A gas refrigeration system using air as the working fluid
Housam Binous and Ahmed Bellagi
A gas refrigeration system using air as the working fluid
Housam Binous and Ahmed Bellagi
Housam Binous and Ahmed Bellagi
Problem 11-79 reproduced from Ref. 1
A gas refrigeration system using air as the working fluid has a pressure ratio of 5. Air enters the compressor at 0°C. The high-pressure air is cooled to 35°C by rejecting heat to the surroundings. The refrigerant leaves the turbine at −80°C and then it absorbs heat from the refrigerated space before entering the regenerator. The mass flow rate of air is 0.4 kg/s. Assuming isentropic efficiencies of 80 percent for the compressor and 85 percent for the turbine and using constant specific heats at room temperature
Part 1 : Let us determine (a) the effectiveness of the regenerator, (b) the rate of heat removal from the refrigerated space, and (c) the COP of the cycle.
Part 2 : Also, determine (d) the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle. Use the same compressor inlet temperature as given, the same turbine inlet temperature as calculated, and the same compressor and turbine efficiencies.
Reference:
[1] Çengel, Yunus A., Boles, Michael A., Kanoğlu, Mehmet, Thermodynamics: An Engineering Approach, Ninth Edition, NY: McGraw-Hill, 2019.
A gas refrigeration system using air as the working fluid has a pressure ratio of 5. Air enters the compressor at 0°C. The high-pressure air is cooled to 35°C by rejecting heat to the surroundings. The refrigerant leaves the turbine at −80°C and then it absorbs heat from the refrigerated space before entering the regenerator. The mass flow rate of air is 0.4 kg/s. Assuming isentropic efficiencies of 80 percent for the compressor and 85 percent for the turbine and using constant specific heats at room temperature
Part 1 : Let us determine (a) the effectiveness of the regenerator, (b) the rate of heat removal from the refrigerated space, and (c) the COP of the cycle.
Part 2 : Also, determine (d) the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle. Use the same compressor inlet temperature as given, the same turbine inlet temperature as calculated, and the same compressor and turbine efficiencies.
Reference:
[1] Çengel, Yunus A., Boles, Michael A., Kanoğlu, Mehmet, Thermodynamics: An Engineering Approach, Ninth Edition, NY: McGraw-Hill, 2019.
Assumptions :
1 - neglect kinetic and potential energy changes
2- air is an ideal gas with constant specific heat: Cp=1.005 kJ/kg*K
3- γ=1.4
4- steady-state operation
1 - neglect kinetic and potential energy changes
2- air is an ideal gas with constant specific heat: Cp=1.005 kJ/kg*K
3- γ=1.4
4- steady-state operation
Part 1
In[]:=
T1=273.15;
In[]:=
PRATIO=5;
In[]:=
γ=1.4;
In[]:=
T2s=T1PRATIO^((γ-1)/γ)(*isentropicrelation*)
Out[]=
432.62
In[]:=
COMPRESSOReff=0.8;
In[]:=
T2=.
In[]:=
T2=T2/.First@NSolve[COMPRESSOReff==(T2s-T1)/(T2-T1),T2](*usingidealgasassumption:ΔH=Cp*ΔT*)
Out[]=
472.488
In[]:=
INVPRATIO=1/5;
In[]:=
TURBINEeff=0.85;
In[]:=
T5=273.15-80
Out[]=
193.15
In[]:=
sol1=Flatten@NSolve[{T5s==T4INVPRATIO^((γ-1)/γ)(*isentropicrelation*),TURBINEeff==(T5-T4)/(T5s-T4)},{T4,T5s}](*usingidealgasassumption:ΔH=Cp*ΔT*)
Out[]=
{T4281.282,T5s177.597}
In[]:=
T4=T4/.sol1
Out[]=
281.282
In[]:=
T5s=T5s/.sol1
Out[]=
177.597
In[]:=
T3=273.15+35
Out[]=
308.15
In[]:=
(*energybalanceonthegenerator*)T6=T6/.First@NSolve[T4-T3==T6-T1,T6]
Out[]=
246.282
In[]:=
EffecitvenessRegenerator=(T3-T4)/(T3-T6)
Out[]=
0.434279
In[]:=
mdot=0.4;
In[]:=
Cp=1.005;
In[]:=
QDOTL=mdot*Cp*(T6-T5)(*inkW*)
Out[]=
21.3591
In[]:=
WDOTCOMP=mdot*Cp*(T2-T1)(*inkW*)
Out[]=
80.1338
In[]:=
WDOTTURB=mdot*Cp*(T4-T5)(*inkW*)
Out[]=
35.4291
In[]:=
WDOTNET=mdot*Cp*(T2-T1)-mdot*Cp*(T4-T5)(*inkW*)
Out[]=
44.7048
In[]:=
COP=QDOTL/WDOTNET(*coefficientofperformance*)
Out[]=
0.477781
Part 2
In[]:=
T1
Out[]=
273.15
In[]:=
T2
Out[]=
472.488
In[]:=
T3
Out[]=
308.15