a. Carnot cycle coefficients of performance:
The energy balance on the cycle is
0=\dot{Q}_{1}+\dot{Q}_{2}+\dot{W}
The entropy balance is
0=\frac{\dot{Q}_{1}}{T_{1}}+\frac{\dot{Q}_{2}}{T_{2}}
since S_{\text {gen }}=0 for the Carnot cycle. Therefore,
\dot{Q}_{2}=\frac{-T_{2}}{T_{1}} \dot{Q}_{1} \quad \text { and } \quad-\dot{W}=\dot{Q}_{2}+\dot{Q}_{1}=\dot{Q}_{1}\left(\frac{T_{1}-T_{2}}{T_{1}}\right)
Winter Operation (Heating Mode)
\text { C.O.P. (winter) }=\frac{-\dot{Q}_{1}}{\dot{W}}=\frac{T_{1}}{T_{2}-T_{1}}=\frac{(5+273.15)}{45}=6.18
Summer Operation (Cooling Mode):
\text { C.O.P. }(\text { summer })=\frac{\dot{Q}_{1}}{\dot{W}}=\frac{T_{1}}{T_{2}-T_{1}}=\frac{(5+273.15)}{30}=9.27
b. Vapor compression cycle coefficients of performance:
The following procedure is used to locate the cycles on the pressure-enthalpy diagram of Fig. 3.3-4, a portion of which is repeated here. First, point 1 is identified as corresponding to the saturated liquid at the condenser temperature. (In this illustration the condenser is the indoor coil at 50°C in the heating mode, and the outdoor coil at 35°C in the cooling mode.) Since the path from point 1 to 2 is an isenthalpic (Joule-Thomson) expansion, a downward vertical line is drawn to the evaporator temperature (corresponding to the outdoor coil temperature of 5°C in the heating mode, or in the cooling mode to the indoor coil temperature, which also happens to be 5°C in this illustration). This temperature and the enthalpy of point 1 fix the location of point 2. Next, point 3 is found by drawing a horizontal line to the saturated vapor at the evaporator pressure. The path from point 2 to 3 describes the vaporization of the vapor-liquid mixture that resulted from the Joule Thomson expansion. The evaporator is the outdoor coil in the heating mode and the indoor coil in the cooling mode.
The path from point 3 to 4 is through the compressor, which is assumed to be isentropic, and so corresponds to a line of constant entropy on the pressure-enthalpy diagram. Point 4 is located on this line at its intersection with the horizontal line corresponding to the pressure of the evaporator. The fluid at point 4 is a superheated vapor. The path from 4 to 1 is a horizontal line of constant pressure terminating at the saturated liquid at the temperature and pressure of the condenser.
The tables below give the values of those properties (read from the pressure-enthalpy diagram) at each point in the cycle needed for the calculation of the coefficient of performance in the heating and cooling modes of the heat pump. In these tables the properties known at each point from the problem statement or from an adjacent point appear in boldface, and the properties found from the chart at each point appear in italics.
