CALIFORNIA STATE UNIVERSITY, NORTHRIDGE
THE DESIGN OF A CONTROL SYSTEM
FOR A SYSTEM OF
PARALLEL CENTRIFUGAL COMPRESSORS
A project submitted in partial satisfaction of the
requirements for the degree of Master of Science in
Engineering
by
Viggen 0. Babayans
January 1983
The Project of Viggen 0. Babayans is approved:
California State University, Northridge
ii
To my Supervisor, Dr. Edward J. Hriber,
my parents, my brother Alfred and my
sister Claudia for their patience and
moral support.
iii
OBJECTIVE
iv
The objective of this project is to design a
control system for five compressors:
three are in
parallel, and this combination is in series with two
others (see Appendix A).
Each of the compressors has a different
capacity for transferring the flow of gas, but the
fourth compressor has larger capacity, and each of
them can operate separately to meet the standard
requirements for system stability.
The main purpose of this project is to
analyze the stability of the three-parallel-compressor
system as well as the overall system, by finding the
mathematical model and signal flow graph of the system.
We will consider two cases:
in Case I, the
input flow rate is constant, and for Case II the input
flow rate is a function of time.
In both cases, one
or more of the compressors operates wi t.h variable
speed.
The state variable method will be used
to analyze the system.
Signal flow graphs and block
diagrams will be constructed initially.
v
Then the
State and Output equations will be derived from these.
Finally the stability, controllability and observability of the system will be determined.
vi
TABLE OF CONTENTS
Page No.
OBJECTIVE
iv
LIST OF FIGURES
ix
LIST OF SYMBOLS
xi
xii
ABSTRACT
CHAPTER ONE - Theory
1
1-1
Introduction
2
1-2
Barrel Compressors
20
1-3
Physical Control of Compressors
22
1-4
Capacity Limitation
30
1-5
Variable Speed Performance
33
1-6
Compressors Surge Control System
36
CHAPTER TWO - Control of Parallel Compressors
42
2-1
Introduction
43
2-2
Block Diagrams
45
2-3
Surging
46
CHAPTER THREE - Stability Analysis of the
System with Constant Input
47
3-1
Introduction
48
3-2
Stability Analysis of Three
Parallel-Compressors
50
Controllability of Three ParallelCompressors
53
Output Controllability of Three
Parallel-Compressors
55
Observability of Three
Parallel Compressors
56
Stability Analysis of Overall System
58
3-3
3-4
3-5
3-6
vii
Page No.
CHAPTER FOUR - Stability Analysis - Constant
Input with Two Compressors
with Variable Speed
61
4-1
Introduction
62
4-2
Controllability of Three
Parallel-Compressors
64
Output Controllability of Three
Parallel-Compressors
66
Observability of the Three
Parallel-Compressors
67
Stability Analysis of the
Overall System
71
4-6
Controllability of Overall System
72
4-7
Calculation of Position Error
Coefficient
77
Efficiency of the Overall System
79
4-3
4-4
4-5
4-8
CHAPTER FIVE - Stability Analysis - Variable
Input Flow with One Compressor
with Variable Speed
81
5-1
Introduction
82
5-2
Controllability of Three
Parallel-Compressors
84
5-3
Transfer Function of Overall System
87
5-4
Controllability of the Overall System
89
5-5
Stability of the Overall System
92
BIBLIOGRAPHY
95
APPENDIXES
96
Schematic Diagram for a Typical
Compressor
97
B.
Typical Simulation Diagram
98
c.
Cofactor 3 x 3 Matrices
99
D.
Table of Laplace Transform Pairs
A.
viii
101
LIST OF FIGURES
Page No.
1.1 - Blade
3
1.2 - Hub or Disc
3
1.3 - Radial Line of Blade
3
1.4 - Semi-open Blade
3
1.5
-
Open Blade
8
1.6
-
Wheel
Sleeve
8
1.7 - Balancing Drum
8
1.8 - Lock Nut
8
1.9
-
&
Internal
&
External Seals
11
1.10 - Three Phase of Compressors
11
1.11 - Eye Seal
13
1.12 - Diffuser
13
1.13 - Return Bend
13
1.14 - Return Channel
13
1.15
-
Diaphragm
13
1.16 - Side Streams
1.17
-
13
Barrel Section
19
1.18 - Discharge Volute
19
Four Wobble Feet
19
1.19 -
1.20 - Single Wobble Foot
19
ix
1.21 - Barrel Compressor with Split
21
1.22 - Barrel Compressor with
Stacked Diaphragms
21
1.23 - Typical Constant Speed
Performance Curves for a
Centrifugal Compressor
25
1.24 - Effect of Changing Gas
Condition on Centrifugal
Compressor Operating at
Constant Speed
26
1.25 - Typical VAriable Speed
Performance Curves for
Centrifugal Compressor
27
1.26 - Typical Compressor Surge
Line Data
35
1.27 - Surge Control Line
39
2.1- Physical Diagram of
Parallel-Compressors
44
2.2 - Block Diagram of Compressors
Control System
45
2.3 - Flow vs. Discharge Pressure
46
3.1 - Simulation Diagram
51
3.2 - Signal Flow Graph
59
4.1 -Block Diagram
62
4.2 - Signal Flow Graph
70
4.3 - Desired Output Flow
78
4.4 - P2
p1
80
vs. time ti
5.1 - Simulation Diagram
X
83
LIST OF SYMBOLS
Q
1
Input Volume, cfm
W
Total Mass Throughput, #/min.
P1
Density at Input Conditions, #/ft
z1
Input Compressibility Factor
H
Polytropic Head
6m
Mean Compressibility Factor
tp
Polytropic Efficiency
Za
Average Suction
p
Gi(s) Transfer Function
es(t) Error Function
Kp
Step Error Coefficient
Kv
Ramp Error Coefficient
Ka
Parabolic Error Coefficient
xi
3
ABSTRACT
THE DESIGN OF A CONTROL SYSTEM
FOR A SYSTEM OF
PARALLEL CENTRIFUGAL COMPRESSORS
BY
Viggen 0. Babayans
Dynamic Compressors, Centrifugal or Axial, are the
heart of many industrial processes:
natural gas pro-
duction and transmission, oil refining, manufacture
of chemicals, steel, etc.
Improving the efficiency and reliability of these
compressors is therefore a matter worthy of serious
attention.
The Centrifugal compressor has become one of the most
common means of gas compression used today.
Its
main advantage is mechanical simplicity compared
to reciprocating and some other forms of rotary
compressors.
This generally reflects lower initial
xii
cost and lower maintenance cost.
Various configurations of compressors can yield
high reliability and efficiency.
Mathematical
models of the given systems permit prediction of
overall system response characteristics for various
configurations, once the mathematical model has
been obtained, several methods of
used.
analy~is
may be
The present study uses the State Variable
approach.
xiii
CHAPTER ONE
THEORY
l
2
1-1:
Introduction
Names and functions of various parts of a
centrifugal compressor are as follows:
The "Blade," (see Figure 1.1)increases the
velocity of a gas by rotating about the center line
and causing the gas to move from the inlet of the
blade (or wheel) to the tip or discharge.
The difference in the distance from the axis
of rotation of the blade inlet the blade discharge
causes the increase in kinetic energy (change of
velocity) as shown in Figure 1.1.
The "Disk" or "Hub" (used interchangeably
for either a closed or a semi-open wheel) serves two
functions ... to drive the blade and to confine the
gas to the blade area.
The "Cover," as with the disk, is used to
confine the gas to the blade area.
The cover is also
useful, for blades which are not radial, as a stiffener opposing the tendency of the blades to take a
radial position due to the forces caused by rotation,
shown on Figure 1.2.
These three parts -- Blade, Cover and Disk
constitute the Impeller, also known as the Wheel.
The wheel illustrated is a closed wheel, as the gas
3
~~~~----~_.BLADES
HUB OR DISC
Figure 1.2
Figure 1.1
Figure 1. 3
SEMI-OPEN
• d
Figure l. 4
4
is confined on both sides of the blade.
The normal
API type compressor wheel is of the closed type with
blades having from five degrees to forty-five degrees
of backward lean.
The most common is a thirty
degree backward lean.
The backward lean of a blade
is defined as the angle between the plane of the
blade at the tip of the wheel, and a radial line
intersecting the tip of the blade, as shown on
Figure 1.3.
The semi-open impeller is one without a
cover but with a full disk therefore confining the
gas on only one side of the blade.
This type of
wheel usually utilizes only radial blades, shown
on Figure 1.4.
The open impeller has no cover or disk,
but has a small hub at the center of the wheel.
The open wheel makes use of only radial blades, and
is often referred to as a "paddle wheel," shown on
Figure 1. 5.
An
Inducer is used on impellers to act
as a scoop to increase the capacity of a wheel.
Inducers change the inlet velocity vector from one
nearly radial to one parallel to the axis.
Inducers
are sometimes used on closed, backward-leaning
wheels; they are extensively used on semi-open and
5
closed radial bladed impellers.
All wheels-open, semi-open or closed;
backward Jean or radial bladed; with or without
inducer -- have been manufactured using many
fabricating techniques.
