jntuk

1. ME 6404 – THERMAL ENGINEERING
UNIT – IV – Air Compressor
by
CH.PAVAN KUMAR
Assistant Professor
Dept. of Mechanical Engg
PRIYADARSHINI INSTITUTE OF TECHNOLOGY AND MANAGEMENT

2. Air Compressors
COMPRESSOR – A device which takes a definite quantity of fluid ( usually gas, and
most often air ) and deliver it at a required pressure.
Air Compressor – 1) Takes in atmospheric air,
2) Compresses it, and
3) Delivers it to a storage vessel ( i.e. Reservoir ).
Compression requires Work to be done on the gas,
Compressor must be driven by some sort of Prime Mover ( i.e. Engine )

3. Classification
Air Compressors
Reciprocating Rotary
Single – acting
Double - Acting
No. of Sides of Piston
in operation
No. of Stages
for Compression
Centrifugal
Single – stage
Multi - stage

4. Reciprocating Compressor - Working
2. Principle of Operation




Fig. shows single-acting piston actions in
the cylinder of a reciprocating compressor.
The piston is driven by a crank shaft via a
connecting rod.
At the top of the cylinder are a suction
valve and a discharge valve.
A reciprocating compressor usually has
two, three, four, or six cylinders in it.

5. Reciprocating Compressor - Working

6. Reciprocating Compressor – Equation for Work
Pressure
P1
P2
3 2’ 2 2”
4 1 (Polytropic)
PV n
C
PV  C
(Isothermal)
PV 
C
(Adiabatic)
V2 V1
Volume
Operations : 4 – 1 : Volume V1 of air aspirated into Compressor, at P1 and T1.
1 – 2 : Air compressed according to PVn = Const. from P1 to P2.
→ Temp increase from T1 to T2.
2 – 3 : Compressed air at P2 and V2 with temperature T2 is delivered.

7. Reciprocating Compressor – Equation for Work
During Compression, due to the excess temperature above surrounding, the air will
exchange the heat to the surrounding.
 Compression Index, n is always less than γ, the adiabatic index.
As Compressor is a work consuming device, every effort is desired to reduce the work.
Work done = Area under P-V curve
 1 – 2” : Adiabatic Compression = Max. Work.
 1 – 2 : Polytropic Compression
 1 – 2’ : Isothermal Compression = Min. Work.

8. Reciprocating Compressor – Equation for Work
Thus, comparison between the Isothermal Work and the Actual Work is important.
Isothermal Efficiency, ηiso =
Isothermal Work
Actual Work
Thus, more the Isothermal Efficiency, more the actual compression approaches to the
Isothermal Compression.
P1
P2
V1V2
3 2’ 2 2”
4
PV  C
(Isothermal)
PV 
C
(Adiabatic)
PV n
C
1(Polytropic)
Actual Work = Wact = Area 4-1-2-3-4
actW = Area (4-1) – Area (1-2) – Area (2-3)











1 1 2 2
2 2
n 1
PV  PV
 P1V1  P2V2  
n 1
PV  P1V1
 P1V1  P2V2  
 P2V2
n 1
P2V2  P1V1
 P1V1 

9. 

 
 


 
 



 1 1 
1 1
1 1 2 2
1 1 2 2
1
PV
P2V2 
PV 1
n1
n
 PV  PV 
n1

n
 PV  PV 
n1
W  1iso
Reciprocating Compressor – Equation for Work
P1
P2
V1V2
3 2’ 2 2”
4
PV  C
(Isothermal)
PV 
 C
(Adiabatic)
PV n
C
1(Polytropic) Now,
nn
P1
V1
V2
P1V1

1/n
 
 P2


 P2V2

 n
Wiso 
P  P 
1/ n

 P1  P2  
 P1V11 2  1 
 n1

10. 
 
 


 
n
n
Wiso 
P  P 
 1/ n

 P1  P1  
P  P 
1/ n

P1  P2  
 P1V11 2  2 
 n1
 P1V11 2  1 
 n1
Reciprocating Compressor – Equation for Work
P1
P2
3 2’ 2 2”
4
PV  C
(Isothermal)
PV 
 C
(Adiabatic)
PV n
C
1(Polytropic) 








  nn
Wiso  
n1

 P1
 P2
 P1V11
 n1
The solution of this equation is always negative.
This shows that Work is done ON the Compressor.






