January 16, 2015 Thermal Lab
Batch 1 - Roll Nos:01 to 06 PERFORMANCE TEST ON TWO-STAGE RECIPROCATING AIR COMPRESSOR Aishwary Mishra,Aman Chauhan,Ananthu M Aji,Ananti Saroj,Ankit Sachan,Ardhendu Barman B.Tech, Fourth semester Department of Aerospace Engineering Indian Institute of Space Science and Technology
ABSTRACT Objective of this experiment is to conduct performance test on two stage reciprocating air compressor and to plot graphs between Volumetric Efficiency Vs pressure ratio, Adiabatic Efficiency Vs pressure ratio and Isothermal Efficiency Vs pressure ratio at various delivery pressure.
V1 - Volume of air actually compressed Vo - Volume of air reduced to NTP Vs - Swept Volume Wact - Actual work done on the compressor Wadia - Adiabatic work done Wiso - Isothermal work done ηiso - Isothermal Efficiency ηadia - Adiabatic Efficiency ηvol - Volumetric Efficiency γw - Specific weight of water ρa - Density of air
NOMENCLATURE A - Area of cross section of LP cylinder a - Area of orifice Cd - Coefficient of discharge D - Diameter of LP Cylinder d2 - Diameter of the orifice Ha - Head causing flow in m of air Hw - Orifice meter manometer reading hw - Stabilization tank manometer reading K - Energy meter constant L - Torque Arm Length lb - Barometer reading in mm of Hg Nm - Speed of motor in rpm Nc - Speed of compressor crank in rpm Pa - Atmospheric pressure Pg - Delivery gauge pressure in kgf/cm2 Pr - Pressure ratio P0 - Pressure of air at NTP P1 - Absolute inlet pressure P2 - Delivery pressure R - Characteristic gas constant S - Spring balance reading T - Torque of the motor T0 - Temperature of air at NTP T1 - Temperature of suction in K T2 - Temperature at delivery t - Time for 20 impulses in sec
INTRODUCTION An air compressor is a device that converts power (usually from an electric motor, a diesel engine or a gasoline engine) into kinetic energy by compressing and pressurizing air, which, on command, can be released in quick bursts [1] The Air Compressor on which experiment was conducted, is a two stage, reciprocating type. Two stage compressor are those air compressor in which compression of air from initial pressure to final pressure is carried out in two cylinders. The air is sucked from atmosphere and compressed in the first cylinder. The compressed air then passes through the air cooler into the second stage cylinder, where the air is further compressed. The air further goes to the air reservoir through safety valve, which operates the electrical switch, when the pressure exceeds the limit. The test unit consists of an air chamber containing an orifice plate, manometer, compressor, and an electrical dynamometer type induction motor. 1
THEORY A reciprocating compressor or piston compressor is a positive-displacement compressor that uses pistons driven by a crankshaft to deliver gases at high pressure. The intake gas enters the suction manifold, then flows into the compression cylinder where it gets compressed by a piston driven in a reciprocating motion via a crankshaft, and is then discharged. Applications include oil refineries, gas pipelines, chemical plants, natural gas processing plants and refrigeration plants. One specialty application is the blowing of plastic bottles made of polyethylene terephthalate. It is also used to start the auxillary and main engines of the ship. Since, the experiment involves measuring the volumetric , adiabatic and isothermal inefficiencies and outlet pressures and temperatures were measured. The inefficiencies are calculated as followsThe atmospheric pressure is calculated using the barometer reading,
Pa =
1.033 × 9.81 × 104 × lb N/m2 760
Density of air, ρa =
From Bernoulli’s equation we can easily see that the volume of air actually compressed, p V1 = Cd A 2gHa
Ha =
Hw ρw ρa
(6)
(7)
where, (1)
ρw =density of water=1000kg/m3 Hw =Orifice meter manometer reading in m of water Cd is the coefficient of discharge = 0.6 (Given) A=Area of the orifice =
πd22 4
Now, swept Volume LANc 3 m /s 60
Vs =
(2)
(8)
where, Nc =speed of compressor crank shaft in rpm
where γw =specific weight of water=9810 N/m3 hw =stabilization manometer reading in m of water
Although the volumetric efficiency could have been calculated as ratio of V1 and Vs it will not be useful to compare efficiency between different compressors operating at different temperatures and pressure thus we need to find the volume with respect to a standard temperature thus we use the NTP conditions. It is important to note that we could have chosen any reference temperature.
