Compressor Control
Whitehouse Consulting
Redway House, East Lane, Merstone, Isle of Wight, PO30 3DJ, United Kingdom
tel:
+44 (0)1983 529931
fax:
+44 (0)1983 530651
e-mail:
mykeking@compuserve.com
web site: www.whitehouse-consulting.com
Contents
• Compressor types
• Polytropic head
• Stonewall
• Surge
• Load control (flow or pressure)
• Surge protection
• Parallel compressors
2
1
Compressor types
• Positive displacement
– reciprocating or rotary
– trap gas and force into smaller volume, raising
pressure
– high pressures, relatively low flows
• Turbo-compressors
– impeller imparts velocity to gas which converts to
pressure rise by passing gas into smaller volume
– centrifugal - gas moves radially
– axial - gas moves parallel to shaft
– multi-stage machines common, with inter-cooling
– moderate pressures, high flow rates
3
Polytropic head
•
•
•
•
Isentropic compression - constant entropy
Polytropic compression - entropy changes
Polytropic head is the work done on the gas
For an ideal gas
Boyle's Law: PV  constant
V
Charles ' Law:
 constant
T
PV
 constant ( R )
T
R = “universal gas constant”
PsVs PdVd

Ts
Td
4
2
Polytropic head
• Reversible adiabatic (isentropic) compression
PV k  constant
k = “adiabatic index”
k
Cp
Cv

specific heat at constant pressure
 1 .4
specific heat at constant volume
PsVsk  PdVdk
5
Polytropic head
• Combining
k
k
Pd  Vs   Td  k 1
   
Ps  Vd   Ts 
• Polytropic compression
Pd  Vs

Ps  Vd
n
  Td
  
  Ts



n
n 1
nk
• Occurs because compression is
– either non-adiabatic (often applies to reciprocating
machines with cylinder coolers)
– or irreversible (often applies to turbo-compressors
due to thermal effect of friction and turbulence)
6
3
Polytropic efficiency
• Polytropic efficiency (p) describes how close
compression is to isentropic
n
k
p
n 1
k 1
Pd  Td

Ps  Ts
 pk
 k 1


7
Polytropic head
• Head is work done per unit weight of gas
d
W   V .dP
s
1
d

1
W  PsnVs  P n .dP
PV n  constant  PsVsn
s
n 
W  Ps Vs
P
n 1 
1
n
n 1
n
d
 Ps
n 1
n


n  Pd

  PsVs
 Ps
1
n






n 1
n

 1


8
4
Polytropic head
• Because gases are not ideal
PV  RTz
R = 8.3144 kJ/kg-mole/K
= 8314.4 Nm/kg-mole/K
= 1.9859 BTU/lb-mole/OF
z = compressibility
• Polytropic head given by
k z s RT s
H p p
k 1 M

