Equations-Review

Switching Overvoltages, Substations

2.8 Equations-Review

Switching Overvoltages, Substations

Table 5 Assumed Values of BSL for Standard Values of BIL for Disconnecting Switches and Bus Support Insulators

BIL, kV S , meters BSL, kV

825 ANSI 850 SEC 900 ANSI 950 IEC 1050 1175 1300 1425 1550 1675 1800 1925 ANSI 1950 IEC 2050 IEC

Chapter 5 SSFOR = - 1 [1 - F(Ze)]

2

and for an extreme value positive skew distribution of SOVs

Example 1. Estimating the SSFOR. Assume that a BSL of 850 kV is used for all apparatus in a 500 kV station. Also assume that the distribution of the SOVs can be approximated by a Gaussian distribution having an E2 of 808 kV (1.8 pu) and a o0 /E2 of 0.10. Also assume a total of 10 parallel insulations or n = 10. Also crf/CF0 = 0.07. From Eq. 4, CFOs = 934 kV, V3s = 738 kV, and V3s/E2 = 0.9134. From Table 1, Kf = 0.91. Then using Eq. 14, Ze = 2.091. From Table 1 of Chapter 3,

0.0183

SSFOR = -

2 ⁼ 0.0091 = 0.91/100 (16)

Using a computer program to perform the numerical integration, the SSFOR is 0.861 100.

To continue this example, assume now that the station is at an elevation of 15OOmeters, so that the relative air density is 6 = 0.840. Also to obtain reasonable values of SSFOR assume a BSLs of 950 kV. Thus the CFOs is 1044 kV. For this BSL, the strike distance per Eq. 6 is 2.827 meters and therefore

Therefore

Then using Eq. 14

Ze = 2.339 SSFOR = -(0.0097) 1 = 0.49/100

2 (19)

Using a computer program to perform numerical integration, the SSFOR is 0.50/100.

Switching Overvoltages, Substations

Example 2. Estimating the BSL and Clearance. Consider the same station as for Example 1, except that the station insulation is to be specified based on a SSFOR of 1/100. First assume sea level conditions.

From Tables 1 and 2, Kf = 0.91 and KG = 1.0, and therefore V3/E2 = 0.91.

Thus V3 = 735 kV, CFO = 931 kV, and the BSL = 847 kV. Thus the required BSL is 847 kV. The next highest BSL, per Table 5, is 895 kV, which is the BSL for a BIL of 11 75 kV. Thus 11 75 kV BIL is selected for the bus support insulators and the disconnecting switches. The BSL for the circuit breaker is 1300 kV, which is more than adequate. Using Eq. 13 with kg = 1.3, the clearance is 2.13 meters.

Using a computer program, the required BSL is 844 kV and the clearance is 2.12 meters. Therefore the same BIL would be selected.

Before continuing this example to include the effect of altitude, consider the requirement that the V3 of the station must be equal to or greater than that of the line. For the line assume that n = 500 and that ES& = 0.9. Therefore na = 100 and for a SSFOR of 1/100, V3/E2 must equal 0.9693, and V3 for the line is 783 kV. Since the V3 for the line, 783 kV, is larger than the V3 for the station, 735 kV, the insulation in the station must be designed for a V3 of 783 kV. Unfortunately, this is always true since for the same SSFOR the KG or Kv for the station and line are approximately equal but the Kf for the line is greater since the number of insulations for the line exceeds that for the station. However, the use of a V3 of 783 kV for the station results in a SSFOR for the station of about 2.12/1000. To expand, from Tables 1 and 2, the required V3/E2 for a SSFOR of 1/1000 is 0.9919, which produces a V3 of 801 kV, which is greater than the 783 kV as required by the line.

Using a required V3 of 783 kV results in a required CFO of 991 kV and a required BSL of 902kV. From Table 5, the next larger BSL is 965 kV, which is the BSL for a BIL of 1300 kV. Thus a BIL of 1300 kV is selected for the bus support insulators and the disconnecting switches. The circuit breaker having a BSL of

1300 kV is again more than adequate.

