Power Transformer Protection

A Seminar on Protection System

Er.Sandeep Yadav , Xen , Protection Division ,

HVPNL,Gurgaon
Important

Over-current Protection
During a fault , the secondary current(s) of the C.T. increases and if the C.T. sec. current increase beyond the setting of the relay , the relay will operates wit
hin a time (as per the fault current)
Generally IDMTL(Inverse Minimum Definite Time Lag) relays are used in HVPNL for protection of primary & secondary side as well as feeder p
rotection of the power transformer. An IDMTL Over current/Earth fault relay has a Aluminum Disc and will take a time to trip after occurrenc
e a fault. The tripping time of the relay is inversely proportional to the fault current.
If fault current is high , operating time is less and if fault current is less , relay will take a long time to operate.
The main drawback of IDMTL relay is to take some minimum operation time even for extreme high fault currents, say as high as 8-10 times the
rating of transformer.
If relay take even a small time to isolate (trip) such heavy fault currents , the power transformer may be in trouble during heavy faults.
So , we need a method by which the IDML relay allow to operate for fault currents below say 5x (5 times the capacity of transformer) and if faul
t current goes beyond 5x then plain over-current relay (non-IDMTL) will trip the transformer instantaneously.
This Non-IDMTL Over-current relay is called High-set or Instantaneous Relay.
Highset relay is always connected in series with the IDMTL relay so that it can see the same current as seen by IDMTL relay as show in picture belo
w.
Highset relays are never connected in incomers i.e. 132KV I/c,66KV I/c,33KV I/c & 11KV I/c and always provided on outgoing circuits i.e. 11KV Out
going feeders , 33KV Outgoings feeders etc so that for heavy outgoing feeder fault , the incomer should not trip and disturb other healthy feeders
connected to it.
In LV side of power transformer , all outgoing feeders are equipped with highset relay to trip the feeder instantaneously during a heavy fault. As al
ready indicated , no highset relay is provided on the LV Incomer panel/VCB of the power transformer. As the LV bus is fed by LV Incomer so if there
is some heavy/serious fault in the LV bus i.e. 11KV bus for a 66/11 or 132/11KV TF , 11KV Incomer panel’s relay will take some time to trip being ID
MTL relay and as bus fault is very nearer to T/F , such delay in tripping can’t be tolerated. So there is a provision of providing the Highset on HV si
de of TF to trip the transformer for LV bus faults.
In a Power Transformer , Restricted Earth Fault (R.E.F.) protection is provided on both H.V. and L.V. side if transformer is of Yy0 type. If primary is D
elta connected (if T/F is Delta-Star type) , no need of R.E.F. on Delta (HV) side. All the Earth Faults on outgoing feeders are looked by their relays. E
arth fault in bus zone will be seen by E/f relay on LV incomer.
Suppose
•an earth fault in a 11KV Incomer cable
•puncture of 11KV or 33KV L.A. of power transformer
•Damage of TF CT on HV side or LV side with a flash with ground
•Damage of LV cable box (LV incomer Indoor/Outdoor)
•Damage of HV or LV winding of transformer after shortening with core/earth
•Bird-age on transformer of HV/LV jack buses etc
Such faults are very nearer to transformer and transformer should be isolated from fault to avoid any mis-happening. We need a relay which will auto
matically sense the zone between HV C.T.s to HV winding/core (for HV R.E.F.) and between LV Winding/core to LV incomer C.Ts (for LV R.E.F.).
This relay needs to sense and isolate the earth faults in the above said zones so called Restricted Earth Fault Relay.As we need immediate tripping
