The document discusses power system protection. It defines the objectives of power system protection as detecting and isolating faults instantaneously while minimizing the number of circuits isolated and restoring the system quickly. It also discusses criteria for proper protection systems, including reliability, selectivity, speed of operation, and discrimination. Detection methods like current transformers and potential transformers are explained. Common protection relays like electromagnetic attraction, balance beam, and electromagnetic induction types are also summarized.

1. ECNG 3015
Industrial and Commercial
Electrical Systems
Lecturer
Prof Chandrabhan Sharma
#3
Power System Protection

2. Chapter 3 - Power System Protection
Figure1: Zoning

3. Objectives of Power System
Protection
- Detect and isolate faults instantaneously.
- Isolate the minimum number of circuits.
- Restore the system to normal configuration ASAP.
(self-clearing faults – Auto reclosure)
- Discriminate between normal and abnormal
conditions (e.g.: load current, overload and fault)

4. Criteria for a Proper Protection System
Reliability must act when called on. Relay is idle most time.
Selectivity operation in its prescribed zone
Speed of operation the longer the fault is ‘on’, the greater
the danger of fire.
Discrimination must be able to distinguish between normal,
overload and fault currents.

5. DETECTION Current, Voltage
Current – using Current Transformer or C.T.
Figure 2: C.T. Characteristic
Knee point:
point at which a 10% increase in secondary voltage produces a
50% increase in current excitation, i.e. saturation occurs.

6. Equivalent Circuit of a Current Transformer (CT)
Figure 3: C.T. equivalent circuit
From the equivalent ckt above
N1
I2 I1 ..........
..........
.....eq.
1
N2
I2 I2 I e ..........
.......... eq. 2
....

7. E2 I2 X 2 X L ..........
..........
..........
.......eq.
3
or
N1
E2 X2 XL I1 - I e ..........
..........
...eq.4
N2
substitute eq.1 in eq. 4 E2 X2 X L I 2 - Ie
I2 - I2
C.T. error
I2

8. Example:
Given a C.T. with ratio of 500:5 and characteristic curve below and
having X2 = 0.5 , find I2 and the C.T. error for XL = 4.5 (burden on
secondary of CT), I1 = 400A and IF = 4000mA.
C.T. Characteristic

9. N1
E2 X2 XL I1 - I e
N2
1
E2 0.5 4.5 400 Ie ; E2 5 4 Ie
100
E2 = 5(4 – Ie)
0.6

10. 0.6
From the graph theC.T. error x 100 15%
4
Note: If a fault occurs beyond the knee point, the CT ratio must be adjusted
such that it operates in the linear region.

11. Voltage – using V.T or commonly called P.T.
Types
- EM P.T.
- Capacitance P.T.
EM P.T.
For high voltages, the cost of insulation of windings is very high
resulting in increased cost and therefore the capacitor voltage
transformer is used.

12. Capacitance P.T. – Used for voltages 132 kV
C1
For capacitance P.T. V2 .V1
C1 C2

13. Protection relays
Types
- EM Attraction
- Balance Beam
- Electromagnetic Induction Type (IDMT relay)
EM Attraction relay – single-quantity relay
Electromagnetic force exerted on the moving element is
proportional to the square of the flux in the air gap. Neglecting the
effect of saturation, the total actuating force may be expressed as
F = K1I2 – K2
At threshold, F = 0 K1I2 = K2
K2
I constant
K1

14. Balance Beam relay
The rod moves in the direction of the side with the greatest
current.

15. Electromagnetic Induction Type relay (IDMT relay)
An IDMT relay is referred to as a split phase induction motor since the
relay uses the principle of induction to develop a torque in the rotor disk.

16. Operation:
- relay detects a current
- this current induces a magnetic flux in the core of the
relay
- the shorting rings split this flux in two (2)
- this causes opposing fluxes to be induced in the disk,
causing rotation (hence the term “split phase”).
See diagram below.

17. The figure shows how the force is produced in a section of a rotor that is
pierced by two adjacent a-c fluxes.
Each flux induces voltage around itself in the rotor and current flows in
the rotor under the influence of the two voltages.
The current produced by one flux reacts with the other flux, and vice
versa, to produce forces that act on the rotor.

18. The quantities involved may be expressed as follows:
1 1 sin t
2 2 sin t
where is the phase angle by which 2 leads 1
Assuming that the paths in which the rotor current flows have negligible
self-inductance then the rotor currents are in phase with their voltages,
i.e.
d 1
i1 α α 1 cos t
dt
d 2
i α α 2 cos t
2
dt

19. Since the twoforcesare opposing
then t net force F F2 F1 α
he 2 i1 i2
1
F α 1 2 sin t cos t sin t cos t
F α 1 2 sin
Also, Fm ax occurs at 90

20. Current Voltage Relays
Receives one actuating quantity from C.T. and the other from
P.T. In terms of actuating quantities, the torque is given by:
T = k1VI cos( - ) – k2
Where
V = rms voltage
I = rms current
= angle between V and I
= angle for maximum torque
At balance point, T = 0
k1VI cos( - ) = k2
Or
VI cos( - ) = k2/k1 = constant

21. GENERAL RELAY EQUATIONS
All relays are comparators, either:-
(a) Amplitude
or
(b) Phase
Amplitude Comparator
Two quantities are opposed and relay operates when
operating quantity exceeds the magnitude of the
restraining quantity irrespective of phase relationship.

