Chapter 6.11, Problem 129RP

A refrigeration system uses a water-cooled condenser for rejecting the waste heat. The system absorbs heat from a space at 25°F at a rate of 21,000 Btu/h. Water enters the condenser at 65°F at a rate of 1.45 lbm/s. The COP of the system is estimated to be 1.9. Determine (a) the power input to the system in kW, (b) the temperature of the water at the exit of the condenser in °F, and (c) the maximum possible COP of the system. The specific heat of water is 1.0 Btu/bm·°F.

To determine

The power input of the refrigerator.

Answer to Problem 129RP

The power input of the refrigerator is $\underset{\_}{\text{3}\text{.239}\text{kW}}$.

Explanation of Solution

Determine coefficient of performance to the refrigerator.

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

Here, the power consumption of the refrigerator is ${\dot{W}}_{\text{in}}$, and the heat removal from the cold medium is ${\dot{Q}}_L$.

Conclusion:

Substitute $21,000\text{Btu}/\text{h}$ for ${\dot{Q}}_L$ and 1.9 for C.O.P in Equation (I).

$\begin{array}{c}{\dot{W}}_{\text{in}}=\frac{21,000\text{Btu}/\text{h}}{1.9}\\ =\frac{21,000\text{Btu}/\text{h}}{1.9}\times \left(\frac{1.055\text{kJ}}{1\text{Btu}}\right)\times \left(\frac{1\text{h}}{3600\text{s}}\right)\\ =3.239\text{kW}\end{array}$

Thus, the power input of the refrigerator is $\underset{\_}{\text{3}\text{.239}\text{kW}}$.

To determine

The exit temperature of the water.

Answer to Problem 129RP

The exit temperature of the water is $\underset{\_}{\text{71}\text{.14}°\text{F}}$.

Explanation of Solution

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

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

Here, the power consumption of the refrigerator is ${\dot{W}}_{\text{in}}$, and the heat removal from the cold medium is ${\dot{Q}}_L$.

Determine exit temperature of the water.

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

Here, the mass rate of the water is $\dot{m}$, the coefficient of pressure is $c_p$, the water enter to the condenser is $T_1$, the exit temperature of the water is $T_2$.

Conclusion:

Substitute $21,000\text{Btu}/\text{h}$ for ${\dot{Q}}_L$ and $3.239\text{kW}$ for ${\dot{W}}_{\text{in}}$ in Equation (II).

$\begin{array}{c}{\dot{Q}}_H=\left(21,000\text{Btu}/\text{h}\right)+\left(3.239\text{kW}\right)\\ =\left(21,000\text{Btu}/\text{h}\right)+\left(3.239\text{kW}\right)\times \left(\frac{1\text{Btu}}{1.055\text{kJ}}\right)\left(\frac{3600\text{s}}{1\text{h}}\right)\\ =32,052.5\text{Btu}/\text{h}\\ \cong 32,053\text{Btu}/\text{h}\end{array}$

Substitute $65°\text{F}$ for $T_1$, $32,053\text{Btu}/\text{h}$ for ${\dot{Q}}_H$, $1.45\text{lbm}/\text{s}$ for $\dot{m}$, and $1.0\text{Btu}/\text{lbm}\cdot °\text{F}$ for $c_p$ in Equation (III).

$\begin{array}{c}{T}_2=65°\text{F+}\frac{32,053\text{Btu}/\text{h}}{\left(1.45\text{lbm}/\text{s}\right)\times \left(1.0\text{Btu}/\text{lbm}\cdot °\text{F}\right)}\\ =65°\text{F+}\frac{32,053\text{Btu}/\text{h}}{\left(1.45\text{lbm}/\text{s}\right)\times \left(\frac{3600\text{s}}{1\text{h}}\right)\times \left(1.0\text{Btu}/\text{lbm}\cdot °\text{F}\right)}\\ =65°\text{F+6}\text{.14}°\text{F}\\ =\text{71}\text{.14}°\text{F}\end{array}$

Thus, the exit temperature of the water is $\underset{\_}{\text{71}\text{.14}°\text{F}}$.

To determine

The maximum possible COP of the refrigerator.

Answer to Problem 129RP

The maximum possible COP of the refrigerator is $\underset{\_}{\text{11}\text{.26}}$.

Explanation of Solution

Determine the average temperature of the water as source temperature.

$T_H=\frac{T_1+T_2}{2}$ (IV)

Determine maximum possible COP of the refrigerator.

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

Here, the system absorb heat from a space is $T_L$.

Conclusion:

Substitute $65°\text{F}$ for $T_1$ and $71.14°\text{F}$ for $T_2$ in Equation (IV).

$\begin{array}{c}{T}_H=\frac{\left(65+71.14\right)°\text{F}}{2}\\ =68.07°\text{F}\end{array}$

Substitute $25°\text{F}$ for $T_L$ and $68.07°\text{F}$ for $T_H$ in Equation (V).

$\begin{array}{c}{\text{COP}}_{\text{rev}}=\frac{25°\text{F}}{68.07°\text{F}-25°\text{F}}\\ =\frac{25°\text{F+460}}{43.07°\text{F}}\\ =11.26\end{array}$

Thus, the maximum possible COP of the refrigerator is $\underset{\_}{\text{11}\text{.26}}$.