MSc Renewable Energy: Air Conditioning Design Project - MP4709

This project delves into the design of air conditioning systems, focusing on thermodynamic principles and the analysis of different refrigerants (R134a, R236ea, and R410A). It includes calculations of entropy, enthalpy, and coefficient of performance (COP) for each refrigerant across various stages of the refrigeration cycle. The project further discusses the heat transfer and work done during evaporation, condensation, compression, and throttling processes. The analysis incorporates EES code for modeling the system and presents temperature-entropy and enthalpy-entropy diagrams. The conclusion emphasizes the importance of energy conservation and sustainable development, comparing the efficiency and environmental impact of the refrigerants, with a particular focus on R134a's lower energy consumption and Global Warming Potential (GWP).

Running head: AIR CONDITIONING DESIGN
Air Conditioning Design
Name of the Student
Name of the University
Author note

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

4AIR CONDITIONING DESIGN
Enthalpy of State 4 H4=108.3 kJ/kg
For R236ea
Entropy of State 1 S1=1.626 kJ/kg-K
Enthalpy of State 1 H1=381 kJ/kg
Entropy of State 2 S2=1.626 kJ/kg-K
Enthalpy of State 2 H2=392.7 kJ/kg
Entropy of State 3 S3=1.171 kJ/kg-K
Enthalpy of State 3 H3=250.2 kJ/kg
Entropy of State 4 S4=1.175 kJ/kg-K
Enthalpy of State 4 H4=250.2 kJ/kg

5AIR CONDITIONING DESIGN
For R410A
Entropy of State 1 S1=1.778 kJ/kg-K
Enthalpy of State 1 H1=425.4 kJ/kg
Entropy of State 2 S2=1.778 kJ/kg-K
Enthalpy of State 2 H2=441.1 kJ/kg
Entropy of State 3 S3=1.221 kJ/kg-K
Enthalpy of State 3 H3=266.2 kJ/kg
Entropy of State 4 S4=1.229 kJ/kg-K
Enthalpy of State 4 H4=266.2 kJ/kg
ii. The temperature entropy diagram
For R134a

6AIR CONDITIONING DESIGN
Entropy-Temperature diagram
Enthalpy-Entropy or Moiller diagram
For R236ea
Entropy-Temperature diagram

7AIR CONDITIONING DESIGN
Enthalpy-Entropy or Moiller diagram

8AIR CONDITIONING DESIGN
For R410A
Entropy-Temperature diagram
Enthalpy-Entropy or Moiller diagram

9AIR CONDITIONING DESIGN
(c)
Between the stage of evaporation and condensing, the heat is reabsorbed or released. On
the other hand, in the compressor and throttle valve no external heat is absorbed or released. In
these stage, only the work is done. For both the heat transfer and work can be calculated by the
changes in the enthalpy of the fluid.
Therefore, the heat absorbed by the evaporation process in the indoor section of the
system is Q_evap = Q41= h1-h4
The heat released by the condensing process at the outdoor section of the system is
Q_con = Q23= h3-h2
The work done by the compressor
W_comp=W12= h2-h1

10AIR CONDITIONING DESIGN
However, the efficiency of the compressor is 85%. To maintain proper heat cycle the
compressor will need 15% more work efficiency. Therefore, the actual work or energy required
by the compressor is W_comp=W12= (h2-h1)*(115/100)=(h2-h1)*1.15
The work done by the throttle valve is W_thr=W34= h4-h3
Refrigerant Stage 1 to 2 Stage 2 to 3 Stage 3 to 4 Stage 4 to 1
R134a Work done (13.83*1.15)=15.9 N/A 0 N/A
Heat transfer N/A 165.5 N/A 151.7
R236ea Work done (11.69*1.15)=13.44 N/A 0 N/A
Heat transfer N/A 142.5 N/A 130.8
R410A Work done (15.66*1.15)=18 N/A 0 N/A
Heat transfer N/A 174.9 N/A 159.8
(d)
COP results for R134a
COP results for R236ea

11AIR CONDITIONING DESIGN
COP results for R410A
e)
Let, the rejected thermal energy at the condenser be Qcon
mcp
dT
dt −As h ( T −T a )=> mcp
dT
dt −As h ( T −T a )=Qcon
¿>∫ mcp dT −∫ As h ( T −T a ) dt =∫ Qcon dt
¿> mcp T − As h ( T −T a )t=Qcon∗t
¿>Qcon∗t−As h ( T −T a ) t=mcp T
=> t= mcp T
(Qcon− As h ( T −Ta ) )
Therefore, the expression for required time to cool down the laboratory is
t= mcp T
(Qcon− As h ( T −Ta ) )