ELE 578 POWER SYSTEM PROTECTION

Course outline

Faults and concept of protection in power systems, symmetrical and unsymmetrical fault

analysis, fuses, circuit breakers, basic principles of relay design, construction, characteristics

and applications, protective relays: distance relay, differential relay, etc, protection of

generators, motors, bus-bars and transformers.

What is power system protection?

Power system protection is the art and science of detecting faults (problems) with power

system components and isolating these components. The protection of the power system

ensures that any part of a system is quickly detected and isolated so that the rest of the

system/equipment and personnel are safe and the remaining system operates satisfactorily.

Thus, the whole protection arrangement must ensure the following:

i. be sufficiently sensitive to detect a fault at its earliest stage

ii. be absolutely reliable in operation, the simpler and more robust the design the better

iii. be able to discriminate between currents fed to the fault points within the protected

section and currents passing through as a result of a fault in another section.

The detection of a fault and disconnection of a faulty section or apparatus can be achieved

by using fuses or relays in conjunction with circuit breakers. However, for high-voltage

circuits, a combination of relays and circuit breakers are employed to serve the desired

function of automatic protective gear, and it shall be our focus.

Protective relays

A protective relay is a device that detects the fault and initiates the operation of the circuit

breaker o isolate the defective element from the rest of the system. Figure shows a typical

single phase of a 3-phase relay circuit system.

1
The relay detects the abnormal conditions in the electrical circuits by constantly measuring

the electrical quantities such as voltage, current, frequency and phase angle, which are

different under normal and fault conditions. The faults signal their presence, type and location

to the protective relays through the changes in one or more of these quantities. Having

detected the fault, the relay operates to close the trip circuit of the breaker. This results in the

breaker's opening and the faulty circuit's disconnection.

When a short circuit occurs at point F on the transmission line, the current flowing in the line

increases to an enormous value. This results in a heavy current flow through the relay coil,

causing the relay to operate by closing its contacts. Consequently, it closes the trip circuit of

the breaker, making the circuit breaker open and isolating the faulty section from the rest of

the system. Thus, the relay ensures the safety of the circuit equipment from damage and the

normal working of the healthy portion of the system.

Essential Requirements of Protective Relaying

Every protective system which isolates a faulty element is required to satisfy the following

essential requirements:

2
 (i) Selectivity and discrimination (ii) speed of operation (iii) sensitivity (iv) reliability

(v) simplicity

Selectivity and discrimination

Selectivity is the ability of the protective system to select correctly that part of the system

in trouble and disconnect the faulty part without disturbing the rest of the system. The

protective system must be able to discriminate between those conditions for which prompt

operation is required and those from which no operation or time-delay operation is required.

Speed of operation

The relay system should disconnect the faulty section as quickly as possible. This is desirable

for many reasons, such as reducing the damage incurred. Improvement in power system

stability results in less annoyance to consumers and a decrease in power outage time, a

decrease in the likelihood of the development of one type of fault into a more severe type,

permits the use of rapid reclosure of circuit breakers to restore service to the consumer.

Sensitivity

A protective system must be sufficiently sensitive to operate reliably when required under

the condition that produces the least operating tendency. The sensitivity of a relay is a

function of the volt-amperes input to the relay's coil necessary to cause its operation. The

smaller the volt-ampere input required to cause relay operation, the more sensitive is the

relay.

Reliability

Reliability is the ability of the relay system to operate under pre-determined conditions.

Without reliability, the protection would be rendered largely ineffective and could even

become a liability.

3
Simplicity. The relaying system should be simple so that it can be easily maintained.

Reliability is closely related to simplicity. The simpler the protection scheme, the greater will

be its reliability.

Unsymmetrical Faults on 3-Phase System

Unsymmetrical Fault Calculations

In a symmetrical fault, the fault currents in the three lines are equal in magnitude and

displaced 120º electrical from one another. Due to the balanced nature of symmetrical faults,

it is sufficient to consider only one phase in calculations since the conditions in the other two

phases are similar.

However, most common faults on the power system are unsymmetrical in nature and when

such fault occurs, it gives rise to unsymmetrical currents (the magnitude of fault currents in

the three lines are different having unequal phase displacement). The calculation procedure

known as method of symmetrical components is used to determine the currents and voltages

on the occurrence of an unsymmetrical fault.

Those faults on the power system which give rise to unsymmetrical fault currents are known

as unsymmetrical faults. On the occurrence of an unsymmetrical fault, the currents in the

three lines become unequal and so is the phase displacement among them. Note: the term

‘unsymmetry’ applies only to the fault itself and the resulting line currents and the system

impedances and the source voltages which are always symmetrical through its main elements

such as generators, transmission lines, and synchronous reactors.

