Mathematical Modeling of Various Systems-Mathematical Modeling and Simulation-Lecture Slides, Slides of Mathematical Modeling and Simulation

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Mathematical Modeling

of Various Systems

Models for Flow of Fluids

First Order Models

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

• A tank contains M (liter) of water

in which are dissolved Q (kg) of

salt.

• About P (liters) of Brine (salted

water) is dissolved into the tank

per minute. Each liter contains S

kg of dissolved salt.

• The mixture is kept uniform by

stirring and it runs out at the

same rate.

Model this system to find the

amount of salt in the water tank.

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) is equal to inflow of salt minus the out flow.

Let us write down both outflow and in-flows:

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

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

We know that the tank contains M (liter) of water in which are dissolved Q (kg) of salt. Then P (liters) of Brian and each liter containing S kg of dissolved salt runs into the tank per minute. The P liters of water leaves the tank.

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

• This is a linear first order ordinary differential equation with non-homogeneous term.
• We can simulate the amount of salt in the tank at any time t using this model.
• The constraints are that the inflow rate and outflow rates are fixed and y( t = 0 ) is equal to a given value Q.
• The fixed parameters in the system are P, S and M respectively.
• Let us do the example using some numerical values.

y t P S M

P dt

dy t  ( )  

( ) Model equation

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

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 salt is 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

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

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

Order: 1;

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

10. 0 0. 025 y ( t ) dt

dy  

y(0) = 40, It is an initial value problem

 0. 025 y ( t ) 10. 0 ; y ( 0 )  40 dt

dy

Its Standard form is given below:

The model equation is given below:

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

• Homogeneity: It is non-homogeneous equation with a force 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 differential equation (ODE) based model.

Some Properties of the Model:

 0. 025 y ( t ) 10. 0 ; y ( 0 )  40

dt

dy

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

Model in SIMULINK: a patch or block diagram

 0. 025 y ( t )  10. 0 ; y ( 0 )  40 dt

dy

Scope

1/s Integrator

-0.

Gain

10 Constant

Add

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

Results from SIMULINK

 0. 025 y ( t ) 10. 0 ; y ( 0 ) 40 dt

dy

0 100 200 300 400 500

0

100

200

300

400

500

600

salt concentration, y(t)

Time(sec)

y(0) = 40

You can see the equilibrium value is 400 from graph and from above equation when dy/dt = 0.

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

50 ( 1 cos t ) 0. 05 y ( t ) dt

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)

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

Example 2: First Order fluid Model

Time rate of change of y = Inflow rate – outflow rate

One liter contains y(t)/1000 kg of salt, then 50 liter outgoing salt will have = 50y(t)/1000;

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

50 ( 1 cos t ) 0. 05 y ( t )

dt

dy

1. 05 y 50 ( 1 cos t ) dt

dy   

Example 2: First Order fluid Model

Solution of the Model and Interpretation:

Using the Integrating factor method we get

y ( t ) exp( 0. 05 t )^ exp( 0. 05 t ) 50 ( 1 cos t ) dt  C 

Then using initial condition , the solution is

1. 05 y 50 ( 1 cos t ) dt

dy   

y ( t ) 1000  2. 494 cos t  49. 88 sin t  802. 5 exp( 0. 05 t )

y(0) = 200,

Example: First Order non-homogeneous Model

Model in MATLAB/SIMULINK

1. 05 y 50 ( 1 cos t ) dt

dy   ^ y(0) = 200.

cos Trigonometric Function

Scope

Ramp

1/s Integrator

50

Gain

-0.

Gain

50 Constant

Add

Example 2: First Order fluid Model