Chapter 22, Problem 94P

(a)

To determine

The heat transfer surface area of the heat exchanger using $LMTD$ method.

Explanation of Solution

Given:

The number of shell passes is $2$.

The number $\left(n\right)$ of tube passes is $8$.

The specific heat $\left(c_{pc}\right)$ of the ethyl alcohol is $2670\text{J}/\text{kg}\text{.K}$.

The inlet temperature $\left(t_{cin}\right)$ of ethyl alcohol is $25°\text{C}$.

The outlet temperature $\left(t_{cout}\right)$ of ethyl alcohol is $70°\text{C}$.

The mass flow rate ${\dot{m}}_c$ of the ethyl alcohol is $2.1\text{kg}/\text{s}$.

The specific heat $\left(c_{ph}\right)$ of water is $4190\text{J}/\text{kg}\text{.K}$.

The inlet temperature $\left(t_{hin}\right)$ of water is $95°\text{C}$.

The outlet temperature $\left(t_{hout}\right)$ of water is $60°\text{C}$.

The overall heat transfer coefficient $\left(U\right)$ is $800\text{W}/{\text{m}}^2.\text{K}$.

Calculation:

Calculate the temperature differences between the two fluids at the two ends of the heat exchanger using the relation.

    $\begin{array}{c}\Delta T_1=\left(t_{hin}-t_{cout}\right)\\ =\left(\left(95°\text{C}+273\right)\text{K}-\left(70°\text{C}+273\right)\text{K}\right)\\ =25\text{K}\end{array}$

    $\begin{array}{c}\Delta T_2=\left(t_{hout}-t_{cin}\right)\\ =\left(\left(60°\text{C}+273\right)\text{K}-\left(25°\text{C}+273\right)\text{K}\right)\\ =35\text{K}\end{array}$

Calculate the logarithmic mean temperature difference ($LMTD$) using the relation.

    $\begin{array}{c}LMTD=\frac{\Delta T_1-\Delta T_2}{\ln \left(\Delta T_1/\Delta T_2\right)}\\ =\frac{25\text{K}-35\text{K}}{\ln \left(25\text{K}/35\text{K}\right)}\\ =29.7\text{K}\end{array}$

Calculate the value of $\left(P\right)$ using the relation.

    $\begin{array}{c}p=\frac{t_{cout}-t_{cin}}{t_{hin}-t_{cin}}\\ =\frac{\left(70°\text{C}+273\right)\text{K}-\left(25°\text{C}+273\right)\text{K}}{\left(95°\text{C}+273\right)\text{K}-\left(25°\text{C}+273\right)\text{K}}\\ =0.64\end{array}$

Calculate the value of $R$ using the relation.

    $\begin{array}{c}R=\frac{t_{hin}-t_{hout}}{t_{cout}-t_{cin}}\\ =\frac{\left(95°\text{C}+273\right)\text{K}-\left(60°\text{C}+273\right)\text{K}}{\left(70°\text{C}+273\right)\text{K}-\left(25°\text{C}+273\right)\text{K}}\\ =0.78\end{array}$

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 “two-shell passes and $4,8,12,etc.$ (any multiple of $4$ ) tube passes” as follows:

    $F=0.93$

Calculate the rate of heat transfer using the relation

    $\begin{array}{c}\dot{Q}={\dot{m}}_c{c}_{pc}\left(T_{hout}-T_{cin}\right)\\ =\left(2.1\text{kg}/\text{s}\right)\times \left(2.670\text{kJ}/\text{kg}\text{.K}\right)\times \left(\left(70°\text{C}+273\right)\text{K}-\left(25°\text{C}+273\right)\text{K}\right)\\ =252.3\text{kW}\end{array}$

Calculate the surface area of heat exchanger using the relation.

    $\begin{array}{c}\dot{Q}=U.A_s.F.LMTD\\ 252.3\text{kW}=\left(800\text{W}/{\text{m}}^2.\text{K}\right)\times A_s\times \left(0.93\right)\times \left(29.7°\text{C}\right)\\ A_s=\frac{252.3\text{kW}}{\left(0.8\text{kW}/{\text{m}}^2.\text{K}\right)\times \left(0.93\right)\times \left(29.7°\text{C}\right)}\\ =11.4{\text{m}}^2\end{array}$

Thus, The heat transfer surface area of the heat exchanger using LMTD method is $11.4{\text{m}}^2$.

