chapter 3-principle of electromechanical energy conversion...electromechanical energy conversion...
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Electric Machines I
2017 Shiraz University of Technology Dr. A. Rahideh
In The Name of God The Most Compassionate, The Most Merciful
2
Table of Contents1. Introduction to Electric Machines
2. Electromagnetic Circuits
3. Principle of Electromechanical Energy Conversion
4. Principle of Direct Current (DC) Machines
5. DC Generators
6. DC Motors
2017 Shiraz University of Technology Dr. A. Rahideh
3
Chapter 3Principle of Electromechanical Energy Conversion
3.1. Electromechanical Relations
3.2. Slow versus Fast Movements
3.3. Mechanical Force Calculations
3.4. Developed Torque in Doubly Excited Systems
2017 Shiraz University of Technology Dr. A. Rahideh
4
Electromechanical RelationsMotoring Case
Electrical System
Coupling Field
Mechanical System
Electrical losses
elecP mechP
Magnetic losses Mechanical losses
Ohmic (copper) losses
Hysteresis losses
Eddy current losses
Bearing losses
Ventilation
Windage
mechfieldelec WWW mechfieldelec dWdWdW
2017 Shiraz University of Technology Dr. A. Rahideh
5
Magnetic System with Single Excitation
irve
i
e
r
v
2irviei
dtirvidteidt 2
mechfieldelec dWdWdW 0mechdW
eidtdWdW fieldelec dtde
N
iddWfield
2017 Shiraz University of Technology Dr. A. Rahideh
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
6
Magnetic Relay with Single ExcitationAssumption 1: The movable part cannot move or is not allowed to move
0mechdW iddWdW fieldelec
2017 Shiraz University of Technology Dr. A. Rahideh
7
Magnetic Relay with Single ExcitationAssumption 1: The movable part cannot move
0mechdW
iddWdW fieldelec
0
idWfield
N
ggcc lHlHNi
B ggcc
field NAdBN
lHlHW
0
0BHg
B
gccfield AdBlBlHW0
0
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
2017 Shiraz University of Technology Dr. A. Rahideh
8
Magnetic Relay with Single ExcitationAssumption 1: The movable part cannot move
where is the core volumeis the air‐gap volume
B
gccfield AdBlBlHW0
0
B
gccfieldBVdBHVW
00
2
2
cVgV
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
Stored magnetic energy in the core
Stored magnetic energy in the air‐gap
2017 Shiraz University of Technology Dr. A. Rahideh
9
Magnetic Relay with Single ExcitationAssumption 1: The movable part cannot move
In the case of linear systems:
0
2
0
2
22 BVBVW g
rcfield
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
Stored magnetic energy in the core
Stored magnetic energy in the air‐gap
constantr
rc
BH0
2017 Shiraz University of Technology Dr. A. Rahideh
10
Magnetic Relay with Single ExcitationExample 1: In the following system if the air‐gap flux density is 1 T and air‐gap length is constant and fringing effect is neglected, calculate:a) The DC source voltage;b) Stored magnetic energy.
1.0
(At/m)H
(T)B
200
0.8
400
1.2
600
0.6
1000800
cm5
i 250N
+ v _
+ e _
5r
cm10
Fixed part
Mov
able
par
t
Spring
mmg 5cm10 Not in scale cm5 cm5
cm10
cm5
2017 Shiraz University of Technology Dr. A. Rahideh
11
Magnetic Relay with Single ExcitationSolution 1: From the curve
1.0
(At/m)H
(T)B
200
0.8
400
1.2
600
0.6
1000800
T1 cg BBAt/m670cH
At/m10795104
1 37
0
g
g
BH
cm5
i 250N
+ v _
+ e _
5r
cm10
Fixed part
Mov
able
par
t
Spring
mmg 5cm10 Not in scale cm5 cm5
cm10
cm5
ggcc lHlHNi 2
m6.0clmm5gl
A4.3321 ggcc lHlH
Ni
V167 rivdc
2017 Shiraz University of Technology Dr. A. Rahideh
12
Magnetic Relay with Single ExcitationSolution 1:
1.0
(At/m)H
(T)B
200
0.8
400
1.2
600
0.6
1000800
T1 cg BB
1
00
2
2BVdBHVW gccfield
33 m10305.01.06.0 cV
35 m10505.01.0005.02 gV
7
253
10421105335103
fieldW
J9.20895.19005.1 fieldW
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13
Energy and CoenergyEnergy
Coenergy
0
idWfield
)A(i
(Wb) Stored Energy
Coenergy
i
field diW0
iWW fieldfield
2017 Shiraz University of Technology Dr. A. Rahideh
14
-i curve in system with changing air-gap
The ‐i curve varies with the air‐gap length
)A(i
(Wb.t)
Increasing air-gap length
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
2017 Shiraz University of Technology Dr. A. Rahideh
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
15
Magnetic Relay with Single ExcitationAssumption 2: The movable part can move but slowly
In this case the current remains constant during the movement.
