Chapter 22, Problem 134DEP

(a)

To determine

The rate of heat transfer.

The steam temperature for this unit.

Explanation of Solution

Given:

The surface area $\left(A_s\right)$ of the heat exchanger is $9{\text{m}}^2$.

The mass flow rate $\left(\dot{m}\right)$ is $10\text{kg}/\text{s}$.

The inlet temperature $\left(T\right)$ is $10°\text{C}$.

The overall heat transfer coefficient $\left(U\right)$ is $600\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\left({1/\dot{m}}_c{}^{0.8}+{2/\dot{m}}_h{}^{0.8}\right)$.

The specific heat $\left(c_p\right)$ of the liquid is $3.15\text{kJ}/\text{kg}\cdot \text{K}$.

Calculation:

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

    $\begin{array}{c}U=\frac{600\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}}{\left({1/\dot{m}}_c{}^{0.8}+{2/\dot{m}}_h{}^{0.8}\right)}\\ =\frac{600\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}}{\left(\frac{1}{8^{0.8}}+\frac{2}{{10}^{0.8}}\right)}\\ =1185\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the rate of heat transfer $\left(\dot{m}\right)$ using the relation.

  $\begin{array}{c}\dot{Q}={\dot{m}}_h{c}_h\left(T_{h,in}-T_{h,out}\right)\\ =\left[\left(10\text{kg}/\text{s}\right)\times \left(3.150\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\left(90°\text{C}+273\right)\text{K}-T_{h,out}\right)\right]\\ \text{=31}\text{.5}\text{kJ}/\text{s}\cdot \text{K}\left(\left(90°\text{C}+273\right)\text{K}-T_{h,out}\right)\end{array}$……$\left(\text{I}\right)$

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

    $\begin{array}{c}\dot{Q}={\dot{m}}_c{c}_c\left(T_{c,out}-T_{c,in}\right)\\ =\left[\begin{array}{l}\left(8\text{kg}/\text{s}\right)\times \\ \left(4.2\text{kJ}/\text{kg}\cdot \text{K}\right)\left(T_{c,out}-\left(10°\text{C}+273\right)\text{K}\right)\end{array}\right]\\ \text{=33}\text{.6}\text{kJ}/\text{s}\cdot \text{K}\left(T_{c,out}-\left(20°\text{C}+273\right)\text{K}\right)\end{array}$         ……$\left(\text{II}\right)$

Calculate the LMTD $\left(\Delta T_{lm}\right)$ using the relation.

    $\begin{array}{c}\dot{Q}=UA\Delta T_{lm}\\ \Delta T_{lm}=UA\frac{\Delta T_1-\Delta T_2}{\ln \left(\frac{\Delta T_1}{\Delta T_2}\right)}\\ =\left(1185\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(9{\text{m}}^{\text{2}}\right)\frac{\left[\begin{array}{l}\left(\left(90°\text{C}+273\right)\text{K}-T_c\right)-\\ \left(T_h-\left(10°\text{C}+273\right)\text{K}\right)\end{array}\right]}{\ln \left(\frac{\left(90°\text{C}-T_c\right)}{\left(T_h-10°\text{C}\right)}\right)}\end{array}$ ……$\left(\text{III}\right)$

Solve the Equation $\left(\text{I}\right)$$\left(\text{II}\right)$ and $\left(\text{III}\right)$ to obtain the value.

    $\begin{array}{l}\dot{Q}=6.2\times {10}^5\text{W}\\ T_{c,out}=29.1°\text{C}\\ T_{h,out}=69.6°\text{C}\end{array}$

(b)

To determine

The rate of heat transfer.

The steam temperature for this unit.

Explanation of Solution

Given:

Calculation:

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

    $\begin{array}{c}U=\frac{600\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}}{\left({1/\dot{m}}_c{}^{0.8}+{2/\dot{m}}_h{}^{0.8}\right)}\\ =\frac{600\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}}{\left(\frac{1}{4^{0.8}}+\frac{2}{5^{0.8}}\right)}\\ =680.5\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the rate of heat transfer $\left(\dot{m}\right)$ using the relation.

  $\begin{array}{c}\dot{Q}={\dot{m}}_h{c}_h\left(T_{h,in}-T_{h,out}\right)\\ =\left[\left(10\text{kg}/\text{s}\right)\times \left(3.150\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\left(90°\text{C}+273\right)\text{K}-T_{h,out}\right)\right]\\ \text{=31}\text{.5}\text{kJ}/\text{s}\cdot \text{K}\left(\left(90°\text{C}+273\right)\text{K}-T_{h,out}\right)\end{array}$       ……$\left(\text{I}\right)$

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

    $\begin{array}{c}\dot{Q}={\dot{m}}_c{c}_c\left(T_{c,out}-T_{c,in}\right)\\ =\left[\left(8\text{kg}/\text{s}\right)\left(4.2\text{kJ}/\text{kg}\cdot \text{K}\right)\left(T_{c,out}-\left(10°\text{C}+273\right)\text{K}\right)\right]\\ \text{=33}\text{.6}\text{kJ}/\text{s}\cdot \text{K}\left(T_{c,out}-\left(20°\text{C}+273\right)\text{K}\right)\end{array}$         ……$\left(\text{II}\right)$

Calculate the LMTD $\left(\Delta T_{lm}\right)$ using the relation.

    $\begin{array}{c}\dot{Q}=UA\Delta T_{lm}\\ \Delta T_{lm}=UA\frac{\Delta T_1-\Delta T_2}{\ln \left(\frac{\Delta T_1}{\Delta T_2}\right)}\\ =\left(680.5\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(2\times 5{\text{m}}^{\text{2}}\right)\frac{\left[\begin{array}{l}\left(\left(90°\text{C}+273\right)\text{K}-T_c\right)-\\ \left(T_h-\left(10°\text{C}+273\right)\text{K}\right)\end{array}\right]}{\ln \left(\frac{\left(90°\text{C}-T_c\right)}{\left(T_h-10°\text{C}\right)}\right)}\end{array}$……$\left(\text{III}\right)$

Solve the Equation $\left(\text{I}\right)$$\left(\text{II}\right)$ and $\left(\text{III}\right)$ to obtain the value.

    $\begin{array}{l}\dot{Q}=4.5\times {10}^5\text{W}\\ T_{c,out}=23.4°\text{C}\\ T_{h,out}=75.7°\text{C}\end{array}$

Thus, the rate of heat transfer is $\dot{Q}=4.5\times {10}^5\text{W}$.