Capacitor Banks 141w43
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- Words: 6,648
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Capacitors Dominik Pieniazek, P.E. – VI Engineering, LLC Nicholas A. Losito Jr. – Castle Power Solutions, LLC
Outline Day 1 • Basic Power Calculations • Capacitor Fundamentals • Capacitor Ratings • Capacitor Application • Capacitor p Protection Day 2 • Harmonics • Capacitor Bank Design Considerations
Shunt Capacitors • Medium Voltage pp • Substation Applications • Power Factor Correction
Basic Power Calculations
Basic Power Calculations C Common Q Questions: ti What are VARs? Wh do Why d we care about b t VARs? VAR ?
Basic Power Calculations Most plant loads (motors, transformers, etc) are INDUCTIVE and require a magnetic field to operate. The magnetic field is necessary, but produces NO USEFUL WORK. The utility must supply the power to produce the magnetic field and the power to produce USEFUL work. The ACTIVE component produces the USEFUL work, the REACTIVE component produces the magnetic field.
VAR
Basic Power Calculations An analogy gy that most can understand.
W
VA
Mug Capacity i off equipment i (i.e. xfmr, cable, swgr, etc) ee Beer Stuff that you want Foam (Head) St ff which Stuff hi h prevents t you from maximizing the amount of beer that you get
Basic Power Calculations
V: Reference f voltage l IR: Resistive load IL: Inductive load IC: Capacitive load
Basic Power Calculations
Vref Ires
Power = Re(V I*)
Basic Power Calculations
Vref Ilag
Power = Re(V I*) Note that due to phase shift (30 degrees) only 86.66% of current is applied to calculate work.
Basic Power Calculations
Vref Ilead
Power = Re(V I*) Regardless whether leading or lagging, power calculation yields similar results
Basic Power Calculations
Vref I90
Power = Re(V I*) In the case where current leads or lags the voltage by 90 degrees, P = 0
Basic Power Calculations Power Factor [ PF ] = Cos
=P/S
A) V k S(
P (W) Power Triangle g The relationship between S, P, and Q. This figure represents a lagging power factor. If Q is negative, leading power factor.
Q (kVAR)
Basic Power Calculations So how do we know if our current is lagging or leading the voltage, and what can we do to correct it? Consider this example, example
Basic Power Calculations Incomingg service is at 99.5% capacity 5500 kW 3400 kVAR 6466 kVA
2000 kW 1000 kVAR
1000 kW 800 kVAR
2500 kW 1600 kVAR
Basic Power Calculations Power Factor [ PF ] = Cos =gP / S 0.85 lag A) V k S(
31.8 deg P (W) 5500 kW Power Triangle g Even though our facility require only 5500 kW to perform Real Work, our incoming service must be sized for 6466 kVA.
Q (kVAR) 3400 kVAR
We have a 6500 kVA mug that is holding 5500 kW and 3400 kVAR.
W
VA
VAR
Basic Power Calculations
With the existing configuration the facilityy cannot add anyy loads without upgrading the incoming service.
Basic Power Calculations Incomingg service is at 85% capacity 5500 kW 0 kVAR 5500 kVA
3400 kVAR
2000 kW 1000 kVAR
1000 kW 800 kVAR
2500 kW 1600 kVAR
Basic Power Calculations Power Factor [ PF ] = Cos = Py / S 1.0 unity A) V k S(
0 deg P (W) 5500 kW Power Triangle g The cap bank is providing 3400 kVAR, so our service is now providing only 5500 kVA (reduction from 6466 kVA.
Q (kVAR) 0 kVAR
Basic Power Calculations COS [ ] = 0.67 COS [ ] = 0.95 Q2 = Q1 + Qc
VA k 00 1,5
Q1 = 1,118 kVAR
VA 1,053 k
Q2 = 330 kVAR Q
1 000 kW 1,000 Required Apparent Power Before and After Adding a Power Capacitor Bank
Qc = 788 kVAR
An example of how to calculate the size of a cap bank based on a target power factor
Basic Power Calculations
But wait, there’s more….
