1

CHAPTER 3RECIPROCATING COMPRESSORS

4.1 Introduction4.2 Working Cycle & p-v Diagram4.3 Indicated Power and Work4.4 Conditions for Minimum Work4.5 Mechanical Efficiency4.6 Isothermal Efficiency

4.8 Volumetric Efficiency4.7 Clearance Volume

4.9 Multistage Compressor

2

4.1 INTRODUCTION

Compressors uses mechanical work to take an amount of fluid and deliver it at a required pressureAn efficient compressor increases pressure with minimum workThe amount of fluid is limited by the volume of the compressor cylinder which is fixedThe reciprocating compressor operates in a cyclic mannerThe properties of the working fluid at inlet and outlet are average values

3

Induction valve

Inlet

Delivery valve

Piston

Connecting rod

Crank

Crank case

Outlet

Schematic Layout

A compressor consist of:crank case encloses the compression volumecrank shaft rotates the crankpiston moves through the cylinder during each cyclecrank and connecting rod connects the crank with the pistonspring loaded induction and delivery valvescylinder 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 opensAir is induced into the cylinderVolume and mass increases Pressure and temperature is constant during this process

(a – b): CompressionInlet valve closesPiston compresses airPressure rises until P2 at (b)Temperature also increases

(b – c): DeliveryDelivery valve opensHigh pressure air is deliveredPressure and temperature is constant during this process

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

6

4.3 WORK & INDICATED POWERThe 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 = constantWork for polytropic process is given by:

gas a ofindex polytropic

11122

=

−−

=

nwhere

nvpvpWin

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 where

1

PowerIndicated

12

=×=

−−

=•

NmNm

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 minuteii. Indicated power

SOLUTION

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

( )( )27315287.0

1100013.1+×

××=m&

min23.1

kg=

RTVpm&

& = so

9

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

nn

−−

=•

• Find T2 first using formula nn

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 graphThe smaller the area the lesser the work and the better the compressorFor reciprocating compressors, the pressure ratio is fixed, so the height of p-v diagram is fixedThe volume of cylinder is also fixed so the line d-a is fixedTherefore the area representing work depends the index n.For n = 1,pV = constant (Isothermal)

For n = γ,pVγ = 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 = CpVn = C

pVγ = C

v2 v1

pV =constant (isothermal)pVγ =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

ad0farea-c0ebarea efab area Work

1

11

ppRTm powerIsothermal

ppmRTW

mRTpVppVp

ppVpW

VpVp

VpVpppVp

in

abin

ba

abb

&=

=

=

==

=

−+=

+=

P2

0 fe

d

c b1

a

V

p

P1

pV = C

13

4.6 ISOTHERMAL EFFICIENCYIsothermal efficiency indicates isothermal work compared to the indicated work.

EXAMPLE 4.2

WorkIndicatedWorkIsothermalEfficiencyIsothermal isoth

, =η

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

SOLUTIONWe know: kPabarPkgm 3.102 @ 023.1,

min23.1 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 nn

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

1lnWppRTmisothermal

&

⎟⎠⎞

⎜⎝⎛×××=

3.102850

ln296287.06023.1

Wisothermal& kW68.3=

iii. Isothermal efficiency can be determine using power indicatedpowerisothermal

=isothη

7.468.3

IP== isothermal

isothW&η %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 frictionTherefore power required to drive the compressor is actually more higher than the indicated powerSo there is need to measure the mechanical efficiency of the cycleMechanical efficiency of the compressor is given by:

Power system[Power required]

Compressor[Indicated power]>

power requiredpowerindicated

=mη

17

• If Indicated power IP = 4.5 kW and mechanical efficiency, ηm 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

ProcessAfter delivery at (c) (volume is VC, pressure is p2 and temperature is T2). So, there are some gas left in the cylinderWhen 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 = CFinally 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 Wcycleper done Work

TTRmmn

n

TTRmn

nTTRmn

nW

cycle

da

da

−−−

=

−−

−−−

=

==

&&

&&&

( )

( )

( ) ⎟⎠⎞

⎜⎝⎛−×=

⎟⎠⎞

⎜⎝⎛×=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=−−

=

=−−

min

min

111

1

1

2112

kgmmNmor

kgmNmwhere

ppRTm

nnTTRm

nnW

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 crankA double-acting compressor completes two compression cycles with one revolutionof the crankSo the mass induce per revolution is twice than a single acting where

