Chapter 7, Problem 110RQ

(a)

To determine

The power input of the refrigerator.

Explanation of Solution

Given:

The coefficient of performance to the refrigerator $\left({\text{COP}}_{\text{R}}\right)$ is 1.9.

The heat removal from the cold medium $\left({\dot{Q}}_L\right)$ is $24,000\text{Btu}/\text{h}$.

The mass rate of the water $\left(\dot{m}\right)$ is $1.45\text{lbm}/\text{s}$.

The coefficient of pressure $\left(c_p\right)$ is $1.0\text{Btu}/\text{lbm}\cdot °\text{F}$

The water enter to the condenser $\left(T_1\right)$ is $65°\text{F}$.

The system absorb heat from a space $\left(T_L\right)$ is $25°\text{F}$.

Calculation:

Calculate the power consumption of the refrigerator $\left({\dot{W}}_{\text{in}}\right)$.

  $\begin{array}{l}{\text{COP}}_{\text{R}}=\frac{{\dot{Q}}_L}{{\dot{W}}_{\text{in}}}\\ {\dot{W}}_{\text{in}}=\frac{{\dot{Q}}_L}{{\text{COP}}_{\text{R}}}\end{array}$

  $\begin{array}{c}{\dot{W}}_{\text{in}}=\frac{24,000\text{Btu}/\text{h}}{1.9}\\ =\frac{24,000\text{Btu}/\text{h}}{1.9}\times \left(\frac{1.055\text{kJ}}{1\text{Btu}}\right)\times \left(\frac{1\text{h}}{3600\text{s}}\right)\\ =3.702\text{kW}\end{array}$

Thus, the power input of the refrigerator is $\underset{\_}{3.702\text{kW}}$.

(b)

To determine

The exit temperature of the water.

Explanation of Solution

Calculate rate of heat rejected in the rejected of the refrigerator.

  ${\dot{Q}}_H={\dot{Q}}_L+{\dot{W}}_{\text{in}}$

  $\begin{array}{c}{\dot{Q}}_H=\left(24,000\text{Btu}/\text{h}\right)+\left(3.702\text{kW}\right)\\ =\left(24,000\text{Btu}/\text{h}\right)+\left(3.702\text{kW}\right)\times \left(\frac{1\text{Btu}}{1.055\text{kJ}}\right)\left(\frac{3600\text{s}}{1\text{h}}\right)\\ =36631.5789\text{Btu}/\text{h}\\ \cong 36632\text{Btu}/\text{h}\end{array}$

Calculate exit temperature of the water $\left(T_2\right)$.

  $\begin{array}{l}{\dot{Q}}_H=\dot{m}{c}_p\left(T_2-T_1\right)\\ T_2=T_1+\frac{{\dot{Q}}_H}{\dot{m}{c}_p}\end{array}$

  $\begin{array}{c}{T}_2=65°\text{F+}\frac{36632\text{Btu}/\text{h}}{\left(1.45\text{lbm}/\text{s}\right)\times \left(1.0\text{Btu}/\text{lbm}\cdot °\text{F}\right)}\\ =65°\text{F+}\frac{36632\text{Btu}/\text{h}}{\left(1.45\text{lbm}/\text{s}\right)\times \left(\frac{3600\text{s}}{1\text{h}}\right)\times \left(1.0\text{Btu}/\text{lbm}\cdot °\text{F}\right)}\\ =65°\text{F}+\text{7}\text{.02}°\text{F}\\ =\text{72}\text{.02}°\text{F}\end{array}$

Thus, the exit temperature of the water is $\underset{\_}{\text{72}\text{.02}°\text{F}}$.

(c)

To determine

The maximum possible COP of the refrigerator.

Explanation of Solution

Calculate the average temperature of the water as source temperature.

  $T_H=\frac{T_1+T_2}{2}$

  $\begin{array}{c}{T}_H=\frac{\left(65+\text{72}\text{.02}\right)°\text{F}}{2}\\ =68.51°\text{F}\end{array}$

Calculate maximum possible COP of the refrigerator.

  ${\text{COP}}_{\text{R}}=\frac{T_L}{T_H-T_L}$

  $\begin{array}{c}{\text{COP}}_{\text{rev}}=\frac{25°\text{F}}{68.51°\text{F}-25°\text{F}}\\ =\frac{\left(25+460\text{}\right)\text{R}}{\left(68.51+460\right)\text{R}-\left(25+460\right)\text{R}}\\ =\frac{485\text{R}}{43.51\text{R}}\\ =11.2\end{array}$

Thus, the maximum possible COP of the refrigerator is $\underset{\_}{11.2}$.