delivery inlet pressure, P2 = (Pg × 104 × 9.81)N/m2
(5)
where, R=characteristic gas constant=287.14 J/kgK T1 =Temperature at suction in K
Where, Ib =Barometer reading in mm of Hg. As the air enters the compressor it has a pressure less than the atmospheric pressure.The pressure is measured by using the manometer reading. Absolute Inlet Pressure, P1 = (Pa − γw hw )N/m2
P1 RT1
(3)
Vo =
where Pg =Delivery pressure gauge reading in kg f /cm2
P1V1 To 3 m /s Po T1
(9)
Vo × 100 Vs
(10)
where To = 273K, Po = 1.03 × 104 × 9.81Pa
Pressure ratio
Volumetric Efficiency, Pr =
P2 P1
η=
(4) 2
now to calculate adiabatic and isothermal work efficiency of the compressor we need to measure actual work done on the compressor this can be done in two ways, 1. using dynamometer- This involves measuring the torque exerted by the motor and multiplying it with the angular velocity. Then we have Wact =
2πNT watts 60
(11)
where, N=speed of the motor in rpm, T=Torque=load X arm length Figure 1.
Two Stage Reciprocating Air Compressor
2. using energy meter
Wact =
20 × 3600 × 1000 watts 20K
OBSERVATION TABLE AND SAMPLE CALCULATION Atmospheric pressure , (12) Pa =
where K=Energy meter constant=1600 impulses/kWh t=time for 20 impulses in seconds
therefore Pa = 99470.56 N/m2
we now calculate isothermal efficiency as isothermal work done, P2 Wiso = P1V1 ln P1
1.033 × 9.81 × 104 × 98157 N/m2 760
inlet pressure P1 = (99470.56 − 9810 × 0.214)N/m2
(13) P1 = 97371.22
ηiso =
Wiso Wact
(14)
delivery pressure, P2 = 99470.56 + 7 × 104 × 9.84N/m2
to calculate the adiabatic efficiency we calculate the adiabatic work,
γ−1 P1V1 P2 γ Wadia = − 1 watts 1−γ P1
P2 = 786170N/m2 (15) Head causing flow
Adiabatic efficiency , Ha = ηadia =
Wadia Wact
18.7 × 1000 m 1.1009
(16) Ha = 169.84 m 3
Sl
Gauge
Orifice side
Suction Side
Spring
Speed
Speed
Temp
Absolute
Absolute
Pressure
No.
Pressure
Manometer
Manometer
balance
of
of
at
inlet
delivery
ratio
Reading
Reading
Reading
motor
compressor
Delivery
pressure
pressure
Pg
h1
h2
Hw
h1
h2
hw
S
Nm
Nc
T2
P1
P2
kgf/cm2
cm
cm
cm
cm
cm
cm
Kg
rpm
rpm
K
Pa
Pa
1
2
0
21.7
21.7
35.8
60.2
24.4
5.2
1463
947.7
347
97077
295671
3.05
2
3
0.8
20.9
20.1
36.8
59.4
22.6
5.4
1452
941.7
356
97254
393771
4.05
3
4
1.1
20.6
19.5
37
59
22
5.7
1455
943.2
367
97312
491871
5.05
4
5
1.3
20.5
19.2
37.2
58.8
21.6
6.8
1447
937.5
376
97352
589971
6.06
5
6
1.4
20.3
18.9
37.3
58.7
21.4
7
1442
934
385
97371
688071
7.07
6
7
1.5
20.2
18.7
37.3
58.7
21.4
7.7
1436
931.4
394
97371
786171
8.07
Sl
Density
Head
Volume of
Volume
Volume
No
of
Causing
air actually
of air
of air
air
flow
compressed
at NTP
Swept
ρ
Torque
Pr
Volumetric
Time
Temp
Actual work
Isothermal
efficiency
for 20
at
done per
work done
impulse
inlet
second
per second
ηvol
t
T1
a
b
Wiso
%
sec
K
Watts
Watts
Watts
Ha
V1
Vo
Vs
kg/m3
m of air
m3 /sec
m3 /sec
m3 /sec
T
1
1.0941
198.3
0.00661
0.00561
0.00619
11.73
90.6
16.09
309
1796.6
2796.8
757.1
2
1.0961
183.4
0.00636
0.00541
0.00615
12.18
87.9
14.66
309
1851.7
3069.6
916.2
3
1.1003
177.2
0.00625
0.00534
0.00616
12.86
86.6
13.88
308
1958.6
3242.1
1046.3
4
1.1008
174.4
0.00620
0.00530
0.00613
15.34
86.5
13.38
308
2323.7
3363.2
1153.3
5
1.1010
171.7
0.00615
0.00526
0.00610
15.79
86.1
12.69
308
2383.8
3546.1
1246
6
1.1010
169.8
0.00612
0.00523
0.00609
17.37
85.9
12.06
308
2611.3
3731.3
1331.1
Volume of Air actually compressed,
swept Volume Vs = 6.08 × 10−3 m3 /s Volumetric Efficiency, Vo Vs
adiabatic work
Isothermal
Adiabatic
Volumetric
ratio
