ASSIGNMENT PROCESS CONTROL 2014LECTURER: DR THAM HENG JINLAB DEMOSTRATOR: MISS HELENSTUDENTS : GROUP 3KEVIN CH'NG JUN YAN, TANG SZU YOU, CHANDRA BINTI RAMLAN, NORAIN BINTI NORDIN, SITI AISHAH BINTI MOHAMED SALLEHASSIGNMENT PROCESS CONTROL 2014LECTURER: DR THAM HENG JINLAB DEMOSTRATOR: MISS HELENSTUDENTS : GROUP 3KEVIN CH'NG JUN YAN, TANG SZU YOU, CHANDRA BINTI RAMLAN, NORAIN BINTI NORDIN, SITI AISHAH BINTI MOHAMED SALLEH
ASSIGNMENT PROCESS CONTROL 2014
LECTURER: DR THAM HENG JIN
LAB DEMOSTRATOR: MISS HELEN
STUDENTS : GROUP 3
KEVIN CH'NG JUN YAN, TANG SZU YOU, CHANDRA BINTI RAMLAN, NORAIN BINTI NORDIN, SITI AISHAH BINTI MOHAMED SALLEH
ASSIGNMENT PROCESS CONTROL 2014
LECTURER: DR THAM HENG JIN
LAB DEMOSTRATOR: MISS HELEN
STUDENTS : GROUP 3
KEVIN CH'NG JUN YAN, TANG SZU YOU, CHANDRA BINTI RAMLAN, NORAIN BINTI NORDIN, SITI AISHAH BINTI MOHAMED SALLEH
QUESTION 1
A system has a second order process with:
Gp s= eθs(5s+1)(3s+1)
Assuming Gm = Gv = 1, design P, PI and PID controller using continuous cycling method (Table 11.4) for case (i) θ = 0, (ii) θ = 2. Comment on the controllers performances with different controller settings. How are the performances if modified IMC tuning relation for SOPTD (Table 11.1) is used to set the controller settings (θ = 2)?
Apply as well ZNFOPTD tuning as given in the table, try comment on the responses and performances of the controller when different cases (i) θ = 5, (ii) θ = 10 (iii) θ = 0.4 are applied.
SOLUTION:
BLOCK DIAGRAM
(i) θ = 0, Continuous Cycling Method
In fact, there are some shortcomings of continuous cycling method :
Time consuming for processes with large time constants
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Process pushed to stability limit
Ultimate gain does not exist for first and secondorder systems without time delays
Generally not applicable to integrating and openloop stable systems because stability is achieved only for intermediate Kv values
Hence, continuous cycling method is not applicable for θ = 0,
(ii) θ = 2, Continuous Cycling MethodZieglerNichols
Find Kcu
When it is continuous cycling
Kcu is 5.05585
Pu is 13.27
Find PID – Kc, I, D Value
Controller P: Kc=0.5Kcu
Kc=0.55.05585=2.5279
Controller PI: Kc=0.45Kcu
Kc=0.45(5.05585)= 2.2751
τI= Pu1.2=13.271.2 = 11.0583
I=KcτI = 2.275111.0583 = 0.2057
Controller PID: Kc=0.6Kcu
Kc=0.65.05585=3.0335
τI=Pu2 = 13.272 = 6.636
I=KcτI =3.03356.636 = 0.4571
τD=Pu8 =13.278 = 1.6588
D=KcτD = (3.0335)(1.6588)
= 5.0320
ZieglerNichols
Kc
τI
I
τD
D
P
2.5279




PI
2.2751
11.0583
0.2057


PID
3.0335
6.636
0.4571
1.6588
5.032
GRAPHRESULT OF ZN CONTROLLER
P Controller:
PI Controller:
PID Controller:
IMC TUNING θ = 2
Modified IMC tuning relation for SOPTD methods:
Model used  K(τ3S+1)eθs(τ1S+1)(τ2S+1),only PID Controller setting available :
τ3=0, K =1,
According to (Skogestad,2003), τc = θ=2.
KCK= τ1+τ2+ τ3 τc+ θ = 5+3+02+2 = 2 , KC=2
τI= τ1+τ2 τ3
= 5 + 3 – 0 = 8
I = KCτI = 28 = 0.25
τD= τ1τ2(τ1+τ2 τ3)τ3τ1+τ2 τ3 = 5×3(5+30)(0)5+30 = 1.875
D = KCτD = 3.75
GRAPHRESULT OF IMC CONTROLLER
PID Controller:
COMPARISON OF IMC AND ZN CONTROLLER
ZN PID IMC PID
DISCUSSION
Compare 2 methods of controller when θ = 2
IMC Controller react faster than ZN Controller in PID controller.
IMPORTANT NOTES
Smaller τI, higher oscillation degree, smaller error, but too high, it will cause slow response of output
Higher Kc, lower offset, faster response to reach steady state, but too high, it will cause undesirable degree of oscillation.
ZNFOPTD – BLOCK DIAGRAM
Taylor's Reduction, Gp= eθs(5s+1)(3s+1)
= e(θ+3)s(5s+1)
ZNFOPTD – θ = 5
For P controller,
KCK = τtd = 58=0.625 , Kc =0.625
For PI controller,
KCK = 0.9τtd = 0.9 ×58=0.5625 , Kc = 0.5625
τI=3.3 td = 26.4 , I = 0.5625/26.4 = 0.02131
For PID controller,
KCK = 1.2τtd = 1.2 ×58=0.75, Kc = 0.75
τI=2 td = 16, I = 0.75/16 = 0.04688
τD=0.5 td = 4, D = 0.75 x 4 = 3.0
GRAPHRESULT OF ZNFOPTD – θ = 5
PCon. PICon PIDCon.
*PICon reach steady state at 430s
*PIDCon reach steady state at 180s
ZNFOPTD – θ = 10
Calculation method same as θ = 5
ZieglerNichols
Kc
τI
I
τD
D
P
0.3846




