Respuesta :
Answer:
a. The work done by the compressor is 447.81 Kj/kg
b. The heat rejected from the condenser in kJ/kg is 187.3 kJ/kg
c. The heat absorbed by the evaporator in kJ/kg is 397.81 Kj/Kg
d. The coefficient of performance is 2.746
e. The refrigerating efficiency is 71.14%
Explanation:
According to the given data we would need first the conversion of temperaturte from C to K as follows:
Temperature at evaporator inlet= Te=-16+273=257 K
Temperatue at condenser exit=Te=48+273=321 K
Enthalpy at evaporator inlet of Te -16=i3=397.81 Kj/Kg
Enthalpy at evaporator exit of Te 48=i1=260.51 Kj/Kg
b. To calculate the the heat rejected from the condenser in kJ/kg we would need to calculate the Enthalpy at the compressor exit by using the compressor equation as follows:
w=i4-i3
W/M=i4-i3
i4=W/M + i3
i4=2.5/0.05 + 397.81
i4=447.81 Kj/kg
a. Enthalpy at the compressor exit=447.81 Kj/kg
Therefore, the heat rejected from the condenser in kJ/kg=i4-i1
the heat rejected from the condenser in kJ/kg=447.81-260.51
the heat rejected from the condenser in kJ/kg=187.3 kJ/kg
c. Temperature at evaporator inlet= Te=-16+273=257 K
The heat absorbed by the evaporator in kJ/kg is Enthalpy at evaporator inlet of Te -16=i3=397.81 Kj/Kg
d. To calculate the coefficient of performance we use the following formula:
coefficient of performance=Refrigerating effect/Energy input
coefficient of performance=137.3/50
coefficient of performance=2.746
the coefficient of performance is 2.746
e. The refrigerating efficiency = COP/COPc
COPc=Te/(Tc-Te)
COPc=255/(321-255)
COPc=3.86
refrigerating efficiency=2.746/3.86
refrigerating efficiency=0.7114=71.14%
