47
Fault Calculations Fault Calculations
• Upload

michaeljmack
• Category

## Engineering

• view

555
• download

0

### Transcript of Fault Calculations

Fault CalculationsFault Calculations

A review of:

Nature of Short Circuit Currents

Fault Types

Per Unit Quantities

Symmetrical Components

Calculation Examples

Fault Calculations

Fault Calculations

Vmaxi(t) =Z

[ sin(ωt + α – θ) – sin (α - θ)e – R t / L]

Vm [ sin(ωt + α ) L didt

] = + Ri

Vmax [sin(ωt + α )]

L R

Faulti(t)

sin (α - θ)e – R t / L = Transient current

sin (ωt + α – θ) = Steady state current

Fault Calculations

Short Circuit Currents

Assume L is constant and R = 0

Time

e(t)

Time of faultoccurance

α

α - θ = 9 0 º th e fau lt occu rs a t V o ltag e ze ro : [s in (ω t + α – θ ) – s in (α - θ )]= s in (ω t + 9 0 º) – s in (9 0 º) = co s (ω t) -1

α = θ th e fau lt o ccu rs a t V o ltag e m ax : [s in (ω t + α – θ ) – s in (α - θ )] = s in (ω t)

Vmax I(t) = [ sin (ωt + α – θ) – sin (α - θ)] Z

Vmax I(t) = [ sin (ωt + α – θ) – sin (α - θ)e – R t / L] Z

Fault Inception Angle

Fault Calculations

Fault Calculations

e(t)i(t)

Time of faultoccurance

Time

i

L is constant an R = 0

Fault at Voltage Maximum

Fault at Voltage Zero

e(t)

i(s)= steadystate current

Time of faultoccurance

Time

L constant and R = 0

i(t)= transientcurrent

i(f) = i(s) + i(t)

i

Fault Calculations

Fault at Voltage Zero R ≠0

e(t)

i(s)= steadystate current

Time of faultoccurance

Time

i(t)= transientcurrent

i(f) = i(s) + i(t)

α - θ = 90º

i

L is constant and R 0≠

Fault Calculations

Transient & Subtransient Reactance

Vmax [sin(ωt + α )]

jXd"jXd'jXd R

Faulti(t)

Fault Calculations

Asymmetrical Fault Current

Time

i''max

R = 0 and L is not constant

i'max

imax

Total Asymmetrical Current

DC Component + AC Component

Fault Calculations

Variation of Current with Time During a Fault

Fault Calculations

Variation of Generator ReactanceDuring a Fault

Fault Calculations

Transient and Subtransient Reactances

Instantaneous units are set with short circuit currents calculated with subtransient reactances, that result in higher values of current.

Time delay units can be set using the same values or the transient reactance, depending on the operating speed of the protection relays.

Transient reactance values are generally used in stability studies.

Fault Calculations

X

X

GBΦCΦ

Short Circuit CalculationFault Types – Single Phase to Ground

Fault Calculations

X

X

GBΦCΦ

Short Circuit CalculationsFault Types – Line to Line

Fault Calculations

Short Circuit CalculationsFault Types – Three Phase

GBΦCΦ

AΦX

X

X

Fault Calculations

Short Circuit CalculationsExample 1– System Impedance

59.5 Ω1.56 Ω

11.2 Ω

Transformer

115kV√3

25kV√3

Fault Calculations

+

V1115kV

√3+

V2 Z2Z1

V1A1 = V2A2 V1A1 =V1

2

Z1

V2A2 =V2

2

Z2

Z1 = Z2 x V12

V22

Short Circuit CalculationsExample 1– Equivalent Impedance

Fault Calculations

Short Circuit CalculationsExample 1– Equivalent Impedance at 25 kV

√3Z25 = Z115 x[25/115/√3

]2

0.53

2.80

11.2

59.5

25kV115 kVSystem —

Xfrm —

Fault Calculations

Short Circuit CalculationsExample 1– Fault Calculation at 25 kV

2.80 Ω 1.56 Ω0.53 Ω25kV√3

25kV

√3 x ΣZ

I fault

25kV√3 x 4.89ΩIF = = = 2952A

Fault Calculations

Convert all system parameters to a common base

All components at all voltage levels are combined

Transformers become “transparent” to calculations

Operating system current and voltage values can bederived as the last calculation

