Chapter 6.11, Problem 106P

(a)

To determine

The rate of heat removal from the refrigerated space.

Answer to Problem 106P

The rate of heat removal from the refrigerated space is $\underset{\_}{4982\text{kJ}/\text{min}}$.

Explanation of Solution

Write the highest thermal efficiency a heat engine between two specified temperature limits.

${\eta }_{\text{th,max}}=1-\frac{T_L}{T_H}$ (I)

Here, the temperature inside the refrigerator is $T_H$, and the temperature outside the refrigerator is $T_L$.

Write the maximum power output of this heat engine.

${\dot{W}}_{\text{net}}={\eta }_{\text{th}}\times {\dot{Q}}_H$ (II)

Here, the rate of heat gain per unit degree is ${\dot{Q}}_H$.

Write the coefficient of performance of the Carnot refrigerator depends on the temperature limits in the cycle.

${\text{COP}}_{\text{R}}=\frac{1}{\left(T_H/T_L\right)-1}$ (III)

Write the rate of heat removal from the refrigerator space.

${\dot{Q}}_{L,R}=\left({\text{COP}}_{\text{R}}\right)\times \left({\dot{W}}_{\text{net}}\right)$ (IV)

Conclusion:

Convert the temperature $°\text{F}$ to $\text{R}$.

$\begin{array}{c}{T}_L=80°\text{F}\\ \text{=}\left(80+460\right)\text{R}\\ \text{=540 R}\end{array}$

Convert the temperature $°\text{F}$ to $\text{R}$.

$\begin{array}{c}{T}_H=1700°\text{F}\\ \text{=}\left(1700+460\right)\text{R}\\ \text{=2160 R}\end{array}$

Substitute $\text{540 R}$ for $T_L$ and $\text{2160 R}$ for $T_H$ in Equation (I).

$\begin{array}{c}{\eta }_{\text{th,max}}=1-\frac{\text{540 R}}{\text{2160 R}}\\ =1-0.25\\ =0.75\end{array}$

Substitute $0.75$ for ${\eta }_{\text{th,max}}$ and $700\text{Btu}/\text{min}$ for ${\dot{Q}}_H$ in Equation (II).

$\begin{array}{c}{\dot{W}}_{net}=\left(0.75\right)\left(700\text{kJ}/\text{min}\right)\\ =525\text{kJ}/\text{min}\end{array}$

Substitute $\text{540 R}$ for $T_L$ and $\text{540 R}$ for $T_H$ in Equation (III).

$\begin{array}{c}{\text{COP}}_{\text{R}}=\frac{1}{\left(\text{540 R}/20°\text{F}\right)-1}\\ =\frac{1}{\left(\text{540 R}/\left(20+460\right)\text{R}\right)-1}\\ =\frac{1}{1.125-1}\\ =8\end{array}$

Substitute 8 for ${\text{COP}}_{\text{R}}$ and $525\text{Btu}/\text{min}$ for ${\dot{W}}_{net}$ in Equation (IV).

$\begin{array}{c}{\dot{Q}}_{L,R}=\left(\text{8}\right)\times \left(525\text{Btu}/\text{min}\right)\\ =4200\text{Btu}/\text{min}\end{array}$

Thus, the rate of heat removal from the refrigerated space is $4200\text{Btu}/\text{min}$.

(b)

To determine

The total rate of heat rejection to the ambient air.

Answer to Problem 106P

The total rate of heat rejection to the ambient air is $4900\text{Btu}/\text{min}$.

Explanation of Solution

Write the total heat rejected by refrigerator.

${\dot{Q}}_{H,\text{R}}={\dot{W}}_{\text{net}}+{\dot{Q}}_L$ (V)

Write the total heat rejected by heat pump.

${\dot{Q}}_{L,\text{HE}}={\dot{W}}_{\text{net}}+{\dot{Q}}_L$ (VI)

Write the total heat rejected to ambient.

${\dot{Q}}_{\text{ambient}}={\dot{Q}}_{L,\text{HE}}+{\dot{Q}}_{H,\text{R}}$ (VII)

Conclusion:

Substitute $700\text{Btu}/\text{min}$ for ${\dot{Q}}_{H,\text{HE}}$ and $525\text{Btu}/\text{min}$ for ${\dot{W}}_{net}$ in Equation (V).

$\begin{array}{c}{\dot{Q}}_{L,\text{HE}}=\left(700\text{Btu}/\text{min}\right)-\left(525\text{Btu}/\text{min}\right)\\ =175\text{Btu}/\text{min}\end{array}$

Substitute $4200\text{Btu}/\text{min}$ for ${\dot{Q}}_{L,\text{R}}$ and $525\text{Btu}/\text{min}$ for ${\dot{W}}_{net}$ in Equation (VI).

$\begin{array}{c}{\dot{Q}}_{H,\text{R}}=\left(4200\text{Btu}/\text{min}\right)+\left(525\text{Btu}/\text{min}\right)\\ =4725\text{Btu}/\text{min}\end{array}$

Substitute $175\text{Btu}/\text{min}$ for ${\dot{Q}}_{L,\text{HE}}$ and $4725\text{Btu}/\text{min}$ for ${\dot{Q}}_{H,\text{R}}$ in Equation (VII).

$\begin{array}{c}{\dot{Q}}_{\text{ambient}}=175\text{Btu}/\text{min}+4725\text{Btu}/\text{min}\\ =4900\text{Btu}/\text{min}\end{array}$

Thus, the total rate of heat rejection to the ambient air is $4900\text{Btu}/\text{min}$.