Question 1. A Refrigerator Uses Refrigerant-134a As The Working Fluid and Operates On An Ideal

Question 1.
A refrigerator uses refrigerant-134a as the working fluid and operates on an ideal

vapor-compression refrigeration cycle between 0.14 and 0.8 MPa. The mass flow rate of the
refrigerant is 0.02 kg/s. Assuming ideal vapour compression refrigeration cycle:
- Draw the p-h diagram, draw and name the different components of the cycle
- Find the COPR, the rate of heat removal from the refrigerated space, the rate of heat
rejection to the environment and the power input to the compressor.
- Find the COPR for a Carnot Cycle operating for the same Tmax and Tmin
- repeat the previous steps for a compressor isentropic efficiency of ηis,C = 75%. What do you
observe?
- calculate the entropy change at the throttle valve.
When applying energy conservation omit kinetic/potential energies.
Solution:
(i)
Isentropic/ideal case:
1: saturated vapour at T1= -18.77oC (p1=140kPa): h1 = 239.19kJ/kg, s1=sg=0.9447kJ/kgK
2: superheated vapour at p2=0.8MPa, s2 = s1 = 0. 9447kJ/kgK => h2=275.41kJ/kg, T2=39oC
3: saturated liquid at p3=0.8MPa => h3= 95.48kJ/kg, T3=31.31oC, s3=0.35408kJ/kgK
4: mixture h4=h3 at T4 = -18.77oC (p4=140kPa): hf= 27.06kJ/kg, hg= 239.19kJ/kg
sf= 0.1108kJ/kg, sg= 0.94467kJ/kg.
h4  h f 95.48  27.06
x4    32.2%
hg  h f 239.19  27.06
s4  xs g  1  x s f  0.322  0.94467  1  0.322  0.1108  0.3797kJ / kgK
Δsvalve = s4-s3=0.3797-0.35408=0.0256kJ/kgK
3 Condenser 2
Valve
Compressor
Evaporator
4
1
Real case - compressor ηis,C = 75%:
1: saturated vapour at T1= -18.77oC (p1=140kPa): h1 = 239.19kJ/kg, s1=sg=0.9447kJ/kgK
2,is: superheated vapour at p2=0.8MPa, s2,is = s1 = 0. 9447kJ/kgK => h2,is =275.41kJ/kg,
T2,is=39oC
h h
2: superheated vapour at p2=0.8MPa, h2= 2,is 1  h1  287.48kJ/kg, T2 ~50oC
is ,C
3: saturated liquid at p3=0.8MPa => h3= 95.48kJ/kg, T3=31.31oC, s3=0.35408kJ/kgK
4: mixture h4=h3 at T4 = -18.77oC (p4=140kPa): hf= 27.06kJ/kg, hg= 239.19kJ/kg
sf= 0.1108kJ/kg, sg= 0.94467kJ/kg.
h4  h f 95.48  27.06
x4    32.2%
hg  h f 239.19  27.06
s4  xs g  1  x s f  0.322  0.94467  1  0.322  0.1108  0.3797kJ / kgK
Δsvalve = s4-s3=0.3797-0.35408=0.0256kJ/kgK
3 Condenser 2
Valve
Compressor
Evaporator
4
1
(ii)
Ideal case:
h h 239.19  95.48
COPR  1 4   3.96
h2,is  h1 275.41  239.19
QL  m  h1  h4   0.02239.19  95.48  2.87kW
QH  m  h2,is  h3   0.02275.41  95.48  3.59kW
W  m h2,is  h1   0.02275.41  239.19  0.72kW
Real case:
h1  h4 239.19  95.48
COPR    2.97
h2  h1 287.48  239.19
 h1  h4   0.02239.19  95.48  2.87kW
QL  m
QH  m h2  h3   0.02287.48  95.48  3.84kW
W m h2  h1   0.02287.48  239.19  0.96kW
(iii)
1 1
COPR ,Carnot    4.4
TH 1 273.15  39 1
TL 273.15  18.77
1 1
COPR ,Carnot    3.69
TH 273.15  50 1
1 273.15  18.77
TL
Question 2. Air enters the compressor of an ideal-gas refrigeration cycle at 250 K and 100
kPa and the turbine at 300 K and 400 kPa. The mass flow rate of air through the cycle is 8.0
kg/s. Assuming constant specific heats for air and isentropic compression and expansion,
determine:
(i) the rate of refrigeration,
(ii) the net power required to run the refrigerator,
Note: cp = 1.005 kJ/kg.K, k = cp/cv = 1.4.
The cycle operates between two pressure levels (reverse Brayton cycle), 100kPa and 400kPa
(rp=4).
1→2: isentropic compression
2→3: heat rejection
3→4: isentropic expansion
4→1: heat absorpion
k 1
T2  p2  k 0.4
  4 1.4
 1.486  T2  250  1.486  371K
T1  p1 
 c p T3  T2   8  1.005  371  300  574.8kW
QH  m
k 1 0.4
T4  p4  k
 1  1.4
      0.673  T4  300  0.673  201.9 K
T3  p3  4
 c p T1  T4   8  1.005  250  201.9  386.8kW

QL  m
W  QH  QL  188kW
Useful formulae
Isentropic efficiency
Compressor
h h
is ,C  2,is 1
h2  h1
Isentropic ideal gas compression
Approximate way
k 1
T2  p2  k
 
T1  p1 
Energy conservation in flowing systems
Steady state forms

 1   1 
Qin  Win   m h  Vel 2  gz   Qout  Wout   m h  Vel 2  gz  in J
in  2  out  2 
 1   1 
Qin  Win   m   h  Vel 2  gz   Q out  Wout   m   h  Vel 2  gz  in J/s
in  2  out  2 
 1   1 
qin  win    h  Vel 2  gz   qout  wout    h  Vel 2  gz  in J/kg
in  2  out  2 
COPR
General formula:
Q kW  QL kJ  qL kJ / kg
COPR  L  
W kW  W kJ  wkJ / kg
Carnot cycle:
1
COPR 
TL / TH  1

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Question 1.

A refrigerator uses refrigerant-134a as the working fluid and operates on an ideal

 c p T1  T4   8  1.005  250  201.9  386.8kW

Isentropic ideal gas compression

Energy conservation in flowing systems

Steady state forms

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