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ESE 470 – Energy Distribution Systems

SECTION 7: FAULT ANALYSIS

K. Webb ESE 470

Introduction2

K. Webb ESE 470

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Power System Faults

Faults in three-phase power systems are short circuits Line-to-ground Line-to-line

Result in the flow of excessive current Damage to equipment Heat – burning/melting Structural damage due to large magnetic forces

Bolted short circuits True short circuits – i.e., zero impedance

In general, fault impedance may be non-zero

Faults may be opens as well We’ll focus on short circuits

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Types of Faults

Type of faults from most to least common: Single line-to-ground faults Line-to-line faults Double line-to-ground faults Balanced three-phase (symmetrical) faults

We’ll look first at the least common type of fault –the symmetrical fault – due to its simplicity

K. Webb ESE 470

Symmetrical Faults5

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Symmetrical Faults

Faults occur nearly instantaneously Lightening, tree fall, arcing over insulation, etc.

Step change from steady-state behavior Like throwing a switch to create the fault at 𝑡𝑡 = 0

Equivalent circuit model:

R: generator resistance L: generator inductance 𝑖𝑖 𝑡𝑡 = 0 for 𝑡𝑡 < 0 Source phase, 𝛼𝛼, determines voltage at 𝑡𝑡 = 0 Short circuit can occur at any point in a 60 Hz cycle

𝑣𝑣 𝑡𝑡 = 2𝑉𝑉𝐺𝐺 sin 𝜔𝜔𝑡𝑡 + 𝛼𝛼

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Symmetrical Faults

The governing differential equation for 𝑡𝑡 > 0 is𝑑𝑑𝑖𝑖𝑑𝑑𝑡𝑡

+ 𝑖𝑖 𝑡𝑡𝑅𝑅𝐿𝐿

=2𝑉𝑉𝐺𝐺𝐿𝐿

sin 𝜔𝜔𝑡𝑡 + 𝛼𝛼

The solution gives the fault current

𝑖𝑖 𝑡𝑡 =2𝑉𝑉𝐺𝐺𝑍𝑍 sin 𝜔𝜔𝑡𝑡 + 𝛼𝛼 − 𝜃𝜃 − sin 𝛼𝛼 − 𝜃𝜃 𝑒𝑒−𝑡𝑡

𝑅𝑅𝐿𝐿

where 𝑍𝑍 = 𝑅𝑅2 + 𝜔𝜔𝐿𝐿 2 and 𝜃𝜃 = tan−1 𝜔𝜔𝐿𝐿𝑅𝑅

This total fault current is referred to as the asymmetrical fault current It has a steady-state component

𝑖𝑖𝑎𝑎𝑎𝑎 𝑡𝑡 =2𝑉𝑉𝐺𝐺𝑍𝑍 sin 𝜔𝜔𝑡𝑡 + 𝛼𝛼 − 𝜃𝜃

And a transient component

𝑖𝑖𝑑𝑑𝑎𝑎 𝑡𝑡 = −2𝑉𝑉𝐺𝐺𝑍𝑍 sin 𝛼𝛼 − 𝜃𝜃 𝑒𝑒−𝑡𝑡

𝑅𝑅𝐿𝐿

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Symmetrical Faults

Magnitude of the transient fault current, 𝑖𝑖𝑑𝑑𝑎𝑎, depends on 𝛼𝛼 𝑖𝑖𝑑𝑑𝑎𝑎 0 = 0 for 𝛼𝛼 = 𝜃𝜃 𝑖𝑖𝑑𝑑𝑎𝑎 0 = 2𝐼𝐼𝑎𝑎𝑎𝑎 for 𝛼𝛼 = 𝜃𝜃 − 90°

𝐼𝐼𝑎𝑎𝑎𝑎 = 𝑉𝑉𝐺𝐺/𝑍𝑍 is the rms value of the steady-state fault current

Worst-case fault current occurs for 𝛼𝛼 = 𝜃𝜃 − 90°

𝑖𝑖 𝑡𝑡 =2𝑉𝑉𝐺𝐺𝑍𝑍 sin 𝜔𝜔𝑡𝑡 −

𝜋𝜋2 + 𝑒𝑒−𝑡𝑡

𝑅𝑅𝐿𝐿

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Symmetrical Faults

Important points here:

Total fault current has both steady-state and transientcomponents – asymmetrical

Magnitude of the asymmetry (transient component) depends on the phase of the generator voltage at the time of the fault

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Generator Reactance

The reactance of the generator was assumed constant in the previous example

Physical characteristics of real generators result in a time-varying reactance following a fault Time-dependence modeled with three reactance values 𝑋𝑋𝑑𝑑′′: subtransient reactance 𝑋𝑋𝑑𝑑′ : transient reactance 𝑋𝑋𝑑𝑑: synchronous reactance

Reactance increases with time, such that𝑋𝑋𝑑𝑑′′ < 𝑋𝑋𝑑𝑑′ < 𝑋𝑋𝑑𝑑

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Sub-Transient Fault Current

Transition rates between reactance values are dictated by two time constants: 𝜏𝜏𝑑𝑑′′: short-circuit subtransient time constant 𝜏𝜏𝑑𝑑′ : short-circuit transient time constant

Neglecting generator resistance and assuming 𝜃𝜃 = −𝜋𝜋2

, the synchronous portion of the fault current is

𝑖𝑖𝑎𝑎𝑎𝑎 𝑡𝑡 = 2𝑉𝑉𝐺𝐺1𝑋𝑋𝑑𝑑′′

−1𝑋𝑋𝑑𝑑′

𝑒𝑒− 𝑡𝑡𝜏𝜏𝑑𝑑′′ +

1𝑋𝑋𝑑𝑑′

−1𝑋𝑋𝑑𝑑

𝑒𝑒− 𝑡𝑡𝜏𝜏𝑑𝑑′ +

1𝑋𝑋𝑑𝑑

sin 𝜔𝜔𝑡𝑡 + 𝛼𝛼 −𝜋𝜋2

At the instant of the fault, 𝑡𝑡 = 0, the rms synchronous fault current is

𝑰𝑰𝐹𝐹′′ =𝑽𝑽𝐺𝐺𝑋𝑋𝑑𝑑′′

This is the rms subtransient fault current, 𝑰𝑰𝐹𝐹′′

This will be our primary metric for assessing fault current

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Symmetrical Three-Phase Short Circuits12

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Symmetrical 3-𝜙𝜙 Short Circuits

Next, we’ll calculate the subtransient fault current resulting from a balanced three-phase fault

We’ll make the following simplifying assumptions: Transformers modeled with leakage reactance only Neglect winding resistance and shunt admittances Neglect Δ-𝑌𝑌 phase shifts

Transmission lines modeled with series reactance only Synchronous machines modeled as constant voltage sources

in series with subtransient reactances Generators and motors

Induction motors are neglected or modeled as synchronous motors

Non-rotating loads are neglected

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We’ll apply superposition to determine three-phase subtransient fault current

