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CHAPTER FIVE

POWER TRANSFORMERS AND CONNECTIONS

1.0Introduction

The transformer is an electro-magnetically coupled circuit, which transforms power from one level of voltage and current to another. It is a vital link in a power system, which has made possible the power generated at lower voltages (11KV) to be transmitted over long distances at higher voltages (330KV, 132KV, etc.)

2.0Theory

In its simplest form, a transformer consists of a laminated core about which are wound two sets of windings; one called the primary and the other the secondary.

When a voltage is applied to the primary, it produces a magnetic flux in the core and the relationship between flux and voltage is given by:

e=- n d(

1

dt

where e and ( are the instantaneous values of voltage and flux and n the number of turns.

This flux lags behind applied voltage by 90oThus if

e=Em Sin(t

(=(m Cos(t

Substituting in eqn. 1 we have:

Em Sin(t= - n d ((m Cos(t)

dt

Em Sin(t=n ( (m Sin(t

Em=2(f n (m

(where ( = 2(f)

(2 E=2(f n (m

(where E=rms value = 1 Em)

(2

E=2 x 3.14 f n (m

(2

=4.44 (m n f volts

=4.44 Bm A n f

(where Bm A=maximum flux density)

Thus if Ep is the voltage applied to the primary, np the number of the turns in

the primary winding, then:

Ep=4.44 Bm A np f

2

This flux produced by voltage Ep links with the secondary winding of ns turns and similarly produces a voltage, i.e.

Es=4.44 Bm A ns f

3

Dividing eqn. 2 by 3 we have:

Ep=4.44 Bm A np fEs

4.44 Bm A ns f

Ep=np

4

Es

ns

There is also a relationship between current and the flux, which is given by:

((nI l

where n=number of turns

l=the length of the magnetic circuit

Thus if the secondary winding delivers a current Is to the load, then a flux (s is produced which is given by:

(s(ns Is

5

l

Thus flux (s links with the primary winding and causes a primary current Ip to be drawn from the source such that:

(p=np Ip

6

l

Equating 6 and 5 we have:

np Ip=ns Is l

l

ornp Ip=ns Is

Ip=nsIs

np

orIs=np

7

Ip

ns

Thus combining eqns. 4 and 7 we have:

Ep=np=IsEs

ns

Ip

This is the equation of an Ideal Transformer.

But in practice if Ip' is the primary current then

Ip=Ip' (primary load current) - Io

where Io is the primary no load current

So thatIp Np=Is Ns

orNp=Is

Ns

Ip

Similarly the secondary load voltage Vs is given by:

Vs=Es - (IsRs + IsXs)

where Es=secondary induced e.m.f

(IsRs + IsXs)=voltage drop due to secondary load current in

secondary windings.

The voltage Es is transformed by primary voltage Ep and

Ep=NpEs

Ns

But the primary applied voltage Vp is given by:

Vp=Ep + (IpRp + IpXp)

where IpRp + IpXp=voltage drop due to primary load current in primary

windings

Hence

Vp=EpVs

Es

And

Vp=NpVs

Ns

The above relationships are explained by the phasor and circuit diagrams shown below

3.0Three-phase unit versus single-phase units:Since the transmission system is 3-phase, transformers may be built as 3-phase single units or as three single-phase units into delta and star combinations or groups.

3.1Advantages of 3 phase units

They occupy less space

No extra support equipment is required to form a 3-phase Delta or Star

connection.

They are cheaper

They can be transported from factory as a compact unit, erected and

commissioned at site quickly

Compact on-load tap changing (OLTC) gear can be provided as a built in unit.

3.2 Disadvantages of 3 phase units Problem of transportation in case of large capacity units weighing more than 100 tons.

Takes time in assembling, erecting and commissioning if parts are dismantled and sent to site.

The cost of one spare 3-phase transformer is more.

Change of connections from star to delta or vice-versa cannot be done.

If reconditioning is undertaken then the complete unit has to be taken out of service and this becomes a problem if no spare capacity is available.

3.3Advantages of Single-Phase Units The cost of a spare transformer is the cost of a single-phase unit, which is comparatively very much less than the cost of a complete spare 3-phase unit.

They can be transported to site as completely assembled units and commissioned quickly.

