Question

A refrigerator uses refrigerant-134a as the working fluid and follows the ideal vapor-compression refrigeration cycle with a non isentropic compressor. The refrigerant enters the evaporator at 120 kPa with a quality of 34 percent and leaves the compressor at 70°C. The compressor consumes 600 W of power.

Determine:

(a) The mass flow rate of the refrigerant.
(b) The condenser pressure, and
(c) The COP of the refrigerator.

Answer 1

Step-by-step explanation:

The pressure at which the saturated liquid-vapor mixture enters the evaporator is, P1=120kPa

P1=120 kPa

Superheated vapor leaves the compressor at temperature

T2=70oC

T2=60oC

To calculate Quality of the refrigerant at the inlet of evaporator,

x=0.3

Power consumed by the compressor,

P=600W

Using the steam table of the refrigerant at P = 120 kPa, the specific enthalpy of the refrigerant at the inlet of evaporator,

h4=hf+xhfgh4=23.32kJ/kg+0.3∗215.54kJ/kgh4=87.982kJ/kg

h4=hf+xhfgh4=21.32 kJ/kg+0.3∗212.54 kJ/kgh4=85.082 kJ/kg

From steam table of the refrigerant at P = 120 kPa,

specific enthalpy of the saturated vapor refrigerant at the inlet of the compressor,

=h1=233.85kJ/kg

s1=0.9355 kJ/kg

For isentropic process, the specific entropy of the refrigerant at the outlet of the compressor S2=S1=0.92545kJ/kg

checking the superheated table of the refrigerant when T=70C

S2=0.9354kJ/kg

The specific enthalpy at the exit of the compressor

h2=284.17kJ/kg

pressure of the refrigerant at the exit

P2=1340.43kPa

a) The mass flow rate of the refrigerant,

P=˙m(h2−h1)600W=˙m(285.17kJ/kg−23.86kJ/kg)˙m=11.92kg/s

b) The condenser pressure:

Pressure of the condenser=P2=1340.43kPa

c) The COP of the refrigerator can be calculated using this formula

COP=h1−h4h1−h2COP=233.86kJ/kg−

85.082

kJ/kg(284.17 kJ/kg−233.86kJ/kg)COP=2.98