Chapter 22, Problem 117RQ

(a)

To determine

The tube length if heat exchanger has one-shell pass and one tube pass.

Explanation of Solution

Given:

The mass flow rate of hot steam $\left(m_h\right)$ is $47\text{kg}/\text{s}$.

The mass flow rate of water $\left(m_c\right)$ is $66\text{kg}/\text{s}$.

The heat capacity of water is $c_{p,c}=4.18\text{kJ}/\text{kg}\cdot \text{K}$.

The heat capacity of hot stream is $c_{p,h}=3.5\text{J}/\text{kg}\cdot \text{K}$.

The hot stream inlet temperature $\left(T_{h,i}\right)$ is $160°\text{C}$.

The hot stream exit temperature $\left(T_{h,o}\right)$ is $100°\text{C}$.

The water inlet temperature $\left(T_{c,i}\right)$ is $10°\text{C}$.

The average heat transfer coefficient $\left(h\right)$ is $220\text{W}/{\text{m}}^2\text{K}$.

The dynamic viscosity of stream $\left(\mu \right)$ is $2.0\text{mPa}\cdot \text{s}$.

The density of stream $\left(\mu \right)$ is $950\text{kg}/{\text{m}}^3$.

Calculation:

Calculate the rate of heat transfer.

    $\begin{array}{c}\dot{Q}=m_h{c}_{p,h}\left(T_{h,i}-T_{h,o}\right)\\ =\left(47\text{kg}/\text{s}\right)\left(3.5\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\begin{array}{l}\left(160°\text{C}-273\right)\text{K}-\\ \left(100°\text{C}-273\right)\text{K}\end{array}\right)\\ =9870\text{kW}\end{array}$

Calculate the outlet temperature of cold water using energy balance.

    $\begin{array}{c}{m}_c{c}_{p,c}\left(T_{c,o}-T_{c,i}\right)=\dot{Q}\\ \left[\begin{array}{l}\left(66\text{kg}/\text{s}\right)\left(4.18\text{kJ}/\text{kg}\cdot \text{K}\right)\times \\ \left(T_{c,o}-\left(10°\text{C}+273\right)\text{K}\right)\end{array}\right]=9870\text{kW}\\ T_{c,o}=\left(318.77\text{K}-273\right)°\text{C}\\ =45.77°\text{C}\end{array}$

Calculate the temperature at the ends.

    $\begin{array}{c}\Delta T_1=\left(T_{h,i}-T_{c,o}\right)\\ =160°\text{C}-45.77°\text{C}\\ =114.23°\text{C}\\ \Delta T_2=\left(T_{h,o}-T_{c,i}\right)\\ =100°\text{C}-10°\text{C}\\ =90°\text{C}\end{array}$

Calculate the LMTD (log mean temperature difference).

    $\begin{array}{c}{T}_{\text{lm}}=\frac{\Delta T_1-\Delta T_2}{\ln \frac{\Delta T_1}{\Delta T_2}}\\ =\frac{\left(114.2°\text{C}+273\right)\text{K}-\left(90°\text{C}+273\right)\text{K}}{\ln \frac{114.2°\text{C}}{90°\text{C}}}\\ =101.6\text{K}\end{array}$

Calculate the Reynolds number.

    $\begin{array}{c}\mathrm{Re}=\frac{VD\rho }{\mu }\\ =\frac{\left(\frac{\dot{m}}{n_{tube}\rho \frac{\pi D^2}{4}}\right)D\rho }{\mu }\\ =\frac{\left(\frac{47\text{kg}/\text{s}}{100\times 950\text{kg}/{\text{m}}^3\frac{\pi {\left(2.5\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2}{4}}\right)\left(2.5\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(950\text{kg}/{\text{m}}^3\right)}{2.0\text{mPa}\cdot \text{s}\frac{{\text{10}}^{-3}\text{kg}/\text{m}\cdot \text{s}}{\text{1}\text{mPa}\cdot \text{s}}}\\ =11968.45\end{array}$

The value of Reynolds number is greater than $10000$. Therefore, the flow is turbulent.

Calculate the value of Prandtl number.

    $\begin{array}{c}\mathrm{Pr}=\frac{\mu c_p}{k}\\ =\frac{\left(0.002\text{kg}/\text{m}\cdot \text{s}\right)\left(3500\text{J}/\text{kgK}\right)}{0.50\text{W}/\text{mK}}\\ =14\end{array}$

Calculate the Nusselt number for turbulent flow.

    $\begin{array}{c}Nu=0.023{\mathrm{Re}}^{0.8}{\mathrm{Pr}}^{0.3}\\ =0.023{\left(11968.45\right)}^{0.8}{\left(14\right)}^{0.3}\\ =92.89\end{array}$

Calculate the heat transfer coefficient on inner surface.