The four most common are:
welded
cast
riveted
forged-milled
The Shaft is used to support the wheels
and to deliver the driving power to the disks.
Methods of fastening the wheels to the shaft are
many and varied.
The most common method is the
combining of shrink fit with keys.
On some
compressors, particularly units of low speed wheels
are fastened to shafts with keys and keyways only.
Some designs use a heavy shrink fit without keys.
However, in some instances this has created a
maintenance problem during disassembly and subsequent
reassembly.
"Built-Up Shafts" come in several variations.
The most common considering all types of dynamic
compressors, is the use of several through bolts, a
given distance from the axis of rotation.
This
type of shaft usually makes use of axial keys in
adjacent elements, primarily for center-line
location.
The axial forces (friction) between the
6
elements due to the through bolts, transmit the
driving torque.
Another type of built-up shaft
utilizes one through bolt on the centerline, and
adjoining elements use radial keys or radial splines
for center-line location and torque transmittal,
see Figure 1. 6.
Shaft Sleeves are used between wheels on
the shaft for axial location of wheels, and to afford
wearing surfaces for interstage seals.
In certain
instances shaft sleeves protect the shaft from
corrosive elements in the gas.
Sleeves are also
used in conjunction with shaft seals and occasionally
for bearing surfaces, as in the case of shafts made
of certain stainless steels, shown on Figure 1.7.
The "Balancing Drum" (balance piston) and
thrust disk or collar are the thrust controlling and
locating elements of the rotor.
Thrust is always
developed in an impeller with a resultant in the
direction of the inlet.
(The one exception to this
is a high inlet pressure, low pressure-rise, singlestage overhung wheel with an axial inlet.
Here, it
is poss:i.ble that the push out thrust may exceed the
normal thrust towards the inlet.)
The majority of the thrust (75% is not
uncommon) is "balanced" by the balancing piston.
7
This is accomplished by subjecting the area on the
side of the balancing piston away from the inlet,
to a lower pressure-most commonly inlet pressure,
thereby creating a pressure differential opposite
in direction to that on the impellers.
The low
pressure is achieved by connecting the area behind
the balance piston to the inlet of machine by a
balance piston line, as shown on Figure 1.8.
Balance pistons are not used on singlestage machines nor on double-flow machines incorporating back-to-back wheels.
Thrust collars are
required to be removable by API and are usually
installed so that they bear against a shoulder on
the shaft and are secured by a lock nut.
In order to
bear against a shaft shoulder, rather than transmitting the thrust through the threads of the lock
nut, the thrust collar is preferably on the "inlet
end" of the shaft.
In order not to "drive through"
the thrust bearing and to avoid the associated losses
of a large thrust, the "drive end" of the shaft is
the "discharge end" and therefore the thrust collar
is located on the outboard end of the compressor.
"Shaft Seals" are used both to contain the
gas within the compressor and to prevent contamination of the gas by external elements.
The seals
8
OPEN
Figure 1. 5
Figure 1.6
Figure 1.7
LOCK NUT
THRUST----.. 1
DISC.
SHAFT
COLLAR
SHAFT THRUST
Figure 1.8
9
on both ends operate at inlet pressure due to the
pressure reduction on the discharge end caused by
the balance piston line, as shown on Figure 1.9.
"Journal Bearings" are the support of the
shaft.
When the bearings are out board of the
shaft seals, the compressor is said to have internal
seals.
Bearings under atmospheric environment are
said to be external.
When the seal is outboard
of the bearings the seals are called external seals
and the bearings are said to be pressurized.
All of the parts mounted on the shaft,
together with the shaft, are known as the " Rotor. "
The rotor is one of the two major elements of a
compressor.
The casing and diaphragm assembly are
the other major elements and shall be considered by
its parts from the inlet to the discharge, shown
on Figure 10.
The inlet flange connects the inlet
nozzle to the piping system.
The inlet nozzle
systematically increases the velocity (by reducing
the pressure) of the gas, guides it to the eye of
the first impeller, and distributes the gas around
the 360° of the impeller eye.
The guiding and
distribution of the gas is aided by guide vanes.
These vanes are at times made movable so they can
change the angle of incidence on the first impeller.
10
This can vary the head efficiency and stability of
a compressor.
The effect of movable guide vanes on
a compressor with more than three stages is questionable and probably is an unnecessary added cost over
an inlet throttle valve, shown on Figure 1.10.
The inlet wall is the first diaphragm and
completes the inlet channel of the compressor as
shown on Figure 10.
The face of the inlet wall
opposite the inlet nozzle is one of the two surfaces
which forms the diffuser.
The inlet wall is position-
ed in the casing using a tongue on the diaphragm and
groove in the casing.
The inlet wall also supports
the eye seal for the first wheel.
This seal is
rolled into the inlet wall using a ."T" slot, tongue
and groove, etc; and the labyrinth is generally made
of aluminum.
At times, stainless steels are used
for these seals, see Figure 1.11.
After the gas passes through the first wheel,
it enters the first diffuser.
One wall of this
channel is formed by a face on the inlet wall, the
other wall of this channel is formed by a face on the
first full diaphragm.
The diffuser systematically reduces the
velocity of the gas as it leaves the wheel by increasing the sectional area.
The type illustrated
11
ATMOSPHERE
VENT AREA
SEAL
AREA
t
GAS
AREA
SHAFT
INTERNAL SEALS
SEAL
AREA
BEARING
GAS
AREA
~
\
SHAFT
EXTERNAL SEALS
Figure 1.9
Figure 1.10
12
is a vaneless parallel wall diffuser and although
the axial dimension remains constant, the flow area
increases in the radial direction.
The change in
velocity converts kinetic energy into pressure,
Figure 1. 12.
Another parallel wall diffuser not quite
as common however, is the vaned diffuser.
The vanes
between the diffuser walls, generally increase the
efficiency but reduces the stability of a stage.
After the diffuser, the gas enters the
"Return Bend" or crossover which guides the gas from
the diffuser to the return channel.
The return bend
is formed by adjacent diaphragms and/or the ID of
the casing, as shown on Figure 1.13. The "Return
Channel" guides the gas from the return bend through
the guide vanes into the next wheel, see Figure 1.14.
The return channel includes vanes which
eliminate the swirl caused by the rotation of the
preceeding stage.
As with the guide vanes on the
inlet, these also direct the flow to a prescribed
incidence on the wheel.
Guide vanes are manufactured so that the
vanes introduce the flow with either an incidence
opposite to the rotation, an incidence with the
rotation, or radially.
Various subtle effects on
13
Figure 1.11
Figure 1.12
Figure 1.13
Figure 1.14
Figure 1.15
Figure 1.16
14
head, efficiency and stability are achieved by using
one or the other of these arrangements.
These
guide vanes are extensions of the return channel
vanes and are either cast integrally with the
diaphragm or in a separate casting which is rolled
into the diaphragms.
Adjustable guide vanes on
other than the first stage are extremely rare as
they present certain control
~nd
maintenance
difficulties, see Figure 1.15.
A full diaphragm then is a stationary
element between two stages which forms half of the
diffuser channel of the earlier stage, part or all
of the return bend, all the retun channel and half of
the diffuser of the later stage.
The diaphragm also
supports the shaft seal following the earlier stage
and the eye seal of the later stage.
"Side Loads," as required by refrigeration
machines, are introduced through the cored section
of a diaphragm (called economizer slots) into the
return channels, if the cored portion of the diaphragm
is in contact with the casing.
In the design which
do not have cored sections of the diaphragm adjacent
to the casing, or when the flow is too large to put
through economizer slots, split diaphragms are used.
In the case of high flows, a gap (blank stage) is
15
left on the shaft which allows sufficient axial
space for the side stream to mix with the main
stream, see Figure 1.16.
The part of the casing between the inlet
and discharge nozzles is known as the "barrel
section" (this is not to be confused with a barrel
compressor) . .The barrel section of the casing
supports the diaphragms.
The diaphragms' tongues
are fitted into grooves in the casings.
In many
designs, these casing diaphragm grooves are of
constant spacing within a given frame, therefore,
permitting the changing of flow conditions with the
minimum of effort, see Figure 1.17.
One wall of the last stage diffuser
channel is formed by the diaphragm between the last
stage and the next to last stage.
The other diffuser
wall is formed by a surface integral with the casing
or by a discharge wall which is fitted into the
casing similar to the inlet wall assembly.
The discharge volute collects the gas from
the last stage diffuser and delivers the gas to the
discharge nozzles.
Portions of many discharge
nozzles particularly with single-stage compressors,
are conical diffuser which reduce the speed of the
gas to the final discharge velocity at the flange.
16
Bearing and seal housings, which can be a
single or two separate sub-assemblies, are fastened
to the "end bell" and complete the casing.
The
casing of the horizontally-split compressor has a
flange which fastens the upper half to the lower
half.
This horizontal split is at the center-line
of the machine, and is usually sealed by a metal
joint, but at higher pressures, gaskets of the string
type ( "0"' ring section) are sometimes used, see
Figure 1 .18.