2017 Shiraz University of Technology Dr. A. Rahideh
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
16
Magnetic Relay with Single ExcitationAssumption 2: The movable part can move but slowly
From o to a: No movement yet, therefore
o )A(i
(Wb.t) Closed
Open a
b
1i
2
1
c
d
oadfieldelec AiddWdW
0mechdW
mechfieldelec dWdWdW
2017 Shiraz University of Technology Dr. A. Rahideh
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
17
Magnetic Relay with Single ExcitationAssumption 2: The movable part can move but slowly
From a to b:
o )A(i
(Wb.t) Closed
Open a
b
1i
2
1
c
d
abcdelec AiiddW 121
oabmech AdW
mechfieldelec dWdWdW
oadobcafieldbfieldfield AAWWdW )()(
2017 Shiraz University of Technology Dr. A. Rahideh
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
18
Magnetic Relay with Single ExcitationAssumption 3: The movable part can move but very fast
In this case the flux linkage remains constant during the movement.
2017 Shiraz University of Technology Dr. A. Rahideh
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
19
Magnetic Relay with Single ExcitationAssumption 3: The movable part can move but very fast
From o to a: No movement yet, therefore
oadfieldelec AiddWdW
0mechdW
mechfieldelec dWdWdW
b
o )A(i
(Wb.t) Closed
Open a
1i
2
1
e
d c
2i
2017 Shiraz University of Technology Dr. A. Rahideh
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
20
Magnetic Relay with Single ExcitationAssumption 3: The movable part can move but very fast
From a to c:
mechfieldelec dWdWdW
b
o )A(i
(Wb.t) Closed
Open a
1i
2
1
e
d c
2i
0 iddWelec
oacmech AdW
oadocdafieldcfieldfield AAWWdW )()(
2017 Shiraz University of Technology Dr. A. Rahideh
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
21
Magnetic Relay with Single ExcitationAssumption 3: The movable part can move but very fast
From c to b: No movement, therefore
cbedfieldelec AiddWdW
0mechdW
mechfieldelec dWdWdW
b
o )A(i
(Wb.t) Closed
Open a
1i
2
1
e
d c
2i
2017 Shiraz University of Technology Dr. A. Rahideh
1‐Why does the current remain constant during the slow movement?
2‐Why does the flux linkage remain constant during the fast movement?
3‐Why should the currents at the open and closed stages be the same?
