Reciprocating Air compressor

1. RECIPROCATING COMPRESSOR
Presented by
Prof. PRABHA CHAND
Department of Mechanical Engineering
National Institute of Technology, Jamshedpur

2. 2
Contents
1. W
orkingCycleandp-vDiagram
2. Indicated Powerandwork
3. Mechanical Efficiency
4. Conditions for minimumwork
5. Isothermal Efficiency
6. Clearance V
olume
7. V
olumetric Efficiency
8. MultistageCompressor

3. 3
What iscompressors?
 Compressor is a mechanical device that increases the
pressureof gasbyreducingitsvolume.
 It usesmechanicalworktotakeanamountoffluidanddeliveritata
requiredpressure.
 Anefficientcompressorincreasespressurewithminimumwork
 The amountof fluid is limited by the volume of the compressor
cylinder whichisfixed.

4. 4
Applications
 Inagasturbineoraircraft/jetengine
 Inrefrigerationandair-conditioningsystem
 Incar
 Inworkshops
 Forpneumaticsystem
 Formedical purposes
 Etc…

5. 5
Typeof compressors

6. Reciprocatingcompressors
Reciprocatingcompressorsalsoknownaspistoncompressorbecauseituse
pistondrivenbyacrankshaft andoperatesinacyclicmanner
.
Theycanbedirect-orbelt-drivenbyelectricmotorsorinternalcombustion
engines,canbeeitherstationaryorportableandcanbesingleormulti-
staged.
Reciprocatingcompressorsaretypicallyusedwherehighcompression ratios
arerequiredperstagewithouthighflowrates,andtheprocessfluidis
relativelydry
.
Smallreciprocatingcompressorsrangefrom5to30hparecommonly
seeninautomotiveapplications.
Largerreciprocatingcompressorswellover1,000hp(750kW)are commonly
foundinlargeindustrialandpetroleumapplications.Discharge pressurescan
rangefromlowpressuretoveryhighpressure(>180MPa).Incertain
applications,multi-stagedouble-actingcompressorsaresaidto bethemost
efficient compressorsavailable,andaretypicallylarger,and morecostlythan
comparablerotaryunits.
6

7. 7
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/receiver,
13. Water drain,

8. Basiccomponentsof reciprocatingcompressor
Compressioncylinder– provide
confinement for theprocessgas
during compression.
Piston – moves through the cylinder in a
reciprocatingactiontocompressthegas.
Intakevalve– tolettheairtodrawninto
thecylinder.
Outlet valve – to let air out of the
cylinder. Crank shaft – connect to the
motorto providetherotarymotion.
Connecting rod – connect the crank and
thepiston– providethereciprocatingups
anddownmovement.
8

9. Workingprincipleofreciprocatingcompressor
9
Refrigerant vapor to air
Whatis:
- topdeadcenter,TDC
- bottomdeadcenter,BDC
- bore,d
- stroke,L
- sweptvolume,Vs
- inducedvolume, Vin

10. 10
Thermodynamicscycleandp-vdiagram
4-1 Induction/intakeprocess
Pistontravel fromTDCtoBDC.
Inductionvalveopens
Airisinducedintothecylinder
Volumeandmassincreases ↑
Pressureandtemperature is
constant duringthisprocessat P1, T1
1 -2 Compressionprocess
Piston travel fromTDCtoBDC
Inlet valvecloses
Pistoncompressesair, Vol ↓
Pressurerises until reachP2 at (2)
T
emperaturealsoincreases
2 -3 Deliveryprocess
Deliveryvalveopensat pt
(2) Highpressureair is
delivered
Note: Compressionprocessis
reversiblepolytropicand
followsthelawpVn =C
Thethermodynamicscycleofreciprocatingcompressorisalwaysshowgraphicallythrough
thep-vdiagram.

11. 1
1
WorkandIndicatedpower
p2Vb  p1Va 
n 1
n
n 1
W
1 a

p2Vb  p1va  (n 1) p2vb  (n 1) p1va
 p2Vb  p1Va
n 1

p2Vb  p V 
Thework doneonair for onecycle is
theareainthegraph(areaabcd)
in  area abcd
cycle
 area abef  area bc0e  area ad0f

2 1
Win
Since p1Va  mRT1 and p2Vb  mRT2
So, work input per cycleis given by
n 1
 mRT T 
mRT2  mRT1 
n 1
n
n
cycle
 where m N  m
and N  rpm
Indicated Power
1

 T 
IP 
n
m R T
n 1 2
0 f
e
d
P2
c b
a
V
p
P1

12. Summary

13. 13
Example 3.1
Singleactingcompressor withoutclearance
Asinglestagereciprocatingcompressoroperatesbyinducing1 m3/minof
airat1.013barand15ºCanddelivers it at 7 bar. Assumethecompression
processbeingpolytropicand the polytropicindexis1.35.
Calculate:
(a) massof airdeliveredperminute, and
(b) indicatedpower.

