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For overhead lines operating at very high voltage (33 KV andabove) use of number of discs connected in series, throughmetal link, is made. The whole unit formed by connecting anumber of discs in series is known as string of insulator. The

line connecting a number of the string and the top disc isconnected to the cross arm of the pole or tower given in fig.

the number of discs connected in series in an insulatorsstring depends upon the line voltage (higher the line

operating voltage, the larger is the number of discs string, asgiven below in tabular form).required for the insulators


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Operating voltage in KV No. of discs in

suspension

Assembly

No. of discs in in tension or dead end assembly

11 1 1

33 2 3

66 5 6

132 9 10

220 14 15

400 21 22

.

Arrangement of Insulators string


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The number of discs indicated in above table is actually the usual number used.However , in case of transmission lines operating at 66 k v or more, one discs lessthan the number indicated is used on about 8 suspension structures near the sub

station . This is accomplished so that in the event of a lightning surge appearingon the line and prevent the surge from traveling to the sub station thus safeguarding the equipment there.It is found that the voltage near the line unit (unitnearest to the line conductor) has the max. Voltage and the figure progressivelydecreasing as the unit nearest the cross arm is approached. This may be moreclearly explained with the help of equivalent circuit of an string.

Each string insulator unit behaves like a capacitor having a dielectric mediumb/w the two metallic parts (via pin and cap) . The capacitance due to these metalplates is mutual capacitance. Further there is also capacitance b/w metal fitting ofeach unit and the earthed pole or tower this is known as shunt capacitance. If thesuspension string is situated very far from the metal work then the shuntcapacitance will become very small as compared to the mutual inductance, thenthe charging current would have been the same through all of the discs beingconnected in series and consequently the voltage across individual units wouldhave been same i.e. applied voltage is divided equally by the number of units inthe string. However practically this is not possible due to nearness of the tower,cross arm and the line. These shunt capacitance is also known as stray capacitance.Due to shunt capacitance the charging current will be different and thus thevoltage across each unit will be different. This unequal potential distribution and

this is expressed as in terms of string efficiency.


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The ratio of voltage across the whole string and the product of the number ofunits and voltage across the unit nearest to the line conductor is known as the

string efficiency i.e.

String efficiency = voltage across the string / n * voltage across the lowest mostunit

Where n is the number of units in the string

Let the mutual capacitance b/w the links be C and shunt capacitance b/w links and

earth be C1, voltage across the first unit (nearest the cross arm) be V1, voltage acrossthe second unit be V2 , voltage across the third unit be V3 , voltage across the fourthunit (nearest to the line conductor) be V4 and voltage b/w conductor and earth volts.


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Applying KCL to node A, we get

I2 = I1 + i1 Or CV2 = CV1 + C1V1 Or CV2 = CV1 + KCV1 V2 = V1 (1 +K)

Applying KCL to node B we get I3= I2 + i2 Or CV3= CV2 + KC(V1+V2)

Voltage across the second capacitanceC1 from the top = V1 + V 2

Or CV3=CV2+K(V1+V2)=KV1+V2(1+K)

Or V3=KV1+V1(1+K)(1+K) From eqution(3.2) V2=V1(1+K)

Or V3=V1(1+3K+K2)

Applying krichoffs first law to nodeC,we get

I4=I3+i3


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Or CV4 = CV3+ C1(V1+V2+V3) Therefore Voltage across the third shunt capacitance C1 from

the top = V1+V2+V3 Or CV4 = CV1(1+3K+k2)+ KC[V1+V1(1+K)+V1(1+3K+K

2)] Or V4 = V1(1+6K+5k

2+K3) Finally voltage b/w line conductor and earth V = V1+V2+V3+V4 = V1 + V1 (1 +K) + V1 (1 +3K + K

2 ) + V1 (1+6K +5K2 + K3)

V1/1 = V2/ (1+K) + V3/ (1+3K+K2) = V4/(1+6K+ 5K

2+ K3) =V/ (4+10K+6K2 + K3)

The greatest voltage will obviously V4 which is given as V4 = (1+6K+5K

2+K3) V (4 +10K + 6K2 + K3 )

and %age string efficiency = V / (n V4) * 10 0 = V * 100/ 4 * (1+6K+5K2 + K3/ 4 + 10K+ 6K2 + K3) V since n = 4 and flash over voltage of one unit = greatest voltage across any unit i.e. V4


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When the number of insulators in the strings is large itbecomes laborious to work out voltage distributionacross each unit, for such standard formula is used.

In this case n is the no. of string unit, V is the max.voltage across the string and all symbols have samemeaning.

The voltage across the mth unit (counted from thetop) is given as

Vm = V [2 sin h (K /2) cos h [ ( m 1/2) K 1/2] sinh(n K 1/2)

Potential adjacent to the line conductor

Vn = V [2sin h(K

2) cos h [(n 1/2) ] K1/2

] sinh(n K 1/2)


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Graphical plot of how voltages aredistributed across the units of an insulatorsstring is given in the figure. The followingpoints may be noted.

The unit nearest to the line conductor is

under maximum electrical stress and islikely to be punctured while the nearest tothe cross arm is under minimum electricalstress.

The voltage distribution across variousunits depends upon the value of K andnumber of discs contained in the string.

The greater the value of K more inequalityin distribution of voltage and lesserefficiency. Non uniform distribution ofvoltage also increases with increase innumber of discs in the string.

When the insulators are wet the value of C(mutual inductance) increases and C1

remains constant so value of K decreasesand string efficiency increases. The value Klow for longer string and high for shorterstring s. in practice K varies from 0.1 to0.1667


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By - satyam jain

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