Chapter 22, Problem 61P

(a)

To determine

The rate of heat transfer in the heat exchanger before any fouling occurs.

Explanation of Solution

Given:

The temperature of glycerin inlet $\left(T_{cin}\right)$ is $65°\text{F}$.

The temperature of glycerin out $\left(T_{cout}\right)$ is $140°\text{F}$.

The temperature of hot water inlet $\left(T_{hin}\right)$ is $175°\text{F}$.

The temperature of hot water out $\left(T_{hout}\right)$ is $120°\text{F}$.

The diameter of tube is $\left(d\right)$ is $0.5\text{in}$.

The length of tube is $\left(l\right)$ is $500\text{ft}$.

The convection heat transfer coefficient of on the glycerin $\left(h_g\right)$ is $4\text{Btu}/{\text{hft}}^{\text{2}}\cdot \text{°F}$.

The convection heat transfer coefficient of on the water $\left(h_w\right)$ is $50\text{Btu}/{\text{hft}}^{\text{2}}\cdot \text{°F}$.

Calculation:

Calculate the temperature differences between the glycerin and water at inlet $\left(\Delta T_1\right)$ using the relation.

    $\begin{array}{c}\Delta T_1=T_{hin}-T_{cout}\\ =175°\text{F}-140°\text{F}\\ =\text{35}°\text{F}\end{array}$

Calculate the temperature differences between the glycerin and water at outlet $\left(\Delta T_2\right)$ using the relation.

    $\begin{array}{c}\Delta T_2=T_{hout}-T_{cin}\\ =120°\text{F}-65°\text{F}\\ =\text{55}°\text{F}\end{array}$

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

    $\begin{array}{c}\Delta T_m=\frac{\Delta T_1-\Delta T_2}{\ln \left(\frac{\Delta T_1}{\Delta T_2}\right)}\\ =\frac{\left(\text{35}°\text{F}\right)-\left(\text{55}°\text{F}\right)}{\ln \left(\frac{\text{35}°\text{F}}{\text{55}°\text{F}}\right)}\\ =44.25°\text{F}\end{array}$

Calculate the ratio P and R using the relation.

  $\begin{array}{c}P=\frac{T_{cout}-T_{cin}}{T_{hin}-T_{cin}}\\ =\frac{120°\text{F}-175°\text{F}}{65°\text{F}-175°\text{F}}\\ =0.5\end{array}$

  $\begin{array}{c}R=\frac{T_{hin}-T_{hout}}{T_{cout}-T_{cin}}\\ =\frac{65°\text{F}-140°\text{F}}{120°\text{F}-175°\text{F}}\\ =1.36\end{array}$

Refer Figure 22-2 “Correction factor F charts for common shell-and-tube and cross-flow heat exchangers”.

Obtain the correction factor $\left(F\right)$  as follows:

    $F=0.50$

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

    $\begin{array}{c}\frac{1}{U}=\frac{1}{\frac{1}{U_1}+\frac{1}{U_2}}\\ =\frac{1}{\frac{1}{50\text{Btu}/{\text{hft}}^{\text{2}}\cdot \text{°F}}+\frac{1}{4\text{Btu}/{\text{hft}}^{\text{2}}\cdot \text{°F}}}\\ U=3.704\text{Btu}/{\text{hft}}^{\text{2}}\cdot \text{°F}\end{array}$

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

    $\begin{array}{c}Q=U\left(n\pi DL\right)F\Delta T_m\\ =\left(3.704\text{Btu}/{\text{hft}}^{\text{2}}\cdot \text{°F}\right)\left(0.5\right)\left(44.25°\text{F}\right)\left(8\pi \left(0.5\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)500\text{ft}\right)\\ =42910\text{Btu}/\text{h}\end{array}$

Thus, the rate of heat transfer in the heat exchanger before any fouling occurs is $42910\text{Btu}/\text{h}$.

(b)

To determine

The rate of heat transfer in the heat exchanger after fouling.

Explanation of Solution

Given:

The temperature of glycerin inlet $\left(T_{cin}\right)$ is $65°\text{F}$.

The temperature of glycerin out $\left(T_{cout}\right)$ is $140°\text{F}$.

The temperature of hot water inlet $\left(T_{hin}\right)$ is $175°\text{F}$.

The temperature of hot water out $\left(T_{hout}\right)$ is $120°\text{F}$.

The diameter of tube is $\left(d\right)$ is $0.5\text{in}$.

The length of tube is $\left(l\right)$ is $500\text{ft}$.

The convection heat transfer coefficient of on the glycerin $\left(h_g\right)$ is $4\text{Btu}/{\text{hft}}^{\text{2}}\cdot \text{°F}$.

The convection heat transfer coefficient of on the water $\left(h_w\right)$ is $50\text{Btu}/{\text{hft}}^{\text{2}}\cdot \text{°F}$.

The fouling factor $\left(R\right)$  is $0.002{\text{hft}}^{\text{2}}\text{°F}/\text{Btu}$

Calculation:

Calculate the thermal resistance of the heat exchanger with a fouling factor $\left(R\right)$ using the relation.

    $\begin{array}{c}R=\frac{1}{h_w{A}_i}+\frac{R}{A_i}+\frac{1}{h_g{A}_o}\\ =\left[\begin{array}{l}\frac{1}{\left(50\text{Btu}/{\text{hft}}^{\text{2}}\cdot \text{°F}\right)\left(\left(8\pi \left(0.5\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)500\text{ft}\right)\right)}+\\ \frac{0.002{\text{hft}}^{\text{2}}\text{°F}/\text{Btu}}{\left(8\pi \left(0.5\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)500\text{ft}\right)}+\\ \frac{1}{\left(8\pi \left(0.5\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)500\text{ft}\right)\left(50\text{Btu}/{\text{hft}}^{\text{2}}\cdot \text{°F}\right)}\end{array}\right]\\ =0.0005195\text{h°F}/\text{Btu}\end{array}$

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

    $\begin{array}{c}Q=U\left(n\pi DL\right)F\Delta T_m\\ =\left(3.676\text{Btu}/{\text{hft}}^{\text{2}}\cdot \text{°F}\right)\left(0.5\right)\left(44.25°\text{F}\right)\left(8\pi \left(0.5\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)500\text{ft}\right)\\ =42585\text{Btu}/\text{h}\end{array}$

Thus, the rate of heat transfer in the heat exchanger after fouling occurs is $42585\text{Btu}/\text{h}$.