done per second
efficiency
efficiency
efficiency
Wadia
ηiso
ηadia
ηvol
Watts
%
%
%
3.04
-841.98
25.569
30.1
90.642
4.04
-1063.27
28.178
34.6
87.872
5.04
-1253.68
30.406
38.7
86.579
6.05
-1422.92
32.348
42.3
86.45
7.05
-1569.34
33.038
44.3
86.102
8.05
-1702.57
33.36
45.6
85.885
Pr
V1 = 6.11 × 10−3 m3 /s
η=
Pressure
× 100
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RESULT AND DISCUSSION The volumetric efficiency of the compressor was determined for varying Pressure ratio. Average Volumetric Efficiency of the compressor is 87.26
η = 85.885% 1. using Dynamometer-
Volumetric efficiency decreases with pressure ratio. Average Isothermal Efficiency of the compressor is 36.6
Wact
2π1436 × 7.7 × 0.23 watts = 60
Average Adiabatic Efficiency of the compressor is 33.7 Isothermal efficiency of the compressor was greater than the adiabatic efficiency.
Wact = 2609.22 W
Both isothermal and adiabatic efficiency was found to be increasing with pressure ratio.
2. using energy meter Wact =
20 3600 × × 1000 watts 1600 12.06
Wact = 3731.34 W isothermal work done, Wiso = 97371.22 × 6.11 × 10−3 ln
786170 watts 97371.22
Wiso = 1244..7889 W Isothermal Efficiency, ηiso = 33.36%
Figure 2.
Graph between Pressure ratio and Thermal,Adiabatic and Vol-
umetric Efficiency
Adiabatic work, 97371.22 × 6.11 × 10−3 Wadia = 1 − 1.4
"
786170 97371.22
1.4−1
#
1.4
−1 INFERENCE 1. Volumetric Efficiency refers to the ratio of the volume entering the conpressor to the volume swept.Now, as Pr increases, compressibility effects dominate and lesser volume enters inside because it faces greater pressure relative to entry pressure.Thus Volumetric efficieny must decrease as Pr increases. 2.Isothermal Efficiency is the ratio of the work done in the isothermal process to the actual work done. Actual work done consists of work done against dissipative forces plus work done
(17) Wadia = −1702.57 W Adiabatic efficiency , ηadia = 45.6 % 5
in the actual process taking place in the compressor. Initially when P2=P1 , Wisothermal=0, even though work done against dissipative forces is considerable.Now as Pr increases Wisothermal increases whereas the actual work done still has major contribution due to dissipative forces which remains nearly thre same.Thus, isothermal efficiency increases. Now, Pr increases further the dissipative forces become larger and larger as well as the actual process path in the compressor shows more deviation from the isothermal path.Thus, isothermal efficiency must decrease.But, it must decreae slowly, as the gap between the actual process path and isothermal process path doesn’t vary much. Similarly we can account for adiabatic efficiency.
REFERENCES [1] klenck,Thomas 30 July 2010,How it works:Air Compressor Popular Mechanics
Appendix A: Error Analysis We have the volumetric efficiency as
η=
Vo × 100 Vs
Now we can calculate the error in volumetric efficiency by,
η=
P1V1 To Po T1 LANc
Now these values can be used to calculate error for each variable and the overall error can be calculated by taking the root mean square of sum of each of the errors. Errors due to each factor has been calculated.
Nc
Hw
P1
T1
m.s.e
0.06131
0.133879
0.002925
0.009402
0.147579
0.059809
0.140105
0.002831
0.009114
0.152636
0.058835
0.142291
0.002787
0.009009
0.154264
0.059102
0.144292
0.002782
0.008995
0.156211
The average uncertainity in volumetric efficiency from this then comes out to be 0.154563.
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