A compressor is a machine that compresses air or gas to pressures over 241.25 KPa. There are three main types: centrifugal for low pressure/high capacity, rotary for medium pressure/low capacity, and reciprocating for high pressure/low capacity. Compressed air has many industrial and specialized uses. Compressors can be analyzed using the steady flow energy equation and isentropic, polytropic, or isothermal processes. Multistage compression saves power by cooling the air between stages to limit temperature and pressure increases.

1. Compressor is a machine used to compressed air or gas to a final pressure exceeding 241.25 KPa gage.
TYPES OF COMPRESSORS
 Centrifugal Compressors - For low pressure and high capacity application
 Rotary Compressors - For medium pressure and low capacity application
 Reciprocating Compressors - For high pressure and low capacity application
USES OF COMPRESSED AIR
 Operation of small engines
 Pneumatic tools
 Air hoists
 Industrial cleaning by air blast
 Tire inflation
 Paint Spraying
 Air lifting of liquids
 Manufacture of plastics and other industrial products
 To supply air in mine tunnels
 Other specialized industrial applications
ANALYSIS OF CENTRIFUGAL AND ROTARY TYPE
W
2
1
Q
Assuming: KE = 0; and PE = 0
Q = h + KE + PE + W (Steady state - steady flow equation)
Q = h + W
For a compressor, work is done on the system; thus
-W = h - Q; Let -W = W (compressor work)
A) Isentropic Compression (PVk = C)
P
2
W = -VdP
PVk = c
dP
V
1
V
For Isentopic compression' Q = 0
W = h ; W = -VdP
W = mCp(T2-T1)
k 1


kmRT  P  k 
1  2
W
 1 KW

k - 1  P 

 1


Q=0
PV1 = mRT1
P,V, and T relationship
k 1
k
P 
2  2
T P 
1  1
T
k 1
V 
 1
V 
 2

2. where
m - mass flow rate of the gas in kg/sec
W - work in KW
P - pressure in KPa
T - temperature in K
R - gas constant in KJ/kg-K
V - volume flow rate in m3/sec
B) Polytropic Compression (PVn = C)
P
2
W = -VdP
PVn =C
dP
V
1
V
For Polytropic compression, Q  0
W = h - Q ; W = -VdP
h = mCp(T2 - T1)
Q = mCn(T2 - T1)
n1


nmRT  P  n 
1  2
W
 1 KW
n - 1  P 
 1 



 k  n
C  Cv

n
 1 n 
P 
2  2
T P 
1  1
T
n1
n
V 
 1
V 
 2
n1
C) Isothermal Compression (PV = C)
P
2
W = -VdP
PV = C
dP
V
1
V
W = -Q
P
W  P V ln 2
1 1 P
1
P1V1 = mRT1

3. valves
cylinder
piston
piston-rod
d
HE
P
P2
P1
CE
L
2
3
4
CVD
D
V1’
1
V
VD
HE - Head end
CE - Crank end
L - length of stroke, m
VD - Displacement volume, m3/sec
D - diameter of bore, m
d - diameter of piston rod, m
A) For Isentropic Compression and RE-expansion process (PVK = C),no heat
is removed from the gas.
W = h ; W = -VdP
W = mCp(T2-T1)
k 1


kmRT1  P2  k 
 
W
 1 KW

k -1  P1 


 

Q=0
PV1' = mRT1
P,V, and T relationship
k 1
k 1
 P  k  V1 

 2 

V 
T1  P1 
 
 2
where:
V1' - volume flow rate measured at intake, m3/sec
m - mass flow rate corresponding V1', kg/sec
B) For Polytropic Compression and Re-expansion process (PVn = C), some
amount of heat is removed from the gas.
W = h - Q ; W = -VdP
h = mCp(T2 - T1)
Q = mCn(T2 - T1)
T2