Reconditioning can be undertaken on individual units with a minimum outage time.

It is possible to obtain different possible pairs of connections between the primary and secondary.

3.4Disadvantages of Single-Phase Units

They occupy more space

They require additional support structure to form 3-phase connections.

Expenditure on civil engineering works is more

The problem of providing on-load tap changing gear and even if provided the cost of providing tap changing gear on each unit works out costlier by at least 50% when compared to a compact unit in a 3-phase transformer.

3.5Considering all the above, there is little argument in favour of the adoption of single-phase units as compared to 3-phase units. Single-phase units are the only choice where 3-phase units cannot be transported because of their weight and dimensions and also if there are no facilities at site for the assembly, preparation and commissioning of the 3-phase units.

4.0Types of Transformers

This is dealt with in reference to units normally installed in a power utility like NEPA.

4.1Power TransformersThese are transformers of high rating of generally not less than 5MVA and 33KV and the rating also increases with the voltage rating. They may be of the step-up type installed at generating stations or of the step-down type installed at substations. They have a high utilisation factor, which means that they are arranged to work at a constant load equal to their rating. Hence their maximum efficiency is designed to be at or near full load. Such power transformers installed in substations are provided with OLTC gear to regulate the voltage to be within permissible limits during peak load and off peak load hours.

However, generator step-up power transformers are provided with only off circuit taps.

4.2Distribution TransformersThese are transformers installed in H.V. distribution feeders to meet consumer voltage requirements. These are generally rated at 11KV and have a rating not exceeding 1000KVA. These transformers are characterised by an intermittent variable load, which is usually considerably less than the full load rating. They are therefore designed to have their maximum efficiency at between half and three quarter of full load. These transformers are not provided with any OLTC gear but with only off circuit taps.

4.3Auto TransformersAn Auto Transformer is a transformer with a common winding for both primary and secondary. They are used in place of two winding power transformers where the ratio of transformation does not exceed 2 as they are cheaper than two winding transformers such as in a 132KV/66KV system or 66KV/33KV system.

They are used in distribution systems for improvement of voltage by boosting or bucking of supply voltage by a small amount. Typical connections of their use are shown below:

4.4Instrument TransformersThis is dealt with exhaustively in a separate chapter.

5.0Three Phase Transformer connections or Vector groupThree phase transformers are divided into four groups depending upon the phase displacement between the primary and secondary terminals. These groups are:

1. No phase displacement (0o)

2. 180o phase displacement

3. - 30o phase displacement

4. + 30o phase displacement

These vector groups; their symbols and connections are shown in the next page

6.0Parallel Operation of Transformers6.1The following conditions must be strictly observed in order that 3-phase transformers may operate in parallel.

(a) The secondaries must have the same phase sequence or the same phase rotation.

(b) All corresponding secondary line voltages must be in phase.

(c) The same inherent phase angle difference between primary and secondary terminals.

(d) Same polarity.

(e) The secondaries must give the same magnitude of line voltages.

In addition, it is desirable that:

(f) The impedances of each transformer, referred to its own rating should be the same, i.e. each transformer should have the same percentage or per unit resistance and reactance.

6.2If conditions (a) to (e) are not complied with, the secondaries will simply short-circuit one another and no output will be possible.

6.3 If condition (f) is not complied with, the transformers will not share the total load in proportion to their ratings and one transformer will become over-loaded before the total output reaches the sum of the individual ratings. It is difficult to ensure that transformers in parallel have identical per unit impedance and this affects the load sharing.

6.4 It follows from the vector group connections indicated in paragraph 5.0, that if a pair of 3-phase transformers belong to the same group provided conditions (a) to (e) are fulfilled, then they can be paralleled with each other by connecting together terminals which correspond both physically and alphabetically. Thus, taking the case of two, 3-phase transformers belonging to vector group 1 with vector symbols Yy 0 and Dd 0, then these can be operated in parallel by connecting the terminals A21, B21 and C21 of the of the first transformer with terminals A21, B21 and C21 of the second transformer and similarly by connecting terminals a21, b21 and c21 of the first transformer with the second.

