Effect of 𝑪𝟏 on 3 - ɸ I.
M (slip ring):
1
The effect of 𝐶1 will be seen when applied to 3 - ɸ I.M and 𝐶1 will
transform the 3- ɸ winding machine into a two phase winding
machine i.e. the old current is ( 𝐼 ) and old voltage ( 𝑉 ) are:
𝐼𝑅 𝑉𝑅
𝐼𝑌 STATOR 𝑉𝑌
𝐼𝐵 𝑉𝐵
𝐼= 𝐼 & 𝑉= 𝑉
𝑟 𝑟
𝐼𝑦 ROTOR 𝑉𝑦
𝐼𝑏 𝑉𝑏
𝐼1 𝑉1
and using partitioning 𝐼= , 𝑉=
𝐼2 𝑉2
The Theory Of The Unified Machines Lecture 6
2 I.e. 𝐼 for stator and for rotor, similarly 𝑉 ′ and 𝐼′ the new voltage and
current.
𝑉𝑂 𝐼𝑂
𝑉𝐴 STATOR 𝐼𝐴
′
𝑉𝐵 ′
𝐼𝐵
𝑉 = 𝑉 & 𝐼 = 𝐼
𝑜 𝑜
𝑉𝛼 ROTOR 𝐼𝛼
𝑉𝛽 𝐼𝛽
and using partitioning
′ ′
𝐼 𝑉
𝐼′ = 1 ′ , 𝑉′ = 1′
𝐼2 𝑉2
The Theory Of The Unified Machines Lecture 6
3 Now using the transformation laws:
𝐼 = 𝐶1 𝐼′ ,𝑉 ′ = 𝐶1𝑡 𝑉 & 𝑍 ′ = 𝐶 ∗1𝑡 𝑍𝐶1
𝑍11 𝑍12
And if we put 𝑍 =
𝑍21 𝑍22
And since 𝐶1 will be applied to stator (1) and to rotor (2):
′ ∗
𝐶1 𝑡 0 𝑍11 𝑍12 𝐶1 0
∴ 𝑍 = 𝐶 1𝑡 𝑍𝐶1 =
0 𝐶1 𝑡 𝑍21 𝑍22 0 𝐶1
𝐶1 𝑡 𝑍11 𝐶1 𝐶1 𝑡 𝑍12 𝐶1 𝑍11 ′ 𝑍12 ′
= =
𝐶1 𝑡 𝑍21 𝐶1 𝐶1 𝑡 𝑍22 𝐶1 𝑍21 ′ 𝑍22 ′
The Theory Of The Unified Machines Lecture 6
4 The 3 - ɸ I.M (slip ring) looks like:
𝜃 is the angle between
Red phase in the stator
and red phase in the
rotor.
The Theory Of The Unified Machines Lecture 6
5 a) Stator to stator:
Since the 3 stator windings are identical
ـــ
𝑅𝑅 = 𝑅𝑌 = 𝑅𝐵 = 𝑅1 , 𝐿𝑅 = 𝐿𝑌 = 𝐿𝐵 = 𝐿1
ـــ
& 𝑀𝑅𝑌 = 𝑀𝑌𝐵 = 𝑀𝐵𝑅 = 𝑀1 (mutual inductance between stator winding phases )
Because of the air gab being uniform 𝐿1 & 𝑀1 doesn’t depend
on the position of the rotor i.e. they are constant w.r.t. 𝜃
So: R Y B
ـــ ـــ ـــ
R 𝑅1 + 𝐿1 𝜌 𝑀1 𝜌 𝑀1 𝜌
ـــ ـــ ـــ
𝑍11 =Y 𝑀1 𝜌 𝑅1 + 𝐿1 𝜌 𝑀1 𝜌
ـــ ـــ ـــ
B 𝑀1 𝜌 𝑀1 𝜌 𝑅1 + 𝐿1 𝜌
The Theory Of The Unified Machines Lecture 6
6 And to get 𝑍11 ′ = 𝐶1 𝑡 𝑍11 𝐶1
ـــ ـــ
𝑅1 + 𝐿1 + 2𝑀1 𝜌 0 0
ـــ ـــ
𝑍11 ′ = 0 𝑅1 + 𝐿1 − 𝑀1 𝜌 0
ـــ ـــ
0 0 𝑅1 + 𝐿1 − 𝑀1 𝜌
ـــ ـــ
if we put 𝐿1 + 2𝑀1 = 𝐿10
ـــ ـــ
and 𝐿1 − 𝑀1 = 𝐿1 0 A B
0 𝑅1 + 𝐿10 𝜌 0 0
We get: 𝑍11 ′ =A 0 𝑅1 + 𝐿1 𝜌 0
B 0 0 𝑅1 + 𝐿1 𝜌
I.e. transforming 𝑍11 into 𝑍11 ′ reduces the matrix.
