SPPU Class-TE-Electrical Subject-Power System-II

1. DEPARTMENT OF ELECTRICAL ENGINEERING
JSPMS
BHIVARABAISAWANTINSTITUTEOFTECHNOLOGYANDRESEARCH,
WAGHOLI,PUNE
A.Y. 2019-20 (SEM-II)
Class: T.E.
Subject: Power System-II
Prepared by Prof. S. D.
Gadekar
Santoshgadekar.919@gmail.com
Mob. No-9130827661

2. Contents
• Types of unsymmetrical faults
• Symmetrical components
• Symmetrical components of three phase system
• Phasor operator
• Evaluation of components or transformation
matrices
• Numerical on Symmetrical Components
• Zero sequence impedance and network of
transformer

3. Types of Unsymmetrical Faults
1.Single‐line to ground (60%‐75%)
2.Double‐line to ground (15%‐25%)
3. Line‐to‐line faults (5%‐15%)

4. Symmetrical Components-
Any unbalance three phase system of currents, voltages or other
sinusoidal quantities can be resolved into three balanced system of
phasor are called as symmetrical components.
Three phase voltage or current is in a balance condition if it has the
following characteristic:
 Magnitude of phase a,b, and c is all the same
 The system has sequence of a,b,c
 The angle between phase is displace by 120 degree
If one of the above is character is not satisfied, unbalanced occur.

5. Symmetrical Components-
 For unbalanced system, power system analysis cannot be
analyzed using per phase as in Load Flow analysis or
Symmetrical fault (Per Unit System).
 Symmetrical component allow unbalanced phase quantities
such as current and voltages to be replaced by three
separate balanced symmetrical components.

6. Symmetrical Components of three phase system
Positive Sequence components (+) 1
Three balance phasors
Equal in magnitude
Displaced from each other by 120ᵒ
Same phase sequence (R-Y-B)
Ib1
Ia1
Ic1
120ᵒ
120ᵒ
120ᵒ
Negative Sequence components (-) 2
Three balance phasors
Equal in magnitude
Displaced from each other by 120ᵒ
Same phase sequence (R-B-Y)
Ic2
Ia2
Ib2
120ᵒ
120ᵒ
120ᵒ
Zero Sequence components 0
Three balance phasors
Equal in magnitude
Zero phase displacement from each other
Ia0
Ib0
Ic0
0ᵒ

7. Phasor Operator
Ib1
Ia1
Ic1
120ᵒ
120ᵒ
120ᵒ
By convention, the direction of rotation of the phasor is taken to be counterclock‐wise.
𝒂 = 𝒆𝒋𝟐𝝅/𝟑
= cos 120 + 𝑗 sin 120
= (-0.5)+j0.866
It indicates that a phasor has a unit length and is oriented
120° or 2π/3 in a positive counter clockwise direction from
the reference axis.
A phasor operated upon by ‘a’ is not changed in magnitude
but is simply rotated in position 120ᵒ in the forward direction.
For Example- 𝐈 𝐜𝟏=𝐈 𝐚𝟏 < 𝟏𝟐𝟎° = 𝐚 𝟏
𝐈 𝐚𝟏
Ic1 is a phasor having the same length as phasor Ia1, but
rotated 120° forward from the phasor Ia1.

8. Phasor Operator
Ib1
Ia1
Ic1
120ᵒ
120ᵒ
120ᵒ
By convention, the direction of rotation of the phasor is taken to be counterclock‐wise.
𝒂 𝟐 = (𝒆𝒋𝟐𝝅/𝟑) 𝟐=𝒆𝒋𝟒𝝅/𝟑=𝒆(−
𝒋𝟐𝝅
𝟑
)
= cos( −120) + 𝑗 sin(−120)
= (-0.5)-j0.866
A phasor operated upon by 𝑎2
is not changed in magnitude
but is simply rotated in position 240ᵒ in the forward direction,
Or 120ᵒ in a negative direction from the reference axis.
For Example- 𝐈 𝐛𝟏=𝐈 𝐚𝟏 < 𝟐𝟒𝟎° = 𝐚 𝟐
𝐈 𝐚𝟏
Ib1 is a phasor having the same length as phasor Ia1, but
rotated 240° forward from the phasor Ia1.
𝒂 𝟑
= 𝟏, 𝒂 𝟒
= 𝒂, 𝒂 𝟓
= 𝒂 𝟐
𝟏 + 𝒂 𝟐 + 𝒂 𝟏 =0 ,
𝟏 + 𝒂 𝟐
+ 𝒂 𝟒
= 𝟎,
𝟏 + 𝒂 𝟑
+ 𝒂 𝟑
= 𝟑

