Air Conditioning Design

Running head: AIR CONDITIONING DESIGN
Air Conditioning Design
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1AIR CONDITIONING DESIGN
Task 1
1. (a)
The testing refrigerants are 134a, 236ea and 410A
Basic consideration of this calculation
PV=RT
Here, P is the pressure of the refrigerant, V is the volume of the refrigerant, T is the temperature
of the refrigerant, and R is the thermodynamic constant
In State 1 the Temperature of the refrigerant T[1]=170 Celsius
In State 2 the compressor efficiency is 85%
Within Step2 and Stap2 the refrigerant will achieve the high temperature TH = 40o Celsius
In State 3 the pressure drops about 30%
Therefore, P[3]= P[2]-30%
In State 4 enthalpy h[4] remains same as State 3
Therefore, h[4]=h[3]
Work process:

2AIR CONDITIONING DESIGN
Work done by compressor
W12=(p[2]*v[2]-p[1]*v[1])/(1-n12)
Work done by heat exchanger
W23=p[3]*v[3]-p[2]*v[2]
Work done by throttle valve
W34=(p[4]*v[4]-p[3]*v[3])/(1-n34)
Total work done by the system
W=W12+W23+W34
The work done by the compressor is: W_comp=h[2]-h[1]
Internal coefficient of performance (COP): COP=Q_evap/W_comp
The Internal model of the Air cooling system is shown bellow

3AIR CONDITIONING DESIGN
(b) i.
Entropy and enthalpy values of each of the process have been presented in the following table:
For R134a
In above the table each row represents each state of the refrigerator system.
Therefore,
Entropy of State 1 S1=0.9235 kJ/kg-K
Enthalpy of State 1 H1=260 kJ/kg
Entropy of State 2 S2=0.9235 kJ/kg-K
Enthalpy of State 2 H2=273.8 kJ/kg
Entropy of State 3 S3=0.3949 kJ/kg-K
Enthalpy of State 3 H3=108.3 kJ/kg
Entropy of State 4 S4=0.4006 kJ/kg-K