1. Automatic Control Theory CSE 322
Lec. 4
Mathematical Modeling
of Dynamic System

2. 2. Mechanical Systems

3. Mechanical Systems Part-I: Translational Mechanical System
Part-II: Rotational Mechanical System
Part-III: Mechanical Linkages

4. 2. Rotational Mechanical Systems

5. Basic Elements of Rotational Mechanical Systems
Rotational Spring

6. Basic Elements of Rotational Mechanical Systems
Rotational Damper

7. 3. Mechanical Linkages

8. Gear
Gear is a toothed machine part, such as a wheel or cylinder, that meshes with another toothed part to transmit motion or to change speed or direction.

9. Fundamental Properties
The two gears turn in opposite directions: one clockwise and the other counterclockwise.
Two gears revolve at different speeds when number of teeth on each gear are different.

10. Gearing Up and Down Gearing up is able to convert torque to velocity.
The more velocity gained, the more torque sacrifice.
The ratio is exactly the same: if you get three times your original angular velocity, you reduce the resulting torque to one third.
This conversion is symmetric: we can also convert velocity to torque at the same ratio.
The price of the conversion is power loss due to friction.

11. Why Gearing is necessary?
A typical DC motor operates at speeds that are far too high to be useful, and at torques that are far too low.
Gear reduction is the standard method by which a motor is made useful.

12. Gear Trains

13. Gear Ratio
You can calculate the gear ratio by using the number of teeth of the driver divided by the number of teeth of the follower.
We gear up when we increase velocity and decrease torque.
Ratio: 3:1
We gear down when we increase torque and reduce velocity.
Ratio= 1:3
Follower
Driver
T1 * S1 = T2 * S2
Let: T1=24, T2 =8 , S1=50rpm find S2?
24*50=8*S2  S2 =150rpm
Gear Ratio = # teeth input gear / # teeth output gear
= torque in / torque out = speed out / speed in

14. Mathematical Modeling of Gear Trains
Gears increase or reduce angular velocity (while simultaneously decreasing or increasing torque, such that energy is conserved). 
Energy of Driving Gear = Energy of Following Gear
Number of Teeth of Driving Gear
Angular Movement of Driving Gear
Number of Teeth of Following Gear
Angular Movement of Following Gear

15. Mathematical Modeling of Gear Trains
In the system below, a torque, τa, is applied to gear 1 (with number of teeth N1, moment of inertia J1 and a rotational friction B1). 
It, in turn, is connected to gear 2 (with number of teeth N2, moment of inertia J2 and a rotational friction B2). 
The angle θ1 is defined positive clockwise, θ2 is defined positive clockwise.  The torque acts in the direction of θ1. 
Assume that TL is the load torque applied by the load connected to Gear-2. B1 B2 N1 N2

16. Mathematical Modeling of Gear Trains
For Gear-1
For Gear-2
Since
therefore
Eq (1) B1 B2 N1 N2
Eq (2)
Eq (3)

17. Mathematical Modeling of Gear Trains
Gear Ratio is calculated as
Put this value in eq (1)
Put T2 from eq (2)
Substitute θ2 from eq (3) B1 B2 N1 N2

18. Mathematical Modeling of Gear Trains
After simplification

19. Mathematical Modeling of Gear Trains
For three gears connected together

20. Electro-Mechanical Systems.
Self Study
Electro-Mechanical Systems.

21. Liquid Level Systems.

23. 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.

24. 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.

25. 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.

26. 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 m3/s
Kl = constant in m2/s
and H = steady-state height in m.
The resistance Rl is

27. 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. h

28. Capacitance of Liquid-Level Systemsh

29. Capacitance of Liquid-Level Systemsh

30. Modelling Example-1

31. 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)
(1)
(2)
(3)

32. Modelling Example -1 From eq. (3) substitute into eq. (1)
After simplifying above equation
Taking Laplace transform considering initial conditions to zero
(3)
(1)

33. Modelling Example -1
The transfer function can be obtained as

34. Modelling Example -1
The liquid level system considered here is analogous to the electrical and mechanical systems shown below.

35. 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).

36. Modelling Example - 2
Tank Pipe 1
Tank Pipe 2

37. Modelling Example - 2 Tank 1 Pipe 1 Tank 2 Pipe 2
Re-arranging above equation

38. Modelling Example - 2
Taking LT of both equations considering initial conditions to zero [i.e. h1(0)=h2(0)=0].
(1)
(2)

39. Modelling Example - 2
(1)
(2)
From Equation (1)
Substitute the expression of H1(s) into Equation (2), we get

40. Modelling Example - 2
Using H2(s) = R2Q2 (s) in the above equation

41. Modelling Example -3
Write down the system differential equations.