Electrical and Electronics Engineering Institute University of the Philippines Diliman

EE 254 – Surge Protection

Switching Transients A switching transient arises whenever an electric circuit is energized or de-energized. There are two types, depending on the magnitude of the over-currents and/or over-voltages that are produced. These are:

1. Simple Switching Transients

2. Abnormal Switching

Simple Switching Transients Simple Switching Transients result in over-voltages or over-currents that do not exceed 2 per unit. These are the most prevalent type of transient in power systems.

There are two types:

1.  Circuit closing transient – transient initiated by the closing of a switch;

2. Circuit opening transient – transient initiated by the opening of a switch.

Example: Circuit Closing Transient

+

L

i t=0

R

v(t) -

In the circuit shown, the switch is closed at t=0 to simulate the occurrence of a fault. Find the fault current i.

The steady-state response can be shown to be

where

and

The transient response is

The complete response is

At t = 0+, i(0+) = 0. We get

or

Thus, the complete response is

Case 1: If θ = ϕ, no transient is initiated. We get

This steady-state fault current persists until the fault is interrupted by circuit breaker action.

Case 2: If , we get the worst transient.

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Note: The DC offset may cause severe CT saturation and relaying problems.

Example: Circuit Opening Transients

230 kV System

Open

50 MVA 230/69 kV

Z=10%

CB x

3 Phase Fault

A three-phase fault occurs at the 69-kV side of a 50 MVA transformer. Assuming a total capacitance to ground of 10 nF at the 69-kV side, find the transient recovery voltage across the first circuit breaker pole to open. For simplicity, neglect the impedance of the system.

The transformer leakage reactance at the 69-kV side

which gives

The three-phase fault current is

Assuming the circuit breaker pole opens at t = 0, we get the equivalent circuit using current injection

L=25.26 mH C=10 nF

ik

v(t) + -

L

Cwhere

From KCL, we get for t ≥ 0,

or

The natural frequency is given by

The natural frequency is 167 times the system frequency. The differential equation can be approximated by

The transient response is

Let the steady-state response, vss = A, a constant. Substitute in the differential equation

or

The complete response

Initial Conditions: From current injection, we know that

and

Thus,

From KCL, we get at t = 0+,

or

Evaluate the Constants: From the solution and its derivative, we get

and

At = 0+,

or

Also at t = 0+,

or

Finally, we get

Substitution gives

Note: This high-frequency transient voltage is referred to as the transient recovery voltage, which is one rating specified for circuit breakers.

The resistance of the system was neglected to simplify the analysis of the problem. With resistance, the transient recovery voltage will exponentially decay to zero. A plot is shown below.

System Voltage

Fault Current

Transient Recovery Voltage

Abnormal Switching Transients

1. Back-to-Back Switching of Capacitor Banks – a very large inrush current results when an incoming capacitor bank is energized on the same bus where a previously-energized capacitor bank is connected

Abnormal Switching Transients result in over-voltages or over-currents that exceed 2 per unit. They involve the release of energy that has been previously trapped in the system.

2. Double Frequency Transients – over-voltages may appear when capacitor banks at different voltage levels are switched

5. Arcing Ground Fault – a single line-to-ground fault on ungrounded or high-impedance grounded power systems may result in large over-voltages

6. Ferroresonance – a series resonant circuit that involves an iron-core inductance

3. Over-voltages Produced by Re-ignition – the de-energization of a capacitive circuit such as a long transmission line or cable, or a capacitor bank may result in large over-voltages due to the trapped charges in the capacitance of the system.

4. Current Chopping – switching-off a vacuum circuit breaker may cause large over-voltages to appear on connected equipment

Example: Back-to-back Switching of Capacitors

The system inductance Ls is 15 mH. Each capacitor bank is wye-grounded and rated 3 MVA, 34.5 kV. At t=0, the system voltage is at its peak value, and the switch is closed. Assuming the incoming bank is uncharged, find the inrush current.

