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 1/38 Process Control Feed Forward Control

Outline 1. Feed forward control 2. Cascade control 3. Cascade control of a web server 2/38 Process Control Feed Forward Control

Ratio Control 3/38 Process Control Feed Forward Control

Ratio Control ◮ Mainly flow rates ◮ Control ratio, with respect to feed ◮ Example: distillation column 4/38 Process Control Feed Forward Control

Ratio Control of Distillation Column PT LT h D Coolant Exit AT Reflux, R Distillate: D , x D Feed Heat ? h B LT AT Bottoms: B , x B ◮ Figure shows manipulating flow rates directly ◮ Instead, manipulate Reflux Ratio and other ratios 5/38 Process Control Feed Forward Control

Ratio control: read from the book 6/38 Process Control Feed Forward Control

1. Feed Forward Control 7/38 Process Control Feed Forward Control

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 = x sp , y = y sp ? ◮ Fz = Dy + (F − D)x ◮ Fz − Fx = D(y − x) ◮ D = F(z − x) / (y − x) ◮ Control law: D = F(z − x sp ) / (y sp − x sp ) 13/38 Process Control Feed Forward Control

Recall the Mixing Process ◮ Mixing of two streams ◮ Variable stream Control Mixture Valve has composition A, B (x 1 ) varying x 1 , w 1 ◮ i.e. x 1 is disturbance Pure A ◮ Want output x 2 = 1 composition w 2 =? constant ◮ Control stream’s x , w flow can be changed 14/38 Process Control Feed Forward Control

Recall: Mass Balance of the Mixing Process ◮ Overflow: w 1 + w 2 = w ◮ w 1 x 1 + w 2 x 2 = wx Control ◮ = (w 1 + w 2 )x Mixture Valve x − x 1 A, B ◮ w 2 = w 1 x 1 , w 1 x 2 − x ◮ Derive feed forward Pure A control law: x 2 = 1 x sp − x 1 (t) w 2 =? ◮ w 2 (t) = w 1 x 2 − x sp x , w 15/38 Process Control Feed Forward Control

Feed Forward Control of Mixing Process AT AC Control Mixture Valve ◮ w 2 (t) = A, B x sp − x 1 (t) x 1 , w 1 w 1 x 2 − x sp Pure A x 2 has been shown x 2 = 1 to be 1 in the fig- w 2 =? ure. This is not used in the above expres- sion. x , w 16/38 Process Control Feed Forward Control

Blending System: Ex. 15.5 of Textbook x sp x 1m w 2 , sp w 2m AT FFC FC FT Mixture p A, B x 1 , w 1 x , w 17/38 Process Control Feed Forward Control

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! 18/38 Process Control Feed Forward Control

2. Cascade Control 19/38 Process Control Feed Forward Control

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 20/38 Process Control Feed Forward Control

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 22/38 Process Control Feed Forward Control

When does cascade control work? ◮ When inner loop is a lot faster than the outer loop 24/38 Process Control Feed Forward Control

Problem from Final Exam, 2009 I This problem is concerned with a scheme, known as cascade control, shown below: Y sp Y 1 1 K c 1 K c 2 s + 1 (2 s + 1)(4 s + 1) − − G m 1. Using Routh-Hurwitz approach, determine the range of proportional controller gain K c1 for which the closed loop system is stable. Take K c2 = 1, G m = 1. 25/38 Process Control Feed Forward Control

Problem from Final Exam, 2009 II 2. Repeat the above for K c2 = 4, G m = 1. 3. Determine the range of K c1 values for which the conventional control scheme, obtained by letting K c2 = 1 and G m = 0, is stable. 4. Compare the three controllers obtained above. Which is preferable and why? 26/38 Process Control Feed Forward Control

Example on Cascade Control Discussed in Slide 20, Lecture 8 Y sp Y 1 1 K c 1 K c 2 s + 1 (2 s + 1)(4 s + 1) − − G m 27/38 Process Control Feed Forward Control

Example on Cascade Control Y sp Y 1 1 K c 1 K c 2 s + 1 (2 s + 1)(4 s + 1) − − G m ◮ Find the range of proportional controller gain K c1 for which the closed loop system is stable. Take K c2 = 1, G m = 1. ◮ Repeat for K c2 = 4, G m = 1. ◮ Find the range of K c1 values for which the conventional control scheme, obtained by letting K c2 = 1 and G m = 0, is stable. 28/38 Process Control Feed Forward Control

Solution to Example 1. Inner loop transfer function = 1 / (s + 2) Characteristic equation of closed loop: 8s 3 +22s 2 +13s+(2+K c1 ) = 0 , 0 < K c1 < 33 . 75 2. Inner loop transfer function = 0 . 8 / (0 . 2s + 1) Characteristic equation: 1 . 6s 3 + 9 . 2s 2 + 6 . 2s + (1 + 0 . 8K c1 ) = 0 , 0 < K c1 < 43 . 31 3. 8s 3 +14s 2 +7s+(K c1 +1) = 0 , 0 < K c1 < 11 . 25 So, answers to the example are 33.75, 43.31, 11.25 29/38 Process Control Feed Forward Control

3. Cascade control of a web server 30/38 Process Control Feed Forward Control

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 31/38 Process Control Feed Forward Control

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 32/38 Process Control Feed Forward Control

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 33/38 Process Control Feed Forward Control

WLAN control 34/38 Process Control Feed Forward Control

WLAN control 35/38 Process Control Feed Forward Control

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) 36/38 Process Control Feed Forward Control

What we learnt today ◮ Ratio control ◮ Feed forward control ◮ Cascade control ◮ A web server application 37/38 Process Control Feed Forward Control

Thank you 38/38 Process Control Feed Forward Control