KHÓA ĐÀO TẠO TÍNH TOÁN ỔN ĐỊNH VÀ HƯỚNG DẪN SỬ DỤNG PHẦN MỀM PSSE CHO KỸ SƯ HỆ THỐNG ĐIỆN (Các tính toán phân tích sự cố trên Phần mềm PSSE): • Symmetrical Components. • Sequence Impedances. • Analysis of Fault Conditions.• Representation of Faults.
Trang 1A Division of Global Power
POWER SYSTEM STABILITY CALCULATION TRAINING
D 5 F lt A l i Day 5 - Fault Analysis
July 10, 2013 Prepared by: Peter Anderson
Trang 33
Trang 4SYMMETRICAL SEQUENCE COMPONENTS
4
SYMMETRICAL SEQUENCE COMPONENTS
Any given set of Unbalanced Three-phase Vectors
Any given set of Unbalanced Three phase Vectors
can be represented by the sum of three sets of
Balanced or Symmetrical vectors
A
B C
Positive Sequence q Negative Sequence g q Zero Sequence q
Trang 5RELATIONSHIPS BETWEEN PHASE VECTORS & SEQUENCE COMPONENTS
A
P iti S Positive Sequence
1 A 1
C 1 A
2 1 B 1
A , I = a I , I = aI I
0 B 2 B 1 B B
0 A 2 A 1 A A
I + I + I
= I
I + I + I
= I
A
B C
Negative Sequence
2
0 C 2 C 1 C C
0 B 2 B 1 B B
I + I + I
= I
I I
I I
2 A
2 2 C 2 A 2
B 2
A , I = aI , I = a I I
2
0 2 1 A
I + aI + I a
= I
I + I + I
= I
Zero Sequence
0 A 0 C 0 A 0 B 0
A , I = I , I = I I
0 2
2 1
C
0 2 1
B
I + I a + aI
= I
I + aI + I a
= I
0 A 0 C 0 A 0 B 0
A , I I , I I I
Trang 6SEQUENCE IMPEDANCES
6
Trang 7voltage at the Point of Fault and the System Zero Sequence Impedance
Trang 9POSITIVE SEQUENCE IMPEDANCE
POSITIVE SEQUENCE IMPEDANCE
Transmission Lines
Ia Ib
Trang 10ZERO SEQUENCE IMPEDANCE
Trang 11ZERO SEQUENCE IMPEDANCE
ZERO SEQUENCE IMPEDANCE
Transformers
E
I I
I I I
Trang 12ZERO SEQUENCE IMPEDANCE
I
I
I n
Trang 13ZERO SEQUENCE TRANSFORMER
Trang 14ZERO SEQUENCE TRANSFORMER
Trang 15ZERO SEQUENCE TRANSFORMER
T Rg
3Rg X0T‐N
T X0T‐N
CORRECT MODEL
Trang 16ANALYSIS OF FAULT CONDITIONS
16
Trang 17ANALYSIS OF SHORT CIRCUIT CONDITIONS
ANALYSIS OF SHORT-CIRCUIT CONDITIONS
Three Phase Fault
0
= I
2 1
0 2
2 1 0
2 1
2 0
‐ Z I a
‐ E a
= V
V
= Z I
‐ Z I
‐ E
= V
V
= V
= V
= V
2 2 1 1 2 2 B
2 2 1 1 A
C B A
0
= I , 0
= ) a + a +
0 2
1
V 3
= 0
= V + V + V
V
= Z I a
‐ Z I a
‐ aE
= V
C B A
2 2
2 1 1 C
2 2 1 1 B
0
= V
= V
=
2 2
3 3 0
=
I
0
= V ) a
‐ 1 (
= Z I ) 1
‐ a (
= ) Z I a
‐ Z I a
‐ E a (
‐ ) Z I
‐ Z I
‐ E (
2 2
2
2 1 1 3 3 2
2 1 1 B
A
Trang 18ANALYSIS OF SHORT CIRCUIT CONDITIONS
18
ANALYSIS OF SHORT-CIRCUIT CONDITIONS
Three Phase Fault
0
= I , 0
= I
, Z
Trang 19ANALYSIS OF SHORT CIRCUIT CONDITIONS
ANALYSIS OF SHORT-CIRCUIT CONDITIONS
Single Phase to Ground Fault
0 2
1
0 2
1
Z + Z + Z
E 3
= I
= I
=
I 0
I
Phase to Phase to Ground Fault
0
2 + Z ) E Z
= I
2 1
0
1 0 0
2 2
1
0
2 1
E Z
‐
= I
Z Z + Z Z + Z Z
)
(
= I
2 0
1 0 0
2 2
1 2
E Z
‐
= I
Z Z + Z Z + Z Z
= I
1 0 0
2 2
1
0
Z Z + Z Z + Z Z
Trang 20REPRESENTATION OF FAULTS
20
Trang 21ANALYSIS OF SHORT CIRCUIT CONDITIONS
ANALYSIS OF SHORT-CIRCUIT CONDITIONS
Fault Positive Negative Zero 3Ф
Ф‐E
F0 F2
F1
Ф‐Ф
F0 F2
F1
Ф‐Ф‐E
F0 F2
F1