Consider the following power system:

Assume there is a balanced three-phase short of bus 1 to ground at 𝑡𝑡 = 0

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The instant of the fault can be modeled by the switch closing in the following line-to-neutral schematic

The short circuit (closed switch) can be represented by two back-to-back voltage sources, each equal to 𝑽𝑽𝐹𝐹

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Applying superposition, we can represent this circuit as the sum of two separate circuits:

Circuit 1 Circuit 2

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Assume that the value of the fault-location source, 𝑽𝑽𝐹𝐹, is the pre-fault voltage at that location Circuit 1, then, represents the pre-fault circuit, so

𝑰𝑰𝐹𝐹1′′ = 0

The 𝑽𝑽𝐹𝐹 source can therefore be removed from circuit 1

Circuit 1 Circuit 2

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The current in circuit 1, 𝑰𝑰𝐿𝐿, is the pre-fault line current Superposition gives the fault current

𝑰𝑰𝐹𝐹′′ = 𝑰𝑰𝐹𝐹1′′ + 𝑰𝑰𝐹𝐹2′′ = 𝑰𝑰𝐹𝐹2′′

The generator fault current is𝑰𝑰𝐺𝐺′′ = 𝑰𝑰𝐺𝐺1′′ + 𝑰𝑰𝐺𝐺2′′

𝑰𝑰𝐺𝐺′′ = 𝑰𝑰𝐿𝐿 + 𝑰𝑰𝐺𝐺2′′

The motor fault current is𝑰𝑰𝑴𝑴′′ = 𝑰𝑰𝑀𝑀1′′ + 𝑰𝑰𝑀𝑀2′′

𝑰𝑰𝑀𝑀′′ = −𝑰𝑰𝐿𝐿 + 𝑰𝑰𝑀𝑀2′′

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Symmetrical 3-𝜙𝜙 Fault – Example

For the simple power system above: Generator is supplying rated power Generator voltage is 5% above rated voltage Generator power factor is 0.95 lagging

A bolted three-phase fault occurs at bus 1 Determine:

Subtransient fault current Subtransient generator current Subtransient motor current

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First convert to per-unit Use 𝑆𝑆𝑏𝑏 = 100 𝑀𝑀𝑉𝑉𝑀𝑀

Base voltage in the transmission line zone is

𝑉𝑉𝑏𝑏,𝑡𝑡𝑡𝑡 = 138 𝑘𝑘𝑉𝑉

Base impedance in the transmission line zone is

𝑍𝑍𝑏𝑏,𝑡𝑡𝑡𝑡 =𝑉𝑉𝑏𝑏,𝑡𝑡𝑡𝑡2

𝑆𝑆𝑏𝑏=

138 𝑘𝑘𝑉𝑉 2

100 𝑀𝑀𝑉𝑉𝑀𝑀= 190.4 Ω

The per-unit transmission line reactance is

𝑋𝑋𝑡𝑡𝑡𝑡 =20 Ω

190.4 Ω= 0.105 𝑝𝑝.𝑢𝑢.

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The two per-unit circuits are

These can be simplified by combining impedances

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Using circuit 2, we can calculate the subtransient fault current

𝑰𝑰𝐹𝐹′′ =1.05∠0°𝑗𝑗0.116

= 9.079∠ − 90° 𝑝𝑝.𝑢𝑢.

To convert to kA, first determine the current base in the generator zone

𝐼𝐼𝑏𝑏,𝐺𝐺 =𝑆𝑆𝑏𝑏

3𝑉𝑉𝑏𝑏,𝐺𝐺=

100 𝑀𝑀𝑉𝑉𝑀𝑀3 ⋅ 13.8 𝑘𝑘𝑉𝑉

= 4.18 𝑘𝑘𝑀𝑀

The subtransient fault current is

𝑰𝑰𝐹𝐹′′ = 9.079∠ − 90° ⋅ 4.18 𝑘𝑘𝑀𝑀

𝑰𝑰𝐹𝐹′′ = 37.98∠ − 90° 𝑘𝑘𝑀𝑀

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The pre-fault line current can be calculated from the pre-fault generator voltage and power

𝑰𝑰𝐿𝐿 =𝑆𝑆𝐺𝐺3𝑽𝑽𝐺𝐺′′

=100 𝑀𝑀𝑉𝑉𝑀𝑀

3 ⋅ 1.05 ⋅ 13.8 𝑘𝑘𝑉𝑉∠ − cos−1 0.95

𝑰𝑰𝐿𝐿 = 3.98∠ − 18.19° 𝑘𝑘𝑀𝑀

Or, in per-unit:

𝑰𝑰𝐿𝐿 =3.98∠ − 18.19° 𝑘𝑘𝑀𝑀

4.18 kA= 0.952∠ − 18.19° 𝑝𝑝.𝑢𝑢.

This will be used to find the generator and motor fault currents

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The generator’s contribution to the fault current is found by applying current division

𝑰𝑰𝐺𝐺2′′ = 𝑰𝑰𝐹𝐹′′0.505

0.505 + 0.15= 7.0∠ − 90° 𝑝𝑝.𝑢𝑢.

Adding the pre-fault line current, we have the subtransient generator fault current

𝑰𝑰𝐺𝐺′′ = 𝑰𝑰𝐿𝐿 + 𝑰𝑰𝐺𝐺2′′

𝑰𝑰𝐺𝐺′′ = 0.952∠ − 18.19° + 7.0∠ − 90°

𝑰𝑰𝐺𝐺′′ = 7.35∠ − 82.9° 𝑝𝑝. 𝑢𝑢.

Converting to kA

𝑰𝑰𝐺𝐺′′ = 7.35∠ − 82.9° ⋅ 4.18 𝑘𝑘𝑀𝑀

𝑰𝑰𝐺𝐺′′ = 30.74∠ − 82.9° 𝑘𝑘𝑀𝑀

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Similarly, for the motor

𝑰𝑰𝑀𝑀2′′ = 𝑰𝑰𝐹𝐹′′0.15

0.505 + 0.15= 2.08∠ − 90° 𝑝𝑝.𝑢𝑢.