Bus inductance is 0.5 µH/ft

34.5 kV Bus

Bank 1 Bank 2

Ls

50 ft

i

vk

+

-

Using the voltage-injection method, the inrush current is the sum of two components:

1.  The current in the incoming capacitor if the switch is not closed – this component is zero.

2. The current in the incoming capacitor due the injected voltage source vk, with all sources and initial conditions set to zero.

vk(t) +

-

C2

Lb

i(t) C1

Ls

Ls=15 mH

Lb=25 µH

Equivalent circuit for t ≥ 0

Let us find the capacitance of a wye-connected bank in terms of its MVAR and kV ratings.

XC

XC XC

VLN

+

-

The reactive power in one phase is

Since VAR3p=3VARp and VLL= VLN, we get

or

or, using MVAR and kV

The given capacitor banks are rated 3 MVAR at 34.5 kV. We get

We expect the natural frequency of the transient response to be very high. This means

1. We can neglect Ls.

2. We can assume vk is constant.

Equivalent Circuit:

Lb=25µH

C1=C2=6.686µF

Vs=28.17kV

Vs +

-

Lb

i(t) C1 C2

We get C = 3.343 µF.

Note: ωn is 290 times the power frequency.

Since the incoming capacitor is uncharged, we get

kA

The rated capacitor current is

A, RMS

= 71 A, peak

Note: The inrush current is 145 times the rated current.

Problems Associated with Back-to-Back Switching of Capacitors:

1. Nuisance capacitor fuse blowing

2. Ruptured capacitor tank

3. Severe duty on switching devices

ANSI C37.06-1987 distinguishes between two types of circuit breakers:

1. General Purpose Circuit Breaker

2. Definite Purpose Circuit Breaker

A Current Limiting Inductance is used to reduce the magnitude and frequency of the inrush current.

Example: What value of inductance should be added to each capacitor if the magnitude of the inrush current is to be limited to 25 times the rated value? Equivalent Circuit:

Lb=25µH

C1=C2=6.686µF

Vs=28.17kV

Vs + -

Lb

i(t)

C1 C2

Lc Lc

The peak value of the inrush current

Ipk = 25(71) = 1,775 Amps

The new surge impedance

From the definition of surge impedance, we get

where CT = 3.343 µF and LT = Lb+2Lc. Substitution gives

We get Lc = 410 µH.

Question: What is the effect of the current-limiting inductor on the reactive power delivered by the capacitor?

At 60 Hz, the impedance of the current-limiting inductor is

At the same frequency, the impedance of the capacitor is 396.75 Ω.

Note: The impedance of the inductor is less than 0.045% of the impedance of the capacitor bank.

General Purpose Circuit Breaker Rated Capacitance Switching Current (1),(2)

Rated Max kV rms

Rated ISC kA rms

Rated ICONT kA rms (3)

Overhead Line, A rms

Isolated I A rms (3)

121 20 1.2 50 50 121 40 1.6, 2.0, 3.0 50 50 121 63 2.0, 3.0 50 50 242 40 2.0, 3.0 160 160 242 50 2.0 160 160 242 63 2.0, 3.0 160 160

550 63 3.0 400 400 550 40 2.0, 3.0 400 400

Part of Table 3A, ANSI C37.06-1987

Definite Purpose Circuit Breaker Rated Capacitance Switching Current (2)

Rated Max kV rms

Overhead Line, A rms

Isolated I A rms (3)

121 160 315 315 16 4250

242 200 400 400 20 4250

Part of Table 3A, ANSI C37.06-1987

Current A rms(3)

Peak I kA

Freq. Hertz

Shunt Capacitor Bank or Cable

Inrush Current(4) Back-to-Back

121 160 315 315 16 4250 121 160 315 315 16 4250

242 200 400 400 20 4250 242 200 400 400 20 4250

Notes for Table 3A:

(1) No ratings for back-to-back shunt capacitor bank or cable switching applications are established for general-purpose circuit breakers.

(2) The capacitance switching current ratings are the highest values that the circuit breaker shall be required to switch at any voltage up to the rated maximum voltage.

