Electrical and Electronics Engineering Institute University of the Philippines Diliman

EE 254 – Surge Protection

Electrical Transients An electrical transient is an outward manifestation of an abrupt change in system configuration, such as when a switch opens or closes in an electric circuit.

1. The transient period is very short;

2. The frequency of the transient is very high;

3. The transient currents and/or voltages are very large; and

3. Power system components will be subjected to excessive stresses if the transient effects are not mitigated.

Definitions A surge is a transient wave of current, potential or power in an electric circuit.

A lightning surge is a transient electrical disturbance caused by lightning.

A switching surge is a heavily damped transient electrical disturbance associated with switching.

An impulse is a surge of unidirectional polarity.

The Standard Lightning Impulse is a full impulse with a front of 1.2 µs and a tail of 50 µs.

The Standard Switching Impulse is a full impulse with a front of 250 µs and a tail of 2500 µs.

Assumptions on Circuit Elements

1.  All circuit elements are assumed to be linear.

R

Resistance

vR

iR

L

Inductance

λ

iL

C

Capacitance

q

vC

2. All circuit elements are assumed to be invariant with time.

3. Whenever possible, all circuit elements will be assumed to be lumped.

Fundamental Circuit Principles

3. Energy conservation must be preserved at all times.

1.  The current through an inductor cannot change instantaneously.

L

+ vL

iL

-

2. The voltage across a capacitor cannot change instantaneously.

C

+ vC

iC

-

Power System Representation At steady state, only power-frequency components are present. In general, capacitances may be neglected. To simplify the analysis of the problem, resistance may be initially omitted, but its effect is included in the interpretation of the final results.

Power Transformer:

At steady state, a loaded transformer is modeled by its leakage reactance while an unloaded transformer is represented by its magnetizing reactance. Under transient condition, transformer capacitance must be included. Typical values of capacitance are given in ANSI/IEEE C37.011-1979.

Example: Given a 25 MVA 115Y–34.5Y kV three-phase transformer with Z=10%. Assume a total capacitance to ground of 2 nF at the 115 kV side. Find the equivalent circuit for loaded condition at 60 Hz and 1 MHz.

At 60 Hz, we get

Equivalent Circuit at the 115 kV side XC

XL

i

Equivalent Circuit at the 115 kV side

XL

i

At 1 MHz, we get

Equivalent Circuit at the 115 kV side XC i

Synchronous Generator:

Immediately after a fault, the machine is modeled by its sub-transient reactance. A few cycles later, the transient reactance is used. At steady state, the machine is represented by its synchronous reactance. For high-frequency transients, the effect of winding capacitance must be included.

For short and medium-length lines, the lumped-parameter model is used. Capacitance is neglected for short lines. For long lines, the line is modeled using distributed parameters.

Transmission Line:

Generator Armature Capacitance

Complete table can be found in ANSI/IEEE C37.011-1979

Generator MVA Winding Capacitance

to Ground, µF

Steam Turbine, 2 poles Conventional Cooling

15 to 30 0.17 – 0.36 30 to 50 0.22 – 0.44 50 to 70 0.27 – 0.52 70 to 225 0.34 – 0.87

Hydro Driven

360-720 rpm, 10 to 30 0.26 – 0.53 85-225 rpm, 25 to 100 0.90 – 1.64

Phase Bus Capacitance

Complete table can be found in ANSI/IEEE C37.011-1979

Amp 15 kV, pF/ft 23 kV, pF/ft 15 kV, pF/ft 1,200 8.9-14.3 8.0-12.4 10 2,000 10.2-14.3 9.0-12.4 10.0-10.2 3,000 10.2-14.3 9.0-12.4 10.0-10.2

4,000 14.0-14.3 12.4-13.5 10.0-12.6 5,000 14.0-19.0 12.7-15.8 12.5-14.9 6,000 14.0-19.0 13.5-15.8 15.0-17.1 7,000 17.3-22.6 14.4-17.6 17.1 8,000 21.7 17.6

