Chapter 6.11, Problem 104P

A Carnot heat pump is to be used to heat a house and maintain it at 25°C in winter. On a day when the average outdoor temperature remains at about 2°C, the house is estimated to lose heat at a rate of 55,000 kJ/h. If the heat pump consumes 4.8 kW of power while operating, determine (a) how long the heat pump ran on that day; (b) the total heating costs, assuming an average price of $0.11/kWh for electricity; and (c) the heating cost for the same day if resistance heating is used instead of a heat pump.

FIGURE P6–111

To determine

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

Answer to Problem 104P

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

Explanation of Solution

Determine the coefficient of performance of the Carnot heat pump depends on the temperature limits in the cycle.

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

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

Determine the total amount of heat lost by the house.

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

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

Determine the work input of a Carnot heat pump.

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

Here the power input required by heat pump is $W_{\text{net,in}}$.

Determine amount of time the heat pump ran.

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

Here, the rate of work input of a Carnot heat pump is ${\dot{W}}_{\text{net,in}}$

Conclusion:

Substitute $2°\text{C}$ for $T_L$ and $25°\text{C}$ for $T_H$ in Equation (I).

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

Substitute $55000\text{kJ}/\text{h}$ for ${\dot{Q}}_H$ and 1 day for per day in Equation (II).

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

Substitute $1,320,000\text{kJ}$ for $Q_H$ and 12.96 for ${\text{COP}}_{\text{HP}}$ in Equation (III).

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

Substitute $101851.852\text{kJ}$ for $W_{\text{net,in}}$ and $4.8\text{kW}$ for ${\dot{W}}_{\text{net,in}}$ in Equation (IV).

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

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

To determine

The total heating cost that day.

Answer to Problem 104P

The total heating cost that day is $\underset{\_}{\$3.11}$.

Explanation of Solution

Determine the total heating cost that day.

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

Conclusion:

Substitute $4.8\text{kW}$ for ${\dot{W}}_{\text{net,in}}$, 5.894 hr for $\Delta t$, $0.11\text{\$}/\text{kWh}$ for price in Equation (V).

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

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

To determine

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

Answer to Problem 104P

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

Explanation of Solution

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

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

Conclusion:

Substitute $1,320,000\text{kJ}$ for $Q_H$, and $0.11\text{\$}/\text{kWh}$ for price in Equation (VI).

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

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