lecture 13&14
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process controlTRANSCRIPT
CHE334 Instrumentation and Process Control Lecture 13 & 14 Chapter 10 : Dynamics of First Order System
By Dr. Maria Mustafa
Department of Chemical Engineering
Dynamic Behavior of first order systems (Chapter 10)
โข First order system โ A first order system is one whose output is
modeled by a first order differential equation. For Linear system (Case 1 )
๐๐๐ ๐
๐ ๐+ ๐๐๐ = ๐๐(๐)
Where f(t) is (forcing ) input function. If ๐๐ is non-zero, then dividing above equation by ๐๐ we have
๐๐๐๐
๐ ๐
๐ ๐+ ๐ =
๐
๐๐๐(๐)
Defining ๐๐๐๐
= ๐๐ ๐๐๐ ๐
๐๐= ๐ฒ๐
Then the equation becomes as
๐๐๐ ๐
๐ ๐+ ๐ = ๐ฒ๐๐(๐)
Where ๐๐ is known as time constant
And ๐ฒ๐ = steady state gain or gain of the
system
If y(t) and f(t) is in deviation variables around a steady state, the initial conditions are
y(0) = 0 and f(0) = 0
Then the transfer function of a first order system or process is given by
๐บ ๐ = ๐ฆ (๐ )
๐ (๐ )=
๐พ๐๐๐๐ + 1
A first order process having transfer function of ๐พ๐
๐๐๐ +1 is also known as first order lag, linear lag,
or exponential transfer lag.
โข Case 2
๐๐๐ ๐
๐ ๐= ๐๐(๐)
Dividing ๐๐ on both sides ๐ ๐
๐ ๐=
๐
๐๐๐ ๐ = ๐ฒโฒ
๐ท๐ ๐
The transfer function of the system is given by
๐ฎ ๐ =๐ (๐)
๐ (๐)=๐ฒโฒ
๐ท
๐
In such case the process is called purely capacitive or pure integrator.
Characteristics of First order System
โข A process that possesses a capacity to tore mass or energy and then acts as a buffer between inflowing and outflowing streams will be ordered as first order system
โข The first order processes are characterized by: โ Their capacity to store mass, energy or
momentum. โ The resistance associated with the flow of
mass. Energy or momentum in reaching the capacity.
Examples
Sr. No. Process Capacity Resistance
1 Dynamic response of the tank
Dynamic response of the tank that have capacity to store liquid or gasses
Resistance to associated with pumps, valves, weirs and pipes attached to either inflowing or outflowing liquids or gasses
2 Dynamic response of temperature of solid, liquid or gaseous system
That can store thermal energy ( thermal capacity Cp)
Resistance is associated with the transfer of heat through walls, liquid or gasses.
Example 10.1: First order system with a capacity for mass storage
โข Consider the tank shown in fig. . The volumetric flow in is Fi and the outlet volumetric flow rate is Fo. In the outlet there is a resistance to flow such as a pipe, valve, pump or weir.
