File Ve Ngan Mach - TA

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fundamental

Short circuit in the power system is the phenomenon which 1 or 2 wires collide

with another wires or ground or neutral wire. Total impedance is reduced when short

circuit occur, another parameter; current and voltage have changed and the circuit in

transient state to maintain state.

Short-term short circuit can be auto repaired and eliminated by instantaneous

effects of circuit breaker and it will not be occur when we connect the circuit breaker

again.

Long-term short circuit which still exist if we don’t have any countermeasure after the circuit breaker turned on.

Short circuit in the electrical system is divided into two types: three-phasesymmetrical short circuit and asymmetrical short circuit. Asymmetrical short circuit

include one-phase-to-ground, line-to-line and double line-to-line to ground short

circuit.

Parameters solved from short circuit calculations used for:

- Select suitable equipment

- Calculate value of relays control circuit breaker

- Solve short circuit current in electrical system

- Study voltage stabilization

CONSIDER ONE-PHASE NO-LOAD SYSTEM

To clarify transient and calculate the initial current when short-circuit occur, we

give AC current into a circuit which consisting of resistance and inductance

component.

Assume voltage input is

u = Umaxsin( , at t=0 là thời điểm khóa S khép kín mạch

u = Umaxsin( , at t=0 is time switch S close



Differential equations represented network

Umaxsin( = Ri + L

Short circuit current is a root of equation above

i N = ( (

i N = ( ( i N = √ I

”[sin( ( ] = iac + idc

which Z = ( và = tg-1

(

Redraw transient current i N

Wave form of short circuit current

First term of this equation is a sinusoid component change-over-time

Second term is a component which reduce to exponential with constant T dc = L/R

called DC current

U

2

2 I "

I x k

A

2

2 I "

= 2

2 I "

t

0 n

i N

i dc

i ac

e _

+

R L

i(1)

S



The term sinusoidal is steady value of current in RL circuit when sinusoidal

voltage input into the circuit. If this value different from 0 when t = 0, DC current

exists to satisfy the physical conditions of the zero line at closing time. Note that DC

current will not exist if the switch is closed when wave voltage

– = 0 or

– =

.

If the switch is closed at – = , DC current has maximum initial value andequal maximum value of sinusoid term.

– = 0

– =

Vmaxsin( at t=0.

DC current can have any value from 0 || , depend on the instantaneous value

of the voltage at the time of switch closing and ratio R/L. When it has input voltage,

DC component and steady component have the same magnitude but opposite in sign

to represent 0 value of instantaneous current.

I. TYPES OF SHORT-CIRCUIT

1. Symmetrical short circuit

For three-phase power system, in the case of three-phase short circuit, the voltage

of the three phases at the point is zero, the currents symmetry and the phase shift is

1200 (regardless the short circuit to-ground or not). Therefore, we calculate short-

circuit current as for symmetrical load.

T ho i gia n

i

T h o i g i an



2. Asymmetrical short circuit

There are many methods to solve the problem asymmetrical short-circuit.

However, due to the simplification of one line diagram for solving symmetrical short-

circuit problem, the method of using symmetrical component to represent an

asymmetrical circuit in form of symmetrical circuit. In this section, we will examine

the method of symmetrical components and how to use it to solve the problem of

asymmetric incident 1-phase model.

3. About symmetrical method

The method of symmetrical components allows the system replaces three-phase

asymmetric quantities such as voltage or current by three symmetrical systems, where

each system consists of the three symmetrical components. In a symmetrical system,

each quantity has the same magnitude and the phase shift is constant.

Note that three phase quantities of initial asymmetrical system is initial quantity

and “phase sequence” is understood as a sequence of three initial quantities whenfollowing a positive rotation on the diagram vector, we have three system of replaced

symmetrical component is described as:

- Positive sequence of three quantities (vector) are equal in magnitude, each

pair of a phase shift about 120

0

and have the same sequence as the originalquantity.

- Negative sequence of three quantities (vector) are equal in magnitude, each

pair of a phase shift about 1200

and have opposite phase sequence of the original

quantity.

- Zero sequence of three quantities (vector) are equal in magnitude and phase

shift.