There are three ways in which unsymmetrical faults may occur in a power system

(i) Single line-to-ground fault (L — G)

(ii) Line-to-line fault (L — L)

(iii) Double line-to-ground fault (L — L — G)

4
 R R

Y Y

B B
L-G Fault
L-L-Fault

B
L-L-G-Fault

Symmetrical Components Method

any unbalanced system of 3-phase currents (or voltages) may be regarded as three separate

sets of balanced vectors

namely: (i) a balanced system of 3-phase currents having positive (or normal) phase sequence.

These are called positive phase sequence components.

(ii) a balanced system of 3-phase currents having the opposite or negative phase sequence.

These are called negative phase sequence components.

(iii) a system of three currents equal in magnitude and having zero phase displacement. These

are called zero phase sequence components

The positive, negative and zero phase sequence components (denoted with subscripts 1, 2 and

0 components, respectively) are called the symmetrical components of the original

unbalanced system.

Suppose an unsymmetrical fault occurs on a 3-phase system having phase sequence abc.

According to symmetrical components theory, the resulting unbalanced currents I a , I b I c can

be resolved into :

(i) a balanced system of 3-phase currents, I a1 , I b1 and I c1 having positive phase

sequence (abc) as shown in Fig. 18.3 (i). These are the positive phase sequence

components

5
 (ii) a balanced system of 3-phase currents I a 2 , I b 2 and I c 2 having negative phase

sequence

(acb) as shown in Fig. 18.3. These are the negative phase sequence components.

(iii) a system of three currents I a 0 , I b 0 and I c 0 equal in magnitude with zero phase

displacement

from each other as shown in Fig. 18.3 (iii). These are the zero phase sequence

components.
I a = I a1 + I a 2 + I a 0

I b = I b1 + I b 2 + I b 0

I c = I c1 + I c 2 + I c 0

The positive phase sequence currents ( I a1 , I b1 and I c1 ), negative phase sequence currents

( I a 2 , I b 2 and I c 2 ) and zero phase sequence currents ( I a 0 , I b 0 and I c 0 ) separately form

balanced system of currents. Hence, they are called symmetrical components of the

unbalanced system.

(ii) The symmetrical component theory applies equally to 3-phase currents and voltages both

phase and line values.

(iii) The symmetrical components do not have separate existence. They are only mathematical

components of unbalanced currents (or voltages) which actually flow in the system.

(iv) In a balanced 3-phase system, negative and zero phase sequence currents are zero.

I c1 I a1 Ib2
Ia2
120o 120 o

Ia0
240 o 240 o
Ib0
Ic0

I b1 Ic2

6
Operator ‘a’

As the symmetrical component theory involves the concept of 120º displacement in the

positive

sequence set and negative sequence set, therefore, it is desirable to evolve some operator

which

should cause 120º rotation. For this purpose, operator ‘a’ is used. It is defined as under

aI = I 120o = I (cos120o + j sin120o ) = I (−0.5 + j0.886)

a = −0.5 + j 0.886

a2 I = I 240o = I (cos 240o + j sin 240o ) = I (−0.5 − j 0.886)

a2 = −0.5 − j 0.886

a3 I = I 360o = I (cos360o + j sin360o ) = I (1)

a3 = 1

Properties of Operator ‘a’

Adding exps. (i) and (ii), we get,

a + a2 = (−0.5 + j 0.886) + (−0.5 − j 0.886) = −1

1 + a + a2 = 0

Subtracting exp. (ii) from exp. (i), we get,

a − a 2 = (−0.5 + j 0.886) − (−0.5 − j 0.886) = j1.732) = j 3

The unbalanced phase currents in a 3-phase system can be expressed in terms of symmetrical

components

as under

I a = I a1 + I a 2 + I a 0

I b = I b1 + I b 2 + I b 0

I c = I c1 + I c 2 + I c 0

7
 Ib = a2 I a1 + aI a 2 + I a 0

Ic = aI a1 + a2 I a 2 + I a 0

I a + Ib + I c = I a1 (1 + a2 + a) + I a 2 (1 + a + a 2 ) + 3I a 0

= I a1 (0) + I a 2 (0) + 3I a 0 = 3I a 0

since (1 + a + a = 0)
2

1
I a0 = ( I a + Ib + Ic )
3
1
I0 = (I a + Ib + Ic )
3

I a + aIb + a2 I c = I a1 (1 + a3 + a3 ) + I a 2 (1 + a 2 + a 4 ) + I a 0 (1 + a + a 2 )

= 3I a1 + I a 2 (0) + I a 0 (0) = 3I a1

1
I a1 = ( I a + aI b + a 2 I c )
3
1
I1 = ( I a + aI b + a 2 I c )
3

I a + a2 Ib + aI c = I a1 (1 + a 4 + a 2 ) + I a 2 (1 + a3 + a3 ) + I a 0 (1 + a 2 + a)

= I a1 (0) + I a 2 (3) + I a 0 (0) = 3I a 2

1
I a 2 = ( I a + a 2 I b + aI c )
3
1
I 2 = ( I a + a 2 I b + aI c )
3

8
 b

aI

120o
I
120o a
120o

a2I
c

Ib2 = a I a 2 Ia2
I c1 = a I a1 I a1
o
120
120o Ia0

120o Ib0
120o 120o
120o
Ic0
I b1 = a 2 I a1
Ic2 = a 2 Ia2

Sequence Impedance

The sequence impedance of the network describes the behaviour of the system under

unsymmetrical fault conditions. All component on power system such as the generators,

transformers and transmission lines, has three values of impedance namely positive sequence

impedance Z1 , negative sequence impedance Z 2 and zero sequence impedance Z o , to each

symmetrical value of current. The concept of impedances of various elements of power system

is essential in determining the fault currents in a 3-phase unbalanced system. Summarily, in

unsymmetrical fault calculations, each piece of equipment will have three values of

impedance corresponding to each sequence current.

