Question

There is a vapor compression type refrigeration cycle using the refrigerant HFC 134 a. In the condenser, it is isostatically cooled, the condensation temperature is 50 ° C., and the condenser outlet is the compressed liquid at 45 ° C. In the evaporator, it is isothermally heated, the evaporation temperature is 10 ° C. and the outlet of the evaporator is heated steam at 15 ° C. When the expansion valve performs isenthalpic expansion, and the adiabatic efficiency of the compressor is 0.70, answer the following questions.

(a) Calculate the specific enthalpy of the compressor outlet temperature, specific enthalpy, condenser outlet and evaporator outlet.
(b) Obtain the cooling operation coefficient of this system.
(c) Obtain the circulation amount of the refrigerant necessary for obtaining the cooling effect of 3.0 kW in this system.

Answer:
(a) Compressor outlet: 68 ° C, 445 [kJ / kg], condenser outlet: 264 [kJ / kg]
Evaporator outlet: 409 [kJ / kg] (b) 4.03 (c) 0.021 [kg / s]

Answer 1

a)

Refrigerant = HFC 134a

Condensation temperature = 50 C.

Condensor outlet is the compressed liquid at 45 C.

From standard tables of R-134 a, the enthalpy at 45 C of condenser outlet = h₃ = 264 KJ/kg

It is said that the evaporator outlet temp is 15 C. Enthalpy at 15 C = 409 KJ/kg.

It is given that condensation temperature is 50 C.

Compressor input temperature = evaporator outlet temp = 15 C.

Isentropic efficiency = 0.7

Compressor outlet temp = 68 C = 445 KJ/kg.

b) COP=?

COP = desired effect / work input

Desited effect = h₁ - h₄

Where h₁ = Entahlphy at exit of evaporator = 409 KJ/kh.

h₄ = enthalphy at end of epnasion or starting of condensor = 264 KJ/kg.

Desired effect = h₁ - h₄ = 409-264 = 145 KJ/kg.

Work input = h₂ - h₁

h₁ = Entahlphy at exit of evaporator =enthalphy at inlet of compressor = 409 KJ/kh.

h₂ = enthalphy at exit of compressor = 445 KJ/kg.

Work input = h₂ - h₁ = 445 - 409 = 36 KJ/kg.

COP = desired effect / work input = 145/ 36 = 4.03.

c) Amount of refrigerant required to obtain the cooling effect of 3 KW.

Desired effect = h₁ - h₄

let mass flow rate be "m"

3 KW = m*(h₁ - h₄)

3 KW = m*(409-264)

mass flow rate = 3/(409-264) = 3/ (145) = 0.02068 kg / sec.