Center-line supports are used for most
all horizontally split compressors.
This means that
the structural members passing the dead weight of
the compressor through its base plate or foundation
to bed rock, are attached to the casing very close
to the casing center-line rather than at the more
convenient bottom of the lower half.
This is done
to allow for expansion in the vertical plane without
disturbing the shaft centerline and thereby destroying the alignment with the driver.
Expansion of compressor casings during
operation is caused by the elevated temperatures
achieved during compression.
On refrigeration
machines the inlet can be considerably below ambient,
while the discharge may be well above ambient, giving
17
an even more complex alighment problem.
Expansion of the casing in the axial
direction is compensated by having a single support
somewhat flexible in the axial direction, while
.considerably rigid in the transverse direction
supporting the compressor on the side opposite the
coupling (driver).
"wobble foot."
This is commonly known as a
When there are two rigid feet
(supports) at the coupling end, any growth due to
differential temperatures would not affect the
drive end (alignment).
Since the differential growth
would "move" only the end opposite the coupling,
shown on Figure 1.19.
Recently a support technique using four
wobble feet, two flexible transversly at the drive
end and two flexible axially at the thrust end with
a "T" locking bar on the thrust end to prevent
"yawing" has allowed for more freedom of casing
expansion, whj_le still maintaining the integrity
of the center-line and, therefore, the surety of
alignment during operation, see Figure 1.20.
Although one design has a horizontallysplit inner casing with the usual grooves to support
to diaphragms, another design uses thru bolts which
held the "stacked" diaphragms together.
The
18
diaphragms of both designs are split on the horizontal center-line.
There are certain maintenance problems
inherent in a machine in which the internals must
be withdrawn from one end.
Care should be exercised
when planning a barrel compressor installation to
insure, if possible, that the compressor (end which
is to be opened) is on the end of the string, and
whether rails to assist the withdrawing of the
intervals are necessary.
19
Figure 1.17
Figure 1.l8
......
I-f--
w//PM
FOUR WOBBLE FEET
Figure 1 .19
SINGLE WOBBLE FOOT Figure 1 · 20
20
1-2:
Barrel Compressors
"Barrel Compressors" are the high-pressure
counter part of the API horizontally-split machines.
These vertically split units, developed as a horizontal joint holding two halves of a cylinder plot
along its axis, will require considerably higher
holding forces than the flanges which hold the top
and bottom onto the same cylinder.
As pressure
increases, or machine size increases, it becomes
progressively more difficult to maintain a gas-tight
seal on the horizontal joint due to the elastic
deformations of the bolting and flanges.
Barrel Compressors are either forged steel,
cast steel, or fabricated steel.
The unit consists
of the barrel section between two end balls.
Some
designs are such that one end ball is either cast
with the barrel section or welded onto the barrel
forging, while other designs bolt both end bells to
barrel flanges, as shown on Figure 1.21 The internals
of a barrel compressor differ from the horizontallysplit unit only in the means of support.
21
BARREL COMPRESSOR
WITH SPLIT:
_r
Figure 1. 21
BARREL COMPRESSOR
WITH STACKED DIAFRAMS
Figure 1.22
. il
__,,
22
1-3:
Physical Control of Compressors
The most universal centrifugal compressors'
performance map would plot capacity versus head and
versus efficiency.
This project will concern itself
with polytropic compression a.nd, therefore, polytropic
head and efficiency.
Figure 1.23 depicts a typical centrifugal
compressor performance map operating at some fixed
speed.
A typical specific performance map plots
capacity in terms of input cfm (cubic feet per minute)
on the abscissa and brake horsepower and discharge
pressure on the ordinate.
The map is usually based
on constant input conditions, such as pressure,
temperature MW and K value of the gas.
In Figure 1.23, the shape of the head curve
is shown for a particular impeller design.
However,
it should be pointed out that centrifugal compressor
impellers can be designed to some extent to achieve
more nearly the sloping characteristics of that
required to suit the system requirements.
For practical reasons, manufacturers
usually will offer standardized impeller design
23
with standardized fram sizes.
A given centrifugal
compressor design ''knows" only its input volume and
its speed.
At a given speed it will develop a
particular head for a given flow throughout.
Consider the input cfm equation:
(1-1)
=
Where:
v1 =
Specific volume at inlet conditions,
3
ft /#
=
R
Universal gas constant
T 1 = Input temperature
=
P1
Input pressure
From equation (1-1) it can be seen that it is possible
to change completely the design inlet gas conditions
and still maintain design inlet cfm.
For all
practical purposes (within the scope of this project)
the polytropic head and efficiency of a given single
stage centrifugal compressor remains the same even
though the design gas and gas conditions have changed
as long as input cfm and speed is obtained.
Consider the polytropic head equation:
K
tp
Hp = ZmRT 1 ( K- 1
P2 oc
) [ (p )
1
Where:
R
Universal gas constant
T
Input. Temperature
1
1]
(1-2)
24
K
=
Specific heat ratio,
p1 =
Input pressure
p2
Discharge pressure
cp cv
Since a given compressor design operating
at design speed will produce design head when passing
design input cfm, the discharge conditions, pressure
and temperature will be determined solely by thermodynamic properties of gas being compressed.
To
illustrate this further, Figure1.24shows the
characteristic curve for a given centrifugal compressor
operating at constant speed but under varying inlet
conditions.
Curve AB represents the characteristic for
a centrifugal compressor designed to handle an input
pressure of 14.5 psia, input temperature is 100°F,
molecular weight of 29.0 (dry air), and K value of
1.40.
This unit would develop 100 percent of
design discharge, pressure at an input capacity of
100 percent design cfm.
If now, the intake air
0
0
temperature is decreased from 100 F to 40 F, all
other conditions remaining the same, the discharge
pressure would increase to 106 percent of design.
25
.......
0
110
H~AO~
100
~ !
:::...,
...,z
- ...
A.-
a:......
......
........
o ...
......
90
0'-'
I
80
I
I
!
...
""
I
60
I
I
110
...a:
...""
a:
i
'
100
i
:
31:
0
"'
0
2:
....z
...
...'-'a:
A.
I
'i
''
!
i
90
I
I
60
so
T
I
l
--7
----,--
! l
I I
I
I
HORSE POW~
r'
.t-
II
I
l
I
80
70
I
I
I
i~i
so
1\
\
I
I
"-z
..,a::
...., ....
a:"-
"\~
./EFFICIENCY
i
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z<-> 70
--~
~
..
v.~
./
i
1
DESIGN
POINT-I--
v
i
zo
30
40
so
60
70
80
90
100
110
120
PERCENT INLET CFM
Figure 1.23 -Typical coastaut speed perlormaace curves for
a ce1Urilugal compressor.
26
130
-
P1•1<4.5 PSIA/
t, •IOO"F
IZO IIW•40.0
K•l.40
~
z 110 P1 •1<4.5 PSIA ~~
2
t 1 : 40"F
\
...c"'
...a:z
...
100
~
u
90
A.
80
iiW:'ZiO
'
K•l.40
'
,
P,•14.5PSIAI
,, :aoo•F
~
I
f A
1
70
1
'1-~ p t•l 2 0 PS-lA'
-,,.IOO"F
IIW• 29.0
~
K• 1.40
r
130
120
......
110
!1:
0
"'a:0
:
..,z
MW•40
100
v
....c
....z
....
'-'
a:
....
A.
90
BO
t 1 •40"F
I
70
I .
v
""
30
40
....,.,..
loo'
y
,/
i
Y:B
YF~,.
[
'
I
I
.
;A
I
!
P1 : 12.0 PSI A
20
~
Yl
I
Xl/1 Yi
~ Y' i / I I
"#("' i
I
i Y·
ry !
:
'
DESIGN AND
I K•I.IO
60
50
i(
!
..,v
--....
y
I
;;;
-~ ~8
I
1<40
....a:
~~
I
'I
IIWO:
29.0
K•LIO
,P, •14.5 PSIA
I\ '• •100°F
IIW•290
K=l.40
l
I
I
I
50
60 . 70
80
90
PERCENT DESIGN INLET CFM
!
I
!
100
110
120
Figure 1 . 24 -Effect of changing gas conditions on centrifugal
compressor operating at constant speed.
27
......,
0
:1:
....
z
...,
u
a:
...,
80
Q.
70
60
50
130
120
....
~
u
70
a:
"'
Q.
60
50
!0~--~~~--~------~----------~--~~~
20
30
40
~0
60
70
80
90
100
110
120
PERCENT INLET CFM
Figure 1. 25.-Typical variable speed perfonnance curves for
centrifugal
compr~ssor.
28
This can be seen by the head equation (1-2)
previously discussed.
The horsepower required under
these conditions would be more than by the ratio of
the absolute temperature
460 + 100 , since by
460 + 40
decreasing temperature we increase density, and,
therefore mass flow for a given volume rate (refer
to equation (1-1).
GHP
( 3300)
1.p
GHP = Gas Horsepower
By lower the input pressure from 14.5 psia to 12.0
psia all other conditions remaining constant, the
discharge pressure would drop to 83 percent of design
by the exact percentage the input pressure dropped,
since the developed pressure ratio has not changed.