To determine the required clearance of strike distance, use Eq. 13, with a gap factor of 1.3 and a relative air density of 1.00. Then S = 2.31 meters.

Now assume that the station is at 1500meters. The required CFO, BSL, and V3 are the same as before except that these values are required at 1500 meters. Thus they are CFOA = 991 kV, VM = 783 kV, and BSLA = 902 kV. A first estimate of the BSLs can be made by assuming that m = 0.5 (8 = 0.840) and therefore that the BSLs = 902/8'" = 984 kV. The iterations are shown in Table 6. Equation 6 is used to find S , then the CFOs is the BSLs/0.91. Go, m, and 5'" can now be found. Then since BSLA = 902kV, from Eq. 10, the BSLs is found. As shown, only a single calculation is necessary, since the iterated BSLs is 981 kV.

Therefore a BSLs of 980 is required. Since the circuit breaker BSL of 1300 kV is greater than this, the circuit breaker is more than acceptable. Entering Table 5 with a

Table 6 Iteration to Find the BSLs for 1500meters

BSLs S , meters CFOs Go m BSLs

172 Chapter 5

required BSL of 980 kV, the next highest is 1032 kV, which translates to a BIL of 1425 kV. If a computer program is used, the BSL required is 981 kV.

As noticed, the required circuit breaker BSL is also obtained using the equations for the post insulators. In explanation, the equations for the post insulators are used only to obtain an estimate of the exponent m, and therefore, since m is not a highly sensitive value, the estimate appears justifiable. Also note that an m of 0.5 is a crude estimate and if used above would require a BSLs of 984 kV. This is not a bad guess.

The remaining task is to determine the strike distance or clearance. This is accomplished in the same manner as in Chapter 3 except that a gap factor of 1.3 is used. From above, a CFOA of 991 kV is required. As an initial guess, with m = 0.5, S = 2.6 meters. The final value from the iterations of Table 7 is 2.69 meters. From a computer program, S = 2.68 meters, much the same value.

Some final comments on this example: In all except one case, BPA, the BIL used for all station insulation except for transformer is 1800 kV. Thus in most cases the BSL employed is much greater than required. As will be shown in Chapter 13, lightning insulation coordination requirements usually exceed the requirements for switching surges.

2.1 1 Phase-Ground Insulation Coordination with Arresters

If the required BSL exceeds the circuit breaker BSL, or the required BILIBSL of the bus support insulators is considered excessive, then line-entrance arresters may be used. As discussed previously, if arresters are used on the line side of the breaker, the requirement that the station insulation strength be equal to or greater than the line insulation strength is eliminated. The arresters essentially isolate the line from the station. As an example, if no closing resistors are used in the breaker, an estimate of the value of E2 is 2.8 pu or 1257 kV. Using the data of the example, the required BSL at 1500meters is about 1400kV, which exceeds the standard BSL of the breaker.

Thus in this case, line-entrance arresters are required.

As is discussed in Chapter 11, line-entrance arresters are sometimes used to protect the breaker when it is in the opened position. This is seldom required at EHV levels but may be necessary for lower voltage systems.

Returning to the use of arresters to mitigate the SOVs, assuming that the arrester discharge voltage is less than the maximum SOV, arresters will not only decrease the SOVs within the station but also decrease the SOVs along a portion of the line. Thus both the line and station apparatus insulation strength may be reduced. Recent failures of the closing resistors within the breaker have stimulated the consideration of the use of line-entrance arresters (or line-end arresters), and some utilities have successfully deleted the resistors and added arresters. However, closing resistors provided a superior means of reducing the SOVs along the entire line, whereas arresters only decrease the SOVs within a relatively short distance of the arrester.

