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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 1 -
Insulation Co-ordination: Numerical Example 1
Case 1: Without Capacitor Switching in Sub.2
Case 2: With Capacitor Switching in Sub.2
New Substation with Un=230kVExisting SubstationUs=Um=245kVPollution level: HeavyAltitude from sea level: 1000m
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 2 -
Insulation Co-ordination: Numerical Example 1
Range I: Um
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 3 -
Step 1:determination of the representative overvoltages (Urp)
-Power Frequency Voltage
Us=Um=245kVminimum Creepage Distance for insulators =25mm/kV (Pollution: Heavy)
-Temporary Overvoltages
earth fault factor
load rejection factor
earth fault and load rejection factor
-Slow-front Overvoltages
surge arrester effect
line entrance equipments
other equipment
-Fast-front Overvoltages
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 4 -
Step 1:determination of the representative overvoltages (Urp)
-Temporary Overvoltages
earth fault factor
3 . .. 2k
1.4k
1.4 1.8k
3 ...1.85k
HV side of power transformerin Sub.1 is solidly earthed.
System study: k=1.5 !(global network simulation)
P-E Overvoltage:
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 5 -
Step 1:determination of the representative overvoltages (Urp)
-Temporary Overvoltages
load rejection factor
System study: k=1.4(overspeed genarator in Sub. 1)
moderately extended systems: < 1.2 p.u. for up to several minutes
widely extended systems: 1.5 p.u. for some seconds
close to turbo generator: 1.3 p.u.
close to salient pole (German: "Schenkelpol") generator: 1.5 p.u.
P-E Overvoltage:
P-P Overvoltage:
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 6 -
Step 1:determination of the representative overvoltages (Urp)
-Temporary Overvoltages
earth fault and load rejection factor
Will be opened after load rejection
Earth Fault factorextremely decreases.(D/Y grounded of Tr.)
k load rejection* k earth fault
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 7 -
Step 1:determination of the representative overvoltages (Urp)
-Power Frequency Voltage
Us=Um=245kVminimum Creepage Distance for insulators =25mm/kV (Pollution: Heavy)
-Temporary Overvoltages
earth fault factor k=1.5; Urp=212kV
load rejection factor k=1.4; Urp=198kV(P-E);
Urp=343kV(P-P)
earth fault and load rejection factor not applicable
Urp=212kV(P-E)
Urp=343kV(P-P)
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 8 -
Step 1:determination of the representative overvoltages (Urp)
-Slow-front Overvoltages
energization and re-energization
Range of 2% slow-front phase-to-earthovervoltages at thereceiving end due to line energization (Approximately Value)
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 9 -
Step 1:determination of the representative overvoltages (Urp)
-Slow-front Overvoltages
energization and re-energization
Range of 2% slow-front phase-to-earthovervoltages at thereceiving end due to line energization (Exact Value)
System study :
Overvoltages originating from Sub.1
Ue2=1.9 p.uUp2=2.9 p.u
Overvoltages originating from Sub.2
Ue2=3.0 p.uUp2=4.5 p.u
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 10 -
Step 1:determination of the representative overvoltages (Urp)
-Slow-front Overvoltages
energization and re-energization
Probability distribution of the representative amplitude of theprospective overvoltage phase-to-earth
Using phase-peak method:
Overvoltages originating from Sub.1
Ue2=1.9 p.uUet=1.25*1.9-0.25=2.125 p.u
Overvoltages originating from Sub.2
Ue2=3.0 p.uUet=1.25*3.0-0.25=3.5 p.u
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 11 -
Step 1:determination of the representative overvoltages (Urp)
-Slow-front Overvoltages
energization and re-energization
Probability distribution of the representative amplitude of theprospective overvoltage phase-to-phase
Using phase-peak method:
Overvoltages originating from Sub.1
Up2=2.9 p.uUpt=1.25*2.9-0.43=3.195 p.u
Overvoltages originating from Sub.2
Up2=4.5 p.uUpt=1.25*4.5-0.43=5.195 p.u
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 12 -
Step 1:determination of the representative overvoltages (Urp)
-Slow-front Overvoltages
Surge Arrester effect
O.V from Sub. 1 O.V from Sub. 2
phase-earth Uet=425kV Uet=700kV
phase-phase Upt=639kV Upt=1039kV
energization and re-energization without S.A:
Surge Arrester Characteristics:
Using Surge Arrester:
Phase-to-earth: Ure= Ups = 410kV < 425 & 700
Phase-to-phase: the lower value of
Urp= 2 Ups= 820kV < 1039 but >639 Urp= Upt = 639kV < 820
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 13 -
Step 1:determination of the representative overvoltages (Urp)
-Power Frequency Voltage
Us=Um=245kVminimum Creepage Distance for insulators =25kV/mm (Pollution: Heavy)
-Temporary Overvoltages
earth fault factor
load rejection factor
earth fault and load rejection factor
-Slow-front Overvoltages
surge arrester effect
line entrance equipments
other equipment
-Fast-front Overvoltages: Simplified Statistical Approach (see step2)
Urp=212kV(P-E)
Urp=343kV(P-P)
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 14 -
