Chapter 6.11, Problem 105P

A Carnot heat engine receives heat from a reservoir at 900°C at a rate of 800 kJ/min and rejects the waste heat to the ambient air at 27°C. The entire work output of the heat engine is used to drive a refrigerator that removes heat from the refrigerated space at −5°C and transfers it to the same ambient air at 27°C. Determine (a) the maximum rate of heat removal from the refrigerated space and (b) the total rate of heat rejection to the ambient air.

To determine

The rate of heat removal from the refrigerated space.

Answer to Problem 105P

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

Explanation of Solution

Determine 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 $T_L$.

Determine 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 ${\dot{Q}}_H$.

Determine 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)

Determine 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:

Substitute $27°\text{C}$ for $T_L$ and $900°\text{C}$ for $T_H$ in Equation (I).

$\begin{array}{c}{\eta }_{\text{th,max}}=1-\frac{\left(27\text{°C}\right)}{\left(900°\text{C}\right)}\\ =1-\frac{\left(27\text{°C+273}\right)}{\left(900°\text{C+273}\right)}\\ =1-0.25575\\ =0.744\end{array}$

Substitute $0.744$ for ${\eta }_{\text{th,max}}$ and $800\text{kJ}/\text{min}$ for ${\dot{Q}}_H$ in Equation (II).

$\begin{array}{c}{\dot{W}}_{net}=\left(0.744\right)\left(800\text{kJ}/\text{min}\right)\\ =595.2\text{kJ}/\text{min}\end{array}$

Substitute $-5°\text{C}$ for $T_L$ and $27°\text{C}$ for $T_H$ in Equation (III).

$\begin{array}{c}{\text{COP}}_{\text{R}}=\frac{1}{\left(\left(27°\text{C}\right)/\left(-5°\text{C}\right)\right)-1}\\ =\frac{1}{\left(\left(27°\text{C+273}\right)/\left(-5°\text{C+273}\right)\right)-1}\\ =\frac{1}{\left(0.11940\text{K}\right)-1}\\ =8.375\end{array}$

Substitute 8.375 for ${\text{COP}}_{\text{R}}$ and $595.2\text{kJ}/\text{min}$ for ${\dot{W}}_{net}$ in Equation (IV).

$\begin{array}{c}{\dot{Q}}_{L,R}=\left(\text{8}\text{.37}\right)\times \left(595.2\text{kJ}/\text{min}\right)\\ =4981.8\text{kJ}/\text{min}\end{array}$

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

To determine

The total rate of heat rejection to the ambient air.

Answer to Problem 105P

The total rate of heat rejection to the ambient air is $\underset{\_}{5782\text{kJ}/\text{min}}$.

Explanation of Solution

Determine the total heat rejected by refrigerator.

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

Determine the total heat rejected by heat pump.

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

Determine the total heat rejected to ambient.

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

Conclusion:

Substitute $800\text{kJ}/\text{min}$ for ${\dot{Q}}_{H,\text{HE}}$ and $595.2\text{kJ}/\text{min}$ for ${\dot{W}}_{net}$ in Equation (V).

$\begin{array}{c}{\dot{Q}}_{L,\text{HE}}=\left(800\text{kJ}/\text{min}\right)-\left(595.2\text{kJ}/\text{min}\right)\\ =204.8\text{kJ}/\text{min}\end{array}$

Substitute $4982\text{kJ}/\text{min}$ for ${\dot{Q}}_{L,\text{R}}$ and $595.2\text{kJ}/\text{min}$ for ${\dot{W}}_{net}$ in Equation (VI).

$\begin{array}{c}{\dot{Q}}_{H,\text{R}}=\left(4982\text{kJ}/\text{min}\right)+\left(595.2\text{kJ}/\text{min}\right)\\ =5577.2\text{kJ}/\text{min}\end{array}$

Substitute $204.8\text{kJ}/\text{min}$ for ${\dot{Q}}_{L,\text{HE}}$ and $5577.2\text{kJ}/\text{min}$ for ${\dot{Q}}_{H,\text{R}}$ in Equation (VII).

$\begin{array}{c}{\dot{Q}}_{\text{ambient}}=\left(204.8\text{kJ}/\text{min}\right)+\left(5577.2\text{kJ}/\text{min}\right)\\ =5782\text{kJ}/\text{min}\end{array}$

Thus, the total rate of heat rejection to the ambient air is $\underset{\_}{5782\text{kJ}/\text{min}}$.