02 - PID Training - Process Dynamics

8/16/2019 02 - PID Training - Process Dynamics

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PROCESS CONTROL

THEORYFUNDAMENTAL PRINCIPLES



• All Processes are dynamic

• i.e. they change with time.

• If a plant were totally static with respect to all it’s variables, then control

would be easy

• But in reality the dynamics are constantly changing

• We will define the basic terms and show how they relate to process response

• The change with respect to time of the process variables such astemperature, pressure, flow, composition etc, due to

• controlled changes e .g. feed rate, temperature, pressure etc

• uncontrolled changes e.g. ambient temperature, feed composition etc

• Understanding process dynamics is fundamental for achieving good control

PROCESS DYNAMICS



• Process Response

• Process Gain

• Process Deadtime

• Process Lag

• Order

• Linearity

• Non-self regulating systems

• Plant tests

PROCESS DYNAMICSTERMS



Feed flow rate

Fuel gas valve position

Time

Time

Coil Outlet temperature

Fuel gas

PID OPEN LOOP



• Dynamic response of a process can usually be characterized by 3parameters :

• Process Gain

• Kp or G and is the Change in the process variable divided by thechange in the manipulated variable. Expressed in Engineering units

• Deadtime

• DT or !, the time between MV changing and a noticeable changein PV. Expressed in minutes

• Lag

• T1 or " Effects the rate at which the PV responds to an MVchange. Expressed in minutes

PROCESS RESPONSECHARACTERIZATION



• Laplace Transforms are simply amathematical technique which can expressequations in the time domain

• This allows straight-forward calculations tobe carried out instead of solving complexdifferential equations

• Format is easily understood

• Let me illustrate by some examples

LAPLACETRANSFORM

Simple Laplace transform techniques shall be detailed here. L(f(t)) = !f(t)e-stdt ; explain simple di" erential equations.Provide introduction to state space representation. Also talk on block diagram and its reductionExample on few simple laplace: A, e-at, t, F’Example on block diagram reduction



LAPLACE FORMAT - 1ORDER

Process Gainof 3.5

Process lag is 1storder and 20 min

Process Deadtimeof 5 min

Laplace on control domain - I order di" erential equation.#dC/dt + Cin = CoutLaplace of this becomesCout = Cin/(1+#s)



Process Gainof 0.5 Process lag is 2nd

order and consists of2 lags each 10 min long

Process Deadtimeof 2 min

Process Leadof 5 min

( )( ) s

e

s s

s 2

110110

155.0

!

++

+

LAPLACE FORMAT - IIORDER

If the 2 time constants di" er much, then this process can be assumed as I order.



Process Gainof 20 Process lag is 2nd

order and consists of2 lags each 10 min long(alternative representation)

No Process Deadtime

Process Leadof 0.5 min

s

e

s s

s 0

2120100

15.00.20

!

++

+

LAPLACE FORMAT - IIORDER

Talk on laplace of P, I and D controllers.For P, G(s) = Kc = Ps/EsFor P&I, G(s) = Kc(1+1/"1s) = Ps/EsFor P&D, G(s) = Kc(1+ "Ds) = Ps/EsFor PID, G(s) = Kc(1+1/"1s+ "Ds) = Ps/EsExplain on closed loop block diagramC/R = GcGp/(1+GcGp) - basis for PID loop tuning



At time, t, h = height of liquid in vessel

h = H exp (-t/ !)

When t = !

h = H.exp(-1) = 0.368 H

i.e. level has fallen by (1-0.368) = 0.632 H

Time

H

H1

1 ORDER SYSTEM



Time

H

Time

H Notice Apparent deadtime

Top tank is 1st order as before

2

1

Critically damped system ("1 = "2)

II ORDER SYSTEMEQUAL TIME CONSTANTS



Over damped system ("1 > "2)

Time

H

Time

H Looks like 1st order

Top tank is much slower

t2 is the dominant time constant

SLOW

FAST2

1

II ORDER SYSTEM

#1 > #2



Under damped system ("1 < "2)

Time

H

Time

H Notice inverse response, level actually rises

initially. This confuses a simple feedback

controller system

Imagine Top tank now empties very fast

FAST

SLOW2

1

II ORDER SYSTEMT1 < T2

Explain typical underdamped process response and its characterOvershoot = exp (-$% / & 1- %2)Settling time = 4"/ % - 98%Decay ratio = OS2



Time

H

No. 6 Looks like

DT+1st order Lag!

due to series of lags

1

2

3

4

6

5

MANY TANKS INSERIES



First order with Deadtime can approximate most of the processFirst order model approximation of 20th order system is shown below

Time

20th order system

Deadtime + 1st order lag

model of system

I ORDER + DT

What does a real plant response look like?

