Chapter 7, Problem 100P

(a)

To determine

The actual running time of the heat pump in a day.

Explanation of Solution

Given:

The temperature inside the house $\left(T_H\right)$ is $25°\text{C}$.

The temperature outside the house $\left(T_L\right)$ is $2°\text{C}$.

The rate of heat gain per unit degree $\left({\dot{Q}}_H\right)$ is $55000\text{kJ}/\text{h}$.

The rate of work input of a Carnot heat pump $\left({\dot{W}}_{\text{net,in}}\right)$ is $4.8\text{kW}$.

The average price is $0.11\text{\$}/\text{kWh}$.

Calculation:

Calculate the coefficient of performance of the Carnot heat.

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

  $\begin{array}{c}{\text{COP}}_{\text{HP}}=\frac{1}{1-\left(\left(2°\text{C}+273\right)/\left(25°\text{C}+273\right)\right)}\\ =\frac{1}{1-\left(\left(275\text{K}\right)/\left(298\text{K}\right)\right)}\\ =12.956\text{K}\\ \cong \text{12}\text{.96}\text{K}\end{array}$

Determine the total amount of heat lost by the house.

  $Q_H={\dot{Q}}_H\left(\text{per}\text{day}\right)$

  $\begin{array}{c}{Q}_H=\left(55000\text{kJ}/\text{h}\right)\times \left(\text{1}\text{day}\right)\\ =\left(55000\text{kJ}/\text{h}\right)\times \left(\text{1}\text{day}\right)\times \left(\frac{24\text{hr}}{1\text{day}}\right)\\ =1,320,000\text{kJ}\end{array}$

Determine the work input of a Carnot heat pump using COP of the heat pump.

  $\begin{array}{l}{\text{COP}}_{\text{HP}}=\frac{Q_H}{W_{\text{net,in}}}\\ W_{\text{net,in}}=\frac{Q_H}{{\text{COP}}_{\text{HP}}}\end{array}$

  $\begin{array}{c}{W}_{\text{net,in}}=\frac{\left(1,320,000\text{kJ}\right)}{\left(12.96\right)}\\ =101851.852\text{kJ}\end{array}$

Calculate amount of time the heat pump ran.

  $\Delta t=\frac{W_{\text{net,in}}}{{\dot{W}}_{\text{net,in}}}$

  $\begin{array}{c}\Delta t=\frac{\left(101851.852\text{kJ}\right)}{\left(4.8\text{kW}\times \left(\frac{1\text{kJ}/\text{s}}{1\text{kW}}\right)\right)}\\ =21219.1\text{s}\times \left(\frac{1\text{hr}}{3600\text{s}}\right)\\ =5.894\text{hr}\\ \cong 5.90\text{hr}\end{array}$

Thus, the actual running time of the heat pump in a day is $\underset{\_}{5.90\text{h}}$.

(b)

To determine

The total heating cost that day.

Explanation of Solution

Calculate the total heating cost that day.

  $\begin{array}{c}\text{Cost}=\text{W}\times \left(\text{price}\right)\\ =\left({\dot{W}}_{\text{net,in}}\times \Delta t\right)\times \left(\text{price}\right)\end{array}$

  $\begin{array}{c}\text{Cost}=\left(\left(4.8\text{kW}\right)\times \left(5.894\text{h}\right)\right)\times \left(0.11\text{\$}/\text{kWh}\right)\\ =\left(28.272\text{kWh}\right)\times \left(0.11\text{\$}/\text{kWh}\right)\\ =\$3.1099\\ \cong \$3.11\end{array}$

Thus, the total heating cost that day is $\underset{\_}{\$3.11}$.

(c)

To determine

The amount of cost if resistance heating is used instead of heat pump.

Explanation of Solution

Calculate the amount of cost if resistance heating is used instead of heat pump.

  $\text{New}\text{Cost}=Q_H\times \left(\text{price}\right)$

  $\begin{array}{c}\text{New}\text{Cost}=\left(1,320,000\text{kJ}\right)\times \left(0.11\text{\$}/\text{kWh}\right)\\ =\left(\left(1,320,000\text{kJ}\right)\times \left(\frac{1\text{kWh}}{3600\text{kJ}}\right)\right)\times \left(0.11\text{\$}/\text{kWh}\right)\\ =\$40.3\end{array}$

Thus, the amount of cost if resistance heating is used instead of heat pump is $\underset{\_}{\$40.3}$.