Subtracting the pre-fault line current gives the subtransient motor fault current

𝑰𝑰𝑀𝑀′′ = −𝑰𝑰𝐿𝐿 + 𝑰𝑰𝑀𝑀2′′

𝑰𝑰𝑀𝑀′′ = −0.952∠ − 18.19° + 2.08∠ − 90°

𝑰𝑰𝑀𝑀′′ = 2.0∠ − 116.9°

Converting to 𝑘𝑘𝑀𝑀𝑰𝑰𝑀𝑀′′ = 2.0∠ − 116.9° ⋅ 4.18 𝑘𝑘𝑀𝑀

𝑰𝑰𝑀𝑀′′ = 8.36∠ − 116.9° 𝑘𝑘𝑀𝑀

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Symmetrical Components26

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Symmetrical Components

In the previous section, we saw how to calculate subtransient fault current for balanced three-phase faults

Unsymmetrical faults are much more common Analysis is more complicated

We’ll now learn a tool that will simplify the analysis of unsymmetrical faults The method of symmetrical components

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Symmetrical Components

The method of symmetrical components: Represent an asymmetrical set of 𝑁𝑁 phasors as a sum of 𝑁𝑁

sets of symmetrical component phasors These 𝑁𝑁 sets of phasors are called sequence components

Analogous to: Decomposition of electrical signals into differential and

common-mode components Decomposition of forces into orthogonal components

For a three-phase system (𝑁𝑁 = 3), sequence components are:

Zero sequence components Positive sequence components Negative sequence components

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Sequence Components

Zero sequence components Three phasors with equal magnitude and

equal phase 𝑽𝑽𝑎𝑎𝑎, 𝑽𝑽𝑏𝑏𝑎, 𝑽𝑽𝑎𝑎𝑎

Positive sequence components Three phasors with equal magnitude and

± 120°, positive-sequence phase 𝑽𝑽𝑎𝑎1, 𝑽𝑽𝑏𝑏1, 𝑽𝑽𝑎𝑎1

Negative sequence components Three phasors with equal magnitude and

± 120°, negative-sequence phase 𝑽𝑽𝑎𝑎2, 𝑽𝑽𝑏𝑏2, 𝑽𝑽𝑎𝑎2

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Sequence Components

Note that the absolute phase and the magnitudes of the sequence components is not specified Magnitude and phase define a unique set of sequence

components

Any set of phasors – balanced or unbalanced – can be represented as a sum of sequence components

𝑽𝑽𝑎𝑎𝑽𝑽𝑏𝑏𝑽𝑽𝑎𝑎

=𝑽𝑽𝑎𝑎𝑎𝑽𝑽𝑏𝑏𝑎𝑽𝑽𝑎𝑎𝑎

+𝑽𝑽𝑎𝑎1𝑽𝑽𝑏𝑏1𝑽𝑽𝑎𝑎1

+𝑽𝑽𝑎𝑎2𝑽𝑽𝑏𝑏2𝑽𝑽𝑎𝑎2

(1)

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Sequence Components

The phasors of each sequence component have a fixed phase relationship If we know one, we know the other two Assume we know phase 𝑎𝑎 – use that as the reference

For the zero sequence components, we have𝑽𝑽𝑎 = 𝑽𝑽𝑎𝑎𝑎 = 𝑽𝑽𝑏𝑏𝑎 = 𝑽𝑽𝑎𝑎𝑎 (2)

For the positive sequence components,𝑽𝑽1 = 𝑽𝑽𝑎𝑎1 = 1∠120° ⋅ 𝑽𝑽𝑏𝑏1 = 1∠240° ⋅ 𝑽𝑽𝑎𝑎1 (3)

And, for the negative sequence components,𝑽𝑽2 = 𝑽𝑽𝑎𝑎2 = 1∠240° ⋅ 𝑽𝑽𝑏𝑏2 = 1∠120° ⋅ 𝑽𝑽𝑎𝑎2 (4)

Note that we’re using phase 𝑎𝑎 as our reference, so𝑽𝑽𝑎 = 𝑽𝑽𝑎𝑎𝑎, 𝑽𝑽1 = 𝑽𝑽𝑎𝑎1, 𝑽𝑽2 = 𝑽𝑽𝑎𝑎2

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Sequence Components

Next, we define a complex number, 𝑎𝑎, that has unit magnitude and phase of 120°

𝑎𝑎 = 1∠120° (5)

Multiplication by 𝑎𝑎 results in a rotation (a phase shift) of 120° Multiplication by 𝑎𝑎2 yields a rotation of 240° = −120°

Using (5) to rewrite (3) and (4)

𝑽𝑽1 = 𝑽𝑽𝑎𝑎1 = 𝑎𝑎𝑽𝑽𝑏𝑏1 = 𝑎𝑎2𝑽𝑽𝑎𝑎1 (6)

𝑽𝑽2 = 𝑽𝑽𝑎𝑎2 = 𝑎𝑎2𝑽𝑽𝑏𝑏2 = 𝑎𝑎𝑽𝑽𝑎𝑎2 (7)

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Sequence Components

Using (2), (6), and (7), we can rewrite (1) in a simplified form


=1 1 11 𝑎𝑎2 𝑎𝑎1 𝑎𝑎 𝑎𝑎2

𝑽𝑽𝑎𝑽𝑽1𝑽𝑽2

(8)

The vector on the left is the vector of phase voltages, 𝑽𝑽𝑝𝑝 The vector on the right is the vector of (phase 𝑎𝑎) sequence

components, 𝑽𝑽𝑠𝑠 We’ll call the 3 × 3 transformation matrix 𝑨𝑨

We can rewrite (8) as

𝑽𝑽𝑝𝑝 = 𝑨𝑨𝑽𝑽𝑠𝑠 (9)

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Sequence Components

We can express the sequence voltages as a function of the phase voltages by inverting the transformation matrix

𝑽𝑽𝑠𝑠 = 𝑨𝑨−1𝑽𝑽𝑝𝑝 (10)

where

𝑨𝑨−1 = 13

1 1 11 𝑎𝑎 𝑎𝑎21 𝑎𝑎2 𝑎𝑎

(11)

So𝑽𝑽𝑎𝑽𝑽1𝑽𝑽2

= 13

1 1 11 𝑎𝑎 𝑎𝑎21 𝑎𝑎2 𝑎𝑎


(12)

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Sequence Components

The same relationships hold for three-phase currents

The phase currents are

𝑰𝑰𝑝𝑝 =𝑰𝑰𝑎𝑎𝑰𝑰𝑏𝑏𝑰𝑰𝑎𝑎

And, the sequence currents are

𝑰𝑰𝑠𝑠 =𝑰𝑰𝑎𝑰𝑰1𝑰𝑰2

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Sequence Components

The transformation matrix, 𝑨𝑨, relates the phase currents to the sequence currents

𝑰𝑰𝑝𝑝 = 𝑨𝑨𝑰𝑰𝑠𝑠𝑰𝑰𝑎𝑎𝑰𝑰𝑏𝑏𝑰𝑰𝑎𝑎

=1 1 11 𝑎𝑎2 𝑎𝑎1 𝑎𝑎 𝑎𝑎2

𝑰𝑰𝑎𝑰𝑰1𝑰𝑰2

(13)

And vice versa

𝑰𝑰𝑠𝑠 = 𝑨𝑨−1𝑰𝑰𝑝𝑝𝑰𝑰𝑎𝑰𝑰1𝑰𝑰2

= 13

1 1 11 𝑎𝑎 𝑎𝑎21 𝑎𝑎2 𝑎𝑎

𝑰𝑰𝑎𝑎𝑰𝑰𝑏𝑏𝑰𝑰𝑎𝑎

(14)

