Engineers often use softwares to perform gas compressor calculations to estimate compressor duty, temperatures, adiabatic & polytropic efficiencies, driver & cooler duty. In the following exercise, gas compressor calculations for a pipeline composition are shown as an example case study.

1. Page | 1
GAS COMPRESSOR
CALCULATIONS TUTORIAL
Engineers often use softwares to perform gas
compressor calculations to estimate compressor
duty, temperatures, adiabatic & polytropic
efficiencies, driver & cooler duty. In the following
exercise, gas compressor calculations for a pipeline
composition are shown as an example case study.
Table 1. Pipeline Gas Composition
Component Mole fraction
Methane 0.9000
Ethane 0.0500
Propane 0.0200
n-Butane 0.0100
n-Pentane 0.0050
n-Hexane 0.0050
Water 0.0100
Total 1.0000
The total fluid flow rate is 7413 kg/hr. The gas
conditions at battery limits (BL) is 2.03 bar (a) &
25.50C & this has to be delivered at a pressure of
3.202 bar(a) & ~500C.
Compressor Performance Curves
To achieve these conditions, the following
compressor maps associated for a chosen
compressor is used.
Figure 1. Head vs. Flow rate Curve
The polytropic head vs. actual volumetric flow rate
curve is indicated in Fig. 1. It is taken in this
exercise, since the surge line is not available; a
surge line is fitted to an equation to obtain the
surge line and used for all calculation purposes. The
polytropic efficiency vs. actual volumetric flow rate
of the gas compressor curves is shown in Fig. 2.
Figure 2. Polytropic Efficiency vs. Flow Rate
Vendor Electric Motor Details
The driver used is an asynchronous electric motor
and its parameters are as follows,
1. Electric Motor = Asynchronous Induction
2. Number of poles = 4
3. Power Source Frequency = 50 Hz (A.C)
4. Motor Slip = 1%
Figure 3. EM Speed vs. Torque characteristics
Motor Characteristics = Fig. 3 & Table 2
Table 2. Speed vs. Torque characteristics
% Speed % Torque
0 55
10 50
20 45
30 50
40 55
50 60
60 70
70 85
80 105
97 150
99 1

2. Page | 2
Preliminary System Estimates
Using the compressor data, a process schematic is
proposed as shown in Fig. 4
Figure 4. Proposed CC system Schematic
For the calculations the following is assumed.
1. Pressure drop from suction side Battery limit
(BL) to compressor suction flange = 0.8 bar
2. Pressure drop from discharge flange to
discharge BL = 0.5 bar
3. Pressure drop across suction scrubber is
assumed to be 0 bar. This causes very little
water to be separated & the final gas
composition at the suction flange is shown in
Table 3.
4. Pressure drop across the discharge side air
cooler is taken to be 0.1 bar
Note: Pressure drop data across equipment is
assumed for sample calculation purposes. In reality
pressure drops can be higher.
Table 3. Gas composition After Suction Scrubber
Component
Mass flow Mass Fraction
[kg/hr] [%]
Methane 5823.61 0.7857
Ethane 606.41 0.0818
Propane 355.72 0.0480
n-Butane 234.44 0.0316
n-Pentane 145.51 0.0196
n-Hexane 173.79 0.0234
Water 72.66 0.0098
Total 7412.14 1.0000
Gas Properties at Flange Conditions
The following properties exist at the flange
conditions of the centrifugal compressor.
Table 4. Gas composition after suction separator
Property Value Unit
Gas MW 18.38 kg/kmol
Suction side Density (1) 0.9153 kg/m3
Discharge side Density (2) 2.03 kg/m3
Suction side Heat capacity 2.138 kJ/kg0C
Discharge side Heat capacity 2.474 kJ/kg0C
Z1 at suction flange 0.9964 -
Z2 at discharge flange 0.9964 -
k1 at suction side 1.2740 -
k2 at discharge side 1.2304 -
Calculations
Polytropic Head
From the above data, in order to arrive at a
discharge pressure of 3.202 bar & ~500C at the
discharge battery limit, a pressure drop 0.498 bar
is added to the battery limit to obtain 3.7 bar(a) at
the compressor discharge flange. Similarly, taking
into consideration the pressure losses across the
suction side of 0.9 bar, the suction flange pressure
is taken to be 1.23 bar(a). Taking into account the
temperature drop across the valves on the suction
side, the suction flange temperature is taken to be
250C. Therefore the head produced (Annexure A) is,

























