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ICE401: PROCESS INSTRUMENTATION
AND CONTROL
Class 21, 22
Summary
Dr. S. Meenatchisundaram
Email: meenasundar@gmail.com
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Control System Components:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Control Loops:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Advanced Control Loops:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Advanced Control Loops:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Mathematical Modeling:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
We can define hydraulic resistance (R) to flow as follows:
Hydraulic Capacitance (A) can be given as:
Liquid-Level System:
0
Potential h
R
Flow q
≡ =
( )
( )
V t Quantity
A
h t Potential
= =
( )
( )
( ) 1i
H s R
Q s RAs
=
+ ( )
0 ( ) 1
( ) 1i
Q s
Q s RAs
=
+
0
( )
( )
H s
Q s
R
=
Mathematical Modeling:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Liquid-Level Systems with Interaction:
Liquid-Level Systems without Interaction:
( )
2
2
1 1 2 2 1 1 2 2 2 1
( ) 1
( ) 1
Q s
Q s RC R C s RC R C R C s
=
+ + + +
( )
2 2
2
1 1 2 2 1 1 2 2 2 1
( )
( ) 1
H s R
Q s RC R C s RC R C R C s
=
+ + + +
1
1
( ) 1
( ) 1
Q s
Q s sτ
=
+
2 2
1 2
( ) 1
( ) 1 1
H s R
Q s s sτ τ
=
+ +
Mathematical Modeling:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Thermal System:
CSTR:
( )
=
H(s) 1 Cs
s R
R
θ∆  
 ∆ + 
( ) ( )A A Ai i A
d d
n C V C F C F rV
dt dt
= = − −
( ) ( )/
0
E RTiA
Ai A A
FdC
C C k e C
dt V
−
= − −
Mathematical Modeling:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Pneumatic System:
Hydraulic System:
0 ( ) 1
( ) ( 1)i
P s
P s RCs
∆
=
∆ +
( )
( ) 1 1i
A A
X s K K
P s RCs sτ
= =
+ +
Dead Time, P&I Diagram:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Process dead time (td) is equal to the distance (D) divided by
the velocity (υ) through the discharge pipe, or td = D/υ.
Direct & Reverse Action:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
A controller operates with direct action when an increasing
value of the controlled variable causes an increasing value of
the controller output.
Reverse action is the opposite case, where an increase in a
controlled variable causes a decrease in controller output.
Classification of Controller Modes:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Controller Action:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Two-Position Mode:
Neutral Zone:
00%
0100%
p
p
e
p
e
<
= 
>
Controller Action:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Multi-Position Mode:
Floating Control Mode – Single Speed:
Multi Speed:
1
1 1
1
100%
50%
0%
p
p
p
e e
p e e e
e e
 >

= − < <
 < −
F p p
dp
K e e
dt
= ± > ∆
Fi p pi
dp
K e e
dt
= ± > ∆
Controller Action:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Proportional Mode:
0p pp K e p= +
100
p
PB
K
=
Controller Action:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Integral Mode:
0
( ) (0)
t
I pp t K e dt p= +∫
1
=I
I
T
K
Controller Action:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Derivative Mode:
Proportional-Integral Control Mode:
Proportional-Derivate Control Mode:
Proportional-Integral-Derivate Control Mode:
( ) p
D
de
p t K
dt
=
0
( ) (0)
t
P P P I p Ip t K e K K e dt p= + +∫
0( )
p
P P P D
de
p t K e K K p
dt
= + +
0
( ) (0)
t
p
P P P I p P D I
de
p t K e K K e dt K K p
dt
= + + +∫
Controller Action:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Proportional-Integral
Controller Action:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Proportional-Derivative
Effects of KP, KI and KD:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
KP KI
Effects of KP, KI and KD:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
KD
Controller Action:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Try:
A PI controller is reverse acting, PB = 20, 12 repeats per
minute. Find (a) the proportional gain, (b) the integral gain,
and (c) the time that the controller output will reach 0% after
a constant error of -1.5% starts. The controller output when
the error occurred was 72%.
Solution: 100
5%
p
PB
K
= =
( )
12%
% min
IK =
−
Controller Action:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015