Basic Power Calculations Odds are that the utilityy is charging you for penalties for a low power factor.
5500 kW 3400 kVAR 6466 kVA
5500 = 0.85 6466 Typically, penalties are applied for power factor less than 0.95%. 2000 kW 1000 kVAR
1000 kW 800 kVAR
2500 kW 1600 kVAR
Basic Power Calculations There are also voltage g considerations. 5500 kW 3400 kVAR 6466 kVA Volt Drop = 5.8%
Assuming typical values: Source impedance = 9% at 6500 kVA The expected voltage drop at the main bus will be close to 6%!
2000 kW 1000 kVAR
1000 kW 800 kVAR
2500 kW 1600 kVAR
Basic Power Calculations Capacitor bank reduces the voltage drop at main bus by 5%! 5500 kW 0 kVAR 5500 kVA Volt Drop = 1.0% 3400 kVAR
2000 kW 1000 kVAR
1000 kW 800 kVAR
2500 kW 1600 kVAR
Basic Power Calculations Note that the capacitor bank can also raise the bus voltage g above the nominal value. 5500 kW -1600 kVAR 5728 kVA Volt Drop = -1.5% 5000 kVAR
2000 kW 1000 kVAR
1000 kW 800 kVAR
2500 kW 1600 kVAR
Basic Power Calculations The voltage continues to rise if the capacitor p bank remains connected and the load is reduced.
3000 kW -3200 kVAR 4386 kVA Volt Drop = -3.6%
5000 kVAR
Out of service
2000 kW 1000 kVAR
1000 kW 800 kVAR
0 kW 0 kVAR
Capacitor Fundamentals
Capacitor Fundamentals
C = eo A / d for a parallel plate capacitor, where eo is the permittivity of the insulating material (dielectric) between plates.
Capacitor Fundamentals
We recall that we can add series capacitances to obtain an equivalent capacitance.
1 / Ceq = 1 / C1 + 1 / C2
Capacitor Fundamentals
Similarly, we can add parallel capacitances to obtain an equivalent capacitance.
Ceq = C1 + C2
Capacitor Fundamentals …but we typically do not have much use for capacitance values. So we convert capacitance to impedance:
1 1 XC C 2fC
Capacitor Fundamentals Z R jX c Z R jX c Z XC 2
V Z S
2
V S XC
Capacitor Fundamentals 2
V [kV ] S [kV kVAR] *1000 000 Z [] Example: The capacitance of a capacitor is 6.22 F and the nameplate voltage is 8000 V. Calculate the power rating.
XC
1 426.7[] 6 2(3.14)(60)(6.22 x10 ) (8) 2 S 1,000 150[kVAR] 426.7
Capacitor Fundamentals
Capacitor Fundamentals
Capacitor Fundamentals
Capacitor Fundamentals
Capacitor Fundamentals
Capacitor Fundamentals Note that IEEE Std 18 requires the discharge resistor to reduce the terminal voltage to 50 V in th time the ti frame f as specified ifi d in the table below.
Discharge resistor
Capacitor Ratings
Capacitor Ratings Medium-voltage Medium voltage capacitors are available in many different styles. styles The main points of differentiation are listed below: • Voltage rating • kVAR rating • Single g bushingg or dual bushingg • Internally fused, externally fused, or fuseless
Capacitor Ratings IEEE 81 defines the ratings for capacitors • Voltage, rms (terminal to terminal) • Terminal-to-case (or ground) insulation class • Reactive power • Number of phases p • Frequency
Capacitor Ratings IEEE 18 provides capacitor tolerances • The capacitance shall not vary more than -0% to +10% of nominal value based on rated kVAR, voltage, and frequency measured at 25 deg C. This means that a new 150 kVAR unit can range g anywhere y from 150 kVAR to 165 kVAR.