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Sequence Components – Balanced System

Before applying sequence components to unbalanced systems, let’s first look at the sequence components for a balanced, positive-sequence, three-phase system

For a balanced system, we have

𝑽𝑽𝑏𝑏 = 𝑽𝑽𝑎𝑎 ⋅ 1∠ − 120° = 𝑎𝑎2𝑽𝑽𝑎𝑎𝑽𝑽𝑎𝑎 = 𝑽𝑽𝑎𝑎 ⋅ 1∠120° = 𝑎𝑎𝑽𝑽𝑎𝑎

The sequence voltages are given by (12) The zero sequence voltage is

𝑽𝑽𝑎 =13 𝑽𝑽𝑎𝑎 + 𝑽𝑽𝑏𝑏 + 𝑽𝑽𝑎𝑎 =

13 𝑽𝑽𝑎𝑎 + 𝑎𝑎2𝑽𝑽𝑎𝑎 + 𝑎𝑎𝑽𝑽𝑎𝑎

𝑽𝑽𝑎 =13𝑽𝑽𝑎𝑎 1 + 𝑎𝑎2 + 𝑎𝑎

Applying the identity 1 + 𝑎𝑎2 + 𝑎𝑎 = 0, we have

𝑽𝑽𝑎 = 0

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The positive sequence component is given by

𝑽𝑽1 =13𝑽𝑽𝑎𝑎 + 𝑎𝑎𝑽𝑽𝑏𝑏 + 𝑎𝑎2𝑽𝑽𝑎𝑎

𝑽𝑽1 =13𝑽𝑽𝑎𝑎 + 𝑎𝑎 ⋅ 𝑎𝑎2𝑽𝑽𝑎𝑎 + 𝑎𝑎2 ⋅ 𝑎𝑎𝑽𝑽𝑎𝑎

𝑽𝑽1 =13𝑽𝑽𝑎𝑎 + 𝑎𝑎3𝑽𝑽𝑎𝑎 + 𝑎𝑎3𝑽𝑽𝑎𝑎

Since 𝑎𝑎3 = 1∠0°, we have

𝑽𝑽1 =13

3𝑽𝑽𝑎𝑎

𝑽𝑽1 = 𝑽𝑽𝑎𝑎

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The negative sequence component is given by

𝑽𝑽2 =13𝑽𝑽𝑎𝑎 + 𝑎𝑎2𝑽𝑽𝑏𝑏 + 𝑎𝑎𝑽𝑽𝑎𝑎

𝑽𝑽2 =13𝑽𝑽𝑎𝑎 + 𝑎𝑎2 ⋅ 𝑎𝑎2𝑽𝑽𝑎𝑎 + 𝑎𝑎 ⋅ 𝑎𝑎𝑽𝑽𝑎𝑎

𝑽𝑽2 =13𝑽𝑽𝑎𝑎 + 𝑎𝑎4𝑽𝑽𝑎𝑎 + 𝑎𝑎2𝑽𝑽𝑎𝑎

Again, using the identity 1 + 𝑎𝑎2 + 𝑎𝑎 = 0, along with the fact that 𝑎𝑎4 = 𝑎𝑎, we have

𝑽𝑽2 = 0

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So, for a positive-sequence, balanced, three-phase system, the sequence voltages are

𝑽𝑽𝑎 = 0, 𝑽𝑽1 = 𝑽𝑽𝑎𝑎, 𝑽𝑽2 = 0

Similarly, the sequence currents are𝑰𝑰𝑎 = 0, 𝑰𝑰1 = 𝑰𝑰𝑎𝑎, 𝑰𝑰2 = 0

This is as we would expect No zero- or negative-sequence components for a

positive-sequence balanced system Zero- and negative-sequence components are only

used to account for imbalance

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41

Sequence Components

We have just introduced the concept of symmetric components Allows for decomposition of, possibly unbalanced,

three-phase phasors into sequence components

We’ll now apply this concept to power system networks to develop sequence networks Decoupled networks for each of the sequence

components Sequence networks become coupled only at the point

of imbalance Simplifies the analysis of unbalanced systems

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Sequence Networks42

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Sequence Networks

Power system components each have their own set of sequence networks Non-rotating loads Transmission lines Rotating machines – generators and motors Transformers

Sequence networks for overall systems are interconnections of the individual sequence network

Sequence networks become coupled in a particular way at the fault location depending on type of fault Line-to-line Single line-to-ground Double line-to-ground

Fault current can be determined through simple analysis of the coupled sequence networks

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44

Sequence Networks – Non-Rotating Loads

Consider a balanced Y-load with the neutral grounded through some non-zero impedance

Applying KVL gives the phase-𝑎𝑎-to-ground voltage

𝑽𝑽𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑦𝑦𝑰𝑰𝑎𝑎 + 𝑍𝑍𝑛𝑛𝑰𝑰𝑛𝑛

𝑽𝑽𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑦𝑦𝑰𝑰𝑎𝑎 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑎𝑎 + 𝑰𝑰𝑏𝑏 + 𝑰𝑰𝑎𝑎𝑽𝑽𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑦𝑦 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑎𝑎 + 𝑍𝑍𝑛𝑛𝑰𝑰𝑏𝑏 + 𝑍𝑍𝑛𝑛𝑰𝑰𝑎𝑎 (15)

For phase 𝑏𝑏:𝑽𝑽𝑏𝑏𝑎𝑎 = 𝑍𝑍𝑦𝑦𝑰𝑰𝑏𝑏 + 𝑍𝑍𝑛𝑛𝑰𝑰𝑛𝑛 = 𝑍𝑍𝑦𝑦𝑰𝑰𝑏𝑏 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑎𝑎 + 𝑰𝑰𝑏𝑏 + 𝑰𝑰𝑎𝑎𝑽𝑽𝑏𝑏𝑎𝑎 = 𝑍𝑍𝑛𝑛𝑰𝑰𝑎𝑎 + 𝑍𝑍𝑦𝑦 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑏𝑏 + 𝑍𝑍𝑛𝑛𝑰𝑰𝑎𝑎 (16)

Similarly, for phase 𝑐𝑐:𝑽𝑽𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑛𝑛𝑰𝑰𝑎𝑎 + 𝑍𝑍𝑛𝑛𝑰𝑰𝑏𝑏 + 𝑍𝑍𝑦𝑦 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑎𝑎 (17)

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Putting (15) – (17) in matrix form

𝑽𝑽𝑎𝑎𝑎𝑎𝑽𝑽𝑏𝑏𝑎𝑎𝑽𝑽𝑎𝑎𝑎𝑎

=𝑍𝑍𝑦𝑦 + 𝑍𝑍𝑛𝑛 𝑍𝑍𝑛𝑛 𝑍𝑍𝑛𝑛𝑍𝑍𝑛𝑛 𝑍𝑍𝑦𝑦 + 𝑍𝑍𝑛𝑛 𝑍𝑍𝑛𝑛𝑍𝑍𝑛𝑛 𝑍𝑍𝑛𝑛 𝑍𝑍𝑦𝑦 + 𝑍𝑍𝑛𝑛


or𝑽𝑽𝑝𝑝 = 𝒁𝒁𝑝𝑝𝑰𝑰𝑝𝑝 (18)

where 𝑽𝑽𝑝𝑝 and 𝑰𝑰𝑝𝑝 are the phase voltages and currents, respectively, and 𝒁𝒁𝑝𝑝 is the phase impedance matrix