, so these are non IDMTL relays i.e. don’t have a rotating disk e.g. CAG14 (Alstom) etc.So following may be cloncluded.
•A H.V. R.E.F. relay will sense any earth fault in H.V. C.Ts , HV Jack Bus, H.V. L.As , TF H.V. Bushing/core and winding.
Similarly
•A L.V. R.E.F. relay will sense any earth fault in L.V.Core/Winding/Bushing of the transformer , LV Jack bus (11KV Incomer cables in case of 132/11 or 6
6/11KV T/F) , LV L.As & L.V. C.T.s ( 33KV I/c C.Ts in case of 132/33KV T/f & 11KV Incomer C.Ts in case of 132/11KV T/F.
 Path of Current during Earth Fault on any outgoing feeder
For an Earth fault , the fault current have to return to neutral bushing of the source transformer through NCT as shown above whatever be the distance of l
ocation of Earth fault from source transformer
 REF relay during Normal Loading condition
During normal condition , the phase currents in all phases will be approx. same and so Ict=0 (as vector sum of three phases currents 120deg apart will b
e zero.The current in NCT secondary will also be zero due to no Earth fault & no current in NCT primary i.e. in normal condition , there won’t be any curr
ent through neutral circuit of Power transformer.
So in normal situation , no current will flow through REF relay and it will not operate.
 REF relay during External Earth Fault Condition
During External Earth Fault condition, the fault current will flow through the NCT as well as through the REF core of the faulty phase CT.As direction of
both the currents are in opposition so nullify both the currents in relay and REF relay won’t operate.
SO, THE REF RELAY WON’T OPERATE FOR A OUT-OF-ZONE EARTH FAULT
 REF relay during In-zone Earth Fault Condition
During In-zone Earth Fault condition, the fault current will only flow through the NCT & no t from the REF core of Incomer CT being fault before Income
r panel.The current from NCT goes to REF relay and trips it being no counter current from REF cores of Incomers CTs to oppose it.
SO, THE REF RELAY SHOULD OPERATE FOR A IN-ZONE EARTH FAULT
 Differential Protection
Differential Protection is most sensitive & fast acting protection of the power transformer. It sense the difference in the currents received from the HV si
de & LV side CT’s differential cores. It covers the area from the differential core of HV CTs to the differential core of LV CT’s including Las , HV/LV jack bu
ses , TF winding , cores & bushings etc
A Buchholz relay is a gas and oil operated device installed in the pipe work between the top of the transformer main tank and the conservat
or. A second relay is sometimes used for the tap changer selector chamber. The function of the relay is to detect an abnormal condition
within the tank and send an alarm or trip signal. Under normal conditions the relay is completely full of oil. Operation occurs when floats
are displaced by an accumulation of gas, or a flap is moved by a surge of oil. Almost all large oil-filled transformers are equipped with a Buch
holz relay, first developed by Max Buchholz in 1921.
PRESSURE RELIEF DEVICE
The pressure relief device (PRD) is designed to open and close automatically when the pressure reaches the operating pressur
e of the PRD.
The pressure relief device will open and remain open until the pressure falls to the reseal pressure. The pressure relief device r
eseals at a positive pressure.
 Magnetic Oil Level Gauges ( Indicator )are most commonly fitted on conservator of Power/ Distribution Transfor
mers. These Gauges give visual indication of oil level and provide Alarm facility at desired level. Float movement d
ue to rise and fall of oil level gives rotary movement to indicating pointer through bevel gears and magnets.
 THANK YOU !!