14. 14
Mechanical Efficiency, ηm
 Because there are movingmechanical parts in the compressor, it is likelythat
losseswill occurduetofriction.
 Therefore power required to drive the compressor is actually more higherthan
theindicatedpower
.
 Sothemechanical efficiencyof thecompressor isgivenby:
ηm
Power system
[Power required]
Compressor
[Indicated power]
>

15. 15
Example 3.2
Mechanical Efficiency
If the compressor of previous example is to be driven at 300
rev/min and is a single-acting, single-cylinder machine, calculate
the cylinder bore required, assumingastroke to bore ratio 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 transmissionis90%.

16. 16
 Weknowthat theworkdoneis equalto theareaunderthegraph.
 Smaller areameanslessworkandthebetter thecompressor.
 For reciprocating compressors, the pressure ratio is fixed, so the
height of p-vdiagramisfixed.
 Thevolumeof cylinder is alsofixed sotheline 4-1is fixed.
 Therefore thearearepresentingwork depends ontheindexfor n.
 For n=1,
pV=constant (isothermal)
 For n=,
pV =constant (isentropic)
 So,theprocesscanbepolytropic, isothermal orisentropic
Conditionsfor MinimumWork

17. 17
o pV
o pV
o pVn
=constant (isothermal)
=constant (isentropic)
=constant (polytropic)
o Fromhereit canbeseenthat the
isothermal processisthebest
becauseit requiresminimumwork.
o Soitisbestthatthegastemperatureis
constantthroughout the compression
cycle.
Conditionsfor MinimumWork

18. Conditionsfor MinimumWork
**Isothermalefficiencyindicatesisothermalworkcomparedtotheindicatedwork.

19. Air is compressedin asingle-stage reciprocating compressor from 1.013
barand15°Cto7bar.Calculatetheindicatedpower requiredforafree
airdeliveryof0.3m3/min,whenthecompressionprocessisasfollows:
(a) isentropic
(b) isothermal
(c) polytropic,withn=1.25
Whatisthedeliverytemperatureateachcase?
Ifthecompressorissingle-actingandhasastroketoboreratioof1.2/1
andistorunat1000rev/min,calculatethesizeofboreandstroke
required.
19
Question3.3
Best conditionfor minimumwork

20. 20
Exercise
Isothermal efficiency
Asingle-stagereciprocatingcompressorinduced1.23kg/minof air at a
pressureof1.023barandtemperature23°Canddeliversit at8.5bar.If
polytropicindexis1.3, determine
(a) Indicatedpower,
(b) Isothermal power, and
(c) Isothermal efficiency

21.  In actual compressors, piston does
not reach the top of wall of the
cylinder head.
 Instead, it reaches maximum stroke
(TDC) at a certain distance from the
cylinder head.
cylinder where piston does
 The remaining volume of the
not
travel through
is call the clearance volume VC.
 The volume where the piston does
travel through is called the swept
volume, VS.
 Purpose of Vc : to give freedom for
working parts and space for valve
operations
ClearanceVolume, Vc

22. (ii) with clearancevolume
(i) without clearancevolume
Cycledifferentbetweenwithandwithoutclearance
volume

23. 23

24.  In actual compressor, the piston cannot
expel all the gases during delivery at
point 3.
 So, there are some compressed gas
trapped and left in the cylinder.
cylinder head, this compressed
expands according to pVn=C until
 When piston moves away from the
gas
the
pressure falls to p1 (reaches point 4).
 At point 4 the inlet valve opens and gas is
drawn in.
 The volume drawn in from point 4 to 1 is
called as induced volume, Vin is smaller
than the swept volume because of the
expansion process of the compressed
gas.
C
V =Clearancevolume
S
V =Swept volume
Thep-Vdiagramfor compressorwith, Vc

25.  Because of the expansion of gas
remaining in the VC, induced volume
is reduced from swept volume VS to
(V1–V4) which is the effective volume
 Mass of air per unit time
m
1  m
2 and m
3  m
4
 Mass delivered per unit time = mass
induced per unit time
m m
2  m
3  m
1  m
4
TheEffect of ClearanceVolume, Vc

26. m
1  m
4  m mass induced per unit time
 

min

 

min

m m  N  kg 







1
1 
 
p
 p2 
2  1 
 1   1 
  
p
n 1
m
1  m
4 RT2  T1 
Indicated powerW 
For compressor with N cycles per unit time (N rpm)
Therefore
with
1 4
m m  m  N  kg 
1
2
n 1
or
n
n1
n
 mRT1
n 1
n
W mRT  T 
V
V1
and      ,
T p
 T2   p2   p2 n
n
n
n1 1
IndicatedPowerfor CompressorwithClearanceVolume

27.  The Free Air Delivery (FAD) is the volume of air drawn into a compressor
fromtheatmosphere.
 Itisastandardizedmeasureofthecapacityofanaircompressor.
 Normally, the pressure inside the compressor during the induction processis
slightlylowerthantheatmosphericpressureoutsideofthecompressor.
 Fore.g:
FreeAir Delivery(FAD)

28. FreeAir Delivery(FAD)

29.  Volumetric efficiency is another definition to measure the
performance of a compressor.
 It can be defined as the ratio of the actual delivered gas volume to
the swept volume of the cylinder or,
 Where,
 If the volumetric efficiency is 100%, it
means the compressor has no clearance
volume.
VolumetricEfficiency, ηv

30.  The volumetric efficiency also can be written as:
 or
pressure ratio and clearance ratio increase.
 This efficiency is made worse if leaks occur past the valves or piston.