We can use (9) and (13) to rewrite (18) as

𝑨𝑨𝑽𝑽𝑠𝑠 = 𝒁𝒁𝑝𝑝𝑨𝑨𝑰𝑰𝑠𝑠 Solving for 𝑽𝑽𝑠𝑠

𝑽𝑽𝑠𝑠 = 𝑨𝑨−1𝒁𝒁𝑝𝑝𝑨𝑨𝑰𝑰𝑠𝑠or

𝑽𝑽𝑠𝑠 = 𝒁𝒁𝑠𝑠𝑰𝑰𝑠𝑠 (19)

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𝑽𝑽𝑠𝑠 = 𝒁𝒁𝑠𝑠𝑰𝑰𝑠𝑠 (19)

where 𝑍𝑍𝑠𝑠 is the sequence impedance matrix

𝒁𝒁𝑠𝑠 = 𝑨𝑨−1𝒁𝒁𝑝𝑝𝑨𝑨 =𝑍𝑍𝑦𝑦 + 3𝑍𝑍𝑛𝑛 0 0

0 𝑍𝑍𝑦𝑦 00 0 𝑍𝑍𝑦𝑦

(20)

Equation (19) then becomes a set of three uncoupled equations

𝑽𝑽𝑎 = 𝑍𝑍𝑦𝑦 + 3𝑍𝑍𝑛𝑛 𝑰𝑰𝑎 = 𝑍𝑍𝑎𝑰𝑰𝑎 (21)

𝑽𝑽1 = 𝑍𝑍𝑦𝑦𝑰𝑰1 = 𝑍𝑍1𝑰𝑰1 (22)

𝑽𝑽2 = 𝑍𝑍𝑦𝑦𝑰𝑰2 = 𝑍𝑍2𝑰𝑰2 (23)

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Equations (21) – (23) describe the uncoupled sequence networks

Positive-sequence network:

Negative-sequence network:

Zero-sequence network:

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We can develop similar sequence networks for a balanced Δ-connected load 𝑍𝑍𝑦𝑦 = 𝑍𝑍Δ/3 There is no neutral point for the Δ-network, so 𝑍𝑍𝑛𝑛 = ∞ - an open circuit




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49

Sequence Networks – 3-𝜙𝜙 Lines

Balanced, three-phase lines can be modeled as

The voltage drops across the lines are given by the following system of equations

𝑽𝑽𝑎𝑎𝑎𝑎′𝑽𝑽𝑏𝑏𝑏𝑏′𝑽𝑽𝑎𝑎𝑎𝑎′

=𝑍𝑍𝑎𝑎𝑎𝑎 𝑍𝑍𝑎𝑎𝑏𝑏 𝑍𝑍𝑎𝑎𝑎𝑎𝑍𝑍𝑏𝑏𝑎𝑎 𝑍𝑍𝑏𝑏𝑏𝑏 𝑍𝑍𝑏𝑏𝑎𝑎𝑍𝑍𝑎𝑎𝑎𝑎 𝑍𝑍𝑎𝑎𝑏𝑏 𝑍𝑍𝑎𝑎𝑎𝑎


=𝑽𝑽𝑎𝑎𝑛𝑛 − 𝑽𝑽𝑎𝑎′𝑛𝑛𝑽𝑽𝑏𝑏𝑛𝑛 − 𝑽𝑽𝑏𝑏′𝑛𝑛𝑽𝑽𝑎𝑎𝑛𝑛 − 𝑽𝑽𝑎𝑎′𝑛𝑛

(24)

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Writing (24) in compact form

𝑽𝑽𝑝𝑝 − 𝑽𝑽𝑝𝑝′ = 𝒁𝒁𝑝𝑝𝑰𝑰𝑝𝑝 (25)

𝒁𝒁𝑝𝑝 is the phase impedance matrix Self impedances along the diagonal Mutual impedances elsewhere Symmetric Diagonal, if we neglect mutual impedances

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We can rewrite (25) in terms of sequence components𝑨𝑨𝑽𝑽𝑠𝑠 − 𝑨𝑨𝑽𝑽𝑠𝑠′ = 𝒁𝒁𝑝𝑝𝑨𝑨𝑰𝑰𝑠𝑠𝑽𝑽𝑠𝑠 − 𝑽𝑽𝑠𝑠′ = 𝑨𝑨−1𝒁𝒁𝑝𝑝𝑨𝑨𝑰𝑰𝑠𝑠𝑽𝑽𝑠𝑠 − 𝑽𝑽𝑠𝑠′ = 𝒁𝒁𝑠𝑠𝑰𝑰𝑠𝑠 (26)

where 𝒁𝒁𝑠𝑠 is the sequence impedance matrix

𝒁𝒁𝑠𝑠 = 𝑨𝑨−1𝒁𝒁𝑝𝑝𝑨𝑨 (27)

𝒁𝒁𝑠𝑠 is diagonal so long as the system impedances are balanced, i.e. Self impedances are equal: 𝑍𝑍𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑏𝑏𝑏𝑏 = 𝑍𝑍𝑎𝑎𝑎𝑎 Mutual impedances are equal: 𝑍𝑍𝑎𝑎𝑏𝑏 = 𝑍𝑍𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑏𝑏𝑎𝑎

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For balanced lines, 𝒁𝒁𝑠𝑠 is diagonal

𝒁𝒁𝑠𝑠 =𝑍𝑍𝑎𝑎𝑎𝑎 + 2𝑍𝑍𝑎𝑎𝑏𝑏 0 0

0 𝑍𝑍𝑎𝑎𝑎𝑎 − 𝑍𝑍𝑎𝑎𝑏𝑏 00 0 𝑍𝑍𝑎𝑎𝑎𝑎 − 𝑍𝑍𝑎𝑎𝑏𝑏

=𝑍𝑍𝑎 0 00 𝑍𝑍1 00 0 𝑍𝑍2

Because 𝑍𝑍𝑠𝑠 is diagonal, (26) represents three uncoupled equations

𝑽𝑽𝑎 − 𝑽𝑽𝑎′ = 𝑍𝑍𝑎𝑰𝑰𝑎 (28)

𝑽𝑽1 − 𝑽𝑽1′ = 𝑍𝑍1𝑰𝑰1 (29)

𝑽𝑽2 − 𝑽𝑽2′ = 𝑍𝑍2𝑰𝑰2 (30)

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Equations (28) – (30) describe the voltage drop across three uncoupled sequence networks




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Sequence Networks – Rotating Machines