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
22
Some Questions (H.W)
2017 Shiraz University of Technology Dr. A. Rahideh
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
23
Magnetic Relay with Single ExcitationAssumption 4: The movable part moves with normal speed
2017 Shiraz University of Technology Dr. A. Rahideh
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
24
Magnetic Relay with Single ExcitationAssumption 4: The movable part moves with normal speed
From o to a: No movement yet, therefore
oadfieldelec AiddWdW
0mechdW
mechfieldelec dWdWdW
b
o )A(i
(Wb.t) Closed
Open a
1i
3
1
f
d
c
2i
e 2
2017 Shiraz University of Technology Dr. A. Rahideh
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
25
Magnetic Relay with Single ExcitationAssumption 4: The movable part moves with normal speed
From a to c:
mechfieldelec dWdWdW
acedelec AiddW
oacmech AdW
oadoceafieldcfieldfield AAWWdW )()(
b
o )A(i
(Wb.t) Closed
Open a
1i
3
1
f
d
c
2i
e 2
2017 Shiraz University of Technology Dr. A. Rahideh
i
N
+ v _
+ e _
r
g
Fixed part
Mov
able
par
t
Spring
26
Magnetic Relay with Single ExcitationAssumption 4: The movable part moves with normal speed
From c to b: No Movement, therefore
mechfieldelec dWdWdW
b
o )A(i
(Wb.t) Closed
Open a
1i
3
1
f
d
c
2i
e 2
cbfefieldelec AiddWdW
0mechdW
2017 Shiraz University of Technology Dr. A. Rahideh
27
Electromechanical Energy ConversionExample 2: The energy conversion cycles of two machines are OABO and OABCO curves shown below. If the energy conversion efficiency is defined as follows, calculate R1 and R2
EnergyElectricalInputEnergyConverted
R
B
O )A(i
(Wb.t)
A
4
3
1
(1)
B
O )A(i
(Wb.t)
A
4
3
1
(2)
C
2
2017 Shiraz University of Technology Dr. A. Rahideh
28
Electromechanical Energy ConversionSolution 2: Part (1)
4.0104
1
11
elec
mech
WW
R
B
O )A(i
(Wb.t)
A
4
3
1
(1)
10443
1
1
01
ddididW B
A
A
Oelec
41 OABOmech AW
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29
Electromechanical Energy ConversionSolution 2: Part (2)
7.0107
2
22
elec
mech
WW
R
10443
3
3
1
1
02
idddidididW C
B
B
A
A
Oelec
72 OABCOmech AW B
O )A(i
(Wb.t)
A
4
3
1
(2)
C
2
2017 Shiraz University of Technology Dr. A. Rahideh
30
Mechanical ForceThe average force is calculated as the mechanical work divided by the displacement
ntDisplacemeWorkMechanical
aveF
g
dx
i
N
+ v _
r
Fixed part
Mov
able
par
t
Spring
x
)A(i
(Wb.t)
a
0x
gx
x dxx
Fully open (no movement yet)
Fully closed
o
c h b
k
d
f
2017 Shiraz University of Technology Dr. A. Rahideh
31
Mechanical Force
• If the flux linkage is constant (fast movement),
)A(i
(Wb.t)
a
0x
gx
x dxx
Fully open (no movement yet)
Fully closed
o
c h b
k
d
f
oabmech AdW
oahmech AdW
0elecdW
mechfieldelec dWdWdW
fieldmech dWdW
fielddWFdx
cte
field
xxW
F
),(
2017 Shiraz University of Technology Dr. A. Rahideh
32
Mechanical Force
• If the current is constant (slow movement),
)A(i
(Wb.t)
a
0x
gx
x dxx
Fully open(no movement yet)
Fully closed
o
c h b
k
d
f
oabmech AdW
oacmech AdW
okafokcdacdfelec AAAdW
mechfieldelec dWdWdW fieldmech WddW
ctei
field
xxiW
F
),(
fieldfieldafieldafieldcfieldcfieldelec WddWWWWWdW )()()()(
fieldWdFdx
2017 Shiraz University of Technology Dr. A. Rahideh
Translational movementForce
Rotational movement Torque
33
Force and Torque Calculation
where is the angular position.
ctei
field
xxiW
F
),(
cte
field
xxW
F
),(
ctei
field iWT
),(
cte
fieldWT
),(
2017 Shiraz University of Technology Dr. A. Rahideh
34
Electromechanical Energy ConversionExample 3: In an electromechanical system for
and . Calculate the force exerted on the movable part if the current is 3 A and air‐gap length is 5 cm.