The impedance offered by an equipment or circuit to positive sequence current is called

positive

9
sequence impedance Z1 . Similarly, impedances offered by any circuit or equipment to

negative and zero sequence currents are respectively called negative sequence impedance Z 2

and zero sequence impedance Z o .

Analysis of Unsymmetrical Faults

In the analysis of unsymmetrical faults, the following assumptions are made :

(i) The generated e.m.f. system is of positive sequence only.

(ii) No current flows in the network other than due to fault i.e. load currents are neglected.

(iii) The impedance of the fault is zero.

(iv) Phase a (R) shall be taken as the reference phase.

In each case of unsymmetrical fault, e.m.f.s’ per phase are denoted by E a , Eb and E c and the

terminal p.d. per phase by Va , Vb and Vc .

Single Line-to-Ground Fault

Consider a 3-phase system with an earthed neutral. Let a single line-to-ground fault occur on

the red phase as shown in Fig. 18.13. It is clear from this figure that

I a = I 0 + I1 + I 2 = 3I 0
a

Ea

N
Eb Va
Ec Ib = 0
c
b
Ic = 0 Vb

Vc

10
 1 1
I0 = ( I a + Ib + Ic ) = I a
3 3
1 1
I1 = ( I a + aI b + a 2 I c ) = I a
3 3

1 1
I 2 = ( I a + a 2 I b + aI c ) = I a
3 3
1
I 0 = I1 = I 2 = I a
3

Fault current

Let Z1 , Z 2 and Z 0 Consider the closed loop NREN. As the sequence currents produce voltage

drops due only to their respective sequence impedances, therefore, we have,

Ia

Z1

3Ea
Z2

Z0

Ea = I1Z1 + I 2 Z 2 + I 0 Z 0 + Va

Va = 0 I = I = I
1 2 0

Ea = I 0 ( Z1 + Z 2 + Z 0 )

Ea
I0 =
Z1 + Z 2 + Z 0

3 Ea
Fault current, I a = 3I 0 = (i)
Z1 + Z 2 + Z 0

11
the equivalent circuit of (i) is shown in fig and the fault current can calculated using the by

connecting the phase sequence impedances in series across an imaginary generator of voltage
3Ea

however, it is assumed is that the fault impedance is zero. Otherwise if the fault impedance is

Z f then expression becomes

3Ea
I a = 3I 0 =
Z1 + Z 2 + Z 0 + 3Z f

Note; if the neutral is not grounded, then zero sequence impedance will be infinite and the

fault current is zero. This is expected because now no path exists for the flow of fault current.

Phase voltages at fault.

For phase voltage at fault, the generated e.m.f. system is of positive sequence only, the

sequence

components of e.m.f. in a-phase are

E0 = 0 E2 = 0 E = Ea
; and 1

The sequence voltages at the fault for a-phase are :

Ea Z1
V1 = Ea − I1Z1 = Ea −
Z1 + Z 2 + Z 0

Z2 + Z0
V1 = Ea
Z1 + Z 2 + Z 0

−Z2
V2 = 0 − Z 2 I 2 = Ea
Z1 + Z 2 + Z 0

−Z2
V0 = 0 − Z 0 I 0 = Ea
Z1 + Z 2 + Z 0

It can be readily seen that V0 + V1 + V2 = 0 . This is expected because a-phase is shorted to

ground.

The phase voltages at fault are :

Va = V0 + V1 + V2 = 0

12
Vb = V0 + a 2V1 + aV2

Vc = V0 + aV1 + a 2V2

Summary of Results. For line (R-phase)-to-ground fault :

3 Ea
I a = fault current = 3I 0 = ; Ib = 0 ; Ic = 0
Z1 + Z 2 + Z 0

Va = 0 Vb = V0 + a 2V1 + aV2 ; Vc = V0 + aV1 + a 2V2

;

Line–to–Line Fault

Consider a line-to-line fault between lines c and b lines as shown in Fig. 18.15. The

conditions created by this fault lead to :

Vb = Vc ; I a = 0; I b + I c =0

Again taking R-phase as the reference, we have

1
I0 = (Ia + Ib + Ic ) = 0
3

Vc = Vb

Expressing in terms of sequence components of red line, we have

V0 + a 2V1 + aV2 = V0 + aV1 + a 2V2

V1 (a 2 − a) = V2 (a 2 − a)

V1 = V2
(i)

13
 Ia = 0
a

Ea

N
Eb

Ec Ib = 0
c
b
Ic = 0

I b + I c =0
Also,
( I 0 + a2 I1 + aI 2 ) + ( I 0 + aI1 + a 2 I 2 ) = 0

I1 + I 2 = 0 (ii)

Fault current. Examination of exp. (i) and exp (ii) reveals

that sequence impedances should be connected as shown in Fig.