PI
0.3462
42.9
0.00807


PID
0.4615
26
0.01775
6.5
3.0
GRAPHRESULT OF ZNFOPTD – θ = 10
PCon. PICon PIDCon.
*PICon reach steady state at 800s
*PIDCon reach steady state at 380s
ZNFOPTD – θ = 0.4
Calculation method same as θ = 5
ZieglerNichols
Kc
τI
I
τD
D
P
1.471




PI
1.3235
11.22
0.1181


PID
1.7647
6.8
0.2595
1.7
3
GRAPHRESULT OF ZNFOPTD – θ = 0.4
PCon. PICon PIDCon.
*PICon reach steady state at 70s
*PIDCon reach steady state at 40s
JUSTIFICATION
PController : Offset present
PIController: Eliminate offset, but more oscillatory
PIDController: Response faster, less oscillatory, but noise present. The random shape of output because the PID controller more depend on Kc and D values.
Different of θ affects the output :
Higher θ(time delay), Slower response of output, Longer time to reach steady state for ZNFOPTD
END OF QUESTION 1
QUESTION 2
Considering a blending system with a feedback control which is used to reduce the effect of disturbances on the inlet feed composition. The transfer functions of the process are represented by:
Gv s= 187.50.0833s+1 Gp s= 2.6 × 1044.71s+1 Gm=32es Gd(s)=0.654.71s+1
Let G = GvGpGm and assuming Km=1, develop the standard block diagram for the feedback control system above. Design P, PI and PID controller using (i) ZieglerNichols ultimategain method (Table 11.4) (ii) Direct synthesis method, with τc=1.57(How does τc value setting affect the response?)
Comment on the performances on these controllers by simulating the closedloop responses to a unit step change in the disturbance. Show the block diagram, all calculations and plots.
SOLUTION:
SCHEMATIC DIAGRAM
* The disturbance is on the inlet feed composition
* Assume the blending System occur in Tank
OutletOutletInletInletDisturbanceDisturbanceMeasurement InstrumentMeasurement InstrumentControllerControllerGv, ValveGv, ValveTANKTANK
Outlet
Outlet
Inlet
Inlet
Disturbance
Disturbance
Measurement Instrument
Measurement Instrument
Controller
Controller
Gv, Valve
Gv, Valve
TANK
TANK
BLOCK DIAGRAM 1
GdGdGcGcGvGv
Gd
Gd
Gc
Gc
Gv
Gv
GpGp
Gp
Gp
GmGm
Gm
Gm
Gc = PID Controller
Gv = Transfer Function of Valve
Gd = Disturbance
Gp = Transfer Function of Process
Gm = Transfer Function of Measurement Instrument
Km = Gain of Gm
However, the question is required to use empirical relation to guess and tune the controller.
What is empirical relation?
Combine Ga,Gp and Gm to G and then introduce disturbance after G transfer function.
Advantages of empirical relation
Quick and easy for guess the PID controller
Disadvantages of empirical relation
Cannot dictate system dynamic response specification.
Controller settings are seldom optimal and most often required field tuning after installation to meet more precise dynamic response specifications.
G = GvGpGmG = GvGpGmBLOCK DIAGRAM OF EMPIRICAL RELATION
G = GvGpGm
G = GvGpGm
GcGc
Gc
Gc
CLOSED LOOP TRANSFER FUNCTION
Mathematic Modeling of Closed Loop
Y=GcG1+GcGYsp+ Gd1+GcG D
Since G=GvGpGm
G =( 187.50.0833S+1) x (2.6×1041+4.71S) x (32es)
= 1.56es0.0833S+1(1+4.71s) = 1.56(1+4.71s) × es(0.0833s+1)
= 1.56es(1+4.71s) 1(0.0833s+1)
*1+GcG cannot equal to 0
(i) ZieglerNichols ultimategain method (Table 11.4)
ZieglerNichols
Kc
τI
τD
P
0.5Kcu


