network fault conditions.ppt 3.pdf

7/28/2019 Network Fault Conditions.ppt 3.pdf

http://slidepdf.com/reader/full/network-fault-conditionsppt-3pdf 1/78

# 1. Network Fault Conditions

# 2. Balanced Faults

# 3. Unbalanced Faults



Network Fault Conditions -Topics

■ Why to Calculate Fault Currents

■ Fault Types

■ Unbalanced Fault Conditions

■ Symmetrical Vectors

■ Phase Sequence Networks



Dimensioning of equipment

Dynamic forces (Idyn)

Network Fault Conditions - Why to Calculate

■ Why it is necessary to calculate fault currents?

erma orces th

Insulation stresses (over voltages)

Breaking capacity of breakers

Setting of protective devicesStability and sensitivity of unit protections

Protection system overall selectivity

Minimum fault current versus maximum load current



Network Fault Conditions - Why to Calculate

■ Why it is necessary to calculate fault currents?

Network stability

Effects the power transfer capacity

Step and touch voltages during earth faults

Control aspects

Allowed and preferred switching combinations



Network Fault Conditions -Topics


■ Fault Types (Active and Passive faults)

■ Balanced Fault Conditions






Network Fault Conditions - Fault Types

FAULT = a state of abnormality within the network that involves

a electrical failure of a primary component

■ Fault Types

Short circuits (Shunt fault)

Open circuits (Series fault)

Simultaneous faults

Winding faults




■ Short Circuits

Faults between phase conductors or between phase conductors

and earth

Three phase Three phase to earth Phase to phase

Single phase to earth Two phase to earth Phase to phase & phase to earth

Three phase fault condition normally used for determination of

system fault levels



Network Fault Conditions -Fault Types

Single phase open circuit Two phase open circuit Three phase open circuit

Failure of one or more phases to conduct

■ Open Circuits

Single and two phase open circuits are causing unbalance in the

system, thus risking a damage with rotating plants

■ Simultaneous (multiple) Faults•~

Presence of two or more similar or different kind of faults

somewhere in the power system

For example: double earth fault



Faults which occur on machine and transformer windings Between windings, winding and earth, within winding or a mixture


■ Winding Faults

Phase to earth Phase to phase Open circuited winding

Short circuited turns



Network Fault Conditions - Topics


■ Fault Types







Network Fault Conditions - Balanced Faults

■ Balanced fault = three phase short circuit

■ Balanced fault = fault current (and voltage) values in all

of the phases are equal in magnitude and symmetrically

° .

■ Balanced values = positive sequence values only



Network Fault Conditions - Unbalanced Faults

■ Unbalanced faults = all other faults except three phase

short circuit

■ Unbalanced fault = fault current (and voltage) values

are different between phases both inma nitude and in hase intervals

■ Unbalance values = positive, negative and zero values





■ Fault Types







Network Fault Conditions - Symmetrical Vectors

■ For proper evaluation of unbalanced fault conditions in asymmetrical three phase network

■ A mathematical method

■ Any unbalanced three phase set of current and/or voltagevectors can be re resented b the sum of three sets of

■ Balanced system vectors:- Positive sequence set

•~ - Negative sequence set•~ - Zero sequence set

balanced symmetrical) vectors

■ Unbalanced system situation is shown as a sum of threebalanced systems



■ Divided into symmetrical current components


■ An unsymmetrical current vector presentation

■ Symmetrical components forming the unsymmetrical currents




■ Positive sequence set of vectors

Equal in magnitude

Spaced with 120° intervals

In positive order

■ Ne ative se uence set of vectors

Equal in magnitude

Spaced with 120° intervals

In reversed (negative) order




Zero sequence set of vectors

◙ Equal in magnitude

◙ Equal in phase

Equations

=a a1 a2 a0

Ib = Ib1 + Ib2 + Ib0

Ic = Ic1 + Ic2 + Ic0

Ia

, Ib

, Ic

denote for any unbalanced three phase vectors in positive

sequence phase order

Second subscripts 1,2 and 0 denote for positive, negative and zero

sequence sets



Unbalanced three-phase voltages and their symmetrical components: (a) unbalanced instantaneous

voltages and their phasors; (b) balanced PPS phasors; (c) balanced NPS phasors and (d) ZPS phasors