The Theory Of The Unified Machines Lecture 6
7 b) Rotor to rotor: 𝒁𝟐𝟐
Similarly:
r y b
ـــ ـــ ـــ
r 𝑅2 + 𝐿2 𝜌 𝑀2 𝜌 𝑀2 𝜌
ـــ ـــ ـــ
𝑍22 =y 𝑀2 𝜌 𝑅2 + 𝐿2 𝜌 𝑀2 𝜌
ـــ ـــ ـــ
b 𝑀2 𝜌 𝑀2 𝜌 𝑅1 + 𝐿2 𝜌
And
𝑍22 ′ = 𝐶1 𝑡 𝑍22 𝐶1
0 𝛼 𝛽
ـــ
ـــ
0 𝑅2 + 𝐿2 + 2𝑀2 𝜌 0 0
ـــ ـــ
𝑍22 ′ =𝛼 0 𝑅2 + 𝐿2 − 𝑀2 𝜌 0 ـــ
ـــ
𝛽 0 0 𝑅2 + 𝐿2 − 𝑀2 𝜌
The Theory Of The Unified Machines Lecture 6
8 Or
0 𝛼 𝛽
0 𝑅2 + 𝐿20 𝜌 0 0
𝑍22 ′ = 𝛼 0 𝑅1 + 𝐿2 𝜌 0
𝛽
0 0 𝑅1 + 𝐿2 𝜌
ـــ ـــ
𝐿20 = 𝐿2 + 2𝑀2
ـــ ـــ
𝐿2 = 𝐿2 − 𝑀2
The Theory Of The Unified Machines Lecture 6
9 C) Rotor to stator: 𝒁𝟐𝟏
The terms of 𝑍21 which as mutual inductance between the rotor
and the stator and it was found that they depend on the angle 𝜃 ,
𝜃2 (𝜃 +120 ) and 𝜃3 (𝜃 + 240). (Fundamental and odd harmonics).
R Y B
ـــ ـــ ـــ ـــ ـــ ـــ
r 𝑀𝜌 cos 𝜃 + 𝑀3 𝜌 cos 3𝜃 𝑀𝜌 cos 𝜃3 + 𝑀3 𝜌 cos 3𝜃 𝑀𝜌 cos 𝜃2 + 𝑀3 𝜌 cos 3𝜃
ـــ ـــ ـــ ـــ ـــ ـــ
𝑍21 = y 𝑀𝜌 cos 𝜃2 + 𝑀ـــ3
𝜌 cos 3𝜃 𝑀𝜌 cos 𝜃 + 𝑀3 𝜌 cos 3𝜃 𝑀𝜌 cos 𝜃3 + 𝑀3 𝜌 cos 3𝜃
ـــ ـــ ـــ ـــ ـــ
b 𝑀𝜌 cos 𝜃3 + 𝑀3 𝜌 cos 3𝜃 𝑀𝜌 cos 𝜃2 + 𝑀3 𝜌 cos 3𝜃 𝑀𝜌 cos 𝜃 + 𝑀3 𝜌 cos 3𝜃
𝑍21 Can be separated into matrix (𝑍21 )1 which represent the
fundamental and (𝑍21 )3 which represent the third harmonic.