Er.Sandeep Yadav
Executive Engineer,
Protection Division,HVPNL
Gurgaon (Haryana)
Certified Energy Auditor from Bureau of Energy Efficiency
Govt. of INDIA

+919310404372
https://www.facebook.com/sanrwr

sandeep.hvpnl@gmail.com
xenmpccggn@hvpn.gov.in
Percentage Impedance:
The percentage impedance of a transformer is the volt
drop on full load due to the winding resistance and
leakage reactance expressed as a percentage of the
rated voltage.
It is also the percentage of the normal terminal voltage
required to circulate full-load current under short circuit
conditions. The impedance is measured by means of a
short circuit test. With one winding shorted, a voltage at
the rated frequency is applied to the other winding
sufficient to circulate full load current. The method of
testing will be discussed later.
 The maximum primary and secondary fault currents
can be calculated from 160MVA.
e.g. Consider a transformer of 16MVA , 66/11kV with
%age impedance of 10%.Calculate the maximum fault
current that will flow during dead short circuit on its LV
bushings.
Fault MVA = 16x10/100=160MVA
Max. HV side fault current = 160x10^6/√3x66x10^3 =
1400A against rated current of 140A
Max. LV side fault current = 160x10^6/√3x11x10^3 =
8400A against rated current of 840A
 A transformer with lower impedance will lead to a
higher fault level (and vice versa)
In practice, the actual fault level will be reduced by the
source impedance, the impedance of cables and
overhead lines between the transformer and the fault,
and the fault impedance itself.
Size MVA Voltage rating in kV
MVA 12kV 33kV 72,5kV 148kV 250kV 300kV 420kV
0.5 4.75% 5.0% 5.5%

0.8 4.75% 5.0% 6.0%

5 4.75% 5.0% 6.0% 7.0%

10 9.0% 9.0% 10.0% 10.0%

20 10% 10.0% 15.0% 13.0%
30 15% 15.0% 12.0% 13.0%
60 12% 12.5% 12.5% 15.0% 15.0% 16%
100 16.0% 17.5% 18%
Way to write Vector Group of Transformer Winding configuration:
First symbol/symbols, capital letters: HV winding connection.
Second symbol/symbols, small letters: LV winding connection.
Third symbol, number: Phase displacement expressed as the clock hour
number.

High Voltage Always capital letters

Delta D
Star Y
Zigzag or Interconnected star Z
Neutral brought out N
Low Voltage Always small letters
Delta d
Star y
Zigzag or Interconnected star z
Neutral brought out n
Phase displacement
Phase rotation is always anti-clockwise. (Internationally adopted
convention).Use the hour indicator as the indicating phase
displacement angle. Because there are 12 hours on a clock,
and a circle consists out of 360°, each hour
represents 30°.
Thus 1 = 30°, 2 = 60°, 3 = 90°, 6 = 180° and 12 = 0°
or 360°.
The minute hand is set on 12 o'clock and replaces
the line to neutral voltage (sometimes imaginary)
of the HV winding. This position is always the
reference point.
Because rotation is anti-clockwise.
Group Phase Clock Description
Number Displaceme Hour
nt number
I 0° 0 No phasor difference

II 180° 6 180° lagging (LV lags HV with 180°)

III -30° 1 30° lagging (LV lags HV with 30°)

IV +30° 11 330° lagging or 30° leading (LV leads HV

with 30°)

Dd0
Delta connected HV winding, delta connected LV
winding, no phase shift between HV and LV.
Dyn11
Delta connected HV winding, star connected LV
winding with neutral brought out, LV is leading HV
with 30°
Group I - (0 o'clock, 0°) - delta/delta, star/star
Group II - (6 o'clock, 180°) - delta/delta, star/star
Group III - (1 o'clock, -30°) - star/delta, delta/star
Group IV - (11 o'clock, +30°) - star/delta, delta/star
1. The voltage ratio must be same.
2. The percentage impedance of the transformers should
be same.(not essential but recommended)
3. The MVA rating of the transformers should be same.
(not essential but recommended)
4. To avoid the circulating currents during parallel
operation, the polarity of the transformers should be
same
5. Phase sequence or phase rotation of the transformers
should be same or we can say that vector group of the
transformers should be same.
 Sr.No. Group Clock Winding Configuration
Number/Angle

1 I 0/0° Yy0 , Dd0 , Dz0

2 II 6/180° Yy6 , Dd6 , Dz6
3 III 1/-30° Dy1 , Yd1 , Yz1
4 IV 11/+30° Dy11 , YD11 , Yz11
The transformers relating to same vector group can be operated
in parallel after ascertaining other conditions/requirements of
the parallel operation of power transformers.
We can make the parallel operation of transformers of group 1 &
2 with suitable internal connections amendments. Group 3 & 4
transformers may also be operated in parallel after a suitable
change of external connections.
In no case the Group 1 or 2 transformers can be paralleled with
group 3 or 4 transformers.
Example1. Equal Voltage Ratio, Equal MVA rating , Equal %Z.
Consider two transformers having MVA rating of 100MVA & 15%
percentage impedance.Voltage ratio of both transformers is same
i.e. 220/66kV.Calculate how a load of 200MVA will be shared by
them.