The horsepower again would be reduced by the ratio
of the input pressure to that of design
12.0
14.5
since, in this case, the inlet density has been
reduced and, therefore, mass flow for a given volume
rate (refer to equation 1-1).
By changing the composition of the gas
such that the molecular weight increases from 29.0
to 40.0, all other conditions remaining the same,
the discharge pressure would increase to 118 percent
of design.
In this case the horsepower would increase
29
from design horsepower by the ratio of molecular
weights (40/29).
Finally, if the ratio of specific heat
value, K, is decreased from 1.40 to 1.10 all other
conditions remaining the same, the discharge
p:.:essure will increase to 102 percent of design
with no charge whatsoever in compressor horsepower.
30
1-4:
Capacity Limitation
There are definite limitations of the
stability range of a centrifugal compressor.
Limiting the minimum capacity of a given centrifugal
compressor is a phenomenon called "surge" which
normally occurs at about 50 percent of the design
input capacity at design speed.
This extremely
complex phenomenon is probably still one of the most
difficult problems in the field of fluid dynamics.
To provide a simplified explanation, consider the
single stage compressor operating at constant speed,
discharging through a throttling valve.
By throttling
on the discharge valve, we increase the system
resistance and, therefore, the head required by the
compressor to over come this resistance.
As we
continue to throttle this valve, less flow will be
capable of flowing through the compressor.
This
continues up to the point of maximum head capability
of the compressor.
At flows below this "surge"
limit, the compressor head characteristic curve takes
a reverse slope, resulting in decreased head
capability.
In this condition, the system back
pressure exceeds that capable of the compressor
delivery, causing a momentary back flow condition.
31
At this time, however, the back pressure has been
lowered, enabling the unit to again be capable of
delivering flow higher than the flow at which the
surge began.
If the obstruction to flow downstream
of the compressor is unchanged, operation follows
back along the head characteristic curve until the
peak head delivery is reached again.
This cyclic
action is what the industry properly calls "surge."
To operate at flows below the surge flow, requires
controls.
While the stability range of a centrifugal
compressor is commonly indicated from the rated
point to the surge point, the unit can operate stabily
to the right of the rated point.
The greater the
load demand on a centrifugal compressor, the greater
the "fall-off" in delivered pressure.
The upper limit of capacity is determined
by the phenomenon of "stonewall."
Stonewall occurs
when the velocity of the gas approached is sonic
velocity somewhere in the compressor, usually at the
impeller input.
Shock waves result which restrict
the flow, causing, "choking" effect and rapid fall
off in discharge pressure for a slight increase in
volume thruput.
Stonewall is usually not a problem
when compressing air and lighter gases; however, in
32
compressing gases heavier than air, the problem
becomes more prevalent as the molecular weight
increases.
In discussing the operating range of
centrifugal compressors, limitation to single stage
compressors was purposely made for simplicity.
As
the number of stages increases, performance maps tend
to show a more sloping curve with lesser stable
range than shown in Figure 1.23 and 1.24 as dictated
by the particular application.
33
1-5:
Variable Speed Performance
A typical variable speed performance map
is shown in Figure 25.
With variable speed, the
compressor easily can deliver constant capacity at
variable pressure, variable capacity at constant
pressure, or a combination variable capacity and
variable pressure.
Basically, the performance of centrifugal
compressors at speeds other than designed, are such
that the capacity will vary directly as the speed,
the head developed as the square of the speed, and
the required horsepower as the cube of the speed.
Each type of compressor exhibits application
and control problems which are unique.
Centrifugal
compressor surge is unique and is a problem in most
applications.
Surge is an unstable condition within the
compressor blades.
A centrifugal compressor head-
flow curve is typified by the 100% speed line in
Figure 1.26. As flow through the compressor decreases
from maximum, the compression ratio increases to a
maximum and falls off again with further decrease in
flow.
This maximum ratio can occur near 50% of
34
design flow for single stage units, or as high as
70% for multistage units.
When the flow through the
compressor falls below this maximum pressure ratio
point the flow through the blades becomes unstable.
This point is called the surge point of the compressor.
For each compressor speed there is a
characteristic head-flow curve with its characteristic surge point.
The line through the surge points
at various compressor speeds is called the surge
line as shown in Figure 1. 26. This line can be closely
approximated for centrifugal compressor by equation(1-1).
(1-3)
A
=
proportionality constant
T = Suction temperature
1
P = Discharge pressure, PSIA.
2
P + Suction pressure PSIA.
1
P~P 1
K =
=
Compression ratio
~v
for the gas
The unstable flow condition in a centrifugal
compressor is usually noisy and frequently violent
enough to damage the compressor or associated piping.
Therefore, in most compressor applications where
the flow may drop below the surge line, it is
35
important to provide a control system which will
In most cases where it is
protect the compressor.
necessary, it is possible for the
~rocess
to force
the compressor into a surge condition very quickly.
This means that the surge control system must be
designed with maximum emphasis on speed of repsonse.
surge
control
= 105%
7.0
Nl.-;
c... c..
0
·r-1
Design point
at P1&T1
6.0
.j..l
...m
c
,..
5.0
0
/
·r-1
1jJ
1jJ
(!)
...
_.
/
/
/
4.0
/surge
0..
E
0
0
3.0
2.0
1.0
0
20
40
60
80
100
120
Q1 % of Design C.F.M.
Figure 1.26:
TYPICAL COMPRESSOR SURGE
LINE DATA
140
36
1-6:
Compressors Surge Control System
The objective of a compressor surge control
system is to maintain the flow through the compressor
at or slightly above the surge line even though the
process demand may be at or below this level.
This
is accomplished by either blowing off the discharge
flow or recirculating it to the compressor suction.
A typical recirculation system is shown on Figure 1.27.
A valve in the recirculation line regulates
the recirculation flow so that the sum of its flow
and the flow to the process which equals the compressor
flow, is at or above the compressor flow,
above the surge line.
control valve.
is at or
The valve is called the surge
Since the blow-off or recirculated
flow represents wasted power, the control system must
not open the control valve when the process demand
is above the surge level nor should it recirculate
more flow than is required to just hold the compressor
flow at a safe level.
Blow-off systems are used only
for gases such as air, which have very little value
and do not create a hazard when released to the
atmosphere.
Normally, a surge control line is established from one to five percent above the actual surge
37
line to provide for control system overshoot and
recovery.
Control systems which have good dynamic
response, and what is commonly called good start-up
characteristics, permit the surge control line to be
placed very close to the actual surge line.
The actual surge line of centrifugal
compressors are provided by curve on Figure 1.26.
The curves shown on such a graph are
usually based on actual test data and the equation
of the actual surge line is not available.
This
means that the determination of the control systerri
computations required to approximate the surge control
line are most easily found graphically.
Data is
frequently presented in the form of discharge pressure
vs. flow with the test suction pressure stated so that
the data can be replotted as pressure ratio vs. flow.
Suction temperature is also given since it is an
important test parameter and is useful in the system
design, as will be shown later~
On a plot of pressure ratio vs. flow for a
specific compressor the surge control line can be
drawn to pr·ovide the desired safe margin between the
actual surge line and the surge control line.
This
is most easily done by picking and multiplying the
flow value for each point by one plus the safety
38
margin (1 + % margin/100) and replotting the new
flow value at the same pressure ratio.
The resultant
curve is still of the same basic form as equation (1-1)
to design the control system which has the desired
function in equation (1-1)must be converted to terms
of the actual measured variables.
Flow is usually
measured by means of an orifice plate.
Equation ( 1-4)
converts the flow to measured variable terms.
Q1 =
Where:
K~h 1 T 1 Z~P 1
(1-4)
K = proportionality constant
1
=
orifice differential pressure in inches
H 0
2
Combining equations (1-1)and (1-4) can be simplified
as follows:
K-1
K
Q 2 = K2 (h1 T 1
1
1]
)
ZV p 1)
Q 2
2
1 = h1 < k1 z1)
p1
T1
LnT
-
K-1
p
K
2
= A Z [ (~)
a P1
p2
1 = LnZa + Ln[ (p-)
1
-1]
( 1-5)
K-1
K
-1]
When the surge control line from Figure La3is replotted
in terms of P2/
p1
vs ~
as in Figure 1.27, all
39
of the effective variables are shown, and the curve
represents the calculations which must be accomplished by the control system.
By picking two points on
the curve in Figure1.27and substituting their values
in
equation(1-5~wo
equations can be written.
Simultaneously solving these two equations yields a
single value representing all of the exponent
K-1
(~).
The control system may be designed based on the
resulting equation.
surge control
line
/
/
//'-/
7.0
6.0
/
/
/
/
K-1
--=1
K
/
5.0
/.
/
4.0
/
3.0
/
/
/
/
/
/
2.0
/
/
/
1.0
500
1000
1500
2000
2500
3000
Figure 1. 27: Surge Control Line Replotted from
Figure 1.26.
40
As previously stated, the control system must have
good dynamic response.