Table 7 Finding the Clearance

S, meters CFOs Go m S, meters

Switching Overvoftages, Substations 173

Returning to the application of arresters, as depicted in Fig. 2 A , assume that the arrester discharge voltage EA can be represented by a single linear line having the equation

Also, per the equivalent circuit of Fig. 2B, the switching overvoltage E is

E = E A + I A Z (21)

where Z is the line surge impedance. Thus

EA = KAE

+

^{(1 - KA)Eo}

where

and

If the original SOV density function or distribution f (V) is Gaussian with a mean, po and a standard deviation on, then the distribution from En to the maximum SOV, EmA, is also a Gaussian distribution with a mean and a standard deviation G A

given by the equations

From E l , the minimum SOV, to Eo, the SOV distribution is equal to the original distribution, i.e., po and 0 0 . See Fig. 3.

1 ^A

Figure 2 Arrester characteristics. Equivalent circuit.

Chapter 5

400 500 600 700\ 800 900 kV E0=714 kV

Figure 3 Arresters alter SOV density.

Alternately, if the original SOV distribution is an extreme value positive skew distribution with the parameters u and b, then the modified distribution from En to EmA is also an extreme value with the parameters UA and

PA

given by

The switching impulse discharge voltage as published in ANSIIIEEE standards (C62.11) [2] is the discharge voltage for the discharge currents of Table 8. Ranges of this discharge voltage are listed in Table 9, and a typical characteristic based on the discharge voltage at 1.0 kA is presented in Table 10. The resultant values of RA and En are given in Table 11. As can be noted, the value of KA and therefore ^OA or

PA

is small. Also PA or UA is primarily determined by EO. The following examples illustrate the procedure.

Table 8 SI Discharge Current Used to Obtain the SI Discharge Voltage

- -

Nominal system

Arrester class voltage, kV SI current, kA

Station 3-150 0.5

15 1-325 1 .0

326-900 2.0

Intermediate 3-1 50 0.5

Switching Overvoltages, Substations

Table 9 SI Discharge Voltage in Per Unit of Crest MCOV

Arrester class Discharge voltage, kV Station

Intermediate

Table 10 Typical Discharge Voltage-Current Characteristics

Discharge voltage per unit Discharge current, kA of voltage, at 1 kA

Table 1 1 Arrester Start Voltage and Resistance in Per unit of Voltage at 1 kA per Table 6

Current range, kA R A Eo

Example 3. With Arresters. Consider the same example as used in Example 2 with the altitude at sea level but with the use of a 318-kV MCOV arrester whose SI discharge voltage is 823 kV. From Table 8, this discharge voltage is for a 2.0 kA discharge current, and therefore the discharge voltage at 1.0 kA is 794 kV; see Table 10. The first task is to select the values of RA and En so that the resultant arrester current at the voltage E2 matches the range of currents per Table 11. Selecting the range of 0.1 to 0.5 kA, RA = 113 and Eo = 714 kV. For E = E2 = 808 kV and for a line surge impedance of 350ohms, from Eq. 23, In = 0.203kA, which is in the selected range of 0.1 to 0.5 kA. For this example, since po = 642kV and oo = 80.8 kV, KA = 0.2441, oA = 19.72 kV, and = 696.4 kV. From these values the value of E2 is reduced to EZA of 737 kV and o A / E Z A = 0.0268. The resultant cumulative SOV distribution is shown in Fig. 4 illustrating the decrease in the SOVs.

Using the new values of E2 and % / E 2 for the arrester, the BSL and the clearance can be determined. From Table 1 for n = 10, Kf = 0.91. From Table 2 for a 0 0 / E 2 of 0.02 and a SSFOR of 1/100, KG = 1.05. Thus the required value of V 3 / E 2 is 0.9555,

Chapter 5

Distz with arrester

400 2 ^I 16 ^I 50 ^I 84 ^I 98 ^I

Probability, %

Figure 4 Arresters effect cumulative SOV distribution.

which in turn results in a required V3 of 0.9555(737) = 704 kV, a required CFO of 891 kV, and a required BSL of 81 1 kV assuming sea level conditions. From Table 5, selecting a BSL of 821 kV gives a BIL of 1050 kV. Therefore for the bus supports and the disconnecting switches, a BIL of 1050 kV can be specified. For the breaker, the BSL of 1300 kV is more than adequate. For the conductor-to-tower-leg gap, kg ⁼ 1.3 and the required clearance is 2.0 m. Using a PC program results in a BSL of 804 kV and a clearance of 2.00m.

