insulation coordination

5/28/2018 Insulation Coordination

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Insulation Coordination

Insulation coordination consists of selecting insulation of various lines and equipment

that have to be interconnected into a system for desired operational requirement. The

system must be reliable and economical.

In high lightning-prone areas or in systems with heavy switching-surge conditions, the

selection of insulation levels will be different from areas with little or no lightning and

with shorter lines.

Normally, insulation systems are designed in a system for no flashovers, or if flashovers

cannot be prevented such flashovers should be restricted to places where damage is not

done, such as air gaps or in gap-type arresters. The flash-over should not disturb normal

system operation and must occur in resalable insulation structures. The over voltages that

can cause damage are due to external origin, namely lightning, and operation of thesystem itself which are at power frequency, earth faults, and switching operations. We

will consider here the insulation co-ordination principles based on lightning. These

insulation levels are known as Basic Impulse Insulation Levels or BIL. Those based on

switching-surge requirements are known as Switching Impulse Levels or SIL.

The lightning arrester is the foundation of protection in e.h.v. ranges, which is selected

for both lightning and switching-surge duty. It is usually of the magnetic blow-out

(current limiting) gap type, or in recent years the gapless ZnO type.

Insulation Coordination:

The process of bringing the insulation strengths of electrical equipment and buses intothe proper relationship with expected over voltages and with the characteristics of the

insulating media and surge protective devices to obtain an acceptable risk of failure.Insulation coordination means the correlation of the insulation of the various

equipments in a power system to the insulation of the protective devices used for the

protection of those equipments against over voltages. In a power system various

equipments like transformers, circuit breakers, bus supports etc. have different

breakdown voltages and hence the volt-time characteristics. In order that all theequipments should be properly protected it is desired that the insulation of the various

protective devices must be properly coordinated.

The basic concept of insulation coordination is illustrated in Fig. 7.27. Curve A is thevolt-time curve of the protective device and B the volt-time curve of the equipment to be

protected. Fig. 7.27 shows the desired positions of the volt-time curves of the protecting

device and the equipment to be protected. Thus, any insulation having a withstandvoltage strength in excess of the insulation strength of curve B is protected by the

protective device of curve A.


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Insulation withstand capacity of any equipment must be higher than the maximumpossible over voltage.

Whatever the insulation withstand of a line, it is not 100% reliable.

Insulation Coordination defines the overvoltage stress and insulation requirementfrom equipment to equipment basis.

It also rates timing coordination and nature of o/v.Basic lightning impulse insulation level (BIL):

The electrical strength of insulation expressed in terms of the crest value of a standardlightning impulse under standard atmospheric conditions.

Basic switching impulse insulation level (BSL):

The electrical strength of insulation expressed in terms of the crest value of a standardswitching impulse.

Necessity of Insulation Coordination:

To ensure the reliability & continuity of service To minimize the number of failures due to over voltages To minimize the cost of design, installation and operation

Requirements of Protective Devices:

Should not usually flash over for power frequency over voltages Volt-time characteristics of the device must lie below the withstandvoltage of the protected apparatus Should be capable of discharging high energies in surges & recover

insulation strength quickly

Should not allow power frequency follow-on current.


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Volt-Time Curve

The breakdown voltage for a particular insulation of flashover voltage for a gap isa function of both the magnitude of voltage and the time of application of the

voltage.

Volt-time curve is a graph showing the relation between the crest flashovervoltages and the time to flashover for a series of impulse applications of a given

wave shape.

Construction of Volt-Time Curve:

o Waves of the same shape but of different peak values are applied to theinsulation whose volt-time curve is required.

o If flashover occurs on the front of the wave, the flashover point gives onepoint on the volt-time curve.

o The other possibility is that the flashover occurs just at the peak value ofthe wave; this gives another point on the V-T curve.

o The third possibility is that the flashover occurs on the tail side of thewave.

To find the point on the V-T curve, draw a horizontal line from the peak value ofthis wave and also draw a vertical line passing through the point where the

flashover takes place

The intersection of the horizontal and vertical lines gives the point on the V-Tcurve.


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Insulation Coordination for L.A(Lightning Arrester)

L.A. plays a major role in insulation coordination L.A. is not used for power frequency o/v, but used for transient o/v Rated voltage of L.A. is maximum temporary o/v (rms)


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Temporary over voltage

VT(rms) = 0.8 x 3 x Vphase

= 1.38 x Vphase

For L.A.:

Temporary o/v is taken as 1pu Switching o/v is taken as 1.6 times of temporary o/v Impulse o/v is taken as 1.8 times of temporary o/v

BIL of L.A.>=Vb

>=VT(rms) x 2 x 1.8 x Margin

(LA)BIL

(min)= VT(rms)x 2 x 1.8 x Margin

(LA)BIL (min)/(LA rating)= 2x1.8x1.15

=2.9217

=Protective Ratio(PR)

Protective ratio may vary from 2.75 t0 3.5


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If PR is less than 2.75, L.A. may operates at temporary overvoltages.

If PR is higher than 2.75 LA may operate for impulse overvoltages that means equipment

insulation level is increased generally which is most costly.

Insulation Coordination for Transformer

Insulation of Transformer is kept slightly greater than L.A.

BIL of Transformer = TPR x BIL of L.A.

TPR : Transformer Protection Ratio

Discharge current through L.A.

