Arefrigerator uses refrigerant-134a as the working fluid and operates on the vapor-compression refrigeration cycle. the evaporator and condenser pressures are 200 kpa and 1400 kpa, respectively. the isentropic efficiency of the compressor is 88 percent. the refrigerant enters the compressor at a rate of 0.025 kg/s superheated by 10.1°c and leaves the condenser subcooled by 4.4°c. determine

(a) the rate of cooling provided by the evaporator, the power input, and the cop. determine
(b) the same parameters if the cycle operated on the ideal vapor-compression refrigeration cycle between the same pressure limits.

a. The COP of the cycle is 2.725 b. The COP of the cycle is 2.886

Explanation:

Working fluid = R-134a

Evaporator pressure P1 = P4 = 200 kPa

Condenser presser P2 = P3 = 1400 kPa

Isentropic efficiency of the compressor ηc = 0.88

Mass flow rate to compressor m = 0.025kg/s

Sub cooled temperature T3’ = 4.4 C

a.State 1

Obtain the saturation temperature at evaporator pressure. Since, the refrigerant enters the compressor in super heated state

Obtain the saturation temperature from the super heated refrigerant R-134a table at P1 = 200kPa and T(sat) = -10.1 C

Calculate the temperature at state 1. As the refrigerant super heated by 10.1 C when it leaves the evaporator.

T1 = (-10.1) + 10.1 = 0 C

Obtain the specific enthalpy and specific entropy at state 1 from the table at T1 = 0 C and P1 = 200 kPa, which is, h1 = 253.05 kJ/kg and s1 = 0.9698 kJ/kg.K

State 2

Obtain the ideal specific enthalpy and saturation temperature at state 1 from refrigerant R-134a table at P2 = 1400 kPa and s1 = s2 = 0.9698kj/kg.K

Using the interpolation

h(2s) = 285.47 + (0.09698 – 0.9389) (297.10 – 285.47)/(0.9733 – 0.9389)

h(2s) = 295.91 kJ/kg

T([email protected]) = 52.40 C

State 3 and State 4

Calculate the temperature at state 3

T3 = T([email protected]) – T3

= 52.40 – 4.4 = 48 C

Obtain the specific enthalpy from the saturated refrigerant R -134a temperature table at T3 = 48 C, which is, h3 = hf = 120.39 kJ/kg

Since state 3 to state 4 is the throttling process so enthalpy remains constant

h4 = h3 = 120.39 kJ/kg

Calculate the actual enthalpy at state 2. Consider the Isentropic efficacy of the compressor

ηc = (h(2s) – h1)/(h2 – h1)

0.88 = (295.91) – (253.05)/h2 – (253.05)

h2 = 301.75 kJ/kg

Calculate the cooling effect or the amount of heat removed in evaporator

Q(L) = m (h1 – h4)

= (0.0025) (253.05 – 120.39)

= 3.317 kW

Therefore, the rate of cooling provided by the evaporator is 3.317 kW

Calculate the power input

W(in) = m (h2 – h1)

= (0.025) (301.75 – 253.05)

= 1.217 kW

Therefore, the power input to the compressor is 1.21 kW

Calculate the Coefficient of Performance

COP = Q(L)/W(in)

= 3.317/1.217

= 2.725

Therefore, the COP of the cycle is 2.725

b.Ideal vapor compression refrigeration cycle

State 1

Since the refrigerant enters the compressor is superheated state. So, obtain the following properties from the superheated refrigerant R-134a at P1 = 200 kPa

X1 = 1, h1 = 244.46kJ/kg, s1 = 0.9377 kJ/kg.K

State 2

Obtain the following properties from the superheated R-134a table at P2 = 1400kPa, which is s1 = s2 = 0.9377kJ/kg.K

Using the interpolation

h2 = 276.12 + (0.9377 – 0.9105) (285.47 – 276.12)/(0.9389 – 0.9105)

= 285.08kJ/kg

State 3

From the saturated refrigerant R-134a, pressure table, at p3 = 1400kPa and x3 = 0

H3 = hg = 127.22 kJ/kg

Since state 3 to state 4 is the throttling process so enthalpy remains constant

H4 = h3 = 127.22 kJ/kg

(hg should be hf because in ideal case it is a should exist as a liquid in state 3)

Calculate the amount of heat removed in evaporator

Q(L) = m (h1 – h4)

= (0.025) (244.46 – 127.22)

= 2.931 kW

Therefore, the rate of cooling provided by the evaporator is 2.931 kW

Calculate the power input to the compressor

W(H) = m (h2 – h1)

= (0.025) (285.08 – 244.46)

= 1.016 kW

Therefore, the power input to the compressor is 1.016 kW

Calculate the COP of the ideal refrigeration cycle

COP = Q(L)/W(in)

= 2.931/1.016 = 2.886

Therefore, the COP of the cycle is 2.886

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