1
1
1
1
21
n
n
avg
p
P
P
n
n
MW
RTZ
H (1)
Where,
pH = Polytropic head produced (ft)
avgZ = Average gas compressibility factor (-)
1T =Gas temperature at suction flange (K)
n= Polytropic volume exponent (-)
1
2
P
P
= Pressure ratio at flange conditions (-)
To estimate the polytropic volume exponent, the
following calculation is made.
p
k
k
n
n



 11
(2)
Where, p = Polytropic Efficiency

3. Page | 3
The specific heat ratio considered is the average
specific heat between suction & discharge flange
2522.1
2
2304.1274.1
2
21





kk
kavg (3)
The polytropic efficiency is taken from Fig. 2, where
the efficiency at which the compressor operates is
~71.84%. Therefore substituting in Eq. (2),
3896.17184.0
12522.1
2522.1
1




n
n
n
(4)
An alternate way to calculate the polytropic
exponent is by using the equation,













1
2
1
2
ln
ln


P
P
n (5)
Or, 3827.1
9153.0
030.2
ln
23.1
7.3
ln
ln
ln
1
2
1
2





























P
P
n (6)
It is seen that the value of ‘n’ calculated using Eq.
(5) is almost equal to the value calculated using Eq.
(2). However, since the value of ‘n’ was estimated
using a graphically calculated value of p by hand as
in Eq. (4), the value obtained using Eq. (6) is
considered for calculations. Using the values
calculated above, the polytropic head produced is
therefore estimated using Eq. (1) as,
  
























1
23.1
7.3
13827.1
3827.1
38.18
67.53635.154529964.09964.0 3827.1
13827.1
pH
(7)
kgkJftHp /92.17217645m36.57890  (8)
Gas Outlet Temperature
The gas discharge temperature is calculated (Eq. 9)
as (Annexure 2),














2
1
1
1
2
1
2
Z
Z
P
P
T
T n
n
(9)
Or,  2515.273
9964.0
9964.0
23.1
7.3 3827.1
13827.1
2 













T (10)
CKT 0
2 2.1314.404  (11)
Adiabatic Efficiency
The adiabatic efficiency is calculated as follows,














































 

1
1
1
1
1
1
2
1
1
2
k
k
n
n
A
p
P
P
k
k
P
P
n
n


(12)
0442.1
233.1
2876.1
1
23.1
7.3
12522.1
2522.1
1
23.1
7.3
13827.1
3827.1
2522.1
12522.1
3827.1
13827.1











































 

A
p


(13)
%78.68
0442.1
7182.0
A (14)
Inlet Volumetric Flow Rate
The inlet volumetric flow rate is calculated as
follows,
1
1

m
Q  (15)
hrmQ 3
1 8100
9153.0
7413
 (16)
Compressor Duty
The power absorbed by the compressor to produce
a discharge pressure of 3.7 bar(a) is therefore,
p
p mH
P


 (17)
hrkJP 54.1784817
7182.0
741392.172


 (18)
Or, kWkJP 496sec/8.495  (19)
Driver Duty
Assuming mechanical losses + margin of 20% of the
absorbed power (Actual value to be confirmed by
Vendor), the estimated power requirements is
2.1496P (20)
rpmatkWP 3000595 (21)
In actual practice, the power value selected as
shown above may or may not meet the required
power to start the compressor from settling out
conditions. In such cases a detailed dynamic

4. Page | 4
simulation must be performed to check - by how
much the compressor loop should be depressurized
prior to re-start. The driver rating is to be chosen
based on standards such as NEMA or other similar
standards. The National Electrical Manufacturer’s
Association (NEMA) is an organization that sets the
standards for manufacturing of electrical
equipment. In the case of electrical induction
motors, NEMA sets 4 types of Torque vs Speed
Characteristics namely, NEMA A, NEMA B, NEMA C
and NEMA D. The characteristics of each type of
NEMA motors are depicted below.
Figure 5. NEMA Torque vs. Speed Types
Table 5. Uses of Different Types of NEMA Motors
Among these, NEMA A & NEMA B are the most
commonly used due to their high breakdown
characteristics and low slip (1% to 5%). For the
purpose of this example, the rated motor selected is
a 600 kW asynchronous induction electric motor to
round–off the calculated value.
Cooler Duty
To cool the gas from the compressor discharge to
say 500C, the cooler duty is calculated as,
 32, TTmCQ avgp  (22)
  15.3234.404
2
220.2473.2
3600
7413


