Question 15.6: Construct the pressure–enthalpy diagram for a system using t......
Construct the pressure–enthalpy diagram for a system using the compressor of Fig. 15-7 with an evaporating temperature of 35 F and a condensing temperature of 115 F. Subcooling and superheat are as specified in Fig. 15-7. Neglect system pressure losses and piping heat losses and gains.
Chart 4 in Appendix E is used to obtain thermodynamic properties and to locate the isotherms for 35 F and 115 F, respectively, as shown in Fig. 15-18. The subcooling is 10 F; therefore, state 1 is determined by the 115 F and 105 F isotherms at the condenser pressure. The superheat is 20 F, and state 3 is located on the 55 F isotherm at the evaporator pressure. As a result of the throttling process, state 2 is located on the 35 F isotherm where i_{2} = i_{1}. Since the data of Fig. 15-7 are for a real compressor, the process from 3 to 4 is not isentropic and some other method will have to be used to locate state 4. This can be done by computing the heat transfer to the refrigerant in process 2–3 and the work done on the refrigerant in process 3–4:
\dot{q}_{23} = \dot{m} (i_{3} – i_{2})
and
\dot{m} = \frac{\dot{q}_{23}}{i_{3} – i_{2}}
From Chart 4, i_{1} = i_{2} = 41 Btu/lbm, i_{3} = 111.5 Btu/lbm, and \dot{q}_{23} = 130,000 Btu/lbm from Fig. 15-7. Then
\dot{m} = \frac{13,000}{111.5 – 41} = 1844 lbm/hr
The compressor power is 12.5 kW or 42,650 Btu/hr from Fig. 15-7. The heat rejected from the refrigerant in the condenser is then
\dot{q}_{41} = \dot{q}_{23} + \dot{w}_{34} = 13,000 + 42,700 = 172,700 Btu/hr
Also
\dot{q}_{41} = \dot{m} (i_{4} – i_{1})
and
i_{4} = i_{1} + \frac{\dot{q}_{41}}{\dot{m}} = 41 + \frac{172,700}{1844} = 135 Btu/lbm
The complete cycle is thus determined in Fig. 15-18. Note an increase in entropy for the compressor of about 0.016 Btu/(lbm-R).
