1

CHAPTER 3RECIPROCATING COMPRESSORS

4.1 Introduction

4.2 Working Cycle & p-v Diagram

4.3 Indicated Power and Work

4.4 Conditions for Minimum Work4.5 Mechanical Efficiency

4.6 Isothermal Efficiency

4.7 Clearance Volume

4.8 Volumetric Efficiency

4.9 Multistage Compressor

2

4.1 INTRODUCTION

Compressors uses mechanical work to take an amount of fluid and deliver it at a required pressure

An efficient compressor increases pressure with minimum work The amount of fluid is limited by the volume of the compressor

cylinder which is fixed The reciprocating compressor operates in a cyclic manner The properties of the working fluid at inlet and outlet are average

values

3

Induction valve

Inlet

Delivery valve

Piston

Connecting rod

Crank case

Crank

Outlet

Schematic Layout

A compressor consist of: crank case encloses the

compression volume crank shaft rotates the crank piston moves through the cylinder

during each cycle crank and connecting rod connects

the crank with the piston spring loaded induction and delivery

valves cylinder where piston travels

The crank shaft is usually driven by an electric motor

Basic Components of a Reciprocating Compressor

4

1. Air intake, 2. Compressor pump, 3.Outlet, 4. Drive belt, 5. Motor, 6. Control switch, 7. Relief valve,

8. Pressure gauge, 9. Manifold, 10. Regulator, 11. Supply line, 12. Air tank, 13. Water drain,

5

4.2 WORKING CYCLE & THE p-V DIAGRAM

p2

0 fe

d

c b

a

v

p

p1

v2 v1

Delivery valve

Induction valve

(d – a): Induction (intake) Induction valve opens Air is induced into the cylinder Volume and mass increases Pressure and temperature is constant

during this process

(a – b): Compression Inlet valve closes Piston compresses air Pressure rises until P2 at (b) Temperature also increases

(b – c): Delivery Delivery valve opens High pressure air is delivered Pressure and temperature is constant

during this process

Compression process is reversible polytropic and follows the law pVn = C

6

4.3 WORK & INDICATED POWER The work done on air for one cycle is the area in the graph (area

abcd) Considering a polytropic process which follows the gas law PVn =

constant Work for polytropic process is given by:

gas a ofindex polytropic

11122

n

wheren

vpvpWin

P2

0 fe

d

c b

a

V

p

P1

7

Work input per cycle

ab

abab

abab

in

VpVpn

nn

vpnvpnvpVp

VpVpn

VpVp

cycleW

12

1212

1212

1

1)1()1(

1

ad0f area bc0e area abef area

abcd area

P2

0 fe

d

c b

a

V

p

P1

12

12in

2211

1

1W

cycleper input work So,

and

TTmRn

n

mRTmRTn

ncycle

mRTVpmRTVpSince ba

R.p.m

and where1

Power Indicated

12

N

mNm

TTRmn

nIP

8

EXAMPLE 4.1

A single stage reciprocating compressor operates by inducing 1m3/min of air at 1.013 bar and 15oC and delivers it at 7bar. Assume the compression process being polytropic and the polytropic index is 1.35. Calculate:

i. Mass of air delivered per minute

ii. Indicated power

SOLUTION

RTmVp i. Mass of air delivered per minute can be determine using

27315287.0

1100013.1

m

min23.1

kg

RTVp

m

so

9

ii. Indicated power can be determine using formula 121IP TTRm

nn

• Find T2 first using formula n

n

PP

TT

1

1

2

1

2

35.1135.1

2 013.17

288

T K4.475

• Indicated power; 2884.475287.06023.1

135.135.1

IP

kW25.4Power Indicated

7

0

2

1

V

P (bar)

1.013

10

4.4 CONDITIONS FOR MINIMUM WORK

We know that the work done is equal to the area under the graph The smaller the area the lesser the work and the better the

compressor

For reciprocating compressors, the pressure ratio is fixed, so the

height of p-v diagram is fixed The volume of cylinder is also fixed so the line d-a is fixed Therefore the area representing work depends the index n. For n = 1,

pV = constant (Isothermal) For n = ,

pV = constant (isentropic) So, the process can be polytropic, isothermal or isentropic

11

P2

0 fe

d

c b1

a

V

p

P1

b b2

pV = C

pVn = C

pV = C

v2 v1

pV =constant (isothermal)

pV =constant (isentropic) pVn =constant (polytropic)

