Chapter 22, Problem 95P

To determine

The surface area of the heat exchanger using both LMTD method and $\epsilon -\text{NTU}$ method.

Explanation of Solution

Given:

The number of shells is $1$.

The number of passes is $8$.

The mass flow rate of water is $50000\text{lbm}/\text{h}$.

The inlet temperature $\left(T_{c,in}\right)$ of water is $70°\text{F}$.

The outlet temperature $\left(T_{c,out}\right)$  of water is $150°\text{F}$.

The specific heat $\left(c_{p,h}\right)$ of the hot air is $0.25\text{Btu}/\text{lbm}\text{.}°\text{F}$.

The inlet temperature $\left(T_{h,in}\right)$ of air is $600°\text{F}$.

The outlet temperature $\left(T_{h,out}\right)$ of air is $300°\text{F}$.

The heat transfer coefficient $\left(h\right)$ on the outer surface of the tube is $30\text{Btu}/\text{h}{\text{.ft}}^2.°\text{F}$.

The possible fouling resistance on water and air side is $0.0015\text{h}{\text{.ft}}^2.°\text{F}/\text{Btu}$ and $0.001\text{h}{\text{.ft}}^2.°\text{F}/\text{Btu}$ respectively.

Calculation:

From Table A-3E, the specific heat $\left(c_{p,c}\right)$ of water at the average temperature of $\left(150+70\right)°\text{F}\text{/}\text{2}=110\text{°F}$ is $1\text{Btu/lbm}\cdot °\text{F}$.

Calculate the heat gained by water from hot air using the relation.

    $\begin{array}{c}\dot{Q}={\dot{m}}_c{c}_{pc}\left(T_{cout}-T_{cin}\right)\\ =\left(50000\text{lbm}/\text{h}\right)\times \left(1\text{Btu}/\text{lbm}\cdot °\text{F}\right)\times \left(150°\text{F}-70°\text{F}\right)\\ =4\times {10}^6\text{Btu}/\text{h}\end{array}$

Calculate the mass flow rate of the hot air using the relation.

    $\begin{array}{l}{\dot{m}}_h{c}_{p,h}\left(T_{h,in}-T_{h,out}\right)={\dot{m}}_c{c}_{p,c}\left(T_{c,out}-T_{c,in}\right)\\ {\dot{m}}_h=\frac{{\dot{m}}_c{c}_{p,c}\left(T_{c,out}-T_{c,in}\right)}{c_{ph}\left(T_{h,in}-T_{h,out}\right)}\\ {\dot{m}}_h=\frac{\left(4\times {10}^6\text{Btu}/\text{h}\right)}{\left(0.25\text{Btu}/\text{lbm}\text{.}°\text{F}\right)\left(600°\text{F}-300°\text{F}\right)}\\ {\dot{m}}_h=53333.33\text{lbm}/\text{hr}\end{array}$

Calculate $\left(\Delta T_1\right)$.

  $\begin{array}{c}\Delta T_1=T_{h,in}-T_{c,out}\\ =600°\text{F}-150°\text{F}\\ =450°\text{F}\end{array}$

Calculate $\left(\Delta T_2\right)$.

  $\begin{array}{c}\Delta T_2=T_{h,out}-T_{c,in}\\ =300°\text{F}-70°\text{F}\\ =230°\text{F}\end{array}$

Calculate the logarithmic mean temperature difference LMTD of a counter flow heat exchanger using the relation.

    $\begin{array}{c}\text{LMTD}=\frac{\Delta T_1-\Delta T_2}{\ln \left(\Delta T_1/\Delta T_2\right)}\\ =\frac{450°\text{F}-230°\text{F}}{\ln \left(\frac{450°\text{F}}{230°\text{F}}\right)}\\ =327.78°\text{F}\end{array}$

Calculate the value of $P$ using the relation.

    $\begin{array}{c}p=\frac{t_2-t_1}{T_1-t_1}\\ =\frac{300°\text{F}-600°\text{F}}{70°\text{F}-600°\text{F}}\\ =0.566\end{array}$

Calculate the value of $R$ using the relation.

    $\begin{array}{c}R=\frac{T_1-T_2}{t_2-t_1}\\ =\frac{70°\text{F}-150°\text{F}}{300°\text{F}-600°\text{F}}\\ =0.266\end{array}$

Refer Figure 22-19 “Correction factor $F$ charts for common shell-and-tube and cross-flow heat exchangers.”