Note : cos3𝜃2 = cos3𝜃 , cos3𝜃3 = cos3𝜃
Angle between Rr=Yy=Bb=𝜃
The Theory Of The Unified Machines Lecture 6
10 𝑍21 ′ = 𝑍21 ′ 1 + 𝑍21 ′ 3
𝑍21 ′ 1 = 𝐶1 𝑡 𝑍21 1 𝐶1
∴ 𝑍21 ′ 1
1 1 1 1
1 0
2 2 2 2
cos 𝜃 cos 𝜃3 cos 𝜃2
2 1 1 1 1 3
= 𝑀𝜌 1 − − cos 𝜃2 cos 𝜃 cos 𝜃3 −
3 2 2 2 2 2
3 3 cos 𝜃3 cos 𝜃2 cos 𝜃 1 1 3
0 − − −
2 2 2 2 2
2 0 0 0
= 𝑀 0 𝜌 cos 𝜃 𝜌sin 𝜃
3 0 −𝜌 sin 𝜃 𝜌cos 𝜃
The Theory Of The Unified Machines Lecture 6
11 Remember:
cos 𝜃 + cos 𝜃2 + cos 𝜃3 = 0
cos 𝜃3 − cos 𝜃2 = 3 sin 𝜃
cos 𝜃 − cos 𝜃3 = 3 sin 𝜃2
cos 𝜃2 − cos 𝜃 = 3 sin 𝜃3
sin 𝜃2 − sin 𝜃3 = 3 cos 𝜃
The Theory Of The Unified Machines Lecture 6
2 ـــ
12 Also we can put M = 3
𝑀
ـــ
1 1 1
𝑍21 ′ 3 = 𝐶1 𝑡 𝑍21 𝐶1 = 𝐶1 𝑡 𝑀3 𝜌 cos 3𝜃 1 1 1 𝐶1
1 1 1
1 0 0
= 𝑀3 𝜌 cos 3𝜃 0 0 0
0 0 0
ـــ
And let: 3𝑀3 = 𝑀3
The Theory Of The Unified Machines Lecture 6
13 Now:
𝑍21 ′ = 𝑍21 ′ 1 + 𝑍21
′
3
0 A B
0 𝑀3 𝜌 cos 3𝜃 0 0
𝑍21 ′ =𝛼 0 𝑀𝜌 cos 𝜃 𝑀𝜌 sin 𝜃
𝛽 0 −𝑀𝜌 sin 𝜃 𝑀𝜌 cos 𝜃
It’s clear that apply of 𝐶1 to 𝑍21 simplify the matrix, four element
equal to zero and 𝜃2 & 𝜃3 are disappear.
The Theory Of The Unified Machines Lecture 6
14 (d) Stator to rotor: 𝒁𝟏𝟐
It was found experimentally: 𝑍12 = 𝑍21 𝑡
𝑍12 ′ = 𝐶1 𝑡 𝑍12 𝐶1
= 𝐶1 𝑡 𝑍21 𝑡 𝐶1
𝑡
= 𝐶1 𝑡 𝑍21 𝐶1
𝑡
= 𝑍21 ′ 𝑡
0 𝛼 𝛽
𝑀3 𝜌 cos 3𝜃
0 0 0
∴ 𝑍12 ′ =𝐴 0 𝑀𝜌 cos 𝜃 −𝑀𝜌 sin 𝜃
𝐵 0 𝑀𝜌 sin 𝜃 𝑀𝜌 cos 𝜃
The Theory Of The Unified Machines Lecture 6
15 The completed transformation 𝒁′ is given:
0 A B 0 𝜶 𝜷
0 𝑅1 + 𝐿10 𝜌 0 0 𝑀3 𝜌 cos 3𝜃 0 0
A 0 𝑅1 + 𝐿1 𝜌 0 0 𝑀𝜌 cos 𝜃 −𝑀𝜌 sin 𝜃
B 0 0 𝑅1 + 𝐿1 𝜌 0 𝑀𝜌 sin 𝜃 𝑀𝜌 cos 𝜃
𝑍′ =
0 𝑀3 𝜌 cos 3𝜃 0 0 𝑅2 + 𝐿20 𝜌 0 0
𝜶 0 𝑀𝜌 cos 𝜃 𝑀𝜌 sin 𝜃 0 𝑅2 + 𝐿2 𝜌 0
𝜷 0 −𝑀𝜌 sin 𝜃 𝑀𝜌 cos 𝜃 0 0 𝑅2 + 𝐿2 𝜌
𝒁′ 0, A, B, 0 , 𝜶 , 𝜷
0
A
B
0
𝛼
𝛽
The Theory Of The Unified Machines Lecture 6
16 Rearranging by moving the fourth column to the second
column and then moving the fourth row to the second row
we get: 0 0 A B 𝜶 𝜷
0 𝑅1 + 𝐿10 𝜌 𝑀𝜌 cos 3𝜃 0 0 0 0
0 𝑀𝜌 cos 3𝜃 𝑅1 + 𝐿20 𝜌 0 0 0 0
′ A 0 0 𝑅1 + 𝐿1 𝜌 0 𝑀𝜌 cos 𝜃 −𝑀𝜌 sin 𝜃
𝑍 =
B 0 0 0 𝑅1 + 𝐿1 𝜌 𝑀𝜌 sin 𝜃 𝑀𝜌 cos 𝜃
𝜶 0 0 𝑀𝜌 cos 𝜃 𝑀𝜌 sin 𝜃 𝑅2 + 𝐿2 𝜌 0
𝜷 0 0 −𝑀𝜌 sin 𝜃 𝑀𝜌 cos 𝜃 0 𝑅2 + 𝐿2 𝜌
𝒁′ ( 0, 0 ,A, B, 𝜶 , 𝜷 ) It’s clear that we have two independent impedance.
