- #1
sci0x
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- Homework Statement
- Brewery operates closed circuit vapour compression refrigeration system using ammonia as refrigerant. The process heat to be removed at the evaporator is 120 kW. The vapour enters the compressor at 300 kPa with 5 deg C superheat. It is compressed isentropically to pressure of 1100 kPa, condensed at constant pressure and sub-cooled by 3 deg C. Finally it is expanded at constant enthalpy to 300 kPa at which pressure evaporation occurs. Using the pressure-enthalpy diagram calculate:
Calculate the required mass flow rate of refrigerant
- Relevant Equations
- M = Q/Cp(T2-T1)
This is a past exam Q, I don't have the attached pressure enthalpy diagram.
Q = 120kW
Cp = Refrigerant enthalpy = I've looked this up, for ammonia enthalpy = 35.06 kJ / mol
T2 = 5 degrees C = (5 deg C) (1100 kPa/300 kPa) = 18.33 deg C +273.15 K = 291.48 K
T1 = 2 degrees C = (2 deg C)(1100 kPa/300 kPa) = 7.33 deg C + 273.15 = 280.48 K
Mass Flow = Q / Cp (T2-T1)
= 120 kW / (35.06 kj / mol)(291.48 K - 280.15 K)
= 0.30 kg /s
Am I on the right track here
The expected answer is 0.115 kg/s
Q = 120kW
Cp = Refrigerant enthalpy = I've looked this up, for ammonia enthalpy = 35.06 kJ / mol
T2 = 5 degrees C = (5 deg C) (1100 kPa/300 kPa) = 18.33 deg C +273.15 K = 291.48 K
T1 = 2 degrees C = (2 deg C)(1100 kPa/300 kPa) = 7.33 deg C + 273.15 = 280.48 K
Mass Flow = Q / Cp (T2-T1)
= 120 kW / (35.06 kj / mol)(291.48 K - 280.15 K)
= 0.30 kg /s
Am I on the right track here
The expected answer is 0.115 kg/s
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