Step 2:determination of the co-ordination withstand voltages (Ucw)
-Temporary Overvoltages
Ucw=Urp (Kc=1)
Urp=212kV(P-E)
Urp=343kV(P-P)
Ucw=212kV(P-E)
Ucw=343kV(P-P)
-Slow Front Overvoltages
S.A Characteristics:
System study :
Overvoltages originating from Sub.1
Ue2=1.9 p.uUp2=2.9 p.u
Overvoltages originating from Sub.2
Ue2=3.0 p.uUp2=4.5 p.u
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 15 -
Step 2:determination of the co-ordination withstand voltages (Ucw)
-Slow Front Overvoltages
Line entrance equipments:
Ups / Ue2 = 410/600 = 0.68 Kcd = 1.10
Ucw= 1.10 * Urp
2Ups/ Up2 = 820/900 = 0.91 Kcd = 1.00
Ucw= 1.00 * Urp
All other equipments:
Ups / Ue2 = 410/380 = 1.08 Kcd = 1.03
Ucw= 1.03 * Urp
2Ups/ Up2 = 820/580 = 1.41 Kcd = 1.00
Ucw= 1.00 * Urp
S.A Characteristics: Ups= 410kV
Line entrance equipments:
Ue2=3.0 p.u=600kVUp2=4.5 p.u=900kV
All other equipments:
Ue2=1.9 p.u=380kVUp2=2.9 p.u=580kV
(a)
(b)
(a)
(b)
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 16 -
Step 2:determination of the co-ordination withstand voltages (Ucw)
-Slow Front Overvoltages
Step. 1:
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 17 -
Step 2:determination of the co-ordination withstand voltages (Ucw)
-Fast Front Overvoltages
Simplified Statistical Approach
S.A Characteristics:
n=2
n: minimum Nos. of lines connected to the Sub.
Upl = 500A = 4500n = 2
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 18 -
Step 2:determination of the co-ordination withstand voltages (Ucw)
-Fast Front Overvoltages
Simplified Statistical Approach
L = 30m for internal insulationL = 60m for external insulation
Lsp= 300m Span Length
Ra= acceptable failure rate = 1 in 400 yearsRkm= outage per 100km per year = 1
La= Ra/ Rkm= 1/400 = 0.25 km = 250 m
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 19 -
Step 2:determination of the co-ordination withstand voltages (Ucw)
-Fast Front Overvoltages
Simplified Statistical Approach
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 20 -
Step 3: determination of the required withstand voltages (Urw)
-Ks: Safety Factor
for any type of Overvoltages (P-E & P-P)
internal insulation : Ks=1.15
external insulation : Ks=1.05
-Ka: Altitude Factor
Only for external insulation !
H=Altitude above sea level = 1000 m
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 21 -
Step 3: determination of the required withstand voltages (Urw)
For Switchingwithstand voltages:
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 22 -
Step 3: determination of the required withstand voltages (Urw)
Step. 2: For Switchingwithstand voltages:
Ucw=451 , curve (a): m=0.94
Ucw=820 , curve (c): m=1.00
For Lightningwithstand voltages:
For Power Frequencywithstand voltages:
m = 1.00
m = 0.50
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 23 -
Step 3: determination of the required withstand voltages (Urw)
For Switchingwithstand voltages:
Ka= e 0.94.(1000/8150)= 1.122 (P-E)Ka= e
1.00 . (1000/8150)= 1.130 (P-P)
For Lightningwithstand voltages:
For Power Frequencywithstand voltages:
Ka= e1.00 . (1000/8150)= 1.130 (P-E & P-P)
Ka= e0.50 . (1000/8150)= 1.063 (P-E & P-P)
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 24 -
Step 3: determination of the required withstand voltages (Urw)
Urw= Ucw . Ks . Ka
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 25 -
Step 4: Conversion to withstand voltages normalized for range I
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 26 -
Step 4: Conversion to withstand voltages normalized for range I
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 27 -
Step 4: Conversion to withstand voltages normalized for range I
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 28 -
Step 5: Selecting of standard withstand voltage values
Summary:
Urw(s): minimum required withstand voltage obtained directlyUrw(c): minimum required withstand voltage obtained by conversion
l l l
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 29 -
Step 5: Selecting of standard withstand voltage values
Short duration power frequency withstand voltages are selected according to direct values Urw(s)
Lightning withstand voltages are selected as a highest value of Lightning Urw(s) orconverted valueof switching withstand voltage into lightning withstand voltage Urw(c)
S 5 S l i f d d i h d l l
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 30 -
Step 5: Selecting of standard withstand voltage values
Results:minimum required withstand voltage
PFWV(kVrms) / LIWV (kVpeak) : 395/950
only P-P minimum required withstand voltage for line enterance equipement is 1127kVwhichis more than 950 kV standardized value and next standard value of 1175kVshall be selected.
Since there is no three phase equipment in line entrance, minimum P-P clearance can bespecified instead of testing. (IEC 71-2: page:101; )
St 5 S l ti f t d d ith t d lt l
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 31 -
Step 5: Selecting of standard withstand voltage values
LIWV (kVpeak) P-P & P-E for other equipments: 950
LIWV (kVpeak) P-P for line equipments: 1175
2.35m P-P clearane for line equipments and 1.9m
P-P and P-E clerance for other equipments.