9. Phasor Operator for Symmetrical Components
Positive Sequence components (+) 1
Ia1=Ia1 < 0° = 𝑎°Ia1
Ib1=Ia1 < 240° = 𝑎2Ia1
Ic1=Ia1 < 120° = 𝑎1Ia1
Ib1
Ia1
Ic1
120ᵒ
120ᵒ
120ᵒ
Negative Sequence components (-) 2
Ia2=Ia2 < 0° = a°Ia2
Ic2=Ia2 < 240° = a2Ia2
Ib2=Ia2 < 120° = a1
Ia2
Ic2
Ia2
Ib2
120ᵒ
120ᵒ
120ᵒ
Zero Sequence components 0
Ia0=Ib0=Ic0
Ia0
Ib0
Ic0
0ᵒ
By convention, the direction of rotation of the phasor is taken to be counterclock‐wise.

10. Evaluation of components or transformation matrices
Let us Express phasor of unbalanced three phase system in terms of their
symmetrical components.
A = A1 + A2 + A0
B = B1 + B2 + B0
C = C1 + C2 + C0
Express all phasor in terms of A1, A2 and A0,
A = A1 + A2 + A0……1
B = 𝑎2A1 + 𝑎1A2 + A0……2
C = 𝑎1
A1 + 𝑎2
A2 + A0……3
Add equation 1, 2 and 3
A+B+C= 1 + 𝑎2
+ 𝑎1
A1 + (1 + 𝑎2
+ 𝑎1
) A2 + 3A0
1 + 𝑎2
+ 𝑎1
=0
A0=
A+B+C
3
…….4

11. Evaluation of components or transformation matrices
A = A1 + A2 + A0……1
B = 𝑎2
A1 + 𝑎1
A2 + A0……2
C = 𝑎1A1 + 𝑎2A2 + A0……3
Multiply equation 2 by a and equation 3 by 𝑎2
.
𝑎B = 𝑎3A1 + 𝑎2A2 + aA0……5
𝑎2
C = 𝑎3
A1 + 𝑎4
A2 + 𝑎2
A0……6
Now add equation 1, 5 and 6
A+ 𝑎B+ 𝑎2
C= 1 + 𝑎3
+ 𝑎3
A1 + 1 + 𝑎2
+ 𝑎4
A2 + 1 + 𝑎1
+ 𝑎2
A0
1 + 𝑎2 + 𝑎1 =0
1 + 𝑎2
+ 𝑎4
= 0
1 + 𝑎3 + 𝑎3 = 3
A1=
A+ 𝑎B+ 𝑎2C
3
…….7

12. Evaluation of components or transformation matrices
A = A1 + A2 + A0……1
B = 𝑎2
A1 + 𝑎1
A2 + A0……2
C = 𝑎1A1 + 𝑎2A2 + A0……3
Similarly Multiply equation 3 by a and equation 2 by 𝑎2
.
𝑎2 B = 𝑎4A1 + 𝑎3A2 + 𝑎2A0……8
𝑎1
C = 𝑎2
A1 + 𝑎3
A2 + 𝑎1
A0……9
Now add equation 1, 8 and 9
A+ 𝑎2
B + 𝑎C = 1 + 𝑎4
+ 𝑎2
A1 + 1 + 𝑎3
+ 𝑎3
A2 + 1 + 𝑎2
+ 𝑎1
A0
1 + 𝑎2 + 𝑎1 =0
1 + 𝑎2
+ 𝑎4
= 0
1 + 𝑎3 + 𝑎3 = 3
A2=
A+ 𝑎2 B + 𝑎C
3
…….10