6.5Sometimes it may be required to operate a 3-phase transformer belonging to one group with another 3-phase transformer belonging to another group. This is only possible with groups 3 and 4 by interchanging the external connections. The -30o phase shift can be corrected to +30o and vice-versa by interchanging the external primary connection of any one of the two transformers. However, this is not possible with groups 1 and 2 or with groups 1 or 2 with 3 and 4.

6.6Phase shift in Delta-Star/Star-Delta Transformations(Vector groups 3 and 4)From triangle ANC, we have:

VA = Va___

Sin 1200

Sin 300

VA=Va Sin 1200 Sin 300=Va Cos 300 Sin 300 = Va___ =Vatan 30o

1(3

VA=(3 Va

Also VA is displaced from Va by 30o as shown

Similarly for the other three phases as follows:

Combining the three phasor diagrams 1, 2 and 3 we have:

Similarly it can be shown for DY 1 group as follows:

The above phase shift can also be explained as follows with reference to DY 11

Delta voltage VA transformed to secondary star voltage Va is given by:

VA=VAB - VCA=(3 Va

Va= VAB - VCA (3

= 1 [VAB - VAB ((240o)]

(3

= 1 [VAB - VAB (- 1 - j (3)]

(3

2 2

= 1 [VAB (1 + + j (3)

(3

2

= 1 VAB (1 + + j (3)

(3

2

= 1 VAB (3 + j (3)

(3 2 2

=(3 VAB (3 + j (3)

3 2 2

= 1 VAB [3(3 + j 3]

3 2 2

=VAB ((3 + j )

2

=VAB ( 30oSimilarly it can be shown for the other phases and vector group DY 1.

6.7Parallel Operation of DY 11 and DY 1 TransformersThe parallel operation of these transformers is done by changing the primary connections to any one of the two transformers as shown.

7.0Procedures in Parallel Operation7.1While paralleling two transformers the following checks are to be conducted:

(a) Measurement of terminal voltages of each transformer - done individually.

(b) Checking the phase sequence of each transformer individually.

(c) Phasing out the terminal voltage between each of the phases of the two transformers.

7.2The following methods are employed for carrying out the above checks

(a) By the use of phasing sticks.

(b) By the use of an external low voltage supply.

(c) By the use of voltage transformers.

7.3Using Phasing Sticks(a) Phasing sticks are high voltage insulated sticks with built in condensers to reduce the voltage to an acceptable value as can be measured by normal indicating instruments.

(b) These sticks are available in ratings of 5 to 33KV. They are also used to indicate if a line is alive or not.

(c) In the diagram shown below A1, B1, C1 and A2, B2, C2 are the three phase secondaries of two transformers 1 and 2 to be paralleled. From the same supply both the transformers are energised keeping the CB or switch open

(d) Three sticks are used to determine the phase sequence. These sticks are labelled (1), (2) and (3) if no colour or other distinguishing marks are available. Two sticks are used to measure the voltages.

(e) The individual voltages are measured and recorded as follows by connecting a voltmeter to the low voltage end of the two phasing sticks (1) and (2).

Phasing Stick connection line endTerminal at which voltage is measuredMagnitude of voltage measured

Stick 1

A1

B1

C1Stick 2

B1

C1

A1A1 B1

B1 C1

C1 A1Say 110V

110V

110V

A2

B2

C2B2

C2

A2A2 B2

B2 C2

C2 A2110V

110V

110V

Adjust the voltage taps of any one of the two transformers if the voltages of transformer (1) are different from those of transformer (2).

(f) The next step is to determine the phase sequence. A phase sequence meter is connected to the low voltage end of the three phasing sticks such that terminals R, Y, B of phase sequence meter are connected to sticks (1), (2) and (3) respectively. The line ends of sticks (1), (2) and (3) are held to terminals A1, B1, and C1 and the phase rotation observed and recorded as positive if anticlockwise and negative if clockwise. Similarly, the phase rotation is observed by holding sticks (1), (2) and (3) to terminals A2, B2 and C2. The phase sequence should be the same in both cases and if not; change any two of the primary connections of any one of the two transformers. Repeat the check and observe phase sequence to be the same.