Consider the following model for a synchronous generator

Similar to the Y-connected load Generator includes voltage

sources on each phase

Voltage sources are positive sequence Sources will appear only in the

positive-sequence network

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Sequence Networks – Synchronous Generator

Sequence networks for Y-connected synchronous generator




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Sequence Networks – Motors

Synchronous motors Sequence networks identical to those for synchronous

generators Reference current directions are reversed

Induction motors Similar sequence networks to synchronous motors,

except source in the positive sequence network set to zero

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Sequence Networks - Y-Y Transformers

Per-unit sequence networks for transformers Simplify by neglecting transformer shunt admittances

Consider a Y-Y transformer

Similar to the Y-connected load, the voltage drops across the neutral impedances are 3𝐼𝐼𝑎𝑍𝑍𝑁𝑁 and 3𝐼𝐼𝑎𝑍𝑍𝑛𝑛 3𝑍𝑍𝑁𝑁 and 3𝑍𝑍𝑛𝑛 each appear in the zero-sequence network Can be combined in the per-unit circuit as long as shunt

impedances are neglected

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Sequence Networks – Y-Y Transformers

Impedance accounting for leakage flux and winding resistance for each winding can be referred to the primary Add together into a single impedance, 𝑍𝑍𝑠𝑠, in the per-unit model

Y-Y transformer sequence networks




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Sequence Networks – Y-ΔTransformers

Y-Δ transformers differ in a couple of ways

Must account for phase shift from primary to secondary For positive-sequence network, Y-side voltage and current lead Δ-

side voltage and current For negative-sequence network, Y-side voltage and current lag Δ-

side voltage and current

No neutral connection on the Δ side Zero-sequence current cannot enter or leave the Δ winding

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Sequence Networks – Y-ΔTransformers

Sequence networks for Y-ΔTransformers




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Sequence Networks – Δ-ΔTransformers

Δ-Δ transformers Like Y-Y transformers, no phase shift No neutral connections Zero-sequence current cannot flow into or out of either

winding

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Sequence Networks – Δ-ΔTransformers

Sequence networks for Δ-ΔTransformers




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Power in Sequence Networks

We can relate the power delivered to a system’s sequence networks to the three-phase power delivered to that system

We know that the complex power delivered to a three-phase system is the sum of the power at each phase

𝑆𝑆𝑝𝑝 = 𝑽𝑽𝑎𝑎𝑛𝑛𝑰𝑰𝑎𝑎∗ + 𝑽𝑽𝑏𝑏𝑛𝑛𝑰𝑰𝑏𝑏∗ + 𝑽𝑽𝑎𝑎𝑛𝑛𝑰𝑰𝑎𝑎∗

In matrix form, this looks like

𝑆𝑆𝑝𝑝 = 𝑽𝑽𝑎𝑎𝑛𝑛 𝑽𝑽𝑏𝑏𝑛𝑛 𝑽𝑽𝑎𝑎𝑛𝑛𝑰𝑰𝑎𝑎∗𝑰𝑰𝑏𝑏∗𝑰𝑰𝑎𝑎∗

𝑆𝑆𝑝𝑝 = 𝑽𝑽𝑝𝑝𝑇𝑇𝑰𝑰𝑝𝑝∗ (28)

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Recall the following relationships

𝑽𝑽𝑝𝑝 = 𝑨𝑨𝑽𝑽𝑠𝑠 (9)

𝑰𝑰𝑝𝑝 = 𝑨𝑨𝑰𝑰𝑠𝑠 (13)

Using (9) and (13) in (28), we have

𝑺𝑺𝑝𝑝 = 𝑨𝑨𝑽𝑽𝑠𝑠 𝑇𝑇 𝑨𝑨𝑰𝑰𝑠𝑠 ∗

𝑺𝑺𝑝𝑝 = 𝑽𝑽𝑠𝑠𝑇𝑇𝑨𝑨𝑇𝑇𝑨𝑨∗𝑰𝑰𝑠𝑠∗ (29)

Computing the product in the middle of the right-hand side of (29), we find

𝑨𝑨𝑇𝑇𝑨𝑨∗ =3 0 00 3 00 0 3

= 3𝑰𝑰3

where 𝑰𝑰3 is the 3×3 identity matrix

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Equation (29) then becomes

𝑺𝑺𝑝𝑝 = 𝑽𝑽𝑠𝑠𝑇𝑇3𝑰𝑰3𝑰𝑰𝑠𝑠∗

𝑺𝑺𝑝𝑝 = 3𝑽𝑽𝑠𝑠𝑇𝑇𝑰𝑰𝑠𝑠∗

𝑺𝑺𝑝𝑝 = 3 𝑽𝑽𝑎 𝑽𝑽1 𝑽𝑽2𝑰𝑰𝑎∗𝑰𝑰1∗𝑰𝑰2∗

𝑺𝑺𝑝𝑝 = 3 𝑽𝑽𝑎𝑰𝑰𝑎∗ + 𝑽𝑽1𝑰𝑰1∗ + 𝑽𝑽2𝑰𝑰2∗

The total power delivered to a three-phase network is three times the sum of the power delivered to the three sequence networks The three sequence networks represent only one of the three

phases – recall, we chose to consider only phase 𝑎𝑎

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Unsymmetrical Faults66

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Unsymmetrical Faults

The majority of faults that occur in three-phase power systems are unsymmetrical Not balanced Fault current and voltage differ for each phase

The method of symmetrical components and sequence networks provide us with a tool to analyze these unsymmetrical faults

We’ll examine three types of unsymmetrical faults Single line-to-ground (SLG) faults Line-to-line (LL) faults Double line-to-ground (DLG) faults

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Unsymmetrical Fault Analysis - Procedure

Basic procedure for fault analysis:

1. Generate sequence networks for the system

2. Interconnect sequence networks appropriately at the fault location

3. Perform circuit analysis on the interconnected sequence networks

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Unsymmetrical Fault Analysis

To simplify our analysis, we’ll make the following assumptions1. System is balanced before the instant of the fault

2. Neglect pre-fault load current All pre-fault machine terminal voltages and bus voltages are equal

to 𝑉𝑉𝐹𝐹3. Transmission lines are modeled as series reactances only

4. Transformers are modeled with leakage reactances only5. Non-rotating loads are neglected6. Induction motors are either neglected or modeled as

synchronous motors

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Unsymmetrical Fault Analysis

Each sequence network includes all interconnected power-system components Generators, motors, lines, and transformers

Analysis will be simplified if we represent each sequence network as its Thévenin equivalent From the perspective of the fault location

For example, consider the following power system:

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Unsymmetrical Fault Analysis – Sequence Networks

The sequence networks for the system are generated by interconnecting the sequence networks for each of the components

The zero-sequence network:

The positive-sequence network: Assuming the generator is operating at the rated voltage at the time of

the fault

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The negative sequence network:

