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CHAPTER 15: FEEDFORWARD CONTROL Outline of the lesson. A process - PowerPoint PPT Presentation

CHAPTER 15: FEEDFORWARD CONTROL Outline of the lesson. A process challenge - improve performance Feedforward design rules Good features and application guidelines Several process examples Analogy to management


  1. CHAPTER 15: FEEDFORWARD CONTROL Outline of the lesson. • A process challenge - improve performance • Feedforward design rules • Good features and application guidelines • Several process examples • Analogy to management principle

  2. CHAPTER 15: FEEDFORWARD CONTROL F 1 L 1 feed T product 1 TC Discuss this PID controller 2 stirred tank heat exchanger. F 2 T 3 heating stream

  3. CHAPTER 15: FEEDFORWARD CONTROL Class exercise: What do IAE = 237.6971 ISE = 758.425 TC 76 we do? F 74 temperature 1 L 1 72 feed minimum T 70 0 20 40 60 80 100 120 140 160 180 200 1 Disturbance = TC feed temperature Let’s use 2 cascade Control performance not acceptable! F 2 T 3 heating stream

  4. CHAPTER 15: FEEDFORWARD CONTROL CASCADE DESIGN CRITERIA FOR T1 Cascade is desired when OK 1. Single-loop performance unacceptable OK 2. A measured variable is available A secondary variable must OK 3. Indicate the occurrence of an important disturbance 4. Have a causal relationship from valve to secondary 5. Have a faster response than the primary NO! Cascade not possible. We need another enhancement!

  5. CHAPTER 15: FEEDFORWARD CONTROL Let’s think about the F 1 L 1 feed process behavior. T product 1 • Causal relationship TC 2 from T 1 disturbance to T 2 (without control) • How can we F 2 T 3 manipulate valve to heating stream compensate? v (valve) → Q → TC T 0 (Feed temperature)

  6. CHAPTER 15: FEEDFORWARD CONTROL CV B (t) = compensation effect CV A + CV B = no deviation 76 74 We want to 72 adjust the T 70 CV A (t) = disturbance effect valve to 68 cancel the 66 0 20 40 60 80 100 120 140 160 180 200 effect of the 60 disturbance. 58 56 v MV(t) = v 54 52 50 0 20 40 60 80 100 120 140 160 180 200 Time T 0 D m (t) = T 0 0 20 40 60 80 100 120 140 160 180 200 Time

  7. CHAPTER 15: FEEDFORWARD CONTROL We use block diagram algebra to determine the form of the calculation [G ff (s)] to achieve the desired performance. Measured disturbance, T 0 CV A (s) D m (s) Controlled G d (s) variable, T CV (s) + Feedforward G ff (s) controller How do we measure CV A ? MV (s) G p (s) CV B (s) Manipulated variable

  8. CHAPTER 15: FEEDFORWARD CONTROL MV ( s ) G ( s ) = = − d G ( s ) ff D ( s ) G ( s ) m p Special case of G p (s) and G d (s) being first order with dead time Please + MV ( s ) T s 1 verify. − θ s = = ld G ( s ) K e ff ff ff + D ( s ) T s 1 m lg Gain Lead-lag Dead time

  9. CHAPTER 15: FEEDFORWARD CONTROL + T s 1 − θ s = ld G ( s ) K e ff ff ff + T s 1 lg Lead-lag = (T ld s+1)/T lg s+1) FF controller gain = K ff = - K d /K p = θ ff = θ d - θ p ≥ 0 controller dead time = T ld = τ p Lead time = T lg = τ d Lag time How do we get values for these parameters?

  10. CHAPTER 15: FEEDFORWARD CONTROL How do we combine feedback with F feedforward? 1 L 1 feed T 1 FF high- lighted TC FF in red 2 TY 1 MV fb + MV ff TY 2 F 2 T 3 heating stream

  11. CHAPTER 15: FEEDFORWARD CONTROL Control Performance Comparison for CST Heater Single-Loop Feedforward with feedback IAE = 237.6971 ISE = 758.425 IAE = 27.772 ISE = 8.0059 76 76 75 temperature 74 74 temperature 73 72 72 71 70 70 0 20 40 60 80 100 120 140 160 180 200 0 20 40 60 80 100 120 140 160 180 200 Much better performance! WHY?

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