Fi, Ti
h
F, T
T
R
Example 10.1: First order system with a capacity for mass storage
โข Consider the tank shown in fig. . The volumetric flow in is Fi and the outlet volumetric flow rate if Fo. In the outlet there is a resistance to flow such as a pipe, valve, pump or weir. Assume that the effluent ( out) flow rate Fo is related to hydrostatic pressure of the liquid h, through resistance R :
๐น๐ =๐
๐ =๐๐๐๐ฃ๐๐๐ ๐๐๐๐๐ ๐๐๐ ๐๐๐๐ค
๐๐๐ ๐๐ ๐ก๐๐๐๐ ๐ก๐ ๐๐๐๐ค
At any time point, the tank has the capacity to store mass. Analyzing the transfer function of the system
โข The total mass balance around the system gives
๐ด๐๐
๐๐ก= ๐น๐ โ ๐น๐ = ๐น๐ โ
๐
๐
Dividing both sides by R we have
๐ด๐ ๐๐
๐๐ก+ ๐ = ๐ ๐น๐
Where A is the cross sectional area
At steady state ๐๐ = ๐ ๐น๐,๐
Subtract the equation B from Equation A
๐ด๐ ๐๐โฒ
๐๐ก+ ๐โฒ = ๐ ๐นโฒ๐
Comparing
๐ด๐ ๐๐โฒ
๐๐ก+ ๐โฒ = ๐ ๐นโฒ๐
And
๐๐๐ ๐
๐ ๐+ ๐ = ๐ฒ๐๐(๐)
We have
๐๐ = ๐จ๐น=time constant of process
๐ฒ๐ = ๐น = Steady state gain
The transfer function of the system is
๐ฎ ๐ = ๐ โฒ(๐)
๐ญ๐โฒ(๐)=
๐ฒ๐
๐๐๐+๐
โข Points to be noted
1. The cross sectional area A , is a measure of its capacitance to store mass. Thus larger the value of A, the larger the storage capacity.
2. Since ๐๐ = ๐จ๐น so we can say that
(time constant)= (storage capacitance ) x (resistance to flow)
Example 10.2 : First Order system with a capacity for Energy Storage
โข The liquid of a tank is heated with saturated steam, which flows through a coil immersed in the liquid. Analyzing the transfer function
Tst
T
Q
Example 10.2 : First Order system with a capacity for Energy Storage
โข The liquid of a tank is heated with saturated steam, which flows through a coil immersed in the liquid. Analyzing the transfer function
โข Energy balance for the system yields
๐[๐๐๐๐ ๐ โ ๐๐๐๐ ]
๐๐ก= ๐ = ๐๐ด๐ก(๐๐ ๐ก โ ๐)
๐๐๐๐๐ ๐
๐๐ก= ๐ = ๐๐ด๐ก(๐๐ ๐ก โ ๐) โฆ. Eqn A
The steady state is given by
0 = ๐๐ด๐ก(๐๐ ๐ก,๐ โ ๐๐ )โฆ.. Eqn B
โข Subtracting Eqn A from Eqn B, we have
๐๐๐๐๐ ๐โฒ
๐๐ก= ๐ = ๐๐ด๐ก ๐
โฒ๐ ๐ก โ ๐โฒ
๐๐๐๐
๐๐ด๐ก
๐ ๐
๐๐ก+ ๐ = ๐๐ ๐ก
The transfer function of the system is
๐ฎ ๐ = ๐ป โฒ(๐)
๐ป๐๐โฒ(๐)=
๐ฒ๐
๐๐๐ + ๐=
๐
๐๐๐๐๐๐ด๐ก
๐ + ๐
Where ๐๐ =๐๐๐๐
๐๐ด๐ก= time constant of the process
and
๐ฒ๐ = steady state gain of process =1
โข Points to be noted
1. The above equation clearly demonstrates that this is a first order lag system
2. The system possesses capacity to store thermal energy and a resistance to the flow of heat characterized by U.
3. The cross sectional area ๐๐๐๐ , is a measure of its
capacitance to store energy and 1
๐๐ด๐ก is the
resistance to the flow of heat from steam to liquid.