Fault Calculations

Short Circuit CalculationsPer Unit System

Establish two base quantities:

Standard practice is to define- Base power – 3 phase- Base voltage – line to line

Other quantities derived with basic power equations

MVA3Φ

kVL-L

Fault Calculations

Short Circuit CalculationsPer Unit System

√3 x kV L-L·baseI base =

x1000MVAbase

Z base =kV2

L-L·base

MVAbase

Fault Calculations

Short Circuit CalculationsPer Unit System

Per Unit Value = Actual QuantityBase Quantity

Vpu = VactualVbase

Ipu = IactualIbase

Zpu = ZactualZbase

Fault Calculations

Short Circuit CalculationsPer Unit System

Short Circuit CalculationsPer Unit System – Base Conversion

Zpu = ZactualZbase

Zbase = kV 2baseMVAbase

Z1base = MVA1basekV 21base X Zactual

Z2base = MVA2basekV 22base

X Zactual

Ratio • Z1baseZ2base

Z2base =Z1base x kV 21base x MVA2base

kV 22base MVA1base

Fault Calculations

Short Circuit CalculationsPer Unit System – Base Conversion

Z2pu = Z1pu x MVA2baseMVA1base

Use if equipment voltage ratings are the same as system base voltages.

Fault Calculations

Short Circuit CalculationsPer Unit System – 25 kV Base

Select MVAbase = 100

Ibase = 100 x 103

√3 x 25

Zbase = 252 = 6.25Ω100

= 2309A

Fault Calculations

Short Circuit CalculationsPer Unit System – Transformers

11.5 / 25kV Delta – Grounded Wye20 / 26.7 / 33.3 MVA Z = 9.0%

Impedance is 0.09 per unit on a 20 MVA base

Z100 pu = 0.09 x 100 = 0.45 pu20

Fault Calculations

Short Circuit CalculationsPer Unit System – Example 1

0.45 pu 0.25 pu0.08 pu25kV√3

1.0

ΣZ

I fault

1.0

0.78IF = = = 1.28 · pu · amperes

I25kV = 1.28pu x 2309A = 2960A

Fault Calculations

Short Circuit CalculationsSymmetrical Components

“Provides a practical technology for understanding and analyzing power system

operation during unbalance conditions”

Protective Relaying Principles and Applications

J. Lewis Blackburn

Fault Calculations

lb2 la2

lc2

NegativePositive

lc1

la1

lb1

lb0

la0

lc0

Zero

Short Circuit CalculationsSymmetrical ComponentsSequence Components

Fault Calculations

Short Circuit CalculationsSymmetrical Components

“a” operator

1∠ 0°

a=1∠ 120°

a2=1∠ 240°

Ib1=1a1∠ 240 = a2Ia1

Ic1=1a1∠ 120 = aIa1

Ia1=1∠ 120

Fault Calculations

Short Circuit CalculationsSymmetrical Components

Network Equations

Ia = I1+I2+I0 I1 = 1 [Ia+aIb+a2Ic]

Ib = a2 I1+aI2+I0 I2 = 1 [Ia+a2Ib+aIc]

Ic = aI1+a2I2+I0 I0 = 1 [Ia+Ib+Ic]

3

3

3

Fault Calculations

Short Circuit CalculationsSymmetrical Component Vectors

I1=1/3 Ia

I2 = 1/3 [Ia+a2Ib+aIc]

Ia=1 ∠0

Ib=1∠ 240

Ic=1∠ 120

I1 = 1/3 [Ia+aIb+a2Ic] aIb= 1∠ 120° 1∠ 240 °x = 1 ∠ 360 °

1∠ 240° 1 ∠ 120°x = 1∠ 360 °a Ic= 2

1∠ 240° 1 ∠ 240°x = 1∠ 120 °a Ib= 2

aIc= 1∠ 120° 1∠ 120 °x = 1 ∠ 240 ° I 2 = 0

a2=1∠ 240°

∠ 120°a=1

Fault Calculations

Symmetrical ComponentsNetwork Equations

In three phase systems, the neutral current is equal to In = (Ia + Ib + Ic) and,therefore, In = 3Io.