    $\begin{array}{c}{h}_i=\frac{0.50\text{W}/\text{mK}}{\left(2.5\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}92.89\\ =1858\text{W}/{\text{m}}^2\text{K}\end{array}$

Calculate the overall heat transfer coefficient.

    $\begin{array}{c}U=\frac{1}{\frac{1}{h_i}+\frac{1}{h_o}}\\ =\frac{1}{\frac{1}{1858\text{W}/{\text{m}}^2\text{K}}+\frac{1}{4000\text{W}/{\text{m}}^2\text{K}}}\\ =1268.69\text{W}/{\text{m}}^2\text{K}\end{array}$

Calculate the tube length.

    $\begin{array}{c}L=\frac{\dot{Q}}{U{n}_{tube}\pi D\Delta T_{\text{lm}}}\\ =\frac{9870\text{kW}\frac{\text{1000}\text{W}}{\text{1}\text{kW}}}{\left(1268.69\text{W}/{\text{m}}^2\text{K}\right)\left(100\pi \right)\left(2.5\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(101.6\text{K}\right)}\\ =9.75\text{m}\end{array}$

Thus, the tube length if heat exchanger has one-shell pass and one tube pass is $9.75\text{m}$.

(b)

To determine

The tube length if heat exchanger has one-shell pass and four tube pass.

Explanation of Solution

Calculation:

Calculate the number of tubes per pass.

    $\begin{array}{c}{n^{\prime }}_{tube}=\frac{n_{tube}}{4}\\ =\frac{100}{4}\\ =25\end{array}$

Calculate the Reynolds number.

    $\begin{array}{c}\mathrm{Re}=\frac{VD\rho }{\mu }\\ =\frac{\left(\frac{\dot{m}}{{n^{\prime }}_{tube}\rho \frac{\pi D^2}{4}}\right)D\rho }{\mu }\\ =\frac{\left(\frac{47\text{kg}/\text{s}}{25\times 950\text{kg}/{\text{m}}^3\frac{\pi {\left(2.5\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2}{4}}\right)\left(2.5\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(950\text{kg}/{\text{m}}^3\right)}{2.0\text{mPa}\cdot \text{s}\frac{{\text{10}}^{-3}\text{kg}/\text{m}\cdot \text{s}}{\text{1}\text{mPa}\cdot \text{s}}}\\ =47873.8\end{array}$

The value of Reynolds number is greater than $10000$. Therefore, the flow is turbulent.

Calculate the Nusselt number for turbulent flow.

    $\begin{array}{c}Nu=0.023{\mathrm{Re}}^{0.8}{\mathrm{Pr}}^{0.3}\\ =0.023{\left(47873.8\right)}^{0.8}{\left(14\right)}^{0.3}\\ =281.60\end{array}$

Calculate the heat transfer coefficient on inner surface.

    $\begin{array}{c}{h}_i=\frac{0.50\text{W}/\text{mK}}{\left(2.5\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}281.60\\ =5632.15\text{W}/{\text{m}}^2\text{K}\end{array}$

Calculate the overall heat transfer coefficient.

    $\begin{array}{c}U=\frac{1}{\frac{1}{h_i}+\frac{1}{h_o}}\\ =\frac{1}{\frac{1}{5632.15\text{W}/{\text{m}}^2\text{K}}+\frac{1}{4000\text{W}/{\text{m}}^2\text{K}}}\\ =2338.89\text{W}/{\text{m}}^2\text{K}\end{array}$

Calculate the two temperature ratio.

    $\begin{array}{c}P=\frac{t_2-t_1}{T_1-t_1}\\ =\frac{100°\text{C}-160°\text{C}}{10°\text{C}-160°\text{C}}\\ =0.4\end{array}$

    $\begin{array}{c}R=\frac{T_1-T_2}{t_2-t_1}\\ =\frac{10°\text{C}-45.8°\text{C}}{100°\text{C}-160°\text{C}}\\ =0.60\end{array}$

Refer figure-22-19, “Correction factor F charts for common shell-and-tube and cross-flow heat exchangers” in book corresponding to calculated value of $P$ and $R$ to obtain the value of correction factor.

    $F=0.96$

Calculate the tube length.

    $\begin{array}{c}L=\frac{\dot{Q}}{UF{n^{\prime }}_{tube}\pi D\Delta T_{\text{lm}}}\\ =\frac{9870\text{kW}\frac{\text{1000}\text{W}}{\text{1}\text{kW}}}{\left(2338.89\text{W}/{\text{m}}^2\text{K}\right)\left(0.96\right)\left(25\pi \right)\left(2.5\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(101.6\text{K}\right)}\\ =5.51\text{m}\end{array}$

Thus, the tube length if heat exchanger has one-shell pass and four tube pass is $5.51\text{m}$.