From here it can be seen that the isothermal process is the best because it requires minimum work

So it is best that the gas temperature is constant throughout the compression cycle

12

ISOTHERMAL WORK

2

1

2

1

2

11

2

12

21

112

12

11

ln

etemperaturconstant the is T where

ln

From

lnln

process isothermalfor

ln

ad0f area - c0eb area efab area Work

1

11

pp

RTm powerIsothermal

pp

mRTW

mRTpV

pp

Vppp

VpW

VpVp

VpVppp

Vp

in

abin

ba

abb

P2

0 fe

d

c b1

a

V

p

P1

pV = C

13

4.6 ISOTHERMAL EFFICIENCY Isothermal efficiency indicates isothermal work compared to the indicated work.

EXAMPLE 4.2 WorkIndicatedWorkIsothermal

EfficiencyIsothermal isoth

,

A single stage reciprocating compressor induce 1.23kg/min of air at pressure 1.023 bar and temperature 23oC and delivers it at 8.5 bar. If its polytropic index is 1.3, determine:

i. Indicated powerii. Isothermal poweriii. Isothermal efficiency

14

SOLUTION

We know: kPabarPkg

m 3.102 @ 023.1,min

23.1 1

kPabarPKCT o 508 @ 5.8 and 296 @ 23 21

i. Indicated power can be determine using 121IP TTRm

nn

• Find T2 first using formula n

n

PP

TT

1

1

2

1

2

3.113.1

2 3.102850

293

T K6.477

• Indicated power; 2936.477287.06023.1

13.13.1

IP

kW7.4Power Indicated

15

ii. Isothermal power can be determine using

2

1lnWpp

RTmisothermal

3.102850

ln296287.06023.1

Wisothermal kW68.3

iii. Isothermal efficiency can be determine using power indicatedpower isothermal

isoth

7.468.3

IP isothermal

isoth

W %78 @ 78.0

16

4.5 MECHANICAL EFFICIENCY, ηm

Because there are moving mechanical parts in the compressor, it is likely that losses will occur due to friction

Therefore power required to drive the compressor is actually more higher than the indicated power

So there is need to measure the mechanical efficiency of the cycle Mechanical efficiency of the compressor is given by:

Power system[Power required]

Compressor[Indicated power]>

power requiredpower indicated

m

17

• If Indicated power IP = 4.5 kW and mechanical efficiency, m is 0.8 the shaft power would be:

kW.kW

625.580

5.4power Shaft

18

4.7 CLEARANCE VOLUME (VC)

In actual compressors, piston does not reach the top of wall of the cylinder.

Instead it reaches maximum stroke at a certain distance from the wall. The remaining volume of the cylinder where piston does not travel

through is call the clearance volume VC. The volume where the piston does travel through is called the swept

volume, VS. Purpose – to give freedom for working parts and space for valve

operations

19

Process After delivery at (c) (volume is

VC, pressure is p2 and temperature is T2). So, there are some gas left in the cylinder

When piston moves downward, this gas expands according to PVn = C until p1 at (d).

Then induction begins (d – a) Then gas is compressed

according to PVn = C Finally there is the delivery (b –

c)

VC VS

a

bc

d

e

f

p2

p1

PVn = C

PVn = C

P

v

VC = Clearance volume

VS = Swept volume

20

Because of the expansion of gas remaining in the VC, induced volume is reduced from swept volume VS to (Va – Vd) which is the effective volume

Mass or air per unit time

Mass delivered per unit time = mass induced per unit time

Effect of VC

dada VVVorVVV

dcba mmandmm

dacb mmmmm

VC VS

a

bc

d

e

f

p2

p1

PVn = C

PVn = C

P

v

21

INDICATED WORK & INDICATED POWER FOR COMPRESSOR WITH CLEARANCE VOLUME

12

1212

1

11 power Indicated

cefd area - abef areaabcd area W

cycleper done Work

TTRmmn

n

TTRmn

nTTRm

nn

W

cycle

da

da

min

min

111

1

1

2112

kgmmNmor

kgmNmwhere

pp

RTmn

nTTRm

nn

W

t timeed per unimass inducmmmbecause

da

nn

da

VC VS

a

bc

d

e

f

p2

p1

PVn = C

PVn = C

P

v

22

We see here that the work done per cycle and indicated power per unit mass is the same whether with or without clearance

23

Double-acting Compressors

A single-acting compressor completes one compression cycle with one revolution of the crank