22. An inherent phase comparator operates when one
input quantity has a defined phase relationship with
the other irrespective of magnitude.
An inherent amplitude comparator acts like a
phase comparator if the i/p quantities are changed to
the sum and difference of the two original quantities.
(vice versa)
e.g. If Amplitude Comparator operates on A B
then A B A - B is only true for a defined phase relationsh ip

23. Relay R is at threshold. A & B supplied to R in any arbitrary
combination.
Using ‘A’ as the reference vector:
i/p # 1 k1 A k 2 B [cos( - θ) j sin( - θ)]
i/p # 2 k3 A k 4 B [cos( - θ) j sin( - θ)]
Where k1, k2, k3 and k4 are design constants.

24. AMPLITUDE COMPARATOR
At threshold, the moduli of both inputs would be equal
irrespective of phase angle.
The locus of the moduli will yield the relay characteristic:
[k1 A k 2 B cos( - θ)]2 [k 2 B sin( - θ)]2
[k3 A k 4 B cos( - θ)]2 [k 4 B sin( - θ)]2
Re - arranging terms
2 2
2 2
(k1 k 3 ) A 2(k1k 2 k 3k 4 ) A B cos( - θ) (k 2 k 2 ) B
2 4 0
2
Divide acrossby (k2 k 2 ) A
2 4
2 2 2
B (k k k 3 k 4 ) B k1 k 3
2 1 22 2
cos( - θ) 0
A k2 k4 A k2 k2
2 4

25. Rearranging in the form of an equation of a circle in the complex
plane:
2
B B
2ζ cos( - θ) ζ r2
A A
Comparing coefficients :-
2 2
k1k 2 k 3k 4 k1 k 3
ζ ζ -r
2 2
k2 k2
2 4 k2 k2
2 4
2 2 2
2 k 1k 2 k 3 k 4 k 1 k 3 2 k 1k 4 k 2 k 3
r r
k2 k2
2 4 k 2
2 k 2
4 k2 k2
2 4

26. This represents a circle with:
k1k 4 k 2 k 3
radius r
k2 k2
2 4
k1k 2 k 3k 4
centrec - ζ θ '
θ'
k2 k2
2 4
i.e. Coordinates = - [cos ’+ j sin ’]

27. PHASE COMPARATOR
Let the input be :
i/p # 1 k1 A k 2 B [cos( - θ) j sin( - θ)]
i/p # 2 k3 A k 4 B [cos( - θ) j sin( - θ)]
For phase comparison, relay will only operate when product of i/p is positive.
Let = phase angle of i/p #1
Let β = phase angle of i/p # 2
∴ for threshold ( - β) = 90º
∴ for threshold tan( - β ) =
tan α - tan β
1 tan α tan β

28. i.e.1 tan α tan β 0
1
Or tan α - .......... .......... .......(1)
tan β
Imag k 2 B sin( - θ)
For i/p # 1 tanα
Real k1 A k 2 B cos( - θ)
k 4 B sin( - θ)
For i/p # 2 an β
t
k3 A k 4 B cos( - θ)
Sub into (1)
2
k 2 k 4 B sin ( - θ) -k1k 3 A
2
k1k 4 A B cos( - θ)
2
k 2 k 3 A B cos( - θ) - k 2 k 4 B
2
Divide acrossby k 2 k 4 A
2
B k1k 4 k 2 k 3 B k1k 3
cos( - θ) 0
A k 2k 4 A k 2k 4

30. Example
Given a percentage differential relay which operates when the
difference of the current entering the circuits exceeds 5% of the
sum of these currents or 10% of the mean through current.
Determine the characteristic equation for the relay.
Solution
The relay operates when one current is 10% > than the other, i.e.
relay is said to have a 10% bias.

31. Relay operates when:
I1-I2 > s{(I1+I2)/2} where s =10%
Let operating winding be supplied with (I1-I2)
Let restraining winding be supplied with s{(I1+I2)/2}
∴ I1-I2 = k1⃒I1⃒ + k2⃒I2⃒(cos +j sin )
s/ (I +I ) = k3⃒I1⃒ + k4⃒I2⃒(cos +j sin )
2 1 2
where is angle between I1 and I2 ( =0)
by comparing coeffs k1 = - k2 = 1
k3= k4 = s/2

32. For an amplitude comparator:
For s = 0.1 → r = 0.1004 and c = 1.004
∴ equation for amplitude comparator is

33. E.g.
Determine the values of k1, k2, k3, k4 for an inherent phase comparator
which will act like the amplitude comparator given before.
For phase comparator :
&
∴ Comparing coeffs:
k1 k4 - k2 k3 = s …………………⑴
k1 k4 + k2 k3 = - {1 + (s/2 )2} …………………⑵
2k2 k4 = 1 - (s/2 )2 …………………⑶
⑴+ ⑵ → 2k1 k4 = - (s/2 ) 2 + s - 1 = -[(s/2 ) – 1]2

34. and 2k2 k4 = - {(s/2) + 1} {(s/2) – 1}
∴ k1 = - 2 k4 = (s/2 ) – 1
k2 = - 2 k3 = (s/2 ) + 1
∴If we incorporate (-2) into k4 and k3 we have
* Amplitude not important for phase comparator

35. ∴ Inputs are:
No. 1. [ I1 {(s/2 ) – 1} + I2 {(s/2 ) + 1}] ⒜
No. 2. [ I1 {(s/2 ) + 1} + I2 {(s/2 ) – 1}] ⒝
For 10% slope → s =0.1
∴ ⒜ → (- 0.95 I1 + 1.05 I2)
⒝ → ( 1.05 I1 - 0.95 I2)
Dividing across by 1.05, inputs become
(- 0.905 I1 + I2) and (I1 - 0.905 I2)
The phase comparator will now act like an amplitude comparator and
the characteristic is as given before.