 


 


n
P
n
Wiso
n1

 1 
 P2 
1mRT 1
n 1
V2 V1
Delivery Temperature,
n  1
 n
 1 
 P 2
T 2  T 1 
P


11. Reciprocating Compressor – Equation for Work
P1
P2
2
5 1
6 3
4
V3 V4 V1
Effective Swept Volume, V1-V4
Swept Volume, V1-V3=Vs
Total Volume, V1
Clearance Volume,
V3=Vc
Clearance Volume :
Volume that remains inside the cylinder
after the piston reaches the end of its
inward stroke.
PV n
 C
Thus, Effective Stroke Volume = V1 – V4
actActual Work = W = Area 1-2-3-4
Wact = Area (5-1-2-6) – Area (5-4-3-6)

12. Reciprocating Compressor – Equation for Work










 
 





n
n
P
Pn
P
m1
 1 
2
1 4
m1

 1 
n   P2
1 1
 PV 1  
n 1
 Wact  PV 1  
n 1





 
act
P  P 
 P V
n
W  
P  P 
1/ n

1 2
2 1
41 1 V  1
 n 1






 



 
nn
act
P
 n
P
n 
W 
 4 
 P3
4 4
m1
 m1

 1 
 P2
1 1  PV 1
n1
PV 1
n1
P1
P2
1
2
5 1
n
PV  C
6 3
4
V3 V4 V
Effective Swept Volume, V1-V4
Swept Volume, V1-V3=Vs
Total Volume, V 1
Clearance Volume,
V3=Vc
But, P4 = P1 and P3 = P2

13. Reciprocating Compressor – Volumetric Efficiency
Volumetric Efficiency :
Ratio of free air delivered to the displacement of the compressor.
Ratio of Effective Swept Volume to Swept Volume.
Effective Swept Volume
Swept Volume
=
c
Vs
=
V1 – V4
V1 – V3
Clearance Volume
Swept Volume
V
Clearance Ratio =
Presence of Clearance Volume
 Volumetric Efficiency less than 1. ( 60 – 85 % )
P1
P2
1
2
5 1
PV n
 C
Volumetric Efficiency =
6 3
4
V3 V4 V
Effective Swept Volume, V1-V4
Swept Volume, V1-V3=Vs
Total Volume, V1
Clearance Volume,
V3=Vc
= γ ( 4 – 10 % )

14. Reciprocating Compressor – Volumetric Efficiency

↑ Pr. Ratio ↑ Effect of ClearanceVolume
…

.Clearance air expansion through greater volume before intake
Cylinder bore and stroke is fixed.
Effective Swept Volume (V1 – V4) ↓ with ↑ Pr. Ratio
↓ Volumetric Efficiency

1 31 3
3 4
31
1 3 3 4
1 3
vol

V3

V4
V1 V3  V3

V3
V1 V3 
1

V4

V3
V1 V3  V3

V3
V1 V3 
1
V V VV 

V

V
1
VV 

V V  V V 
V V

V1V4

P1
P2
1V4 V
2
5 1
6 3
4
V3
Effective Swept Volume,
V1-V4
Swept Volume, V1-V3=Vs
Total Volume, V1
Clearance Volume,
V3=Vc

15. Reciprocating Compressor – Volumetric Efficiency








 4 
1 3  4 
3  4 
 1
1 3  4 
1

1/ n

1/ n
vol
P3

P3  1
V V  P
V3
V3  1  
V V V
V3
 V3  1  1  
V V V
V3
 1   


P
 1
vol
  1 
vol
vol
P1
P2
V4 V1
2
5 1
6 3
4
V3
Effective Swept Volume,
V1-V4
Swept Volume, V1-V3=Vs
Total Volume, V 1
Clearance Volume,
V =V3 c

16. Reciprocating Compressor – Actual P-V Diagram
P1
P2
2
1
3
4
Valve Bounce
Intake Depression
Atmospheric Pressure
Receiver Pressure 1-2-3-4-1 : Theoretical P-V Diagram.
At 4, inlet valve does not open due to :
1. There must be a pressure difference across the valve to open.
2. Inlet valve inertia.
Pr. Drop continues till sufficient level
for valve to force its seat.
Some valve bounce is set (wavy line).
Eventually, the pressure sets down at a level lower
than atmospheric pressure. This negative pressure
difference is known as Intake Depression.
Similar situation appears at 2, i.e. at the start of the delivery.
Pressure rise, followed by valve bounce and then pressure settles at a level higher than
the delivery pressure level.
Air delivery to a tank / receiver, hence, generally known as Receiver Pressure.

17. Reciprocating Compressor – F.A.D.
t
t
T  150
C  288K
P  101.325KN / m2
Free Air Delivery (F.A.D.) : If the volume of the air compressor is reduced to atmospheric
temperature and pressure, this volume of air is called FAD (m3/min)
Delivered mass of air = intake mass of air
PtVt
 P1 V1  V4   P2 V2  V3 
Tt T1 T2
If clearance volume is neglected
PtVt

P1V1

P2V2
Tt T1 T2
Where

18. Reciprocating Compressor – Multistage
High Pressure required by Single – Stage :
 1. Requires heavy working parts.
2. Has to accommodate high pressure ratios.
3. Increased balancing problems.
4. High Torque fluctuations.
5. Requires heavy Flywheel installations.
This demands for MULTI – STAGING…!!