Heating (winter operation)
| Point |
T (°C) |
P (MPa) |
Condition |
\hat{H}\left(\frac{ kJ }{ kg }\right) |
\hat{S}\left(\frac{ kJ }{ kg K }\right) |
| 1 |
50 |
1.319 |
Saturated liquid |
\begin{matrix} 271.9 \\ \downarrow \end{matrix} |
| 2 |
5 |
\begin{matrix} 0.3499 \\ \downarrow \end{matrix} |
Vapor-liquid mixture |
271.9 |
|
| 3 |
5 |
0.3499 |
Saturated vapor |
401.7 |
\begin{matrix} 1.7252 \\ \downarrow \end{matrix} |
| 4 |
58 |
1.319 |
Superheated vapor |
432 |
1.7252 |
Therefore,
\begin{gathered}\frac{-\dot{Q}_{23}}{\dot{M}}=\hat{H}_{3}-\hat{H}_{2}=(407.1-271.9) \frac{ kJ }{ kg }=129.8 \frac{ kJ }{ kg } \\\frac{\dot{W}_{34}}{\dot{M}}=\hat{H}_{4}-\hat{H}_{3}=(432-401.7) \frac{ kJ }{ kg }=30.3 \frac{ kJ }{ kg }\end{gathered}
and
\text { C.O.P. }=\frac{129.8}{30.3}=4.28
Cooling (summer operation)
| Point |
T (°C) |
P (MPa) |
Condition |
\hat{H}\left(\frac{ kJ }{ kg }\right) |
\hat{S}\left(\frac{ kJ }{ kg K }\right) |
| 1 |
35 |
0.8879 |
Saturated liquid |
\begin{matrix} 249.2 \\ \downarrow \end{matrix} |
| 2 |
5 |
\begin{matrix}0.3499 \\ \downarrow \end{matrix} |
Vapor-liquid mixture |
249.2 |
|
| 3 |
5 |
0.3499 |
Saturated vapor |
401.7 |
\begin{matrix}1.7252 \\ \downarrow \end{matrix} |
| 4 |
45 |
0.8879 |
Superheated vapor |
436 |
1.7252 |
Consequently, here
\begin{aligned}&\frac{\dot{Q}_{23}}{\dot{M}}=\hat{H}_{3}-\hat{H}_{2}=(401.7-249.2) \frac{ kJ }{ kg }=152.5 \frac{ kJ }{ kg } \\&\frac{\dot{W}_{34}}{\dot{M}}=\hat{H}_{4}-\hat{H}_{3}=(436-401.7) \frac{ kJ }{ kg }=34.3 \frac{ kJ }{ kg }\end{aligned}
and
\text { C.O.P. }=\frac{152.5}{34.3}=4.45
Note that in both the heating and cooling modes, the heat pump in this illustration has a lower coefficient of performance (and therefore lower efficiency) than a reverse Carnot cycle operating between the same temperatures. Finally, it should be mentioned that the thermodynamic properties listed above were obtained from a detailed thermodynamic properties table for HFC-134a, akin to the steam tables for water in Appendix A.III; values cannot be obtained from Fig. 3.3-4 to this level of accuracy.
As in Illustration 5.2-2 Aspen \text { Plus }{ }^{\circledR} with the Peng-Robinson equation of state will be used. To use the folder Aspen Illustration>Chapter 5>5.2-6 on the Wiley website for this book it is necessary to compute the vapor pressures of HFC-134a as predicted by equation of state at 5°C, 35◦C and 50°C. These are computed using Properties > Pure option in Aspen \text { Plus }{ }^{\circledR} which predicts the following
\begin{array}{cr}T\left({ }^{\circ} C \right) & P( MPa ) \\5 & 0.3484 \\35 & 0.8864 \\50 & 1.3205\end{array}
Winter
\begin{aligned}&\begin{array}{llclrr}\text { Point } & & T\left({ }^{\circ} C \right) & P( MPa ) & Q \text { (Watts) } & W \text { (Watts) } \\1 & \text { Condenser } & 50 & 1.3205 & -158276 & \\2 & \text { Valve } & 5 & 0.3484 & & \\3 & \text { Boiler } & 5 & 0.3484 & 130762 & \\4 & \text { Compressor } & 56.4 & 1.3205 & & 2796.4\end{array}\\&\end{aligned}
\text { C.O.P }=\frac{130762}{27964}=4.676
Summer
\begin{aligned}&\begin{array}{llcllrr}\text { Point } & & T\left({ }^{\circ} C \right) & P( MPa ) & Q(\text { Watts }) & W(\text { Watts }) \\1 & \text { Condenser } & 35 & 0.8864 & -174316 & \\2 & \text { Valve } & 5 & 0.3484 & & \\3 & \text { Boiler } & 5 & 0.3484 & 154692 & \\4 & \text { Compressor } & 37.3 & 0.8864 & & 19625\end{array}\\&\end{aligned}