■ Fault Types







■Three phase unbalanced network fault case shown withthree equivalent single phase balanced networks

▲ Symmetrical (balanced) current and voltage vectors lead to phasesequence networks

▲ Balanced faults: positive sequence network only Unbalanced faults:

positive, negative and zero sequence networks

Network Fault Currents - Phase Sequence Networks

▲Connections between the sequence networks (serial, parallel or

combination) depend on the fault case under consideration

▲ E.m.f.s produced by generators are positive sequence voltagesonly, there being no generated negative or zero sequencevoltages in the system




Example how to measure sequence impedance

of a transmission line

Positive and negative

sequence impedance

Zero sequence

circuit

Zero sequence

impedance



Overhead lines Z1 = Z2

Zo ≈ 2...3*Z1 (depends on soil, ground wires, etc.)

Cables

Z1≈

Z2


o

Transformers

Z1 ≈ Z2 (typically)

Zo

depends highly on connection group (earthing impedance)

Synchronous machines Z1 > Z2

Zo varies largely, Ro > R1, Xo « X1



#1 . Network Fault Conditions

#2 . Balanced Faults

#3 . Unbalanced Faults



Balanced Faults - Topics

■ Fundamentals

■ Network Components

■ Calculation Methods

■ Ехam le #1

■ Ехample #2



Network laws

Ohm's law

U = I x Z

Vector equation

Complex values

Kirchoff's first law

Balanced Faults – Fundamentals

Ia+Ib+Ic+Id = 0

Kirchoff's second law

The vector sum of source voltages effecting a closed loopis equal to the vector sum of voltage drops in the loop

E1+E2+E3 = I1Z1+I2Z2+I3Z3



Balanced Faults - Fundamentals

■ Different stages during the fault

Initial stageI” k initial short circuit current

X” d initial axial reactance of a synchronous machine

Presence of DC-component

Determines the Idyn level (dynamic short circuit level)

Transient stage I’ k transient short circuit current

Suppressing DC-component

X’ d transient axial reactance of a synchronous machine

Steady stageIk steady stage short circuit current

Xd axial synchronous reactance of a synchronous machine

No DC-component present

Determines the Itherm level (thermal short circuit level)




■ DC - component in the short circuit current

Short circuit current with Short circuit current withoutDC-component (DC offset) DC-component

What further away the generating points in the network are, the lessdifference there will be between initial, transient and steady stage currents




■ DC-component in the short circuit current

Magnitude depends on the moment when the fault occurs The suppression time of the DC-component depends the R/X

ratio of the network

■ Peak Current

Peak value hi hest value of the short circuit current

k = 1,02 + 0,98e-~R ~~

Factor varies between 1,0 and 2,0. Usually with HV-networksvalue 1,8 is used

ip = k kx√2 x I"k (ip = peak current)

If I"k ≈ Ik ,then ip = 1,8 x√ 2 x Ik ≈ 2,5 x Ik

with equipment the Idyn level is usually 2,5 times the Itherm leve




■ Fundamentals



■ Ехample #1

■ Ехample #2



Balanced Faults - Network Components

■ One phase representation of the fault case Driving voltage (E)

Nominal phase voltage of the selected level

IEC recommends following factors for the driving voltage whencalculating the maximum current (factor "c"):

UN > 1kV→ E = 1,1 x UN /√3 = , ... ,

Prefault load situation

To be considered as impedances (usually ignored when calculating by hand)

Network components

To be considered as impedances Calculation with complex values

Z = 78Ω∟65° = 32,96+j70,69Ω




■

Presentation of network components Synchronous generators

Resistive component usually ignored (typically X/R > 10)

Axial reactances used

Effect of X"d , X'd and Xd to be considered case by case

Xd = xd (%) /100 *U2N / SN

Power transformer

Zk = uk (%) /100 *U2N / SN

R k =copper losses = PkN / SN * U2N / SN

Xk = (Z2k – R 2k )

Position of tap changer can be ignored, effects mainly to the load or fault

current division between parallel in feeds




■ Presentation of network components

Network feeder

Resistive component usually ignored

Often the initial short circuit level is given (Sk3)

X"s = U2s / S"k3

If the initial short circuit reactance of the network is less than half from the short

•~

circuit reactance of the power transformer in between the network and the fault

point, the initial level can be regarded the same as the steady stage level (Xs < 0,5 * Xk )

Overhead line or cable circuits

Resistive component recognized

With lower voltages the effect of conductor heating has to be recognized. Usuallywhen calculating the maximum fault current the temperature of 40°C is used.