Solution: 1st method (Coenergy) 2nd method (Energy)
209.0xi 40 i cm103 x
i
field diW0
ctei
field
xxiW
F
),(
cte
field
xxW
F
),(
0
idWfield
2017 Shiraz University of Technology Dr. A. Rahideh
35
Electromechanical Energy Conversion
Solution 3: 1st method (Coenergy)
For and
209.0xi 40 i cm103 x
2/3
00 3209.009.0 i
xdi
xidiW
ii
field
2/32 3
209.0),(i
xxxiW
Fctei
field
A3i cm5x ?F
xi09.0
A3i cm5x N1243209.0 2/3
2 ix
F
2017 Shiraz University of Technology Dr. A. Rahideh
36
Electromechanical Energy Conversion
Solution 3: 2nd method (Energy)
For and
209.0xi 40 i cm103 xA3i cm5x ?F
A3i cm5x
309.009.0
3
2
2
0
2
0
xdxidWfield
3
2
3
2
09.031
09.02
309.02),(
x
ixxx
xWF
cte
field
N7.12409.031
09.02
3
2
xixF
2017 Shiraz University of Technology Dr. A. Rahideh
37
Electromechanical Energy ConversionExample 4: In the following system assume flux is constant during the movement, calculate the average force.
g
Fixed part A
Movable part N
xxg
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38
Electromechanical Energy ConversionSolution 4: Since the permeability of the core goes to infinity, the system is linear.
g
Fixed part A
Movable part N
x xg
Li
2NL
221
2
21
212
21
LiLiWfield
N
2
021
AxgWfield
AxxW
Fcte
field
0
221 1),(
2017 Shiraz University of Technology Dr. A. Rahideh
39
Electromechanical Energy ConversionExample 5: In an electromechanical system the characteristics is defined as for , calculate the average force when .
i 22/3 15.2 xi m10 xm6.0x
0
223
0)1(5.2 dxidWfield
)1(245),( 2
xx
xWF
cte
field
224525
52 )1( xWfield
26.0 xF
2017 Shiraz University of Technology Dr. A. Rahideh
40
Developed Torque in Doubly Excited Systems
Consider the following electric motor with double excitation.
wherestator currentrotor currentstator flux linkagerotor flux linkagestator self‐inductancerotor self‐inductancemutual inductance
rrsselec dididW
is
vs
2sN
2sN
Stator
vr
rN
ir
rrssrr
rsrsss
iLiMiMiL
sr
risi
sLrLsrM
2017 Shiraz University of Technology Dr. A. Rahideh
41
Developed Torque in Doubly Excited Systems
The inductances are defined as follows
wherestator number of turnsrotor number of turnsreluctance seen by stator fluxreluctance seen by rotor fluxreluctance seen by resultant flux
s
ss
NL
2
is
vs
2sN
2sN
Stator
vr
rN
ir
s
r
rNsN
sr
r
rr
NL
2
sr
rssr
NNM
2017 Shiraz University of Technology Dr. A. Rahideh
42
Developed Torque in Doubly Excited Systems
Assumption 1: The rotor cannot rotate
is
vs
2sN
2sN
Stator
vr
rN
ir
rrsselec dididW
rrssrr
rsrsss
iLiMiMiL
rrssrrrsrssselec iLiMdiiMiLdidW
rrrsrsrrssrssselec diiLdiiMdiiMdiiLdW
mechfieldelec dWdWdW
(1)
rrrrssrsssfield diiLiidMdiiLdW
rrss i
rrr
ii
rssr
i
sssfield diiLiidMdiiLW000
0
2212
21
rrrssrssfield iLiiMiLW
2017 Shiraz University of Technology Dr. A. Rahideh
43
Developed Torque in Doubly Excited Systems
Assumption 2: The rotor can rotate
Taking differentiation from the relationyields
is
vs
2sN
2sN
Stator
vr
rN
ir
rrsselec dididW
rrssrr
rsrsss
iLiMiMiL
rrssrrrsrssselec iLiMdiiMiLdidW
rrrrrsrrssrsr
srsrrssrssssselec
dLidiiLdMiidiiM
dMiidiiMdLidiiLdW2
2
rrrrrsrsr
srsrrssrsssssfield
dLidiiLdiiM
dMiidiiMdLidiiLdW2
21
221
2212
21
rrrssrssfield iLiiMiLW
2017 Shiraz University of Technology Dr. A. Rahideh
44
Developed Torque in Doubly Excited Systems
Assumption 2: The rotor can rotate
is
vs
2sN
2sN
Stator
vr
rN
ir
rrsselec dididW
rrssrr
rsrsss
iLiMiMiL
rrrrrsrrssrsr
srsrrssrssssselec
dLidiiLdMiidiiM
dMiidiiMdLidiiLdW2
2
rrrrrsrsr
srsrrssrsssssfield
dLidiiLdiiM
dMiidiiMdLidiiLdW2
21
221
mechfieldelec dWdWdW
rrsrrsssmech dLidMiidLidW 2212
21
2017 Shiraz University of Technology Dr. A. Rahideh
45
Developed Torque in Doubly Excited Systems
Assumption 2: The rotor can rotate
Since the torque is defined as
it yields
is
vs
2sN
2sN
Stator
vr
rN
ir
rrsrrsssmech dLidMiidLidW 2212
21
ddWT mech
ddL
id
dMiiddL
iT rr
srrs
ss
2212
21
Electromagnetic torque
Reluctance torque Reluctance torque
2017 Shiraz University of Technology Dr. A. Rahideh
46
Developed Torque in Doubly Excited Systems
Case 1: Both rotor and stator have salient structures:, and are function of .
