3. A refrigeration system utilizing refrigerant R134a as the working fluid is considered. The following engineering data are provided for your convenience: i. The system is operated to maintain the refrigerated space at -30°C ii. The inlet state of the compressor is 60 kPa and -34°C while gaining a net heat of 0.450 kW (Qin) from the surrounding iii. The exit temperature and pressure of the compressor are 65°C and 1.2 MPa iv. The cooling water enters and exits the condenser at 18°C and 26°C with a mass flow rate of 0.25kg/s v. The exit temperature of the condenser is 42°C vi. The h₁ = 230.04 kJ/kg, h₂ = 295.18 kJ/kg, h3 =111.28 = h4 vii. The water inlet and outlet enthalpy to cool the condenser are hw1,18°c = 75.47 kJ/kg and hw2,26°C 108.94 kJ/kg Determine (a) the actual enthalpy at state 2 with aid of interpolation, (b) the enthalpy at state 4, (c) the quality of the refrigerant at the evaporator inlet and the enthalpy at state 4, (d) the mass flow rate of the refrigerant, (e) the waste heat transferred from the refrigerant, the compressor power input, and the refrigeration load of the thermal system using an energy balance, and (f) the COP of the refrigerator. Calculation of enthalpy and internal energy at state 1 From table At P₁ = 60 kPa, T₁ = - 34°C h₁ =230 kJ/kg, u₁ = 211.100 kJ/kg Calculation of enthalpy and internal energy at state 2 From table At P2=1.2 MPa, T₂ = 65°C h₂ = 295.18 kJ/kg, v₂ = 0.019 m³/kg h₂ = u₂+Pv2 ⇒ 295.18 kJ/kg = u2+1200×0.019 kJ/kg = ➡u₂ 272.38 kJ/kg Calculation of enthalpy and internal energy at state 3 P3= 1.2 MPa, T3 = 42°C T₂
3. A refrigeration system utilizing refrigerant R134a as the working fluid is considered. The following engineering data are provided for your convenience: i. The system is operated to maintain the refrigerated space at -30°C ii. The inlet state of the compressor is 60 kPa and -34°C while gaining a net heat of 0.450 kW (Qin) from the surrounding iii. The exit temperature and pressure of the compressor are 65°C and 1.2 MPa iv. The cooling water enters and exits the condenser at 18°C and 26°C with a mass flow rate of 0.25kg/s v. The exit temperature of the condenser is 42°C vi. The h₁ = 230.04 kJ/kg, h₂ = 295.18 kJ/kg, h3 =111.28 = h4 vii. The water inlet and outlet enthalpy to cool the condenser are hw1,18°c = 75.47 kJ/kg and hw2,26°C 108.94 kJ/kg Determine (a) the actual enthalpy at state 2 with aid of interpolation, (b) the enthalpy at state 4, (c) the quality of the refrigerant at the evaporator inlet and the enthalpy at state 4, (d) the mass flow rate of the refrigerant, (e) the waste heat transferred from the refrigerant, the compressor power input, and the refrigeration load of the thermal system using an energy balance, and (f) the COP of the refrigerator. Calculation of enthalpy and internal energy at state 1 From table At P₁ = 60 kPa, T₁ = - 34°C h₁ =230 kJ/kg, u₁ = 211.100 kJ/kg Calculation of enthalpy and internal energy at state 2 From table At P2=1.2 MPa, T₂ = 65°C h₂ = 295.18 kJ/kg, v₂ = 0.019 m³/kg h₂ = u₂+Pv2 ⇒ 295.18 kJ/kg = u2+1200×0.019 kJ/kg = ➡u₂ 272.38 kJ/kg Calculation of enthalpy and internal energy at state 3 P3= 1.2 MPa, T3 = 42°C T₂
Refrigeration and Air Conditioning Technology (MindTap Course List)
8th Edition
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Chapter44: Geothermal Heat Pumps
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Problem 13RQ: If 14 gpm of water is flowing through an open-loop heat pump with a temperature difference of 7F,...
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Transcribed Image Text:3. A refrigeration system utilizing refrigerant R134a as the working fluid is considered. The following
engineering data are provided for your convenience:
i. The system is operated to maintain the refrigerated space at -30°C
ii. The inlet state of the compressor is 60 kPa and -34°C while gaining a net heat of 0.450 kW (Qin)
from the surrounding
iii. The exit temperature and pressure of the compressor are 65°C and 1.2 MPa
iv. The cooling water enters and exits the condenser at 18°C and 26°C with a mass flow rate of
0.25kg/s
v. The exit temperature of the condenser is 42°C
vi. The h₁ = 230.04 kJ/kg, h₂ = 295.18 kJ/kg, h3 =111.28 = h4
vii. The water inlet and outlet enthalpy to cool the condenser are hw1,18°c = 75.47 kJ/kg and
hw2,26°C 108.94 kJ/kg
Determine (a) the actual enthalpy at state 2 with aid of interpolation, (b) the enthalpy at state 4, (c) the
quality of the refrigerant at the evaporator inlet and the enthalpy at state 4, (d) the mass flow rate of the
refrigerant, (e) the waste heat transferred from the refrigerant, the compressor power input, and the
refrigeration load of the thermal system using an energy balance, and (f) the COP of the refrigerator.
Transcribed Image Text:Calculation of enthalpy and internal energy at state 1
From table
At P₁ = 60 kPa, T₁
= - 34°C
h₁ =230 kJ/kg, u₁ = 211.100 kJ/kg
Calculation of enthalpy and internal energy at state 2
From table
At P2=1.2 MPa, T₂ = 65°C
h₂ = 295.18 kJ/kg, v₂ = 0.019 m³/kg
h₂ = u₂+Pv2
⇒ 295.18 kJ/kg = u2+1200×0.019 kJ/kg
=
➡u₂ 272.38 kJ/kg
Calculation of enthalpy and internal energy at state 3
P3= 1.2 MPa, T3 = 42°C
T₂<T=46.31°C (subcooled liquid))
=
h3 111.28 kJ/kg, u3 = 110.2 kJ/kg
Calculation of enthalpy at state 4
T= -30°C
h = 12.65 kJ/kg, u₁ = 12.59 kJ/kg
hg=232.2 kJ/kg, ug = 213.1 kJ/kg
=
=
h4 h3 111.28 kJ/kg (throttling process)
Calculation of dryness fraction at state 4
h₁ = h₂+x4 (hg-h₂)
⇒ 111.28
12.65+x(232.2-12.65)
⇒ x4 = 0.449237
Calculation of internal energy at state 4
U₁₁ = U₁ + × 4 (ug-u₁)
= 12.59+0.449237 (213.1-12.59)
⇒u4=102.6665 kJ/kg
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