lec 7 first order fluid

7/27/2019 Lec 7 First Order Fluid

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Mathematical Modeling

of Various Systems

Nasir M. Mirza


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Models for Flow of Fluids

First Order Models


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Hydraulic or Fluid Systems

When a system is based on flow of incompressible fluids, we

have a hydraulic system. This appears in chemical processes, automatic control

systems, drive motors and actuators.

A turbine driven by water to generate electricity is anillustration of a system in which hydraulic, mechanical and

electrical sub-systems interact with each other. Variables used to describe a hydraulic system are

flow rate (m3/s),

volume (m3),

height (m) and pressure (N/m2).

These systems have capacitance, resistance to flow andinertia.

Let us consider some simple hydraulic systems throughexamples.


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Example 1: Modeling of Flow of Salt in a Water Tank

A tank contains M (liter)of water

in which are dissolved Q (kg)ofsalt.

About P (liters)of Brine (saltedwater) is dissolved into the tank

per minute . Each liter contains Skg of dissolved salt.

The mixture is kept uniform bystirring and it runs out at thesame rate.

Model this system to find theamount of salt in the water tank.


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Let us define the total amount of salt in the

tank at any time t equal to y(t).

Then the time rate of change of y(t) isequal to inflow of salt minus the out flow.

Let us write down both outflow and in-flows:

The in-flowof salt = P( liters ) S (kg of salt/liter ).

One liter in tank contains y(t)/M of salt, then P liters outgoingwill have = P y(t)/M ;

We know that the tank contains M (liter)of water in whichare dissolved Q (kg)of salt. Then P (liters)of Brian and eachliter containing Skg of dissolved salt runs into the tank perminute. The P litersof water leaves the tank.



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which may be rearranged as

The initial amount of salt is y( t = 0 ) = Q.

)t(yM

P

SPdt

)t(dy

SP)t(yM

P

dt

)t(dy

The differential equation model for the salt

in the tank is given by following balance offlow rates:


Rate of Change of salt in tank = in-flow rate out-flow rate


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This is a linear first order ordinary differential equation withnon-homogeneous term.

We can simulate the amount of salt in the tank at any time tusing this model.

The constraints are that the inflow rate and outflow rates arefixed and y( t = 0 )is equal to a given value Q.

The fixed parameters in the system are P, S and Mrespectively.

Let us do the example using some numerical values.

SPtyMP

dttdy )()( Model equation



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Consider y(t) is the total amount of salt in tank.

Inflow rate = 5 x 2 kg per min = 10 kg/min

The tank shown in Figure contains 200 liter of

water in which an initial amount of 40 kg of saltis dissolved.

Five liters of brine and each liter contain 2 kg of

salt and run into the tank per minute.

Time rate of change of y = Inflow rate

outflow rate


The mixture is kept uniform by stirring. It runs out at the same rate.


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)(025.00.10 tydt

dy

One liter contains y(t)/200 of salt, then 5

liters outgoing will have

= 5y(t)/200 = 0.025y(t) ;

y(0) = 40, (initial condition)

Salt outflow rate = 0.025y(t)

Inflow rate = 5 x 2 kg per min = 10 kg/min

Time rate of change of y = Inflow rate outflow rate

Mathematical Model



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Order: 1;

Linearity:It is a linear equation as it has no product term

)(025.00.10 tydt

dy

y(0) = 40,

It is an initial value problem

40)0(;0.10)(025.0 ytydt

dy

Its Standard form is given below:

The model equation is given below:



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Homogeneity:It is non-homogeneous equation with aforce term (10.0).

Conditions:Initial conditions are given.

Coefficients:There are constant coefficients

Driving term type:It has analytical term as opposed to a

tabular form.

Model Equation Type:It is a single ordinary differentialequation (ODE) based model.

Some Properties of the Model:

40)0(;0.10)(025.0 ytydt

dy



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Analytical Solution and Interpretation:

0.1040

)(

ty

dt

dy

0.10)(025.0 tydtdy

dty

dy025.0

400

Integrating we get

Cty 025.0|400|ln

)025.0exp(400 tCy

Using initial condition, C = -360 ; and the solution is

y(0) = 40

)025.0exp(360400 ty



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Model in SIMULINK: a patch or block diagram

40)0(;0.10)(025.0 ytydt

dy

Scope

1/s

Integrator

-0.025

Gain

10

Constant

Add



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Results from SIMULINK

40)0(;0.10)(025.0 ytydt

dy

0 100 200 300 400 5000

100

200

300

400

500

600

saltcon

centration,y(t)

Time(sec)

y(0) = 40You can see the

equilibrium value is

400 from graph and

from above equation

when dy/dt = 0.



