Electromechanical Energy Conversion

Unit 3

Energy in magnetic systems – field energy, coenergy and mechanical force – singly and multiply excited systems.

Syllabus

IntroductionUnit 3. Electromechanical Energy Conversion

Electrical Energy

Advantages: High Efficiecncy Suitable for long distance transmission Can be able to link with other forms of energy

Electromechanical System: Electrical Enenrgy ↔ Mechanical Energy

Electromechanical System

Electromechanical Energy Conversion Devices

• Category I– Signal Producing Devices like microphones,

sensors, speakers etc.,

• Category II– Force producing devices like solenoids, relays,

electromagnets etc.,

• Category III– Continuous energy conversion equipment like

motors, generators.

Electromechanical Energy Conversion Concept

• Motor:

• Generator:

• Energy Balance Equation:

• Energy Flow:

heat into

ConvertedEnergy

field) magnetic(in

StoredEnergy Total

Output

Energy Mechanical

InputEnergy

Electrical Total

heat into

ConvertedEnergy

field) magnetic(in

StoredEnergy Total

Output

Energy Electrical

InputEnergy

Mechanical Total

heatfieldmagneticoutputinput dWdWdWdW

Magnetic System

Types of Magnetic System:i. Single excited systems

Ex: Electromagnetic Relay, Reluctance Motor, Toroid Coil, Hysteresis Motor, Solinoid coil, etc.,

ii. Multiple excited systems Ex: Synchronous motor, Alternators, DC Shunt

Machines, Loud Speakers, etc.,

Energy in Magnetic System

i

v

R

Hinge

Fluxϕ

e N

Core

Armature

x

Energy in Magnetic System• Flux linkage, ---(1)

• EMF induced due to flux linkage, ---(2)

• By applying KVL in the circuit, ---(3)

• Electrical Energy applied, ---(4) • Assuming the applied energy stored in magnetic field,

----(5)

where, dWf is the change in field energy in time dt.

• Substituting ‘e’ from Eqn.(2),

• where, F=Ni is the MMF.

N

dt

de

dt

diReiRv

dteidWe

dteidWdW ef

fe dWFdiNdiddtdt

didW

Energy in Magnetic System

• Energy absorbed by the field for finite change in flux linkages,

--- (6)• Energy absorbed by the magnetic system to set up flux ϕ,

---- (7)

• Practically, ‘λ’ may vary according to ‘i’ or ‘i’ may vary according to ‘λ’ . So, mathematically,

i=i(λ,x) λ-Independent variable

λ= λ(i,x) i – Independent variable• Depending upon the independent variable, the stored field energy is also

the function of i,x or λ,x.

i.e., Wf=Wf(λ,x) or Wf(i,x) --(8)

2

1

2

1

dFdiW f

00

dFdiW f

Energy in Magnetic System

i-λ Relation ship

• i-λ Relation ship is linear for non-saturated magnetic system.

• As per Eqn (7),

• While,

• The co-energy has no physical significance; but it is importanct in obtaining magnetic forces.

Concept of Co-energy

Energy FieldOABO Area0

diW f

energyCo' OABO, Area0

diW f

Energy in Magnetic System

For i-λ linear relationship without magnetic saturation,

Area OABO=Area OACO

i.e., Wf=Wf’

or Wf+Wf’ =Area OABO+Area OACO= iλ

where,

)9(2

1

2

1

2

1 2 SFiW f

F

S

)13(2

1,' 2 ixLxiW f

)12(2

1,

2

xL

xW f

Energy in Magnetic System

Self inductance,

The co-energy,

From the equation (10), it is found that Wf is a function of independent variables λ and x.

The Co-energy is the function of two independent variables ‘i’ and ‘x’,

LiorLi

iL

)10(2

1

2

1)(

2

1

2

1

2

12

2 JoulesL

iWorLiLiiiW ff

FPJoulesPFPFFFiW f

)11(2

1

2

1

2

1

2

1' 2

Mechanical Force

• Magnetic field produces a mechanical force, Ff .

• The force Ff, drives the mechanical system consisting of active and passive mechanical elements.

