Chapter 22, Problem 118RQ

(a)

To determine

The rate of heat transfer and the mass rates of water and hydrocarbon streams.

Explanation of Solution

Given:

The specific heat of hydrocarbon stream $\left(c_p\right)$ is $2.0\text{kJ}/\text{kg}\cdot \text{K}$.

The number of tube $\left(n\right)$ is $160$.

The diameter of tube $\left(d\right)$ is $2.0\text{cm}$.

The length of tube $\left(l\right)$ is $1.5\text{m}$.

The heat transfer coefficient of the tube side $\left(h_i\right)$ is $1.6\text{kW}/{\text{m}}^2\cdot \text{K}$.

The heat transfer coefficient of the shell side $\left(h_o\right)$ is $2.5\text{kW}/{\text{m}}^2\cdot \text{K}$.

The inlet temperature of hydrocarbon stream is $\left(T_{cin}\right)$ is $20°\text{C}$.

The outlet temperature of hydrocarbon stream is $\left(T_{cout}\right)$ is $50°\text{C}$.

The inlet temperature of water stream is $\left(T_{hin}\right)$ is $80°\text{C}$.

The outlet temperature of water stream is $\left(T_{hout}\right)$ is $40°\text{C}$.

Calculation:

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

    $\begin{array}{c}\Delta T_1=T_{hin}-T_{cout}\\ =80°\text{C}-50°\text{C}\\ =\text{30}°\text{C}\end{array}$

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

    $\begin{array}{c}\Delta T_2=T_{hout}-T_{cin}\\ =40°\text{C}-20°\text{C}\\ =\text{20}°\text{C}\end{array}$

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

    $\begin{array}{c}\Delta T_m=\frac{\Delta T_1-\Delta T_2}{\ln \left(\frac{\Delta T_1}{\Delta T_2}\right)}\\ =\frac{\left(30°\text{C}+273\right)\text{K}-\left(20°\text{C}+273\right)\text{K}}{\ln \left(\frac{30°\text{C}}{20°\text{C}}\right)}\\ =24.66\text{K}\end{array}$

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

    $\begin{array}{c}U=\frac{1}{\frac{1}{h_i}+\frac{1}{h_i}}\\ =\frac{1}{\frac{1}{1.6\text{kW}/{\text{m}}^2\cdot \text{K}}+\frac{1}{2.5\text{kW}/{\text{m}}^2\cdot \text{K}}}\\ =0.976\text{kW}/{\text{m}}^2\cdot \text{K}\end{array}$

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

  $\begin{array}{c}Q=UAF\Delta T_m\\ =\left(0.976\text{kW}/{\text{m}}^2\cdot \text{K}\right)\left[\left(160\pi \left(2\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(1.5\text{m}\right)\right)\right]\times 0.90\times 24.66\text{K}\\ =326.5\text{kW}\end{array}$

Calculate the mass flow rate of cold fluid stream $\left(m_c\right)$ using the relation.

    $\begin{array}{c}{m}_c=\frac{Q}{c_p\left(T_{cout}-T_{cin}\right)}\\ =\frac{326.5\text{kW}}{\left(2.0\text{kJ}/\text{kg}\cdot \text{K}\right)\left[\left(50°\text{C+273}\right)\text{K}-\left(20°\text{C+273}\right)\text{K}\right]}\\ =5.44\text{kg}/\text{s}\end{array}$

Calculate the mass flow rate of hot fluid stream $\left(m_c\right)$ using the relation.

    $\begin{array}{c}{m}_c=\frac{Q}{c_p\left(T_{cout}-T_{cin}\right)}\\ =\frac{326.5\text{kW}}{\left(4.18\text{kJ}/\text{kg}\cdot \text{K}\right)\left[\left(80°\text{C+273}\right)\text{K}-\left(40°\text{C+273}\right)\text{K}\right]}\\ =1.95\text{kg}/\text{s}\end{array}$

Thus, The rate of heat transfer and the mass rates of water is $5.44\text{kg}/\text{s}$ and hydrocarbon streams is $1.95\text{kg}/\text{s}$.