(3) When applied to shunt capacitor banks, the current rating shall be selected to include the effects of tolerance in capacitance, system and capacitor bank grounding, and additional current magnitude and heating due to harmonics.

(4) The rated transient inrush current peak is the highest magnitude that the circuit breaker shall be required to to close at any voltage up to the rated maximum voltage, and shall be as determined by the system and unmodified by the circuit breaker. The rated transient inrush current frequency is the natural frequency that the circuit breaker is required to close at 100% of its rated back-to-back shunt capacitor bank or cable switching current.

For application at less than 100 % of rating, the product of the inrush current peak and natural frequency shall not exceed the product of the rated transient current peak and the rated transient inrush current frequency.

ANSI/IEEE C37.012-1979: IEEE Application Guide for Capacitance Current Switching for AC High-Voltage Circuit Breakers Rated on a Symmetrical Current Basis

4.7.1 Capacitor Bank Current. The actual capacitance current should include the following effects:

(1) Voltage Factor. This factor can be as large as 1.1 since capacitors can be operated continuously up to 10% above rated voltage.

(2) Capacitor Tolerance. A multiplier in the range of 1.05 to 1.15 should be used to take into account the allowable capacitor tolerance.

(3) Harmonic Component. Capacitor banks provide a low-impedance path for the flow of harmonic currents. When capacitor banks are ungrounded, no path is provided for the zero-sequence harmonics (3rd, 6th, 9th, etc.). A multiplier of 1.1 is generally used for grounded neutral bank and 1.05 for ungrounded neutral.

Note: In the absence of specific information on multipliers for the above factors, it will usually be conservative to use a total multiplier of 1.25 times the nominal capacitor current at rated capacitor voltage for ungrounded neutral operation and 1.35 times the nominal current for grounded neutral operation.

4.7.2 Isolated Capacitor Bank. A capacitor bank is considered isolated if the maximum rate of change, with respect to time, of transient inrush current on energizing an uncharged capacitor bank does not exceed the maximum rate of change of the symmetrical interrupting current at the voltage at which the circuit breaker is supplied. The limiting value is

Ir = rated rms short-circuit current, Amps

ωs = system frequency, Hz

4.7.3 Back-to-Back Capacitor Bank. A capacitor bank is considered switched back-to-back if the highest rate of change of inrush current on closing exceeds that specified for isolated capacitor banks.

4.7.4. Inrush Current. The energization of a capacitor bank by the closing of a circuit breaker will result in a transient inrush current. The magnitude and frequency of this inrush current is a function of the following: applied voltage (point on the voltage wave at closing), capacitance of the circuit, inductance of the circuit (amount and location), any charge on the capacitor bank at the instant of closing, and damping due to the resistors in the circuit.

4.7.4. Inrush Current, cont. When capacitor banks are switched back to back, transient currents of prospective high magnitude and with a high natural frequency may flow between the banks on closing of the switching device or in the event of a restrike on opening. This oscillatory current is limited only by the impedance of the capacitor bank and the circuit between the energized bank or banks and the switched bank. This transient current usually decays to zero in a fraction of a cycle of the system frequency. In the case of back-to-back switching, the component supplied by the source is at a lower frequency and so small it may be neglected.

Inrush Current and Frequency for Switching Capacitor Banks

Condition Quantity When Using Current

Energizing an isolated bank

Table 1 of ANSI/IEEE C37.012-1979

Energizing a bank with another on the same bus

imax pk, A

f, Hz

imax pk, A

f, kHz

Inrush Current and Frequency for Switching Capacitor Banks, cont.

Condition Quantity When Using Current

Energizing a bank with an equal bank on the same bus

imax pk, A

f, kHz

fs = system frequency, Hz

Leq = total equivalent inductance per phase between capacitor banks, µH Isc = symmetrical rms short-circuit current, Amps

I1,I2 = currents of bank being switched and of bank already energized, respectively. Capacitor bank being switched is assumed uncharged, with closing at a voltage crest of the source voltage. The current should include the effect of operating the bank at voltages above nominal and the effect of capacitor tolerance. In the absence of specific information, 1.15 times nominal capacitor current is recommended.

imax pk = A peak value calculated without damping. In practical circuits, it will be about 90 percent of this value.

