chapter three - philadelphia university · 2016. 5. 4. · laith batarseh mathematical modeling...

5/4/2016

1

Chapter three

Laith Batarseh

Mathematical modeling

Home

Nex

t

Pre

vio

us

End


Example [1]: mass-spring-damper system

Find the transfer function for the mechanical system shown in the fig. assumezero initial conditions and the desired output is v1

Solution:To find the mathematical model, youcan apply Newton’s 2nd law to eachmass. This procedure produce two DE

0.

)(

0

21212

2

211211

1

t

dttvkvvbdt

tdvM

trvbtvbbdt

tdvM

Solved examples

5/4/2016

2


Now , use Laplace transform to transform these equations into the s-domain.Remember, we assumed zero initial conditions (i.e. v1(0) = v2(0) = 0.

0 )( 2121222112111

s

sVksVsVbssVMsRsVbsVbbssVM

Rearrange these equations

0- )( 21211211211

sV

s

kbsMsVbsRsVbsVbbsM

These are two equations with tow unknowns: V1(s) and V2(s). To solve theseequations we can use Cramer’s rule. First represent this system of equations bymatrix notation

02

1

121

1211 sR

V

V

s

kbsMb

bbbsM



The solution of V1(s) is calculated as:

To find the transfer function (G(s)): G(s) = V1(s) / R(s)

2

112211

121

bskbsMbbsM

sRskbsMsV

2

112211

12

bskbsMbbsM

skbsMsG

We can represent the transfer function in terms of the displacement x(t) byrepresenting the velocity (v) in terms of (x) : v(t)= dx(t)/dt. Now, transfer it to s-domain: V1(s) =sX1(s). Remember, we assume zero initial conditions. Apply this toG(s) expression:

s

sG

ssR

sV

sR

sX 11


5/4/2016

3


Example [2]: DC motor

Find the transfer function for the mechanical system shown in the fig. assumezero initial conditions and the desired output is v1



Solution:

The DC motor converts direct current (DC) electrical energy into rotationalmechanical energy. The transfer function of the DC motor will be developed for alinear approximation to an actual motor, and second-order effects, such as hysteresisand the voltage drop across the brushes, will be neglected. The air-gap flux ф of themotor is proportional to the field current, provided the field is unsaturated, so that:

ф = Kfif ---- (1)

The motor torque (Tm) is related to air-gap flux (ф) by:

Tm = K1 ф ia(t) = K1Kfif(t)ia(t) ---- (2)

as you can see, there are two controlling variables (if and ia). So, we can chose one of them as a controlling input.

5/4/2016

4



Solution:

Assume the controlling current is the filed current (if) and so the armature current(ia) is constant. Now, Eq(2) become:

Tm = K1 ф ia(t) = [K1Kfia] if(t) = Kmif(t)---- (3)

Transfer Eq(3) using Laplace transformation to the s-domain

Tm(s) = KmIf(s)---- (4)Where: Km is the motor constant.The motor torque (Tm) is the summation of torque delivered to the load (TL) and thedisturbance torque (Td), mathematically:

Tm(s) = TL(s) + Td(s) ---- (5)

Where the load torque (TL) can be calculated as: . Note that the output in our case is Ө or ω. Assume zero initial conditions and transform it to s-domain:

TL(s) = Js2Ө(s)+bsӨ(s)---- (6)

... bJTL



Solution:

We can relate the field current to the field voltage as: . Again

assume zero initial conditions and transform this relation to s-domain:

dt

diLiRv

f

ffff

7

ff

f

fffffsLR

sVsIsIsLRsV

Assume that the disturbance torque is neglected (i.e. Td = 0). Rearrange equations 1-7 to find relation between the input (If(s)) and the output (Ө(s)). This relation is the required transfer function (G(s)):

8//

/

ff

fm

ff

m

LRsJbss

JLK

RsLbJss

K

sV

ssG

5/4/2016

5



Solution:

Now we can represent this mathematical process by the following block diagram

ff sLR

1Km bJs

1

s

1Vf(s) If(s) Tm(s)

_

+

Td(s)

TL(s)

speed

ω(s) Ө(s)

Field Load

This step can be done if take thespeed (ω) as the desired output. Themain difference in calculations willbe using instead of bJTL

.

