Chapter 7, Problem 101P

(a)

To determine

The rate of heat removal from the refrigerated space.

Explanation of Solution

Given:

The temperature $\left(T_{H\left(\text{HE}\right)}\right)$ of the reservoir (heat engine) is $900°\text{C}$.

The temperature $\left(T_L\right)$ of the sink is $27°\text{C}$.

The rate of heat gain per unit degree $\left({\dot{Q}}_H\right)$ is $800\text{kJ}/\text{min}$.

The temperature $\left(T_{H\left(R\right)}\right)$ of the sink is $27°\text{C}$.

Calculation:

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

  ${\eta }_{\text{th,max}}=1-\frac{T_L}{\left(T_{H\left(\text{HE}\right)}\right)}$

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

Calculate the maximum power output of this heat engine.

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

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

Calculate the coefficient of performance of the Carnot refrigerator.

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

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

Calculate the rate of heat removal from the refrigerator space.

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

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

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

(b)

To determine

The total rate of heat rejection to the ambient air.

Explanation of Solution

Calculate the total heat rejected by refrigerator.

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

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

Calculate the total heat rejected by heat pump.

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

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

Calculate the total heat rejected to ambient.

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

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

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