So load MVA shared by both the transformers

equally.
Example2. Equal Voltage Ratio, Unequal MVA rating, Equal %Z.
Consider two transformers having MVA rating of 100MVA & 50MVA.%
percentage impedance is same 15% for both transformers. Voltage
ratio of both transformers is same i.e. 220/66kV.Calculate how a load
of 200MVA will be shared by them.

So if the %age impedance is same , load will be shared

by transformers in proportion to their MVA rating.
Example3. Equal Voltage Ratio, Equal MVA rating, Differnent %Z.
Consider two transformers having MVA rating of 100MVA each,%
percentage impedance is 10% & 15%. Voltage ratio of both
transformers is same i.e. 220/66kV.Calculate how a load of 200MVA
will be shared by them.
Alternative way to calculate load sharing:
Example : Suppose we need the parallel operation of 3
transformers having following details:

Sr.No. TF MVA Rating %age

Impedance
1 T1 100 9%
2 T2 75 10%
3 T3 50 8.5%
4 T4 150 12%
Sr.No. TF MVA Rating %age A=MVA/%age impedance
Impedance
1 T1 100 9% 11.1
2 T2 75 10% 7.5
3 T3 50 8.5% 5.88
4 T4 150 12% 12.5
Total 375 36.98
Load Shared by T1 TF 375x11.1/36.98= 112.56MVA
Load Shared by T2 TF 375x7.5/36.98= 76.05MVA
Load Shared by T3 TF 375x5.88/36.98= 59.62MVA
Load Shared by T4 TF 375x12.5/36.98= 126.75MVA
Calculation of max. load which can be shared without overloading any of the transformer

X factor for T1 TF 112.56/100 = 1.1256

X factor for T2 TF 76.05/75 =1.014
X factor for T3 TF 59.62/50=1.1924
X factor for T4 TF 126.75/150=0.845
Maximum X Factor 1.1924
Max. load which can be shared without over 375/1.1924= 314.5MVA
loading any transformer
 Conclusion:
1. TF T1 will be ,over-loaded i.e. share 112.56MVA against rated
capacity of 100MVA
2. TF T2 will be badly over-loaded i.e. will share 76.05MVA against
rated capacity of 75MVA.
3. TF T3 will be over-loaded i.e. share 59.62MVA against installed
capacity of 50MVA
4. TF T3 will also be under-load and share 126.75MVA during
parallel operation against installed capacity of 100MVA.
5. If we want to avoid over loading of any transformer then we
share a load up to 315MVA (approx) i.e. under utilization of
60MVA (375-315=60MVA) installed capacity.
Cooling of Power Transformer:
The cooling systems of a transformer can be divided in to natural
or forced type. Natural cooling works on the principle of
convection. In forced cooling, fans and pumps are used. Different
methods used for cooling of power transformers are listed below:
O - Oil, A – Air, N – Natural, F – Forced
Sr.No. Symbols Meaning Description
1 ONAN Oil Natural Air Natural The transformer is fitted with panel type
radiators for natural oil cooling & air natural
cooling.
2 ONAF Oil Natural Air Forced Air if forced by cooling fans on the radiators. Oil
is circulated by natural convection.
3 OFAN Oil Forced Air Natural Oil immersed transformers with oil pumped
around the cooling system. No air cooling/ fan
arrangements are used.
4 OFAF Oil Forced Air Forced Both oil & air are forced by oil pump & air
cooling fans.
5 OFWF Oil Forced Water Oil is pumped around the cooling system by oil
Forced pump, which then further cooled by a heat
exchange pumped-water system.
•Voltage Ratio or Transformer-Turn-Ratio (TTR) test :
TRANSFORMER TURN RATIO (TTR) TEST
SHORT CIRCUIT TEST
VECTOR GROUP TEST
FAULTS FACED BY A TRANSFORMER
PHASE-TO-PHASE FAULT
In case of phase to phase fault or three phase fault, the
fault current will not flow through the neutral or
earthing. So the fault current will be limited by the
impedance of the transformer. The winding resistance
is deliberately designed to limit the maximum short
circuit current of the transformer. This can be achieved
by a properly designed leakage reactance or flux.
However the increase of the leakage reactance may
drop the efficiency of the transformer considerable in
loading conditions.
FAULTS FACED BY A TRANSFORMER

