Topic 4

Controller Actions And Tuning

Chemical Processes

Self-regulating Integrating

Process Dynamics

SS Gain, Kp

Deadtime, θLag, τ

Process Dynamics

Gain, Kp,integrate

Deadtime, θ

In the last lecture

What We Will Cover

Topic 1

Introduction To Process Control

Topic 2

Introduction To Process Dynamics

Topic 3

Plant Testing And Data Analysis

Topic 5Enhanced

Regulatory Control Strategies

Topic 6

Process Control Hardware Systems

Topic 4

Controller Actions And Tuning

Topic 7

Control Valves

Topic 8

Process Control Troubleshooting

In This Lecture…

Controller Actions

Controller Modes

Proportional Control

Problems of Proportional-Only Control

Feedback Controller PID controller is the most common type

Sole purpose is to adjust an MV in order to bring a CV (PV) as close to SP in as short a time as possible

The extent at which the controller adjusts the MV depends on the PID tuning constants– Kc – Proportional Term

– τI – Integral Term (min)

– τD – Derivative Term (min)

Controller Modes Manual (MAN)

– Operator changes OP “manually”– SP not used for control

Auto (AUTO)– Operator changes SP– Controller adjusts OP “automatically” to bring PV to SP

Cascade (CAS)– Secondary controller on CAS and Primary controller on AUTO– Primary’s OP “cascaded” down to secondary’s SP– Secondary adjusts own OP to bring PV to SP

Consider this....

Fin fluctuates; SP fixed at 50% When level = 10%, we want valve full close (OP = 0%) When level = 90%, we want valve full open (OP = 100%) Level in between 10 and 90%, valve will be partially open This is an example of proportional-only control

Fin

LC

OP

SP = 50%

Essence Of P-Action

Take drastic action when far away from SP

Go easy when close to SP

OP = Kc x Error + Bias– Kc = controller gain– Error = SP - PV (depends on manufacturer)– Bias = some constant

Good? Yes, but….

An example

DeltaP fluctuates so flow fluctuates if loop is on MAN Let’s say we now have a flow rate of PV=SP=500 BD, and at that

flow rate, OP = 40% (i.e. valve is 40% open)

OP = Kc x Error + Bias

40 = Kc x 0 + Bias

Bias = 40

FCInstrument range

0~1000 BD

Pressure Drop,Delta P

SP = 500 BD(Barrels per Day)

An example We now want to control the flow at 600 BD (Operator

increase SP from 500 to 600)

Assume Kc = 0.5, so OP = 0.5(Error) + Bias

The controller detects an error of (600-500)/1000 = 10%

New OP = 0.5(10)+40 = 45%, so valve opens to 45%

What will be the new flow rate?– 3 possibilities: 500 < Flow < 600; 600 exactly; >600– Unlikely to get exactly 600 BD

An example Let’s say at 45%, the PV= 562.5 BD

– Error = (600 – 562.5)/1000 = 3.75%

Error has now decreased from 10% to 3.75%– New OP = 0.5(3.75%) + 40% = 41.88%– New flow = 523.4 BD

Recalculate Error and OP, and observe flow– Error = (600 – 523.4)/1000 = 7.66%– New OP = 0.5(7.66%) + 40% = 43.83%– New flow = 547.9 BD

This cycle will repeat itself Finally it will settle at a steady value, BUT

P-Only Control ResponseFlow Rate

400

450

500

550

600

650

0 5 10 15 20 25 30

Time (Sec)

Flow

(BD

)

39.0%

44.0%

49.0%

54.0%

59.0%

SP

PV

OP

Problem With P-Only Control

It will not settle at 600 BD

There will always be an Offset (SP-final value that PV settles at)

Offset can be reduced by higher controller gain, Kc

But that can result in more drastic cycling before the PV settles down

In This Lecture…

Controller Actions

Proportional Control

Problems of Proportional-Only Control

In The Next Lecture… Integral Control

– Equation– How it works– Interaction with Proportional Action– Problems with Integral Action

Derivative Control– Equation– How it works– Problems with Derivative Action