Now, let’s assume there is some sort of fault at bus 1 Determine the Thévenin equivalent for each sequence network from the

perspective of bus 1 Simplifying the zero-sequence network to its Thévenin equivalent

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The positive-sequence network simplifies to the following circuit with the following Thévenin equivalent

Similarly, for the negative-sequence network, we have

Next, we’ll see how to interconnect these networks to analyze different types of faults

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Unsymmetrical Fault Analysis – SLG Fault

The following represents a generic three-phase network with terminals at the fault location:

If we have a single-line-to-ground fault, where phase 𝑎𝑎 is shorted through 𝑍𝑍𝑓𝑓 to ground, the model becomes:

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The phase-domain fault conditions:

𝑰𝑰𝑎𝑎 = 𝑽𝑽𝑎𝑎𝑎𝑎𝑍𝑍𝑓𝑓

(1)

𝑰𝑰𝑏𝑏 = 𝑰𝑰𝑎𝑎 = 0 (2)

Transforming these phase-domain currents to the sequence domain

𝑰𝑰𝑎𝑰𝑰1𝑰𝑰2

= 13

1 1 11 𝑎𝑎 𝑎𝑎21 𝑎𝑎2 𝑎𝑎

𝑽𝑽𝑎𝑎𝑎𝑎/𝑍𝑍𝑓𝑓00

= 13

𝑽𝑽𝑎𝑎𝑎𝑎/𝑍𝑍𝑓𝑓𝑽𝑽𝑎𝑎𝑎𝑎/𝑍𝑍𝑓𝑓𝑽𝑽𝑎𝑎𝑎𝑎/𝑍𝑍𝑓𝑓

(3)

This gives one of our sequence-domain fault conditions

𝑰𝑰𝑎 = 𝑰𝑰1 = 𝑰𝑰2 (4)

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We know that

𝑰𝑰𝑎𝑎 = 𝑽𝑽𝑎𝑎𝑎𝑎𝑍𝑍𝑓𝑓

= 𝑰𝑰𝑎 + 𝑰𝑰1 + 𝑰𝑰2 (5)

and𝑽𝑽𝑎𝑎𝑎𝑎 = 𝑽𝑽𝑎 + 𝑽𝑽1 + 𝑽𝑽2 (6)

Using (5) and (6) in (1), we get

𝑰𝑰𝑎 + 𝑰𝑰1 + 𝑰𝑰2 = 1𝑍𝑍𝑓𝑓

𝑽𝑽𝑎 + 𝑽𝑽1 + 𝑽𝑽2

Using (4), this gives our second sequence-domain fault condition

𝑰𝑰𝑎 = 𝑰𝑰1 = 𝑰𝑰2 = 13𝑍𝑍𝑓𝑓

𝑽𝑽𝑎 + 𝑽𝑽1 + 𝑽𝑽2 (7)

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The sequence-domain fault conditions are satisfied by connecting the sequence networks in series along with three times the fault impedance

We want to find the phase domain fault current, 𝑰𝑰𝐹𝐹

𝑰𝑰𝐹𝐹 = 𝑰𝑰𝑎𝑎 = 𝑰𝑰𝑎 + 𝑰𝑰1 + 𝑰𝑰2 = 3𝑰𝑰1

𝑰𝑰1 = 𝑽𝑽𝐹𝐹𝑍𝑍0+𝑍𝑍1+𝑍𝑍2+3𝑍𝑍𝐹𝐹

𝑰𝑰𝐹𝐹 = 3𝑽𝑽𝐹𝐹𝑍𝑍0+𝑍𝑍1+𝑍𝑍2+3𝑍𝑍𝐹𝐹

(8)

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SLG Fault - Example

Returning to our example power system

The interconnected sequence networks for a bolted fault at bus 1:

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SLG Fault - Example

The fault current is

𝑰𝑰𝐹𝐹 = 3𝑽𝑽𝐹𝐹𝑍𝑍0+𝑍𝑍1+𝑍𝑍2+3𝑍𝑍𝐹𝐹

𝑰𝑰𝐹𝐹 = 3.𝑎∠3𝑎°𝑗𝑗𝑎.487

= 6.16∠ − 60° 𝑝𝑝.𝑢𝑢.

The current base at bus 1 is

𝑰𝑰𝑏𝑏 = 𝑆𝑆𝑏𝑏𝑉𝑉𝑏𝑏1

= 15𝑎 𝑀𝑀𝑉𝑉𝑀𝑀3 23𝑎 𝑘𝑘𝑉𝑉

= 376.5 𝑀𝑀

So the fault current in kA is𝑰𝑰𝐹𝐹 = 2.32∠ − 60° 376.5 𝑀𝑀

𝑰𝑰𝐹𝐹 = 2.32∠ − 60° 𝑘𝑘𝑀𝑀

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Unsymmetrical Fault Analysis – LL Fault

Now consider a line-to-line fault between phase 𝑏𝑏 and phase 𝑐𝑐 through impedance 𝑍𝑍𝐹𝐹

Phase-domain fault conditions:

𝑰𝑰𝑎𝑎 = 0 (9)

𝑰𝑰𝑏𝑏 = −𝑰𝑰𝑎𝑎 = 𝑽𝑽𝑏𝑏𝑎𝑎−𝑽𝑽𝑐𝑐𝑎𝑎𝑍𝑍𝐹𝐹

(10)

Transforming to the sequence domain𝑰𝑰𝑎𝑰𝑰1𝑰𝑰2

= 13

1 1 11 𝑎𝑎 𝑎𝑎21 𝑎𝑎2 𝑎𝑎

0𝑰𝑰𝑏𝑏−𝑰𝑰𝑏𝑏

= 13

0𝑎𝑎 − 𝑎𝑎2 𝑰𝑰𝑏𝑏𝑎𝑎2 − 𝑎𝑎2 𝑰𝑰𝑏𝑏

(11)

So, the first two sequence-domain fault conditions are

𝑰𝑰𝑎 = 0 (12)

𝑰𝑰2 = −𝑰𝑰1 (13)

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To derive the remaining sequence-domain fault condition, rearrange (10) and transform to the sequence domain

𝑽𝑽𝑏𝑏𝑎𝑎 − 𝑽𝑽𝑎𝑎𝑎𝑎 = 𝑰𝑰𝑏𝑏𝑍𝑍𝐹𝐹

𝑽𝑽𝑎 + 𝑎𝑎2𝑽𝑽1 + 𝑎𝑎𝑽𝑽2 − 𝑽𝑽𝑎 + 𝑎𝑎𝑽𝑽1 + 𝑎𝑎2𝑽𝑽2

= 𝑰𝑰𝑎 + 𝑎𝑎2𝑰𝑰1 + 𝑎𝑎𝑰𝑰2 𝑍𝑍𝐹𝐹

𝑎𝑎2𝑽𝑽1 + 𝑎𝑎𝑽𝑽2 − 𝑎𝑎𝑽𝑽1 − 𝑎𝑎2𝑽𝑽2 = 𝑎𝑎2 − 𝑎𝑎 𝑰𝑰1𝑍𝑍𝐹𝐹

𝑎𝑎2 − 𝑎𝑎 𝑽𝑽1 − 𝑎𝑎2 − 𝑎𝑎 𝑽𝑽2 = 𝑎𝑎2 − 𝑎𝑎 𝑰𝑰1𝑍𝑍𝐹𝐹

The last sequence-domain fault condition is

𝑽𝑽1 − 𝑽𝑽2 = 𝑰𝑰1𝑍𝑍𝐹𝐹 (14)