(b)

To determine

The magnitude of fouling factor

Explanation of Solution

Given:

The specific heat of hydrocarbon stream $\left(c_p\right)$ is $2.0\text{kJ}/\text{kg}\cdot \text{K}$.

The number of tube $\left(n\right)$ is $160$.

The diameter of tube $\left(d\right)$ is $2.0\text{cm}$.

The length of tube $\left(l\right)$ is $1.5\text{m}$.

The heat transfer coefficient of the tube side $\left(h_i\right)$ is $1.6\text{kW}/{\text{m}}^2\cdot \text{K}$.

The heat transfer coefficient of the shell side $\left(h_o\right)$ is $2.5\text{kW}/{\text{m}}^2\cdot \text{K}$.

The inlet temperature of hydrocarbon stream is $\left(T_{cin}\right)$ is $20°\text{C}$.

The outlet temperature of hydrocarbon stream is $\left(T_{cout}\right)$ is $50°\text{C}$.

The inlet temperature of water stream is $\left(T_{hin}\right)$ is $80°\text{C}$.

Calculation:

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

  $\begin{array}{c}Q=m_c{c}_p\left(\left(T_{cout}-5°\text{C}\right)-{\text{T}}_{cin}\right)\\ =\left(5.44\text{kg}/\text{s}\right)\left(2.0\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\begin{array}{l}\left(\begin{array}{l}\left(50°\text{C}+273\right)\text{K}-\\ \left(5°\text{C}+273\right)\text{K}\end{array}\right)-\\ \left(20°\text{C}+273\right)\text{K}\end{array}\right)\\ =272\text{kW}\end{array}$

Calculate the outlet temperature of hot fluid is $\left(T_{hout}\right)$ is $40°\text{C}$.

  $\begin{array}{c}Q=m_h{c}_p\left({\text{T}}_{hin}-{\text{T}}_{hout}\right)\\ 272\text{kW}=\left(1.95\text{kg}/\text{s}\right)\left(4.18\text{kJ}/\text{kg}\cdot \text{K}\right)\left(80°\text{C}-{\text{T}}_{hout}\right)\\ \frac{272\text{kW}\left(\frac{1\text{kJ}/\text{s}}{1\text{kW}}\right)}{\left(1.95\text{kg}/\text{s}\right)\left(4.18\text{kJ}/\text{kg}\cdot \text{K}\right)}=\left(\left(80°\text{C}+273\right)\text{K}-{\text{T}}_{hout}\right)\\ T_{hout}=\left(319.6\text{K}-273\right)°\text{C}\\ =46.6°\text{C}\end{array}$

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

    $\begin{array}{c}\Delta T_1=T_{hin}-\left(T_{cout}-5°\text{C}\right)\\ =80°\text{C}-\left(50°\text{C}-\text{5}°\text{C}\right)\\ =\text{35}°\text{C}\end{array}$

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

    $\begin{array}{c}\Delta T_2=T_{hout}-T_{cin}\\ =46.6°\text{C}-20°\text{C}\\ \text{=26}\text{.6}°\text{C}\end{array}$

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

    $\begin{array}{c}\Delta T_m=\frac{\Delta T_1-\Delta T_2}{\ln \left(\frac{\Delta T_1}{\Delta T_2}\right)}\\ =\frac{\left(35°\text{C}+273\right)\text{K}-\left(26.6°\text{C}+273\right)\text{K}}{\ln \left(\frac{35°\text{C}}{26.6°\text{C}}\right)}\\ =30.61\text{K}\end{array}$

Calculate the ratio P and R using the relation.

  $\begin{array}{c}P=\frac{T_{cout}-T_{cin}}{T_{hin}-T_{cin}}\\ =\frac{\left(50°\text{C}-\text{5}°\text{C}\right)-20°\text{C}}{80°\text{C}-20°\text{C}}\\ =0.42\end{array}$

  $\begin{array}{c}R=\frac{T_{hin}-T_{hout}}{T_{cout}-T_{cin}}\\ =\frac{80°\text{C}-46.6°\text{C}}{45°\text{C}-20°\text{C}}\\ =1.34\end{array}$

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

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

    $F=0.97$

Thus, The magnitude of fouling factor is $0.97$.