Chapter 22, Problem 130RQ

(a)

To determine

The outlet temperatures of both streams in steady operation using the LMTD method.

Explanation of Solution

Given:

The mass flow rate of steam $\left(m\right)$ is $2700\text{kg}/\text{h}$.

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

The temperature of steam inlet $\left(T_{h,in}\right)$ is $120°\text{C}$.

The mass flow rate of other steam $\left(m\right)$ is $1800\text{kg}/\text{h}$.

The specific heat of other steam $\left(c\right)$ is $4.2\text{kJ}/\text{kg}\cdot \text{K}$.

The temperature of other steam inlet $\left(T_{c,in}\right)$ is $20°\text{C}$.

The overall heat transfer coefficient $\left(U\right)$ is $2.0\text{kW}/{\text{m}}^2\text{K}$.

The area of heat exchanger $\left(A\right)$ is $0.5{\text{m}}^2$.

Calculation:

Calculate the rate of heat transfer $\left({\dot{Q}}_1\right)$ using the relation.

    $\begin{array}{c}{\dot{Q}}_1=\dot{m}{c}_p\left(T_{h,in}-T_{h,out}\right)\\ =\left(2700\text{kg}/\text{h}\left(\frac{1\text{h}}{3600\text{s}}\right)\right)\left(2.0\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\left(120°\text{C}+273\right)\text{K}-T_{h,out}\right)\end{array}$

Calculate the rate of heat transfer $\left({\dot{Q}}_2\right)$ using the relation.

    $\begin{array}{c}{\dot{Q}}_2=\dot{m}{c}_p\left(T_{c,out}-T_{c,in}\right)\\ =\left(1800\left(\frac{\text{kg}}{\text{h}}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)\right)\left(4.2\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\begin{array}{l}{T}_{c,out}-\\ \left(20°\text{C}+273\right)\text{K}\end{array}\right)\end{array}$

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

  $\begin{array}{c}\Delta T_m=\frac{\Delta T_1-\Delta T_2}{\ln \left(\frac{\Delta T_1}{\Delta T_2}\right)}\\ =\frac{\left(T_{h,in}-T_{c,in}\right)-\left(T_{h,out}-T_{c,out}\right)}{\ln \left(\frac{T_{h,in}-T_{c,in}}{T_{h,out}-T_{c,out}}\right)}\\ =\frac{\left(120°\text{C}-20°\text{C}\right)-\left(T_{h,out}-T_{c,out}\right)}{\ln \left(\frac{120°\text{C}-20°\text{C}}{T_{h,out}-T_{c,out}}\right)}\end{array}$

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

    $\begin{array}{c}\dot{Q}=\left(2.0\text{kW}/{\text{m}}^2\text{K}\right)\left(0.5{\text{m}}^2\right)\frac{\left(120°\text{C}-20°\text{C}\right)-\left(T_{h,out}-T_{c,out}\right)}{\ln \left(\frac{120°\text{C}-20°\text{C}}{T_{h,out}-T_{c,out}}\right)}\\ =\frac{100°\text{C}-\left(T_{h,out}-T_{c,out}\right)}{\ln \left(\frac{100°\text{C}}{T_{h,out}-T_{c,out}}\right)}\end{array}$

Solve the above equation to obtain the outlet temperature of hot steam $\left(T_{hout}\right)$ and outlet temperature of cold steam $\left(T_{cout}\right)$ as follows:

    $\begin{array}{c}{T}_{hout}=80.3°\text{C}\\ T_{cout}=48.4°\text{C}\end{array}$

Thus, the outlet temperatures of hot steam is $80.3°\text{C}$ and cold steam is $48.4°\text{C}$ in steady operation using the LMTD method.

(b)

To determine

The outlet temperatures of both streams in steady operation using the effectiveness–NTU method.

Explanation of Solution

Calculation:

Calculate the product of mass flow rate of hot stream and specific heat $\left(m_h{c}_{ph}\right)$ using the relation.