 Pd
 Ps

k 1

 k p
  1



M = molecular weight
9
Compressor capacity limits
• Stonewall
– upper capacity limit - gas velocity (relative to the
impeller) approaching speed of sound
– no harm in operating at this point
• Surge
– lower capacity limit - impeller discharge pressure
temporarily drops below discharge piping pressure
– large flow and pressure oscillations
– risk of mechanical damage
10
5
Compressor performance
11
Compressor performance curves
6000
8000 rpm
polytropic head (kJ/kg)
5000
7300 rpm
surge
6600 rpm
4000
5900 rpm
3000
process curve
5200 rpm
surge
2000
stonewall
1000
stonewall
0
0
50
100
150
200
250
300
350
400
suction flow (actual m3/min)
12
6
Compressor
mol wt
13
Effect of discharge control valve p
35
30
pressure (barg)
25
compressor discharge
control
valve p
20
15
process inlet
10
process exit
5
0
0
100
200
300
400
500
600
700
800
900
flow (F nm /min)
3
14
7
Use of linear discharge control valve
900
800
flow (F nm3/min)
700
600
actual p
500
fixed p
400
300
200
100
0
0
10
20
30
40
50
60
70
80
90
100
valve position (V % open)
15
Compensation for compressor curve
100
90
80
signal to valve (%)
70
or use an "equal percentage"
control valve
60
50
40
include in control system as continuous
function or as a look-up table
30
20
10
0
0
10
20
30
40
50
60
flow controller output (%)
70
80
90
100
16
8
Equal percentage valve
100
90
flow (% of maximum)
80
70
same increase in valve opening also gives 50% increase in flow
60
50
40
30
increase in valve opening gives 50% increase in flow
20
10
0
0
10
20
30
40
50
60
70
80
90
valve position (% open)
100
17
Equal percentage valve
• For equal percentage valve, equal increments of valve
position (V) produce equal percentage changes flow
(F)
F
 V
F
ln( F )  k .V  A
dF
 k .dV
F
F  e ( k .V  A)
Fmax  e ( k .100 A)
F
e ( k .V  A)
 ( k .100 A)  e k (V 100)
Fmax e
• k is valve constant
18
9
Equal percentage valve
• Characterisation adjusted slightly to ensure F is zero
when V is zero
• Alternatively equal percentage behaviour can be
approximated by conditioning the controller output
(M) before applying it to a linear valve, e.g.
V
100kM
100  (1  k ) M
19
Control valve types
100
90
quick opening
% of maximum flow
80
70
k=0.05
60
k=0.02
k=0.01
50
k=0 (linear)
40
k=0.01
30
k=0.02
20
k=0.05
10
equal percentage
0
0
10
20
30
40
50
60
70
80
90
100
valve position (V % open)
20
10
Case Study Exercise 10.1
Plot the process curve on the compressor performance map. Ensure
all controls are on manual and the recycles shut. Fully open the
discharge valve (HC2OP). Vary the speed and record the suction flow.
Repeat with HC2OP set at 20% and at 30%.
speed
FI17
HC2OP=100%
FI17
HC2OP=20%
FI17
HC2OP=30%
5200
5900
6600
7300
8000
What problem do you foresee in terms of tuning FC18 to manipulate
discharge throttling?
21
At constant speed, vary discharge valve position over its feasible range
and plot FC18PV against HC2OP. Under what conditions would it best
to tune FC18?
Switch the discharge valve to non-linear, enter a value for k and again plot
FC18PV against HC2OP. Does this help? Try different values for k .
At constant discharge valve position, vary speed over its feasible range
and plot FC18PV against SC1PV. Would there be a similar problem in
tuning FC18 to manipulate speed?
22
11
Compressor performance curves
6000
8000 rpm
polytropic head (kJ/kg)
5000
7300 rpm
6600 rpm
4000
5900 rpm
3000
5200 rpm
surge
2000
1000
stonewall
0
0