18.16. It is clear from the figure that :

Ea
I1 = − I 2 =
Z1 + Z 2

 Ea   − Ea 
Fault current, I b = I 0 + a 2 I1 + aI 2 = 0 + a 2   + a 
 Z1 + Z 2   Z1 + Z 2 

 Ea  − j 3Ea
I b = (a 2 − a)  = = − Ic
 Z1 + Z 2  Z1 + Z 2

Phase voltages. Since the generated e.m.f. system is of positive phase sequence only, the

sequence

components of e.m.f. in a-phase are

E0 = 0; E2 = 0; E1 = Ea

14
The sequence voltages at the fault for R-phase are

Ea Z1
V1 = Ea − I1Z1 = Ea −
Z1 + Z 2

Z 2 Ea
V1 =
Z1 + Z 2

Z2
V2 = 0 − Z 2 I 2 = Ea
Z1 + Z 2

V0 = 0 − Z 0 I 0 = 0

The phase voltages at fault are

Z 2 Ea Z E
Va = V0 + V1 + V2 = 0 + + 2 a
Z1 + Z 2 Z1 + Z 2

2 Z 2 Ea
Va =
Z1 + Z 2

 Z E   Z E 
Vb = V0 + a 2V1 + aV2 = 0 + a 2  2 a  + a  2 a 
 Z1 + Z 2   Z1 + Z 2 

 Z E 
= (a 2 + a)  2 a 
 Z1 + Z 2 
Z 2 Ea
 Vb = −
Z1 + Z 2

 Z E   Z E 
Vc = V0 + aV1 + a 2V2 = 0 + a  2 a  + a 2  2 a 
 Z1 + Z 2   Z1 + Z 2 

 Z E 
= (a 2 + a)  2 a 
 Z1 + Z 2 
Z 2 Ea
Vc = −
Z1 + Z 2

Summary of Results. For line-to-line fault (Blue and Yellow lines)

− j 3 Ea
I a = 0; I b = − I c =
Z1 + Z 2

15
 Z 2 Ea
Vb = Vc = −
Z1 + Z 2

2 Z 2 Ea
Va =
Z1 + Z 2

I1
I2

V1 Z1 Z2 V2

Ea

Double Line-to-Ground Fault

Consider the double line-to-ground fault involving Y–B lines and earth as shown in Fig. 18.17.

Then conditions created by this fault lead to

I a = 0; Vb = Vc = 0

(ii)

Fault current. Examination of exp. (i) and exp. (ii)

reveals that sequence impedances should be *connected as

shown in Fig. 18.18. It is clear that

16
 Ia = 0
a

Ea

N
Eb

Ec Ib = 0
c
b
Ic = 0 Fault current
= Ib + I c

Vb = Vc = 0

1
V1 = V2 = V0 = Va
3 (i)

Also, I a = I1 + I 2 + I 0 = 0 (ii)

Ea
I1 =
Z Z
Z1 + 2 0
Z2 + Z0

Z0
I 2 = − I1
Z2 + Z0

Z2
I 0 = − I1
Z2 + Z0

 Z2 
Fault current, I f = I b + I c = 3I 0 = 3  − I1 
 Z2 + Z0 

3Z 2 Ea 3Z 2 Ea
=  =
Z2 + Z0 Z2 Z0 Z 0 Z1 + Z 0 Z 2 + Z1Z 2
Z1 +
Z2 + Z0

Phase Voltages. The sequence voltages for phase a are3

17
V1 = Ea − I1Z1; V2 = 0 − I 2 Z 2 ; V0 = 0 − I 0 Z 0

1
V1 = V2 = V0 = Va
3
Va = V1 + V2 + V0 = 3V2

Vb = a 2V1 + aV2 + V0 = (a 2 + a + 1)V2 (V1 = V2 = V0 )

= 0  V2 = 0
( a 2 + a + 1 = 0)

Vc = aV1 + a 2V2 + V0 = (a 2 + a + 1)V2 = 0

I1
I2 I0

Z1 V1 Z2 V2 Z0 V0

Ea

:This is a wonderful part of the method of symmetrical components

and makes the analysis easy and interesting. In fact, this method if the made in arriving at

exp. (i) permits to bring any unsymmetrical fault into a simple circuit of

The following points may be noted :

Transducers in protection schemes

A current Transformer (CT), is a transducer/device which reproduces a primary current as

the secondary current with a reduced value or level. That is, the primary current is

proportional to the secondary current. It reduces high voltage currents to a much lower value

and provides a convenient way of safely monitoring the actual electrical current flowing in

an AC transmission line using a standard ammeter. A typical CT is shown in Figure

18
.