10,000 21.7 18.1 12,000 23.7 20.5

Isolated Phase Bus Segregated Phase Bus

Outdoor Bushing Capacitance

Complete table can be found in ANSI/IEEE C37.011-1979

kV Rating Amp Rating Range in µF 15 600 160 – 180 1,200 190 – 220 34.5 1,200 170 – 390 2,000 240 – 360

69 1,200 160 – 290 2,000 210 – 320 115 1,200 250 – 420 1,600 250 – 430 138 1,200 250 – 420 1,600 250 – 460

Principle of Superposition In a linear electric circuit containing N independent sources, the current (or voltage) in any branch is equal to the algebraic sum of N components, each of which is due to one independent source acting alone.

Note: Reducing an independent source to zero:

1. For a voltage source, remove the source and replace with a short circuit;

2. For a current source, remove the source and replace with an open circuit.

Thevenin’s Theorem

Consider a linear circuit which can be represented by two networks, A and B. Suppose we want to simplify network A.

Linear Network

A y

x

Linear Network

B

Thevenin’s Theorem states that Network A can be replaced by a single voltage source Vth which is connected in series with a single impedance Zth.

y

x

Linear Network

B

Zth

Vth

+

-

Vth= open-circuit voltage from terminal x to terminal y, with network B removed

Zth= the equivalent impedance from terminal x to terminal y, looking into network A, with all independent sources reduced to zero.

Voltage Injection Method In a linear circuit, the closing of a switch may be simulated by injecting a voltage source vk across the terminals of the switch, which is equal to the negative of the voltage that would appear across the switch if it had remained open. The voltage (or current) at any point in the network is equal to the sum of two components:

1.  The voltage (or current) that would appear at the point if the switch had remained open; and

2. The voltage (or current) at the point due to the injected voltage source vk, with all sources and initial conditions set to zero.

Equivalent network after closing

RLC, Sources,

IC’s

vk

vk - +

+ -

Network before closing

RLC, Sources,

IC’s vk

+

-

Network after closing

v=0

RLC, Sources,

IC’s

+

-

Equivalent network after closing

RLC, Sources,

IC’s vk

+ -

RLC only

vk - +

+

Equivalent network after closing

RLC only

vk - +

+ RLC,

Sources, IC’s

vk

+

-

Example: Find all currents and voltages.

We get

4Ω

24v

+

-

I1

8Ω

V1 V2

+

-

+ -

I2 4Ω

24v

+

-

I1

8Ω

V1 + -

I2

16v

+

-

We get

Example: Back-to-back Switching of Capacitors

At t=0, the switch is closed to energize bank 2. Assume that at the instant of switching, the system voltage is at its peak value and that the incoming capacitor is uncharged. Find the equivalent circuit for t ≥ 0 using the voltage injection method.

Bus inductance between the two banks is 10 µH

34.5 kV Bus

Bank 1

5 µF

Bank 2

5 µF vk

+

-

The peak value of the system voltage is

If the switch does not close, the voltage across the switch for t ≥ 0 is

Note: i(t) is the current in the incoming bank.

Equivalent circuit for t ≥ 0

vk(t) +

-

5 µF 5 µF

10 µH

i(t)

Current Injection Method In a linear circuit, the opening of a switch may be simulated by injecting a current source ik through the terminals of the switch, which is equal to the negative of the current that would flow through the switch if it had remained closed. The voltage (or current) at any point in the network is equal to the sum of two components:

1.  The voltage (or current) that would appear at the point if the switch had remained closed; and

2. The voltage (or current) at the point due to the injected current source ik, with all sources and initial conditions set to zero.

