unit 3: electromechanical energy conversion
DESCRIPTION
Class Notes on Electrical Machines I (Anna University Chennai Syllabus)TRANSCRIPT
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Electromechanical Energy Conversion
Unit 3
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Energy in magnetic systems – field energy, coenergy and mechanical force – singly and multiply excited systems.
Syllabus
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IntroductionUnit 3. Electromechanical Energy Conversion
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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
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Electromechanical System
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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.
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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
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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.,
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Energy in Magnetic System
i
v
R
Hinge
Fluxϕ
e N
Core
Armature
x
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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
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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
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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
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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
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)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
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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
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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
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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
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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
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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
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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
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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.
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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).
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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
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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
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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
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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
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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
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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