MODULE-1: System Modelling

Part-a: Control Terminology and Transfer Functions;

System Block diagram - simplifying/reduction approach;

Part-b: Modelling simple systems - deriving Transfer Functions - Mechanical

- Electrical

- Electromechanical

- Level control

- Thermal

- Pressure difference

- Hydraulic

- Pneumatic

- Others

Lecture 3c

Simple Pressure difference System

Gas flow Resistance:

Gas flow Capacitance:

q = Gas flow rate = dm/dt

m = Mass of the gas in the vessel (gas-stored) C= capacitance of the gas in the vessel V= Volume of the vessel

---(1)

---(2)

At steady state:

Applying Laplace Transform (assuming zero initial conditions):

---(3)

Substituting in (3)

Large multiplication of force

Hydraulic systems are preferred for continuous control of motion

with significant external loads.

Pressure is transmitted undiminished in an enclosed static fluid!

Pstatic fluid = ρ.g.h

ρ = m/V Fluid density

g = acceleration of gravity

h = depth of fluid

PASCAL’s Principle

Standard symbols used & SI units

Gases:

Liquids:

oi qqdt

dv

oq

hR

h

vC

Constraints R and C in the output flow

Hydraulic Systems

Study of Incompressible fluids such as oil and water

Fluid’s density remains constant despite changes in fluid pressure.

Conservation of mass is equivalent to conservation of volume

Main System Variables:

Pressure,

Mass (m) and

Mass flow rate (qm)

The volume flow rate:

(ρ, the density = constant )

qqm . ---(1)

If q1 and q2 are the total volume inflow and outflow rates in the container, such that

2

1

qq

qq

out

in

21 qqqq outin

V

qq

)( 21

For a container of volume V holding the incompressible fluid of mass m,

qin and qout are the inflow and outflow rates

The conservation of the mass is given by:

---(2) Vqqm outin .)(

---(3)

Simple Hydraulic System

At lower piston, the oil flow-rate is

given by :

Pilot Valve

Power Cylinder

e will cause a displacement in x: e (and x) will move to the right:

Ports I and II will open as shown;

dt

dyAq ..

dtA

qdy

xKq 1

K1 is a constant

At Port-II, the oil flow rate would be:

Find the Transfer function Y/X:

is oil density

A is the piston cross-section area Motion of the Piston:

---(4) ---(5)

Displacement x as a result of adding two small displacements

For the Flapper movement:

yba

ae

ba

bx

Feedback link:

b

yx

ba

ye

xKq 1

dt

dyAq ..

dt

dyAxK ..1

)()(1 sAsYsXK

Hence, equating the two, we get:

Transfer Function: sA

K

sX

sY 1

)(

)(

sA

K1 Y(s) X(s)

---(4)

---(5)

Applying Laplace Transform (assuming zero initial conditions):

---(6)

Proportional Controller

Transfer Function:

ba

a

s

Ks

K

ba

b

sE

sY

1)(

)(

Feedback link: b

yx

ba

ye

Under normal operating conditions, we can

write (8) as |Ka/[s(a+b)]| >>1 pKa

b

sE

sY

)(

)(

)(.)(1.)()(

baasK

sK

ba

bsEsY

---(7)

---(8)

---(9)

Working medium compressible fluid (air/gas);

Slower response than that of hydraulic systems;

Forces are greater than those available from electrical

drive systems;

Quantities used:

— Mass (m),

— Volume (V),

— Pressure (P)

— Temp. (T)

The main variables of pneumatic system are:

Mass flow rate q (a through variable)

pressure P

Pneumatic Systems

(These are analogues of current and voltage in electrical networks)

Ideal Gas Law:

Linearizing

R is similar to turbulent flow resistance;

∆ p is the pressure drop across the component

The compressible Flow/pneumatic Resistance is modeled as,

p is the absolute pressure of the gas with volume V,

m is the mass,

T its absolute temperature, and

Rg the gas constant that depends on the type of gas

TmRpV g

pmqR 2

---(1)

---(2)

---(3)

Pneumatic Proportional Controller

For the Flapper movement:

yba

ae

ba

bx

Bellows act like springs:

zKp

zKp

xKp

c

b

b

3

2

1

xK

KKp

K

Kb

2

31

2

3

ykpA sc .

A is the effective area of bellows

and

kS is the equiv. spring constant

Diaphragm Valve ---(4)

---(5)

---(6)

Transfer Function could be obtained by dividing eqn. (4) by (5) and

applying Laplace transform

s

c

k

A

ba

aK

Kba

b

sE

sP

..1

.

)(

)(

xK

KKpc

2

31

yb

ax

b

bae

ykpA sc .

From eqn. (4)

From eqn. (6)

From eqn. (5)

pc K

sE

sP

)(

)(

Exercise: Aircraft Elevator Control

Automobile systems

-such as hydraulic actuators as the

brake on an automobile

Elevators systems

-such as hydraulic jacks and lifts;

lifting heavy loads in the construction

and mining industry.

Aircraft systems

-such as control flaps of airplanes

Automation and Industrial Robots

Automatic controllers

Guided Missiles etc…

Applications of Hydraulic and Pneumatic systems

Rescue Robot

Next coming up:

- Others combinations…

Hydraulic & Pneumatic Systems

Large forces and torques;

High sped of response;

Availability of both Linear and Rotary Actuators (Hydraulic)

Power is not readily available;

Contaminated oil may cause system failure (Hydraulic);

Leakage;

Expensive;