Question 14.2: Modified Vapor Compression Cycle Consider the same example a......

We start with the fact that the vapor at state 1´ is superheated by 6°C. Hence, for a temperature of (−14 + 6 =) −8°C and pressure of 0.1707 MPa, the vapor enthalpy h´_{1} = 395.24  kJ/kg. With the condition that the  process is isentropic from 1´ to 2´, i.e., s´_{1}  = s´_{2}, we can determine the enthalpy of point 2´.
We determine h´_{2} = 427.5  kJ/kg

We can deduce the enthalpy of the subcooled uid leaving the condenser (Equation 14.13)

(h_{3}  –  h´_{3}) = (h´_{1}  –  h_{1})       (14.13)

= 236.74  kJ/kg

Next, we compute the refrigerant mass flow from Equation 14.6 with h´_{4} = h´_{3}

= 0.0326  kg/s         (earlier it was 0.0336)

We rearrange Equation 14.14 to get the enthalpy of the vapor leaving the compressor:

\eta_{isen} = \frac{h´_{2}  –  h´_{1}}{h"_{2}  –  h´_{1}}          (14.14)

= 435.565  kJ/kg

This allows the compressor power input to be determined (from Equation 14.7)

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

= 1.473  kW           (earlier it was 1.05)

The COP of the modied VC cycle:

COP_{mod} = \frac{5.0}{1.473} = 3.40        (earlier it was 4.76)

From Equation 14.10, the kW/ton of the machine = 3.516/3.39 = 1.04 kW/ton (earlier it was 0.739) The Carnot efficiency remains unaltered and is the same as that of Example 14.1, i.e., COPCarnot = 5.886. The refrigerating efficiency from Equation 14.11 is finally determined:

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

Comments
Due to the isentropic efficiency of the compressor, the COP and the kW/ton have become worse than those for the standard VC cycle of Example 14.1. In fact, multiplying the COP found in Example 4.1 by \eta_{R} =0.8 (the value assumed in Example 14.2) yields a COP for the modified cycle of 3.81. The further difference in COP between 3.81 and 3.40 found earlier is because of the superheat effect.