VLL = rated maximum voltage, kilovolts

Example: Energizing an Isolated Capacitor Bank

Bus The switch is closed when the system voltage is at its peak value.

Equivalent circuit using voltage injection vk

+ -

C

Lc i(t)

Ls Ls = system inductance Lc = bank inductance

If Vp is the rms line-to-neutral voltage, we get

Solving for the peak value of the current, we get

or

Isc = system short-circuit current, A rms

Ic = capacitor current, A rms

where

Solving for the natural frequency, we get

where fs = system frequency in Hz, and fn is the natural frequency, also in Hz.

or

Example: Back-to Back Switching of Equally-Sized Capacitor Banks

The switch is closed when the system voltage is at its peak value.

Bus

C C

Equivalent circuit using voltage injection

C

Lc

vk + -

i(t)

Ls

C

Lc

Ls = system inductance Lc = bank inductance

Note: since Ls>>Lc, we can neglect Ls at the transient frequency.

Equivalent circuit vk

+ -

i(t) C

Lc

C

Lc LT = 2Lc

CT = C/2

Solving for the peak value of the current, we get

Vp is the rms line-to-neutral voltage in volts. Let us use line-to-line kV. We get

In this expression, LT is expressed in Henry. Let us use the inductance in µH.

or

Solving for the natural frequency, we get

or

Simplifying, we get

in Hz

4.7.4.1 Method for Calculating Transient Inrush Current. A typical circuit for back-to-back switching is shown.

Lbus C1 LC1 L1 C2 LC2 L2

Bank 2 Bank 1

SW1 SW2

Source

Notes:

(1)  The inductance of the bus can be calculated similar to a transmission line.

(2) The inductance within the capacitor bank itself is not easy to obtain, but in general it is of the order of 10 µH for banks above 46 kV, and 5 µH for banks below 46 kV.

(3) Inherent resistance of the circuit causes rapid decay of the transient current so that the first peak actually may only reach 90%

to 95% of the maximum value calculated.

Typical Values of Inductance Between Capacitor Banks

≤15 0.214 10 – 20 38 0.238 15 – 30 48.3 0.256 20 – 40 72.5 0.256 25 – 50 121 0.261 35 – 70 145 0.261 40 – 80 169 0.268 60 – 120 242 0.285 85 – 170

Rated Max. Voltage, kV

Inductance/ phase, µH/ft

Typ. Inductance Bet. Banks,* µH

Table 2 of ANSI/IEEE C37.012-1979

4.7.4.2 Considerations for Transient Inrush Current. Consider the 115 kV system shown.

Bank 1 12 MVAR

Bank 2 12 MVAR

Bank 3 12 MVAR

CB1

CB2

Ls

L1’

L2’

L3’

144 ft

Lbus

I2

I1 35 ft

35 ft

35 ft

25 ft

25 ft

Figure 5 of ANSI/IEEE C37.012-1979

The capacitor banks have a nominal rating of 12 MVAR. The nominal current per bank is 60 Amps. The maximum voltage is 121 kV. The system inductance is 10 mH or a reactance of 3.77 Ω. The short-circuit current at the bus is 18,600 Amps at 121 kV.

Note: In Figure 5, Ls is the system inductance, while L1’, L2’, and L3’ are the inductances between the circuit breakers and capacitor banks, including the inductances of the capacitor banks, and Lbus is the inductance of the bus between the switching devices.