(g) The last step is to phase out the two supply voltages. Stick (1) is held to source A1 and stick (2) is held to source terminals A2, B2, and C2 in succession and the voltages are recorded as follows:

Phasing Stick connection line endTerminal at which voltage is measuredMagnitude of voltage measured

Stick 1Stick 2

A1

B1

C1A2

B2

C2

A2

B2

C2

A2

B2

C2A1 A2

A1 B1

A1 C2

B1 A2

B1 B2

B1 C2

C1 A2

C1 B2

C1 C20

190

190

190

0

190

190

190

0

From the above it indicates that terminals A1, B1 and C1 correspond to terminals A2, B2, C2 and the CB or switch can now be safely closed to parallel the two sources. However, during the above test, if A1 - B2, B1 - C2 and C1 - A2 show zero voltages as against A1 - A2, B1 - B2 and C1 - C2 respectively then, the phases B2, C2, and A2 must be paralleled with A1, B1, and C1 respectively by interchanging the secondary terminals.

7.4By the method of an external supply source(a) This method is employed where phasing sticks are not available and also if V.T.s are not available.

(b) The supply used is generally 400 volts, 3-phase supply from which both the transformers are energised keeping the CB or paralleling switch open.

(c) Checks as mentioned in paragragh.7.3 (e), (f) and (g) are conducted for paralleling.

7.5By the use of Voltage Transformers(a) This is by far the method always employed in 330KV, 132KV and other substations for paralleling of transformers and for paralleling of two different sources of supply.

(b) Two sets of V.T.s are essentially required for this method. The checks are explained with reference to the diagram appended below.

(c) Transformer 1 is energised by closing breaker 52HT1 and keeping 52LT1, 52LT2 and 52HT2 open. The phase sequence and voltages at the secondary of VT1 are measured and recorded as stated in paragraph 7.3 (e) and (f). Next 52LT2 is closed to energise VT2. The phase sequence and voltages at secondary of VT2 are measured and recorded as per paragraph 7.3 (e) and (f). The V.T. secondaries of VT1 and VT2 are also phased out as per paragraph 7.3 (g).

This test is to ensure that both the V.T.s have the same polarity, connecting secondary voltages, ratio and phase sequence. Breakers 52LT2, 52LT1 and 52HT1 are now opened out.

(d) The test as per paragraph (c) above is repeated for transformer (2) by closing 52HT2 and keeping 52LT2, 52LT1 and 52HT1 open. This test is to ensure again that both the V.T.s have the same polarity, connections, secondary voltages, ratio and phase sequence.

(e) If the V.T.s have a difference in phase sequence, polarity etc., then these have to be suitably corrected and tests (c) and (d) repeated.

(f) The last step is phasing out the two secondary voltages. For this test, breakers 52LT1 and 52LT2 are kept open. Both the transformers are energised through breakers 52HT1 and 52HT2 and the voltages phased out through the secondaries of the two VTs as enumerated in paragraph 7.3 (g).

(g) If there is a duplicate bus system provided with bus V.T.s for each bus, then each bus is charged from the secondary of each transformer with the bus coupler breaker open for conducting the necessary checks before paralleling.

8.0Case studies on paralleling of Transformers

8.1Paralleling of Transformers of unequal ratings and same percentage impedances

Data

(KVA) 1 -Rating of Transformer No.1

(KVA) 2 -Rating of Transformer No.2

Z1

-% impedance of Transformer No.1

Z2

-% impedance of Transformer No.2

It

-Total load current

I1

-Load current shared by Transformer No.1

I2

-Load current shared by Transformer No.2

Basically the problem is one of two impedance connected in parallel as shown.

Here:

V=I1 Z1=I2 Z2

1

And:

It=I1 + I2From eqn. (1)

I1=I2 Z2

Z1

It=I2 Z2 + I2 Z1

=I2 (Z2 + Z1) Z1

Or I2=It (Z1)____

2

(Z1 + Z2)

Similarly I1=It (Z2)___

3

(Z1 + Z2)

Multiply eqn. (2) on both sides by V the secondary load voltage

We now have

VI2=VIt (Z1) (Z1 + Z2)

Or VI2 = VIt (Z1)___ 1000

1000 (Z1 + Z2)

(KVA) 2=(KVA) t (Z1)___ (Z1 + Z2)

Similarly (KVA) 1=(KVA) t (Z2)___ (Z1 + Z2)

8.2 Problem: To find the load shared by 2Nos. of 132/33KV Transformers of rating 15 MVA and 25 MVA with % impedances of 10% and 6% respectively.

