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Examples_Ch 2_Compressor

S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines

Compressor

Examples

Applied Thermodynamics & Heat Engines

S.Y. B. Tech.

ME0223 SEM - IV

Production Engineering



Example 1A single-stage reciprocating compressor takes 1m3 of air per minute at 1.013 bar and 15 0C and delivers it at 7 bar. Assuming that the law of compression is PV1.35 = constant, and the clearance is negligible, calculate the indicated power.

Mass of air delivered per min. :

min/226.1)288()./287.0(

)1()/10013.1( 325

1

11 kgKKkgkJ

mmNX

TR

VPm

Delivery Temperature :

Kbar

barK

P

PTT

n

n

2.425013.1

7)288(

35.1

135.11

1

212

Indicated Work :

kWkJ

KKCkgkJkgTTRmn

nW

23.460

254min/254

)2882.475()./287.0(min)/226.1(35.1

135.1)(

112

….ANS


S. Y. B. Tech. Prod Engg.

Example 2If the compressor in last example is driven at 300 rpm and is a single acting, single-cylinder machine, calculate the cylinder bore required, assuming a stoke to bore ration of 1.5:1.Calculate the power of the motor required to drive the compressor if the mechanical efficiency of the compressor is 85 % and that of the motor transmission is 90%.

ME0223 SEM-IV Applied Thermodynamics & Heat Engines

Volume dealt with per min. at inlet = 1 m3/min

5.1

400333.0)5.1(

400333.0

4222

X

XDDDLD

Hence, Vol. drawn in per cycle = 1/300 = 0.00333 m3/cycle = Cylinder Volume

Thus, cylinder bore, D = 0.1414 m or 141.4 mm ….ANS

Power input to the compressor = kWkW

53.585.0

23.4 ….ANS



An air compressor takes in air at 1 bar and 20 ºC and compresses it according to law PV1.2

= C. It is then delivered to a receiver at a constant pressure of 10 bar. R = 0.287 kJ/kg.K. Determine :1.Temperature at the end of compression.2.Work done and heat transferred during compression per kg of air.

Example 3

P1

P2

V1V2

3 2

4 1

CVP 2.1

Delivery Temperature :

CKbar

barK

P

PTT

n

n

1574301

10)293(

2.1

12.11

1

212

Work done :

airofkgkJ

KCkgkJkg

P

PTRm

n

nW

n

n

/236

1

101)293()./287.0(min)/1(

2.1

12.1

11

2.1

12.1

1

1

21

….ANS



Example 3….contd

P1

P2

V1V2

3 2

4 1

CVP 2.1

Heat Transferred :

airofkgkJ

K

n

RCTT

n

TTRTTC

n

VPVPTTC

WUQ

V

V

V

/)23.98(

12.1

287.0718.0)293430(

1)(

1

)()(

1)(

12

1212

112212

….ANS



Example 4Following data relate to a performance test of a single-acting 14 cm X 10 cm reciprocating compressor:Suction Pressure =1 bar Suction temperature = 200 CDischarge Pressure = 6 bar Discharge temperature = 180 0C Speed of compressor = 1200 rpm Shaft Power = 6.25 kWMass of air delivered =1.7 kg/min

Calculate the following:1. The actual volumetric efficiency. 4. The indicated power.2. The isothermal efficiency. 5. The mechanical efficiency.3. The overall isothermal efficiency.


P1

P2

V1V2

3 2

4 1

CVP n min/8473.1)1200()1.0()14.0(

4

4

32

2

mrpmXmm

NXLDVd

….for single-acting compressor

min/4295.1

101

)293()./287.0()7.1(

3

51

1

m

X

KKkgkJkg

P

TRmFAD



Example 4….contd


P1

P2

V1V2

3 2

4 1

CVP n

%38.77100min/8473.1

min/4295.1100

3

3

Xm

mX

V

FAD

dvol ….ANS

32.1

)1/6(ln

293/453ln1

1

)/(ln

/ln1

12

12

1

1

2

1

2

nnPP

TT

n

n

P

P

T

T n

n

Indicated Power :

kW

KCkgkJkg

P

PTRm

n

nW

n

n

346.5

1

61)293()./287.0(sec/

60

7.1

32.1

132.1

11

32.1

132.1

1

1

21

….ANS



Example 4….contd


P1

P2

V1V2

3 2

4 1

CVP n

Isothermal Power :

kW

bar

barKCkgkJkg

P

PTRmWiso

269.4

1

6ln)293()./287.0(sec/

60

7.1

ln1

21

%85.79100346.5

269.4100 X

kW

kWX

PowerIndicated

PowerIsothermaliso

….ANS

%5.8510025.6

364.5100 X

kW

kWX

PowerShaft

PowerIndicatedmech

….ANS

%3.6810025.6

269.4100)( X

kW

kWX

PowerShaft

PowerIsothermaloveralliso

….ANS



Example 5A single-stage double-acing air compressor is required to deliver 14 m3 of air per min measured at 1.013 bar and 15 OC. the delivery pressure is 7 bar and the speed 300 rpm. Take the clearance volume as 5 % of swept volume with the compression and expansion index of n = 1.3. Calculate:1. Swept volume of the cylinder 2. Indicated power 3. Delivery temperature.