Capacitor Ratings IEEE 18 states the capacitor is intended to operate at or below rated voltage. Capacitors shall be capable of continuous operation given that none of the following limitations are exceeded: • 110% of rated rms voltage (temporary overvoltage parameters will be discussed later)) • 120% of peak voltage, including harmonics but excluding transients • 135% of nominal rms current based on rated kVAR and rated voltage • 135% of rated kVAR
Capacitor Ratings
* Impulse tests shall be applied between terminals and case, with the terminals connected together. For capacitors having bushings with two different BIL ratings, this test shall be based on the bushing with the lower BIL. The nameplate shall show both BIL ratings, e.g. 150/95 kV BIL. ** Not applicable to indoor ratings
Capacitor Application
Capacitor Application Power factor correction capacitor banks are typically installed in the following ways : • Pole top • Metal-Enclosed / Pad-Mount • Ope Open rack ac • Terminal end at equipment
Capacitor Application
P l Top Pole T Installation I t ll ti
Capacitor Application
Transient inrush reactors
P l Top Pole T Installation I t ll ti
Capacitor Application
Pad Mounted Installation Pad-Mounted
Capacitor Application
Metal-Enclosed Metal Enclosed Substation Installation
Capacitor Application
Three-phase iron core harmonic filter reactor Metal-Enclosed Metal Enclosed Substation Installation
Capacitor Application
Open Rack, Medium-Voltage Substation Installation
Capacitor Application
Open Rack, High-Voltage Substation Installation
Capacitor Application
Installation in Equipment
Capacitor Application Power factor correction capacitor banks can be configured in the following ways : • Delta • Wye - Solidly Grounded • Wye y - Ungrounded g A common misconception is that the capacitor bank should be connected Delta since it is being applied to a delta or high highimpedance grounded system. This is NOT true.
Capacitor Application The driving factor which determine the configuration for the given application is COST. Voltage considerations IEEE 1036 suggests that only banks rated 2400 V and below should be Delta connected. connected This is mainly because standard voltage ratings for wye connected banks may not be available. Cost of phase-to-phase vs phase-to-neutral rated capacitors at higher voltages tends to point installations towards wye connected banks for larger bank installations.
Capacitor Application Delta Lower voltages (<= 2400 V) • Standard capacitors are typically not available at 1380 V Distribution systems (pole top) • Units are configured with a single series group of capacitors with capacitors rated phase-to-phase. phase to phase Therefore Therefore, unbalance detection is not required.
Capacitor Application Wye – Solidly Grounded • Initial cost of the bank may be lower since the neutral does not have to be insulated from ground. ground • Capacitor switch recovery voltages are reduced • High inrush currents may occur in the station ground system • The Th grounded-wye d d arrangement provides id a low-impedance l i d fault f l path which may require revision to the existing system ground pprotection scheme. Typically yp y not applied pp to ungrounded g systems. y • When applied to resistance-grounded systems, difficulty in coordination between capacitor fuses and upstream ground protection relays (consider coordination of 40 A fuse with a 400 A grounded system). • Typical for smaller installations (since auxiliary equipment is not required) i d)
Capacitor Application The most common capacitor bank configurations for larger substation applications are Wye-Ungrounded
Three of the most common unbalance protection schemes are shown. Discussion of the protection schemes will be presented later.
Capacitor Protection Fusing • Fuseless • Internally I t ll Fused F d • Externally Fused
Capacitor Protection
B k Protection Bank P t ti Summary S
Capacitor Protection Fuseless Capacitors Constr cted of small capacitor elements which Constructed hich are arranged in series and parallel. The elements are constructed of aluminum foil with a dielectric of electrical grade polypropylene. This design provides a safe failure mode. In the event that the dielectric fails, the energy in the resulting small arc punctures many layers of the thin film and foil within the element. The arc causes the film layer to receded allowing many layers of the aluminum foil electrodes to touch and weld together forming an electrically stable electrical t. t This results in an entire series section being shorted.
Capacitor Protection Example of Fuseless Installation
Capacitor Protection Internally Fused Capacitors Constr cted such Constructed s ch that each element is protected with a series connected current limiting fuse. The design is such that isolated fusing prevents potential damage to the adjacent elements and fuses. The current limiting mode chops the fault current to prevent the energy stored in the parallel connected elements from being discharged into the faulted element.