Of all of the terms repre-
sented in equation (1-5).
The orifice differential
pressure is the most sensitive to process load
changes and most directly related to the compressor
surge characteristics.
Since most analog computation
steps add time constants which may be determined by
the control loop response, the orifice differential
pressure is directly used as the process input to
the controller and all of the other terms applied as
the remote set-point to this controller.
An
additional advantage to this arrangement is that the
controller can be operated using its local set-point
to achieve some surge protection when the more
complex remote set point system is being serviced.
Rearranging the terms of equation (1-3) to
reflect the computational form required for the
set-point to the surge controller yields equation (1-4):
(1-6)
2
which we assume (A Za
K1 2Z1
) = K constant.
The h 1 computed in equation (1-6) is the set-point
value of the controller which adjusts the control
value to hold h , measured in the process equal to
1
41
the computed h .
1
The resulting control system is
shown in Appendix A.
CHAPTER TWO
CONTROL OF PARALLEL COMPRESSORS
42
43
2-1:
Introduction
Every process or unit operation is entitled
to some design attention of its own, no matter how
many times an apparently identical system has been
implemented.
This is true in the design of control
systemsfor centrifugal compressors and doubly so when
those compressors operate in parallel.
Many types of parallel compressor applications exist in the processing industries.
Most
common, perhaps, are refrigeration compressors in
ethylene and propylene plants.
Here, three or more
compressors frequently operate in parallel sharing a
common suction and a common discharge.
These
compressors usually provide suction to a second set
of parallel compressors, in Figure 1, which also share
a common suction and discharge.
In each set, one
compressor remains in standby as a spare with automatic start-up controls.
This type of configuration
contains all the control problems which are uniquely
associated with parallel compressors.
44
spillback valve
Knockout
Drum
Spare
Condensate
Figure (2-1).
Physical Diagram of Parallel-Compressors
45
2-2:
Block Diagram
The follwoing diagram shows the inter-
connection of the three parallel compressors.
There
are three inlet valves which transfer the flow to
the compressors, with a pressure of P. related to
l
the capacity and the specification of the compressors.
These input pressures are P , P and P .
1
3
2
The output
pressure of the three parallel compressors would be
Y , Y , Y .
1
2
3
The input flow for the system is U.
c
u
Flow
Figure (2.2):
Block Diagram of Compressors
Control System
46
2-3:
Surging
Starting up a spare compressor and taking
a compressor out of service both create problems.
Bringing a compressor up to speed when it must pump
against the head established by a parallel compressor
is not only difficult, it is dangerous.
If the
coming up compressor is tied into the discharge manifold too early, it can be thrown into violent surge:
If a compressor is failing or being shut down, bringing up a new compressor too late may cuase the process
to starve for gas because of the failing compressor's
inability to pump.
Conversely, bringing up the spare
and tying it in too early may cause reverse rotation
in the failing compressor.
Controlling input pressure is a common
method for controlling compressor discharge.
In
Figure 29 throttling input pressure by 4.6 psi changes
the output flow from point A to point B without
effecting the discharge pressure .
........~----,output pressure P 2
· nput pressure P1
Flow
Figure (2.3):
Flow vs. Discharge Pressure
CHAPTER THREE
STABILITY ANALYSIS OF THE SYSTEM
WITH CONSTANT INPUT
47
48
3-1:
Introduction
For parallel compressors, the control system
also must be able to bring another compressor on line
in conjunction with operating compressors.
In Figure (2-2) consider three parallel
compressors in series with the fourth compressor.
The transfer function of each compressor in the· open
loop system is first order and type zero and has the
form, G(s)=
K
+ a
s
i.e., U(t)=U 0
If the input flow is constant,
there is no need for a feedback element,
,
but if the input flow U(t) changes with time a feedback element H. must be involved.
l
The addition of the feedback element permits
stable operation and permits the system to follow a
U(t) which changes with time.
The gain of each compressor transfer function
depends on the speed of the compressor, assuming first
that one compressor has speed, K
1
1, and the three
~
others all have unity gain.
For the second case considered, two of the
compressors have variable speeds K1 and K2 which are
If the flow changes in this
case, there is a feedback element H., which is a
l
fraction of
~
U.
49
Case I
G (s)
1
G (s)
3
=
- Flow is constant, U(t)::ouo
K1
s + 0.1
G (s)
2
=
1
s + 0.5
G (s)
4
=
s
s
1
+ 0.2
1
+ 0.25
The following are the impulse responses of the
compressors:
g1(t)
-0.1t
= K1 e
(3-1)
g2(t)
-0.2t
= e
(3-2)
g3(t)
-0.5t
= e
(3-3)
g4(t)
-0.25t
= e
(3-4)
50
3-2:
Stability Analysis of Three Parallel Compressors
The following simulation diagram Figure (3-1)
shows physical variables which appear in the system.
x 1 , x2
In this diagram
x3
,
and
x4
represent the
output pressure of each compressor.
x1
The signal
which represents the input pressure of each compressor
is integrated once.
The summation of the output
pressures of the three parallel compressors is
R
which
is also the input pressure for the fourth compressor.
The simulation diagram.
;J,.~p.ds
.of
Differential Equation
to the Vector Matrix
state for the three parallel
compressors:
•
B u
......
X
_,
A ......
X +
.....
U(t)
Uo u(t)
(3-6)
c
(3-7)
R
-
R
.....
--
X + ....,
D u
.....
(3-5)
,...J
Xl+X2+X3
(3-8)
x1
x1
x2
x2
----;o.
Flow
Transfer
Elements
K'=l
x3
L---
Figure (3.1).
~
I
R
+
x4
>
X4
I
-Y
'------;O.c:>>~-----
x3
o.s,·~------~
Simulation Diagram
Ul
I-'
52
(3-9)
=
(3-10)
=
(3-11)
-0.1
=
0
0
D
R
=
0
0
-0.2
0
1
-0.5
0
Zero Matrix
X
[1,1,1]
l
X~ J
x3
+
0
u
.-
(3-13)
53
3-3:
Controllability of Three Parallel Compressors
To check the controllability of the
three parallel compressors we have to find the
State Controllability Matrix (S.C.M.) which is
indicated as matrix P.
2
This matrix is defined asP= [B,AB,A B ... ,A
n-1
"n" is the dimension of Matrix A.
For the three parallel compressors:
0
-0.1
A
=
-0.2
0
A B
=
-0.5
1
0.01 K
1
-0.2
0.04
-0.5
0.25
-0.1K
p
B =
0
0
0
-0.1 K
0
=
1
1
O.OK
-0.2
0.04
-0.5
0.25
1
1
Now we have to find the determinant of Matrix P;
if it has rank 3 the three parallel-compressor
system is controllable.
B]
54
=
-0.2K (0.25)
1
=
-0.022K
1
(3-14)
This shows that the system is controllable if all
three compressors work with speed
v1
~
0.
If one of the compressors stops or fails, K
1
=
the system is not controllable.
At this point the entire system should be shut
down and all necessary repairs made.
0;
55
3-4:
Output Controllability of Three Parallel
Compressor-System
Definition of Output Controllability:
If there exists a piecewise continuous input flow
U(t) such that it drives R(t
0
)
to any final R (tf)
during the interval which is tf - t
system has controllable output.
0
>
o, then the
By definition:
so the output controllability matrix should have
rank of "n".
2
CAn-1 B, D)
[CB, CAB, CA B
!'1 =
"n" is the dimension of matrix A
. .
c
(3-15)
= [1,1,1]
f.§=
[1,1,1]
=
-0.1
~~=
0
0
-0.2
0
1
0
0.04
0
0
-
0.01K
-0.5
-0.7
Ol
0
A2 B
-0.2
-0.5
-0.1K
0.01
c
,...,
0
0
CA.B
---
2
A B=
0
0~1
~Kll
I1
" ') L·
Vo"-
1
+ 0.29
_l.JI
ra.01KJ
I
0 .04
I 0 ,...,~
L..L:)
I
J
56
=
=
if K
1
0
p1 = [2,
and P
(3-16)
-0.7, 0.29]
has rank 3
1
The above result shows the system is output
controllable; therefore if we shut down one of the
compressors the other two compressors will continue
to operate properly.
3-5(1):
Observability of the Three Parallel
Compressors
By definition, the state X(t ) is observo
able if for any input flow U(t), for an interval at
tf
~
t
0
the knowledge of input U(t), the matrices
,
A, B, C, D and the output R is sufficient to
determine x(tf).
The observability matrix is P
2
which is defined as follows:
P
2
=
[ c' '
A' c' '
(A' ) 2c' '
A' and C' are the transpose matrices of A and C.
0
-0.1
A'
=
0
-0.2
0
In this case A = A'
0
0
0
-0.5
C'=
[~]
57
The observability of the system will be assured
if the matrix P
has rank 3, where P
2
2
is defined
as follows:
-0.1
A' C'
-0.2
-0.5
0.01
(A') 2=A2=
0
0
0
0
0
0.04
0.25
0
0.01
2
(A') C'=
0.04
0.25
1
-0.1
0.01
1
-0.2
0.04
1
-0.5
0.25
-0.012
The matrix P
2
(3-17)
has rank 3; therefore the three
parallel-compressor
system is observable.
