Known A heat pump cycle operates with Refrigerant 134a. The states of the refrigerant entering and exiting the compressor and leaving the condenser are specified. The refrigerant mass flow rate and interior and exterior temperatures are given.
Find Determine the compressor power, the isentropic compressor efficiency, the heat transfer rate to the building, the coefficient of performance, and the cost to operate the electric heat pump for 80 hours of operation.
Schematic and Given Data:
Engineering Model
1. Each component of the cycle is analyzed as a control volume at steady state.
2. There are no pressure drops through the evaporator and condenser.
3. The compressor operates adiabatically. The expansion through the valve is a throttling process.
4. Kinetic and potential energy effects are negligible.
5. Saturated vapor enters the compressor and saturated liquid exits the condenser.
6. For costing purposes, conditions provided are representative of the entire week of operation and the value of electricity is 15 cents per kW ⋅ h.
Analysis Let us begin by fixing the principal states located on the accompanying schematic and T–s diagram. State 1 is saturated vapor at −8°C; thus h_{1} \text { and } s_{1} are obtained directly from Table A-10. State 2 is superheated vapor; knowing T_{2} \text { and } p_{2}, h_{2} is obtained from Table A-12. State 3 is saturated liquid at 10 bar and h_{3} is obtained from Table A-11. Finally, expansion through the valve is a throttling process; therefore, h_{4}=h_{3}. A summary of property values at these states is provided in the following table:
| State |
T(°C) |
p(bar) |
h(kJ/kg) |
s(kJ/kg ⋅K) |
| 1 |
-8 |
2.1704 |
242.54 |
0.9239 |
| 2 |
50 |
10 |
280.19 |
– |
| 3 |
– |
10 |
105.29 |
– |
| 4 |
– |
2.1704 |
105.29 |
– |
a. The compressor power is
\dot{W}_{ c }=\dot{m}\left(h_{2}-h_{1}\right)=0.2 \frac{ kg }{ s }(280.19-242.54) \frac{ kJ }{ kg }\left|\frac{1 kW }{1 kJ / s }\right|=7.53 kW
b. The isentropic compressor efficiency is
\eta_{ c }=\frac{\left(\dot{W}_{ c } / \dot{m}\right)_{ s }}{\left(\dot{W}_{ c } / \dot{m}\right)}=\frac{\left(h_{2 s }-h_{1}\right)}{\left(h_{2}-h_{1}\right)}
where h_{2 s } is the specific entropy at state 2s, as indicated on the accompanying T–s diagram. State 2s is fixed using p_{2} \text { and } s_{2 s }=s_{1}.
Interpolating in Table A-12, h_{2 s }=274.18 kJ/kg. Solving for compressor efficiency
\eta_{ c }=\frac{\left(h_{2 s }-h_{1}\right)}{\left(h_{2}-h_{1}\right)}=\frac{(274.18-242.54)}{(280.19-242.54)}=0.84(84 \%)
c. The heat transfer rate provided to the building is
\dot{Q}_{\text {out }}=\dot{m}\left(h_{2}-h_{3}\right)=\left(0.2 \frac{ kg }{ s }\right)(280.19-105.29) \frac{ kJ }{ kg }\left|\frac{1 kW }{1 kJ / s }\right|
= 34.98 kW
d. The heat pump coefficient of performance is
\gamma=\frac{\dot{Q}_{\text {out }}}{\dot{W}_{ c }}=\frac{34.98 kW }{7.53 kW }=4.65
e. Using the result from part (a) together with the given cost and use data
[electricity cost for 80 hours of operation]
=(7.53 kW )(80 h )\left(0.15 \frac{\$}{ kW \cdot h }\right)=\$ 90.36
Skills Developed
Ability to…
• sketch the T–s diagram of the vapor-compression heat pump cycle with irreversibilities in the compressor.
• fix each of the principal states and retrieve necessary property data.
• calculate the compressor power, heat transfer rate delivered, and coefficient of performance.
• calculate isentropic compressor efficiency.
• conduct an elementary economic evaluation.
Quick Quiz
If the cost of electricity is 10 cents per kW⋅ h, which is the U.S. average for the period under consideration, evaluate the cost to operate the heat pump, in $, keeping all other data the same. Ans. $60.24.