Positive sequence Negative sequence Zero sequence

Symmetrical component

Each initial asymmetrical phase quantity is total of symmetrical components, so

we can rewrite: = + +

= + + =

= + + =

=

=

Sequence by phase quantity

=

(

(

(

II. METHOD OF CALCULATIONS SHORT-CIRCUIT

- The short-circuit in the unit named

- The short-circuit in percentage units

- The short-circuit in relative units

- The short-circuit matrix by the relative unit

Here we examine short-circuit calculations using matrix in relative units

1. Ma trận tổng dẫn nút: - Admittance matrix

In this section, we will present a common method of calculating short-circuit in

multiple nodes network. By using the impedance matrix element busbar (nodes), fault

current as well as fault voltage nodes would be calculated easily and more convenient.



Calculate admittance between nodes: y*ij

Calculate elements on diagonal of admittance matrix Ynode : the diagonal elements

corresponding to each node is the sum of admittance connect to this node.

∑

Element outside the diagonal has the same magnitude but opposite sign with

admittance between 2 nodes:

= = - ( )

Infer admittance matrix YBUS

YBUS =

2. Impedance matrix

For a large power system with multiple nodes, the impedance matrix will have

zero elements and sometimes the inverse matrix can not be calculated. Another way to

find Znode matrix is build it from the simple impedance matrix of one branch then

adding each network element step by step until solve Znode of full network.

Build direct ZBUS: build impedance matrix by adding each branch until a full

system has been built. Each time adding a branch would correspond to a expansion

step of new impedance matrix and so on until the last branch has been added.

Each time adding a branch will fall into one of following case:

- Add a branch from the standard bus to a new bus

- Add a branch from the standard bus to an old bus

- Add a radial branch (line or transformer) from an old bus to a new bus.

- Add lines or transformer between two old bus



equivalent rake circuit

Four types above will be mentioned below. It should be emphasized that

impedance matrix and equivalent rake circuit are based on one phase equivcalent

three-phase system. Assume that the process of adding branches until now beingimplemented matrices of 3x3, ie it represents a system with three standard buses. The

equivalent rake circuit has shown above.

Type 1: Adding a transformer or lines from a standard bus to a new bus. This type

expand impedance matrix of n n to (n +1) (n +1). Thus, the new equivalent rake

circuit has been built by adding impedance Znh of new branch from standard bus to

new bus.

A new loop is established and any current in this loop (equivalent to load

current or generator at bus 4) no interactive with other branches so that explains why

has zero element in the new matrix.



[

]

Type 2: Adding a branch from a standard bus to an old bus.

Assuming a branch with impedance from standard bus to bus 3 as shown in

Figure 3.

Figure 3



Table of Contets

fundamental................................................................................................................................................... 1

CONSIDER ONE-PHASE NO-LOAD SYSTEM ....................................................................................... 1

I. TYPES OF SHORT-CIRCUIT.......................................................................................................... 3

1. Symmetrical short circuit ............................................................................................................... 3

2. Asymmetrical short circuit............................................................................................................. 4

3. About symmetrical method ............................................................................................................ 4

II. METHOD OF CALCULATIONS SHORT-CIRCUIT ............................................................... 5

1. Ma trận tổng dẫn nút: - Admittance matrix................................................................................. 5

2. Impedance matrix ........................................................................................................................... 6

3. Impedance matrix in asymmetrical fault .................................................................................... 14

III. SHORT-CIRCUIT CALCULATION ......................................................................................... 15

1. 3-phase faults ................................................................................................................................. 15

2. Phase-to-ground fault (( ...................................................................................................... 16

3. Line-to-Line faults ((............................................................................................................ 18

4. Double line-to-ground (( .................................................................................................. 20

A new loop is formed entirely within the passive power system.

To solve this case, we need:

- Find impedance loop matrix.

- Remove inner loop by dividing matrix and simplify matrix to find new

[Z and equivalent rake circuit of 3 busbar with standard bar

Impedance matrix:

[

]

=

Where



is an old bus matrix rank 3x3. Note that and repeat row 3 and column 3 of

the old matrix. This is reasonable because the loop 1,2 and 3 have the same mutual

impedance with loop 4 as well as main loop. Consider that element of and have

negative sign if I4 in opposite direction. However, loop 4 will be eliminated so we can

infer that direction of current I4 will not affect to the final result. Assume thatdirection of is counter-clockwise and any element of and have positive sign.

Impedance of loop 4 is total impedance of branch push impedance of loop 3

To eliminate new inner loop 4, we apply compact matrix formula by loop removal

method.

=

Equivalent rake circuit of matrix is drawn in the picture below

Type 3: adding a branch between an old bus and a new bus.

