Lecture 30 Ratio, Feed Forward, Cascade Control Process Control - - PowerPoint PPT Presentation

Lecture 30 Ratio, Feed Forward, Cascade Control

Process Control

• Prof. Kannan M. Moudgalya

IIT Bombay Monday, 28 October 2013

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Outline

• 1. Feed forward control
• 2. Cascade control
• 3. Cascade control of a web server

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Ratio Control

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Ratio Control

◮ Mainly flow rates ◮ Control ratio, with respect to feed ◮ Example: distillation column

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Ratio Control of Distillation Column

? LT PT Feed Coolant Exit Reflux, R Bottoms: B, xB Distillate: D, xD Heat hD hB AT AT LT

◮ Figure shows manipulating flow rates directly ◮ Instead, manipulate Reflux Ratio and other

ratios

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Ratio control: read from the book

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• 1. Feed Forward Control

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FF Control Law (SS) Derivation

◮ F = D + B ◮ Fz = Dy + Bx ◮ What should D be for changes in F, z,

given that we want x = xsp, y = ysp?

◮ Fz = Dy + (F − D)x ◮ Fz − Fx = D(y − x) ◮ D = F(z − x)/(y − x) ◮ Control law: D = F(z − xsp)/(ysp − xsp)

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Recall the Mixing Process

x, w Valve Control Pure A x2 = 1 w2 =? Mixture A, B x1, w1

◮ Mixing of two

streams

◮ Variable stream

has composition (x1) varying

◮ i.e. x1 is

disturbance

◮ Want output

composition constant

◮ Control stream’s

flow can be changed

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Recall: Mass Balance of the Mixing Process

x, w Valve Control Pure A x2 = 1 w2 =? Mixture A, B x1, w1

◮ Overflow: w1 + w2 = w ◮ w1x1 + w2x2 = wx ◮ = (w1 + w2)x ◮ w2 = w1

x − x1 x2 − x

◮ Derive feed forward

control law:

◮ w2(t) = w1

xsp − x1(t) x2 − xsp

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Feed Forward Control of Mixing Process

AC Pure A x2 = 1 w2 =? x, w AT Mixture A, B x1, w1 Valve Control

◮ w2(t) =

w1 xsp − x1(t) x2 − xsp x2 has been shown to be 1 in the fig-

• ure. This is not used

in the above expres- sion.

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Blending System: Ex. 15.5 of Textbook

w2m FFC xsp x1m w2,sp

p

AT Mixture A, B x1, w1 x, w FT FC

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Optional Assignment

◮ Example 15.5 of the textbook works out in

detail feed forward and feedback control strategies of the blending system

◮ Compares the efficacy of these strategies ◮ You will have to work these calculations out in

detail, including Scilab code, simulation results, plots, etc.

◮ A surprise!

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• 2. Cascade Control

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Motivation for cascade control

◮ Some times the response may be slow ◮ Do not want to introduce feed forward control

• may not want to model the disturbance

transfer function exactly

◮ Possible to overcome this through another

control loop - called slave control

◮ The set point for this is given by the outer

control

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Application from furnace control

◮ Furnace is used to heat oil ◮ By changing the fuel gas ◮ One option is to change the gas flow rate

directly to obtain correct heating

◮ Fuel gas supply pressure could upset the

calculations

◮ Tell what the desired pressure is ◮ Use it to regulate the fuel gas flow rate

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When does cascade control work?

◮ When inner loop is a lot faster than the outer

loop

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Problem from Final Exam, 2009 I

This problem is concerned with a scheme, known as cascade control, shown below:

Ysp −

Y

Kc1 Kc2 Gm 1 s + 1

1 (2s + 1)(4s + 1)

−

• 1. Using Routh-Hurwitz approach, determine the

range of proportional controller gain Kc1 for which the closed loop system is stable. Take Kc2 = 1, Gm = 1.

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Problem from Final Exam, 2009 II

• 2. Repeat the above for Kc2 = 4, Gm = 1.
• 3. Determine the range of Kc1 values for which

the conventional control scheme, obtained by letting Kc2 = 1 and Gm = 0, is stable.

• 4. Compare the three controllers obtained above.

Which is preferable and why?

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Example on Cascade Control Discussed in Slide 20, Lecture 8

Ysp −

Y

Kc1 Kc2 Gm 1 s + 1

1 (2s + 1)(4s + 1)

−

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Example on Cascade Control

Ysp −

Y

Kc1 Kc2 Gm 1 s + 1

1 (2s + 1)(4s + 1)

− ◮ Find the range of proportional controller gain

Kc1 for which the closed loop system is stable. Take Kc2 = 1, Gm = 1.

◮ Repeat for Kc2 = 4, Gm = 1. ◮ Find the range of Kc1 values for which the

conventional control scheme, obtained by letting Kc2 = 1 and Gm = 0, is stable.

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Solution to Example

• 1. Inner loop transfer function = 1/(s + 2)

Characteristic equation of closed loop: 8s3+22s2+13s+(2+Kc1) = 0, 0 < Kc1 < 33.75

• 2. Inner loop transfer function = 0.8/(0.2s + 1)

Characteristic equation:

1.6s3 + 9.2s2 + 6.2s + (1 + 0.8Kc1) = 0, 0 < Kc1 < 43.31

• 3. 8s3+14s2+7s+(Kc1+1) = 0, 0 < Kc1 < 11.25

So, answers to the example are 33.75, 43.31, 11.25

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• 3. Cascade control of a web server

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Application from CS

◮ Title: Feedback based distributed admission

control in 802.11 WLANs

◮ Authors: Preetam Patil, Vipul Mathur, Varsha

Apte, and Kannan Moudgalya

◮ Conference: ◮ The 34th Annual IEEE Conference on Local

Computer Networks (LCN)

◮ Place: Zurich, Switzerland ◮ Date: 20-23 Oct 2009

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Connection admission control in WLAN

◮ Wireless LAN is an ubiquitous entity ◮ It should admit a large number of

applications/people (flows) to be active at any time

◮ Quality of service should be maintained

◮ Response time should be reasonable ◮ Should not crash because of admittance of

excessive number of applications on the LAN

◮ It is better restrict the number of applications,

rather than letting it crash - restart time could be enormous

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Connection Admission Control

◮ Ultimate control action is to regulate the

number of admitted flows at a station

◮ A measure of admitted flows is provided by the

utilisation threshold, θ

◮ Depending on the difference between the set

point of this threshold and the actual value, flows are increased or decreased

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WLAN control

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WLAN control

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Connection Admission Cascade Control

◮ Master controller K1 decides the desired

threshold value

◮ Slave controller K2 changes the number of

flows and helps achieve this threshold

◮ G2 denotes the transfer function between flows

and threshold

◮ Inner loop (K2, G2) is faster than outer loop

(K1,G1)

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What we learnt today

◮ Ratio control ◮ Feed forward control ◮ Cascade control ◮ A web server application

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Thank you

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