Real plants exhibit very high order dynamics (i.e. many lags in series) We can approximate a high order system such as the 20 order one above as a 1st order system with

deadtime with a pretty good fit

Work out an example: Y/X = 3/(0.1s+1)(0.5s+1)(s+1)(3s+1)reduced form: Y/X = 3exp(-1.6s)/(3s+1)plot the response and try compare it.

Pade approximation for dead time shall also be explained here: 1 order: (1+tds/2)/(1-tds/2)Example:Y/X = exp(-3s)/(10s+1)

with first order pade, it is = (1.5s+1)/(10s+1)(1-1.5s), plot the response and see



Time

100%

63.2%

0%

High-order system

Approximated to first order

deadtime plus lag

% of finalvalue

DERIVING PROCESSDYNAMICS

The dynamics may be derived in two ways

The first is to approximate the curve to one of deadtime and 1st order lag and to draw a tangent at the point of steepest slope. The point of intersection on the X-axis marks

the end of the deadtime and start of the lag period



Time

100%

0%

20%

80%

T20 T80

Deadtime = t20 - 0.161 (t80-t20)

Lag = 0.721 (t80-t20)

Steady-state value

DERIVING PROCESS DYNAMICSWITH THE 20%/80% METHOD

Another and more easy method is the 80/20 method developed by Honeywell Hi-Spec.

It uses a simple empirical formula to calculate the dynamics directly.

Simply measure 20% and 80% of the final value on the Y-axis and then determine the corresponding times. Then use theformula above



EFFECT OF VARYING

DYNAMIC PARAMETERS

Increasingdeadtime

Increasing Gain

IncreasingTime Constant

IncreasingOrder

Order = 0

! = 0



INVERSE RESPONSE (1)

PC



INVERSE RESPONSE (2)

Time

Time

Valve

Position

Pressure

Inverse Response



STABLE / INTEGRATINGPROCESS

Ti

Fuel gas

Self-regulating Process Non-Self Regulating Process

If increase opening of fuel gas control valve,

then coil outlet temperature will rise and move

to a new steady-state value

If increase opening of outlet valve, then

level will fall and continue to fall without

reaching a new steady-state value

LI

FC



LINEARITY

• A linear system has a constant process gain

• Most processes in refining, chemicals etc are atleast slightly non-linear

• This has implications for controller tuning(need to tune for different conditions)

• Implications for plant tests or step tests

• Some processes are known to be highly non-linear e.g. pH control



LINEAR / NON LINEAR

Process Variable, PV

Manipulated

Variable, MV

Linear

Highly non-lineare.g. pH

pH = -log10 [H+]

(Note a system is linear if

dPV/dMV is constant)



PLANT TESTS,GUIDELINES 1

• Choose the right time

- Not at a shift changeover

- Not during a feed switch

- Not at dawn or dusk

- Not on Monday morning!

- Not on Friday afternoon!

• Set up sampling/lab analysis if required

• Talk about test with operations and get approval for test

• Ensure instrumentation is OK

• Ensure process is steady

Open loop response is required for getting accurate tuning parameters. With out step test, is it possible deriving open loop response from closed loop data?Yes...explain it through bock diagram.



PLANT TESTS,GUIDELINES 2

• Open required loop, make step change of say 2-5% valvemovement

• Record trend of OP and PV

• Reach new steady state some people use !+5"• Carry out test again in opposite direction, double amount if

possible

• Ensure Manipulated variable really changes

• Document results

• Do tests under all modes of operations

• Step away from any constraints on the unit for safety



TYPICAL PLANT TEST

Time

MV CV

Time



CONCLUSION

• Understanding dynamics is the key to

controlling any process

• Simple dynamics can be represented by

gain, deadtime and lag

• These may be determined by carrying out

plant step tests

Documents

02 - PID Training - Process Dynamics