A double-acting compressor completes two compression cycles with one revolution of the crank

So the mass induce per revolution is twice than a single acting where

Delivery Delivery

InductionInduction

min 2

min 2

kgmmNmor

kgmNm da

24

EXAMPLE 4.3 A single stage, double-acting compressor is required to deliver 8m3/min of

air measured at pressure of 1.013 bar and 15oC. Delivery pressure is 6 bar and crank speed is 300rpm. The clearance volume is 5% of swept volume and the compression index is 1.3. Calculate

i. Swept volume, VS

ii. Delivery temperature, T2

iii. Indicated power

25

SOLUTION

We know: rpmNbarPbarPKCT o 300 and 6 ;013.1 and 288 @ 15 211 Since it is double acting, per minute, it will

have 300 x 2 = 600 cycle that induces 8 m3. It means for one cycle it will induce;

30133.06008

mVV da

i. Swept volume can be determine using the information of the induced air volume per cycle

VC VS

a

bc

d

6

1.013

PV1.3 = C

PV1.3 = C

P

v

cas VVV

• From the diagram

sas VVV 05.0

sa VV 05.1 (1)

26

• From polytropic equation nc

nd VPVP 21

3.111

1

2

013.16

05.0

s

n

cd VPP

VV

sd VV 196.0 (2)

• Insert (1) and (2) in equation 30133.0 mVV da

30133.0196.005.1 mVs

litre 15.6or 0156.0 3mVs

ii. Delivery temperature, T2 can be determine usingn

n

PP

TT

1

1

2

1

2

3.113.1

2 013.16

288

T CK o161.6or 6.434

VC VS

a

bc

d

PV1.3 = C

PV1.3 = C

P

v

27

iii. Indicated power can be determine using 121IP TTRm

nn

• First, find mass induce per cycle

288287.0

0133.0100013.1

1

1

RT

VVPm da kg0163.0

• Since it is double acting, mNm 2 0163.03002 min

78.9kg

NOTE: we can straight away obtain using the value of mmin

83m

V

288287.0

8100013.1

1

1

RT

VPm

min8.9

kg

2882.434287.06078.9

13.13.1

1IP 12

TTRmn

n

kW64.29IP

28

4.5 VOLUMETRIC EFFICIENCY, ηv

Volumetric efficiency is another definition to measure the performance of a compressor.

The are two ways how to define volumetric efficiency: 1st definition:

The ratio of the actual induced mass (mactual) in the cylinder with ideal induced mass at free air condition (mideal). Free air condition is basically the ambient condition

1

1

RTVVP

m daactual

and

o

soideal RT

VPm

Where Po is the ambient pressure

To is the ambient temperature

29

So by first definition,

s

da

s

da

v VPRT

RTVVP

RTVP

RTVVP

0

0

1

1

0

0

1

1

1

0

0

1

TT

PP

VVV

s

dav

If assume , and 11 oo TTPP

s

dav V

VV

VC VS

a

bc

d

e

f

p2

p1

PVn = C

PVn = C

P

v

s

d

s

c

s

s

s

dcsv V

VVV

VV

VVVV

1111

c

d

s

c

c

d

s

cv V

VVV

VV

VV (1)

30

Since n

cn

d VPVP 21

n

c

d

n

c

d

PP

VV

PP

VV

1

1

2

1

2 therefore and

Insert the above equation to equation (1) and we get

11

1

1

2n

s

cv P

PVV

NOTE: The above equation is only true when Po=P1 and To=T1

31

2nd definition:

The ratio of the actual volume (Vactual) in the cylinder that is measured at free air condition with swept volume (Vs)

VOLUMETRIC EFFICIENCY, ηv

sV

conditionair freeat actualv

V

We know that actual mass induced is

1

1

RTVVP

m daactual

If we measure actual mass induced at free air condition, it will be

o

actualoactual RT

VPm

32

Combining the two mathematical definition, we get

1

1

RTVVP

RTVP da

o

actualo

1

0

0

1

TT

PP

VVV daactual (1)

sV

conditionair freeat actualv

V Insert equation (1) into

1

0

0

1

TT

PP

VsVV da

v

Note that the equation above is the same the one in the first definition.

33

FREE AIR DELIVERY (FAD)

The actual volume of air induced or delivered that is measured at free air temperature & pressure is called free air delivery (FAD).