4. Since ๐๐ =๐๐๐๐
๐๐ด๐กso we can say that
(time constant)= (storage capacitance ) x (resistance to flow)
Example 10.3: Pure Capacitive System
โข Consider the tank shown in fig. . The volumetric flow in is Fi and the outlet volumetric flow rate is Fo which remain constant for all time. Fo is determined by a constant โ displacement pump and not by the hydrostatic pressure of the liquid level โh Fi
h
Fo
Analyzing the transfer
function of the system
โข The total mass balance around the system gives
๐ด๐๐
๐๐ก= ๐น๐ โ ๐น๐
Dividing both sides by A we have ๐๐
๐๐ก= (
1
๐ด)๐น๐โ(
1
๐ด)๐น๐
Where A is the cross sectional area
At steady state
0 = (1
๐ด)๐น๐,๐ โ(
1
๐ด)๐น๐
Subtract the equation B from Equation A ๐๐โฒ
๐๐ก= (
1
๐ด)๐นโฒ๐
Comparing ๐๐โฒ
๐๐ก= (
1
๐ด)๐นโฒ๐
And ๐ ๐
๐ ๐= ๐ฒโฒ๐๐(๐)
We have
๐ฒโฒ๐ = (๐
๐จ) = Steady state gain
The transfer function of the system is
๐ฎ ๐ = ๐ โฒ(๐)
๐ญ๐โฒ(๐)=
๐ฒโฒ๐
๐
10.3: Dynamic Response of a Pure Capacitive system
The transfer function of the pure capacitive system is given by
๐ฎ ๐ =๐ (๐)
๐ (๐)=๐ฒโฒ
๐ท
๐
โข Let y(t) changes with time, when f(t) undergoes a unit step change :
๐ ๐ก = 1 ๐๐๐ ๐ก > 0
The Laplace of f(t) is given by
๐ ๐ =๐
๐
โข Then y(s) becomes as follows :
๐ ๐ = ๐ ๐๐ฒโฒ
๐ท
๐=๐ฒโฒ
๐ท
๐๐
And after inversion y(t) is given by ๐ฆ ๐ก = ๐พโฒ๐๐ก
This means that the output grows linearly wih time in an unbounded fashion. Thus
๐ฆ ๐ก โ โ ๐๐ ๐ก โ โ
Such type of response , characteristic of a pure capacitive process, lends the name pure integrator because it behaves as if there were an integrator between its input and output.
โข A pure capacitive process can not balance itself, loose control and is known as non-self regulation process. It will cause serious control problems. As in example 10.3, a small change in inlet Flow will make the tank either flood or run dry (empty).
Dynamic Behavior of First Order Lag system
โข The transfer function of first order lag system is given by:
๐บ ๐ = ๐ฆ (๐ )
๐ (๐ )=
๐พ๐
๐๐๐ + 1
โข Lets assume that the f(t) undergoes a unit step change with time
๐ ๐ก = 1 ๐๐๐ ๐ก > 0
Then ๐ฆ ๐ ๐๐ ๐๐๐ฃ๐๐ ๐๐ฆ ๐พ๐
(๐๐๐ +1)
๐
๐ and
By taking inverse the y(t) will given as follows
๐ฆ ๐ก = ๐พ๐(1 โ ๐โ๐ก๐๐)
โข Lets assume that the f(t) undergoes a step change of magnitude A with time
๐ ๐ก = ๐ด ๐๐๐ ๐ก > 0
Then ๐ฆ ๐ ๐๐ ๐๐๐ฃ๐๐ ๐๐ฆ ๐ด๐พ๐
(๐๐๐ +1)
๐
๐ and
By taking inverse the y(t) will given as follows
๐ฆ ๐ก = ๐ด๐พ๐(1 โ ๐โ๐ก๐๐)
In terms of the dimensionless coordinator
๐ฆ ๐ก /๐ด๐พ๐ = (1 โ ๐โ๐ก๐๐)
Plotting ๐ฆ ๐ก
๐ด๐พ๐๐ฃ๐ (1 โ ๐
โ๐ก
๐๐)
t = ๐๐ t/๐๐ Output response
๐ฆ ๐ก /๐ด๐พ๐ = (1 โ ๐โ๐ก๐๐)
% output response
0 0 0 0
๐๐ 1 0.632121 63.2
2๐๐ 2 0.864665 86.466
4๐๐ 4 0.981684 98.17
6๐๐ 6 0.997521 99.75
0
0.2
0.4
0.6
0.8
1
0 1 2 3 4 5 6 7
(1 โ ๐โ๐ก๐๐)
๐๐/๐จ
๐ฒ๐
t/๐๐
Characteristics of First order Lag system by analyzing graph
1. First order Lag system is self regulating . After disturbances, the system output response reaches to new steady state.