Obtain the value of correction factor $\left(F\right)$ from “one-shell passes and $4,8,12,etc.$ (any multiple of $4$ ) tube passes” as follows:

    $F=0.96$

Calculate the overall heat transfer rate using the relation.

    $\begin{array}{c}\frac{1}{U}=\frac{1}{h}+R_{f,water}+R_{f,air}\\ =\frac{1}{30\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot °\text{F}}+0.0015\text{h}{\text{.ft}}^2.°\text{F}/\text{Btu}+0.001\text{h}{\text{.ft}}^2.°\text{F}/\text{Btu}\\ U=27.9\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot °\text{F}\end{array}$

Calculate the heat transfer rate using the relation.

    $\begin{array}{c}\dot{Q}=U\cdot A_s\cdot F\cdot \text{LMTD}\\ A_s=\frac{\dot{Q}}{U\cdot F\cdot \text{LMTD}}\\ =\frac{\left(4\times {10}^6\text{Btu}/\text{h}\right)}{\left(27.9\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot °\text{F}\right)\left(0.96\right)\left(327.78°\text{F}\right)}\\ =455.5{\text{ft}}^2\end{array}$

Calculate the heat capacity rate of the cold fluid using the relation.

    $\begin{array}{c}{C}_c={\dot{m}}_c{c}_{pc}\\ =\left(50000\text{lbm}/\text{h}\right)\left(1\text{Btu}/\text{lbm}.°\text{F}\right)\\ =50000\text{Btu}/\text{h}.°\text{F}\end{array}$

Calculate the heat capacity rate of the hot fluid using the relation.

    $\begin{array}{c}{C}_h={\dot{m}}_h{c}_{ph}\\ =\left(53333.33\text{lbm}/\text{h}\right)\left(0.25\text{Btu}/\text{lbm}.°\text{F}\right)\\ =13333.33\text{Btu}/\text{h}.°\text{F}\end{array}$

Calculate the capacity rate ration using the relation.

    $\begin{array}{c}c=\frac{C_{\min }}{C_{\max }}\\ =\frac{\left(13333.33\text{Btu}/\text{h}.°\text{F}\right)}{\left(50000\text{Btu}/\text{h}.°\text{F}\right)}\\ =0.266\end{array}$

Calculate the effectiveness of the heat exchanger using the relation.

    $\begin{array}{c}\epsilon =\frac{\dot{Q}}{{\dot{Q}}_{\max }}\\ =\frac{C_h\left(T_{hin}-T_{hout}\right)}{C_{\min }\left(T_{hin}-T_{cin}\right)}\\ =\frac{\left(600°\text{F}-300°\text{F}\right)}{\left(600°\text{F}-70°\text{F}\right)}\\ =0.566\end{array}$

Calculate the $\text{NTU}$ using the relation.

    $\begin{array}{c}\text{NTU}=-\frac{1}{\sqrt{1+c^2}}\ln \left(\frac{2/\epsilon -1-c-\sqrt{1+c^2}}{2/\epsilon -1-c+\sqrt{1+c^2}}\right)\\ =-\frac{1}{\sqrt{1+{0.266}^2}}\ln \left(\frac{2/0.566-1-0.266-\sqrt{1+{0.266}^2}}{2/0.566-1-0.266+\sqrt{1+{0.266}^2}}\right)\\ =-\frac{1}{1.034}\ln \left(\frac{1.233}{3.034}\right)\\ =0.952\end{array}$

Calculate the surface area of the heat exchanger using the relation.

    $\begin{array}{c}\text{NTU}=\frac{U{A}_s}{C_{\min }}\\ A_s=\frac{\text{NTU}\times C_{\min }}{U}\\ =\frac{\left(0.952\right)\left(13333.33\text{Btu}/\text{h}\text{.}°\text{F}\right)}{27.9\text{Btu}/\text{h}{\text{.ft}}^2.°\text{F}}\\ =454.95{\text{ft}}^2\end{array}$

Thus, the surface area of the heat exchanger using both $LMTD$ method and $\epsilon ,NTU$ method is $455.5{\text{ft}}^2$ and $454.95{\text{ft}}^2$ respectively.