4. n 1


nmRT1  P2  n 
 
W
 1 KW

n -1  P1 
 




P1V1' = mRT1
k  n 
Cn  Cv
 KJ/kg-K
1n 
n 1
n 1
 P  n  V1 

 2 

V 
T1  P1 
 
 2
C) For Isothermal compression and re-expansion process (PV = C), an
amount of heat equivalent to the compression work is removed from the
gas.
W = -Q
P
W  P1 V1'ln 2
P1
P1V1' = mRT1
T2
PERCENT CLEARANCE
Clearance Volume
C 
Displacement Volume
V
C  3 x 100%
VD
For compressor design, values of percent clearance C ranges from
3 to 10 %.
V3 = CVD
where: V3 - clearance volume
VOLUMETRIC EFFICIENCY
v = Volume flow rate at intake x 100%
Displacement Volume
V
ηv  1' x 100%
VD
A) For Isentropic Compression (PVk= C)
1/k

 P2  
ηv  1  C  C   x 100%
P  

 1 

B) For Polytropic Compression (PVn = C)
1/n

 P2  
ηv  1  C  C   x 100%
P  

 1 

C) For Isothermal Compression (PV = C)

 P 
ηv  1  C  C 2  x 100%
 P 

 1 


5. DISPLACEMENT VOLUME
A) For single acting
VD = LD2Nn' m3/sec
4(60)
B) For Double acting without considering the volume of piston rod
VD = 2LD2Nn' m3/sec
4(60)
C) For Double acting considering volume of piston rod
VD = LNn'[2D2 - d2] m3/sec
4(60)
where: L - length of stroke, m
D - diameter of bore, m
d - diameter of piston rod, m
n' - no. of cylinders
ACTUAL VOLUMETRIC EFFICIENCY
V
ηva  a x 100 %
VD
Va - actual volume of air or gas drawn in
MEAN EFFECTIVE PRESSURE
W
Pm 
KPa
VD
W in KJ, KJ/kg, KW
VD in m3, m3/kg, m3/sec
PISTON SPEED
PS = 2LN m/min
PS = 2LN m/sec
60
EFFICIENCY
A) COMPRESSION EFFICIENCY
cn = Ideal Work
x 100%
Indicated Work
B) MECHANICAL EFFICIENCY
m = Indicated Work x 100%
Brake Work
C) COMPRESSOR EFFICIENCY
c = cn = m = Ideal Work x 100%
Brake work
MULTISTAGE COMPRESSION
Multi staging is simply the compression of air or gas in two or more cylinders in place of a single cylinder compressor. It is
used in reciprocating compressors when pressure of 300 KPa and above are desired, in order to:
 Save power
 Limit the gas discharge temperature
 Limit the pressure differential per cylinder
 Prevent vaporization of lubricating oil and to prevent its ignition if the temperature becomes too high.
It is common practice for multi-staging to cool the air or gas between stages of compression in an intercooler, and it is this
cooling that affects considerable saving in power.

6. A) 2 - Stage Compression without pressure drop in the intercooler
1
suction
2
Qx
3
4
discharge
Intercooler
1stStage
2nd Stage
For an ideal multistage compression, with perfect inter-cooling and minimum work, the cylinder were properly designed so
that:
 the work at each stage are equal
 the air in the intercooler is cooled back to the initial temperature
 no pressure drop occurs in the intercooler
 the pressure at each stage are equal
W1 = W2 ;
T1 = T3 ; P2 = P3 = Px
where: W1 - work of the LP cylinder (1st stage)
W2 - work of the HP cylinder (2nd stage)
Px = ideal intercooler pressure, optimum pressure
Assuming polytropic compression and expansion processes:
P
P4
Px
P1
5
4
6 7
3 2
8
W1 = W2
T1 = T3
P2 = P3 = P x
W = W1 + W2
Px - the ideal intercooler pressure or optimum pressure
Work 1st Stage:
n 1