2. The slope of the response at t=0 is equal to 1. This implies that , the output response would reach its final value in one time constant. Also we can conclude:
โ the smaller the value of time constant , the steeper the initial output response of the systemโ. Or โThe time constant of a process is a measure of the time necessary for the process to adjust to a change in it input.โ
Characteristics of First order Lag system by analyzing graph
3. The value of output response y(t) reaches 63.2 % of its final value when the time elapsed is equal to one time constant. [ see above Table ]
4. As we know that ๐ฆ ๐ก = ๐ด๐พ๐ 1 โ ๐โ
๐ก
๐๐ , When t โ , then output response ๐ด๐พ๐. Since for any step change in input , the steady state change in output is given by โ(output) = ๐พ๐โ(output). This tells us how much change we have to made in input in order to have desired value of output response for a process with given gain ๐พ๐ . Thus to effect the same change in output, we need
A small change in the input if ๐พ๐ is large ( very sensitive systems )
A large change in the input if ๐พ๐ is small.
Example 10.4: Effect of Parameters on the response of first order system
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 2 4 6 8 10
Consider two systems (System 1 and system 2) of liquid storage ( as shown in Example 10.1) of different two areas A1 and A2 resp. such that A1 > A2 . This means that time constant of systems 1 is greater than system provided resistance to flow R is same in both system. So the System 2 initial response is steeper than that of system 1 as shown in this Plot.
System2(A2)
System1(A1)
0
0.02
0.04
0.06
0.08
0.1
0.12
0 1 2 3 4 5 6 7
Consider two systems (System 1 and system 2) of liquid storage ( as shown in Example 10.1) of different two areas A1 and A2 resp. and different two resistances R1 and R2 resp. such that A1 > A2, R2>R1 and A1RI =A2R2 . This means that gain of systems 2 is greater than system 1 provided time constants are same in both system. So the System 2 initial response is same to that of system 1 but as time goes on, the system 2 output response caused due to change in same input will rise to higher steady state value as compare to that of system 1 as shown in this Plot.
For total mass balance
๐(๐)
๐๐ก= ๐ ๐
๐:๐๐๐๐๐ก
โ ๐ ๐๐:๐๐ข๐ก๐๐๐ก
Mass Balance on component
๐(๐๐ด)
๐๐ก=๐(๐ฅ๐ด๐)
๐๐ก= ๐ฅ๐ด๐๐ ๐
๐:๐๐๐๐๐ก
โ ๐ฅ๐ด๐๐ ๐๐:๐๐ข๐ก๐๐๐ก
Total energy balance ๐(๐ธ)
๐๐ก=๐(๐ + ๐พ + ๐)
๐๐ก
= ๐ ๐๐๐๐:๐๐๐๐๐ก
โ ๐ ๐๐๐๐:๐๐ข๐ก๐๐๐ก
ยฑ ๐ ยฑ๐๐
For total mass balance
๐(๐๐)
๐๐ก= ๐๐๐น๐
๐:๐๐๐๐๐ก
โ ๐๐๐น๐๐:๐๐ข๐ก๐๐๐ก
Mass Balance on component
๐(๐๐ด)
๐๐ก=๐(๐๐ด๐)
๐๐ก= ๐๐ด๐๐น๐
๐:๐๐๐๐๐ก
โ ๐๐ด๐๐น๐๐:๐๐ข๐ก๐๐๐ก
ยฑ ๐๐
Total energy balance ๐(๐ธ)
๐๐ก=๐(๐ + ๐พ + ๐)
๐๐ก
= ๐๐๐น๐๐๐๐:๐๐๐๐๐ก
โ ๐๐๐น๐๐๐๐:๐๐ข๐ก๐๐๐ก
ยฑ ๐ ยฑ๐๐