PI
0.45Kcu
Pu1.2

PID
0.6Kcu
Pu2
Pu8
STEP 1 : Set Kc as smaller value (e,g.0.5), I=0, D=0
STEP 2 : Introduce a small change to Kc until continuous cycling occurs. Hence this value of Kc is called as ultimate gain, Kcu. The period is ultimate period.Pu
STEP 3: Calculate PID Controller using the ZieglerNichols (ZN) tuning relations I table above
STEP 4 : Evaluate
Find Kcu
When it is continuous cycling
Kcu is 4.895
Pu is 4.0
Find PID – Kc, I, D Value
Controller P: Kc=0.5Kcu
Kc=0.54.895=2.4475
Controller PI: Kc=0.45Kcu
Kc=0.45(4.895)= 2.2028
τI= Pu1.2=4.01.2 = 3.3333
I=KcτI = 2.20283.3333 = 0.661
Controller PID: Kc=0.6Kcu
Kc=0.64.895=2.937
τI=Pu2 = 4.02 = 2.0
I=KcτI =2.9372.0 = 1.4685
τD=Pu8 =4.08 = 0.5
D=KcτD = (2.9202)(0.5) = 1.4685
ZieglerNichols
Kc
τI
I
τD
D
P
2.4475




PI
2.2028
3.3333
0.6610


PID
2.9370
2.0
1.4685
0.5
1.4685
GRAPHRESULT OF ZN CONTROLLER
Controller P:
Controller PI:
Controller PID:
Comparison of 3 controller setting:
PCon. PICon PIDCon.
DISCUSSION
Controller P :
1) Offset happen and could not able to eliminate.
2) It is because it use the errors between inputs at 2 different time to improve the output performance, hence once input reach steady state, there is no different between in inputs no matter how the time going. As a result, the output has a steady state error all the time. It required integral function or differential function to eliminate steady state error.
3) We can try to increase the Kc in order to get greater result (nearer to 1), however, there is a limit for it since it will go for diverge at certain point.
Controller PI:
Larger overshoot
Slower than PID and P to reach steady state, but higher accuracy than P
Smaller oscillation than PID during steady state, but error bigger than PID during oscillation [if consider noise of PID as oscillation]
Controller PID:
Fastest response to reach steady state
Noise.
(1) and (2) because the differential function improve the output by providing prediction with the linear extrapolation over time Td.
RECOMMENDATION
Nodisturbance – PI controller better, since less oscillation, relatively stable during steady state
Disturbance – PID controller, response to disturbance change faster and better.
How if we didt use Empirical Relation Method, we use block diagram 1(pg ) to tune controller?
The kcu u found will be slightly different with E.R., hence controller setting is also slight different even apply same theory (NZ method). However, it is believed more precise than E.R. in dynamic response observation.
(ii) Direct synthesis method
It is considered as a second plus time delay model
K eθsτ1s+1(τ2s+1)=1.56 es0.0833s+1(4.71s+1)
K = 1.56, θ= 1, τ1 = 0.0833, τ2 = 4.71
Kc=1K τ1+ τ2τc+ θ =11.56 0.0833+4.711.57+ 1 = 1.196
τI= τ1+ τ2= 0.0833 + 4.71 = 4.793
I=KcτI= 1.1964.793= 0.2495
τD= τ1τ2τ1+ τ2= 0.0833 ×4.714.793=0.0819
D=Kc×τD=1.196 x 0.0819=0.09795
Guess τc = 0, 1.57 and 3.0 to analyze the different
τc
Kc
τI
I
τD
D
0
3.073
4.793
0.6411
0.0819
0.2517
1.57
1.196
4.793
0.2495
0.0819
0.09795
3.0
0.768
4.793
0.1602
0.0819
0.06290
GRAPHRESULT OF DIRECTSYNTHESIS METHOD
For τc =0,
P Controller:
PI Controller:
PID Controller:
Comparison of P,PI,PID Controller :
For τc =1.57,
P Controller:
PI Controller:
PID Controller:
Oscillating pattern is same as PID controller of τc =0
Comparison of P,PI,PID Controller :
PCon. PICon PIDCon.
For τc =3.0,
P Controller
PI Controller
PID Controller
Oscillating pattern is same as PID controller of τc =0
Comparison of P,PI,PID Controller :
Comparison of 3 different τc of PID Controller:
PCon. PICon PIDCon.
Justification:
The closedLoop responses in Figure above are more sluggish and less oscillatory for τc =3.0. It will make the reaction to slow down. Again, the overshoot for τc =3.0 is smaller for the setpoint change. Among all the 3 τcvalue, the PI and PID controller will not show any changes. In this system, blending process acts as an integrator.
When τI increase, integral, I value will be decreased, the ability of the integral to accumulate the error decrease. This will make the controller become less effective. Decreasing τI tends to speeds up the response. When τI is held constant, the effect of Kc on the higherorder system can be summarized as follow:
Values of Kc
ClosedLoop Response
Small
Oscillatory
Moderate or large
Overdamped
Very large
Oscillatory or unstable
When τd increase, the time of the derivation action will be increased, and the number of oscillation will be decreased. So, the reaction time is slower to eliminate error. However, when τdis held constant, the D value decreased when τcincrease, the oscillatory becomes less and more stable.
END OF QUESTION 2