PERFORMANCE TEST ON TWO-STAGE RECIPROCATING AIR COMPRESSOR

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

4

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.

6