50
100
150
200
250
300
350
400
suction flow (actual m3/min)
23
Effect of discharge valve
900
800
flow (FC18PV nm3/min)
700
600
500
400
300
200
100
0
0
10
20
30
40
50
60
70
80
90
100
discharge valve position (HC2OP %)
24
12
Effect of compressor speed
900
800
flow (FC18PV nm3/min)
700
600
500
400
300
200
100
0
5200
5600
6000
6400
6800
7200
7600
8000
speed (SC1OP rpm)
25
Case Study Exercise 10.1
Plot the process curve on the compressor performance map. Ensure
all controls are on manual and the recycles shut. Fully open the
discharge valve (HC2OP). Vary the speed and record the suction flow.
Repeat with HC2OP set at 20% and at 30%.
speed
FI17
HC2OP=100%
FI17
HC2OP=20%
FI17
HC2OP=30%
5200
195.7
5900
237.1
117.7
162.2
6600
265.7
158.0
206.5
7300
287.2
190.3
237.7
8000
302.6
213.7
258.5
102.4
What problem do you foresee in terms of tuning FC18 to manipulate
discharge throttling?
26
13
Compressor performance curves
6000
8000 rpm
5000
HC2OP = 20%
7300 rpm
polytropic head (kJ/kg)
HC2OP = 30%
6600 rpm
4000
5900 rpm
3000
5200 rpm
HC2OP = 100%
surge
2000
1000
stonewall
0
0
50
100
150
200
250
300
350
400
suction flow (actual m3/min)
27
Effect of discharge valve
900
SC1SP = 8000
800
SC1SP = 6600
700
flow (FC18PV nm3/min)
SC1SP = 7300
SC1SP = 5900
600
SC1SP = 5200
500
400
300
surge
200
100
0
0
10
20
30
40
50
60
70
80
90
100
discharge valve position (HC2OP %)
28
14
Use of non-linear valve
900
800
600
500
k = 0.95
400
k = 0.90
k = 0.75
k = 0.50
k=0
300
200
SC1SP = 6600 rpm
100
0
0
10
20
30
40
50
60
70
80
90
100
discharge valve position (HC2OP %)
29
Use of non-linear valve
100
80
valve position
flow (FC18PV nm3/min)
700
60
40
20
0
0
20
40
60
controller output
80
100
30
15
Effect of compressor speed
900
800
HC2OP = 100%
700
flow (FC18PV nm3/min)
HC2OP = 30%
600
HC2OP = 20%
500
400
surge
300
200
100
0
5200
5600
6000
6400
6800
7200
7600
8000
speed (SC1OP rpm)
31
Load control
• Load control may be FC or PC, several MVs possible
turbo-machines
discharge throttling
suction throttling
inlet guide-vanes
recycle manipulation
speed control
reciprocating machines
suction throttling
cylinder unloading
recycle manipulation
speed control
• Each has different rangeability and power
requirements
• Common to have combination of schemes, e.g. if
operating turbo-machines close to surge
32
16
Discharge throttling
33
Suction throttling
34
17
Inlet guide-vanes
35
Recycle manipulation
36
18
Speed control
37
Reciprocating machines
• Reciprocating machines not prone to surge
• Some load control schemes similar to turbo-machines
– discharge throttling has no effect on flow (just
reduces efficiency)
– suction throttling effective
– inlet guide-vanes not applicable
– recycle manipulation effective but inefficient
– speed control effective
38
19
Reciprocating machines
• Can also adjust cylinder loading
– on multi-cylinder machines, keep one or more of
inlet valves open (for all or part of compression
stroke)
– vary stroke length
– adjust compression volume
• opening/closing chambers within cylinder
• adjustable fixed piston
39
Reciprocating machines
discharge
D
C
pressure of gas in cylinder
A
inlet valve opens
B100 inlet valve closes (max load)
B25
A
TDC
B50
B75
C
exhaust valve opens
D
exhaust valve closes
B100
suction
volume of gas in cylinder
BDC
40
20
Case Study Exercise 10.2
Obtain the dynamic behaviour of FC18 with respect to each
manipulated variable and tune each controller as P+I only.
Kp


