Lecture-13: Mathematical Modelling of Liquid
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Transcript Lecture-13: Mathematical Modelling of Liquid
Feedback Control Systems (FCS)
Lecture-13
Mathematical Modelling of Liquid Level Systems
Dr. Imtiaz Hussain
email: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
Laminar vs Turbulent Flow
• Laminar Flow
– Flow dominated by viscosity
forces is called laminar flow and
is characterized by a smooth,
parallel line motion of the fluid
• Turbulent Flow
– When inertia forces dominate,
the flow is called turbulent flow
and is characterized by an
irregular motion of the fluid.
Resistance of Liquid-Level Systems
• Consider the flow through a short pipe connecting two
tanks as shown in Figure.
• Where H1 is the height (or level) of first tank, H2 is the
height of second tank, R is the resistance in flow of liquid
and Q is the flow rate.
Resistance of Liquid-Level Systems
• The resistance for liquid flow in such a pipe is defined as the change
in the level difference necessary to cause a unit change inflow rate.
Resistance
change in level diff erence
change in flow rate
R
( H 1 H 2 )
Q
m
3
m /s
m
3
m /s
Resistance in Laminar Flow
• For laminar flow, the relationship between the steady-state flow
rate and steady state height at the restriction is given by:
Q kl H
• Where Q = steady-state liquid flow rate in m/s3
• Kl = constant in m/s2
• and H = steady-state height in m.
• The resistance Rl is
Rl
dH
dQ
Capacitance of Liquid-Level Systems
• The capacitance of a tank is defined to be the change in quantity of
stored liquid necessary to cause a unity change in the height.
h
Capacitanc e
change in liquid stored
change in height
m
3
m
• Capacitance (C) is cross sectional area (A) of the tank.
or m
2
Capacitance of Liquid-Level Systems
h
Rate of change of fluid volume in the tank flow in flow out
dV
dt
qi qo
d( A h)
dt
qi qo
Capacitance of Liquid-Level Systems
h
A
dh
dt
C
dh
dt
qi qo
qi qo
Modelling Example#1
Modelling Example#1
• The rate of change in liquid stored in the tank is equal to the flow in
minus flow out.
C
dh
dt
qi qo
(1)
• The resistance R may be written as
R
dH
dQ
h
(2)
q0
• Rearranging equation (2)
q0
h
R
(3)
Modelling Example#1
C
dh
dt
qi qo
q0
(1)
h
R
• Substitute qo in equation (3)
C
dh
dt
qi
h
R
• After simplifying above equation
RC
dh
dt
h Rq i
• Taking Laplace transform considering initial conditions to zero
RCsH ( s ) H ( s ) RQ i ( s )
(4)
Modelling Example#1
RCsH ( s ) H ( s ) RQ i ( s )
• The transfer function can be obtained as
H (s)
Qi ( s )
R
( RCs 1)
Modelling Example#1
• The liquid level system considered here is analogous to the
electrical and mechanical systems shown below.
RC
de o
dt
eo ei
b dx o
k dt
RC
dh
dt
h Rq i
xo xi
Modelling Example#2
• Consider the liquid level system shown in following Figure. In this
system, two tanks interact. Find transfer function Q2(s)/Q(s).
Modelling Example#2
• Tank 1
• Tank 2
C1
C2
dh 1
dt
dh 2
dt
q q1
q1 q 2
Pipe 1
Pipe 2
R1
h1 h 2
q1
R2
h2
q2
Modelling Example#2
• Tank 1 C 1
dh 1
• Tank 2 C 2
dh 2
q
h1 h 2
dt
R1
dt
h1 h 2
R1
h2
R2
q1
Pipe 1
h1 h 2
R1
q2
Pipe 2
h2
R2
• Re-arranging above equation
C1
dh 1
dt
h1
R1
q
h2
R1
C2
dh 2
dt
h2
R1
h2
R2
h1
R1
Modelling Example#2
C1
dh 1
dt
h1
R1
q
h2
R1
C2
dh 2
dt
h2
R1
h2
R2
h1
R1
• Taking LT of both equations considering initial conditions to zero
[i.e. h1(0)=h2(0)=0].
1
1
C1 s
H 1( s ) Q( s )
H 2(s)
R1
R1
(1)
1
1
1
C2s
H 2(s)
H 1( s )
R1
R2
R1
(2)
Modelling Example#2
1
1
C1 s
H 1( s ) Q( s )
H 2(s)
R
R
1
1
(1)
1
1
1
C2s
H 2 (s)
H 1( s )
R1
R2
R1
• From Equation (1)
H 1( s )
R1 Q ( s ) H 2 ( s )
R1 C 1 s 1
• Substitute the expression of H1(s) into Equation (2), we get
1
1
1 R1 Q ( s ) H 2 ( s )
C2s
H 2(s)
R
R
R
R
C
s
1
1
2
1
1 1
(2)
Modelling Example#2
1
1
1 R1 Q ( s ) H 2 ( s )
C2s
H 2(s)
R
R
R
R
C
s
1
1
2
1
1 1
• Using H2(s) = R2Q2 (s) in the above equation
R 2 C 2 s
Q2 (s)
Q(s)
1 R1 C 1 s 1 R 2 C 1 s Q 2 ( s ) Q ( s )
1
R 2 C 1 R1 C 2 s
2
R1 C 1 R 2 C 2 R 2 C 1 s 1
Modelling Example#3
• Write down the system differential equations.
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END OF LECTURES-13