19. Reciprocating Compressor – Multistage
Intercooler :
Compressed air is cooled
between cylinders.
Series arrangement of cylinders, in which the compressed air from earlier cylinder
(i.e. discharge) becomes the intake air for the next cylinder (i.e. inlet).
L.P. = Low Pressure
I.P. = Intermediate
Pressure
H.P. = High Pressure
L.P.
Cylinder
I.P.
Cylinder
H.P.
Cylinder
Intercooler
Intercooler
Air Intake
Air Delivery

20. Reciprocating Compressor – Multistage
Intake Pr.
P1 or Ps
P3 or Pd
2
9 3 5
1
PV n
C
8
Delivery Pr. 6
Intermediate Pr. 7 4
P2 PVC
Without Intercooling
Perfect Intercooling
L.P.
H.P.
Overall Pr. Range : P1 – P3
Single – stage cycle : 8-1-5-6
Without Intercooling :
L.P. : 8-1-4-7
H.P. : 7-4-5-6
With Intercooling :
L.P. : 8-1-4-7
H.P. : 7-2-3-6
Volume
Perfect Intercooling : After initial compression in L.P. cylinder, air is cooled in the
Intercooler to its original temperature, before entering H.P. cylinder
i.e. T2 = T1 OR
Points 1 and 2 are on SAME Isothermal line.

21. Reciprocating Compressor – Multistage
Ideal Conditions for Multi – Stage Compressors :
A. Single – Stage Compressor :
2 PV C
3 5
4
1
PV n
C
8
7
6 9
L.P.
H.P.
Single – stage cycle : 8-1-5-6






 
 1 
1 1
n

P
P5n1
PV 1
n1
n
W 
Delivery Temperature,
n  1
 n
 
P 1 
 P 5
T 5  T 1

22. Reciprocating Compressor – Multistage
2 PV C
3 5
4
1
PV n
C
8
7
6 9
L.P.
H.P.
B. Two – Stage Compressor (Without Intercooling) :
Without Intercooling :
L.P. : 8-1-4-7
H.P. : 7-4-5-6







 




 
n
P
n
 P 
Pn
W 
n1

 4 
 P5  n
4 4
n1

 
 1 
4
1 1 

 P V 1 
n 1
P V 1 
n 1
Delivery Temperature also remains SAME.
Without Intercooling  This is SAME as that of Work done in Single – Stage.

23. Reciprocating Compressor – Multistage
2 PV C
3 5
4
1
PV n
C
8
7
6 9
L.P.
H.P.
C. Two – Stage Compressor (With Perfect Intercooling) :
With Intercooling :
L.P. : 8-1-4-7-8
H.P. : 7-2-3-6-7







 







P
n
P
n
W  P V
n1

 2 
 P3  n
2 2
n1

 
 1 
 P4  n
1 1
 P V 1 
n 1
1 
n 1
123
PP
nn
, as T2  T1 
 2 
 T 
 2 
 P3   P3
T  T
n1
Delivery Temperature,
n1

24. Reciprocating Compressor – Multistage
2 PV C
3 5
4
1
n
PV  C
8
7
6 9
L.P.
H.P.
C. Two – Stage Compressor (With Perfect Intercooling) :
With Intercooling :
L.P. : 8-1-4-7-8
H.P. : 7-2-3-6-7






     
PP
n
W  P V
n1

 P3  n
n 1
 1   2 
 P2  n
1 1 2 
n 1
Now, T2 = T1
P2V2 = P1V1
Also P4 = P2
Shaded Area 2-4-5-3-2 : Work Saving due to Intercooler…!!

25. Reciprocating Compressor – Multistage
Condition for Min. Work :
2 PV C
3 5
4
1
PV n
C
8
7
6 9
L.P.
H.P.
Intermediate Pr. P2 → P1 : Area 2-4-5-3-2 → 0
Intermediate Pr. P2 → P3 : Area 2-4-5-3-2 → 0
2
 There is an Optimum P for which Area 2-4-5-3-2
is maximum,
i.e. Work is minimum…!!





 nnn
n1
n1
 1   2 
 P2  P3
2 
P
 
P
W  PV
n1 1 1


  0


  

dP2
 P2  P1 
d  
dW
dP2
 P3  n
n 1

 P2  n
n 1
For min. Work,

26. Reciprocating Compressor – Multistage
Condition for Min. Work :
232  0



 

 
 
 
   
n
 
n11
n
 n1
n
 n1
1
n1
n
P 
n
  n1
  P P 
n
1  n1
P1


  0


 

dP2
 P2
 P3
 P1
 P2
d 
dW
dP2
n
n1
n1
n

 

 
n
n
 P P 
P 
P   n1
1 3 2n1
2
1/n
2
2
2P   P1P3
P P2
 3
P1 P2
P2  P1 P3 OR
PVC2
1
PV n
C
8
7 4
6 9 3 5
L.P.
H.P.