Total load current = 700 A

Transformer No.1 - 15MVA, 132/33KV; 10% impedance

Ifl=15 x 103 =262.5 A

(3 x 33

Transformer No.2 - 25MVA, 132/33KV; 6% Impedance

Ifl=25 x 103 =437.5 A

(3 x 33

Assume MVA base = 100.

The per unit impedances of the transformers is given by:

Zpu (1)=0.1 x 10015

=0.67 p.u

Zpu (2)=0.06 x 100 25

=0.24 p.u

Load current shared by Transformer No.1

I1=It (Z2)___

(Z1 + Z2)

=700 (0.24)____ 0.67 + 0.24

=700 x 0.240.91

=184.6A

Load current shared by Transformer No.2

I2=It (Z1)__ (Z1 + Z2)

=700 (0.67) 0.91

=515.4A

It can be observed that transformer No.2 is already overloaded while transformer No.1 is lightly loaded.

This shows that with unequal % impedances, the load will not be shared in proportion to their ratings.

8.3Paralleling of Transformers with unequal % ImpedancesProblem: To find the load shared by 2 Nos. of 132/33KV 15 MVA Transformer of equal rating but with unequal % impedances of 10% and 10.2%.

Total load current = 500 A

Load current shared by Transformer No.1 (Impedance 10%)

=500 x (10.2)_____(10 + 10.2)

=500 x 10.2

20.2

=252.5 A

Load current shared by Transformer No.2 (Impedance 10.2%)

=500 x 10.020.2

=247.5 A

This shows that the transformer with higher impedance shares less load than the transformer with lower impedance. In such a case, the loading should be such as not to exceed the full load current.

8.4Paralleling of Transformers with unequal secondary voltages

LetE1=secondary phase voltage of Transformer No. (1)

E2=secondary phase voltage of Transformer No. (2)

Z1=impedance of Transformer No. (1)

Z2=impedance of Transformer No. (2)

The unequal secondary voltages will cause a circulating current Ic to flow.

The magnitude of this current is given by the equation:

Ic=E1 - E2_(E1 > E2)

Z1 + Z2

The current in transformer (1) will be (I1 + Ic) and that in transformer (2) will be (I2 - Ic). Since Z1, Z2 are small in magnitude, the difference (E1 - E2) must also be small as otherwise a large circulating current will flow overloading the transformers.

Problem - To find the load shared by 2 Nos. of 5MVA; 33/11KV transformers of equal % impedance of 6% but with unequal secondary voltages of 11.2KV and 11.0KV. Total load current = 500 A

Secondary full load current of each transformer will be:

= 5 x 106 __(3 x 11 x 103

Ifl=262.5 A

Impedance Z of each transformer will be:

=Vph x % ZIfl x 100

=11000 x 6

(3________262.5 x 100

Z1 = Z2 = 1.45 ohms

Ic =E1 - E2Z1 + Z2

=(11.2 - 11.0) x 103 (3__

1.45 + 1.45

=39.8 A

Current in transformer 1=I1 + Ic=It (Z2) + Ic (Z1 + Z2)

=500 (6) + 39.8

12

=250 + 39.8

=289.8 A

This is greater than the full load current of 262.5 A. Hence it is not safe to operate the two transformers in parallel with unequal secondary voltages. But the transformers may be operated in parallel provided that the current in each transformer does not exceed the full load current.

9.0Three Winding Power Transformers9.1In a large EHV substation there will be at least three high voltage systems from the low-tension auxiliary supplies. In some substations there may be even four or five high voltage systems. Although transformers with four high voltage windings are being manufactured, such transformers are not extensively used, because there is no advantage in having four different voltage systems in the same tank as the risk of a fault on any one voltage system involves all the voltage systems.

9.2An example of an EHV substation having three different voltages is a 330KV substation with voltages at 330KV, 132KV and 11KV.

9.3A comparison is now made as whether to have two winding transformers of 330/132KV and 132/11KV or three winding transformers of 330/132/11KV in an EHV substation with three voltages. The 11KV load in such a substation is to meet the local loads around the substation and also for the requirements of the station auxiliary supplies. This load may be around 10 to 15MVA.