Reluctance torque is independent of current direction.
is
vs
2sN
2sN
Stator
vr
rN
ir
ddL
id
dMiiddL
iT rr
srrs
ss
2212
21
Electromagnetic torque
Reluctance torque due to rotor saliency
Reluctance torque due to stator
saliency
sL rL srM
2017 Shiraz University of Technology Dr. A. Rahideh
47
Developed Torque in Doubly Excited Systems
Case 2: Rotor has salient structures but stator is cylindrical:and are function of .
ddL
id
dMiiddL
iT rr
srrs
ss
2212
21
Electromagnetic torque
Reluctance torque due to rotor saliency
sL srM
0
ddMii
ddL
iT srrs
ss 2
21
is
vs
Stator
vr
rN
ir
2017 Shiraz University of Technology Dr. A. Rahideh
48
Developed Torque in Doubly Excited Systems
Case 3: Stator has salient structures but rotor is cylindrical:and are function of .
ddL
id
dMiiddL
iT rr
srrs
ss
2212
21
Electromagnetic torque
Reluctance torque due to rotor saliency
rL srM
0
ddL
id
dMiiT rr
srrs
221
ir
is
vs
2sN
2sN
Stator
vr
rN
2017 Shiraz University of Technology Dr. A. Rahideh
49
Developed Torque in Doubly Excited Systems
Case 4: Both rotor and stator are cylindrical (non‐salient):only is a function of .
ddL
id
dMiiddL
iT rr
srrs
ss
2212
21
Electromagnetic torque
srM
0
ddMiiT sr
rs
0
is
ir
Stator
vr
rN vs
2017 Shiraz University of Technology Dr. A. Rahideh
50
Developed Torque in Doubly Excited Systems
Example 6: In the following figure the rotor has no winding and the self‐inductance is assumed to be where is the rotor position. If stator current is .a) Calculate the torque exerted on the rotor.b) Find the conditions in which the average torque is not zero if
where is the angular velocity of the rotor and is the initial position of the rotor.