Capacitor Protection
Group Fusing
Individual d v du Fusing us g
Capacitor Protection Group Fusing– Considerations for Selecting Fuse (typical for distribution pole mounted racks)
• Continuous Current • Transient Current • Fault Current • Tankk Rupture Curve C C di i Coordination • Voltage on Good Capacitors
Capacitor Protection Continuous Current • For wye-solidly grounded systems: Fuse > = 135% of rated capacitor current (includes overvoltage, capacitor tolerances, and harmonics). • For wye-ungrounded y g systems: y Fuse > = 125% of rated capacitor current (includes overvoltage, capacitor tolerances, and harmonics). Care should be taken when using NEMA Type T and K tin links which are rated y 1.50. 150%. In this case, the divide the fuse ratingg by
Capacitor Protection Transient Current • Capacitor switching (specifically back-to-back switching) • Lightning surges Back-to-back is typically yp y not a factor for ppole mounted capacitors p banks. High frequency lightning surges: Use NEMA T tin links for ampere ratings up to 25 A. Use NEMA K tin links for ampere ratings above 25 A.
Capacitor Protection Fault Current Ensure that the fuse can interrupt the available fault current Tank Rupture Coordination Ensure that the fuse maximum clearingg TCC curve for the fuse link is plotted below the capacitor tank rupture curve. In cases of high fault currents, the tank rupture curve should be compensated for asymmetry. asymmetry Voltage on Good Capacitors For an ungrounded d d system, a fault f l on one phase h results l in i a 1.73 times overvoltage on the un-faulted phases. Ensure that the fault p failure. is cleared before the second capacitor
Capacitor Protection Problems with Fusing of Small Ungrounded Banks Consider a 12.47 kV, 1500 kVAR cap bank made of three (3) 500 kVAR single-phase i l h units. i 1500[kVAR] 69.44[ A] 69.44 1.5 104[ A] 100[ A]Fuse 3 12.47[kV ]
If a capacitor fails, we will expect approximately 3x line current. pp y 500 seconds to clear this It will take a 100 A fuse approximately fault (3 x 69.44 A = 208.32 A). The capacitor case will rupture long before the fuse clears the fault. The solution is using smaller units (explanation to follow).
Capacitor Protection Individual Fusing– Considerations for Selecting Fuse (typical for substation capacitor banks)
• Continuous Current • Transient Current • Fault Current • Tank Rupture Curve Coordination • Voltage on Good Capacitors • Energy Discharge into Faulted Unit • Outrush Current • Coordination with Unbalance Detection System
Capacitor Protection Continuous Current • Fuse > = 135% of rated capacitor current (includes overvoltage, capacitor tolerances, and harmonics) Care should be taken when using NEMA Type T and K tin links which are rated 150% In this case, 150%. case the divide the fuse rating by 1.50. 1 50
Capacitor Protection Transient Current • Lightning surges • Capacitor switching (specifically back-to-back switching) High g magnitude, g , high g frequency q y lightning g g surges g are typically yp y not a concern for substation installations. Back-to-back Back to back switching is typically controlled with pre-insertion pre insertion closing resistors or current limiting reactors. By the h nature off installation, i ll i the h parallel ll l fuses f will ill share h the h transient current and will not be a factor.
Capacitor Protection Fault Current Ensure that the fuse can interrupt the available fault current. In substation banks with multiple series groups, fault current will not flow through g a failed capacitor p unit unless other units experience a simultaneous failure. For this reason expulsion fuses are commonly used rather than current limiting fuses Tank Rupture Coordination Ensure that the fuse maximum clearing TCC curve for the fuse li k is link i plotted l d below b l the h capacitor i tankk rupture curve. In cases off high fault currents, the tank rupture curve should be compensated y y for asymmetry.
Capacitor Protection Example of a Definite Tank Rupture Curve. The time Th i between b the h rupture curve and the fuse maximum clear curve is the coordination margin.