The lightning current passing through the arrester material is calculated as follows:Consider a travelling wave of voltage Vw, crest, which is accompanied by a current wave

Ion a line with surge impedance Z, Fig. 9.11. They strike an arrester whose duty is to

hold the voltage across it constant at the protective level V. Now, by using Thevenin'stheorem, with the arrester terminals open, the incident travelling wave will give a voltage

2Vpdue to total reflection. The Thevenin's impedance looking through the open arrester

terminals is equal to the surge impedanceZof the line.


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Therefore, with the arrester connected, the current through it will be

Ia= (2VwV)/ZThe maximum value the travelling-wave voltage Vpcan reach is the flashover voltage of

the line insulation. Also, it is assumed that Vpstays fairly constant at all current values

discharged by the arrester.

Example :For a 750 kV line, take Vw= 3000 kV, crest, travelling on the line and V

=1700kV. The line surge impedance is Z = 300 ohms. Calculate and discuss (a) thecurrent flowing in the line before reaching the arrester, (b) the current through the

arrester, and (c) the value of arrester resistance for this condition and verify the reflection

and refraction coefficients giving rise to the voltage and current conditions.

Solution.(a)Iw= Vw/Z= 3000/300 = 10 kiloamperes.(b)

Ia=(2Vw -V)/2Z= (60001700)/300 = 14.33 kA.

(c)The reflected current in the line is + 4.33 kA.

This gives rise to a reflected voltage of 4.33 300 = 1300 kV. Under theseconditions, the arrester resistance is

Ra= Vp/Ia= 1700 kV/14.33 kA=118.6 ohms. (120 is taken ohms).


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Example: For the above example , if an 80% arrester is used, calculate the protective

ratioNp= Vp/V.

.

Solution.For rated line-to-line voltage of 750 kV, arrester rating is

Va= 0.8 750 = 600 kV (R.M.S.).

Protective ratioNp= 1700/600 = 2.83.

Voltage Across Equipment Protected by ArresterIn the ideal case, the arrester must be located adjacent to the equipment which is usually a

large transformer or shunt reactor. In practice, however, there may be a length of the linebetween the two extending to 20 to 40 meters. This results in a slightly higher voltage

across the equipment due to repeated reflections. The high inductance of a transformer or

reactor represents nearly an open-circuit to a surge. The excess voltage experienced is

given by an empirical equation and depends on the line length and the rate of rise of thevoltage, thus:

V= (dVw/dt). (l/150) kV

Where l= length of line in meters anddVw /dt= steepness of wave front in kV/ms of the incoming wave. This can be taken as

approximately 500 kV/ms for lines with overhead ground wiresand 1000 kV/ms when a

line conductor is hit.(A line without earth wires).

Example .A transformer is connected by a length of 20 meters of line to an arrester.

The rate of rise of voltage is 700 kV/microsecond. The arrester voltage is 1700 kV.

Calculate the voltage across the transformer.

Solution.V= (700) (20/150) = 93 kV.Transformer voltage = 1700+93 kV=1793 kV, impulse crest.

Very important Numerical:

For 400 kV line having effectively grounded system with L.A. of protection ratio 3.

L.A. is put 30 m from transformer and transformer protection ratio is 1.25, surge

impedance Zc=350 . Lightning current equals to 10 kA at an arrester with peak

time of 1 s. Check whether the transformer insulation is properly co-ordinated

with the arrester or not. If not, also find maximum possible separation between

Transformer & L.A. for proper co-ordination.


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Solution:

Temporary o/v=400*0.8*1.05=336kV=LA rating

BIL of L.A. Va =PR x LA rating=3*336 kV=1008kV

BIL of Transformer =TPR x BIL of L.A. =1.25*1008 kV=1260kV

Excess voltage experienced in protected device

V=(dv/dt)*(l/150)=(di/dt)*Zc*(l/150)=700kV

Vb=Va + V=1008 kV+700 kV=1708kV

Here BIL of transformer is less than the voltage experienced by it Vb= 1708 kV

thats why it is not properly co-ordinated.

Now boundary condition for proper co-ordination is

BIL of transformer= voltage experienced by it

Vb=1260 kV

Va +(dv/dt)*(l/150)=1260

1008+(di/dt)*Zc*(l/150)=1260

(di/dt)*Zc*(l/150)=252

10*350*(l/150)=252

Therefore l=10.8 m is the maximum possible separation between L.A and transformer forproper co-ordination. Also for safe operation distance should be less than 10.8 m for safe

operation.


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Example:

In 132 kV transmission line having Zc= 400, effectively grounded system, the

proposed L.A. has the following specification:

Residual voltage Option1 Option2At discharging current 0f 10 kA 300 kV 300 kV

At discharging current 0f 20 kA 360 kV 320 kV

Residual voltage: voltage across LA when discharging current flows

i. Compute BIL withstand voltage if maximum current that could strike atthe arrester is 20 kA and 50 kA

ii. Compute the lightning maximum permissible over voltage if themaximum residual voltage of a arrester is to be limited to that of 20 kA

discharge current.

Solution:

Arrester current I=

For lightning arrester:

Residual voltage V=

For option 1:

300= k10x

360= k20x

Solving we get,

x=0.263

k=163.727

Therefore V=163.727 I0,263

For 20 kA lightning current discharge current of L.A.

L.A. withstand capacity = kIx= 163.727 * 38.10.263=429.366 kV

For 50 kA discharge current

Va=163.727 * 500.263= 458.05 kV


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BIL of LA= 163.727 * 98.8550.263= 547.987 k

Do same for option 2[ ans: for 20kA, Va=340.512 kV, for 50 kA Va= 371.22 kV]

(ii)I=

Option 1: 20=

V=4180 kV

Option 2:

V=4160 kV

insulation coordination

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