ddL
id
dMiiddL
iT rr
srrs
ss
2212
21
2cos10 LLLs tIi ms sin
tm m
si
N
Stator
Rotor
m
2017 Shiraz University of Technology Dr. A. Rahideh
51
Developed Torque in Doubly Excited Systems
Solution 6:Part a)
Since rotor has no winding: d
dLi
ddMii
ddL
iT rr
srrs
ss
2212
21
2cos10 LLLs tIi ms sin
si
N
Stator
Rotor
m
0ri
ddL
iT ss2
21
2cossin 102
21 LL
ddtIT m
tLIT m 21
2 sin2sin
2017 Shiraz University of Technology Dr. A. Rahideh
52
Developed Torque in Doubly Excited Systems
Solution 6:Part b)
To have non‐zero average torque the coefficient of t in one of the above sin terms should be zero:
or or
2cos10 LLLs tIi ms sin
si
N
Stator
Rotor
m
0ritLIT m 2
12 sin2sin
tm
2
2cos12sin12 ttLIT mm
tttLIT mmmm 2sin2sin2sin 21
21
12
21
0m m m
2017 Shiraz University of Technology Dr. A. Rahideh
53
Developed Torque in Doubly Excited Systems
Solution 6:Part b)
1) If
2) If
si
N
Stator
Rotor
m
tttLIT mmmm 2sin2sin2sin 21
21
12
21
0m
m
2sin12
21 LIT mave
2sin12
41 LIT mave
2017 Shiraz University of Technology Dr. A. Rahideh
54
Electromechanical Energy ConversionExample 7: In the following system assume the movable part is fixed and the air‐gap length is .1) Calculate the current and voltage if flux density in air‐gap is 0.5 T.2) Calculate the stored energy.
mm1d
3) Obtain the force exerted on the movable part.
4) Compute the inductance of the winding.
1.0
(At/m)H
(T)B
200
0.8
400
1.2
600
0.6
1000800
0.4 0.2
4r
V
Movable
cm5
i
500N
cm5 Spring
cm5
cm10
Not in scale cm5
d
cm10
cm5
2017 Shiraz University of Technology Dr. A. Rahideh
55
Electromechanical Energy ConversionSolution 7:Part 1)
From the curve
mm1 gld
T5.0 cg BB At/m350cH
1.0
(At/m)H
(T)B
200
0.8
400
1.2
600
0.6
1000800
0.4 0.2
NllH
NlHlH
i gB
ccggccg
0
m6.0cl 70 104
A215.1i
V86.4 riV
?i ?V
4r
V
Movable
cm5
i
500N
cm5 Spring
cm5
cm10
Not in scale cm5
d
cm10
cm5
2017 Shiraz University of Technology Dr. A. Rahideh
56
Electromechanical Energy ConversionSolution 7:Part 2)
mm1 gldT5.0 cg BBB At/m350cH
1.0
(At/m)H
(T)B
200
0.8
400
1.2
600
0.6
1000800
0.4 0.2
m6.0cl 70 104
B
gccfieldBVdBHVW
00
2
2
33 m105.105.005.06.0 cV36 m105.205.005.0001.0 gV
J38.0fieldW
?fieldW
4r
V
Movable
cm5
i
500N
cm5 Spring
cm5
cm10
Not in scale cm5
d
cm10
cm5
2017 Shiraz University of Technology Dr. A. Rahideh
57
Electromechanical Energy ConversionSolution 7:Part 3)
mm1 gld
T5.0 cg BBB
m6.0cl 70 104
B
ggcccfieldBAldBHAlW
00
2
2
xlg
0
2
2
BAx
WF g
cte
field
23 m105.205.005.0 gA
N7.248F
?F
4r
V
Movable
cm5
i
500N
cm5 Spring
cm5
cm10
Not in scale cm5
d
cm10
cm5
2017 Shiraz University of Technology Dr. A. Rahideh
58
Electromechanical Energy ConversionSolution 7:Part 3)
mm1 gld
T5.0 cg BBB
m6.0cl 70 104
iNAB
iN
iL
23 m105.205.005.0 gA
H514.0L
?L
215.15.0105.2500 3
L
4r
V
Movable
cm5
i
500N
cm5 Spring
cm5
cm10
Not in scale cm5
d
cm10
cm5
2017 Shiraz University of Technology Dr. A. Rahideh
59
Electromechanical Energy ConversionExample 8: In the following system assume the current is 1.215 A and the air‐gap is fully closed.1) Calculate the flux density.2) Calculate the force exerted on spring.3) Calculate the stored energy