Rating of Definite Purpose Circuit Breaker:

Rated maximum voltage = 121 kV Rated Continuous Current = 1,600 Amps Rated Short Circuit Current = 40 kAmps Rated Isolated and Back-to-Back Capacitance Switching Current = 160 Amps Rated Inrush Current = 10 kAmps, peak Rated Inrush Frequency = 5.3 kHz

Multipliers for Capacitor Current: Voltage Factor = 121/115 = 1.05 Capacitor Tolerance = 1.1 Harmonic Component = 1.1

Total Multiplier for Isolated and Back-to-Back Capacitor Current:

1.05 x 1.1 x 1.1 = 1.27

For Capacitor 1: I1 = 1.27 x 60 = 76 Amps For Capacitor 2: I2 = 76 Amps or 152 Amps

Bus Inductances:

The bus is an extruded aluminum tube with 8-foot spacing between phases. The section between breakers is 3 inch iron pipe size (IPS) while the section between breakers and capacitors is 2 inch IPS. L (3 in) = 0.261 µH/ft

L (2 in) = 0.285 µH/ft

Capacitor Bank Inductance:

Lbank = 10 µH

Calculate Inductances:

Lbus = 0.261 x 144 = 37.6 µH L1’ = 0.285 x 35 + 10 = 20 µH L2’ = L3’ = 0.285 x (35+25) + 10 = 27.1 µH

Multiplier for Capacitor Current in Table 1:

1.05 x 1.1 = 1.15

For Capacitor 1: I1 = 1.15 x 60 = 69 Amps For Capacitor 2: I2 = 69 Amps or 138 Amps

Case 1: Energization of Capacitor Bank 1 (isolated switching)

Calculate the rate-of-rise of the inrush current

Compare with the rate-of-rise of the rated short circuit current of 40 kAmps at 60 Hz.

Since the rate-of-rise of the inrush current is less than the rate-of-rise of the rated short-circuit current, there is no problem wit isolated switching.

Note: Given the sinusoidal function

Case 2: Energization of Capacitor Bank 1 with Bank 2 Energized on the Bus

The equivalent inductance between the banks is L1’ +Lbus +L2’ = 84.7 µH.

Note: The circuit breaker ratings of 10 kA peak and 5.3 kHz are exceeded.

Current-Limiting Inductor: Let a 0.6 mH inductor be added to each capacitor bank. The equivalent inductance between the banks is 84.7 µH + 600 µH.

Case 3: Energization of Bank 1 with Banks 2 and 3 Energized on the Bus

The equivalent inductance between the banks is L1’ +Lbus +0.5L2’ = 71.2 µH.

Note: Again, the circuit breaker ratings of 10 kA peak and 5.3 kHz are exceeded.

ANSI/IEEE C37.99-1980: IEEE Guide for the Protection of Shunt Capacitor Banks

The magnitude and consequent effects of inrush current to a switched capacitor bank may be greatly reduced by the use of series reactors or a capacitor switching device furnished with pre-insertion resistors.

Example: Double Frequency Transient in Capacitor Bank Energization

115 kV

Industrial Plant

10 MVA 115/34.5 kV

Z=6%

Bank 1 Bank 2

10 MVAR 115 kV

1 MVAR 34.5 kV

The three-phase short circuit MVA is 1,420 at the 115-kV bus. At t=0, the system voltage is at its peak value and the switch is closed. Find the voltage that appears at the terminals of bank 2. Assume bank 1 is initially uncharged.

Equivalent Circuit: For the system, we get

kA, RMS

For the transformer,

For the 115-kV capacitor bank

For the 34.5-kV capacitor bank

, referred to the 34.5-kV side

the 115-kV side , referred to

Prior to switching, the equivalent circuit is

Xs = 9.3 Ω

Xt = 79.4 Ω

Xt

+ i vC2

-

+ vs

- X’C2

Xs X’C2 = 13,225.4 Ω

vs = 66.4 kV, rms

fs = 60 Hz

Since X’C2 >> (Xs+Xt), then . If the switch is not closed, the voltage across C2 is equal to the system voltage.

Using the voltage injection method, we get the equivalent circuit for t ≥ 0.

Ls = 24.7 mH

C1 = 2 µF vk

+

- C1

Lt

C’2

Ls

v’C2

+

-

Lt = 210.5 mH

C’2 = 0.2 µF

Note: We expect the transient frequency to be high, thus vk can be assumed constant.