The two schemes are shown by single line diagrams as follows:

Scheme (A)

Scheme (B)

Two winding transformers

Three winding transformers

9.4Comparing scheme (B) with scheme (A) we have the following merits and demerits

Merits

(a) The number of transformers, circuit breakers, CTs, isolators and control panels is reduced to a minimum. There is therefore a considerable saving in the cost of equipment required.

(b) There is considerable saving in the cost of civil engineering and structural works because of the fewer equipment.

(c) The layout is simple and occupies less space because of the fewer equipment

and operation is also simple.

(d) There is saving in energy because of the reduced transformation losses.

(e) Besides, it is inevitable to provide a third winding in a star-star connected power transformer. This third winding in such transformers is also called a `Stabilizing Winding' or Tertiary Winding. This winding is connected in a closed delta to provide a circulating path for the third harmonic voltages and zero sequence currents or ground fault currents.

It is pertinent to note here that a star-star connection is almost always resorted to in the case of EHV transformers of 132KV and above such as in 330/132KV transformers. The reason being that the cost of such a transformer is cheaper because the windings need be insulated for only 1/(3 times of the line voltage instead of for the full line voltage of (3 times the star voltage with a delta winding. Such a closed delta winding can be made use of for the third voltage, without the necessity of having a separate transformer.

Demerits(a) The main disadvantage is the increased fault level at 11KV because the voltage is directly transformed from 330 to 11KV. Hence 11KV switchgear of adequately higher rupturing capacity will have to be installed.

The cost of such switchgear may be much more than that of such switchgear installed in the secondary of a 132/11KV transformer.

(b) Since the third winding is a closed delta, an artificial neutral has to be necessarily

created by the use of earthing transformers. This is a disadvantage as it adds to the initial cost.

(c) The other disadvantage is that the units are exposed directly to the short circuit stresses because of faults on 11KV lines. The 11KV overhead networks, particularly if carried into rural areas are quite long and extensive. These lines are carried on pin insulators and are therefore susceptible to frequent faults. Such frequent faults, stress the windings and reduces the life of the transformer. If the 11KV winding feeds an urban network through an underground cable system, then this arrangement would prove to be the best and the fear of the transformer being exposed to short circuit stresses is not there.

(d) The capacity of the third winding is generally limited to 1/3rd of the capacity of the main transformer. Hence if there is a rapid increase in the growth of the 11KV load, augmentation of the 11KV capacity to meet this load becomes a problem unless another two winding 132/11KV transformer is added at the substation. This creates problems in load sharing and parallel operation because the impedance of the third delta winding is very much low when compared to the impedance of a similar voltage in a two winding transformer.

10.0Cooling of transformers and Cost Comparison of the cooling methods

10.1The B.S.S. recognises three cooling methods for transformers namely Air, Mineral oil and Synthetic liquid. Since almost all of the power transformers are mineral oil cooled, the method of cooling by mineral oil is only dealt with here. The methods of cooling with oil immersed transformer is classified as follows:

10.2Oil Immersed Natural Cooled - Type ONCooling is by circulation of oil under natural thermal heat only. In large transformers the surface area is not sufficient for dissipation of heat by radiation. As such additional surface area is provided for the cooling fins; also called radiators.

10.3Oil Immersed Air Blast - Type OBCooling is similar to type ON except that air circulation is done by external fans mounted below the radiators.

The advantage is the reduction in the size of the transformer for the same rating and consequently a saving in the cost.

10.4Oil Immersed Water Cooled - Type OWAn internal cooling coil or tubing is mounted through which water is circulated. This requires a free and abundant supply of water. Cooling is by convection.

10.5Forced Oil Natural Air Cooled - Type OFN

It is similar to type ON except that a cooling pump is installed in the oil circuit for better circulation of oil.

10.6Forced Oil Air Blast Cooled - Type OFBIt is a combination of type OB and type OFN.

10.7Forced Oil Water Cooled - Type OFWIt is similar to type OW except that a cooling pump is added in the oil circuit for forced oil circulation into a heat exchanger in which water is allowed to flow.

10.8It must be noted here that transformers with type OFB and type OFW cooling will carry no load if air or water supply is cut off.

