Thermodynamic Property Methods

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Ref.1: Kumar, Gas Production Engineering, Gulf Publishing Com.,
1987, Chapter 9.
Ref.2: GPSA Electronic Data Book, Gas Processors Association,
1998, Chapter 13.
Ref.3: Mokhatab et al, Handbook of Natural Gas Transmission and
Processing, Gulf Publishing Com., 2006, Chapter 8.
1
Gas Compression
Reciprocating Compressors
Number of Stages:
1. Calculate the overall compression ratio (rt). If the
compression ratio is under 4, consider using one stage. If it is
not, select an initial number of stages (ns) based on the
following equation (3% pressure drop was assumed in each
coolers):
r  rt1/ ns / 0.97
2. Calculate the discharge gas temperature for the first stage. If
the discharge temperature is too high (more than 300 oF),
either increase the number of stages or reduce the suction
temperature through precooling.
Gas Compression
Reciprocating Compressors
BHP of each stage:
Calculate the ideal head of each stage based on the Isentropic
assumption, then calculate the GHP and BHP with using
isentropic and mechanical efficiency:
is  0.83 0.93, m  0.88  0.95
Gas flow rate at suction conditions:
Gas flow rate at suction conditions is required for capacity
calculation of each stage.
 Psc
q g1 (Mcfm )  q g sc 
 P1
 T1

 Tsc
o
z
T
(
R )q g sc (Mscfd )

5 1 1
 z1  1.9671 10
P1 (psia)

Gas Compression
Reciprocating Compressors
Capacity (or speed) of each stage:
For single acting:
Piston diameter
Stroke length
q1  ( / 4)d p2 Ls Sv
 q1 (Mcfm)  5.454106 d p2 (in ) Ls (ft) S (rpm)v
For double acting:
q1 (Mcfm)  5.45410 (2d  d )Ls (ft)S (rpm)v
6
Rod diameter
2
p
2
r
Gas Compression
Reciprocating Compressors
Volumetric efficiency:
The term “volumetric efficiency” refers to the actual
pumping capacity of a cylinder compared to the piston
displacement.
 z1 1/ k 
v  0.96  0.01r  C  r  1
 z2

Where:
ClearanceVolum e
C
StrokeVolum e
Subtract about 0.05 from volumetric efficiency when a nonlubricated compressor is used.
Gas Compression
Centrifugal Compressors
Number of Stages:
The total polytropic head (Hpt)
is an indication of the number
of stages required for
centrifugal compression:
ns  H pt / H pstage ,
 lb  ft 
H pstage  f   15000 1500M g0.35
 lb m 
Or from Fig 13-39:
The discharge temperature
must be less than 350 oF
 ft  lbf
1.0 
 lb m
 Nm 


  3.0 
 k 

 g 
Gas Compression
Centrifugal Compressors
BHP of all stages (one unit):
Calculate the ideal head of all stages based on the polytropic
assumption, then:
1- Calculate the GHP
with using polytropic
efficiency:
2- Calculate the BHP
with using
mechanical losses:
MechanicalLosses  0.75(GHP) 0.4
BHP  GHP  M .L.
Gas Compression
Centrifugal Compressors
Capacity (or speed):
Gas Compression
Centrifugal Compressors
Performance Curves:
Generally, the performance of a centrifugal compressor follows
the “affinity laws.”According to the affinity laws, as the
rotational speed of the centrifugal compressor is changed, the
inlet flow and head vary as the speed and the square of the speed,
respectively
qg  S , H  S 2  HP  S 3
q g new
 S new 
 S new 



 q g old 
, H new  H old 

 S old 
 S old 
2
As the speed deviates from the design speed, the error of
“affinity laws” will increase.
Gas Compression
Centrifugal Compressors
Gas Compression
Centrifugal Compressors
Surge Line Control: In all modern centrifugal compressors, a
recycle line with a control valve that allows the flow to increase
through the compressor is used for surge control. The flow
through the compressor must be 10-20% greater than surge limit.
SC
Gas Compression
Example
Pressure of a gas stream (30 oC, 200 MMscfd, 90% C1, 10% C2)
must increase from 600 psia to 1100 psia.
a) Design a reciprocating compressor (dp= 25 cm, Ls=1.2 m,
dr=3 cm, single acting, Clearance fraction = 0.15)
b) Design a Centrifugal compressor
Solution:
T1  546 o R, r  1.833, M g  17.45,  g  0.602
Ppc  674.3 psia, Tpc  355.5 o R, Tpr  1.54, Ppr  0.89  z1  0.93
q1  3.33 Mcfd, k  1.271
Gas Compression
Example: Solution
a) Reciprocating: The values in the parenthesis are calculated by HYSYS
r  4  ns  1, assume:is  0.83  T2  637 o R  177 o F  300
 ft  lbf
 lbm
v  0.85 (0.88), S (rpm)  1825(1763), H is  29070



GHP  6783(6647), assumem  0.88  BHP  7708
b) Centrifugal:
 p  0.715, H p  29848 ns  3, T2  237o F  350
GHP  8085(7935), M . L.  27  BHP  8112, S  11000rpm
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