mathematical modeling of liquid-level systems

ICE401: PROCESS INSTRUMENTATION
AND CONTROL
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal,Aug – Nov 2015
Class 7: Mathematical Modeling of
Liquid-Level Systems
Dr. S. Meenatchisundaram
Email: meenasundar@gmail.com

Liquid-Level Systems:
Fluid-Flow Basics:
• In analysing systems involving fluid flow, it is necessary to
divide flow regimes into laminar flow and turbulent flow,
according to the magnitude of the Reynolds number.
• If the Reynolds number is greater than about 3000 to 4000,
then the flow is turbulent.
• The flow is laminar if the Reynolds number is less than about
2000.
• In the laminar case, fluid flow occurs in streamlines with no
turbulence.
• Systems involving laminar flow may be represented by linear
differential equations.

• Industrial processes often involve flow of liquids through
connecting pipes and tanks.
• The flow in such processes is often turbulent and not laminar.
• Systems involving turbulent flow often have to be represented
by nonlinear differential equations.
• If the region of operation is limited, however, such nonlinear
differential equations can be linearized.
• We shall discuss such linearized mathematical models of liquid-
level systems in this section.
• Note that the introduction of concepts of resistance and
capacitance for such liquid-level systems enables us to describe
their dynamic characteristics in simple forms.

Hydraulic Resistance:
• Figure shows liquid flow in a pipe, with a restricting device
(a valve) providing a hydraulic resistance (Rh) to the flow.
• Note that the walls of the pipe will also provide a small
amount of resistance to flow, depending on how rough they
are.

• When turbulent flow occurs from a tank discharging under its
own head or pressure, the flow is found by the following
equation:
(5.1)
• Where q0 is the flow rate (ft3/s), K is a flow coefficient, A is the
area of the discharge orifice (ft2), g is gravitation constant (ft/s2),
and h is pressure head of liquid (ft).
• We can define hydraulic resistance (R) to flow as follows:
(5.2)
q0  KA 2gh
R 
Potential

h
Flow q0

• Therefore, the instantaneous rate of change of hydraulic
resistance to flow is
(5.3)
• Rearranging Equation 5.1, we arrive at:
(5.4)
(5.5)
0
hi
R
dq

dh
q0
KA 2g
h 
dq0
KA 2gh
• Differentiating Equation 5.4 with respect to q0 gives us,
dh

2h

(5.6)
• The hydraulic resistance is analogous to electrical resistance in
that it is inversely proportional to flow q but directly
proportional to two times the differential pressure, h, or the
driving potential.
• The difference lies in the fact that turbulent flow involves the
square root of the driving potential or head h.
0 0
hi
dq q
• According to Equation 5.1, the denominator of Equation 5.6 is
q, so substituting q into Equation 5.6 gives us, the instantaneous
hydraulic resistance as
dh
 R 
2h

Hydraulic Capacitance:
• In a tank being filled with a liquid, the equation for the
volume (V) of the liquid in the tank is given by the following
equation:
(5.7)
where
V(t) = the volume of liquid as a function of time
h(t) = height of liquid
A = the surface area of the liquid in the tank
• Note that the volume V of the tank and the liquid height or
head are a function of time. The flow of liquid into the tank,
qi , and the flow liquid out of the tank, qo , vary with time.
V(t)  Ah(t)

(5.8)
to the equation for electrical
• Comparing this
capacitance (i.e., C=q/V) clearly shows that liquid
capacitance Cl
tank, or Cl
is simply the surface area of the liquid in the
= A. Furthermore, taking the derivative of
Equation 5.8 with respect to time yields
(5.9)
h(t) Potential
equation
• Rearranging equation 5.7,
A 
V(t)

Quantity
dV(t)
 A
h(t)
dt dt

• Consider the liquid-level system shown in Figure.
• Assume that the system consists of a tank of uniform cross-
sectional area A, which is attached a flow resistance R such as
valve, a pipe, or a weir.

• The volumetric output flow-rate qo, through the resistance R,
is related to the head h and can be given by a linear
relationship as
(5.10)
• The transient mass balance equation can be given as
Mass flow-in – Mass flow-out = Rate of accumulation of mass
in the tank
• In terms of variables used in this analysis, the mass balance
equation becomes
(5.11)
0
q
R

h
i 0
i 0
dt dt dt
q (t)  q (t) 
d(V)

d(Ah)
⇒ q (t)  q (t)  A
dh

• Substituting eqn. 5.10 into 5.11,
(5.12)
• Rearranging eqn. 5.12,
(5.13)
(5.14)
• Taking Laplace Transform of eqn. 5.14 by assuming zero
initial conditions,
i
R dt
q (t) 
h
 A
dh
i
dt R
q (t)  A
dh

h
i
dt
Rq (t)  RA
dh
 h

RAs 1H(s)  RQi (s)
Qi (s) RAs 1
Qi (s)
RAs 1 0
R
Q (s) 
H(s)
(5.15)
• where,
H(s) = L[h(t)] and Qi(s) = L[qi(t)]
H (s)

R
(5.16)
• If however, q0 is taken as the output, the input being the
same, then the transfer function is
Q0 (s)

1
(5.17)

• To obtain the differential equation in terms of variable qo,
dh/dt must be expressed in terms of dqo/dt.
• We know from the equation 5.11 and 5.6 that
i 0
dt
q (t) q (t)  A
dh(t)
0
h
R
dq

dh

Therefore, the system equation becomes
Rearranging and Taking Laplace Transform both sides,
h
dt dt
• or dh(t) = Rh*dq0(t). Taking the derivative of this equation
with respect to time yields
dh(t)
 R
dq0
i 0 h
dt
q (t)  q (t)  AR
dq0 (t)
 
0 h
i
o
o h i
i h
dt
Q (s) 1
Q (s)
q (t)  AR
dq0 (t)
 q (t)
Q (s) 1 AR s  Q (s) ⇒ 
1 AR s

Ex1: Determine the differential equation for flow (qo) out of the
process tank as shown earlier. Find the system time constant if
the operating head is 5m, the steady state flow is 0.2m3/s, and
the surface area of the liquid is 10m2.
The system time constant ‘τ’
  ARh  500s
0
h
q m2
R 
2h
 50
s

Assignment 1.3: A tank having a time constant of 1 min and a
resistance of (1/9) ft/cfm is operating at steady state with an
inlet flow of 10ft3/min. At time t=0, the flow is suddenly
increased to 100ft3/min for 0.1 min by adding an additional 9 ft3
of water to the tank uniformly over a period of 0.1 min. Plot the
response in tank level and compare with the impulse response.

References:
• Modern Control Engineering, 5th Edition, by Katsuhiko Ogata
• Measurement and Control Basics, 3rd Edition, by Thomas A.
Hughes.
• Process Control: Concepts Dynamics And Applications by Shio
Kumar Singh

mathematical modeling of liquid-level systems

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