Q (23)
hrkJkWQ 6
10414.16.392  (24)
Summary
The following table shows a summary of the
calculated values,
Table 6. Summary of Preliminary Estimates
Parameter Value Unit
Inlet Flange Pressure 1.23 Bar(a)
Outlet Flange Pressure 3.70 Bar(a)
Polytropic Head 172.92 kJ/kg
Discharge Temperature 131.2 0C
Adiabatic Efficiency 68.78 %
Inlet Volume Flow 8100 Act_m3/hr
Compressor Duty 496 kW
Driver Rated Duty 600 kW
Cooler Duty 392.6 kW
Electric Motor Sizing
The induction type electric motor available is a 4
pole asynchronous model with a power source at
the plant site operating at 50Hz. Therefore, the
synchronous speed of the motor is,
4
50120120 



Poles
frequency
N ssynchronou (25)
rpmN ssynchronou 1500 (26)
From the steady calculations, a 600 kW induction
type electric motor has been proposed to be
installed. The torque required to sustain the
compressor at the rated conditions, the torque
required is,
kW
NT
P
100060
2



(27)
 
14852
100060496
100060
14852
496








T
mNTrpm
kW (28)
mNT  53.3189 (29)
Hence the electric motor has to provide a torque of
3189.53 kg-m2 to sustain the compressor at 3000
rpm to meet the rated suction & discharge
conditions. The maximum torque that can be
provided by the electric motor before it breaks
down is calculated at the breakdown torque’s
corresponding speed at 97% (Figure 3)
rpmN 45.1440148597.0  (30)

5. Page | 5
The breakdown torque is estimated as,
mNT downbreak 


 45.59665.1
45.14402
100060600

(31)
Hence the power absorbed to generate 5966.5 N-m
is 900 kW at a speed of ~1440 rpm during start-up.
The inertia offered by the electric motor is
calculated by the empirical relationship as,
48.1
0043.0 






N
P
I (32)
Where,
P = Power [kW]
I = Inertia [kg-m2]
N = Speed [rpm/1000]
Therefore, the inertia offered by the motor is,
48.1
1000
1485
600
0043.0








I (33)
2
97.30 mkgI  (34)
Therefore the inertia offered by the electric motor
(EM) is 30.97 kg-m2. In order to scale up the speed
of the compressor, a gearbox is installed. The gear
ratio is,
0202.2
1485
3000
GR (35)
The gear ratio calculated is 2.0202. Note that the
total inertia required to be overcome by the
Electric Motor (EM) is the sum of compressor rotor
inertia, EM inertia, Gear box Inertia & Gas inertia
during start-up & settling out conditions.
Annexure A: Compressor Head
Derivation
The general energy balance for compressors may
be written in differential form as
gdz
c
ddhdqdy 






2
2
(A.1)
Where,
y = Specific mass for compressor work input
q = Heat flow through the compressor walls [kJ]
h = Enthalpy of Gas [kJ/kg]
c = Absolute velocity of gas [m/s]
g = Gravitational acceleration [m/s2]
z = Elevation [m]
From the above expression it is seen that the
specific mass referenced work input and the heat
flow to the compressor is equal to the enthalpy
change, change in kinetic energy & static heat
difference. Neglecting the velocity terms, static
head contributions & the heat input through the
walls of the compressor, we obtain,
dhdy  (A.2)
The change in enthalpy of gas is given as,

dp
dh  (A.3)
Where,
 = Density of Gas (kg/m3)
p = Pressure of Gas (bara)
The specific compressor mass referenced work
input is calculated as,

2
1
p
p
dp
y

(A.4)
The actual work is found by dividing the mass
referenced work input by efficiency

y
W  (A.5)
Where,
W = Actual work applied to the compressor
The integral (Eq. A.4) can be solved in ways
equivalent to different compression paths such as,
1. Isentropic compression (reversible & adiabatic) -
Entropy is constant - 0;0;0  STQ
2. Isothermal compression (reversible &
diathermic) - Temperature is constant - 0T
3.Polytropic Compression (irreversible &
adiabatic) – Efficiency is constant
Isentropic Process
As an isothermal process is not feasible in real
world applications, this is neglected. However an
Isentropic compression can be considered to be an
idealistic situation, as it can exist when the process
is completely adiabatic & not heat transfer takes
place. Therefore for isentropic compression,

6. Page | 6
constvp k
. (A.6)
Where, 
v
p
C
C
k Ratio of Specific Heats
Rewriting eq. (A.6), we get
const
p
constp k
k







1
(A.7)
kkk
k
kk
p
p
p
ppp
1
1
1
1
1
1
1












(A.8)
k
p
p
1
1
1 





  (A.9)
Applying Eq. (A.9) in Eq. (A.4),
 







2
1
2
1
2
1
1
1
1
1
1
1
1
1
p
p
k
kp
p
k
p
p p
dpp
dp
p
p
dp
y



(A.10)
2
1
2
1
11
11
1
1
1
1
1
1
1
p
p
kkp
p
k
k
k
pp
ydpp
p
y












 
(A.11)
2
1
1
1
1
1
1
p
p
k
k
k
k
k
pp
y

















 