PHASE-EARTH FAULT
The flow of the earth fault current depends upon the flow of zero
sequence component of the current. The AT (ampere-turn) balance
between primary & secondary windings should be maintained. The
magnitude of the earth fault current is depends upon the method of
the earthing , the impedance of the winding and the transformer
configuration (Start-star , Delta-Star etc).It also depends upon the
position of fault in the winding.
Solidly earthed transformer windings have only their winding
impedance to limit the earth fault currents.
We can use earthing transformers & resistors to add further
impedance to the path of zero sequence currents.
FAULTS FACED BY A TRANSFORMER
INTER-TURN FAULTS
Degradation of the winding insulation is the major
cause of inter-turn faults. Such faults cause localized
over heating & formation of hotspots as the faulted
current carry more currents.
Inter-turn faults may cause very heavy currents to
flow within the shorted turns. Such heavy inter-turn
fault current in the shorted turns when referred to
primary side will reflect a very small amount of
current, which may be quite less to initiate any
protective device.
Resistance of faulted turns = 1mΩ (assume)
Voltage drop across faulty turns = 4V (assume)
Fault current = 4/0.001=4000A

Power dissipated in faulty turns = I^2.R =

4000x4000x0.001 = 16kW
As IpVp=IsVs
Ip x 220000/√3 = 4000 x 4
Ip = 0.125A or 125mA
As seen from the calculations, a current of 125mA will
flow through primary circuit for inter-turn fault of 4000A
on the secondary side. So such inter-turn faults are not
easy to be detected by the over-current or differential
relays. But such heavy fault currents may cause the
formation of hot-spots on the shorted turns.
Such faults cause generation of heat & decomposition
of the oil which can be sensed by Buchholz relay.
CORE FAULTS

Core faults may occur mainly due to following reasons:

 Breakdown in the insulation between CRGO core
laminations.
 Failure of core bolt insulation.
Both caused flow of excessive eddy currents. Hot
spot will form due to excessive eddy currents.
Such faults are generally incipient in nature and
cause gradual deterioration in the transformer
performance before developing into electrical fault
due to insulation failure.
Core faults are detected by Oil & DGA.
FAULTS DUE TO IMPROPER COOLING SYSTEM

A degradation of insulation in the tank & further

leakage of the transformer oil can occasionally cause
the flashover between windings and end connections.
The cooling system of the transformer should be in
good condition. Failure of cooling system may leads
to reduced capacity of the transformer.
OVER FLUXING
Over fluxing is often caused by poor regulation of
voltage and frequency on the power system.
Consider the basic emf equation of the transformer:
V α 4.44x f x N x ϕ
Where
V = rms value of the voltage
f = frequency
N = number of turns in the winding
Φ = FLUX
So Flux , Φ = V/(4.44.f.N)
 Φ = V/(4.44.f.N)
It is obvious from the above equation that an Over fluxing
condition can occur if the ratio of voltage to frequency
exceeds a predetermined value. Transformers are rated for
certain values of over fluxing.
Whenever there is an overvoltage (with constant
frequency) , the core of the transformer may be subjected
to higher values of flux to justify the increased voltage.