 From the definition, the volumetric efficiency decreases as




  1
1
s  1 
v
P
V
Vc  P2 n
 1
VolumetricEfficiency, ηv

31.  A single-acting compressor completes one compression cycle with one
revolutionofthecrank
 Adouble-acting compressor completes two compression cycles with onerevolutionofthe
crank
 Sothemassinduceperrevolutionistwicethanasingleactingwhere
 
 

min


min
 a d
m 2  N  m kg  or m 2  N  m  m  kg 
Delivery Delivery
Induction
Induction
DoubleActingCompressors

32. Gas is compressed in a reciprocating compressor from 1 bar to 6
bar. The FAD is 0.013 m3/s. The clearance ratio is 0.05. The
expansion processofthecyclefollowsthelawofpV1.2=C.Thecrank
speedis360 rev/min. Calculate,
(a) theswept volume, and
(b) thevolumetricefficiency
Question3.4
FAD

33. FAD
Asingle-stage, single-acting air compressor running at 1000 rev/min delivers air at 25bar
.For this
purpose the induction and freeairconditions can be taken as1.013barand 15°C,and theF
ADas
0.25 m3/min. Theclearance volume is 3%of the swept volume and the stroke/bore ratio is 1.2/1.
Calculate:
(a) theboreand stroke
(b) thevolumetricefficiency
(c) theindicatedpower
(d) theisothermalefficiency
T
aketheindexofcompressionand re-expansionas1.3
(Answers: 73.2mm;87.8mm;67.7%
; 2kW;67.7)
Question3.5

34. Question3.6
FAD
Thecompressorof Question3.5hasactual induction conditions of
1 barand 40°C,andthedeliverypressureis25bar.Takingthe
boreas calculatedinQuestion3.5, calculatetheFADreferred
to1.013bar and 15°C,and theindicatedpower
required. Calculatealsothevolumetric efficiencyand
compareit withthat of Question3.5.
(Answers: 0.226m3/min; 1.98kW; 61.4%)

35. Whendeliverypressureisincreasedtoa
highervalue, several weaknesseswill
occur:
i. inducedvolumewill beless
ii. increaseindeliverytemperature
iii. decreaseinvolumetricefficiency(Vin
become less were else no change in
Vs)
Toovercomethosematter,multi-staging
compressorisintroduced.
MultistageCompressors

36. Pi,Ti Pi,T1 P2,T2
P1,T1
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.
MultistageCompressors

37. e-f :
a-b : PVn=Ccompression
b-e : Qfromairtosurrounding
TemperaturedropsfromTbtoTe.
IdeallyTe=Ta
PVn=Ccompression
Advantages:
i. Slight increaseintemperature ii.
iii. Increaseinvolumetricefficiency
Savinginwork (shadedarea)
NOTES:
 Sincenomassisallowtoescapeduringitstravel, mLPC =mHPC
 If pressureratio andtheratio of Vc/Vsis the same,volumetric efficiency for both
compressoristhesame.
a
b
e
f
g
c h
d
Vc
p
V
Vs
P2
Pi
P1
LP CPMPRESSOR
HP
CPMPRESSOR
MultistageCompressors

38. TheIdeal IntermediatePressure

39. IndicatedPowerfor z-StagesCompressor

40. A single acting air compressor runs at 700 rpm and compresses air in two-
stages to 80 bar from an induction pressure of 1 bar with an ideal
intermediate pressureandcompleteintercooling. Thefree air delivery is 0.06
m3/s at the pressure 1.013 bar and 20oC with the clearance volume is 3%of
the swept volume in each cylinder. The index of the compression and re-
expansion is 1.3 in both cylinders and the temperature at the end of the
induction stroke in each cylinder is 30oC. The mechanical efficiency of the
compressor is85%. Calculate:
i. theindicatedpower required,
ii. thesavinginpower over single-stagecompressionbetweenthesame
pressure,
iii. theswept volumeforeachstage, and
iv. therequiredpoweroutputofthedrivemotor.
Exercise3.7 MultistageandIntercooling

41. 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. Bothstages have the same pressureratio, and the law
of compressionandexpansionin both stages is PV1.3=C. The clearance volume
ofbothstagesare5%oftheirrespectivesweptvolumesand itrunsat 300rpm.If
intercoolingiscompleteandacoolingwatercoolstheairto32oC calculate:
i. indicatedpower,
ii. volumetricefficiency,
iii. swept volumefor thehighpressurecompressor, and
iv. themassflowrateofthecoolingwaterifwaterenterstheintercooler
at20oCandexitat46oC.
v. Alsoshowsthepowersavingonp-vdiagramandgetthevaluein
percentage.
Exercise3.8 MultistageandIntercooling