Swept Vol. Vs = V1 – V3 = V1 – Vc = V1 – 0.05 Vs

P1

=1.013 bar

P2

=7 bar

V1V4

6 2

5 1

CVP 3.1

3

4

V3

Swept Volume, V1-V3=Vs

V3=Vc

= 0.05 Vs

288 K

Vol. induced per cycle = (V1 – V4 )

V1 = 1.05 Vs

lecycmsidesXrpm

mVV /0233.0

2300

min/14)( 3

3

41

423.4013.1

705.1

3.1

11

1

2

3

41

bar

bar

P

P

V

VandVV

n

S



Example 5….contd

V4 = 4.423 V3 = 4.423 X 1.05 Vs = 0.221 Vs (V1 – V4 ) = 1.05 Vs – 0.221 Vs = 0.0233 m3

33

0281.0)221.005.1(

0233.0m

mVS

….ANSSwept Volume,

Delivery Temperature : Kbar

barK

P

PTT

n

n

450013.1

7)288(

3.1

13.11

1

212

….ANS

Indicated Power :

kW

bar

bar

X

mmNX

P

PVVP

n

nW

n

n

56.57

0132.1

71

6010

min)/14()/10013.1(

3.1

13.1

11

3.1

13.1

3

325

1

1

2411

….ANS



Example 6A four-cylinder double-acting compressor is required to compress 30 m3 / min of air at 1 bar and 27 0C to a pressure of 16 bar. Determine the size of motor required and cylinder dimensions if the following data is given :Speed of compressor = 320 rpm. Clearance volume = 4 % Stroke : Bore ratio = 1.2 Mechanical Efficiency = 82 %Index = 1.32Assume no pressure change in suction valves and air gets heated by 12 0C during suction stroke.

Net Work done :

kW

bar

bar

X

mmNX

P

PVVP

n

nW

n

n

64.197

1

161

6010

min)/30()/101(

32.1

132.1

11

32.1

132.1

3

325

1

1

2411

….ANS

Motor Power : kWWmech

24182.0

64.19764.197

….ANS



Example 6….contd


Vol. Efficiency :

%6.68)39273(

)27273(.)1(

)1(

1

1604.004.01

.1

32.1

1

1

1

2

K

K

bar

barX

bar

bar

T

T

P

PX

P

Pkk

inside

atm

atm

insiden

vol

Swept Vol. per cylinder :

33

01708.0686.0

1.)320()2(

1.)4(

min)/30(m

rpmsidescylinders

m

)2.1(4

01708.04

22 DDLD

mmmXL

mmmD

6.315263.02.1

263263.0

….ANS

….ANS



Example 7 Air at 103 kPa and 27 0C is drawn in L.P. cylinder of a two-stage air compressor and is isentropically compressed to 700 kPa. The air is then cooled at constant pressure at 37 0C in an intercooler and is then again compressed isentropically to 4 MPa in the H.P. cylinder, and is deliverer at this pressure. Determine the power required to run the compressor if it has to deliver 30 m3 of air per hour measured at inlet conditions.

P1 = 103 kPa

P3 = 4 MPa

3

2

1

CVP

P2 = 700 kPa

L.P.

H.P.

Volume

Mass of air delivered per min. :

hrkgKKkgkJ

hrmmNX

TR

VPm /89.35

)300()./287.0(

)/30()/10103( 323

1

11

Temperature for compression process 1-2’ :

KkPa

kPaK

P

PTT 7.518

103

700)300(

4.1

14.11

1

21'2

Temperature for compression process 2-3 :

KkPa

MPaK

P

PTT 1.510

700

4)310(

4.1

14.11

2

323



Work required to run the compressor :

kW

KKKKCkgkJhrkg

TTTTRm

TTRmTTRmW

194.4

)3101.510()3007.518()./287.0()/89.35(4.1

14.1

)()(1

)()(1

231'2

231'2

….ANS

Example 7….contd



Example 8

P1 = 1 bar

P3 = 40 bar

3

2

1

CVP 3.1

P2 = 6 bar

L.P.

