Feedback Control Systems (FCS)
Lecture-13
Mathematical Modelling of Liquid Level Systems
Dr. Imtiaz Hussain
email: imtiaz.hussain@faculty.muet.edu.pk
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