13. Evaluation of components or transformation matrices
From equation 4 ,7 and 10, A0, A1 and A2 can be written as
A0=
A+B+C
3
A1=
A+ 𝑎B+ 𝑎2C
3
A2=
A+ 𝑎2 B + 𝑎C
3
A0
A1
A2
=
1 1 1
1 𝑎 𝑎2
1 𝑎2 𝑎
A
B
C
*
1
3

14. Evaluation of components or transformation matrices
From equation 1,2 and 3, A, B and C can be written as
A = A1 + A2 + A0
B = 𝑎2
A1 + 𝑎1
A2 + A0
C = 𝑎1
A1 + 𝑎2
A2 + A0
A0
A1
A2
=
1 1 1
1 𝑎2 𝑎
1 𝑎 𝑎2
A
B
C
*

15. Example 1-Obtain the symmetrical components for the set of unbalanced
voltages
𝑉𝑎 = 176 − 𝑗132, 𝑉𝑏 = −128 − 𝑗96 , 𝑉𝑐 = −160 + 𝑗100.
𝐴0=
A+B+C
3
=
(176−𝑗132)+(−128−𝑗96)+(−160+𝑗100)
3
𝐴0=(-37.33-j42.47)
𝐴1=
A+ 𝑎𝐵+ 𝑎2 𝐶
3
=
(176−𝑗132)+(−0.5+j0.866 )∗(−128−𝑗96)+(−0.5−j0.866) ∗(−160+𝑗100)
3
𝐴1=(163.24-j35.42)
𝐴2=
A+ 𝑎2 𝐵 + 𝑎𝐶
3
=
(176−𝑗132)+(−0.5−j0.866 )∗(−128−𝑗96)+(−0.5+j0.866) ∗(−160+𝑗100)
3
𝐴2=(50-j50.9)

16. Example 2-Compute the symmetrical components of the three phase voltages.
Va = 100 < 0° Volts,
Vb = 110 < −100° Volts,
Vc = 115 < 110° Volts.
Symmetrical Components are given by,
A0=
A+B+C
3
=
Va+Vb+Vc
3
A1=
A+ 𝑎B+ 𝑎2C
3
=
Va+ 𝑎Vb+ 𝑎2Vc
3
A2=
A+ 𝑎2 B + 𝑎C
3
=
Va+ 𝑎2Vb+ 𝑎Vc
3
Va0 =
100<0° + 110<−100° +(115<110°)
3
Va0 =13.86< −0.37° Volts
V𝑎1 =
100<0° + 1<120° ∗ 110<−100° + 1<240° ∗(115<110°)
3
V𝑎1 = 105.7< 3.19° Volts
V𝑎2 =
100 < 0 + 1 < 240° ∗ 110 < −100° + 1 < 120° ∗ (115 < 110°)
3
Va2=20.24< −163.35° Volts

17. Example 3-A single phase load of 100 kVA is connected across lines
bc of a three phase supply of 3.3 kV. Determine symmetrical
components of line current.
c
b
a
𝐼 𝑎 = 0
100 kVA Load
𝐼 𝑏 = −𝐼𝑐 =
100 ∗ 1000
3.3 ∗ 1000
=30.30 Ampere
Symmetrical Components are given by,
A0=
A+B+C
3
=
Ia+Ib+Ic
3
A1=
A+ aB+ a2C
3
=
Ia+ aIb+ a2Ic
3
A2=
A+ a2 B + aC
3
=
Ia+ a2Ib+ aIc
3
Ia0 =
Ia+Ib+Ic
3
=
0+30.30−30.30
3
=0 Ampere
Ia1 =
Ia+ aIb+ a2Ic
3
=
0+30.30∗(𝑎−𝑎2)
3
=17.49< 90° Ampere
Ia2 =
Ia+ 𝑎2Ib+ 𝑎Ic
3
=
0+30.30∗(𝑎2−𝑎)
3
=17.49< −90° Ampere

18. Sequence Impedances and networks of transformers-
The positive and negative sequence impedances of transformer is equal to its
leakage impedance.
Z1 = Z2 = ZLeakage
The zero sequence currents can flow through the winding connected in star only,
if star point is grounded.
General Circuit to determination of zero sequence network of transformer is given below,
Primary Secondary
1 2
3 4
𝑍0 is the zero sequence impednace of transformer.
The series switch of particular side is closed, if it is star grounded.
The shunt switch is closed if that side is delta connected.
Z0

