Air Conditioning Design

This document provides an in-depth analysis of air conditioning design, focusing on the thermodynamics and efficiency of different refrigerants. It discusses the working principles of air conditioning systems and provides solutions to common design problems. The document also highlights the importance of energy conservation and sustainable development in the field of air conditioning.

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 ) )

12AIR CONDITIONING DESIGN
Task 2
a) Introduction
Refrigeration systems have advanced over the past few decades owing to rapid
development in technology (7). Unlike the traditional traditions, today’s refrigeration systems
employees the use of electricity, magnetism and laser to facilitate their operation. It is a highly
developing industry with the primary role of keeping indoors households, offices and other
buildings comfortable and healthy. Its applications have also been extended to industrial freezers,
household refrigerators, air conditioning among others (1).
b) Methods
EES Code:
F$='R134a' or F$='R236ea' or F$='R410A'
//State1
x[1]=1
T[1]=17
h[1]=enthalpy(F$, x=x[1], T=T[1])
s[1]=entropy(F$, x=x[1], T=T[1])
P[1]=pressure(F$, x=x[1], T=T[1])
//State3
x[3]=0
T[3]=40
h[3]=enthalpy(F$, x=x[3], T=T[3])
s[3]=entropy(F$, x=x[3], T=T[3])
P[3]=(pressure(F$, x=x[3], T=T[3]))*(97/100)

13AIR CONDITIONING DESIGN
//State2
P[2]=pressure(F$, x=x[3], T=T[3])
s[2]=s[1]
h[2]=enthalpy(F$, P=P[2], s=s[2])
T[2]=temperature(F$, P=P[2], s=s[2])
PR=P[2]/P[1]
//State4
h[4]=h[3]
P[4]=P[1]
s[4]=entropy(F$, h=h[4], P=P[4])
T[4]=temperature(F$, h=h[4], P=P[4])
x[4]=quality(F$, h=h[4], P=P[4])
Q_evap=h[1]-h[4]
W_comp=h[2]-h[1]
COP_en=Q_evap/W_comp
Ex|Q_evap=(1-(T[1]+273)/(T[3]+273))*Q_evap
COP_ex=Ex|Q_evap/W_comp
In the above code the F$ function declaration denotes the use of refrigerant in that
system. In State 3 the pressure was lost 3% due to mechanical inefficiency. Therefore, the
pressure formulae has been coded as P[3]=(pressure(F$, x=x[3], T=T[3]))*(97/100)
c) Results
The relation between enthalpy and entropy:
As per the

14AIR CONDITIONING DESIGN
dS=dQrev/T
=> ∆entropy = ∆heat added or subtracted at constant temperature/ Temperature
As per the first law of thermodynamics
∆enthalpy = change in heat content of the body based on internal energy and flow work = ∆u+
∆(pv)
=> H=U+PV
=> dH=dU+PdV+VdP
Therefore,
dH = T dS + V dP;
Where dH=change in enthalpy, dS=change of entropy, T=temperature, dP=change Pressure, V=
volume.
Tabulated results of entropy and enthalpy:
Refrigerant Stage 1 Stage 2 Stage 3 Stage 4
R134a Entropy Si 0.9235 0.9235 0.3949 0.4006
Enthalpy hi 260 273.8 108.3 108.3
R236ea Entropy Si 1.626 1.626 1.171 1.175
Enthalpy hi 381 392.7 250.2 250.2
R410A Entropy Si 1.778 1.778 1.221 1.229
Enthalpy hi 425.4 441.1 266.2 266.2

15AIR CONDITIONING DESIGN
d) Discussion:
Clausius statement for the second thermodynamics law forms the fundamental basis for
the stable refrigeration cycle. On the other hand, the Rankine cycle is based on the condenser,
compressor, evaporator and expansion device required for a stable refrigeration cycle. Therefore,
for compressor additional 15% energy has been taken under consideration. As a result, the most
energy taken refrigerant R410A has shown its energy requirement of compressor as 174.9 kJ/kg
The target of this refrigeration systems should be energy conservation and sustainable
development. While considering the heat absorption and rejection R410A Refrigerant has been
found as most efficient. However, R410A has the heist Global Warming Potential, which is even
higher than Carbon dioxide. it can have several applications in air conditioning in households,
buildings and food preservations.. Generally, it uses renewable sources of energy like electricity
thus reducing the overdependence on fossils fuels such as coal. This has led to reduction in air
pollution that will also result to reduce global warming at long run.
e) Conclusion:
It can be concluded that, the refrigeration cycle consists of the condenser, compressor,
evaporator and expansion device that enables its stability for efficient operation. The study
showed that indeed efficiency with respect to other two Refrigerant R410A shows inefficient
heat release during the Stage 3 and in terms of computability with traditional R22 mechanism of
refrigerator R410a is more effective than other two refrigerants. However, R134a consumed
lower energy and has very low GWP. Therefore, It can be also concluded that the R134a
refrigeration systems has the ultimate technological benefits that will help promote energy saving
and sustainable growth.

16AIR CONDITIONING DESIGN
References and Bibliography:
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2. Moran M, Moran M, Shapiro H, Boettner D, Bailey M. Fundamentals of engineering
thermodynamics.
3. Pitzer K. Thermodynamics. New York: MacGraw-Hill; 1995.
4. Sweetser R. The fundamentals of natural gas cooling. Lilburn, GA: Fairmont Press, Inc.; 1996.
5. Davis T. The Technical Arts and Sciences of the Ancients (Neuberger, Albert). Journal of
Chemical Education. 1933;10(5):319.
6. Macchi E, Astolfi M. Organic Rankine cycle power systems. Duxford, UK: Woodhead
Publishing is an imprint of Elsevier; 2016.
7. Annamalai K, Puri I, Jog M. Advanced thermodynamics engineering. Boca Raton, FL: CRC
Press; 2011.
8. Öhman H, Lundqvist P. Theory and method for analysis of low temperature driven power
cycles. Applied Thermal Engineering. 2012; 37:44-50.