So increase of flux due to increased voltage may drive the

operating region of the transformer in to saturation zone. A
saturated transformer will draw more magnetizing current and
cause over-excitation .Core losses will increased and the
overheating of the transformer occur.
 Differential Protection

Ib + Ic + Id = Ia
 Factors affecting Differential Protection Relay

1. CT Saturation: Excessive fault currents when

circulating through CTs may lead to CT saturation. The
waveform of the secondary current of a saturated CT
may be distorted with presence of harmonics. The
distorted wave-shape may cause the false operation of
differential relay even for external faults. The harmonics
contents present in the output of a saturated CT can
also cause a time delay of operation of differential
protection for in-zone faults.
1. Inrush Currents: During the sudden energization of
the transformer, the transformer may draw a current
even up to 5-20 times of the rated current value.
This current may persists for 10 cycles i.e. 200ms
depending upon the system impedance, type of
transformer, remanance flux & the position of
difference voltages (phase) while energization.
This excessive current drawn by the transformer
on primary side from source while no secondary
side current is there. A huge difference in primary
& secondary current will reflect as a fault
scenario for the differential relay and the relay
may mal-operate .
Over-excitation: The line capacitance & light load
conditions on a transformer can lead to high voltage on
the transformer. High voltage can results in more flux
and more eddy current losses. Transformers are
typically designed to operate just below the saturation
level. Any more increase in V/f (Voltage/frequency ratio)
can lead to saturation of core. After the saturation of
core , more excitation current will be drawn by the
transformer from the source.
Wrong ratio selection of
primary & secondary CTs
 TRANSFORMATION RATIO VARIATION
For a 132/33kV, 16MVA transformer, the voltage ratio is
132/33=4.It means that the secondary voltage is 1/4th of
primary voltage. Similarly, the secondary current will be 4
times the primary side current.
This condition is true if the transformer is running on principal
tap. The ratio of the secondary to primary currents will not be
exactly 4 at tap positions other than principal tap as indicate in
table

Tap Vp Vs Ip Is Ratio
1 138600 33000 66.65134 279.9356 4.200
5 132000 33000 69.9839 279.936 4.000
17 112200 33000 82.334 279.9356 3.400
TRANSFORMATION RATIO VARIATION

Transformation ratio at Tap 1 = 4.2

LV side CT Secondary current =6000x1/300=20A
HV side CT secondary current = (6000/4.2)x(1/75) =19.04A
Difference of current = 20-19.04 = 0.96A or 960mA
So even for a external (out-zone) fault , differential relay will
operate.
 Transformation ratio at Tap 5 = 4.2
(Principal Tap)
LV side CT Secondary current =6000x1/300=20A
HV side CT secondary current = (6000/4)x(1/75) =20A
Difference of current = 20-20A = 0
No current will flow through differential relay.
Transformation ratio at Tap 1 = 3.4
LV side CT Secondary current =6000x1/300=20A
HV side CT secondary current = (6000/3.4)x(1/75) =23.529A
Difference of current = 23.529-20 = 3.529A or 3259mA
So even for a external (out-zone) fault , differential relay
will operate.
CT CONNECTIONS:
Harmonic Restraint:
As already discussed, at the time of energization of the
transformer, the transformer can draw excessive current
from the source called inrush. This current is sensed by
the differential relay as HV side current. As no current is
there in LV side of the transformer , the differential relay
assume that it is the internal fault due to which
transformer is drawing excessive current in HV side and
no current is on LV side.
An over-excitation of the transformer during light load
conditions (less frequency or high voltage) can cause false
operation of differential relay
A typical inrush waveform of a transformer contains
the Harmonic components as listed in table
Sr.No. Harmonic Magnitude
1 Fundamental 100%
2 DC components 50-60%
3 2nd Harmonic 50-70%
4 3rd Harmonic 15-30%
5 4th Harmonic <5%
6 5th Harmonic <5%
7 6th & 7th Harmonic <5%

During the energization of the transformer, the

operation of differential relay is blocked called
‘Harmonic Inhibit’
 HARMONIC RESTRAIN FEATURE

As the 2nd harmonics form the major harmonic component in

the wave-shape of inrush current of the transformer, the
harmonic restrain is designed to be provided by 2nd
harmonics. Some relays also uses 4th harmonic in addition to
2nd harmonic for differential restrain feature.