Electric Motors
M = Km⋅Φ⋅I
from electrcal overview
Lorentz force...
Ref: Chapter 9
Km = constant_for_given_motor
M = torque
N⋅ m
(ref: 2.93)
Φ = magnetic_flux
Wb = 1 weber
I = current
A = 1 amp
(9.1)
1Wb⋅ 1A = 1 N⋅
when rotating, electromotive force induced in rotor given by ..
from electrical overview Faraday's force ...
E = KE⋅Φ⋅n
(ref: 2.96)
KE = constant_for_given_motor
E = induced_electromotive_force
V = 1 volt
Φ = magnetic_flux
Wb = 1 weber
rpm = 6.283
n = rotation_speed
1
(9.2)
Wb⋅ rpm = 0.105 V
min
model motor as resistance in series with EMF generator (opposing applied voltage)
U = E + I⋅ R
M = Km⋅Φ⋅I
with ...
I :=
n collect , Φ →
(9.1)
M
KE⋅Φ
−
M
⋅
R
n=
2
Km
KE⋅ Φ
(9.4)
I
E = KE⋅Φ⋅n
and ...
E := (U − I⋅ R)
Km⋅Φ
U
2
P = E⋅ I + I ⋅R
(9.3)
n :=
U
KE⋅Φ
(9.2)
RA
Um
E
E
KE⋅Φ
separately
excited
field
M⋅ R
−
2
KE⋅Km⋅Φ
(9.5)
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to see an example of DC motor behavior assume a set of reasonable parameters. Not all are independent.
for fixed magnetic field Φ and rpm at maximum power
, maximum current Im and maximum torque Mm
M = Km⋅Φ⋅I
set Φ, n, R and applied voltage U
Φ := 1Wb
derived
n := 100rpm
R := 2Ω
U = E + I⋅ R
E
E⋅ Im = M m⋅n⋅ 2⋅ π
M
M m = Km⋅Φ⋅Im
Km
E = KE⋅Φ⋅n
KE
maximum current
Um := 400V
Im := 10A
E := Um − Im⋅R
E = 380 V
M m :=
Km :=
KE :=
E⋅ Im
M m = 57.753 N⋅ m assuming EMF*I converted
into mechanical power
n ⋅ 2⋅ π
2
Mm
Km = 5.775
Φ⋅Im
E
2
P = U⋅ I = E⋅ I + I ⋅R = M ⋅ n⋅ 2⋅ π + I ⋅R
KE = 36.287
Φ⋅n
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1
11/15/2006
M = U⋅
a :=
Km⋅Φ
R
Km⋅Φ
− n⋅
KE⋅Km⋅Φ
Um⋅
R
b :=
R
2
Km KE⋅Φ
R
− n⋅
KE⋅Km⋅Φ
2
R
= 57.753 N⋅ m
2
M ( U , n ) := ( U⋅ a − b ⋅ n)
R
nn( U , M ) :=
M 0 := 0N⋅ m
Km⋅Φ
M m = 57.753 N⋅ m
(
calculate M when U and n known ...