• Let the armature moves a distance of dx in positive direction.• The mechanical work done by the magnetic field is,

• Based on energy balance equation,

• In such electromechanical systems the independent variables can be (i,x) or (λ,x)

)14( dxFdW fm

fem dW Energy, Stored

in Change

dW Input,

Energy Electrical

dW Output,Energy

Mechanical

)15( ff dWiddxF

Mechanical Force

• Case I: Independent variables are (i,x) i.e, current constant.• Thus λ changes as i and x. Hence,

λ= λ(i,x)

From Eqn (8),

Using Eqns. (15), (16) and (17),

)16(

dxx

dii

d

xiWW ff ,

17

dxx

Wdi

i

WdW ff

f

)18(

dii

W

iidx

x

W

xidxF

dxx

Wdi

i

Wdx

xdi

iidxF

fff

fff

Mechanical Force

• As there is no term of di on LHS, it should be zero on RHS.

• This is the expression for the mechanical force developed by the magnetic coupling field.

0

dii

W

ii f

dxx

W

xidxF f

f

)19(

,,

x

xiW

x

xiiF f

f

)20(,,

xiWxiix

F ff

Mechanical Force

• From the fig,

• Thus for independent variables (i,x),

• Current is kept constant. Such a system is current excited system.

ff WiW '

)22(

,'

x

xiWF f

f

21,,,' xiWxiixiW ff

Mechanical Force

• Case II: Independent variables are (λ,x) i.e., λ is constant.• Thus i changes as λ and x hence,

• Using Eqn.(23) and (15),

• No dλ on LHS, so

),( xii ),( xWW ff

)23(

dxx

Wd

WdW ff

f

dxx

Wd

WiddxF ff

f

)24(

d

Widx

x

WdxF ff

f

xW

idW

i ff ,0

Mechanical Force

• This is the expression for system in which λ is independent variable. i.e., flux producing voltage is constant. Such a system is voltage controlled system.

• In rotational Systems, the force is replaced by torque and linear displacement dx is replaced by angular displacement dθ.

And

)25(

,

x

xWF f

f

)26(

,'

iWT f

f

)27(

,

ff

WT

Multiple Excited System

• For continuous energy conversion devices like alternators, synchronous motors etc, multiple excited magnetic systems are used. Practically, doubly excited systems are widely used.

• Fig shows doubly excited system with two independent sources.

Multiple Excited System

• i1= Current due to source 1

• i2= Current due to source 2

• λ1=Flux linkages due to i1

• λ2=Flux linkages due to i2

• θ- Angular displacement of rotor

• Tf=Torque developed

• Due to two sources, there are two sets of three independent variables i.e., (λ1, λ2, θ) or (i1, i2, θ).

Multiple Excited System

• Case I: Independent variables are (λ1, λ2, θ). i.e., λ1and λ2 are constants.

• From Eqn(27),

• Currents are variables.• While the field energy is,

)1(

,, 21

f

f

WT

)2(,,21

0

22

0

1121

didiW f

Multiple Excited System

• Let,

L11-Self inductance of rotor

L22-Self inductance of stator

L12=L21-Mutual inductance between stator and rotor

λ1=L11i1+L12i2 ---> (3)

λ2=L21i1+L22i2 ----> (4)

Multiply Eqn. (3) by L12 ,

Multiply Eqn. (4) by L11 ,

(3) – (4), 22211112112

21211211211112 iLLiLLiLiLLLL

2221111211211 iLLiLLL

22211212222112

212211112 iLLLiLLiLLL

)5(22211222211

212

111

2211212

122

LLL

L

LLL

Li

221211211112 iLiLLL

Multiple Excited System

• Similarly,

• Using i1 and i2 in Equation (2),

)6(2121111 i

2122211

121221

2122211

1122

2122211

2211,

LLL

L

LLL

L

LLL

Lwhere

21

0

2222112

0

121211121 ,,

ddW f

Multiple Excited System

• Integrating the above equation, we get,

• The self and mutual inductances of the coils are dependent on the angular position θ of the rotor.

)7(2

1

2

1,, 2

2222112211121 fW

Multiple Excited System

• Case II: Independent of variables i1,i2,θ. i.e, i1,i2 are constants.

• The co-energy is given by,

• Using

)8(

,,' 21

iiW

T ff

)9(,,'21

0

22

0

1121 ii

f didiiiW

22211222121111 and iLiLiLiL

21

0

2222112

0

121211121 ,,'ii

f diiLiLdiiLiLiiW

)10(2

1

2

1,,' 2

2222112211121 iLiiLiLiiW f

Multiple Excited System

• Force in a doubly excited system:

where, i1 and i2 are constants which are the stator and rotor currents respectively.

,,'

21 iiW

F f

22222112

211121 2

1

2

1,,' iLiiLiLiiWF f

2222

1221

1121 2

1

2

1 Li

Lii

LiF