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Compressor Design
General Equations
If the suction pressure and the discharge pressures of the
compressor are Ps and Pd respectively,then the specific work done
by the compressor can be found using:
w=Ps v s(k
k1)((
PdPs
)( kk1 )1)
Here k is the index of compression. For the refrigerant R134a,
it can be approximated to 1.15 undernormal operating conditions.
However, if more accuracy is required, k can be found from
theequation:
k=ln(
PsPd
)
ln(vdvs)
Based on the mass flow rate of the refrigerant, the compressor
power consumption and the volumetric flow rate of refrigerant
through the compressor can be found out sing:
P=mwV r=mv S
Once we find the value of V r , the actual swept volume of the
compressor can be found, if we knowthe volumetric efficiency of the
compressor. The actual swept volume of the compressor is:
V s=V rv
Assume a suitable stroke to bore ratio, based on the following
guidelines: Vacuum pumps and high speed compressors:
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Calculations and Results
The pressure drops at the suction and discharge valves have been
assumed to be 0.2 bar and 0.4 bar respectively. Based on the
properties of refrigerant at the exit of evaporator ( T e=8
o C ) and at the entry of condenser ( T c=40
o C ), the following following pressures have been
found:P1=349.9kPa and P2=1017 kPa
Hence the suction and the discharge pressures
are:Ps=P120kPa=329.9kPa and Pd=P2+40 kPa=1057 kPa
Further, the specific volumes at these states are:v
s=0.06327m
3/kg and vd=0.02032 m3/kg
Based on these properties of refrigerant, the index of
compression can be found out to be:
k=ln (
PsPd
)
ln(vdvs)=
ln( 329.91057
)
ln ( 0.020320.06327
)=1.025
The specific work of the compressor is:
-
w=Ps v s(k
k1)((
PdPs
)( kk1 )1)=329.9kPa0.02032m3 /kg( 1.025
1.0251)(( 1057
329.9)( 1.0251.0251
)1)=24.65kJ /kg
The actual power consumed by the compressor and the volumetric
flow rate of the refrigerant are:P=mw=0.02361 kg/ s24.65kJ
/kg=0.5821kW
V r=mv s=0.02361 kg /s0.06327 m3/kg=0.001494 m3/s
The compression ratio of the compressor is:
rC=PdP s
= 1057329.9
=3.204
For the above compression ration, the volumetric efficiency is
estimated to be v=0.795 . Hence theswept volumetric flow rate of
the refrigerant is:
V s=V rv =
0.0014940.795
=0.001879 m3/s
The refrigerant chosen here is R134a, which is a fluorocarbon
refrigerant. Hence the stroke-to-boreratio, is assumed to be 0.8.
The rotational speed of the compressor, N is taken to be
1000 rpm which equals 104.7 rad /s .
Hence the bore diameter of the compressor is:
D= 3 240V sN = 3 2400.001879 m3/ s0.8104.7 =0.1109mHence the
length of stroke is:
L=D=0.80.1109=0.08868 m
From the cycle analysis, the ideal work done by the compressor
is:Pideal=m(h3h2)=0.02361 kg /s (281.5256.6)kJ /kg=0.5863kW
Hence the efficiency of the compressor is:
= PPideal
=0.58210.5863
=0.9832
Critical parameters of the compressor:
Bore diameter 110 mm
Length of stroke 90mm
Power consumed 0.6kW