(A.12)
Applying Limits & Re-arranging Eq. (A.12)




























k
k
k
k
pp
k
kp
yp
k
kp
y
k
p
p
k
kk 112
1
12
1
1
1
1
1
1
1
11 
(A.13)
Taking out P1 as a common term from the brackets,


















 

1
1
1
11
1
2
1
1
1
1
k
k
k
k
k
k
p
p
k
kpp
y
k

(A.14)




















































1
1
1
1
1
1
2
1
11
1
1
1
2
1
1
1
1
1
k
k
k
k
kk
k
k
p
p
k
kp
y
p
p
k
kpp
y
k
k

(A.15)

























1
1
1
1
2
1
11
1
k
k
k
k
p
p
k
kp
y

(A.16)

























1
1
1
1
2
1
1
k
k
p
p
k
kp
y

(A.17)
Using Ideal gas equation the pressure & density
terms are rewritten using the expression (A.18),
nZRTpv  (A.18)
ZRT
p
v
n
 (A.19)
The above expression gives the density in kmol/m3
& is written in terms of kg/m3 by dividing with
molecular weight of the gas.
ZRT
p
MWv
M


(A.20)
Where M = mass of gas
MW
ZRTp
ZRT
MWp
v
M




 (A.21)
Hence for the inlet conditions of the gas into the
compressor Eq. (A.21) becomes,
MW
ZRTp 1
1
1


(A.22)
Substituting Eq. (A.22) into Eq. (A.17),

























1
1
1
1
21
k
k
gas p
p
k
k
MW
ZRT
y (A.23)
The above expression gives the adiabatic head
produced by the compressor & can be written as,
aHy  (A.24)
Where, Ha = Adiabatic Head [m]

























1
1
1
1
21
k
k
gas
a
p
p
k
k
MW
ZRT
H (A.25)
Polytropic Process
In all real world compression applications, the
polytropic process is predominant & hence the
exponent in the ideal gas equation becomes ‘n’
(polytropic volume exponent) which is  n1 &
Eq. A.6 becomes,
constvp n
. (A.26)
Similarly, performing the above set of calculations
using the polytropic exponent, the polytropic head
is expressed as follows,

7. Page | 7

























1
1
1
1
21
n
n
gas
p
p
p
n
n
MW
ZRT
H (A.27)
The power absorbed by the compressor or the
power that is needed at the compressor shaft is
estimated as,
p
gasinp
actual
mH
P

,
 (A.28)
Where,
p = Polytropic Efficiency [-]
gasinm , =Mass flow rate at compressor inlet [kg/s]
actualP =Required power at compressor shaft [kW]
Annexure B: Compressor Discharge
Temperature Derivation
The temperature rise at compressor discharge after
compression is calculated from Eq. (A.26) as,
const
p
ZRT
pconstvp
n
n






. (B.1)
Taking the value of gas constant to the right hand
side of Eq. (B.1),
  constZTp nn
1
(B.2)
For inlet conditions (suffix 1) to outlet conditions
(suffix 2), Eq. (B.2) can be written as,
   nnnn
TZpTZp 22
1
211
1
1

 (B.3)
 
 
nn
n
n
n
n
TZ
TZ
p
p
TZ
TZ
p
p
















22
11
1
1
2
22
11
1
1
1
2
(B.4)

























1
2
1
1
2
2
1
22
11
1
1
2
Z
Z
p
p
T
T
TZ
TZ
p
p n
n
n
n
(B.5)
Annexure C: Electric Motor (EM)
Implementation in ASPEN HYSYS
Dynamics
Aspen HYSYS Dynamics provides the option of
simulating centrifugal compressors with an Electric
Motor (EM). In order to setup the electric motor,
the calculations made to arrive at the Electric
Motor Input data is entered into Aspen HYSYS
Dynamics.
Figure 6. Electric Motor in Aspen HYSYS Dynamics
References
1. Gas Processors Association. Gas Processors
Suppliers Association (1998) P.13-1 to P.13-20
2. HYSYS 2004.2, Operations Guide































 


2
1
1
1
2
1
2
1
1
2
1
2 1
Z
Z
p
p
Z
Z
p
p
T
T n
n
n
n
(B.6)


















 
2
1
1
1
2
1
2
Z
Z
p
p
T
T n
n
(B.7)
From the above, the discharge temperature at the
compressor outlet is calculated from Eq. (B.7).