Capacitor Protection Example of a 10% and 50% Rupture Curve for a 100 kVAR Capacitor. Probability y based tank rupture p curves are developed when there is too much variance in rupture test data. Based on the 10% and 50% curves, one can extrapolate the curves for f any probability. b bili
Capacitor Protection Voltage on Good Capacitors When a short-circuit on one unit occurs, an overvoltage results on the un-faulted phases. Ensure that the fault is cleared before the second capacitor failure. A table summarizes this voltage rise on the un-faulted units
Per Unit Voltage on Un-failed Capacitors
Capacitor Protection Energy Discharge Into a Failed Unit When a capacitor failure occurs, the stored energy in the parallel connected capacitors can discharge thro through gh the failed capacitor and its fuse. The total calculated parallel stored energy should not exceed the energy capability (Joule rating) of the capacitor and fuse. If the energy capabilities are exceeded, a failure of the fuse and/or rupture of the capacitor tank can result. Typical rating of film capacitors is 15,000 Joules (4650 kVAR in parallel) and 10,000 Joules (3100 kVAR in parallel) for paperfilm capacitors. capacitors Expulsion fuses are typically rated 30 30,000 000 Joules. Current limiting fuses are required if ratings are exceeded. (1 Joule = 1 W x sec sec, use 00.2 2 cycle clearing time for calculation)
Capacitor Protection Outrush Current When a capacitor failure occurs, the parallel connected capacitors can discharge high frequency current into the failed capacitor. The fuses of the un-failed capacitors should be able to withstand the high g frequency q y discharge g currents. These calculations and measurements are complex and are determined by the manufacturer. Coordination with Unbalance Detection Scheme The individual fuse must clear the fault before the unbalance protection i scheme h trips i the h entire i capacitor i bank. b k
Capacitor Protection Fusing Recommendations by McGraw Edison
Capacitor Protection Recall Problem with Fusing of Small Ungrounded Banks 12.47 kV, 1500 kVAR cap bank made of three (3) 500 kVAR units 1500[kVAR] 69.44[ A] 69.44 1.5 104[ A] 100[ A]Fuse 3 12.47[kV ]
It will take a 100 A fuse approximately 500 seconds to clear this fault (3 x 69.44 A = 208.32 A). The capacitor case will rupture long before the fuse clears the fault. The solution is using smaller units with individual fusing. Consider five ((5)) 100 kVAR capacitors p pper phase, p , each with a 25 A fuse. The clear time for a 25 A fuse @ 208.32 A is below the published capacitor rupture curve.
Capacitor Protection Why is the Current 3 x Nominal Line Current for a Phase-toNeutral Fault on a Wye-Ungrounded Capacitor Bank? B
A N
A N
VNG Since V = I*Z, where Z is constant t t (assuming ( i f = 60 hz) h ) If voltage across capacitor is increased by 1.732, the current also increases byy factor of 1.732
C
IA = 3.0 3 0 p.u.
B
C
Capacitor Protection
Minimum Conductor Size It was noted that capacitors are rated 135% of rating. This requires the conductor to be sized 135% of the nominal capacitor rating. rating
Capacitor Protection Unbalance Protection As single-phase units in a multiple unit/phase installation fail and are removed from service, service the remaining units are experience an overvoltage condition. IEEE Standard 1036 provides overvoltage limitations. Duration
Max Voltage (x rated RMS)
6 cycles
2.20
15 ccycles cles
2 00 2.00
1s
1.70
15 sec
1.40
1 min
1.30
An unbalance protection scheme must by implemented to prevent g units. the failure of the overvoltaged
Capacitor Protection
Capacitor Protection
Neutral Voltage Unbalance with Unbalance Compensation
Capacitor Protection
VA
VNG
Normal Conditions VN = VG VAN = VBN = VCN
VG VN VC
VB
Capacitor Protection Ungrounded or Impedance Grounded System
VA
VNG
Normal Conditions VN = VG VAN = VBN = VCN = 1.0 .0 p.u.
VG VN VC
VB
Capacitor Protection
VNG Phase to Neutral Fault VNG = VLN VAN = VBN = VLL = 1.732 p.u.
Capacitor Protection
VA
VNG VG One Can Removed VNG = 0.2 p.u. VCN = 1.2 1 2 p.u. pu
VC
VN VB
Capacitor Protection Ungrounded or Impedance Grounded System
VA
VNG
Normal Conditions VN = VG VAN = VBN = VCN = 1.0 .0 p.u.