Looking back at, FAD is 1

0

0

1

TT

PP

VVFADV daactual

Where Po is the ambient pressure

To is the ambient temperature

For a single acting compressor, if N rpm, FAD can be defined as

NTT

PP

VVFADV daactual 1

0

0

1

For a double acting compressor,

NTT

PP

VVFADV daactual 21

0

0

1

34

EXAMPLE 4.4 A single stage, single-acting compressor delivers 3m3/min of air measured

at pressure of 1.014bar and 23oC. During induction, pressure and temperature or air is 0.98 bar and 43oC respectively. Delivery pressure is 6.5 bar and crank speed is 358 rpm. The clearance volume is 5% of swept volume and the compression index is 1.3. Calculate

i. Indicated power

ii. Volumetric efficiency

35

SOLUTION

We know:01.4kPa1 @ 014.1 and 296 @ 23 00 barPKCT o

8kPa9 @ 98.0 and 163 @ 43 11 barPKCT o kPa 506 @ 5.6 and 358N ,min3 2

3barP rpmmFAD

VC VS

a

bc

d

6.5

0.98

PV1.3 = C

PV1.3 = C

P

v

i. Indicated power can be determine using

121IP TTRm

nn

We know:

o

o

RTFADP

m

296287.034.101

min

58.3kg

T2 can be determine usingn

n

PP

TT

1

1

2

1

2

3.113.1

2 98.05.6

316

T K489

36

121IP TTRm

nn

316489287.06058.3

13.13.1

kW84.12

ii. Volumetric efficiency can be determine usings

@ V

FADVactualv

We know: 1

1

RTVVP

m da

and min3,min

58.33mFAD

kgm

1

1

PTRm

VV da

98

316287.058.3 min31.3

3m

Since N = 358 rpm,358

31.3 da VV 300925.0 m

cas VVV

From the diagram

cas VVV 05.0

sa VV 05.1 (1)

VC VS

a

bc

d

P

v

37

• From polytropic equation nc

nd VPVP 21

n

cd PP

VV

1

1

2

sd VV 214.0 (2)

• Insert (1) and (2) in equation 300925.0 mVV da

sV214.005.1

litre 11or 011.0 3mVs

Since N = 358 rpm, 358011.0 sV min938.33m

sV @

FADVactual

v 3.9383

%76 @ 76.0

3.11

98.05.6

05.0

sV

300925.0 m VC VS

a

bc

d

P

v

P2

P1

38

4.6 MULTI-STAGING COMPRESSOR• When delivery pressure is increased to a

higher value, several weaknesses were found:

1. Induce volume will become lesser

2. Increase in delivery temperature

3. Decrease of volumetric efficiency (FAD becomes lesser were else no change in Vs)

VC VS

P

V

p1

p2

p3

p4

d d’ d”

b

b’

b”

c

c’

c”

a

• To overcome those matter, multi-staging compressor is introduced

39

Pi,Tb Pi,Ta P2,TcP1,Ta

Coolant in Coolant out

Intercooler

LP Compressor

HP Compressor

It consist of more than one compressor where the air passes through an intercooler before entering the next compressor.

The size of the next compressor is smaller to compromise Vs. In the intercooler, heat is transferred to the surrounding and temperature will

decreased. It will be brought back to its inlet temperature (before induction process). It is assumed that all compressors will have the same polytropic index.

40

a

be

fg

c h

d

Vc

p

VVs

P2

Pi

P1

LP CPMPRESSOR

HP CPMPRESSOR

a-b : PVn=C compression

b-e : Q from air to surrounding

Temperature drops from Tb to Te. Ideally Te=Ta

e-f : PVn=C compression

Advantages:

a. Slight increase in temperature

b. Increase in volumetric efficiency

c. Saving in work ( shaded area)

***NOTES:

a. Since no mass is allow to escape during its travel, mLP = mHP

b. If pressure ratio and the ratio of Vc/Vs is the same, volumetric efficiency for both compressor is the same.

41

EXAMPLE 4.5 In a single acting, two-stage reciprocating air compressor, 4.5 kg/min of air

is compressed from 1.013 bar and 15oC surrounding conditions through a pressure ratio of 9 to 1. Both stages have the same pressure ratio, and the law of compression and expansion in both stages is PV1.3=C. The clearence volume of both stages are 5% of their respective swept volumes and it runs at 300 rpm. If intercooling is complete, calculate:

i. Indicated power

ii. Volumetric efficiency

iii. Cylinder swept volumes required.