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Class 21 22 - summary

  • 1. ICE401: PROCESS INSTRUMENTATION AND CONTROL Class 21, 22 Summary Dr. S. Meenatchisundaram Email: meenasundar@gmail.com Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
  • 2. Control System Components: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
  • 3. Control Loops: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
  • 4. Advanced Control Loops: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
  • 5. Advanced Control Loops: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
  • 6. Mathematical Modeling: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015 We can define hydraulic resistance (R) to flow as follows: Hydraulic Capacitance (A) can be given as: Liquid-Level System: 0 Potential h R Flow q ≡ = ( ) ( ) V t Quantity A h t Potential = = ( ) ( ) ( ) 1i H s R Q s RAs = + ( ) 0 ( ) 1 ( ) 1i Q s Q s RAs = + 0 ( ) ( ) H s Q s R =
  • 7. Mathematical Modeling: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015 Liquid-Level Systems with Interaction: Liquid-Level Systems without Interaction: ( ) 2 2 1 1 2 2 1 1 2 2 2 1 ( ) 1 ( ) 1 Q s Q s RC R C s RC R C R C s = + + + + ( ) 2 2 2 1 1 2 2 1 1 2 2 2 1 ( ) ( ) 1 H s R Q s RC R C s RC R C R C s = + + + + 1 1 ( ) 1 ( ) 1 Q s Q s sτ = + 2 2 1 2 ( ) 1 ( ) 1 1 H s R Q s s sτ τ = + +
  • 8. Mathematical Modeling: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015 Thermal System: CSTR: ( ) = H(s) 1 Cs s R R θ∆    ∆ +  ( ) ( )A A Ai i A d d n C V C F C F rV dt dt = = − − ( ) ( )/ 0 E RTiA Ai A A FdC C C k e C dt V − = − −
  • 9. Mathematical Modeling: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015 Pneumatic System: Hydraulic System: 0 ( ) 1 ( ) ( 1)i P s P s RCs ∆ = ∆ + ( ) ( ) 1 1i A A X s K K P s RCs sτ = = + +
  • 10. Dead Time, P&I Diagram: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015 Process dead time (td) is equal to the distance (D) divided by the velocity (υ) through the discharge pipe, or td = D/υ.
  • 11. Direct & Reverse Action: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015 A controller operates with direct action when an increasing value of the controlled variable causes an increasing value of the controller output. Reverse action is the opposite case, where an increase in a controlled variable causes a decrease in controller output.
  • 12. Classification of Controller Modes: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
  • 13. Controller Action: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015 Two-Position Mode: Neutral Zone: 00% 0100% p p e p e < =  >
  • 14. Controller Action: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015 Multi-Position Mode: Floating Control Mode – Single Speed: Multi Speed: 1 1 1 1 100% 50% 0% p p p e e p e e e e e  >  = − < <  < − F p p dp K e e dt = ± > ∆ Fi p pi dp K e e dt = ± > ∆
  • 15. Controller Action: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015 Proportional Mode: 0p pp K e p= + 100 p PB K =
  • 16. Controller Action: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015 Integral Mode: 0 ( ) (0) t I pp t K e dt p= +∫ 1 =I I T K
  • 17. Controller Action: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015 Derivative Mode: Proportional-Integral Control Mode: Proportional-Derivate Control Mode: Proportional-Integral-Derivate Control Mode: ( ) p D de p t K dt = 0 ( ) (0) t P P P I p Ip t K e K K e dt p= + +∫ 0( ) p P P P D de p t K e K K p dt = + + 0 ( ) (0) t p P P P I p P D I de p t K e K K e dt K K p dt = + + +∫
  • 18. Controller Action: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015 Proportional-Integral
  • 19. Controller Action: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015 Proportional-Derivative
  • 20. Effects of KP, KI and KD: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015 KP KI
  • 21. Effects of KP, KI and KD: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015 KD
  • 22. Controller Action: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015 Try: A PI controller is reverse acting, PB = 20, 12 repeats per minute. Find (a) the proportional gain, (b) the integral gain, and (c) the time that the controller output will reach 0% after a constant error of -1.5% starts. The controller output when the error occurred was 72%. Solution: 100 5% p PB K = = ( ) 12% % min IK = −
  • 23. Controller Action: Process Instrumentation and Control (ICE 401) Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015