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basic principles for feedback and feedforward controller to any process in the field of engineering
Typology: Lecture notes
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(^) What is process control and why do we need control?What is process control and why do we need control? (^) “What” is easy to explain, but “why” is more“What” is easy to explain, but “why” is more important from an engineering point of view. Toimportant from an engineering point of view. To understand “why”, let’s design a process withoutunderstand “why”, let’s design a process without considering any control.considering any control.
(^) Your boss comes and says: “Here’s all the money thatYour boss comes and says: “Here’s all the money that
you need. Build me a factory to produce diesel fuel.”you need. Build me a factory to produce diesel fuel.” (^) How are you going to do that? More importantly, how’sHow are you going to do that? More importantly, how’s
your four years of chemical engineering educationyour four years of chemical engineering education related with this end?related with this end? (^) Let’s make a very brief summary of what you haveLet’s make a very brief summary of what you have
learned.learned.
CO SO 2 H 22 N 2
CO SO 2 2
S (^17)
coal powder in air ash
diesel fuel
N 2 coal powder
H 2
S H 2 H 2 S
NH 3
solids liquids
gases
H 2 O O 2
N 2
coal lumps in air
air
coal powder
coal powder N 2 N 2
N CH 2 H 2 S^4
(^1 2 3 4 )
6 7
8 10 11 12
13 14 15
CO SO 2 N 2
CO SO 2 H N 2 2
H N 2 H N 2 2 2 H N 2 2
H 2
N O 2 2
N CH 2 4
16
9
Let’s use an existing design of the process, where coal is Let’s use an existing design of the process, where coal is converted to fuel. converted to fuel.
(^) Let’s say we have agreed on which unit operations to use.Let’s say we have agreed on which unit operations to use. The next step is decide on the capacity of the process.The next step is decide on the capacity of the process. (^) Why? Because we have to know how large reactors,Why? Because we have to know how large reactors, distillation columns, heat exchangers, pipes, etc. are going todistillation columns, heat exchangers, pipes, etc. are going to be.be. (^) So comes the concept ofSo comes the concept of operatingoperating (or(or designdesign ) conditions!) conditions! Each equipment is designed to operate at a certain set ofEach equipment is designed to operate at a certain set of conditions.conditions. (^) Ex. Your boss wants a fuel production of 1 mEx. Your boss wants a fuel production of 1 m^33 /min. You design/min. You design the equipment in such a way to handle this capacity andthe equipment in such a way to handle this capacity and make the pipe connections to maintain this production rate.make the pipe connections to maintain this production rate. (^) The important point is that most of the chemical processes isThe important point is that most of the chemical processes is continuous (like the one described here), so it seems thatcontinuous (like the one described here), so it seems that unsteady balances are not required in design. Steady stateunsteady balances are not required in design. Steady state balances will be sufficient.balances will be sufficient.
To control a process means to affect theTo control a process means to affect the behavior of the system in a desirable way.behavior of the system in a desirable way.
(^) Product qualityProduct quality (^) Profit optimizationProfit optimization (^) Equipment protectionEquipment protection (^) Environmental protectionEnvironmental protection
Easier to understand on a real example: Easier to understand on a real example:
A reactor with cooling jacket (no rxn for the sake A reactor with cooling jacket (no rxn for the sake of simplifying the process)of simplifying the process)
Steady-state energy balance around the CSTR:Steady-state energy balance around the CSTR: (^) FF (^) ffcc (^) pp(T(T (^) ff - T- T ssss) =) = UA(TUA(T ssss^ – T– T (^) jjssss)) Steady-state energy balance around the jacket:Steady-state energy balance around the jacket: (^) (^) jjFF (^) jfjfcc (^) pjpj(T(T (^) jjssss^ - T- T (^) jfjfssss) =) = UA(TUA(T ssss^ – T– T (^) jjssss))
Two equations, two unknowns. Solve by anyTwo equations, two unknowns. Solve by any method you’d like (such as MATLAB)method you’d like (such as MATLAB) fsolve('jacket_design', [5 -10], optimset('fsolve'), 15);
j
jf
If process is to be operated at T = 14If process is to be operated at T = 14 ooC, what should be FC, what should be F (^) jfjf??
fsolve('jacket_design', [5 -20], optimset('fsolve'), 14)
j
jf
3. If there is a disturbance? (Unsteady State Simulation) Ff suddenly increases to 2.3 L/s
If neither any variable in the process changes, nor my manager changes his/her mind about the process, then this process will work fine without a controller, as has been taught in other chemical engineering courses.
So what should be done to bring T to 15 oC? (The concept of feedback control )
Ff suddenly increases to 2.3 L/s (at t = 100 s), but this time there is automatic control!
(^) Process:Process: CSTR with cooling jacketCSTR with cooling jacket (^) Controlled variable:Controlled variable: Output of interest, variable that we want to controlOutput of interest, variable that we want to control (T).(T). (^) Manipulated variable:Manipulated variable: Input to the process, on which we have the powerInput to the process, on which we have the power to manipulate, so that the controlled variable be kept at the desired valueto manipulate, so that the controlled variable be kept at the desired value (T(T^ jfjf or For F^ jfjf).). (^) Disturbance:Disturbance: Variables which affect the controlled variable, but overVariables which affect the controlled variable, but over which we do not have manipulative power (Twhich we do not have manipulative power (T^ ff, F, F^ f,f, stirring rate,stirring rate, temperature of the surroundings, impurities in the feeds, etc.).temperature of the surroundings, impurities in the feeds, etc.).
Process Control Terminology Continued
Openloop Process
Closedloop Process