H.P.

Volume

A trial on a two-stage single-acting reciprocating compressor gave the following data :Free air delivered =6 m3/min Delivery Pr. = 40 barAtm. Pr. and Temp. = 1 bar 27 ºC Speed = 400 rpm Intermediate Pr. = 6 bar Temp. at inlet to 2nd stage = 27 ºC Law of Compression =PV1.3 = C Mechanical Efficiency = 80 %Stroke of L.P. = Stroke of H.P. = Diameter of L.P.

Assuming Vol. Efficiency of 100 %,

mmLD

XDXDRPMXLXDFAD

PLPL

PLPLPLPL

3.267

4004

64

....

..2....

2..

….ANS

Swept Vol. of H.P. cylinder = Vol. of air at 6 bar, 27 ºC

mmD

XXD

RPMXLXDm

PH

PH

PHPH

109

4002673.04

4min/1

..

2..

..2

..3

….ANS



Example 8….contd

Indicated Work :

kW

bar

bar

bar

bar

X

mmNX

ngIntercooliPerfectTTP

P

P

PVP

n

nW

n

n

n

n

42.58

6

40

1

62

6010

min)/6()/101(

3.1

13.1

......21

3.1

13.1

3.1

13.1

3

325

13

1

2

3

1

1

211

….ANS



A two-stage single-acting reciprocating air compressor takes in air at the rate of 0.2 m3/sec. The intake pressure and temperature are 0.1 MPa and 16 ºC. The air is compressed to a final pressure of 0.7 MPa. The intermediate pressure is ideal and intercooling is perfect. The compression index in both stages is 1.25 and the compressor runs at 600 rpm. Neglecting clearance, determine :1.Intermediate Pr. 2. Total Vol. of each cylinder.3. Power required to drive compressor 4. Rate of heat rejection in intercooler

Example 9

Intermediate Pr., P2 : ngIntercooliPerfectMPaPPP ....2646.0)7.0()1.0(312 ….ANS

Vol. of cylinder 1, Vs1 :3

1111 02.02.060

600

60mVXVV

NXV SSS ….ANS

Vol. of cylinder 2, Vs2 :

32

3

2

2

1122211

00756.0

2646.0

)02.0()1.0(

mV

MPa

mMPaV

P

VPXVVPVP

S

S

SSSS

….ANS



Power Required :

kW

MPa

MPammNXX

ngIntercooliPerfectP

PVP

n

nP

n

n

96.42

1.0

7.01

10

sec)/2,0()/101.0(

125.1

25.12

......11

2

25.1

125.1

3

326

2

1

1

311

….ANS

Example 9….contd

Mass of air handled : sec/241.0)289()./287.0(

sec)/2.0()/101.0( 326

1

11 kgKKkgkJ

mmNX

TR

VPm

Delivery Temperature : KMPa

MPaK

P

PTT

n

n

1.3511.0

2646.0)289(

25.1

125.11

1

212

Heat rejected in Intercooler :

kWKKKkgkJkg

TTCm P

4.152891.351)./005.1(sec)/241.0(12

….ANS



Example 10A two-stage air compressor with complete intercooling delivers air to the mains at a pressure of 30 bar, the suction conditions being 1 bar and 15 0C. if both cylinders have same stroke, find the ratio of cylinder diameter for the efficiency of compression to be maximum. Assume index of compression to be 1.3.

P1

P3 3

2

1

CVP 3.1

P2

L.P.

H.P.

Volume

2’5

6

42

1

..

..

2..

2..

2

1

4

4V

V

D

D

LD

LD

V

V

PH

PL

PH

PL

3.1/1

1

2

'2

13.1'22

3.111

3.1

P

P

V

VVPVPCVP

For max. efficiency : barPPP 48.5)30()1(312

7.348.5 3.1/1

'2

1 V

V



Example 10

DAm /4

Delivery Temperature : Kbar

barK

P

PTT

n

n

4.4261

48.5)288(

3.1

13.11

1

21'2

For Const. Pr. Process 2 – 2’ : 48.1290

4.426

2

'2

2

'2

'2

'2

2

2 K

K

T

T

V

V

T

V

T

V

Thus, we get : 476.548.17.32

'2

'2

1

2

1 XV

VX

V

V

V

V

Hence : 34.2476.52

1

..

.. V

V

D

D

PH

PL ….ANS

Se prod thermo_examples_compressor

Education

compressor examples

contd me0223 sem

stage air compressor

ans motor power

contd p

ans power input

singleacting compressor

volume mass of air