Chapter 22, Problem 103P

(a)

To determine

The mass flow rate of ethylene glycol.

Explanation of Solution

Given:

The temperature $\left(T_{h,in}\right)$ of glycol is $110°\text{C}$.

The initial temperature $\left(T_{c,in}\right)$ to heat oil is $10°\text{C}$.

The exit temperature $\left(T_{c,out}\right)$ to heat oil is $70°\text{C}$.

The dimensions $\left(D\right)$ of the tube is $0.5\text{m}\times 0.5\text{m}$.

The length $\left(L\right)$ of the tube is $25\text{mm}$.

The outer diameter $\left(D_o\right)$ of the tube is $0.5\text{m}$.

The mass flow rate $\left(\dot{m}\right)$ is $4.05\text{kg}/\text{s}$.

The heat transfer coefficient $\left(h\right)$ is $2500\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}$.

The exit temperature $\left(T_{h,out}\right)$ is $90°\text{C}$.

The properties of ethylene glycol are:

$\begin{array}{c}\rho =1062\text{kg}/{\text{m}}^3\\ \mu =2.499\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\\ c_p=2742\text{J}/\text{kg}\cdot \text{K}\\ k=0.262\text{W}/\text{m}\cdot \text{K}\\ \mathrm{Pr}=26.12\\ {\mathrm{Pr}}_s=96.97\end{array}$

Calculation:

Calculate the bulk mean temperature $\left(T_m\right)$ using the relation.

    $\begin{array}{c}{T}_M=\frac{T_s+T_a}{2}\\ =\frac{70°\text{C}+10°\text{C}}{2}\\ =40°\text{C}\end{array}$

Refer table $\text{A-19}$ “properties of oil”.

Obtain the following properties of oil corresponding to the temperature of $50°\text{C}$.

$k=01964\text{W}/\text{m}\cdot \text{K}$

Calculate the mass flow rate $\left(\dot{m}\right)$ using the relation.

    $\begin{array}{c}{\dot{m}}_h=\frac{{\dot{m}}_c{c}_{pc}\left(T_{c,out}-T_{c,in}\right)}{c_{ph}\left(T_{h,in}-T_{h,out}\right)}\\ =\frac{\left(4.05\text{kg}/\text{s}\right)\left(\left(1964\text{J}/\text{kg}\cdot \text{K}\right)\right)\left(\left(70°\text{C}+273\right)\text{K}-\left(10°\text{C}+273\right)\text{K}\right)}{\left(\left(2742\text{J}/\text{kg}\cdot \text{K}\right)\right)\left(\left(110°\text{C}+273\right)\text{K}-\left(90°\text{C}+273\right)\text{K}\right)}\\ =8.7\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of ethylene glycol is $8.7\text{kg}/\text{s}$.

(b)

To determine

The number of tube rows.

Explanation of Solution

Given:

Calculation:

Calculate the maximum velocity $\left(V_{\max }\right)$ using the relation.

    $\begin{array}{c}{V}_{\max }=\frac{S_T}{S_T-D_o}V\\ =\frac{S_T}{S_T-D_o}\frac{{\dot{m}}_h}{\rho A}\\ =\frac{0.035}{\left(0.035-0.025\right)}\frac{8.7\text{kg}/\text{s}}{\left(1062{\text{kg/m}}^3\right)\left(0.5\times 0.5\right){\text{m}}^2}\\ =0.114\text{m}/\text{s}\end{array}$

Calculate the Reynolds number $\left(\mathrm{Re}\right)$ using the relation.

    $\begin{array}{c}\mathrm{Re}=\frac{\rho V_{\max }{D}_o}{\mu }\\ =\frac{\left(1062{\text{kg/m}}^3\right)\left(0.114\text{m}/\text{s}\right)\left(0.025\right)\text{m}}{2.499\times {10}^{-3}\text{kg/m}\cdot \text{s}}\\ =1211\end{array}$

Calculate the Nusselt number $\left(\text{Nu}\right)$ using the relation.

Assume that the tube rows are greater than $16$.

Thus, from table $7\text{-}2$ obtain the value of Nusselt number.