ts
Kc
Ti
HC2
HC1
SC1
Test each of your controllers, particularly at low gas rates.
Adjust tuning by trial and error as necessary.
Investigate the impact each controller has on power consumption.
Initialise and put all the controllers on manual. Increase the speed
to maximum, reduce the guide vane angle to minimum and fully
open the discharge valve. Configure FC18 to manipulate the
the discharge valve. Put it on automatic and vary the flow.
Plot power consumption versus flow.
41
Compressor power
6
5
power (MW)
4
3
2
1
0
0
100
200
300
400
500
600
700
800
900
1000
flow (nm3/min)
42
21
Repeat the exercise with FC18 manipulating speed.
Repeat again with FC18 manipulating guide vanes.
Test recycle as a possible manipulated variable. Put FC18 on
manual. Run the compressor at maximum speed, with the
discharge valve fully open and the inlet guide-vanes at the
minimum angle. Vary the anti-surge recycle valve position
and plot power versus FC18.
Identify the advantages and disadvantages of each flow control
scheme.
43
Case Study Exercise 10.2
Obtain the dynamic behaviour of FC18 with respect to each
manipulated variable and tune each controller as P+I only.
Kp


ts
Kc
Ti
HC2
2.90
0
0.49
2
0.8
0.4
HC1
0.25
0
1.02
2
10.6
0.7
SC1
0.75
0
1.50
2
3.4
1.1
Test each of your controllers, particularly at low gas rates.
Adjust tuning by trial and error as necessary.
Investigate the impact each controller has on power consumption.
Initialise and put all the controllers on manual. Increase the speed
to maximum, reduce the guide vane angle to minimum and fully
open the discharge valve. Configure FC18 to manipulate the
the discharge valve. Put it on automatic and vary the flow.
Plot power consumption versus flow.
44
22
Compressor power
6
5
discharge
throttling
3
recycle
manipulation
inlet
guide-vanes
2
speed
control
1
0
0
100
200
300
400
500
600
700
800
900
1000
flow (nm3/min)
45
Impact of discharge throttling
6
6
5
Discharge throttling causes
Pd to rise and so increases
Hp - but not linearly
4
4
3
3
Power consumption (Hp x gas flow)
therefore can pass through a maximum
2
2
1
power consumption (MW)
5
polytropic head (Hp MJ/kg)
power (MW)
4
1
0
0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
gas flow (kg/s)
46
23
Anti-surge control
• Unnecessary recycling is costly
• Objective is to predict where surge will occur based
on available measurements and recycle only as
necessary
• Relatively simple for constant speed machines
compressing gas of constant molecular weight
– either keep flow above pre-determined minimum
– or keep discharge pressure below pre-determined
maximum
47
Anti-surge control (constant speed machine)
48
24
Anti-surge control (constant speed machine)
49
Anti-surge control
• Many schemes available for more complex situations
but performance very machine dependent
• Surge line can usually be described by
H p  f (F 2 )
• Since F  f ( orifice dP ) then H p  f (dP )
• Schemes based on simplification of
H p  m.dP  c
• Common to use PID controller with
SP  H p
PV  m.dP  c
50
25
Anti-surge control
• If surge is approached controller will recycle to keep
operation on the surge line
• Usual to build in a safety margin equivalent to about
15% on flow
• Achieved by basing m on 1.152dP
• Requires collection of operating data when machine
is close to surge
51
Anti-surge control
• Schemes include
SP
PV
Pd  Ps
m.dP  c
Pd
Ps
m
dP
c
Ps
speed
m dP  c
power
m dP.Ps  c
52
26
Anti-surge control
53
Anti-surge control
• Need to know PV range for controller tuning
PVmin  m.dPmin  c
PVmax  m.dPmax  c
PVrange  PVmax  PVmin  m(dPmax  dPmin )
• Remember to use absolute units for pressure and
temperature
– calculation of m and c
– calculation of PV range
54
27
Anti-surge control
• More rigorous approach
• For orifice
F  cd
d 2
2dP
4

• At compressor suction
Fs2 
• Where
dP
s
Fs = actual volumetric flow
dP = orifice pressure drop
s = gas density
55
Anti-surge control
s 
 Fs2 
MPs
Ts z s
dP.Ts z s
M .Ps
• Polytropic head

k z s RTs  Pd

H p p
k  1 M  Ps




k 1
k p

 1


56
28
Anti-surge control
• Remember
Td  Pd 
 
Ts  Ps 
T
log d
k 1
 Ts


 pk
P
log d
 Ps
k 1
 pk
• Surge line slope
P
log d
Hp
 Ps

Fs2
T
log d
 Ts








 Ps  Td  1


 dP  Ts 


57
Case Study Exercise 10.3
Determine tuning for:
Kp


ts
Kc
Ti
dP
Pd
Put FC18 on manual and close the recycles. At a number of speeds
close the discharge valve as far as possible, avoiding surge. Record
the operating conditions on Table 1 and calculate the parameters.
Plot H p versus F s 2 for each approach and determine slope (m) and
intercept (c) . Obtain the dynamics for each and tune.
2
Hp
Fs
Pd -Ps
1.15 dP
m
P d /P s
1.15 dP/P s
c
Kp


ts
Kc
Ti
Kp


ts
Kc
Ti
2
2
0.5
speed
1.15dP
power
1.15(dP.P s )
0.5
Determine tuning for:
H p /F s
2
58
29
Table 1
speed
5200
5600
6000
6400
6800
7200
7600
8000
HC2OP
FI17 (dP)
PC3 (P s )
TI6 (T s )
PI4 (P d )
TI7 (T d )
power
Pd -Ps
2
1.15 dP
P d /P s
2
1.15 dP/P s
1.15dP
0.5
1.15(dP.P s )
H p /F s
0.5
2
59
Surge line
50
45
40
Pd - Ps
35
30
25
20
15
10
5
10
12
14
16
18
20
22
24
1.152dP
60
30
Surge line
16
14
Pd/Ps
12
10
8
6
4
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
1.152dP/Ps
61
Surge line
8500
8000
7500
speed
7000
6500
6000
5500
5000
4500
3.0
3.2
3.4
3.6
3.8
4.0
4.2
4.4
4.6
4.8
5.0
1.15dP0.5
62
31
Surge line
3.0
2.5
power
2.0
1.5
1.0
0.5
5.5
6.0
6.5
7.0
7.5
8.0
8.5
1.15(dP.Ps)0.5
63
Set up FC18 so that it manipulates the discharge valve. Test each
anti-surge scheme in turn by reducing FC18 below surge flow and
ensuring the anti-surge recycle opens properly. Adjust parameters as
necessary. Vary compressor speed and ensure operation tracks close to
the surge line. Repeat the same test at different suction and reactor
pressures.
Commission RC18 with a setpoint of 2. Vary the gas molecular weight by
increasing FC15 to about 90 sm3/min. Again check whether the anti-surge
schemes cope with the change.
64
32
Case Study Exercise 10.3
Determine tuning for:
Kp


