Chapter 22, Problem 89P

(a)

To determine

The overall heat transfer coefficient of this exchanger using the LMTD method.

Explanation of Solution

Given:

The specific heat $\left(c_{ph}\right)$  of the oil is $0.525\text{Btu}/\text{lbm}\text{.}°\text{F}$.

The inlet temperature $\left(t_{hin}\right)$ of oil is $300°\text{F}$.

The outlet temperature $\left(t_{hout}\right)$ of the oil is $105°\text{F}$.

The mass flow rate ${\dot{m}}_h$ of the oil is $5\text{lbm}/\text{s}$.

The specific heat $\left(c_{pc}\right)$ of water is $1.0\text{Btu}/\text{lbm}\text{.}°\text{F}$.

The inlet temperature $\left(t_{cin}\right)$ of water is $70°\text{F}$.

The mass flow rate ${\dot{m}}_c$ of the oil is $3\text{lbm}/\text{s}$.

The diameter $\left(D\right)$  of the tube is $5\text{inch}$

The length $\left(L\right)$  of the tube is $200\text{ft}$.

Calculation:

Calculate the rate of heat transfer using the relation

    $\begin{array}{l}\dot{Q}={\dot{m}}_h{c}_{ph}\left(T_{hin}-T_{hout}\right)\\ =\left(5\text{lbm}/\text{s}\right)\times \left(0.525\text{Btu}/\text{lbm}\text{.}°\text{F}\right)\times \left(300°\text{F}-105°\text{F}\right)\\ =511.9\text{Btu}/\text{s}\end{array}$

Calculate the outlet temperature of the cold fluid using the relation.

    $\begin{array}{c}\dot{Q}={\dot{m}}_c{c}_{pc}\left(T_{cin}-T_{cout}\right)\\ T_{cout}=T_{cin}+\frac{\dot{Q}}{{\dot{m}}_c{c}_{pc}}\\ =70°\text{F}+\frac{511.9\text{Btu}/\text{s}}{\left(3\text{lbm}/\text{s}\right)\left(1.0\text{Btu}/\text{lbm}\text{.}°\text{F}\right)}\\ =240.6°\text{F}\end{array}$

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

    $\begin{array}{c}\Delta T_1=T_{hin}-T_{hout}\\ =300°\text{F}-240.6°\text{F}\\ =\text{59}\text{.4}°\text{F}\end{array}$

    $\begin{array}{c}\Delta T_2=T_{hout}-T_{cin}\\ =105°\text{F}-70°\text{F}\\ =\text{35}°\text{F}\end{array}$

Calculate the logarithmic mean temperature difference using the relation.

    $\begin{array}{c}LMTD=\frac{\Delta T_1-\Delta T_2}{\ln \left(\Delta T_1/\Delta T_2\right)}\\ =\frac{\text{59}\text{.4}°\text{F}-\text{35}°\text{F}}{\ln \left(\text{59}\text{.4}°\text{F}/\text{35}°\text{F}\right)}\\ =46.1°\text{F}\end{array}$

Calculate the overall heat transfer coefficient using the relation.

    $\begin{array}{c}\dot{Q}=U.A_s.LMTD\\ U=\frac{\dot{Q}}{A_s.LMTD}\\ =\frac{511.9\text{Btu}/\text{s}}{\pi \left(5\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)\left(200\text{ft}\right)\left(46.1°\text{F}\right)}\\ =0.0424\text{Btu}/\text{s}{\text{.ft}}^2.°\text{F}\end{array}$

Thus, the overall heat transfer coefficient of this exchanger using the LMTD method is $0.0424\text{Btu}/\text{s}{\text{.ft}}^2.°\text{F}$.

(b)

To determine

The overall heat transfer coefficient of this exchanger using the $\epsilon ,NTU$ method.

Explanation of Solution

Calculation:

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

    $\begin{array}{c}{C}_h={\dot{m}}_h{c}_{ph}\\ =\left(5\text{lbm}/\text{s}\right)\times \left(0.525\text{Btu}/\text{lbm}\text{.}°\text{F}\right)\\ =2.625\text{Btu}/\text{s}.°\text{F}\end{array}$

    $\begin{array}{c}{C}_c={\dot{m}}_c{c}_{pc}\\ =\left(3\text{lbm}/\text{s}\right)\times \left(1.0\text{Btu}/\text{lbm}\text{.}°\text{F}\right)\\ =3.0\text{Btu}/\text{s}.°\text{F}\end{array}$

Here $C_h=2.625\text{Btu}/\text{s}.°\text{F}$ is minimum.

Calculate the capacity rate ratio using the relation.

    $\begin{array}{c}c=\frac{C_{\min }}{C_{\max }}\\ =\frac{2.625\text{Btu}/\text{s}.°\text{F}}{3.0\text{Btu}/\text{s}.°\text{F}}\\ =0.875\end{array}$

Calculate maximum heat transfer rate using the relation.

    $\begin{array}{c}{\dot{Q}}_{\max }=C_{\min }\left(T_{cin}-T_{cin}\right)\\ {\dot{Q}}_{\max }=\left(2.625\text{Btu}/\text{s}\text{.}°\text{F}\right)\left(300°\text{F}-70°\text{F}\right)\\ =603.75\text{Btu}/\text{s}\end{array}$

Calculate the actual rate of heat transfer using the relation.

    $\begin{array}{c}\dot{Q}=C_h\left(T_{cin}-T_{hout}\right)\\ =\left(2.625\text{Btu}/\text{s}\text{.}°\text{F}\right)\left(300°\text{F}-105°\text{F}\right)\\ =511.9\text{Btu}/\text{s}\end{array}$

Calculate the effectiveness of the heat exchanger using the relation.

    $\begin{array}{c}\epsilon =\frac{\dot{Q}}{{\dot{Q}}_{\max }}\\ =\frac{511.9\text{Btu}/\text{s}}{603.75\text{Btu}/\text{s}}\\ =0.85\end{array}$

Calculate the number of transfer units $\left(NTU\right)$ using the relation.

    $\begin{array}{c}NTU=\frac{1}{c-1}\ln \left(\frac{\epsilon -1}{\epsilon c-1}\right)\\ =\frac{1}{0.875-1}\ln \left(\frac{0.85-1}{0.85\times 0.875-1}\right)\\ =4.28\end{array}$

Calculate the heat transfer surface area using the relation.

    $\begin{array}{c}{A}_s=\pi DL\\ =\pi \left(5/12\text{ft}\right)\left(200\text{ft}\right)\\ =261.8{\text{ft}}^2\end{array}$

Calculate the overall heat transfer coefficient using the relation.

    $\begin{array}{c}NTU=\frac{U{A}_s}{C_{\min }}\\ U=\frac{NTU.C_{\min }}{A_s}\\ =\frac{\left(4.28\right)\left(2.635\text{Btu}/\text{s}\text{.}°\text{F}\right)}{261.8{\text{ft}}^2}\\ =0.0429\text{Btu}/\text{s}{\text{.ft}}^2\text{.}°\text{F}\end{array}$

Thus, the overall heat transfer coefficient of this exchanger using the $\epsilon ,NTU$ method is $0.0429\text{Btu}/\text{s}{\text{.ft}}^2.°\text{F}$.