    $\begin{array}{c}\text{Nu}=0.35{\mathrm{Re}}^{0.6}{\mathrm{Pr}}^{0.36}{\left(\frac{\mathrm{Pr}}{{\mathrm{Pr}}_s}\right)}^{0.25}\\ =0.35{\left(1211\right)}^{0.6}{\left(26.12\right)}^{0.36}{\left(\frac{26.12}{96.97}\right)}^{0.25}\\ =57.76\end{array}$

Calculate the heat transfer coefficient $\left(h_o\right)$ using the relation.

    $\begin{array}{c}{h}_o=\frac{\text{Nuk}}{D_o}\\ =\frac{\left(57.76\right)\left(0.262\text{W}/\text{m}\cdot \text{K}\right)}{\left(25\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)}\\ =605.3\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

Calculate the overall heat transfer coefficient $\left(U\right)$ using the relation.

    $\begin{array}{c}\frac{1}{U}=\frac{D_o}{h_i{D}_i}+\frac{D_o\left(\ln D_o/D_i\right)}{2k}+\frac{1}{h_o}\\ \frac{1}{U}=\left[\begin{array}{l}\frac{\left(25\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)}{\left(2500\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(23\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)}+\\ \frac{\left(25\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)\ln \frac{\left(25\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)}{\left(23\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)}}{2\left(250\text{W}/{\text{m}}^2\cdot \text{K}\right)}\end{array}\right]+\frac{1}{605.3\text{W}/{\text{m}}^2\cdot \text{K}}\\ \frac{1}{U}=\left(4.34\times {10}^{-4}+4.169\times {10}^{-6}+1.652\times {10}^{-3}\right){\text{m}}^2\cdot \text{K}/\text{W}\\ U=478.4\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

Calculate the heat capacity rate $\left(C_{\min }\right)$ using the relation.

    $\begin{array}{c}{C}_{\min }={\dot{m}}_c{c}_{pc}\\ =\left(4.05\text{kg}/\text{s}\right)\left(\left(1964\text{J/kg}\cdot \text{K}\right)\right)\\ =7954.2\text{W}/\text{K}\end{array}$

Calculate the heat capacity rate $\left(C_{\max }\right)$ using the relation.

    $\begin{array}{c}{C}_{\max }={\dot{m}}_h{c}_{ph}\\ =\left(8.7\text{kg}/\text{s}\right)\left(\left(2742\text{J/kg}\cdot \text{K}\right)\right)\\ =23855.4\text{W}/\text{K}\end{array}$

Calculate the capacity ratio $\left(c\right)$ using the relation.

    $\begin{array}{c}c=\frac{C_{\min }}{C_{\max }}\\ =\frac{7954.2\text{W}/\text{K}}{23855.4\text{W}/\text{K}}\\ =0.33\end{array}$

Calculate the effectiveness $\left(\epsilon \right)$ using the relation.

    $\begin{array}{c}\epsilon =\frac{\dot{Q}}{{\dot{Q}}_{\max }}\\ \epsilon =\frac{C_c\left(T_{c,out}-T_{c,in}\right)}{C_c\left(T_{h,in}-T_{c,in}\right)}\\ =\frac{\left(70-10\right)°\text{C}}{\left(100-10\right)°\text{C}}\\ =0.6\end{array}$

Calculate the $\left(\text{NTU}\right)$ using the relation.

Refer table $22\text{-5}$ to obtain the expression of $\left(\text{NTU}\right)$.

    $\begin{array}{c}\text{NTU}=-\ln \left[1+\frac{\ln \left(1-\epsilon c\right)}{c}\right]\\ =-\ln \left[1+\frac{\ln \left(1-\left(0.6\right)\left(0.33\right)\right)}{0.33}\right]\\ =1.106\end{array}$

Calculate the surface area $\left(A_s\right)$ using the relation.

    $\begin{array}{c}{A}_s=\frac{\text{NTU}{C}_{\min }}{U}\\ =\frac{\left(1.106\right)\left(7954.2\text{W}/\text{K}\right)}{\left(478.4\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)}\\ =18.38{\text{m}}^2\end{array}$

Calculate the number of rows $\left(N_T\right)$ using the relation.

    $\begin{array}{c}{N}_L=\frac{A_s}{\pi D_{o}L{N}_T}\\ =\frac{18.38{\text{m}}^2}{\pi \left(23\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)\left(0.5\text{m}\right)\left(\frac{0.5\text{m}}{0.035}\right)}\\ =33\end{array}$

Thus, the number of rows are $33$.