Ia = I1+I2+I0

Ib = a2 I1+aI2+I0

Ic = aI1+a2I2+I0

Fault Calculations

Symmetrical ComponentsNetwork Equations

The same equations apply to the voltages:

Va0 = 1/3(Va + Vb + Vc)

Va1 = 1/3(Va + aVb + a2Vc)

Va2 = 1/3(Va + a2Vb + aVc)

Fault Calculations

Short Circuit CalculationsSymmetrical ComponentsNetwork Representations

I2

Z2

NegativeSequence

I0

Z0

Zero Sequence

1 pu I1

Z1

PositiveSequence

Fault Calculations

Short Circuit CalculationsSymmetrical Components

Three Phase Fault

1 pu I1

Z1

Positive Sequence

X

Fault Calculations

Short Circuit CalculationsSymmetrical Components

Phase to Phase Fault

1 pu I1

Z1

Positive Sequence

X

XI2

Z2

Negative Sequence

Fault Calculations

Short Circuit CalculationsSymmetrical Components

Phase to Ground Fault

1 pu I1

Z1

Positive Sequence

X

XI2

Z2

Negative Sequence

XI0

Z0

Zero Sequence

Fault Calculations

Short Circuit CalculationsSymmetrical ComponentsOpen Conductor Condition

1 puX

X

X

Z2

Z1

Z0

I1

I2

I0

Positive Sequence

Negative Sequence

Zero Sequence

Fault Calculations

Short Circuit CalculationsSymmetrical Components

Transformer Representations

Grounded Wye - Grounded Wye

LZ1 or Z2

H

N1 or N2

H LZ0

N0

Fault Calculations

Short Circuit CalculationsSymmetrical Components

Transformer Representations

Z1 or Z2

Z0

N0

H L

N1 orN2

H LDelta - Grounded Wye

Fault Calculations

Z1 or Z2

H L

N1 or N2

R

Z0

N0

H L

3R

G

Delta-Grounded Wyewith Grounding Resistor

Short Circuit CalculationsSymmetrical Components

Transformer Representations

In = (Ia + Ib + Ic)In = 3I0

Fault Calculations

Short Circuit CalculationsSymmetrical Components

Three Phase Fault

0.45 pu 0.25 pu0.08 pu1 pu I1

I2 = I0 = 0

Ia = I1 + I2 + I0 = I1= 1.28pu A

Ib = a2I1 Ic = aI1

I25kV = 1.28pu x 2309A = 2960A

Fault Calculations

Short Circuit CalculationsSymmetrical Components

Phase to Ground Fault

1 pu

X

X

X

Zero Sequence

0.27 pu 0.45 pu 0.95 pu

Negative Sequence

Positive Sequence

0.45 pu0.08 pu 0.25 pu

0.45 pu0.08 pu 0.25 pu

I1 = I2 = I0

I1 = 1.0 = 1pu = 0.35puA

ΣZ 2.86pu

Ia = I1 + I2 + I0 = 1.05puA

I25kV = 1.05 x 2309 = 2424A

Fault Calculations

Short Circuit CalculationsSymmetrical Components

Phase to Ground Fault

1 pu

X

X

X

Zero Sequence

0.27 pu 0.45 pu

Negative Sequence

Positive Sequence

0.45 pu0.08 pu

0.45 pu0.08 pu

I1

I2

I0

Transformer LowSide Faults

I1 = I2 = I0

I1 = 1.0 = 1pu = 0.66puA

ΣZ 1.51pu

Ia = I1 + I2 + I0 = 1.99puA

I25kV = 1.99 x 2309 =4595A

Fault Calculations

ReferencesReferencesBlackburn, J. I., Protective Relaying Principles and Applications, Marcel Dekker, Inc., copyright 1987

Blackburn, J. I., Symmetrical Components for Power Systems Engineering, Marcel Dekker, Inc., copyright 1993

ABB Power T&D Co., Protective Relaying Theory and Application, Marcel Dekker, Inc., copyright 1994

Stevenson, W.D., Elements of Power System Analysis, McGraw-Hill Book Company, Inc., copyright 1962

IEEE Std 242-1986, IEEE Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems

Cooper Power Systems, Electrical Distribution System Protection, copyright 1990, Third Edition

Fault Calculations

©2008 Beckwith Electric Co., Inc.