The figure below shows adding branch impedance Znh between bus 2 and bus 4



Assume that no mutual coupling between new line and another one. Total

impedance and mutual of loop 1, 2 and 3 unchanged. We need to expand the system to

four bus and standard bus; the rank of new impedance matrix is 4x4.

[

]

Element of the new bus matrix equal plus total impedance of the old one.

We also notice that the new impedance , , of new matrix need to be in

pair with

,

,

because loop 4 and loop 2 have the same impedance with another

one. Particularly is the common element of loop 2 and loop 4 because two loops has

the same mutual impedance.

The rank of new matrix is 4x4 and it is equivalent to the rake circuit 4 bus

with a standard bus as this picture below.



Type 4: Adding a branch between two old bus

Assume that we connect the branch between bus 2 and bus 3 as this picture.

The drawing inside dashed line is the rake circuit of

. Amount of buses are not

change but the equivalent rake circuit need to be changed to get the new

matrix

By adding the new branch creates a new inner loop in the rake diagram.

Impedance loop matrix 4x4 can be re-written by observed in the figure.

[

]=

Where

The rank of old submatrix is 3x3. The final purpose is get four rows and four

column of . We observe the picture above

Element represent the impedance of loop 4. This impedance consists of plus mutual of on its own loop. We saw that current through ,

this current reflects the branch voltage on proportional to . The sign of this mutual

voltage opposite to voltage generated by run upward at branch (current through

and in opposite directions) so that the impedance has negative sign. Summary:





3. Impedance matrix in asymmetrical fault

In the generalize system n nodes, we used impedance matrix contain positive

sequence to determine the voltage and current when the 3phase symmetrical faults. This

method can easily expand to apply in asymmetrical faults due to negative sequence as

well as zero sequence. For positive sequence, the impedance matrix will be written:

=

[

]

Impedance matrix in negative and zero sequence network

=

[

]

=

[

]

(1) (2) (k) (n)

(1)

(2)

(n)

(k)

(1)

(2)

(n)

(k)

(1) (2) (k) (n)

(1) (2) (k) (n)

(1)

(2)

(n)

(k)



III. SHORT-CIRCUIT CALCULATION

1. 3-phase faults

Short circuit through intermediate station resistor Z N is shown in the picture

below:

Sự cố ba pha

3-phase faults

Calculate symmetrical 3-phase fault current: (breakdown point N is note as node

K)

= ( =

Inferred short-circuit current running on the branch and the node voltage when the

fault occurred

Voltage at break-down point K and at any node i.

[ ]

=

[ ( ( ( ( ]

= ( = (

c

I Na ZN

a

b

ZN

ZN

I Nb

I Nc



Short-circuit current in branches when fault at node K. Calculate current in

branches i-j with branch impedance is

=

2. Phase-to-ground fault ((

Phase to ground fault is a problem most often occurs in the system, usually caused

by lightning strikes or by line contact with the ground. This picture below represents one

phase to ground fault through impedance Z N and phase fault is phase a. Note that K is

breakdown point, three current on 3-phases flowing out of node K when the fault occur is , and . In this case, is current flowing out of node K to ground.

Breakdown phase a to ground through impedance ZN

Z N

I Na

N k

I Nb

I Nc

a

b

c



Equivalent diagram

=

Short circuit current on the fault phase at breakdown node K

=

= 1. + 1.

= 1. + a2.

= 1. + a.

= 0 = 0

= 3 = 3 =

Short circuit sequence voltage at node K



= 0

=

= 0

Short circuit voltage on phases at node K

=

= 1. + 1.

= 1.

+ a

2.

= 1. + a.

3. Line-to-Line faults ((

Assume line-to-line faults occur at phase b and phase c through impedance Z N

as follow picture

Line-to-line fault through impedance ZN

Equivalent diagram

Z N

I Na

N k

I Nb

I Nc

a

b

c

k

k



From diagram inferred the equation of current NM sequence at node K

= 0

= =

Short circuit current on the phase fault at breakdown node K

=

= 1. + 1. = 1. + a2.

= 1. + a.

= 0

=(

= (

Short circuit voltage at breakdown node K

= 0



=

= 0

Short circuit voltage on the phase fault at breakdown node K

=

= 1. + 1.

= 1.

+ a

2.

= 1. + a.

4. Double line-to-ground ((

Assume short circuit double line to ground occur at phase b and phase c at

node K through impedance Z N shown as this picture below.

Double line-to-ground fault



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