... bJTL



Solution:

If the armature current used to control the system:

Tm = K1 ф ia(t) = [K1Kfif] ia(t) = Kmia(t)---- (9)

Transfer Eq(9) using Laplace transformation to the s-domain

Tm(s) = KmIa(s)---- (10)The motor torque (Tm) is the summation of torque delivered to the load (TL) and thedisturbance torque (Td), mathematically:

Tm(s) = TL(s) + Td(s) ---- (10)

Where the load torque (TL) can be calculated as: . Note that the output in our case is Ө or ω. Assume zero initial conditions and transform it to s-domain:

TL(s) = Js2Ө(s)+bsӨ(s)---- (11)

... bJTL

5/4/2016

6



Solution:

We can relate the armature current to the armature voltage as:

Again assume zero initial conditions and transform this relation to s-domain:

Vb(s) is the back electromotive force and is given as :Vb(s) = Kbω(s). Rearrange to find

G(S)

12

b

aaaaa v

dt

diLiRv

13

aa

baabaaaa

sLR

sKsVsIsVsIsLRsV

14

mbaa

m

KKbJssLRs

K

sV

ssG



Solution:

14

mbaa

m

KKbJssLRs

K

sV

ssG

fa

m

sLR

K

Kb

bJs

1

s

1Va(s)

Back electromotive force

speed

ω(s) Ө(s)

Armature

Tm(s)

_

+

Td(s)

+

-

5/4/2016

7


Example [3]: Liquid level system

The governing equations can be derived as:

R

hqo

dtqqCdh oi

iRqhdt

dhRC

Laplace

)()()( sRQsHssHRC

1)(

)(

RCs

R

sQ

sH

i

1

1

)(

)(

RCssQ

sQ

i

o

Where:C: tank capacitance h: head R: pipe resistance q : flow rate

= change in liquid stored/change in head=cross-sectional area of the tank


Exercise

Governing equations

1

211

R

hhq

11

1 qqdt

dhC

2

2

2 qR

h

212

2 qqdt

dhC

Find :1. Q1(s)/Q(s)2. Q2(s)/Q(s)3. Q2(s)/Q1(s)

5/4/2016

8


Exercise

1

211

R

hhq

11

1 qqdt

dhC

2

2

2 qR

h

212

2 qqdt

dhC


Exercise

5/4/2016

9


Exercise


Example [4]: thermal system

A simple thermal system is shown in the figure below

For conduction or convection heat transfer

convectionfor

conductionfor

HAK

x

kAK

Where:q: heat flowK: coefficient ∆Ө = temperature difference

Thermal Resistance and Thermal Capacitance.

KR

1

kcal/sec rate, flowheat in change

C,difference turein tempera change o

The thermal resistance for conduction or convection heat transfer is given by

The thermal capacitance C is defined by

mcC C, turein tempera change

kcal stored,heat in changeo

specific heat of substance, kcal/kg °C

5/4/2016

10



Consider the system shown in Figure. If

kcal/sec rate,input heat state-steadyH

Ckcal/ e,capacitanc thermalC

sec/kcal C ,resistance thermalR

C kcal/kg liquid, ofheat specificc

kg in tank, liquid of massM

kg/sec rate, flow liquid state-steadyG

C liquid, outflowing of re temperatustate-steady

C liquid, inflowing of re temperatustate-steady

o

i

Assume that the temperature of the inflowing liquid is kept constant and that the

heat input rate to the system (heat supplied by the heater) is suddenly changed

from to where hi represents a small change in the heat input rate. The

heat outflow rate will then change gradually from to .The temperature of

the out-flowing liquid will also be changed from to .

H ihH H ohH

o o



For the pre-described case, ho, C, and R are obtained, respectively, as:

GchR

McC

Gch

o

o

1

The heat-balance equation for this system is

ioioi Rhdt

dRChh

dt

dCdthhCd

Apply Laplace: 1

RCs

R

sH

ssRHssRCs

i

i

chapter three - philadelphia university · 2016. 5. 4. · laith batarseh mathematical modeling...

Documents