Chapter 22, Problem 124RQ

(a)

To determine

The outlet temperature of the cold water.

Explanation of Solution

Given:

The overall heat transfer coefficient $\left(U\right)$ is $1400\text{W}/\text{K}$.

The specific heat of hot water $\left(c_p\right)$ is $4180\text{J}/\text{kg}\cdot \text{K}$.

The temperature of hot water inlet $\left(T_{hin}\right)$ is $95°\text{C}$.

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

The mass of hot water $\left(m_h\right)$ is $1.5m$.

The mass of cold water $\left(m_c\right)$ is $m$.

Calculation:

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

    $\begin{array}{c}\left(m_h{c}_p\right)=\left(1.5m\right)\left(4180\text{J}/\text{kg}\cdot \text{K}\right)\\ =6.27m\end{array}$

Calculate the product of mass flow rate of hot water and specific heat $\left(m_c{c}_p\right)$ using the relation.

    $\begin{array}{c}\left(m_c{c}_p\right)=\left(m\right)\left(4180\text{J}/\text{kg}\cdot \text{K}\right)\\ =4.18m\end{array}$

Calculate the ratio of minimum and maximum specific heat $\left(c\right)$ using the relation.

    $\begin{array}{c}c=\frac{\left(m_h{c}_p\right)}{\left(m_c{c}_p\right)}\\ =\frac{4.18m}{6.27m}\\ =0.667\end{array}$

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

    $\begin{array}{c}Q=\left(m_c{c}_p\right)\left(T_{cout}-T_{cin}\right)\\ =4.18m\left(T_{cout}-20°\text{C}\right)\end{array}$

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

  $\begin{array}{c}Q=\left(m_h{c}_p\right)\left(T_{hin}-T_{hout}\right)\\ =6.27m\left(95°\text{C}-\left(T_{hout}+15°\text{C}\right)\right)\\ =6.27m\left(80°\text{C}-\left(T_{hout}\right)\right)\end{array}$

Calculate the outlet temperature of the cold water $\left(T_{cout}\right)$ using the relation.

  $\begin{array}{c}4.18m\left(T_{cout}-20°\text{C}\right)=6.27m\left(80°\text{C}-\left(T_{cout}\right)\right)\\ T_{cout}=56°\text{C}\end{array}$

Thus, the outlet temperature of the cold water is $56°\text{C}$.

(b)

To determine

The effectiveness of the heat exchanger.

Explanation of Solution

Given:

The overall heat transfer coefficient $\left(U\right)$ is $1400\text{W}/\text{K}$.

The specific heat of hot water $\left(c_p\right)$ is $4180\text{J}/\text{kg}\cdot \text{K}$.

The temperature of hot water inlet $\left(T_{hin}\right)$ is $95°\text{C}$.

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

The mass of hot water $\left(m_h\right)$ is $1.5m$.

The mass of cold water $\left(m_c\right)$ is $m$.

Calculation:

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

$\begin{array}{c}\epsilon =\frac{4.18m\left(T_{cout}-T_{cin}\right)}{4.18m\left(T_{hin}-T_{cin}\right)}\\ =\frac{4.18m\left(56°\text{C}-20°\text{C}\right)}{4.18m\left(95°\text{C}-20°\text{C}\right)}\\ =0.480\end{array}$

Thus, The effectiveness of the heat exchanger is $0.48$.

(c)

To determine

The mass flow rate of the cold water.

Explanation of Solution

Given:

The overall heat transfer coefficient $\left(U\right)$ is $1400\text{W}/\text{K}$.

The specific heat of hot water $\left(c_p\right)$ is $4180\text{J}/\text{kg}\cdot \text{K}$.

The temperature of hot water inlet $\left(T_{hin}\right)$ is $95°\text{C}$.

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

The mass of hot water $\left(m_h\right)$ is $1.5m$.

The mass of cold water $\left(m_c\right)$ is $m$.

Calculation:

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

$\begin{array}{c}NTU=\frac{1}{c-1}\ln \left(\frac{\epsilon -1}{\epsilon \epsilon -1}\right)\\ =\frac{1}{0.667-1}\ln \left(\frac{0.48-1}{0.48\times 0.6667-1}\right)\\ =0.8048\end{array}$

Calculate the mass flow rate of the cold water $\left(m\right)$ using the relation.

    $\begin{array}{c}NTU=\frac{UA}{\left(m_c{c}_p\right)}\\ 0.8048=\frac{1400\text{W}/\text{K}}{4.18m}\\ m=0.4162\text{kg}/\text{s}\end{array}$

Thus mass flow rate of the cold water is $0.4162\text{kg}/\text{s}$.

(d)

To determine

The heat transfer rate.

Explanation of Solution

Given:

The overall heat transfer coefficient $\left(U\right)$ is $1400\text{W}/\text{K}$.

The specific heat of hot water $\left(c_p\right)$ is $4180\text{J}/\text{kg}\cdot \text{K}$.

The temperature of hot water inlet $\left(T_{hin}\right)$ is $95°\text{C}$.

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

The mass of hot water $\left(m_h\right)$ is $1.5m$.

The mass of cold water $\left(m_c\right)$ is $m$.

Calculation:

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

$\begin{array}{c}Q=\left(m_c{c}_p\right)\left(T_{cout}-T_{cin}\right)\\ =0.4162\text{kg}/\text{s}\left(4180\text{J}/\text{kg}\cdot \text{K}\right)\left(\begin{array}{l}\left(56°\text{C}+273\right)\text{K}-\\ \left(20°\text{C}+273\right)\text{K}\end{array}\right)\\ =\left(62629.7\text{J}/\text{s}\times \frac{1\text{kW}}{1000\text{J}/\text{s}}\right)\\ =62.6\text{kW}\end{array}$

Thus the heat transfer rate is $62.6\text{kW}$.