27. Reciprocating Compressor – Multistage













n
P

PP

n1
1
1/2
1 3
1 1P V 1
n1
W 
2n









 n
n1

 1 
 P2
1  
P
W  PV
n 1 1 1
2n




 n1

 1 
 P3 2n

1 
P
W  PV
n 1 1 1
2n
P2 obtained with this condition (Pr. Ratio per stage is equal) is the Ideal Intermediate
Pr. Which, with Perfect Intercooling, gives Minimum Work, Wmin.
 Equal Work per cylinder…!!

28. Reciprocating Compressor – Efficiency
Isothermal work done / cycle = Area of P – V Diagram
= P1V1 loge(P2/P1)
Isothermal Power = P1V1 loge(P2/P1)N kW
60 X 1000
Indicated Power : Power obtained from the actual indicator card taken during a
test on the compressor.
Compressor Efficiency = Isothermal Power
Indicated Power
Isothermal Efficiency = Isothermal Power
Shaft Power
NOTE : Shaft Power = Brake Power required to drive the Compressor.

29. Reciprocating Compressor – Efficiency
Adiabatic Efficiency : Ratio of Power required to drive the Compressor; compared
with the area of the hypothetical Indicator Diagram; assuming
Adiabatic Compression.
Brake Power required to drive the Compressor
adiabatic









 1

 1 
 P2 

1  
P

 1
P1V1
Mechanical Efficiency : Ratio of mechanical output to mechanical input.
Mechanical Efficiency, ηmech = Indicated Power
Shaft Power

30. Reciprocating Compressor – Efficiency
How to Increase Isothermal Efficiency ?
A. Spray Injection : Assimilation of water into the compressor cylinder towards the
compression stroke.
Object is to cool the air for next operation.
Demerits : 1. Requires special gear for injection.
2. Injected water interferes with the cylinder lubrication.
3. Damage to cylinder walls and valves.
4. Water must be separated before delivery of air.
B. Water Jacketing : Circulating water around the cylinder to help for cooling the
air during compression.

31. Reciprocating Compressor – Efficiency
How to Increase Isothermal Efficiency ?
C. Inter – Cooling : For high speed and high Pr. Ratio compressors.
Compressed air from earlier stage is cooled to its original
temperature before passing it to the next stage.
D. External Fins : For small capacity compressors, fins on external surfaces are useful.
E. Cylinder Proportions : Short stroke and large bore provides much greater surface
for cooling.
Cylinder head surface is far more effective than barrel
surface.

32. Reciprocating Compressor – Efficiency
Clearance Volume : Consists of two spaces.
1. Space between cylinder end & the piston to allow for wear.
2. Space for reception of valves.
High – class H.P. compressors : Clearance Vol. = 3 % of Swept Vol.
: Lead (Pb) fuse wire used to measure the gap
between
Low – grade L.P. compressors : CcylelianrdaenrcenVdoal.n=d6p%istoonf.SweptVol.
: Flattened ball of putty used to measure the gap
between cylinder end and piston.
Effect of Clearance Vol. :
↑ Size of compressor
↑ Power to drive compressor.
Vol. taken in per stroke < Swept Vol. 

33. P1
P2
1V4 V
2
5 1
6 3
4
V3
Effective Swept Volume,
V -V1 4
1 4Swept Volume, V -V =V s
Total Volume, V1
Clearance Volume,
V =V3 c
Reciprocating Compressor – Work Done




 



 
nn
 P 
Pn
 P 
Pn
n1

 
 4 
3
4 4 

n1

 
 1 
2
1 1 

 P V 1 
n 1
W  P V 1 
n 1
Assumption : Compression and Expansion follow same Law.
Work / cycle = Area 1-2-3-4-1
P3 = P2 and P4 = P1







  






  
P
 P  n
P V
n
P
n
n 1
 1 
2
1 a
n1

 1 
 P2  n
4
1 
n 1
V ) 1 W  P (V 
n 1 1 1

34. P1
P2
V4 V1
2
5 1
6 3
4
V3
Effective Swept Volume,
V1-V4
Swept Volume, V1-V4=Vs
Total Volume, V1
Clearance Volume,
V3=Vc
Reciprocating Compressor – Work Done





 nn
n1

 1 
 P2
1 
P
W  m RT
n 1 1 1
m1 is the actual mass of air delivered.
Work done / kg of air delivered :







nn
W 
n1

 P1
 P2
RT1 1
n1