nmRT  P  n 
1  2
W 
 1 KW
1
n - 1  P 
 1 



Work 2nd Stage:
n 1


nmRT  P  n 
3  4
W 
 1 KW
2
n - 1  P 
 3 



PVn = C
1

7. Pressure Ratio:
P2 P4

P1 P3
but P2 = P3 = Px
Px P4

P1 Px
then
Px  P1P4
Since W1 =W2, the total work W is;
n1


2nmRT1  P2  n
 
W
 1
 P 

n -1  1 




KW
substituting Px to W, it follows that
n1


2nmRT1  P4  2n
 
W
 1
KW

n -1  P1 
 




By performing an energy balance on the inter-cooler
Qx = mCP(T3 - T2)
T-S Diagram:
T
P4
T2 = T4
T1 = T3
Px
P1
Qx
4
2
3
1
S
B) 2 stage compressor with pressure drop in the intercooler For 2 stage compression with pressure drop in the intercooler,
P2  P3.The air in the intercooler may or may not be cooled to the initial temperature, and the work at each stage may
or may not be equal, thus the work W = W1 + W2
Work 1st Stage:
n 1


nmRT1  P2  n 
 
W1 
 1 KW

n -1  P1 
 




Work 2nd Stage:
n 1


nmRT3  P4  n 
 
W2 
 1 KW

n -1  P3 


 


8. The total work W is;
W = W1 + W2
The pressure, P2  P3, but the 1st stage may compress the air or gas to the optimum intercooler pressure P x, but a
pressure drop will occur in the inter-cooler.
P
5
4
P4
P2
P3
7
2
6
3
P1
8
1
V
Heat Rejected in the inter-cooler
Qx = mCp(T3 - T2)
C. Three-Stage compressor without pressure drop in the intercooler
Qx
Qy
suction
1
2
3
4
LP Intercooler
1st Stage
5
6
discharge
HP Inercooler
2nd stage
3rd stage
Considering Polytropic compression and expansion processes and with perfect inter-cooling;
Work of 1st stage cylinder:
n 1


nmRT1  P2  n 
 
W1 
 1 KW

n -1  P1 


 

Work of the 2nd stage cylinder:
n 1


nmRT3  P4  n 
 
W2 
 1 KW

n -1  P3 
 




Work of the 3rd stage cylinder:
n 1


nmRT5  P6  n 
 
W3 
 1 KW

n - 1  P5 
 



For perfect inter-cooling:
W1 = W2 = W3
T1 = T3 = T5

9. and
P2
P1
But
Thus

P4
P3

P6
P5
P2 = P3 = Px (Ideal LP Intercooler pressure)
P4 = P5 = Py (Ideal HP Intercooler pressure)
Px
P1

Py
Px

P6
Py
By expressing Px and Py in terms of P1 & P6:
2
Px  3 P1 P6
Py  3 P1P6
2
The total compressor work is equal to:
W = W1 + W2 + W3
but: W1 = W2 = W3 ;therefore
W = 3W1
n 1


3nmRT1  P2  n 
 
W
 1 KW

n -1  P1 


 

then substituting Px andPy then simplify, the result is:
n 1


3nmRT1  P6  3n 
 
W
 1 KW

n -1  P1 
 




For multistage compression with minimum work and perfect inter-cooling and no pressure drop in the inter-coolers
between stages, the following conditions apply:
1. the work at each stage are equal
2. the pressure ratio between stages are equal
3. the air temperature in the inter-coolers are cooled to the
original temperature T1
4. the total work W is equal to

10. n 1


SnmRT1  P2S  Sn 


W
 1 KW

n -1  P1 






where S - number of stages
Example
An ideal 3-stage air compressor with intercoolers handles air at the rate of 2 kg/min. The suction pressure is 101 Kpa,
suction temperature is 21C, delivery pressure is 5000 KPa. Assuming perfect inter-cooling and minimum work,
calculate total power required if compressor efficiency is 60% and both compression and expansion processes are
PVn=C, where n = 1.2. (21 KW)
Given
m  2 kg / min
P1  101 KPa
T1  21  273  294 K
P6  5000KPa
ec  0.60
PV n  C
n  1 .2
3nmRT1
W
n -1
 P  n 13n

 6 
 1
 P1 

 

KW
1. 2  1



3(1.2)(2)(0.287)(294) 5000 3(1.2)
W
 1


60(1.2 - 1)
 101 



W  20.41 KW