A measuring current transformer is designed and constructed to operate satisfactorily over

the range of the rated normal current of the CT and has specified errors at the operating

value. A protection CT on the other hand operates over a range of currents many times the

rating of the circuit the CT is protecting and it is very often subjected to conditions greatly

exceeding these to which a measuring CT is subjected. The corresponding flux density leads

to the saturation of the core and the transient period of the short circuit current is usually very

important to the correct performance of the CT.

The design of CTs is the same as in power or distribution transformers in which the emf

equation is given as

Erms = 4.44 f BAc N (1)

 = BAc Ac = w l

f = frequency, Hz; B -= flux density; Ac = core area cross-section m2 ; N= number of secondary

turns; l ,w = length and width of the core in m.

19
The CT Knee Point

The accuracy of the CT's measurement depends on its linearity, which is the degree to which

the secondary current is proportional to the primary current. Ideally, a CT would provide a

linear output for any primary current, but in practice, CTs have a limit to their linearity,

known as the knee point. The knee point is the point at which the CT core begins to saturate

due to the magnetic flux density becoming too high. As the core saturates, the output of the

CT becomes non-linear, which means that the ratio of primary current to secondary current

is no longer linear. This can result in inaccurate measurements and can lead to errors in the

monitoring and protection of the electrical system.

knee point curve

so, in design, once the core saturation has been fixed, then the flux density is also limited to

the knee point value of about 1.5 wb/m2 or Tesla. The normal design criterion is to limit the

flux to the value

20
For instance, consider a CT ratio of 400/1 with a core section of l= 40mm and w = 30mm then

the flux , at the knee-point flux density of 1.5 Tesla is

 = 1.5  40  30  10−6 = 1.8  10−3 wb and

Erms = 4.44  1.8  10−3  400  50 = 159.8 V

In this design, if the CT is used to protect a 150 kVA transformer of 11/0.415 kV at the

secondary, then the primary current of the CT at full load is

150  10 3
= = 208.7 Amps.
3  0.415

Accordingly, this CT can be used to protect this transformer and the secondary current of the

CT at the full primary load current of the protected circuit is

208.7
=  1 = 0.522 Amps.
400
Burden of a Current Transformer.

The burden of a current transformer is the value of the impedance/load (largely resistive)

connected across the secondary winding of the transformer. It is expressed as the output in

volt-amperes (VA). For a 5 VA burden on a 1 Amp CT, the impedance will be 5 Ω. Burdens

are usually and always connected in series with the CT and the increase in the impedance

increases the burden of the transformer.

The current transformer is of two types in make:

The bar primary CTs for which the primary is a centrally situated straight conductor,

extending some distance on both sides of the CT which is a component of the power system.

In all cases, the primary is simply one primary turn. The secondary on the other hand is

wound uniformly over the whole periphery of the core, which is usually of the ring type (or

annular rings of steel). In this design, the leakage flux of both the primary and secondary are

negligible.

21
The primary wound types: the primary of this type consists of one or more turns; and because

of the small number of turns, they cannot be distributed uniformly over the whole core. These

types, therefore, tend to have high leakage flux, and are usually built with cores of E, T, I or

L-shaped laminations steel stampings. A typical type is shown in figure

Current Transformer Errors

The error of a protective CT is specified at the rated current and the accuracy limit current.

ratio error

A common error in current transformers is the ratio error. It emerges when the transformer

fails to convert current precisely according to a predefined ratio. Often, the non-linear

characteristics of the core and changes in coil impedance contribute to this error.

It can be best understood going back to the equivalent and phasor diagram figures

22
 Ip I s' Is
Rp Xp I Xs Ib
Rs
Io Im Rb
Vs
Ep Es
Xb

b Es = secondary induced voltage

−Ep
Ep = primary induced voltage
Ip
 c I p = primary current
aI s
a
I s = secondary current

 nI s = secondary current referred to the


Io  I primary
at = turn ratio
o 
Im  I = excitation current
 I m = magnetising current
 I o = working component
Is
 = main flux
Vs Es
 = Phase angle
Rb I s  = Burden Angle
X s Is

Ns
turn ratio , a =
Np

That is
nominal ratio-actual ratio
Current ratio error = (5)
actual ratio

a − at
=  100%
at

where

Ip
Transformation Ratio ( at ) at = (6)
Is

23
 To evaluate I p , we extend the line of aI s from a to b and trace to c from Figure

looking at obc

I p2 = ob2 + bc 2 (7)
2 2
=  aI s + I cos(90 − ( +  )) +  I sin(90 − ( +  ))  (8)
2 2
=  aI s + I sin ( +  )  +  I cos ( +  ) 

= a 2 I s2 + 2aI s I o sin ( +  ) + I2 sin 2 ( +  ) + I2 cos 2 ( +  )

