Self and Mutual Inductances for
Fundamental Harmonic in
Synchronous Machine with
Round Rotor (Cont.)
Double Layer Lap Winding
on Stator
Round Rotor Field Winding (1)
d axis
sr = 2 nr
even
Define Sr is the number of rotor slots.
Define sr is the number of rotor slots
between two poles, or the number of slots
per pole.
Sr
sr =
P
Round Rotor Field Winding (2)
sr
odd
MMF from Round Rotor When sr is Even (1)
F f from all conductors
(2υ − 1)γ r υ = 1, 2, ...,
with N υ turns
sr
2
Nυ i f
π
− Nυ i f
d axis
∑
h =1,3,5...
θ de=
= θd
F
fh
F fh =
2π
θ de
2
Ff =
P
2
π
F fh
F=
F fh cos( h
fh cos( hθ de )
P
θd )
2
sr
2
 π − (2υ − 1)γ r 
sin
N
∑
υ
h

2
π h υ =1


4i f
Rotor winding turns are typically all connected in series.
MMF from Round Rotor When sr is Even (2)
sr
2
N f = P ∑ Nυ
Define:
υ =1
total number of series turns of rotor


sr


2
4  N f if   1
 π − (2υ − 1)γ r   4i f  Nˆ fh 
Nυ sin  h
=


∑
 sr



πh  P  2
π
h
P
2


υ =1


 ∑ Nυ

 υ =1

⇒ F fh
where
Nˆ fh = N f k wf ,h
k wf ,h =
1
sr
2
Nυ
∑
υ
=1
sr
2
effective number of series turns for hth harmonic
 π − (2υ − 1)γ r 
N
sin
∑
υ
h

2


υ =1
winding factor for hth harmonic
MMF from Round Rotor When sr is Odd (1)
Fa from conductors
with N1 turns
N1i f
2
−
N1i f
2
π
π
2π
2
θ de
Fa from all conductors
with N υ turns
2(υ − 1)γ r υ = 2, ...,
Nυ i f
− Nυ i f
=
Ff
∑
h =1,3,5...
π
sr + 1
2
π
2π
2
F fh =
where F fh F=
F fh cos( h
fh cos( hθ de )
sr +1

4i f  N1
 π 2
 π − 2(υ − 1)γ r
=
F fh
sin  h  + ∑ Nυ sin  h
πh  2
2
 2  υ =2


P
θd )
2





θ de
MMF from Round Rotor When sr is Odd (2)
Define:
sr +1


2
N
N f P  1 + ∑ Nυ 
=
 2 υ =2



total number of series turns of rotor
sr +1


2
π
υ
γ
N
i
−
−
2(
1)
4i f  Nˆ fh

4  f f 
1
f 
 N1 sin  h π  +
=

 ∑ Nυ sin  h


  π h  P
sr +1

πh  P 
2
2
2

 υ =2


2

N1


+ ∑ Nυ
2 υ =2
F fh



where
Nˆ fh = N f k wf ,h
k wf ,h
effective number of series turns for hth harmonic
sr +1


2
1
N
π
π
2(
υ
1)
γ
−
−

r 
 1 sin  h  +
N
sin
h
∑
υ



  winding factor
sr +1

2
2
2




2
=
υ
2
N1

 for hth harmonic
+ ∑ Nυ
2 υ =2
Mutual Inductance between Stator and
Rotor Field Winding
If we apply current in rotor field winding, then when rotor is moving,
the magnetic field in airgap from rotor field winding is: θ = θ − θ
4 µ 0  Nˆ f
Bf =
πg eff  P
d
a

i f cos( P θ d ) Define: Nˆ = k N
f
wf
f

2

Now, we can calculate flux in Phase A winding from field current.
λa |from field winding = Nˆ a Φ f , pk cos θ me
m
θ me =
Effective number
of turns on field
winding.
β = −θ me
 Nˆ f 
i f

B f , pk
where
Φ f , pk =
 P 
P


 Nˆ a Nˆ f 
 Nˆ f 
8
µ
µ
2
4
Dl
Dl
0
0
 cos θ mei f

i f =

λa = Nˆ a cos θ me
2
π g eff  P 
P π g eff  P 
λa 8µ 0 Dl  Nˆ a Nˆ f 
cos θ me = Lsf cos θ me
⇒ Laf =
=
2


if
π g eff  P 
2 B f , pk Dl
where
4µ0
=
πg eff
8µ 0 Dl  Nˆ a Nˆ f
Lsf =
π g eff  P 2