58
3-S(II):
Signal Flow Graph
Observability can also be determined
from a signal flow graph of the system.
The block diagram of Figure (3-1) can be
used to construct the signal flow graph of Figure (3-2).
From the signal flow graph it is apparent, by
inspection that the system is completely observable.
Since R is a function of all three state variables.
3-6:
Stability Analysis of Overall System
The transfer function for our first part of
(3-18)
R
-( s)
+
1
1
u
S+0.1
G(s)
(K +2)s +(0.7K +0.9)s+0.1K +0.07
1
1
1
(s+0.1)(s+0.2)(s+0.5)
s + 0.2
+ ( S+ 0. 5)
2
(3-19)
G(s) is the transfer function of the three parallelcompressors system.
system and type zero.
Note that it is third order
""
R
Figure (3.2):
Signal Flow Graph of the Three Parallel-Compressor
System.
Ul
lO
60
The transfer function of the overall system
in the case of one compressor with variable speed
G. ( s)
l
y
= -(s)
u
R
U
-(s)
l
u
y
-( s)
R
(3-20)
Transfer function of the three parallel-
compressor
G. ( s)
R
= -(s)
is:
=
system.
2
(K +2)s + (0.7K +0.9)s + 0.1K 1 +0.07
1
1
(s+O.l)(s+0.2)(s+0.25)(s+0.5)
(s+O.l)(s+0.2)(s+0.25)(s+0.5) = 0
(3-21)
(3-22)
The equation (3-22) shows it is independent of K.;
l
therefore the speed of the compressor has no effect
on the stability of the overall system.
The overall system is stable because all the
poles are located on the right hand side of the S
plane.
CHAPTER FOUR
STABILITY ANALYSIS-CONSTANT INPUT,
TWO COMPRESSORS WITH VARIABLE SPEED
61
62
4-1:
Introduction
In this chapter the controllability,
observability and output controllability of the
overall system will be discussed based on two
compressors operating with variable speeds and the
remaining two compressors operating with constant
speeds.
The following simulation diagram leads
to Vector Matrix Differential Equation of state for
the overall system:
+
U(t)
Flow
K'
Pressure
'----i
Figure (4.1):
0. 5 ! - - - - '
Simulation Diagram of the Overall System
63
x1
=
-0.1X
+ K1Uo
(4-1)
x2
=
-0.2X 2 + K2Uo
(4-2)
=
-0.5X
=
X +X +X -0.24X 4
1 2 3
.
.
x3
x4
1
2
+ u0
(4-3)
(4-4)
x1
y
= [0,0,0,1]
(4-5)
x2
x3
x4
(4-6)
y
Since
(4-7)
= X1+X2+X3
R
Output of the three parallel-compressors system .
.
(4-8)
-0.24Y
•
R - x4
0.24
R
x4
y
(4-9)
Output of the overall system.
x1
f-o.l
0
x2
=
0
0
0
x1
K1
-0.2
0
0
x2
K2
-0.5
0
x3
1
-0.25
x4
0
+
x3
0
0
x4
1
1
1
L
u 0 (4-10)
64
4-2:
Controllability of the Three ParallelCompressors System.
The state equations for the given system
are:
-0.1
A
0
-0.2
0
=
2
[B, AB, A B]
-0.1K
AB =
-0.2K
B
0
= State Controllability Matrix
0.01 K1
1
2
A B
2
0.04 K2
=
-0.5
p
=
=
-0.5
0
0
P
0
0.25
K1
-O.OK 1
0.01K
K2
-0.2K
0.04K
1
-0.5
2
1
2
0.25
The matrix P shows that if one of the compressors
stops (no flow through compressors K or K = 0), the
2
1
matrix P is not of rank 3, and the system is
uncontrollable.
Now assume K
1
~
0, K
2
~
0 find
the determinant of matrix P.
=
+
65
=
(4-11)
The determinant of P is not equal to zero if neither
K nor K is equal to zero.
1
2
Under these conditions
the rank of P is equal to three, and the system
is controllable.
66
4-3:
Output Controllability of the Three ParallelCompressors
If there exists a piecewise continuous input
flow U such that it drives R(t 0 ) to any final R(tf)
......
during the interval, is tf-t >o; then the system
0
is output controllable by definition.
X
=
AX
+
BU
R
=
CX
+
DU
so the output controllability matrix P
1
should have a
rank of n and we will consider this matrix as follows:
[CB, CAB, CA2B ... CAn-1 B ] ,
=
"n" is the dimension of matrix A.
Hint:
-0.1
A
=
0
0
c =
[1,1,1]
=
[1,1,1]
CB
0
0
-0.2
0
K: 1J
[ 2
0
-0.5
=
K1
B
=
K2
1
67
-0.1
AB=
0
-0.2
0
0
0
0
-0.1K 1
0
-0.2K 2
-0.5
-0.5
CAB= (-0.1 K , -0.2K , -0.5)
2
1
0.01
0
0
P1
0
0.04
0
0
K1
0
K2 = 0.01K 1 +0.04K +0.25
2
0.25
1
[K 1 +K 2 +1, -0.1K 1 -0.2K -0.5, 0.1K +0.04K +0.25]
2
1
2
(4-12)
Observing Matrix P :
1
If two of the compressors fail
to operate properly (zero speed at transferring the
flow, K =K =0), the third compressor would be capable
1 2
of maintaining the transmission of flow.
Meanwhile
we would be able to adjust the two failed compressors
to make them functional again.
4-4:
Observability of the Three Parallel-Compressors
System:
Now, consider the observability of the three
parallel-compressors.
By definition, the state X(t )
""
0
is observable if for any input flow U(t), the matrices
A,B,C,D and the output R is sufficient to determine
68
Now, define the observability matrix P
2
as follows:
A' and C' are the transpose matrices of A and C.
0
-0.1
A' =
0
0
-0.2
0
0
C' =
0
-0.5
:1
1J
in our case, A' = A
Our system is observable if rank of
matrix P
is n, which is the dimension of matrix A.
2
.o
0
1
0
0.2
0
1
0
0
-0.1
A'C'=
-0.01
0
0
-0.5
0
0.04
0
1
0
0
0.25
-0.1
=
-0.2
-0.5
69
0.01
0
0
1
0
0.4
0
1
0
0
=
0.25
1
-0.1
0.01
1
-0.2
0.04
1
-0.5
0.25
1
0.01
=
0.04
0.25
The matrix shows the observability of the system
is independent of the spped of the compressors.
Since
!PI =
-0.012
~
P
0,
has rank 3; therefore
the three parallel-compressors system is observable.
The signal flow graph gives the same result
without calculation.
x1
=
-0.1X 1
+
K1 U0
x2 =
-0.2X 2
+
K2 U0
+
u0
The signal flow graph of Figure (4.2) shows R is a
linear combination of all three states x , x 2 and x
1
3
and therefore the system is observable.
R
Figure (4.2):
Signal Flow Graph of the Three Parallel-Compressors System.
-....1
0
71
Stability Analysis of Overall System
4-5:
In this case the transfer function of the
overall system is G. which is defined as follows:
l
G.
l
y
= -(s)
u
+
G.
l
G.
l
(4-13)
=
=
1
1
S+0.5 ]. S+0.25
2
(K +K +1)S +(0.7K +0.6K +0.3)S+0.01K +0.05K +0.02
2
1
1
2
1 2
(S+0.1)(S+Q2)(S+0.25)(S+0.5)
(S+0.1)(s+0.2)(S+0.25)(S+0.5)
=
0
(4-14)
The equation (4-14) shows the stability of the system
is independent of K and K ; therefore the speed of
1
2
the compressors has no effect on the stability of
the overall system.
72
4-6:
Controllability of the Overall System with Two
Compressors having Variable Speeds.
The Differential Equations and State Matrices
of the Overall System are defined as follows:
..
=
-O.SX
3
+ U
0
=
-0.1
x1
0
0
-0.2
x3
0
0
x4
1
1
x2
=
0
0
x1
0
0
X
0
x3
1
-0.25
x4
0
-0.5
1
2
K1
+
K2
u0
73
.
X
=
y
A
X + B U
0
[0,0,0,1]
To check the controllability of our
system in this case, we determine the matrix P 1 :
2
3
P = [B, AB, A B, A B]
1
-0.1
0
0
-0.2
A.B =
0
0
1
1
0.01
2
0
0
K1
-0.1 Kl
0
0
K2
-0.2 K2
0
1
-0.25
0
-0.5
1
0
0
0.04
0
0
0
0
0
0
=
-0.5
I
J
A =A.A=
3
2
A =A.A=
0.25
0
-0.35
-0.45
-0.75
-0.001
0
0
0
0
0
0
0
0.0975
-0.008
0
0 .. 1525
0.0625
-0.125
0
0.4375 -0.0156
K +K +1
1
2
"'"
0.01K
0.04K
2
A B=
-0.001 K
1
-0.008 K
2
3
A B=
2
-0.125
0.25
-0.35K
P.=
'"'l
1
-0.45K
1
2
0.0975K +0.1525K 2 +0.4375
1
-0.75
0.01 K
-0.001 K
K1
-0.1 K
K2
-0.2 K
2
0.04 K
2
-0.008 K
-0.5
0.25
-0.125
I1
0
K +K +1
2
1
1
1
-0.35K -0.45K -0.75
2
1
1
2
0.0975K +0.1525K 2 +0.4375
1
-...]