U⋅ a − M
calculate n when U and M known useful at ends of torque range 0 - M m
b
)
nn Um , M m = 100 rpm
derived check
(
)
nn Um , 0 = 105.26316 rpm
for example at U = 0.25 U m, calculate n at 0 and maximum torque
(
(
)
)
n11 := nn Um⋅0.25, M m n11 = 21.053 rpm
n12 := nn Um⋅0.25, M 0 n12 = 26.316 rpm
maximum torque
0 torque
n1 := n11 , n11 + 0.1rpm .. n12
Mm
plot M vs n for U = 0.25*U m
(
)
N⋅ m
50
M U m⋅0.25, n1
N⋅ m
and if develop similar data for 0.5 * U m, 0.75 * Um and Um
0
obtain the following plot
0
10
20
30
n1
rpm
plot data
torque N*m
60
0.25 * Um
0.5 * Um
0.75 * Um
1.0 * Um
Mm max torque
40
20
0
0
20
40
60
80
100
120
140
rpm
2
11/15/2006 additional operating envelope is available beyond design rpm by reducing the field strength Φ. But the region is limited
by the maximum power available
3
Pmax := M m⋅ 2⋅ π⋅n
beyond a M limited by Pmax
Pmax = 3.8 × 10 W
above base sped and torque with power consatant at P max, torque is limited inversely with rpm
M max⋅nn⋅ 2⋅ π = Pmax
Pmax
M max_n( nn) :=
nn⋅ 2⋅ π nn1 := n , n + 1rpm .. 2 ⋅
n
60
Mm
using .. again
M max_n(nn1)
40
N⋅ m
n=
20
100
150
200
U
M⋅ R
−
KE⋅Φ
KE⋅Km⋅Φ
2
(9.5)
250
nn1
rpm
Pmax
n=
U
KE⋅Φ
M⋅ R
−
nn =
KE⋅Km⋅Φ
2
U
KE⋅Φ
nn⋅ 2⋅ π
−
Pmax
⋅R
U
2
nn =
KE⋅Km⋅Φ
KE⋅Φ
2
⋅R
2⋅π
⋅nn −
KE⋅Km⋅Φ
2
1
solved symbolicly on blank sheet
nnmax( Φ ) :=
2
1
4⋅π⋅KE⋅Km
⋅
2
2 ⋅ Um⋅π⋅Km + 2 ⋅ ⎛ Um ⋅π ⋅Km − 2⋅π⋅KE⋅Km⋅ Pmax⋅R⎞
⎝
⎠
2
2
Φ
calculate n when U and M known - useful at ends of torque range 0 - M m and in this application a and b are functions of Φ
a( Φ ) :=
Km⋅Φ
R
b ( Φ ) :=
M ( U , n) = U⋅ a − b ⋅ n
Km KE⋅Φ
2
R
calculate M when U and n known ...
so with Φ = 0.75 Φ base
n51 := nnmax( Φ⋅0.75)
(
n52 := nn 0.75⋅Φ , M 0
nn( Φ , M ) :=
)
(
b(Φ )
MM( Φ , n ) := Um⋅ a( Φ ) − b ( Φ ) ⋅n
)
and Φ = 0.6 of Φ base
n51 = 133.333 rpm
n52 = 140.351 rpm
n5 := n51 , n51 + 1rpm .. n52
MM( Φ⋅0.75, n51 ) = 43.315 N⋅ m
MM( Φ⋅0.75, n52 ) = 0 N⋅ m
Um⋅a( Φ ) − M
n61 := nnmax( Φ⋅0.6)
n61 = 166.667 rpm
(
n62 = 175.439 rpm
n62 := nn 0.6⋅Φ , M 0
)
n6 := n61 , n61 + 0.1rpm .. n62
MM( Φ⋅0.6, n61 ) = 34.652 N⋅
m
MM( Φ⋅0.6, n62 ) = 0 N⋅ m
3
11/15/2006
60
0.25 * Um
0.5 * Um
0.75 * Um
1.0 * Um
Mm
FI = 0.75
FI = 0.6
Pmax
50
torque N*m
40
30
20
10
0
0
50
100
150
200
250
rpm
now if we plot this data in terms of power,
nrpm ⎞
load( nrpm) := Pmax⋅ ⎛⎜
⎟
1.5⋅ n
⎝
power = torque⋅ rpm⋅ 2 ⋅ π
and superimpose a cubic load curve
reaching max power at 1.5 base rpm
0 ⎞
⎛ 0
⎜
Pmax ⎟
Pmax_plot :=
⎜ n
⎟
⎝
rpm W ⎠
3
⎠
nrpm := 0rpm , 1rpm .. 1.5⋅ n
4000
0.25 * Um
0.5 * Um
0.75 * Um
1.0 * Um
M max
FI = 0.75
FI = 0.6
P max
load
power (W)
3000
2000
1000
0
0
20
40
60
80
100
120
140
160
180
200
220
rpm
So ... as observed in the text: "The operational envelopes show that a DC motor is very suited to drive a propeller for ship
propulsion."
4
11/15/2006
practical aspects ...
IA
shunt motor: field in parallel with armature
IF
operates same as separate excitation ...
field excitation constant
RA
shunt
field
Um
E
I
series motor: field in series with armature
now the field current and armature current is the same
using the same relationships from above ...