42

SOLUTION

We know:

a

be

fg

c h

d

Vc

p(bar)

VVs

P2

Pi

1.013

LP COMPRESSOR

HP COMPRESSOR

3.1 and 288300min

54 1 nKrpm,T, Nkg

.m

i. Indicated power can be determine using

LPHP IPIPIP

ei

i

TTPP

PP

PP

11

2

1

2 and , 9

9 2

11

2

1

2

PP

PP

PP

PP ii

i

39 1

PPi

11TTRm

nn

IP iLP

43

nn

ii

PP

TT

1

11

3.113.1

3288

iT K371

a

be

fg

c h

d

Vc

p(bar)

VVs

P2

Pi

1.013

LP COMPRESSOR

HP COMPRESSOR

11TTRm

nn

IP iLP

288371287.060

5.413.1

3.1

LPIP kW74.7

eHP TTRmn

nIP

21

nn

ie PP

TT

1

22

3.113.1

3288

iT K371

1 and TTe

eHP TTRmn

nIP

21

288371287.060

5.413.1

3.1

kW74.7

LPHP IPIPIP 274.7 kW48.15

44

a

be

fg

c h

d

Vc

p(bar)

VVs

P2

Pi

1.013

LP COMPRESSOR

HP COMPRESSOR

ii. Since pressure ratio for and the ratio of Vc:Vs is the same for both stages,

HPvLPv

We know that air is induced at free air condition, so

oo TTPP 11 and

Vs

VVTT

PP

VsVV dada

v

1

0

0

1

We know

Nm

cyclem

300

5.4 kg015.0

1

1

PTRm

VV da

100013.1288287.0015.0

301224.0 m

45

cas VVV

From the diagram

cas VVV 05.0

sa VV 05.1 (1)

a

be

fg

c h

d

Vc

p(bar)

VVs

P2

Pi

1.013

LP COMPRESSOR

HP COMPRESSOR

From polytropic equation nci

nd VPVP 1

ni

cd PP

VV

1

1

sd VV 1164.0 (2)

Insert (1) and (2) in equation 301224.0 mVV da

sV1164.005.1

litresmV LPs 13or 013.0 3

3.11

305.0 sV

301224.0 m

Vs

VV dav

013.001224.0

%94or 94.0

46

ii. We already calculated Vs for LP compressor. Since volumetric efficiency for both stages is the same

a

be

fg

c h

d

Vc

p(bar)

VVs

P2

Pi

1.013

LP COMPRESSOR

HP COMPRESSOR

94.0

VsVV he

v

We know barPPmm iHPLP 039.33 and 1

i

ehe P

TRmVV

100039.3288287.0015.0

300408.0 m

1 TTe

v

hes

VVV

94.0

00408.0 litresm 34.4or 00434.0 3

***NOTES:

Easier steps are shown in McConkey page 399-400

47

IDEAL INTERMEDIATE PRESSURE

The value chosen for the intermediate pressure pi influences the work to be done on the air and its distribution between the stages.

Minimum power happen when 0idP

Wd

11

11

1

2

1

11

nn

ie

nn

i

PP

RTmn

nPP

RTmn

nW

We know 1 TTe

a

be

fg

c h

d

Vc

p(bar)

VVs

P2

Pi

1.013

LP COMPRESSOR

HP COMPRESSOR

21

1

2

1

11

nn

i

nn

i

PP

PP

RTmn

nW

0121

21

nn

in

n

i

PPPdPWd

48

nn

in

n

PPP121

21

i

i

PP

PP 2

1

221 iPPP or (pressure ratio is the same for each stage)

The total minimum work can be written as

21

1

2

1

11

nn

i

nn

i

PP

PP

RTmn

nW

11

22

1

1

21

nn

PP

RTmn

nW

So for compressor with Z stages, total minimum work is

11

1

1

21

Znn

PP

RTmn

nZW

49

EXAMPLE 4.6 A three stage, single acting compressor running in an atmosphere at 1.013

bar and 15oC has an FAD of 2.83 m3/min. The induced pressure and temperature is 0.98 bar and 32oC respectively. The delivery pressure is 70 bar. Assuming complete intercooling, n =1.3 and that the machine is design for minimum work, calculate the indicated power required.

SOLUTION

0

0

RTFADP

m

27315287.0

83.2100013.1

min

47.3kg

11

1

1

21

Znn

PP

RTmn

nZW

198.0

70288287.0

6047.3

13.13.1

33.1313.1

kWW 2.24