~
75
Now, to find the determinant of P..
l
If it
has rank 4, the dimension of Matrix A, the overall
system is controllable.
pl
p2
P.
p3
p4
l
11
+p
1
pl
p2
pl
2
p2
2
3
3
pl
p2
4
4
=
l
P.=p
1
a -p
31
12
i -p
1
1
p3
p4
p3
2
p4
2
3
3
c.
p4
4
4
b +pl. c -p d -p e +P2 f -p g +P2 h
14
21
23
3
2
4
32
j
+P3 k -p l
34
3
-p
41
m +p4 n -p4 q +p4 r
2
3
4
For specific details of matrices a,b,c.
Appendix
p3
.
r·
'
see
76
-t
I P.l I
=
If either K or K
2
1
=
0,
IP.l I
=
0; under these
conditions, matrix P. does not have rank 4, and the
l
overall system is uncontrollable.
But if K and
1
K are both non-zero, Pi has rank 4, and the system
2
is controllable.
77
4-7:
Calculation of Position Error Coefficient
The input flow in our system for each period
of time 6t
tf-t
=
0
is constant, and therefore can be
represented in terms of the step function u(t), so
U(t) = U
u(t).
0
The Laplace Transform of u(t) would be:
u(t) -.·
... U(s)
1
U u ( t ) - U (-)
0
0 s
U
has unit
0
The error transfer function of the overall
system is E(s).
E(s)
=
Taking the Laplace Transform of e(t):
U(s)
1 + G(s)
=
u0
s[1+G(s)]
The steady-state error.
u
e
s
(t)
= lim s
[~
s
1
u0
1+G( s) ] = 1+limG(s)
~0
K
K
p
p
Kp
=
lim G(s)
s---.o
=
0.01K +0.05K +0.02
2
1
0.0025
=
2K +10K +4
1
2
0.5
lim G(s), position error coefficient
s~o
=
u0
1+K
p
78
e (t)
s
Steady-State Error.
The above implies that our system is capable of
following a constant input with a constant error
in the steady-state.
u0
. ·.·
e
Uo
s = l+Kp
Kp
l+Kp
Figure (4.3):
y
s
=
Uo
The desired output for the overall
system.
2K +10K +4.5
2
1
0.5
u0
2K +10K +4
1
2
2K +10K +4.5
1
2
u
0
.
79
4-8:
Efficiency of the Overall System
In the previous discussion we found that
if K and K are non-zero, the overall system is
2
1
both stable and controllable.
Now we can assume
different amounts of variable speeds for two compressors of our system and ascertain speed effects
on our output flow.
2K
y
s
=
1
+10K 2 +4
u0
Steady-State output
I
No. of
Cases
Ys
K1
K2
1
0
0
.0.8888Uo
2
1
0
0.9230Uo
3
0
1
0.9655Uo
4
1
1
0.9696Uo
5
5
1
0.9795Uo
6
1
5
0.9911Uo
7
10
0
0.9795Uo
8
0
10
0.9952Uo
9
10
10
0.9959Uo
10
100
100
0.9995Uo
TABLE 1.
Table 1 shows the best output flow for our
system is when both compressors are operating with
This would lead to Y
s
=
0.9696U
0
which
gives the smoothest operation, seeK 1 =1 curve on
80
Figure (4.4) below.
This variable speed does not have a real
effect on the output flow, but it has a direct effect
on input pressure and output pressure.
The following
figure will show the effect of this factor (K. ).
l
The curves on Figure (4.4) are sketched
based on three values of K . This Figure shows that
1
when the compressors operate with low speed K =1,
1
the corresponding curve drops less rapidly than the
curves for K
>
1 -
10.
p2
p1
8
7
6
5
K
4
1
=5
3
2
1
0
Figure
( 4. 4):
1
2
p2
p1
3
4
vs. time t ..
l
5
6
t
CHAPTER FIVE
STABILITY ANALYSIS - VARIABLE INPUT FLOW
WITH ONE COMPRESSOR HAVING VARIABLE SPEED
81
82
5-l:
Introduction
In this case, assume that our input flow
is not constant.
This input flow could be a function
of time, which is defined as U(t)=(t)-nuo,
U(t)=e -nt Uo, u(t)=nt -nUo or ... etc.
Reducing the
input flow at a given time, reduces the input
p2
pressure P , and the ratio PI .
1
an undesirable situation.
Therefore, this is
This could damage the
compressor, in this situation feedback elements
can be used to bring back the input flow to the
normal level, which takes the system out of the
surge point.
A fraction of the output flow is
fed back to the input to stabilize operating condition
and, is particular to remove the danger of reaching
to the surge point.
"'"
xl
+
xl
Pressure
xz
U(t)
Flow
+
Figure (5.1):
x3
xz
y
x3
Simulation Diagram- State Equations.
():)
(N
84
5-2:
Controllability of the Three ParallelCompressor System
In this case assume that one of the
compressors has a variable speed and the rest of
the compressors are operating with constant speed.
The transfer functions of the compressors
are:
G(s)
G. ( s) =
l
1+GH(s)
G ( s) =
1
K1
S+0.1
K1
1+
S+0.1
=
G (s) =
2
1
S+0.2
0.5
1+
S+0.2
=
G (s)
3
G (s) =
4
1
S+0.7
1
S+2
1
S+1.25
U(t) = The Input Flow
Rate
The stateequation of the system can be obtained by
Simulation Diagram Figure (S.1).
=
X2
-0.5X 2 + U(t)
=
=
-l.SX~
~
+ U(t)
85
Thus, the State Equations are:
-1
x1
0
0
x2
•
-0.5
x3
0
0
x4
+1
1
.
y
x4 =
0
0
x1
0
0
x2
-1.5
1
[o,o,0,1]
+ 0
K1
1
+
(5-1)
0
x3
1
-1
x4
0
U(t)
(5-2)
Output Equation
Now we have to consider the controllability for the
three parallel compressors.
- --
P1= [B
~
AB
~
-1
A=
0
0
-0.5
0
AB=
0
[~:~~
-1.5
K1
2
A B=
6-213]
lI
0.25J
2.25
controllability state matrix.
0
0
-1.5
B=
1:1
I
L1
86
Therefore,
p1 =
K1
-K
1
-0.5
0.25
1
-1.5
2.25
K1
1
and
=
(5-3)
=
If K1 the speed of one of the compressors,
approaches zero, the three parallel-compressor
system will not be controllable.
Therefore, in this
case we can use one of the compressors as a spare.
To check the controllability of the
three parallel-compressor
system, the P. matrix must
l
first be evaluated.
2
P. = (CB, CAB, CA B]
l
-1
A
=
0
0
-0.5
0
c
[1' 1' 1]
0
0
0
-1.5
B
87
CB
=
K +2
1
CAB =
-K 1 -2
2
CA B=
[1,1,1]
r Kl ~
l0.25
2.25
=
K +2.5
1
(5-4)
The matrix shows that even if K = 0 and one of the
1
three compressors completely stops, the system would
still be controllable.
P.
1
=
Then,
[2, -2, 2.5]
and P. has rank 3.
1
5-3:
Transfer Function of the Overall System
The following calculations lead to the
transfer function of the overall system with
variable input flow U(t):
G. ( s) = R ( s)
u
J
Transfer Function of the ThreeParallel-Compressor
(5-5)
G. ( s) = G (s)+G (s)+G (s)
1
2
3
J
K1
G. ( s) =
s+0.1+K
J
+
1
1
+
S+0.7
1
S+2
88
G .( s)
=
J
2
2
K1 (s +2.7s+1.4)[s +(K 1 +0.3)s+0.2+2K 1 ]
(s+0.l+K )(s+0.7)(s+2)
1
2
s +(0.8+K )s+0.07+0.7K
1
1
+
(s+0.l+K )(s+0.7)(s+2)
1
2
(K +2)s +(2.7K +K +0.3+0.8+K )s+1.4K
1
1
1 1
1
(s+0.1+K )(s2+2.7s+1.4)
1
+ 0.2+2K 1 +0.07+0.7K 1
2
(5-6)
(s+0.l+K )(s +2.7s+1.4)
1
GT(s) =
GT(s) =
y
-( s)
Transfer Function of the
Overall System
(5-7)
u
y
R
(s )
R
.
(s ) =
u
y
R
(s )
.
G jC s)
(5-8)
Where Gj ( s) is defined in equations (5-5) and
(5-6) .