(b)

To determine

The heat transfer surface area of the heat exchanger using $\epsilon ,NTU$  method.

Explanation of Solution

Calculation:

Calculate the rate of heat transfer using the relation.

    $\begin{array}{c}\dot{Q}={\dot{m}}_c{c}_{pc}\left(T_{cout}-T_{cin}\right)\\ =\left(2.1\text{kg}/\text{s}\right)\left(2.670\text{kJ}/\text{kg}\text{.K}\right)\left(70°\text{C}-25°\text{C}\right)\\ =\left(2.1\text{kg}/\text{s}\right)\left(2.67\text{kJ}/\text{kg}\text{.K}\right)\left(\begin{array}{l}\left(70°\text{C}+273\right)\text{K}\\ -\left(25°\text{C}+273\right)\text{K}\end{array}\right)\\ =252.3\text{kW}\end{array}$

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

    $\begin{array}{c}\dot{Q}={\dot{m}}_h{c}_{ph}\left(T_{hin}-T_{hout}\right)\\ {\dot{m}}_h=\frac{\dot{Q}}{c_{ph}\left(T_{hin}-T_{hout}\right)}\\ =\frac{\left(252.3\text{kW}\right)}{\left(4.19\text{kJ}/\text{kg}\text{.K}\right)\times \left(\begin{array}{l}\left(95°\text{C}+273\right)\text{K}\\ -\left(60°\text{C}+273\right)\text{K}\end{array}\right)}\\ =1.72\text{kg}/\text{s}\end{array}$

Calculate the heat capacity rates of the hot and cold fluids using the relation

    $\begin{array}{c}{C}_h={\dot{m}}_h{c}_{ph}\\ =\left(1.72\text{kg}/\text{s}\right)\times \left(4.19\text{kJ}/\text{kg}\text{.K}\right)\\ =7.21\text{kW}/\text{K}\end{array}$

    $\begin{array}{c}{C}_c={\dot{m}}_c{c}_{pc}\\ =\left(2.1\text{kg}/\text{s}\right)\times \left(2.67\text{kJ}/\text{kg}\text{.K}\right)\\ =5.61\text{kW}/\text{K}\end{array}$

Calculate the capacity rate ratio using the relation.

    $\begin{array}{c}c=\frac{C_{\min }}{C_{\max }}\\ =\frac{C_c}{C_h}\\ =\frac{5.61\text{kW}/\text{K}}{7.21\text{kW}/\text{K}}\\ =0.78\end{array}$

Calculate the maximum heat transfer rate using the relation.

    $\begin{array}{c}{\dot{Q}}_{\max }=C_{\min }\left(T_{hin}-T_{cin}\right)\\ =\left(5.61\text{kW}/\text{K}\right)\times \left(\left(95°\text{C}+273\right)\text{K}-\left(25°\text{C}+273\right)\text{K}\right)\\ =392.7\text{kW}\end{array}$

Calculate the effectiveness of heat exchanger using the relation.

    $\begin{array}{c}\epsilon =\frac{Q}{Q_{\max }}\\ =\frac{252.3\text{kW}}{392.7\text{kW}}\\ =0.64\end{array}$

Refer Figure 22-27 “effectiveness of heat exchangers”.

Obtain the value of $\left(NTU\right)$ from “two-shell passes and $4,8,\mathrm{12.....}$ tube passes” as follows:

    $NTU=1.7$

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

    $\begin{array}{c}NTU=\frac{U{A}_s}{C_{\min }}\\ A_s=\frac{NTU.C_{\min }}{U}\\ =\frac{\left(1.7\right)\times \left(5.61\text{kW}/\text{K}\right)}{0.8\text{kW}/{\text{m}}^2\text{.K}}\\ =11.9{\text{m}}^2\end{array}$

Thus, The heat transfer surface area of the heat exchanger using $\epsilon ,NTU$ method is $11.9{\text{m}}^2$.