    $\begin{array}{c}\left({\dot{m}}_h{c}_p\right)=\left(2700\text{kg}/\text{h}\times \left(\frac{1\text{h}}{3600\text{s}}\right)\right)\left(2000\text{J}/\text{kg}\cdot \text{K}\right)\\ =\left(1500\text{J}/\text{s}\cdot \text{K}\times \frac{1\text{kW}}{1000\text{J}/\text{s}}\right)\\ =1.5\text{kW}/\text{K}\end{array}$

Calculate the product of mass flow rate of cold steam and specific heat $\left(m_c{c}_{pc}\right)$ using the relation.

    $\begin{array}{c}\left({\dot{m}}_c{c}_{pc}\right)=\left(1800\text{kg}/\text{h}\times \left(\frac{1\text{h}}{3600\text{s}}\right)\right)\left(4200\text{J}/\text{kg}\cdot \text{K}\right)\\ \left(2100\text{J}/\text{s}\cdot \text{K}\times \frac{1\text{kW}}{1000\text{J}/\text{s}}\right)\\ =2.1\text{kW}/\text{K}\end{array}$

Calculate the heat capacity ratio $\left(c\right)$ using the relation.

    $\begin{array}{c}c=\frac{C_{\min }}{C_{\max }}\\ =\frac{1.5\text{kW}/\text{K}}{2.1\text{kW}/\text{K}}\\ =0.7143\end{array}$=

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

  $\begin{array}{c}NTU=\frac{UA}{C_{\min }}\\ =\frac{\left(2.0\text{kW}/{\text{m}}^2\text{K}\right)\left(0.5{\text{m}}^{\text{2}}\right)}{1.5\text{kW}/\text{K}}\\ =0.667\end{array}$

Calculate the effectiveness of this parallel-flow heat exchanger $\left(\epsilon \right)$  using the relation.

    $\begin{array}{c}\epsilon =\frac{1-\exp \left[-NTU\left(1+c\right)\right]}{1+c}\\ =\frac{1-\exp \left[-\left(0.6667\right)\left(1+0.7143\right)\right]}{1+0.7143}\\ =0.3973\end{array}$

Calculate the maximum heat transfer rate $\left(Q_{\max }\right)$ using the relation.

    $\begin{array}{c}Q=C_{\min }\left(T_{h,in}-T_{c,in}\right)\\ =\left(1.5\text{kW}/{\text{m}}^2\text{K}\right)\left(\left(120°\text{C}+273\right)\text{K}-\left(20°\text{C}+273\right)\text{K}\right)\\ =150\text{kW}\end{array}$

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

    $\begin{array}{c}Q=\epsilon Q_{\max }\\ =\left(0.3973\right)\left(150\text{kW}\right)\\ =59.60\text{kW}\end{array}$

Calculate the outlet temperature of cold steam $\left(T_{c,out}\right)$ using the relation.

    $\begin{array}{c}{T}_{c,out}=T_{c,in}+\frac{Q}{C_c}\\ =\left(20°\text{C}+273\right)\text{K}+\frac{59.6\text{kW}}{2.1\text{kW}/\text{K}}\\ =\left(321.38\text{K}-273\right)°\text{C}\\ =48.4°\text{C}\end{array}$

Calculate the outlet temperature of hot steam $\left(T_{h,out}\right)$ using the relation.

    $\begin{array}{c}{T}_{h,out}=T_{h,in}-\frac{Q}{C_h}\\ =\left(120°\text{C}+273\right)\text{K}-\frac{59.6\text{kW}}{1.5\text{kW}/\text{K}}\\ =\left(353.3\text{K}-273\right)°\text{C}\\ =80.3°\text{C}\end{array}$

Thus, the outlet temperatures of hot steam is $80.3°\text{C}$ and cold steam is $48.4°\text{C}$ in steady operation using the NTU method.