19. Primary Secondary
3 4
1. Star-Star with isolated neutral
2. Star-Star with primary
neutral grounded
1 2Z0
Z0
1 2Z0
1 2
3. Star-Star with both
neutral grounded
1 2Z0
4. Star-Delta with grounded star
neutral grounded 1 2Z0
4
Reference Bus

20. Primary Secondary
3 4
5. Star-Delta with isolated Star
6. Delta-Delta
1 2
Z0
Z0
1 2Z0
1 2
4
4
3
Reference Bus

21. 7. Star-Star with primary
neutral grounded and Delta tertiary
1 2Z0
Tertiary
Primary Secondary
3 4
1 2
Z0
Reference Bus

22. Example 4-The line to neutral voltage in a three phase system are,
Van = 200 < 0° Volts,
Vbn = 600 < 100° Volts,
Vcn = 400 < 270° Volts.
Determine symmetrical components of voltages.
Symmetrical Components are given by,
A0=
A+B+C
3
=
Van+Vbn+Vcn
3
A1=
A+ 𝑎B+ 𝑎2C
3
=
Van+ 𝑎Vb𝑛+ 𝑎2Vcn
3
A2=
A+ 𝑎2 B + 𝑎C
3
=
Va𝑛+ 𝑎2Vb𝑛+ 𝑎Vcn
3
Va0 =
200<0° + 600<100° +(400<270°)
3
Va0 = 71.19 < 63.34° Volts
V𝑎1 =
200<0° + 1<120° ∗ 600<100° + 1<240° ∗(115<110°)
3
V𝑎1 = 211.28 < −162.96° Volts
V𝑎2 =
200 < 0° + 1 < 240° ∗ 600 < 100° + 1 < 120° ∗ (400 < 270°)
3
Va2=370.79< −0.2689° Volts

23. Example 5-The delta connected is connected to a three phase supply. One
line of supply is open. The current in other two lines is
20 < 0° A , 20 < 180° A,
Determine symmetrical components of line currents.
Ir = 20 < 0°, Iy = 20 < 180°,Ib=0
Symmetrical Components are given by,
A0=
A+B+C
3
=
Ir+Iy+Ib
3
A1=
A+ 𝑎B+ 𝑎2C
3
=
I 𝑟+ 𝑎Iy+ 𝑎2I 𝑏
3
A2=
A+ 𝑎2 B + 𝑎C
3
=
Ir+ 𝑎2I 𝑦+ 𝑎Ib
3
Ia0 =
20<0° + 600<100° +(0)
3
Ia0 = 0 A
Ia1 =
20<0° + 1<120° ∗ 20<180° + 1<240° ∗(0)
3
I 𝑎1 = 11.54 < −30° A
Ia2 =
20 < 0° + 1 < 240° ∗ 20 < 180° + 1 < 120° ∗ (0)
3
Ia2=11.34< 30° A

24. Example 4-The line to neutral voltage in a three phase system are,
Van = 200 < 0° Volts,
Vbn = 600 < 100° Volts,
Vcn = 400 < 270° Volts.
Determine symmetrical components of voltages.
Symmetrical Components are given by,
A0=
A+B+C
3
=
Van+Vbn+Vcn
3
A1=
A+ 𝑎B+ 𝑎2C
3
=
Van+ 𝑎Vb𝑛+ 𝑎2Vcn
3
A2=
A+ 𝑎2 B + 𝑎C
3
=
Va𝑛+ 𝑎2Vb𝑛+ 𝑎Vcn
3
Va0 =
200<0° + 600<100° +(400<270°)
3
Va0 = 71.19 < 63.34° Volts
V𝑎1 =
200<0° + 1<120° ∗ 600<100° + 1<240° ∗(115<110°)
3
V𝑎1 = 211.28 < −162.96° Volts
V𝑎2 =
200 < 0° + 1 < 240° ∗ 600 < 100° + 1 < 120° ∗ (400 < 270°)
3
Va2=370.79< −0.2689° Volts