It can be shown that the voltage across bank C2 due to the injected voltage source is

where

The total voltage across bank C2 is the sum of two components:

1.  The voltage across C2 if the switch is not closed. This is approximately constant and equal to the peak line-to-neutral voltage.

2. The voltage across C2 due to the injected voltage source vk, with all sources and initial conditions set to zero.

We get

Substitution gives

where Vpk = 28.17 kV. A plot is shown below.

vC2

Vpk

2Vpk

3Vpk

4Vpk

-Vpk

-2Vpk

0.8 1.6 2.4 t, msec

. . . . . . . . . . . . .

. . . . . .

.

. . . . . .

. . . . . . . . . . . .

. . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Note: The voltage reaches up to 4 p.u.

Further analysis shows that:

1.  The highest voltage appears when ω1 is of the same order as ω2.

2. The over-voltage increases as the surge impedance of the second loop becomes large compared to the surge impedance of the first loop.

The problem can be avoided by:

1.  Switching the high-voltage bank in sections

2. Using a pre-insertion resistor in the high-voltage capacitor bank

3. Synchronous switching

Example: Over-voltages Produced by Re-ignition

The system inductance Ls is 1 mH. The capacitor bank is wye-grounded and rated 5 MVA, 34.5 kV. The switch is opened to de-energize the capacitor. Sketch the voltages and current before and after the switch is opened.

34.5 kV Bus

Ls

+ i vC

-

Ls = 1 mH

Equivalent circuit for t < 0

Vs = 34.5 kV, rms line-to-line

fs = 60 Hz

Ls

+ i vC

-

+ vs

- C

The reactance of the system is negligible compared to the capacitor’s reactance. This means that

(2) the current leads the voltage by 90o

(1)

Note: The current is not immediately interrupted when the circuit breaker contact opens. Initially, an arc jumps the gap to sustain the current. The arc is extinguished when the current crosses its normal zero value.

A, rms

A sketch of the current and voltages is shown below.

vs, vC

i 0

vs

vC

A B

Assume that the arc is extinguished at point A. The peak value of the system voltage is trapped in the capacitor. One half cycle (8.33 ms) later at point B, the voltage across the open breaker contact reaches twice the peak line-to-neutral value.

If the dielectric strength of the open contact is not sufficient to overcome the magnitude of the voltage, an arc will jump the gap and the circuit closes anew. This phenomenon is called a restrike.

Equivalent Circuit: Assuming there is a restrike, we get the circuit below. For convenience we reset time so that the restrike occurs at t=0.

Ls = 1 mH Ls

+ i vC

-

+ vs

- C

C = 11.14 µF

vC(0+) =-28.17 kV

vs = 28.17 cos ωst kV

The natural frequency of the transient response is

The natural frequency is 25 times the power frequency. We get the circuit below.

Ls = 1 mH C = 11.14 µF

vC(0+) =-28.17 kV

E = 28.17 kV

Ls

+ i vC

-

+

E -

C

Note: This is a series LC circuit with the voltage source approximated by a constant.

The capacitor voltage and current are described by

where

Substitution gives

Plots of vc and i are shown below.

-28.17

i

vC

28.17

56.34

84.51 kV

C D

The rated capacitor current is

A, RMS

Since the transient current is 50 times the rated current, the circuit breakers will trip to interrupt the current.

(1)  If the arc is extinguished at point D, the capacitor voltage will remain at –28.17 kV.

(2) If the arc is extinguished at point C, the capacitor voltage will be at 84.51 kV. This is 3 times the peak line-to-neutral voltage.

Assume that the current is interrupted at point C. One half cycle (8.33 ms) later, the voltage across the open breaker contact reaches 4 times the peak line-to-neutral voltage. The possibility of a restrike is even bigger.