109It is quite common to select large power transformers of 15MVA and above with two or more systems of cooling namely ON/OFB or ON/OB or ON/OB/OFB. These determine the type of cooling and permissible loading and as soon as the loading exceeds a preset value, fans/pumps are switched on automatically. An indication of the operation of the fans/pumps is given in the Transformer control panel. The rating of such a transformer with ON/OB cooling will be written as for example 45/60MVA, which means that up to 45 MVA load, the fans will not be working. The fans will be switched on automatically when the load exceeds 45MVA.

10.10The type of cooling has a bearing on the cost of the Transformer. The approximate relationship on the cost with different methods of cooling is mentioned below.

Type of coolingONOFNON/OBON/OFBOFBOFW

% Cost

100 95 90

85

80 75

10.11The ON cooling is the simplest method of cooling with no fans or pumps or auxiliary motors. It is used in all distribution transformers and in power transformers up to 15MVA. The saving in cost in power transformers of up to 15MVA in changing the cooling from ON to other types is negligible.

10.12The OFW cooling is only employed in transformers installed at hydroelectric power stations where an abundant supply of cooling water is assured. But at other stations, special arrangements have to be made for water supply and disposal of hot water, which may increase the cost of the transformer.

11.0Requirements and characteristics of insulating oil

11.1The mineral oil in transformers is used not only as an insulating medium but also as a heat-transferring medium to dissipate the heat produced in the windings and the core. The life of a transformer is dependent on the quality of the insulating oil and as such it is very necessary to use insulating oil of a high quality or standard.

11.2The essential qualities required of the insulating oil are:

(a) High dielectric strength

(b) Permits good transfer of heat

(c) Low specific gravity - suspended particles settle at bottom of tank rapidly.

(d) Low viscosity - better cooling rate

(e) Low pour point

(f) High flash point - prevents vaporisation of oil

(g) Chemical stability

11.3 There are various national and international standards on characteristics of oil. These are characterised by:

(a) Sludge value (Max) - %

(b) Acidity after oxidation - mg KOH/g

(c) Flash point (min) - oF or oC

(d) Viscosity at 70 oF or 21.1oC (Max) centistakes or ... secs.Redwood

(e) Pour point - oF or oC

(f) Electric strength - KV rms for 1 minute

(g) Acidity Neutralisation value

Total - mg KOH/g

Inorganic

(h) Saponification value (max) - mg KOH/g

(i) Copper discoloration - +ve or ve

(j) Specific Gravity

(k) Volume resistivity - ohms/cm3

(l) Water content ppm

(m) Tan delta or loss angle

12.0Tests on TransformersThese are governed by various national and international standards. Most of these standards recommend the following tests.

12.1Routine tests(a) Measurement of winding resistance

(b) Ratio, polarity and phase relationship

(c) Impedance voltage

(d) No load losses and no load current.

(e) Load losses

(f) Insulation resistance

(g) Separate source voltage withstand test

12.2Type tests(a) Impulse voltage withstand test

(b) Temperature rise test

13.0Field tests and CommissioningThese tests are conducted at the time of commissioning on a completely assembled transformer after necessary drying out of the winding core and filtering of oil.

(a) Measurement of insulation resistance and Polarisation Index

(b) Ratio test on all the tap positions

(c) Open circuit test, no load current and no load losses

(d) Short circuit test and load losses

(e) Oil test

(f) Operation of tap changer manually and electrically on local and remote

(g) Operation of cooling fans/pumps and motors

(h) Measurement of earth resistance of transformer grounds namely; neutral and body

(i) Operation of Bucholtz relay for alarm/tripping

(j) Measurement of loss angle of EHV bushings

14.0Maintenance of Power Transformers

Normally, every manufacturer lists out the maintenance procedures to be followed during the lifetime of a transformer in service. However, the commonly recommended measures in almost all power transformers are the following:

Hourly: Recording readings of:

(i) Load current

(ii) Load KW

(iii) Temperature

(iv) Voltage

Half Yearly/Yearly:

(i) Insulation resistance

(ii) Oil test for breakdown voltage, water content and acidity.

Periodically: Changing the silica gel when the colour has changed from blue to pink.

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