0
0
A
B
𝛼
𝛽 The Theory Of The Unified Machines Lecture 6
17 The zero sequence impedance 𝑍0 , the zero sequence voltage 𝑉0
and the zero sequence current 𝐼0 .
𝑉0 = 𝑍0 𝐼0
𝑉10 𝑅1 + 𝐿10 𝜌 𝑀𝜌 cos 3𝜃 𝐼10
Or: =
𝑉20 𝑀𝜌 cos 3𝜃 𝑅2 + 𝐿20 𝜌 𝐼20
𝐼20 can flow only in the rotor if there is an external slip ring
connection to the start point which doesn’t happen in a practical
machines, also 𝐼10 can flow only if there is a 3 - ɸ four wire supply
also it’s not a normal case.
𝐼10 & 𝐼20 Cannot flow and there for 𝑉0 = 𝑍0 𝐼0 can be neglected.
The Theory Of The Unified Machines Lecture 6
18 The voltage equation will reduce to: 𝑉 ′ = 𝑍 ′ 𝐼′
𝑉𝐴 𝑅1 + 𝐿10 𝜌 0 𝑀𝜌 cos 𝜃 −𝑀𝜌 sin 𝜃 𝐼𝐴
𝑉𝐵 0 𝑅1 + 𝐿10 𝜌 𝑀𝜌 sin 𝜃 𝑀𝜌 cos 𝜃 𝐼𝐵
𝑉𝛼 = 𝑀𝜌 cos 𝜃 𝑀𝜌 sin 𝜃 𝑅2 + 𝐿20 𝜌 0 𝐼𝛼
𝑉𝛽 −𝑀𝜌 sin 𝜃 𝑀𝜌 cos 𝜃 0 𝑅2 + 𝐿20 𝜌 𝐼𝛽
A B 𝜶 𝜷
𝑅1 + 𝐿10 𝜌
A 0 𝑀𝜌 cos 𝜃 −𝑀𝜌 sin 𝜃
0 𝑅1 + 𝐿10 𝜌 𝑀𝜌 sin 𝜃 𝑀𝜌 cos 𝜃
𝑍 ′ =𝜶B
𝑀𝜌 cos 𝜃 𝑀𝜌 sin 𝜃 𝑅2 + 𝐿20 𝜌 0
𝜷 −𝑀𝜌 sin 𝜃 𝑀𝜌 cos 𝜃 0 𝑅2 + 𝐿20 𝜌
The Theory Of The Unified Machines Lecture 6
19 The torque equation :
1 𝑑𝐿
𝑇 = 𝐼𝑡 𝐼
2 𝑑𝜃
𝐼𝑅
𝐼𝑌
𝐼𝐵
Where: 𝐼 = 𝐼
𝑟
𝐼𝑦
𝐼𝑏
1 𝑑𝐿 1 ′ 𝑑𝐿 ′ 1 ′ 𝑑𝐿
𝑇 = 𝐼𝑡 𝐼= 𝐼𝑡 𝐶𝑡 𝐶𝐼 = 𝐼𝑡 𝐶𝑡 𝐶 𝐼′ = 𝑇 ′
2 𝑑𝜃 2 𝑑𝜃 2 𝑑𝜃
𝑑𝐿′ 𝑑𝐿
′
And since 𝑇 = 𝑇 ======> = 𝐶𝑡 𝐶 from 𝑍 ′ find the Torque 𝑇 ′
𝑑𝜃 𝑑𝜃
The Theory Of The Unified Machines Lecture 6
20
The Theory Of The Unified Machines Lecture 6
