of 22
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
1/22
Feedback Control Systems (FCS)
Dr. Imtiaz Hussainemail: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
Lecture-13Mathematical Modelling of Liquid Level Systems
mailto:[email protected]:[email protected]7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
2/22
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
3/22
Laminar vs Turbulent Flow
Laminar Flow Flow dominated by viscosity
forces is called laminar flow andis characterized by a smooth,
parallel line motion of the fluid
Turbulent Flow
When inertia forces dominate,the flow is called turbulent flowand is characterized by anirregular motion of the fluid.
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
4/22
Resistance of Liquid-Level Systems
Consider the flow through a short pipe connecting twotanks as shown in Figure.
Where H1 is the height (or level) of first tank, H2 is theheight of second tank, R is the resistance in flow of liquidand Q is the flow rate.
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
5/22
Resistance of Liquid-Level Systems
The resistance for liquid flow in such a pipe is defined as the changein the level difference necessary to cause a unit change inflow rate.
sm
m
flow ratechange in
erencelevel diffchange in
Resistance /3
sm
m
Q
HHR
/
)(3
21
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
6/22
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:
Where Q= steady-state liquid flow rate in m/s3
Kl= constant in m/s2
andH = steady-state height in m.
The resistance Rlis
HkQl
dQ
dHRl
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
7/22
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.
Capacitance (C) is cross sectional area (A) of the tank.
2
3
morm
m
heightchange in
storedliquidchange ineCapacitanc
h
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
8/22
Capacitance of Liquid-Level Systems
outflowinflowtanktheinvolumefluidofchangeofRate
h
oi qqdt
dV
oi qq
dt
hAd
)(
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
9/22
Capacitance of Liquid-Level Systems
h
oi qq
dt
dhA
oi qqdt
dhC
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
10/22
Modelling Example#1
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
11/22
Modelling Example#1
The rate of change in liquid stored in the tank is equal to the flow in
minus flow out.
The resistance R may be written as
Rearranging equation (2)
oi qqdt
dhC (1)
0q
h
dQ
dHR (2)
R
hq 0 (3)
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
12/22
Modelling Example#1
Substitute qo in equation (3)
After simplifying above equation
Taking Laplace transform considering initial conditions to zero
oi qqdt
dh
C (1) R
h
q 0 (4)
Rhq
dtdhC i
iRqhdt
dhRC
)()()( sRQsHsRCsH i
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
13/22
Modelling Example#1
The transfer function can be obtained as
)()()( sRQsHsRCsHi
)()(
)(
1
RCs
R
sQ
sH
i
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
14/22
Modelling Example#1
The liquid level system considered here is analogous to the
electrical and mechanical systems shown below.
i
Rqhdt
dhRC
ioo ee
dt
deRC
ioo xx
dt
dx
k
b
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
15/22
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).
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
16/22
Modelling Example#2
Tank 1 Pipe 1
Tank 2 Pipe 2
1
1
1 qqdt
dhC
1
211q
hhR
21
2
2 qqdt
dh
C 2
2
2 q
h
R
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
17/22
Modelling Example#2
Tank 1 Pipe 1
Tank 2 Pipe 2
1
211
1R
hhqdt
dhC
1
211R
hhq
2
2
1
212
2 R
h
R
hh
dt
dh
C
2
2
2 R
h
q
Re-arranging above equation
1
2
1
11
1R
hq
R
h
dt
dhC
1
1
2
2
1
22
2R
h
R
h
R
h
dt
dhC
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
18/22
Modelling Example#2
Taking LT of both equations considering initial conditions to zero
[i.e. h1(0)=h2(0)=0].
1
2
1
11
1R
hqR
h
dt
dhC
1
1
2
2
1
22
2R
h
R
h
R
h
dt
dhC
)()()( sHR
sQsHR
sC2
1
1
1
1
11
(1)
)()( sHR
sHRR
sC1
1
2
21
2
111
(2)
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
19/22
Modelling Example#2
From Equation (1)
Substitute the expression of H1(s) into Equation (2), we get
)()()( sHRsQsHRsC 21
1
1
1
11
)()( sHRsHRRsC 11
2
21
2
111
(1) (2)
111
21
1 sCR
sHsQRsH )()()(
1
111
11
21
1
2
21
2sCR
sHsQR
RsH
RRsC
)()()(
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
20/22
Modelling Example#2
Using H2(s) = R2Q2 (s) in the above equation
1
111
11
21
1
2
21
2sCR
sHsQR
RsHRRsC)()(
)(
)()( sQsQsCRsCRsCR 2121122
11
11
122211
2
2112
2
sCRCRCRsCRCRsQ
sQ
)(
)(
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
21/22
Modelling Example#3
Write down the system differential equations.
7/28/2019 Lecture 13-Mathematical Modelling of Liquid Level Systems
22/22
END OF LECTURES-13
To download this lecture visit
http://imtiazhussainkalwar.weebly.com/
http://imtiazhussainkalwar.weebly.com/http://imtiazhussainkalwar.weebly.com/