ANSI/IEEE C37.99-1980: IEEE Guide for the Protection of Shunt Capacitor Banks

5.2 Transient Overvoltage. An important considera-tion for application of circuit breakers for capacitor switching is the transient overvoltage which may be generated by restrike during the opening operation. At current zero, the capacitor is left charged to nearly full peak line voltage… The capacitance current arc is usually interrupted at the first current zero after the circuit breaker contacts open. One half cycle later, the voltage across the breaker contacts will reach nearly twice the peak line voltage and a breakdown may occur.

The possibility of restrike may be eliminated by using a circuit breaker with pre-insertion resistor.

R

C1 C2

C

The circuit breaker has two contacts. During the opening operation, contact C1 opens first, putting the resistor in series with the capacitor. Then contact C2 is opened. The sequence is reversed during the closing operation. The resistor brings the current and capacitor voltage closer in phase and also acts as a voltage divider.

Example: Current Chopping

The transformer has a 2% magnetizing current. At t=0, the current is at its peak value and the circuit breaker is opened to de-energize the transformer. Assume that the total capacitance at the 34.5 kV side is 6,000 pF. Find the voltage across the transformer for t ≥ 0.

34.5 kV Bus No Load

Vacuum CB

5 MVA 34.5-4.16 kV

The rated current of the transformer

A, RMS

The magnetizing current and inductance

A, RMS

= 2.37 A, pk

The capacitive reactance

Note: When the magnetizing current is chopped at its peak value, the system voltage is zero.

Equivalent Circuit: Using current injection, we get the equivalent circuit for t ≥ 0.

CT = 6 nF

ik

v(t) +

-

Ls

CT Lm

Lm = 31.57 H

ik = 2.37 cos ωst A

ωs = 60 Hz

Note: The natural frequency is 6 times the power frequency. To simplify the analysis of the problem, let us assume ik is constant. We get for t ≥ 0

The natural frequency of the transient response is

or

Solving for v(t), we get

whose derivative is

From current injection, we know that vC(0+)=0 and iL(0+)=0. Thus, we get at t=0+,

and

or

At t=0+, we get

or

Thus, we get

or

The total voltage across the transformer is the sum of two components:

2. The voltage across the transformer due to the injected current source ik.

1.  The voltage across the transformer if the switch is not opened. This is given by

The second component is the dominant term.

Note: The voltage across the transformer is 6 times the normal value.

The over-voltage can be reduced by connecting a surge capacitor across the transformer. Suppose we want to limit the over-voltage to twice the rated line-to-neutral voltage.

CT Lm Cs

At 60 Hz,

Arcing Ground Fault Arcing ground fault is common in high-impedance grounded systems where the single line-to-ground fault current is small.

Ls

CL

CL

CL

Ls

Ls Zg

x Fault

+

-

Xs

XC Xg

x Vs i

Equivalent Circuit: Assuming the grounding impedance is a reactor, we get the equivalent circuit of the faulted phase.

Assumption:

Xg>>Xs or XL

This means:

1.  The fault current i is very small and the arc may extinguish by itself;

2. The fault current i lags the system voltage vs by 90o.

vs i

0

A B

Assume that the arc extinguishes by itself at point A. At this time,

A plot of the fault current and system voltage is shown.

1.  The voltage across the line capacitance XC is nearly zero;

2. The system voltage vs is at peak value.

+

-

Ls

CL Lg

Vpk +

- vC

Equivalent Circuit:

vC(0)=0

A plot of the capacitor voltage is shown, where vC

Vpk

C 2Vpk

vC

D

2Vpk

-2Vpk

The arcing ground fault is usually caused by a puncture in the line insulation. When the voltage vC is at peak value, the fault may re-ignite. The equivalent circuit is shown. This time the capacitor

CL x i

voltage oscillates between 2Vpk and –2Vpk.

The arc may still extinguish since the fault current is small. If this happens at point D, a voltage –2Vpk will be trapped in the capacitor. This time the capacitor voltage will oscillate between –2Vpk and +4Vpk.

vC

2Vpk

4Vpk

-2Vpk

Vpk

Note: The magnitude of the capacitor voltage is progressively increasing. This phenomenon is called “ringing”.