= a 2 I s2 + 2aI s I  sin ( + ) + I 2

= a 2 I s2 + 2 aI s I sin ( +  ) + I2 (9)


1

Ip a2 I s2 + 2aI s I o sin( +  ) + I2  a2 I s2 + 2aI s I sin( +  ) + I2  2

at = = = 
Is Is Is

In design I  aI s thus, if I sin ( +  ) used instead of I  the value of at is not affected. Then

we can write
1

 s s   (
 a2 I 2 + 2aI I sin( +  ) + I sin( +  ) 2  2
 )
at = 
Is

aI s + I sin ( +  )
at
Is

I
at a+ sin( +  ) (10)
Is

Looking at the phasor diagram equation 10 can further be written as

I sin  cos  + I sin  cos 

at a+
Is

I m sin  + I o cos 
at a+ (11)
Is

If  = 0; cos  = 1; sin  = 0

24
 Io
at a+ (12)
Is

Phase Angle Error – In an ideal current transformer the vector angle between the primary

and reversed secondary current is zero. But in an actual current transformer, there is a phase

difference between the primary and the secondary current because the primary current has to

supply core loss and magnetizing components of the CT for which its losses some phase angle.

Thus, the difference between the two phases is termed phase angle error . The phase angle

error can understand by the below phasor diagram of a current transformer.

From the phase diagram,

ac I cos( +  )
tan  = = (13)
ob aI s + I sin( +  )

Note that I sin( +  )  I s for a small value of  ,tan  = 

I cos( +  ) I cos  cos  − I sin  sin 

= = (14)
aI s aI s

 I cos  − I o sin  
=   radian
 aI s
 
(15)

 I cos  − I o sin   180

 =    degrees
 aI s  
Effect of Open Secondary Windings of a CT

Under normal operating conditions, the secondary winding of a CT is connected to its burden,

and it is always closed. When the current flows through the primary windings, it always flows

through secondary windings and the amperes turns of each winding are subsequently equal

and opposite. The secondary turns will be 1% and 2% less than the primary turns and the

difference being used in the magnetising core. Thus, if the secondary winding is opened and

the current flows through the primary windings, then there will be no demagnetizing flux

due to the secondary current.

25
Due to the absence of the counter ampere-turns of the secondary, the unopposed primary

MMF will set up an abnormally high flux in the core. This flux will produce core loss with

subsequent heating, and a high voltage will be induced across the secondary terminal.

This voltage caused the breakdown of the insulation and also the loss of accuracy in the future

may occur because the excessive MMF leaves the residual magnetism in the core. Thus, the

secondary of CT may never be open when the primary is carrying the current.

Specific Items of CT

Rated secondary current

CTs are designed to have rated secondary current of 0.5 A, 1.0A or 5A. the burden should

have a definite VA at the rated CT current value. Since Z = VA I 2 that is the impedance varies

inversely to the rated CT current. Many CT burdens are situated at some distance from the

CT. Accordingly, the connected leads resistances do introduce an additional burden on the

CT; and the higher the current rating, the more the burden introduced by the leads. Consider

that the lead resistance is 2 ohms and at 1A rating the lead burden is 2 VA, but at 5 A rating,

the VA will be (2  5)  5 = 50 A. Therefore, in all cases where the leads burden is appreciable,

there is a great advantage in using lower rated CTs. For this reason, modern practice tends

towards using 1-ampere secondary windings.

Primary windings

It is necessary to achieve a satisfactory voltage output from a current transformer. The basic

output voltage equation is

d d r  o
N2 = N 2 ( N1 I p − N 2 I s ) Ac
dt dl lc

Hence, for cases of I p = 80 A and below and N1 = 1 , the core size sectional area, Ac will be large.

It is therefore more economical to increase the number of primary turns Np to two or more in

26
certain designs. The number of secondary turns will also increase thereby increasing the knee-

point voltage for a given cross-section Ac. Figure 2.10. the transformer may be specially

constructed or bypass the primary and secondary turns through the ring-type CT design and

construction.

Secondary winding impedance

To limit the secondary winding losses, because of the secondary winding current as a result

of the CT resistance, it is best to reduce the winding resistance value. in this way, copper losses

and temperature rise are reduced. In core design using ring-type CT, there are no leakage

reactances hence the impedance is mainly resistive.

Effect of CT Magnetizing Current on Relay Setting

The overall setting of a protection scheme/system is affected by the magnetizing current of

the current transformers. The effect may not be much in overcurrent relays but it does have

an effect on the overall setting of the earth-fault relay; and sometimes a profound effect on the

differential protection system. This effect is particularly so when a large number of current

transformers are connected as in the case of busbar zone protection. The error arises from the

fact that the primary operating current (POC) of a protection system = sum of relay setting

currents and magnetizing currents of all the connected current transformers at the voltage

across the relay setting multiplied by the CT ratio.