P
θm
2
Self Inductance of Rotor Field Winding
For the magnetic field from rotor field winding is:
4µ0
Bf =
πg eff
 Nˆ f

 P


i f cos( P θ d )

2

Now, we can calculate flux in field winding by integrating on θ d .
λ f = Nˆ f Φ f , pk
where
Φ f , pk
 Nˆ f 
i f

 P 


8µ 0 Dl  Nˆ f
µ 0  Nˆ f 
if =


π g eff  P
g eff  P 
2
ˆ


8µ 0 Dl  N f 
=
π g eff  P 
4µ0
2 B f , pk Dl B
=
f , pk
=
πg eff
P
2 Dl 4
ˆ
λf = k f N f
P π
⇒ Lmf =
λf
if
L f = Llf + Lmf
2

 if


Steady State Analysis of Synchronous
Machine with Round Rotor
Terminal Voltage for Round Rotor Motor
 va   Rs
v   0
 b=
 vc   0
  
v f   0
When

 λa  
λ  
 b =
 λc  
  
λ f   L
 sf

0
Rs
0
0
0
0
Rs
0
0   ia 
 λa 
0   ib  d  λb 
+
0   ic  dt  λc 
 
 
R f  i f 
λ f 
ia + ib + ic = 0
Ls
0
0
cos θ me
Lsf cos θ me 
2π   ia 
Ls
Lsf cos(θ me − )   
0
3  ib
2π   
0
Ls
Lsf cos(θ me + )  ic 
3  
 i f 
2π
2π
Lsf cos(θ me − ) Lsf cos(θ me + )
Lf

3
3

0
0
Round Rotor Motor at Steady State
va Rsia +
=
d λa
dt
=
λa Lsia + Lsf cos θ mei f
dθ me
ωet + φr
= ωe θ=
me
At steady state
dt
di
va =
Rsia + Ls a − Lsf I f ωe sin (ωet + φr )
dt
π

j  φr + 


dia
2  jωe t

=
+ Re  Lsf I f ωe e
Rsia + Ls
e 


dt


π

+
j
φ


r
jωe t
jωe t
2

=
V
v
e
Re
ia = Re I Ae
Let
E A = Lsf I f ωe e
φ
a
(
)
(
⇒ Vφ = Rs I A + jX s I A + E A
X s = ωe Ls
8µ 0 Dl  Nˆ a Nˆ f
Lsf =
π g eff  P 2




)
Round Rotor Generator at Steady State
Motor
Generator
Vφ =Rs I A + jX s I A + E A
Vφ =
− Rs I A − jX s I A + E A
d λa
dt
λa = Ls ( −ia ) + Lsf cos θ mei f
va =
− Rsia +
π

j  φr + 


dia
2  jωe t

− Rsia − Ls
+ Re  Lsf I f ωe e
va =
e 
 

dt


EA
va = Re Vφ e jωet
ia = Re I Ae jωet
8µ 0 Dl  Nˆ a Nˆ f 
Lsf =
π g eff  P 2 
(
)
(
)
Open Circuit Voltage (1)
Assume the armature windings are open circuit, the magnetic field in the air gap
comes from the field winding only.
P
P
−
=
+
Bg |from =
B
ω
t
θ
φ
B
θ a − ωet − φr )
cos(
)
cos(
f , pk
e
a
r
f , pk
field winding
2
2
2 B f , pk Dl
(
)
β
−
ω
t
+
φ
=
Φ f , pk =
From Notes Flux Linkage in Phase Winding
e
r
P
⇒ λa |from field winding = Nˆ a Φ f , pk cos β = Nˆ a Φ f , pk cos(ωet + φr )
d λa |from field winding
d Φ f , pk
ˆ
ˆ
E A (t ) =
=
− N a Φ f , pk ωe sin(ωet + ϕ r ) + N a cos(ωet + ϕ r )
dt
dt
At steady state, we have
dΦ f , pk
dt
=0
E A (t ) = − Nˆ a Φ f , pk ωe sin(ωet + φr )
π
= Nˆ a Φ f , pk ωe cos(ωet + φr + )
2
Phasor of EA(t) is:
=
EA
ωe Nˆ a Φ f , pk e
π
j (ϕ r + )
2
Open Circuit Voltage (2)
E A = ωe Lsf I f e
8µ Dl
Lsf = 0
π g eff
π
j (φ r + )
2
 Nˆ a Nˆ f