St 5 S l ti f t d d ith t d lt l
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 32 -
Step 5: Selecting of standard withstand voltage values
Results (cont.):
Refeinment in system study leads to lower value of PFWV for external insulations. So 360kVfor PFWVand 850kVfor LIWVmay be possible to select as an insulation levels.
for internal insulation, it is economic to select 750and 850kVlightning withstand voltageforPhase to Earthand Phase to Phase, respectively. But PFWVshall be at least 395kV.
I l i C di i N i l E l 1
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 33 -
Insulation Co-ordination: Numerical Example 1
Case 2: With Capacitor Switching in Sub.2
With Capacitor Switching
C s 2: C p cit r S itchin in Sub 2 (St p 1: U )
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 34 -
Case 2: Capacitor Switching in Sub. 2 (Step.1: Urp)
Having a capacitor switching in Sub.2 affects only in Slow-front Overvoltages.
System study shows the following maximum amplitude of O.V in new Sub. Due to capacitorswitching in Sub. 2:
Ups< Uet and 2 Ups< Upt
Surge Arrester Characteristics:
Urp= 410 kV (P-E)Urp= 820 kV (P-P)
Case 2: Capacitor Switching in Sub 2 (Step 2: U )
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 35 -
Case 2: Capacitor Switching in Sub. 2 (Step.2: Ucw)
Ups= 410 kV (P-E)2 . Ups= 820 kV (P-P)
IEC 60071-2 page: 215, Mistake!!!Kcd= 1.075 , Ucw= 441 kV;
Case 2: Capacitor Switching in Sub 2 (Step 3: U )
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 36 -
Case 2: Capacitor Switching in Sub. 2 (Step.3: Urw)
Ucw= 441 kV (P-E)Ucw= 820 kV (P-P)
Ucw=441 , curve (a): m=0.94Ucw=820 , curve (c): m=1.00
Ka= e 0.94(1000/8150)= 1.122 (P-E)Ka= e
1.0(1000/8150)= 1.130 (P-P)
Safety Factor :internal insulation : Ks=1.15external insulation : Ks=1.05
-External insulation:(P-E): Urw= 441 * 1.05 * 1.122 = 519 kV(P-P): Urw= 820 * 1.05 * 1.13 = 973 kV
-Internal insulation:(P-E): Urw= 441 * 1.15 = 507kV(P-P): Urw= 820 * 1.15 = 943 kV
Case 2: Capacitor Switching in Sub 2 (Step 4: Conversion)
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 37 -
Case 2: Capacitor Switching in Sub. 2 (Step.4: Conversion)
-Conversion to short-duration power-frequency withstand voltage (SDW):-External Insulation:(P-E): SDW = 519 * (0.6 + 519 / 8500) = 343 kV;(P-P): SDW = 973 * (0.6 + 973 / 12700) = 658 kV;-Internal Insulation:(P-E): SDW = 507 * 0.5 = 254 kV;(P-P): SDW = 943 * 0.5 = 472 kV.
-Conversion to lightning impulse withstand voltage (LIW):-External Insulation:(P-E): SDW = 519 * 1.3 = 675 kV;(P-P): SDW = 973 * (1.05 + 973 / 9000) = 1127 kV;-Internal Insulation:(P-E): SDW = 507 * 1.1 = 558 kV;
(P-P): SDW = 943 * 1.1 = 1037 kV.
Case 2: Capacitor Switching in Sub 2 (Step 5)
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 38 -
Summary:
Urw(s): minimum required withstand voltage obtained directlyUrw(c): minimum required withstand voltage obtained by conversion
Urw
- kV rms for PFWV
-kV peak for SIWV or LIWV
External Insulation Internal
Insulation
Urw(s)
Urw(c)
Urw(s)
Urw(c)
PFWV
P-E 237 343 243 254
P-P 383 658 395 472
SIWV
P-E 519 - 507 -
P-P 973 - 943 -
LIWV
P-E 884 675 715 558
P-P 884 1127 715 1037
Case 2: Capacitor Switching in Sub. 2 (Step.5)
Case 2: Capacitor Switching in Sub 2 (Step 5)
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 39 -
Case 2: Capacitor Switching in Sub. 2 (Step.5)
Results:
minimum required withstand voltagePFWV(kVrms) / LIWV (kVpeak) : 395/950
only P-P minimum required withstand voltage for all equipments(not onlyline enteranceequipement) is 1127kVwhich is more than 950 kV standardized value and next standard valueof 1175kVshall be selected.
2.35m P-P clearane for all external equipments needed if no P-P test would like to beprovided.
for internal insulation, 460/1050 kV (due to capacitor switching in Sub.2)
Case 2: Capacitor Switching in Sub 2 (Comparision)
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Hochspannungstechnik Overvoltage Protection and Insulation Coordination / Chapter 6
Numerical Example 1 - 40 -
Case 2: Capacitor Switching in Sub. 2 (Comparision)