Voltage Transformer (VT)

A voltage or potential transformer fulfils one requirement, that the secondary voltage must

be an accurate replica of the primary voltage in both magnitude and phase. To achieve this

requirement, they are designed to operate at fairly low flux densities so that the magnetizing

current and hence the turn ratio and phase angle errors are small if any. The core area for a

given output is therefore larger than that of a power transformer. Therefore, the overall size

27
of a comparison unit is large. The stringent performance characteristics call for five limbs

design of the cores, figure to avoid magnetic interference between phases. For protection

purposes accuracy of measurement is important particularly during fault conditions when

the voltage is suppressed. Therefore, a VT for protection must meet the extended range of

requirements over a range of 5% to 80% of rated voltages, and for certain applications, the

range is between 120% to 190% of the rated voltage.

Protection of VT

VTs are usually protected on the primary side of HRC fuses while other types of fused or

miniature circuit breakers are used on the secondary side. Since they are designed to operate

at low flux densities, their impedances are low and so the secondary side short circuit will

produce a fault current of many times the rated current of its normal burden.

Residual connection

Correct voltage magnitude and phase angle must be presented to directional earth fault relays

the earth fault elements of impedance relays. But an earth fault can occur on any of the three

phases of the system, and it is therefore not possible to derive a voltage conventionally. The

solution is to use the residual or broken delta connection as shown in figure

28
Under balanced conditions, the three winding voltages sum up to zero. But if one voltage is

absent or reduced because of an earth fault on a phase, then the difference between that

voltage and the normal voltage is delivered to the relay and hence it trips the appropriate

circuit.

Capacitor voltage transformers

At a voltage of 132kV and more, the cost of an electromagnetic voltage transformer is high.

Therefore, a more economical approach is the use of a capacitor voltage transformer.

Essentially this is a capacitor voltage divided into a turning inductance and an auxiliary

transformer as shown in Figure

Any simple voltage divider system causes wide voltage variations and this affects the burden.

But by turning c2 with L the burden can be varied over a wide range with very low regulation.

The capacitors connected in series act like potential dividers provided the current taken by

the burden is negligible compared with the current passing through the series-connected

capacitors. However, the burden current becomes relatively larger and ratio error and also

phase error is introduced. Compensation is carried out by ‘tuning’.

The reactor connected in series with the burden is adjusted to such a value that at supply

frequency it resonates with the sum of two capacitors. This eliminates the error.

29
Fuses

A fuse is a safety device that protects electrical circuits from the effects of excessive currents.

It interrupts a circuit whenever that circuit carries a current larger than that for which it is

intended. Figure shows a typical fuse consisting of a low-resistance metallic wire (element)

enclosed in a non-combustible material. The element is generally made of materials having a

low melting point, high conductivity and least deterioration due to oxidation. Under normal

operating conditions, the element is at a temperature below its melting point and carries the

normal current without overheating. However, when a fault occurs, the current through the

fuse increases beyond its rated value thereby raises the temperature and the fuse element

melts (or blows out), thus, disconnecting the protected circuit. In this way, a fuse protects the

machines and equipment from damage due to excessive currents.

Advantages

i. It is the cheapest form of protection available.

ii. It requires no maintenance.

iii. Its operation is inherently completely automatic unlike a circuit breaker which requires

an

elaborate equipment for automatic action.

iv. It can break heavy short-circuit currents without noise or smoke.

v. The smaller sizes of fuse elements impose a current limiting effect under short-circuit

conditions.

30
vi. The inverse time-current characteristic of a fuse makes it suitable for overcurrent

protection.

vii. The minimum time of operation can be made much shorter than with the circuit

breakers.

Disadvantages

i. Considerable time is lost in rewiring or replacing a fuse after operation.

ii. On heavy short circuits, discrimination between fuses in series cannot be obtained

unless

there is sufficient difference in the sizes of the fuses concerned.

iii. The current-time characteristic of a fuse cannot always be co-related with that of the

protected apparatus.

Important Terms

The following terms are much used in the analysis of fuses :

i. Current rating of fuse elements. It is the current that the fuse element can normally

carry

without overheating or melting. It depends upon the temperature rise of the contacts of the

fuse holder, fuse material and the surroundings of the fuse.

ii. Fusing current. It is the minimum current at which the fuse element melts and thus

disconnects

the circuit protected by it. For a round wire, the approximate relationship between fusing

current I and diameter d of the wire is I = k d 3 2 , where k is the fuse constant.

iii. Fusing factor. It is the ratio of the minimum fusing current to the current rating of the

fuse element.
Minimum fusing current
Fusing factor=
Current rating of fuse

31
 iv. Prospective Current. It is the r.m.s. value of the first loop of the fault current obtained

if the fuse is replaced by an ordinary conductor of negligible resistance

v. Cut-off current. It is the maximum value of fault current reached before the fuse

melts.

vi. Pre-arcing time. It is the time between the commencement of fault and the instant

when cut off occurs.

vii. Arcing time. This is the time between the end of pre-arcing time and the instant when

the arc is extinguished.

viii. Total operating time. It is the sum of pre-arcing and arcing times.

ix. Breaking capacity. It is the r.m.s. value of a.c. component of maximum prospective

current that a fuse can deal with at rated service voltage

Types of Fuses

Generally, fuses are classified into:

i. Low voltages fuse ii. High voltage fuses

It is a usual practice to provide isolating switches in series with fuses where it is necessary to

permit fuses to be replaced or rewired with safety.