Table 12 compares the results of the examples plus the use of arresters at 15OOmeters. (A PC program gives a BSL of 881 kV and a clearance of 2.33m.) Interestingly, the use of arresters at 15OOm results in a lower BIL than required at sea level when no arresters are used, 1300 vs. 1175 kV BIL and 2.31 vs. 2.33 m. As mentioned previously, the normal BIL used at a nominal system voltage of 500 kV is 1800 kV. As will be evident in Chapter 13, the BIL required from a lightning stand- point is larger than that required by switching overvoltages.

In this example, E2* was less than E2, and thus arresters were beneficial. If E2*

is equal to or greater than E2, then the design should be based on the original value of E2.

Table 12 Comparison of BSLs and Clearances

Altitude, Req'd BSL, Selected BIL, Clearance,

meters Criterion kV kV m

0 w/o considering line 844 1175 2.12

0 Considering line 902 1300 2.31

0 With arresters 904 1050 2.00

1500 Considering line 98 1 1425 2.68

1500 With arresters 881 1175 2.33

Switching Overvoltages, Substations 177

Also note that in Table 2, KG and KE vary significantly dependent on the values of 9 / E 2 or PIE2 and the design value of SSFOR. For the Gaussian distribution of SOVs, KG increases as the 0 0 / E 2 decreases. Thus the required value of V3 increases.

2.1 3 Transformer and Transformer Bushing

Almost universally, an arrester is located immediately adjacent to the transformer, either at the transformer or on the transformer bus, and therefore the arrester switching impulse discharge voltage is compared to the BSL. Since the transformer and the internal insulation of the transformer bushing is non-self-restoring, a deter- ministic insulation coordination method must be used with a minimum safety margin of 1%. Therefore

where EsI is the arrester switching impulse discharge voltage as published by the manufacturer for the discharge current as specified in Table 8. Using the previous example, where this voltage EsI was 823 kV, the required BSL is 946kV. From Table 4, the next highest transformer BSL is 975 kV, which translates to a BIL of 1175 kV. Again using Table 4 for the transformer bushing, the next highest BSL is 1050 kV, which is for a bushing BIL of 1300 kV. Thus the required BILs of the transformer and of the internal insulation of the bushing are 1175 and 1300kV, respectively. Examining Table 4, it is noted that the minimum BIL used at 500 kV is 1300 kV. Thus, considering only switching overvoltages, a BIL of 1300 kV would be specified.

For the external transformer bushing insulation, the effect of altitude must be considered. In addition, the use of a safety margin is debatable. In selecting the BSL of other self-restoring insulations in the stations, a safety margin was not considered.

It appears better to decrease the desired or design SSFOR than to use a safety margin. Thus, for sea level conditions, for the external bushing, Eq. 27 could be used without a margin. Not using a safety margin, the required BSL of the external bushing would be 823 kV, which translates to a BIL of 1175 kV. However, since the internal insulation BIL of the transformer is 1300 kV, the bushing would normally be specified as a 1300 kV BIL.