2
P

=
EA



ωe Nˆ a Φ f , pk e
Φ f , pk
4 µ0
=
Br
π g eff
2 Br , pk Dl
=
P
 Nˆ f

 P

P
θ a − θ me )
I
cos(
 f
2

We can find out that they are the same.
EA
λ f , pk= Nˆ a Φ f , pk
π
π
j (φr + )
j (φr + )
8µ0 Dl  Nˆ a Nˆ f 
2
2
ω=
I
e
ω
λ
e


e
e f , pk
π g eff  P 2  f
E A, rms
Dl  Nˆ a Nˆ f 
8=
2 f e µ0

 I f
2
g eff  P 
π
j ( φr + )
2
1
ωeλ f , pk
2
Volt-Second Balance
E=
A,rms
2π f e Nˆ a Φ f , pk ≈ 4.44 f e Nˆ a Φ f , pk
E A,rms
=
or:
fe
λ f , pk
2π Nˆ a Φ f , pk ≈ 4.44 Nˆ a Φ f , pk
Example: if a 60Hz generator is to be operated at 50 Hz, then the
operating voltage must be derated to 50/60 of its original value.
Induced Phase Voltage
For the net magnetic field, at steady state:
P
Bnet = B pk cos(ωet − θ a + φnet )
2
⇒ λa ,net =
Nˆ a Φ pk cos (ωet + φnet )
Φ pk =
2 B pk Dl
P
The steady state phase voltage in armature phase A winding is:
Vφ (t ) =
dλa
π
= − Nˆ a Φ pk ωe sin(ωet + φnet ) = Nˆ a Φ pk ωe cos(ωet + φnet + )
dt
2
π
j (ϕ net + )
2
ˆ
Vϕ
ωe N a Φ pk e
Phasor =
The rms phase voltage is
=
Vφ ,rms
2π f e Nˆ a Φ pk ≈ 4.44 f e Nˆ a Φ pk
λ pk
Y Connection (Generator)
∆ Connection (Generator)
Example 1
For a 3 phase, 4 pole, 24 slot, 5/6 pitch machine with double layer lap
winding, the peak magnetic filed intensity in the airgap is 0.45 T. There
is one slot skew. The machine shaft speed is 8000 rpm. The stator
inner diameter is 0.5 m. The machine length is 0.3 m. There are 10
turns per coil. All the turns are connected in series. The three phase
coils are Y connected.
(1)What is the rms phase voltage of the machine?
(2)What is the rms terminal voltage of the machine?
=
Vϕ ,rms
2π f e Nˆ a Φ pk ≈ 4.44 f e Nˆ a Φ pk
Φ pk
2 B pk Dl
=
P
P=4
λ pk
Nˆ a = N a k w
N a = PqN c
q=
S
mP
ACmachine1.m
Example 2
For a simple 2 pole, 3 phase, Y connected machine (single layer
winding) shown in the figure, the peak magnetic field intensity in the
airgap is 0.2T. There is no skew. The machine shaft speed is 3600 rpm.
The stator inner diameter is 0.5 m. The machine length is 0.3 m. There
are 15 turns in the coil.
(1)What is the rms phase voltage of the machine?
(2)What is the rms terminal voltage of the machine?
=
Vϕ ,rms
2π f e Nˆ a Φ pk ≈ 4.44 f e Nˆ a Φ pk
Φ pk
λ pk
2 B pk Dl
=
P
P=2
For this example:
Nˆ a = N c
(Note: This is single layer winding.)
ACmachine2.m
Voltage and Speed Regulation
Voltage regulation:
VR =
Vnl − V fl
V fl
× 100%
Speed regulation:
SR =
Or:
nnl − n fl
n fl
× 100%
ω nl − ω fl
SR =
× 100%
ω fl
AC Machine Efficiency
=
η
Pout
× 100%
Pin
Pout
= Pin − Ploss
AC Machine Loss Mechanism
1.
2.
3.
4.
Electrical or copper losses (I2R losses)
Core losses
Mechanical losses
Stray or miscellaneous losses
Electrical or Copper Loss
Stator Copper Loss (SCL):
PSCL = 3I A2 Rs
Rotor Copper Loss (RCL):
PRCL = I F2 RF
Core, Mechanical and Stray Losses
AC Generator Power Flow
power converted from mechanical to electrical
Pconv = Temωm
AC Motor Power Flow
power converted from electrical to mechanical
Pconv = Temωm