M := Km⋅Φ⋅I
(9.1)
E = KE⋅Φ⋅n
(9.2)
U = E + I⋅
R
(9.3)
Φ := KF⋅II
Um
I :=
U− E
2
M := Km⋅KF⋅I
I
→
R
(
I⋅ R = U − KE⋅KF⋅I⋅n
2
M → Km⋅KF⋅I
E := KE⋅Φ ⋅n
M := Km⋅KF⋅I
E → KE⋅ KF⋅ I⋅ n
E := KE⋅ KF⋅ I⋅ n
U − KE⋅KF⋅I⋅ n
)
I=
U − KE⋅ KF⋅ I⋅ n I =
R
I⋅ R + KE⋅KF⋅n = U
R
U
I :=
R + KE⋅KF⋅n
2
M → Km⋅ KF⋅
U
R + KE⋅KF⋅n
2
U
(R + KE⋅KF⋅n)
(9.7)
I = current
2
M := Km⋅Φ⋅I
U = E + I⋅ R
series
field
E
KF = constant_for_given_motor
eliminate I ...
RA
2
M := Km⋅ KF⋅
U
(R + KE⋅KF⋅n)
(9.8)
2
some numerical values for a plot ...
Um := 100V KE := 1
Km := 1
2
M ( U , n ) := Km⋅KF⋅
U
(R + KE⋅KF⋅n)
2
Wb
KF := 1
A
R := 4Ω
(
)
M Um , 100rpm = 47.747 N⋅ m
Imax := 9A
2
M max := Km⋅KF⋅Imax
M max = 81 N⋅
n := 1rpm , 2rpm .. 200rpm
5
11/15/2006
100
Mmax
Torque (N*m)
80
N⋅ m
60
40
20
0
Um
0.5*Um
0
50
100
150
200
250
rpm
motor suitable for traction purposes - high torque at low rpm
Induction motors (AC)
IF := 1
ω :=
1
s
t := FRAME⋅
4⋅π
100
⋅sec
t to go from 0 to 4*π in
100 steps
current sinusoidal (cos) with time
π⎞
⎛
Ib ( t) := IF⋅ cos⎜ ω⋅t − 2 ⋅ ⎟
3 ⎠
⎝
Ia( t) := IF⋅ cos( ω⋅t)
π⎞
⎛
Ic( t) := IF⋅ cos⎜ ω⋅t − 4 ⋅ ⎟
3 ⎠
⎝
field vector displaced by 2*π/3 and 4*p/3, and current at appropriate phase shift applied
IF⋅ cos( ω⋅t)
= 1
Bza( t) := IF⋅ cos( ω⋅t)
π⎞ ⎛ ⎛ π⎞
⎛
⎛ π⎞ ⎞
Bzb ( t) := IF⋅ cos⎜ ω⋅t − 2 ⋅ ⎟ ⋅ ⎜ cos⎜ 2 ⋅ ⎟ + sin⎜ 2 ⋅ ⎟ ⋅i⎟
3
3
⎝
⎠⎝ ⎝ ⎠
⎝ 3⎠ ⎠
π⎞
⎛
IF⋅ cos⎜ ω⋅t − 2 ⋅ ⎟ = −0.5
3 ⎠
⎝
π⎞ ⎛ ⎛ π⎞
⎛
⎛ π⎞ ⎞
Bzc( t) := IF⋅ cos⎜ ω⋅t − 4 ⋅ ⎟ ⋅ ⎜ cos⎜ 4 ⋅ ⎟ + sin⎜ 4 ⋅ ⎟ ⋅i⎟
3⎠ ⎝ ⎝ 3⎠
⎝
⎝ 3⎠ ⎠
(
)
(
)
(
Re_sum ( t) := Re Bza( t) + Re Bzb ( t) + Re Bzc( t)
)
π⎞
⎛
IF⋅ cos⎜ ω⋅t − 4 ⋅ ⎟ = −0.5
3 ⎠
⎝
(
)
(
)
(
Im_sum( ) := Im Bza( t) + Im Bzb ( t) + Im Bzc( t)
Re_sum ( t) = 1.5
Im_sum( t) = 0
2
)
2
Re_sum( t) + Im_sum( t) = 1.5
6
11/15/2006 1.5
0
⎛
⎞
⎜Im Bz ( t) ⎟
⎝
( a )
⎠
1
0
⎛
⎞
⎜Im Bz ( t) ⎟
⎝
( b )
⎠
0.5
0
0
⎛
⎞
⎜Im Bz ( t) ⎟
⎝
( c )
⎠
0.5
0
⎞
⎛
⎜
⎟
⎝
Im_sum( t) ⎠
1
1.5
1.5
1
0.5
0
0.5
1
1.5
0
0
0
⎞ ⎛
⎞ ⎛
⎞ ⎛
⎛
0
⎞
⎟
⎜Re Bz ( t) ⎟ , ⎜Re Bz ( t) ⎟ , ⎜Re Bz ( t) ⎟ , ⎜
⎝
( a )
⎠
⎝ ( b ) ⎠
⎝
( c )
⎠
⎝
Re_sum( t) ⎠
t1 :=
π
3
IF⋅ cos( ω⋅t1) = 0.5
sec
(
)
(
)
(
)
Re Bza( t1)
π⎞
⎛
IF⋅ cos⎜ ω⋅t1 − 2 ⋅ ⎟ = 0.5
3 ⎠
⎝
Re Bzb ( t1)
imaginary part of field vector
π⎞
⎛
IF⋅ cos⎜ ω⋅t1 − 4 ⋅ ⎟ = −1
3 ⎠
⎝
Re Bzc( t1)
2
(
)
2
= 0.5
(
)
2
= 0.5
(
)
2
=1
+ Im Bza( t1)
2
+ Im Bzb ( t1)
2