ts
Kc
Ti
dP
0.30
0.0
0.2
2
5.4
0.12
Pd
1.09
0.0
0.2
2
1.6
0.15
Put FC18 on manual and close the recycles. At a number of speeds
close the discharge valve as far as possible, avoiding surge. Record
the operating conditions on Table 1 and calculate the parameters.
Plot H p versus F s 2 for each approach and determine slope (m) and
intercept (c) . Obtain the dynamics for each and tune.
2
Hp
Qs
Pd -Ps
1.15 dP
P d /P s
1.15 dP/P s
2
2
0.5
speed
1.15dP
power
1.15(dP.P s )
0.5
m
c
Kp


ts
Kc
Ti
2.86
-22.3
0.36
0.0
0.2
2
4.6
0.12
2.86
-6.40
0.12
0.0
0.2
2
13.7
0.12
2186
-2585
0.54
0.0
0.2
2
3.0
0.12
0.77
-3.62
0.32
0.0
0.2
2
5.3
0.12
Determine tuning for:
H p /Q s
2
Kp


ts
Kc
Ti
0.52
0.0
0.2
2
3.3
0.13
65
Table 1
speed
5200
5600
6000
6400
6800
7200
7600
8000
HC2OP
28.2
21.4
17.7
15.3
13.6
12.3
11.4
10.8
FI17 (dP)
9.1
10.6
12.0
13.3
14.4
15.3
16.1
17.0
PC3 (P s )
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
TI6 (T s )
30.0
30.0
30.0
30.0
30.0
30.0
30.0
30.0
PI4 (P d )
15.7
19.8
24.2
28.7
33.2
37.6
41.6
45.1
TI7 (T d )
219.2
250.9
280.0
306.5
330.2
351.0
368.7
383.2
power
1.07
1.35
1.63
1.90
2.15
2.36
2.56
2.74
Pd - Ps
13.7
17.8
22.2
26.7
31.2
35.6
39.6
43.1
1.15 dP
12.0
14.0
15.9
17.6
19.0
20.2
21.3
22.5
P d /P s
5.5
6.9
8.4
9.9
11.4
12.8
14.1
15.3
3.99
4.65
5.27
5.84
6.32
6.72
7.07
7.46
3.47
3.74
3.98
4.19
4.36
4.50
4.61
4.74
6.02
6.50
6.92
7.28
7.58
7.81
8.01
8.23
0.730
0.731
0.731
0.730
0.731
0.737
0.738
0.729
2
2
1.15 dP/P s
1.15dP
0.5
1.15(dP.P s )
H p /F s
2
0.5
66
33
Surge line
50
45
40
Pd - Ps
35
30
25
20
15
y = 2.864x - 22.3
10
5
10
12
14
16
18
20
22
24
1.15 2dP
67
Surge line
16
14
Pd/Ps
12
10
8
y = 2.864x - 6.4007
6
4
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
1.152dP/Ps
68
34
Surge line
8500
8000
7500
speed
7000
6500
6000
y = 2186.3x - 2584.8
5500
5000
4500
3.0
3.2
3.4
3.6
3.8
4.0
4.2
4.4
4.6
4.8
5.0
1.15dP0.5
69
Surge line
3.0
2.5
power
2.0
1.5
y = 0.7662x - 3.6178
1.0
0.5
5.5
6.0
6.5
7.0
7.5
8.0
8.5
1.15(dP.Ps)0.5
70
35
Anti-surge control
71
Anti-surge control
• Anti-surge control will interact with other controls
– opening recycle will reduce discharge pressure
and hence flow to process
– flow (or pressure) controller will compensate and
interact with anti-surge control
– variety of modifications which detune flow (or
pressure) control when anti-surge control acts
• Opening recycle around first stage of multi-stage
machine will starve other stages of gas
– can feedforward to compensate
72
36
Parallel compressors
• Compressor curves never identical
• If curves are flat then small pressure changes will
cause large flow changes
– one machine may take the full load while another is
close to surge
– load may swing between machines
– difficulty in starting machines
• Need to balance machines (approximately equal
distance from surge and equal power consumption)
73
Compressor balancing
74
37