Chapter 7, Problem 123RQ

To determine

The maximum money saved by using the lake water instead of outside air as the heat source.

Explanation of Solution

Given:

The lower temperature of the reversible heat pump ${\left(T_L\right)}_{\text{Air}}$ is $0°\text{C}$.

The lower temperature of the reversible heat pump ${\left(T_L\right)}_{\text{Lake}}$ is $25°\text{C}$.

The higher temperature of the reversible heat pump $\left(T_H\right)$ is $25°\text{C}$.

The rate of heat reject by the heat pump $\left({\dot{Q}}_H\right)$ is $140,000\text{kJ}/\text{h}$.

The cost of electricity is $\text{\$0}\text{.105}/\text{kWh}$.

The number of operating hours is 100 hours.

Calculation:

Initially calculate for outside air:

Calculate coefficient of performance for the reversible heat pump.

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

  $\begin{array}{c}{\text{COP}}_{\text{HP,rev}}\text{=}\frac{1}{1-\left(0°\text{C}/25°\text{C}\right)}\\ \text{=}\frac{1}{1-\left(0+273\text{K}/25+273\text{K}\right)}\\ =11.92\end{array}$

Calculate the minimum power input required to operate the heat pump.

  ${\dot{W}}_{\text{in,min}}=\frac{{\dot{Q}}_H}{{\text{COP}}_{\text{HP,rev}}}$        (II)

  $\begin{array}{c}{\dot{W}}_{\text{in,min}}=\frac{\left(140,000\text{kJ}/\text{h}\right)}{\left(11.92\right)}\\ =\frac{\left(140,000\text{kJ}/\text{h}\right)\times \left(\frac{1\text{kW}}{3600\text{kJ}/\text{h}}\right)}{\left(11.92\right)}\\ =\frac{38.88889\text{kW}}{11.92}\\ =\text{3}\text{.2624}\text{kW}\end{array}$

Calculate the cost of the energy in a heat pump $\left({\text{Cost}}_{\text{air}}\right)$.

  $\text{Cost}=\left[\begin{array}{l}\left({\dot{W}}_{\text{in,min}}\right)\times \left(\text{number of hours operates by heat pump}\right)\\ \times \left(\text{cost of electricity}\right)\end{array}\right]$        (III)

  $\begin{array}{c}{\text{Cost}}_{\text{air}}=\left(\text{3}\text{.2624}\text{kW}\right)\times \left(\text{100}\text{h}\right)\times \left(\text{\$0}\text{.105}/\text{kWh}\right)\\ =\$34.255\\ \cong \$34.26\end{array}$

Similarly,

For lake water:

Substitute ${\left(T_L\right)}_{\text{Lake}}=10°\text{C}$ and $T_H=25°\text{C}$ in Equation (I).

  $\begin{array}{c}{\text{COP}}_{\text{HP,rev}}\text{=}\frac{1}{1-\left(10°\text{C}/25°\text{C}\right)}\\ \text{=}\frac{1}{1-\left(10+273\text{K}/25+273\text{K}\right)}\\ =19.866\\ \cong 19.87\end{array}$

Substitute ${\dot{Q}}_H=140,000\text{kJ}/\text{h}$ and ${\text{COP}}_{\text{HP,rev}}=19.87$ in Equation (II).

  $\begin{array}{c}{\dot{W}}_{\text{in,min}}=\frac{\left(140,000\text{kJ}/\text{h}\right)}{\left(19.87\right)}\\ =\frac{\left(140,000\text{kJ}/\text{h}\right)\times \left(\frac{1\text{kW}}{3600\text{kJ}/\text{h}}\right)}{\left(19.87\right)}\\ =\frac{38.88889\text{kW}}{19.87}\\ =\text{1}\text{.957}\text{kW}\end{array}$

Substitute ${\dot{W}}_{\text{in,min}}=\text{1}\text{.957}\text{kW}$, 100 h for number of hours operates by heat pump, and $\text{\$0}\text{.105}/\text{kWh}$ for cost of electricity in Equation (III).

  $\begin{array}{c}{\text{Cost}}_{\text{lake}}=\left(\text{1}\text{.9571}\text{kW}\right)\times \left(\text{100}\text{h}\right)\times \left(\text{\$0}\text{.105}/\text{kWh}\right)\\ =\$20.549\\ \cong \$20.55\end{array}$

Calculate the maximum money saved by using the lake water $\left({\text{Cost}}_{\text{lake}}\right)$ instead of outside air as the heat source.

  $\text{Money}\text{saved}={\text{Cost}}_{\text{air}}-{\text{Cost}}_{\text{lake}}$        (IV)

  $\begin{array}{c}\text{Money}\text{saved}=\left(\$34.26\right)-\left(\$20.55\right)\\ =\$13.71\\ \cong \$13.7\end{array}$

Thus, the maximum money saved by using the lake water instead of outside air as the heat source is $\underset{\_}{\text{\$13}\text{.7}}$.