For higher altitudes, the insulation strength of the external bushing insulation decreases so that the required BSL increases. That is, the required BSL at an altitude A , BSLA, is

where BSLs is the standard value (at sea level). To obtain the value of m, the strike distance across the bushing and the CFO must be known. If the strike distance is not known, it can be estimated from the equation for a rod-plane gap. However, first the CFO is determined by Eq. 4. The strike distance is then obtained from the equation for a rod-plane gap, i.e.,

178 Chapter 5

Table 13 Finding the Bushing BSL

BSLs CFOs S Go m BSLs

997 1096 3.80 0.5763 0.271 992

992 1090 3.78 0.5774 0.272 992

where S is the strike distance in meters. Then the value of m and Go are found from the previous equations. To demonstrate, assume as before a required BSL of 946 kV at an altitude of 15OOmeters, 6 = 0.840, i.e., BSLA = 946kV. To start, assume m = 0.3, resulting in a BSLs of 94610.9165 = 997 kV. Proceeding as in Table 13 results in a BSLs of 992 kV.

Normally, the value of m ranges from about 0.3 to 0.5, and to obtain a quick and easy estimate, a conservative value of 0.3 is recommended.

Applying no safety margin, the required BSL is 992 kV. From Table 4, the next highest BSL is 1050 kV, which translates to a BIL of 1300 kV. If a safety margin of 15% is used, the BIL would increase to 1550 kV.

Comparing this to previous results with arresters at 15OOmeters, the BSL required for other equipment was only 881 kV, which for a transformer bushing would lead to a BIL of 1300 kV.

3 PHASE-PHASE INSULATION COORDINATION 3.1 Insulation Strength

As explained in Chapter 4, two methods exist for determining the strength of p h a s e phase insulations. To review,

The Alpha Method

A positive voltage ctVp is applied to one phase and a negative voltage (1 - ct)Vp is applied to the other phase, where Vp is the phase-phase voltage. With rx held con- stant, the voltage Vp is varied to determine the phase-phase CFO denoted as CFOp.

Per this method, cx is defined as

The equation for the CFOp is

CFO ^- k 3400

- gp 1

+

(81s)

where S is the phase-phase spacing and k is the gap factor for this method. This method is employed in the IEC and used with alpha equal to 0.5, i.e. gp. V+ = V-, to estimate the strike distances (clearances). The phase-phase BSL, BSLp, in IEC stan- dards [3] is defined for an alpha of 0.5 and therefore

Switching Overvoltages, Substations

with u = 0.5

The V+-V- Method

With a constant negative polarity voltage V - applied to one phase, the positive polarity voltage on the other phase is varied to obtain the positive polarity CFO'.

This CFO' may be related to the negative polarity voltage by the equation

where KL is a constant dependent on the gap configuration and CFOo is the CFO when the negative polarity component is zero (one of the two electrodes is grounded) and is given by the equation

Because the alpha method was first employed, most if not all test results are available using this method. However, these test results may be transcribed or changed so as to apply to the other method. That is,

Using this equation, the gap factors for the two methods are shown in Table 14, as obtained from Chapter 4, for alternate gap configurations. As discussed in Chapter 4, the CFOo is the phasephase CFO when one of the electrodes is grounded. Thus the CFOo is similar to the C F O ~ since this CFO is also the CFO when one of the electrodes is grounded. Therefore of/CFO or O~/CFO; should be identical to oFG/CFO0, and also KGp should be equal to kg. It is true that the CFOo is used differently from C F O ~ , but the gap factors are measured or determined in the same manner or by the same type of test, i.e., with one electrode grounded. Therefore in Table 14, for the v+-V- method, the nomenclature for the gap factor has been changed to k g and that for the standard deviation has been changed to of/CFOo. In addition, since ofg/CFOp is also equal to of/CFO, this nomenclature has been changed to of/CFOp. This should make the presentation and use of these variables clearer and simpler.

Referring to Fig. 1, the length of the bus upon which the switching surge impinges is relatively short, less than the span length of the line. Thus the conduc- tor<onductor gap configuration whose length is large is not given in Table 14.

Rather the 10-meter conductor~onductor configuration will be used for setting the phase-phase clearance of the bus. The ring-ring or large smooth electrode con- figuration is applicable to large-diameter grading rings such as used at EHV on bushings. At lower voltages where the grading rings are considerably smaller, the rod-rod configuration is more applicable. Referring to the rod-rod configuration,