Delivery Delivery

InductionInduction

[ ] ( )[ ] ⎟⎠⎞

⎜⎝⎛−××=⎟

⎠⎞

⎜⎝⎛××=

min 2

min 2

kgmmNmorkgmNm da&&

24

EXAMPLE 4.3A 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 VPPVV

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 m&min

83mV =&

( )288287.0

8100013.1

1

1

×××

==RT

VPm&

&min

8.9kg

=

( ) ( )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

RTVVPm da

actual−

= ando

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

×−

=

−

=η

( )1

0

0

1

TT

PP

VVV

s

dav ××

−=η

If assume , and 11 oo TTPP ==( )

s

dav V

VV −=η

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

−+=−+

=η

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−+= 1111

c

d

s

c

c

d

s

cv V

VVV

VV

VVη (1)

30

Since nc

nd 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

PVVη

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

sVconditionair freeat actual

vV

=η

We know that actual mass induced is( )

1

1

RTVVPm da

actual−

=

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

PPVVV daactual ××−= (1)

( )sV

conditionair freeat actualv

V=ηInsert equation (1) into

( )1

0

0

1

TT

PP

VsVV da

v ××−

=η

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

PPVVFADV 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

PPVVFADV daactual ×××−==

1

0

0

1&

For a double acting compressor,

( ) NTT

PPVVFADV daactual 2

1

0

0

1 ×××−==&

34

EXAMPLE 4.4A 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 powerii. 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

RTFADPm ×

=•

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 &

&=η

We know: ( )1

1

RTVVPm da&& −

=•

and min3,min

58.33mFADkgm ==&

1

1

PTRmVV da

××=−

&&&98

316287.058.3 ××= min31.3

3m=

Since N = 358 rpm,358

31.3=− da VV 300925.0 m=

cas VVV −=

cas VV

From the diagram

V 05.0−=

sa VV 05.1= (1)

VC VS

a

bc

d

P

v

37

• From polytropic equation nc

nd VPVP 21 =

n

cd PPVV

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 @

&

& FADVactualv =η 3.938

3= %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,TfP1,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 inlettemperature (before induction process).It is assumed that all compressors will have the same polytropicindex.

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 4.5In 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 clearencevolume of both stages are 5% of their respective swept volumes and it runs at 300 rpm. If intercooling is complete, calculate:

i. Indicated powerii. Volumetric efficiencyiii. 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&

( ) ( )LPHP IPIPIP +=

i. Indicated power can be determine using

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

nnIP 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

nnIP 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& ( )288371287.0

605.4

13.13.1

−×⎟⎠⎞

⎜⎝⎛×

−= 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 ηη =

We know that air is induced at free air condition, so oo TTPP == 11 and

( ) ( )Vs

VVTT

PP

VsVV dada

v−

=××−

=1

0

0

1η

We know

Nm

cyclem &

=300

5.4= kg015.0=

( )1

1

PTRmVV da

××=−

100013.1288287.0015.0

×××

= 301224.0 m=

45

cas VVV −=

cas VV

From the diagram

V 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 PPVV

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

−=η

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

vη

We know ( ) ( ) barPPmm iHPLP 039.33 and 1 =×==

( )i

ehe P

TRmVV ××=− 100039.3

288287.0015.0×

××=

300408.0 m=

1 TTe =

( )v

hes

VVVη−

=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 0=idP

Wd &

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

−−

11

11

1

2

1

11

nn

ie

nn

i

PPRTm

nn

PPRTm

nnW &&&

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

PPRTm

nnW &&

( )( )

0121

21 =⎟⎠⎞

⎜⎝⎛−=

−−nn

inn

i

PPPdPWd &

48

( )( )

⎟⎠⎞

⎜⎝⎛=

−−nn

inn

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

PPRTm

nnW &&

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

×=

−

11

22

1

1

21

nn

PPRTm

nnW &&

So for compressor with Z stages, total minimum work is

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

×=

−

11

1

1

21

Znn

PPRTm

nnZW &&

49

EXAMPLE 4.6A 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

RTFADPm ×

=&( )

( )27315287.083.2100013.1

+×××

=min

47.3kg

=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

×=

−

11

1

1

21

Znn

PPRTm

nnZW && ( )

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−⎟⎠⎞

⎜⎝⎛×⎟

⎠⎞

⎜⎝⎛

−×=

−

198.0

70288287.0

6047.3

13.13.1

33.1313.1

kWW 2.24=&