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Sequence-domain fault conditions𝑰𝑰𝑎 = 0 (12)𝑰𝑰2 = −𝑰𝑰1 (13)𝑽𝑽1 − 𝑽𝑽2 = 𝑰𝑰1𝑍𝑍𝐹𝐹 (14)

These can be satisfied by: Leaving the zero-sequence network open Connecting the terminals of the positive- and negative-sequence

networks together through 𝑍𝑍𝐹𝐹

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The fault current is the phase 𝑏𝑏 current, which is given by

𝑰𝑰𝐹𝐹 = 𝑰𝑰𝑏𝑏 = 𝑰𝑰𝑎 + 𝑎𝑎2𝑰𝑰1 + 𝑎𝑎𝑰𝑰2

𝑰𝑰𝐹𝐹 = 𝑎𝑎2𝑰𝑰1 − 𝑎𝑎𝑰𝑰1

𝑰𝑰𝐹𝐹 = −𝑗𝑗 3 𝑰𝑰1 = −𝑗𝑗 3 𝑽𝑽𝐹𝐹𝑍𝑍1+𝑍𝑍2+𝑍𝑍𝐹𝐹

𝑰𝑰𝐹𝐹 = 3 𝑽𝑽𝐹𝐹∠−9𝑎°𝑍𝑍1+𝑍𝑍2+𝑍𝑍𝐹𝐹

(15)

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LL Fault - Example

Now consider the same system with a bolted line-to-line fault at bus 1

The sequence network:

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LL Fault - Example

𝑰𝑰1 = 1∠3𝑎°𝑗𝑗𝑎.363

= 2.75∠ − 60°

The subtransient fault current is given by (15) as

𝑰𝑰𝐹𝐹 = 3 2.75∠ − 60° ∠ − 90°

𝑰𝑰𝐹𝐹 = 4.76∠ − 150° 𝑝𝑝.𝑢𝑢.

Using the previously-determined current base, we can convert the fault current to kA

𝑰𝑰𝐹𝐹 = 𝐼𝐼𝑏𝑏1 ⋅ 4.76∠ − 150°

𝑰𝑰𝐹𝐹 = 4.76∠ − 150° 376.5𝑀𝑀

𝑰𝑰𝐹𝐹 = 1.79∠ − 150° 𝑘𝑘𝑀𝑀

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Unsymmetrical Fault Analysis – DLG Fault

Now consider a double line-to-ground fault Assume phases 𝑏𝑏 and 𝑐𝑐 are shorted to

ground through 𝑍𝑍𝐹𝐹 Phase-domain fault conditions:

𝑰𝑰𝑎𝑎 = 0 (16)

𝑰𝑰𝑏𝑏 + 𝑰𝑰𝑎𝑎 = 𝑽𝑽𝑏𝑏𝑎𝑎𝑍𝑍𝐹𝐹

= 𝑽𝑽𝑐𝑐𝑎𝑎𝑍𝑍𝐹𝐹

(17)

It can be shown that (16) and (17) transform to the following sequence-domain fault conditions (analysis skipped here)

𝑰𝑰𝑎 + 𝑰𝑰1 + 𝑰𝑰2 = 0 (18)

𝑽𝑽1 = 𝑽𝑽2 (19)

𝑰𝑰𝑎 = 13𝑍𝑍𝐹𝐹

𝑽𝑽𝑎 − 𝑽𝑽1 (20)

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Sequence-domain fault conditions𝑰𝑰𝑎 + 𝑰𝑰1 + 𝑰𝑰2 = 0 (18)𝑽𝑽1 = 𝑽𝑽2 (19)

𝑰𝑰𝑎 = 13𝑍𝑍𝐹𝐹

𝑽𝑽𝑎 − 𝑽𝑽1 (20)

To satisfy these fault conditions Connect the positive- and negative-sequence networks together directly Connect the zero- and positive-sequence networks together through 𝑍𝑍𝐹𝐹

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The fault current is the sum of the phase 𝑏𝑏 and phase 𝑐𝑐currents, as given by (17) In the sequence domain the fault current is

𝑰𝑰𝐹𝐹 = 𝑰𝑰𝑏𝑏 + 𝑰𝑰𝑎𝑎 = 3𝑰𝑰𝑎

𝑰𝑰𝐹𝐹 = 3𝑰𝑰𝑎 (21)

𝐼𝐼𝑎 can be determined by a simple analysis (e.g. nodal) of the interconnected sequence networks

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DLG Fault - Example

Now determine the subtransient fault current for a bolted double line-to-ground fault at bus 1

The sequence network:

Here, because 𝑍𝑍𝐹𝐹 = 0, 𝑽𝑽𝑎 = 𝑽𝑽1 = 𝑽𝑽2

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DLG Fault - Example

To find 𝑰𝑰𝐹𝐹, we must determine 𝑰𝑰𝑎 We can first find 𝑉𝑉𝑎 by applying voltage division

𝑽𝑽𝑎 = 𝑽𝑽𝐹𝐹𝑍𝑍2||𝑍𝑍𝑎

𝑍𝑍1 + 𝑍𝑍2||𝑍𝑍𝑎

𝑽𝑽𝑎 = 1.0∠30°𝑗𝑗0.194||𝑗𝑗0.124

𝑗𝑗0.169 + 𝑗𝑗0.194||𝑗𝑗0.124

𝑽𝑽𝑎 = 0.309∠30°

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DLG Fault - Example

Next, calculate 𝑰𝑰𝑎

𝑰𝑰𝑎 =−𝑽𝑽𝑎𝑍𝑍𝑎

=−0.309∠30°

𝑗𝑗0.124= 2.49∠120° 𝑝𝑝.𝑢𝑢.

The per-unit fault current is𝑰𝑰𝐹𝐹 = 3𝑰𝑰𝑎 = 7.48∠120° 𝑝𝑝.𝑢𝑢.

Using the current base to convert to kA, gives the subtransient DLG fault current

𝑰𝑰𝐹𝐹 = 7.48∠120° 376.5 𝑀𝑀

𝑰𝑰𝐹𝐹 = 2.82∠120° 𝑘𝑘𝑀𝑀

Documents

SECTION 7: FAULT ANALYSIS - web.engr.oregonstate.eduweb.engr.oregonstate.edu/~webbky/ESE470_files/Section 7 Fault Analysis.pdf · subtransient fault current for balanced three -phase