+ Im Bzc( t1)
1
0
1
1
0
1
real part of field vector
7
11/15/2006 speed of rotation of this machine = frequency of the supplied AC. as shown, there are two poles (one pair) N-S
with multiple pairs the speed of rotation is reduced proportional to the number of poles
ns = f =
with AC frequency ω
and two poles
ω
n s = rotation_speed
2⋅π
f = frequency
rpm
Hz
1
ω = frequency
for p poles
ns =
2⋅ f
Hz = 1
1
s
Hz = 9.549 rpm
s
Hz
( 9.10 )
assumes radians
2⋅π⋅Hz = 60 rpm
p
one stator winding ...
d
E = −N⋅ Φ
dt
d
Φ = −constant⋅Φ⋅f
dt
(2.95)
small ...
U = IF⋅R + E
R<1
U=E=
Φ⋅f
E=
=>
Φ⋅f
KF
U
Φ = KF⋅
f
=>
KF
KF = constant_for_given_motor
now consider the rotor,if it is turning at the same speed as the rotating magnetic field of the stator, there is no EMF
the current induced in the rotor is strongly dependent on the relative speed
define ...
at low slip
s=
0%
n s − n
(9.14)
n
s
to ..
10%
s = slip
n s = rotation_speed_stator
fR_EMF = n s − n
n = rotation_speed_rotor
fR_EMF = frequency_of_rotor_EMF
(9.15)
because the rotor induced current will be at slip frequency, EMF is low so reactance (L) will be low. I A depends
primarily (only) on rotor (armature) resistance R A and is in phase with the flux pattern. The net result is torque is ~
directly proportional to slip
M = K⋅ s
(9.16)
8
11/15/2006 some typical curves from text for discussion
M,I
pull out torque
I
M
increasing
resistance
load curve
0
n
s
load curve
ns
0
0
n
ns
0
s
1
fig. 9.20A Torque - speed curve varying rotor resistance from Woud
1
fig. 9.18 Torque - speed curve induction motor from Woud
Um
M
I
increasing
resistance
0.75*Um
0.5*Um
load curve
load curve
0.25*Um
0
n
s
ns
0
0
1
fig. 9.20B Current - speed curve varying rotor resistance from Woud
n
s
ns
0
1
fig. 9.21 Torque - speed curve induction motor
varying supply voltage, from Woud
starting has challenges and text reviews some alternatives
There is much more to this subject and text covers quite well. Next lecture will review ship applications.
One comment regarding opinion in text regarding DC motor drive. Dc drives not typical as commutation brushes require
significant maintenance. DDX motor has innovative new brush technology.
see: http://www.globalsecurity.org/military/library/report/2002/mil-02-04-wavelengths02.htm
9
11/15/2006