■ Presentation of network components

Unsynchronous motor

supplies fault current during the first 3-5 cycles

usually recognised only when calculating the initial short circuit

current => switchgears Idyn level




■ Fundamentals■ Network Components


■ Ехample #1 ■ хamp e



Balanced Faults - Calculation Methods

Basic Methods to solve the circuit

Thevenin's theorem

applicable to any linear network

sum of driving voltages can replaced with a single

driving voltage acting in series with a single impedance




■ Basic Methods to solve the circuit

Superposition theorem

applicable to any linear network

a current flowing in any branch of a network as a result of several driving

voltages is a vector sum of currents driven by each individual voltage

. .



■ "Aid" Methods

Star to delta transformation and vice versa

Combination of equal driving voltages

Combination of series or parallel branches





■ Fundamentals



■

Ехample #1 ■



Balanced Faults - Example #1

Fault Case

Calculate the steady stage three phase short circuit current in 24kV switchgear

busbars (using Thevenin's theorem)

How to proceed step by step:

1. Select the voltage level2. Draw the one phase substitute circuit3. Calculate Impedance values4. Convert the impedance values to the chosen voltage level5. Calculate the current




3. Calculate impedance values

X”s= Reactance of the network feeder

X”s = U”s2 / S”k3

Xs" = 110kV2/2300MVA = 5,26Ω

Xl= Reactance of the overhead line

Xl = 130km * 0,4Ω

/km = 52Ω

X = Resistance of the overhead line

Zs= 0+j5,26Ω = 5,26 Ω∟90

Zl= 26+j52 Ω = 58,14 Ω∟63,4

Xl = 130km * 0,2Ω/km = 26 Ω

Zs= 0+j5,26 Ω = 5,26Ω∟90

Zk= Impedance of the power transformer

Zk

= uk

(%) /100 * U2

N

/ SNZk= 11%/100* 113kV2/25MVA=56,18Ω

Rk= Resistance of the power transformer

Rk=PkN/SN* U2N/SN

Rk = 56kW/25MVA * 113kV2/25MVA = 1,14 Ω




4. Convert the impedance values to the chosen voltage level

Zs = Impedance the network feeder

Zs1=Zs * (U2/U1)2

Zs1 = 5,26Ω L90 *(20kV/ 110kV)2 = 0,174Ω L90

Zl = Impedance of the overhead line Zl1 = 58,14Ω L63,4 * (20kV 110kV)2 = 1,922Ω L63,4

Zk = Impedance of the power transformer

Zk1 = 56,18Ω L88,8 * (20kV/ 110kV)2 = 1,857Ω L88




5. Calculate the fault current

I k3 = Three phase fault current

Ik3 = c* Ef/(Zs1 +Zl1+Zk1)

Ik3 = 1,1 * 11,55kV ∟00 / (0,174 Ω∟90 + 1,922 Ω∟63,4 + 1,857Ω∟88,8)

I k3 = 3300 A∟-76,5




■ Fundamentals



■ Ехample #1

■ Ехample #2




Fault Case

Calculate:

a) the steady stage three phase short circuit current

b) the initial three phase short circuit current



#1. Network Fault Conditions

#2. Balanced Faults

#3 . Unbalanced Faults



Unbalanced Faults - Topics

Equations

Power Transformer Connection Groups

Example



Unbalanced Faults -Equations



Unbalanced Faults - Equations




Equations




Unbalanced Faults - PT Connection Groups




U b l d F l PT C i G



Unbalanced Faults - Example

Equations


Example




U b l d F lt E l




Unbalanced Faults Example




U b l d F lt E l

network fault conditions.ppt 3.pdf

Documents