ =0.01083\text{K}/\text{W}\end{array}$

Calculate the overall heat transfer based on the inner surface.

  $\begin{array}{c}\frac{1}{U_i{A}_i}=\frac{1}{U{A}_s}\\ \frac{1}{U_i}=\left(0.01083\text{K}/\text{W}\right)\left(0.0942{\text{m}}^{\text{2}}\right)\\ U_i=979.4\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the overall heat transfer based on the outer surface.

  $\begin{array}{c}\frac{1}{U_o{A}_o}=\frac{1}{U{A}_s}\\ \frac{1}{U_i}=\left(0.01083\text{K}/\text{W}\right)\left(0.106{\text{m}}^{\text{2}}\right)\\ U_i=871.09\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Thus, overall heat transfer based on the outer surface is $871.09\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}$ and overall heat transfer based on the inner surface is $979.4\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}$.