2
(K +2)s +(4.7K +l.l)s+4.1K +0.27
1
1
1
2
(s+0.1+K )(s +2.7s+1.4)
1
1
S+l.25
2
(K +2)s +(4.7K +1.1)s+4.1K 1 +0.27
1
1
2
(s+0.1+K )(s +2.7s1.4)(s+1.25)
1
The transfer function of the overall system is type
zero and fourth order.
• d
89
5-4:
Controllability of the Overall System
The controllability matrix for the overall
system is defined as follows:
(5-9)
=
-1
0
0
AT =
-0.5
0
0
1
1
0
0
K1
0
0
1
-1.5
1
BT =
0
1
0
-1
The overall system is controllable if the matrix PT
is of rank 4.
-K
1
1
-0.5
ATBT=
-1.5
K +2
1
AT
2
0
0
0.25
0
0
0
0
=
0
0
0
-1.5
2.25
-2.5
0
1
90
-1
0
0
-4
-K
1
0
-0.125
0
0
0
0
-3.375
0
1. 75
4.75
0
-1
1
-0.125
-3.375
(5-10)
-K
-K 1
1
1
-0.5
0.25
-0.125
1
-1.5
2.25
-3.375
0
-4
Evaluation of the determinant of matrix PT
yields the following:
±IPTj
= -1.125K
1
(K +6.5)-0.843K CK +2)-0.75K
1
1
1
1
+0.281K (K +2)
1
1
1
1
+ 0.375~ (K +6.5)+6.75K
1
91
1
4
IPTI=
2
K1
2
-1.125Kl -7.3125
-0.843Kl -1.686Kl-0.75Kl
+0.281K
2
1
2
+0.562K +0.375K
+2.4375K +6.75K
1
1
1
1
2
2
+2.25K 1 +14.625K +0.5K +0.25K
+1.625K
1
1
1
1
2
2
-13.5K -1.5K -9.75K -0.l25K
-0.25K +
1
1
1
1
1
+0.5K
2
1
2
+3.25K +3.375K
+6.75K +6K +0.25K
1
1
1
1
1
2
2
+0.5K 1 -2K -2.25K -4.5K
1
1
1
1.438K
2
1
+3.251K
1
The determinant shows if all compressors operate
with some finite, non-zero speed, the overall
system is controllable, but if one of the compressors
stops (K =0) the overall system is not controllable.
1
It is better to have one compressor as a spare for
this situation.
When one of the compressors stops
or fails, the spare compressor can then be used.
92
5-5:
Stability Analysis of the Overall System
Routh's Criterion:
The characteristic equation for the overall
system, can be found from the transfer function of
the overall system.
Characteristic Equation
2
(K +2)s + (4.7K +1.1)s +4.1K + 0.27
1
1
1
(5-12)
= 0
The following tables show the stability
of the overall system in three different cases
based on three values of K :
1
(I) :
2s
s
s
s
(II) :
s
s
s
2
+ 1.1s + 0.27 = 0
2
1.1
0
2
1
0
0.27
2
1
K1
0
=
K1
stable
0.1485
1
=
3
5.8
8.44
4.37
Stable
93
(III):
K
1
12s
s
s
s
2
1
0
=
2
10
+ 48.1s + 41.27 = 0
12
41.27
48.1
Stable
165.42
The above results show the stability of the overall
system is indepenC'ent of the compressor speed and
if the speed of one of the compressors changes
during the operation it would have no effect on the
stability of the overall system.
Since our overall system is type zero
we have:
Step Error
System·
Type
Coefficient Kp
0
K
Ramp Error
Coefficient Kv
0
0
0
Parabolic
Error
Coefficient Ka
TABLE 2.
2
K
p
Kp
=
=
limGT(s)
s-o
=
lim
s-o
4.1K +0.27
1
1.75K +0.175
1
(K +2)s +(4.7K +1.1)s+4.1K +0.27
1
1
1
2
(s+0.1+K )(s +2.7s+1.4)(s+1.25)
1
(5-13)
94
The step error coefficient Kp is related directly
to the speed of compressors.
Speed
Factor
K1
Error
Coefficient
(See equation (5-13)).
Steady-State
Error
e ( t)
s
K
p
0
1.542
1
2.270
5
2.327
0.3005
10
2.335
0.2998
50
2.341
0.2993
100
TABLE 3.
2.342
=
Uo
1 + Kp
0.3933
Low Speed
High Speed
0.3058
0.2992
As Table 3 shows at low speed and high speed
the value of the error coefficient does not change
much except when the compressor stops; therefore
in this case, the value of error coefficient of the
overall system drops to its lowest value, but the
steady-state error e
s
(t) has the greatest value when
one of the compressors operates with the speed of
zero.
95
BIBLIOGRAPHY ·
Ayres, Frank Jr., Schaum's Outline Series, MATRICES,
1962, pp. 7-10, pp. 20-32, pp. 39-45, pp. 55-63.
Cameron, Joseph and Danowski, Frank M. Jr.,
Elliott Compressor-Materials for Centrifugal
Compressors, April 1976.
D'Azzo and Houpis, Linear Control System Analysis
and Design, 1975, pp. 29-30, p. 68, p. 83,
pp. 92-94, pp. 103-105, pp. 123-125,
pp. 148-153, pp. 169-174, pp. 184-188,
pp. 190-196, p. 210.
Gupta, Someshwar C. and Hasdorff, Lawrence,
Fundamentals of Automatic Control, 1970,
pp. 52-112, pp. 247-285, pp. 474-504.
Ogata, Katsuhiko, State Space Analysis of Control
System, 1976, pp. 8-16, pp. 45-65, pp. 175-203,
pp. 370-418, pp. 460-470.
Tezekjian, E. A., How to Control Centrifugal
Compressors-Elliott Company, 1963 .
• d
APPENDIX
96
97
A.
Schematic Diagram for
a typical Compressor
Feedback line
Input
Flow
y
u
Pressure Transmitter
Flow Transmitter
P (x)
= output pressure of compressor
P 1 ( x)
= input pressure of compressor
y
= output flow
FCV
= Feedback controller
2
ply & p2y & P3Y & p4y & PSY
input & output pressure process
This process has six important quantities.
1.
h
2.
P2
1
P1
input flow
= f(x)
98
3.
[f(x) ]n
n
= (P2)K-1
P1 K
K-1
=I<
power reducer
4.
(P2)K-1 - 1
P1 K
5.
p1 [f 1 (x)]
6.
2
(A Za ) f (x)
K 2z
2
1
1
B.
Typical Simulation Diagram
= [f)x) ]n - 1 = f 1 (x)
= f2 ( x)
= Kf 2 (x)
Before drawing the overall system and
writing down the state equations for each compressor
and later for the overall system we have to consider
the following:
U(t)
1
y
T
S±a
Block Diagram
Integration Element
99
C.
Cofactor 3 x 3 Matrices
P2
a=
p3
p4
P2
C=
p3
p4
p1
e=
p3
p4
p1
g=
p3
p4
i=
2
2
2
1
1
1
2
2
2
1
1
1
P2
p3
3
3
p4
P2
p3
p4
3
P2
p3
p4
p1
p3
p4
p1
p3
p4
~12
p1
I P22
P2
~42
p4
2
2
2
3
3
3
2
2
2
3
P2
p3
p4
p1
p3
p4
p1
p3
p4
P2
4
b=
4
p3
p4
4
P2
4
d=
4
p3
p4
4
p1
4
f=
4
p3
p4
4
p1
4
h=
4
p3
p4
4
p1
p1:l
4 '
1
1
1
1
P2
p3
p4
P2
3
3
3
2
P~
1
1
1
1
1
1
1
1
1
-.)2
p4
p1
2
3
p3
P241
3
p4J
j=
p2
p4
1
1
p3
p4
P2
p3
p4
p1
p3
3
p4
p1
p3
p4
p1
3
2
2
2
3
I
3
P2
P2
3
p4
p4
p1
p3
4
4
3
3
3
4
4
4
3
3
p43J
p14l
p24J
p4
3
4
4
100
p1
k=
p2
p4
1
1
1
p1
p2
p4
2
2
2
p1
p2
p4
p1
4
l=
p2
4
p4
4
1
1
1
p1
p2
p4
2
2
2
p1
p2
p4
3
3
3
r
p1
m=
p2
p3
p1
q=
p2
p3
2
2
2
1
1
1
p1
p2
p3
p1
p2
p3
3
3
3
2
2
2
p1
p2
p3
p1
p2
p3
4
n=
4
I p1 1
p1
p2
p2
p3
4
1
1
p·1
4
1
r=
4
4
p2
p3
1
1
p3
p1
p2
p3
3
3
3
2
2
2
p1
p2
p3
p1
p2
p3
4
4
4
3
3
3
101
D.
Table of Laplace Transform Pairs
f(t)
F(s)
6(t)
1
-1
1 or u(t) unit step
s
1
s
tu(t), ramp function
2
1
s
1
1
'(-n--~~)~!
n
1
s
unit impulse
e
-as
1
-s + a
1
(s+a)n
tn-1 u(t)' n lS
. a pos1·t·1ve
integer
u(t-a), unit step starting at t =a
a
-at
exponential decay
1
n-1 -at
(n-1)! t
e
n is a positive
integer