Ferroresonance

1.  Ferroresonance is a no-load problem.

2. It involves a series RLC circuit where L is an iron-core inductance.

3. The capacitance is due to the presence of shunt capacitor banks, cable circuits, overhead lines and the internal capacitance of equipment.

4. It is prevalent where single-pole switching is used.

5. The problem is aggravated by an ungrounded transformer primary.

When ferroresonance is present in a distribution system, it usually causes one or more of the following abnormalities which are easily measured or observed:

1.  High voltage phase-to-phase, phase-to-ground, or both with peak voltages which may be five or more times the normal system voltage;

2. Extremely jagged and irregular voltage and current wave shapes; and

3. Excessively loud noise in the transformer due primarily to magnetostriction at high flux densities.

Source: ANSI/IEEE C57.105-1978

Effectively Grounded

Source

A typical ferroresonant circuit is shown below.

Shielded Cable

Unloaded Transformer

Figure 13 of ANSI/IEEE C57.105-1978

CL

CL

CL

Single-Pole Switches Xm

Xm

Xm

A

B

C

“When the single-pole switch for phase A is closed, two phases of the transformer are energized by a path through the cable capacitances from phases B and C to ground. At the instant the switch in phase A is closed, the capacitance to ground on phases B and C appears as a short circuit, and the transformer windings of legs A-B and A-C start to draw normal inrush current or exciting current.”

“The transformer iron during the first cycle of applied voltage may saturate due to closing at or near voltage zero, or due to residual flux in the transformer core or both. Saturation results in a large current pulse through the transformer windings and capacitances of phases B and C.”

“Next the transformer iron drops out of saturation leaving a substantial trapped charge (voltage) on the cable capacitance. In subsequent cycles the transformer iron may go into saturation in the opposite direction, thereby changing the polarity of the trapped voltage on the capacitance.”

“If the transformer continues to go into and out of saturation in either a cyclical or random fashion, high sustained overvoltages will occur phase-to-phase and phase-to-ground. These sustained overvoltages can cause overexcitation of the transformer, surge arrester failure, and even failure of major insulation in the transformer or system.”

The path of the current, after one pole is closed, is shown.

CL

CL

CL

Lm Lm

Lm

Va

+

-

Va

+

-

The Equivalent Ferroresonant Circuit

Connections Susceptible to Ferroresonance

Single-Pole Switch

Open Wire or Cable Circuit

Transformer Bank

No Load H X

Part of Table 2 of ANSI/IEEE C57.105-1978

H X

H X

H X

H X

H X

H X

“For the connections shown, ferroresonance does not usually occur if the primary voltage is 15 kV or lower and the lateral circuit is overhead line (open wire) provided the length of line between the single-pole devices and transformer bank does not exceed ordinary lengths.”

“In 25 kV and particularly in 35 kV multi-grounded neutral overhead systems, ferroresonance is most likely with the connections shown in Table 2.”

“Ferroresonance is also likely if the primary voltage is 12 kV or higher (effectively-grounded system) and the lateral circuit is single-conductor shielded cable or three-conductor shielded cable.”

The following measures can be applied to minimize the probability of ferroresonance with ungrounded primary connection:

1.  The use of properly adjusted 3-pole switches at the source-end of the cable to prevent single-phasing. However, this is a costly solution.

2. A sufficient load at the secondary of the transformer. In general, a 10% resistive load (based on the transformer rating) will prevent ferroresonance.

3. Switching at the transformer

Source: ANSI/IEEE C57.105-1978

4. Resistance grounding of wye-connected primary

Single-Pole Switch

Open Wire or Cable Circuit

Transformer Bank

No Load H X

Part of Table 2 of ANSI/IEEE C57.105-1978

Connections Which Prevent or are Least Susceptible to Ferroresonance

H X

H X

H X

H X

•  The Lecture Notes are prepared by the Faculty of Electrical and Electronics Engineering Institute initiated by Prof. Artemio P. Magabo

•  With contributions from Dr. Allan C. Nerves and Prof. Rowaldo D. del Mundo