Question 14.1: Standard VC Cycle A refrigeration cycle uses refrigerant 134......

Part of the solution is embodied in the property values assembled in Table 14.2.
(i) The saturated pressure at a temperature T_{evap} = −14°C is the low-side pressure
equal to 0.1707 MPa (24.8 psia) and the high-side pressure at T_{cond} = 30°C is 0.7701 MPa (113.3 psia).
(ii) Next, we begin at point 1 where the enthalpy and entropy are rst found
from the property table.

(iii) We then use the fact that entropies at points 1 and 2 are equal. So, using the entropy value from point 1 and the known high-side pressure, properties at 2 can be found from the superheat Table A8. Alternatively, one could simply use the p–h diagram (Figure 12.7), and find the intersection of a constant entropy line drawn from point 1 and the constant pressure horizontal line; this is point 2.
(iv) Since point 3 represents saturated liquid at p_{2}, the properties at 3 can be read from the saturated-liquid column of the property table.
(v) Finally, the conditions at 4 can be found since the pressure and enthalpy are both known (i.e., h_{3} = h_{4}). (The entropy values at points 3 and 4 are not needed for the cycle calculation and need not be determined unless one is concerned about the irreversibility inherent in the throttling process 3–4.)
The compression ratio is given by Equation 14.12:

CR = (p_{cond}/p_{evap})            (14.12)

The refrigerant mass flow is (Equation 14.6)

\dot{m}_{r} = \dot{Q}_{cool}/(h_{1}  –  h_{4})             (14.6)

The compressor power input is (Equation 14.7)

\dot{W}_{comp} = \dot{m}_{r}  (h_{2}  –  h_{1})           (14.7)

The COP of the standard VC cycle is thus

The Carnot refrigeration cycle efficiency is (Equation 12.4)

COP_{Carnot,c} = \frac{Q_{l}}{W_{net}} = \frac{T_{l}}{T_{h}  –  T_{l}}            (12.4)

From which the refrigerating efficiency is (Equation 14.11)

\eta_{R} = (COP_{std}/COP_{Carnot})        (14.11)

Following Equation 14.10, the kW/ton of the machine = 3.516/4.76 = 0.739 kW/ton.

(\dot{W}/\dot{Q}_{cool}) = 3.516/COP           (14.10)
Comments
The careful reader will observe that the throttling process 3–4 represents an irreversible process in the ideal vapor compression cycle as indicated by the entropy increase. This irreversibility is forced on us by the difficulties of trying to build a wetmixture turbine that could survive in a refrigeration machine. However, if such a turbine could be made,* some of the power input to the compression process could be avoided since it would be provided by this hypothetical turbine.
Though using the p–h diagram is convenient to analyze refrigeration cycles, the state properties cannot be determined accurately. Even small differences can result in large changes in certain quantities, such as the COP. Hence, for any sort of accuracy, it is urged that points 1 and 3 be read off the saturated tables while state 2 be determined from the superheat tables (such as Tables A7 and A8 for R-22 and R-132a respectively). Alternatively, software programs (there are several available) such as the online HCB software accompanying this textbook should be used for determining thermodynamic properties to achieve the needed accuracy. You are encouraged to try the software for the problems in this chapter and for generally analyzing such vapor compression cycles and system

* A few small-scale chiller manufacturers have indeed developed such turbines.