Network before opening

RLC, Sources,

IC’s

ik

Network after opening

RLC, Sources,

IC’s i=0

Equivalent network after opening

RLC, Sources,

IC’s ik ik

i=0

Equivalent network after opening

+ RLC,

Sources, IC’s

ik RLC only

ik

Equivalent network after opening

RLC only

ik

RLC, Sources,

IC’s

ik +

Example: Current Chopping

At t=0, the vacuum circuit breaker is opened to de-energize an unloaded transformer. Assume that at the instant of switching, the magnetizing current is at its peak value of 4.74 Amps. The total capacitance at the 34.5 kV side is 6,000 pF. Find the equivalent circuit for t ≥ 0 using the current injection method.

34.5 kV Bus No Load

Vacuum CB

If the circuit breaker does not open, the current through the switch for t ≥ 0 is

Equivalent circuit for t ≥ 0

ik

v(t)

+

-

Ls

6nF Lm

Note: v(t) is not the total voltage across the transformer.

First-Order Transients

Consider the homogeneous differential equation

with initial condition x(0)=X0.

Substitution gives

The solution can be shown to be an exponential of the form

After canceling the exponential term, we get

or

Thus the solution is

At t=0, we get

The final solution is

RL Network Consider the circuit shown. Let i(0) = I0. From KVL, we get for t ≥ 0

Note: The transient response of an RL network is a decaying exponential with a time constant τ=L/R.

Substitution gives

Li

R

RC Network Consider the circuit shown. Let vC(0) = V0. From KCL, we get for t ≥ 0

Note: The transient response of an RC network is a decaying exponential with a time constant τ=RC.

Substitution gives

i

R

vC

+

- C

Second-Order Transients Consider the homogeneous differential equation

The solution can be shown to be an exponential of the form

Substitution gives

After canceling the exponential term, we get the characteristic equation

Using the quadratic formula, we get two roots

Thus the solution is

Note: The roots may be real or complex numbers.

Series RLC Circuit

C

L

i

t=0

R

Consider the circuit shown. From KVL, we get for t ≥ 0

Differentiating once, we get

Note: The circuit is described by a homogeneous second-order differential equation.

Example: Real and Distinct Roots (Over-damped Case)

E=12 V L=1 H R=10Ω

vc(0)=0

+ + C

L

i

t=0 R

E vC - -

For t ≥ 0, we get

Differentiating once, we get

or

Substitution gives and the roots are s1=-2 and s2=-8. The total response is

The characteristic equation is

Using the quadratic formula, we get the two roots

Initial Conditions: At t = 0+,

0 0

From (2), we get

Evaluate Constants: Differentiate the total response. We get

At t=0+

Solving simultaneously, we get K1=2 and K2=-2. Thus,

Example: Real but Repeated Roots (Critically- Damped Case)

E=12 V L=1 H R=8Ω

vc(0)=0

+ + C

L

i

t=0 R

E vC - -

The characteristic equation is whose roots are s1=s2=-4. When the roots are real but repeated, the transient response can be shown to be given by

Substitution gives

and

At t=0+

We get K1=12. The complete response is

Example: Complex Roots (Under-damped Case)

E=12 V L=1 H R=6Ω

vc(0)=0

+ + C

L

i

t=0 R

E vC - -

The characteristic equation is whose roots are s1,s2=-3 ± j2.65. When the roots are complex conjugates, i,e. s1,s2=-α ± jωd, the transient response is given by

whose derivative is

Substitution gives

At t=0+

We get

R=8Ω

R=10Ω

R=6Ω

Plot of the Currents

Series LC Circuit

vC

+

- E C

L

i + -

t=0

or

Consider the circuit shown. From KVL, we get for t ≥ 0

The characteristic equation is whose roots are imaginary. We get s1,s2=± jωn and the transient response is given by

Substitute in

We get

The initial conditions are

and

where

= surge impedance, Ω

= natural frequency, rad/sec

Substitution gives

We can also show that

Plots of vc for various values of vC(0+) are shown.

Note: The voltage across the capacitor oscillates symmetrically above and below the source.

E

2E

3E

4E

-E

-2E

vC(0+)=0

vC(0+)=-E

vC(0+)=-2E

•  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