Low Voltage Fuses

Low voltage fuses can be subdivided into two classes viz: (i) semi-enclosed rewireable fuse

(ii) high rupturing capacity (H.R.C.) cartridge fuse.

i. Semi-enclosed rewireable fuse is used where low values of fault current are to be

interrupted. It consists of (i) a base and (ii) a fuse carrier. The base is of porcelain and

carries the fixed contacts to which the incoming and outgoing phase wires are

connected. The fuse carrier is also of porcelain and holds the fuse element between its

terminals. The fuse carrier can be inserted in or taken out of the base when desired.

Advantages.

32
 (i) the detachable fuse carrier permits the replacement of fuse elements without any

danger of coming in contact with live parts.

(ii) the cost of replacement is negligible.

Disadvantages

(i) There is a possibility of renewal by the fuse wire of the wrong size or by improper

material.

(ii) It has a low-breaking capacity and hence cannot be used in circuits of high fault levels.

(iii) The fuse element is subjected to deterioration due to oxidation through the

continuous heating up of the element. Therefore, after some time, the current rating of

the fuse is decreased.

(iv) The protective capacity of such a fuse is uncertain as it is affected by ambient

conditions.

(v) Accurate calibration of the fuse wire is not possible because the fusing current very

much depends upon the length of the fuse element.

High-Rupturing capacity (H.R.C.) cartridge fuse.

Figure shows a typical H.R.C. cartridge fuse. It consists of a heat resisting ceramic body

having metal end-caps to which is welded silver current-carrying element. The space within

the body surrounding the element is completely packed with a filling powder which may be

chalk, plaster of paris, quartz or marble dust and acts as an arc quenching and cooling

medium. Under normal load conditions, the fuse element carries the normal current without

overheating. When a fault occurs, the current increases and the fuse element melts before the

fault current reaches its first peak. The heat produced in the process vapourises the melted

silver element. The chemical reaction between the silver vapour and the filling powder results

in the formation of a high resistance substance which helps in quenching the arc.

33
Advantages

(i) They are capable of clearing high as well as low fault currents.

(ii) They do not deteriorate with age.

(iii) They have a high speed of operation.

(iv) They provide reliable discrimination.

(v) They require no maintenance.

(vi) They are cheaper than other circuit interrupting devices of equal breaking capacity.

(vii) They permit consistent performance.

Disadvantages

(i) They have to be replaced after each operation.

(ii) Heat produced by the arc may affect the associated switches.

High Voltage Fuses

High-voltage fuses are used in high-voltage circuits. Some of the high-voltage fuses are:

(i) Cartridge type. This is similar in general construction to the low-voltage cartridge type

except that special design features are incorporated. Some designs employ fuse elements

wound in the form of a helix to avoid corona effects at higher voltages. In some designs, there

are two fuse elements in parallel; one of low resistance (silver wire) and the other of high

resistance (tungsten wire). Under normal load conditions, the low-resistance element carries

34
the normal current. When a fault occurs, the low-resistance element is blown out and the high-

resistance

element reduces the short-circuit current and finally breaks the circuit. High voltage cartridge

fuses are used upto 33 kV with a breaking capacity of about 8700 A at that voltage.

(ii) Liquid type. These fuses are filled with carbon tetrachloride and have the widest range of

applications to h.v. systems. They may be used for circuits up to about 100 A rated current on

systems up to 132 kV and may have breaking capacities of the order of 6100 A. Fig. 20.5 shows

the essential parts of the liquid fuse. It consists of a glass tube filled with carbon tetrachloride

solution and sealed at both ends with brass caps. The fuse wire is sealed at one end of the tube

and the other end of the wire is held by a strong phosphor bronze spiral spring fixed at the

other end of the glass tube. When the current exceeds the prescribed limit, the fuse wire is

blown out. As the fuse melts, the spring retracts part of it through a baffle (or liquid director)

and draws it well into the liquid. The small quantity of gas generated at the point of fusion

forces some part of the liquid into the passage through the baffle and there it effectively

extinguishes the arc.

(iii) Metal clad fuses. Metal clad oil-immersed fuses have been developed with the object of

providing a substitute for the oil circuit breaker. Such fuses can be used for very high voltage

circuits and operate most satisfactorily under short-circuit conditions approaching their rated

capacity.

Current Carrying Capacity of Fuse Element

The current carrying capacity of a fuse element mainly depends on the metal used and the

cross-sectional area. It is also affected by the length, the state of the surface and the

surroundings of the fuse.

When the fuse element attains a steady temperature,

35