1.0
(At/m)H
(T)B
200
0.8
400
1.2
600
0.6
1000800
0.4 0.2
0d
4r
V
Movable
cm5
i
500N
cm5 Spring
cm5
cm10
Not in scale cm5
d
cm10
cm5
2017 Shiraz University of Technology Dr. A. Rahideh
60
Electromechanical Energy ConversionSolution 8:Part 1)
From the curve
0 gld
T21.1cB
1.0
(At/m)H
(T)B
200
0.8
400
1.2
600
0.6
1000800
0.4 0.2
ggcc lHlHNi
m6.0cl 70 104
?B
cclHNi At/m1012c
c lNiH
A215.1i
4r
V
Movable
cm5
i
500N
cm5 Spring
cm5
cm10
Not in scale cm5
d
cm10
cm5
2017 Shiraz University of Technology Dr. A. Rahideh
61
Electromechanical Energy ConversionSolution 8:Part 2)
0 gldT21.1cB At/m1012cH
1.0
(At/m)H
(T)B
200
0.8
400
1.2
600
0.6
1000800
0.4 0.2
m6.0cl 70 104
B
gccfieldBVdBHVW
00
2
2
33 m105.105.005.06.0 cV
0gV
J92.0fieldW
?fieldW
B
ccfield dBHVW0
4r
V
Movable
cm5
i
500N
cm5 Spring
cm5
cm10
Not in scale cm5
d
cm10
cm5
2017 Shiraz University of Technology Dr. A. Rahideh
62
Electromechanical Energy ConversionSolution 8:Part 3)
0 gld
T21.1 cg BBB
m6.0cl 70 104
B
ggcccfieldBAldBHAlW
00
2
2
xlg
0
2
2
BAx
WF g
cte
field
23 m105.205.005.0 gA
N4.1456F
?F
4r
V
Movable
cm5
i
500N
cm5 Spring
cm5
cm10
Not in scale cm5
d
cm10
cm5
2017 Shiraz University of Technology Dr. A. Rahideh
63
Developed TorqueExample 9: In the following figure the rotor has no winding and the self‐inductance is assumed to be where is the rotor position. If stator current is 10 Aac with frequency of 60 Hz: .a) In which rotational velocity the average torque is not zero if
where is the angular velocity of the rotor and is the initial position of the rotor.
b) In the velocity obtained above calculate the maximum torque and mechanical power.
c) Obtain maximum torque at zero velocity.
ddL
id
dMiiddL
iT rr
srrs
ss
2212
21
4cos2.02cos3.01.0 sL
tm m
si
N
Stator
Rotor
m
2017 Shiraz University of Technology Dr. A. Rahideh
64
Developed TorqueSolution 9:Part a)
ddL
iT ss2
21
4cos2.02cos3.01.0 sL tm
ti ss cos10
si
N
Stator
Rotor
m
4cos2.02cos3.01.0cos10 221
ddtT s
120602 s
4sin8.02sin6.0cos50 2 tT s
)(4sin8.0)(2sin6.02
2cos150
tttT mm
s
2017 Shiraz University of Technology Dr. A. Rahideh
65
Developed TorqueSolution 9:Part a)
or or
4cos2.02cos3.01.0 sL tm
ti ss cos10
si
N
Stator
Rotor
m
120602 s
)(4sin8.0)(2sin6.02
2cos150
tttT mm
s
422sin4.0422sin4.022sin3.022sin3.0
)(4sin8.0)(2sin6.025
tttt
ttT
smsm
smsm
mm
)sin()sin(cossin 21 bababa
0m 120 sm 6021 sm
2017 Shiraz University of Technology Dr. A. Rahideh
66
Developed TorqueSolution 9:Part b)
at
at
4cos2.02cos3.01.0 sL tm
si
N
Stator
Rotor
m
120602 s
120 sm
6021 sm
2sin5.7aveT
Nm5.7max aveT 4 k
4sin10aveT
W28271205.7max maveTP
Nm10max aveT82 k
W18856010max maveTP
2017 Shiraz University of Technology Dr. A. Rahideh
67
Developed TorqueSolution 9:Part c)
4cos2.02cos3.01.0 sL tm
si
N
Stator
Rotor
m
120602 s
0m 4sin202sin15 aveT
Nm31max aveT
04cos802cos300 aveT
012cos282cos3 2
082cos32cos16 2 62.02cos
81.02cos o8.25o9.71 Nm10max aveT
2017 Shiraz University of Technology Dr. A. Rahideh