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This second example has a tank shown in

Figure and it contains 1000 liters of water inwhich an initial amount of 200 kg of salt is

dissolved.

Fiftyliters of brine each contain (1 + cos t) kgof

salt and run into the tank per minute. The

deriving force term has time dependence now.


The mixture is kept uniform by stirring.

It runs out at the same rate.

Model this nonhomogeneous system.

Example 2: First Order fluid Model


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)(05.0)cos1(50 tytdt

dy

Inflow rate = 50 x (1 - cos t) kg per min

Let us say y(t) is the total amount of salt in tank.

Salt outflow rate = 0.05y(t)

y(0) = 200, It is an initial value.Mathematical Model



One liter contains y(t)/1000 kg of salt, then 50 liter

outgoing salt will have = 50y(t)/1000;


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Order: first;

Dependent variable = y(t);

independent Variable = t

Linearity: It is a linear equation as it has no product term for y

and its derivatives.

y(0) = 200, let us say.

Standard form

)(05.0)cos1(50 tytdt

dy

)cos1(5005.0 ty

dt

dy



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Conditions:Initial conditions are given.

Coefficients:There are constant coefficients

Driving term type: It has analytical term as

opposed to a tabular form.

Model Equation Type: It is a single ordinary

differential equation (ODE) based model.

Some Properties of the Model

Homogeneity:It is non-homogeneous equation

with a time dependent force term.



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Solution of the Model and Interpretation:

Using the Integrating factor method we get

Cdttttty )cos1(50)05.0exp()05.0exp()(

Then using initial condition , the solution is

)cos1(5005.0 tydt

dy

)05.0exp(5.802sin88.49cos494.21000)( tttty

y(0) = 200,

Example: First Order non-homogeneous Model


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Model in MATLAB/SIMULINK

)cos1(5005.0 tydtdy y(0) = 200.

cos

TrigonometricFunction

Scope

Ramp

1/s

Integrator

50

Gain1

-0.05

Gain

50

Constant

Add



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0 20 40 60 80 100 120 140 160 180 2000

100

200

300

400

500

600700

800

900

1000

1100

saltconcentratio

n,y(t)

Time(sec)

y(0) = 200

Results from simulation

What is differentabout theequilibrium state of

this system?



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Example 3: Flow of Water Through a Hole in a Tank

A cylindrical tank 150 cm high stands on its circular base

of diameter 100 cm and it is initially filled with water. Atthe bottom of the tank, there is a hole of diameter onecm, which is opened at some instant, so that the waterstarts draining under the influence of gravity.

GIVEN:

According to the Bernoullis law, waterflows out of the hole with velocityproportional to the square root of theheight at that time. Take proportionalityconstant as 5 for this system.

The velocity increases when heightdecreases.

Assume height of water in the tank at anyinstant of time t is h(t).

Develop the mathematical model.

h


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System:A cylindrical tank with maximum waterheight of 150 cm and diameter of 100 cm.

Variable:Water level in the tank at any instant is h(t).

h

Conditions:The initial height of the water

in the tank is 150 cm (i.e. h(t) = 150 cm at

time t= 0). The volume of the tank is fixedand the diameter of the hole does not

change with the passage of time. There is

no internal source.

Fixed Parameters: Acceleration due to gravity g = 980

cm/sec2 , and area of the hole in the tank are fixed

parameters.


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Here we must keep in view that final velocity is high ascompared to initial velocity. Then equation reduced tofollowing first order differential equation:

)()(

thkdt

tdh

The negative sign indicates that high

decreases the velocity increases.h

)()(

thdt

tdh

Such a model has analytical solution for given

initial condition.


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Model in MATLAB/SIMULINK :

k = 5; h(0) = 200 m.h


)()(

thkdt

tdh

simout

To Workspace

Scope

sqrt

MathFunction

1/s

Integrator

-5

Gain

h(0) = 200


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Results from SIMULINK

You can see the equilibrium

value is 400 from graph and

from above equation when

dy/dt = 0.


h0 2 4 6 8 10

0

50

100

150

200

Heightofwater(m)

time(min)

using ode45 in matlab k = 5; h(0) = 200 m.

)()(

thkdt

tdh


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End of slides ----

lec 7 first order fluid

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