VG VN VC
VB
Capacitor Protection Ungrounded or Impedance Grounded System
VA Gnd VNG
Ground Fault VNG = VLN VAG = VBG = VLL = 1.732 1 732 p.u. pu
VN VGC VC =V Ground Fault at Cap Bank or A Anywhere h on the h S System
VB
Capacitor Protection
Wye-Ungrounded: Voltage Between Capacitor Bank Neutral and Ground vs. Percentage of Capacitor Units Removed from Series Group
Capacitor Protection
Wye-Ungrounded: Voltage on Remaining Capacitor Units in Series Group vs. Percentage of Capacitor Units Removed from Series Group
Capacitor Protection
Wye-Grounded: Neutral Current vs Percentage of Capacitor Units Removed from Series Group
Capacitor Protection
Wye-Grounded or Delta: Voltage on Remaining Units in Series Group vs. Percentage of Capacitor Units Removed from Series Group
Capacitor Protection
Double Wye-Ungrounded, Neutrals Tied Together: Neutral Current vs. Percentage of Capacitor Units Removed from Series Group
Capacitor Protection
Double Wye-Ungrounded, Neutrals Tied Together: Voltage on Remaining Capacitor Units in Series vs. Percentage of Capacitor Units Removed from Series Group
Capacitor Protection A phase
B phase
C phase P: Number of units in group (P=6)
S: Number of series groups (S=4)
Reference Figure for Calculations to Follow
Capacitor Protection # of Series Groups
Grounded Y or Delta
Ungrounded Y
Split Ungrounded Y (equal sections)
1
-
4
2
2
6
8
7
3
8
9
8
4
9
10
9
5
9
10
10
6
10
10
10
7
10
10
10
8
10
11
10
9
10
11
10
10
10
11
11
11
10
11
11
12 and over
11
11
11
Minimum recommended number of units in parallel per series Group to limit voltage on remaining units to 110% with one unit out
Capacitor Protection
Many more configurations and calculations shown in IEEE C37.99
Day 2
Capacitor Fundamentals Further discussion on capacitor voltage ratings: On a ungrounded or impedance grounded system, a ground fault on one phase will cause the other two phases will be elevated by 1 732 1.732. Does this mean that capacitors must be rated phase-to-phase? Certainly nothing wrong with this, but cost will be significantly higher. g
Capacitor Fundamentals Recall:
V2 S XC
This means that a 150 kVAR, kVAR 12470 V unit applied at 7200 V will provide only 50 kVAR. S NEW S NEW
7200[V ]2 150[kVAR] 2 12470[V ] 50[kVAR]
Capacitor Fundamentals Using 12470 V capacitors on a 12470 V Ungrounded or Resistance-Grounded System y will require q 3x more cans. It should be noted that the 12470 V cans will also be larger than the 7200 V cans. cans Results in a much larger and more costly installation. This solution would be required if a ground fault could be maintained for extended pperiods of time.
Capacitor Fundamentals Perhaps a 150 kVAR or 200 kVAR, 7620 V or 7960 V units applied at 7200 V would be a better solution. 7200[V ] 150[kVAR ] 2 7620[V ] 134[kVAR ]
S NEW S NEW
2
S NEW OR
OR 2
7200[V ] 150[kVAR ] 2 7960[V ] 123[kVAR ]
S NEW S NEW
S NEW
7200[V ]2 200[kVAR ] 2 7620[V ] 178[kVAR ]
S NEW S NEW
7200[V ]2 [kVAR] 2 7960[V ] 163[kVAR ]
Note that the 7620 V unit provides an additional 6% The 7960 V unit provides an additional 11%
Capacitor Fundamentals Explusion Fuses: Provides a means of disconnecting a failed capacitor from the circuit by melting a tin-lead low current link. The shorted capacitor unit causes a large increase in the current through the fuse. The current is limited only by the power system reactance and the other capacitor units in series with the failed capacitor unit. The pressure is generated by the hot arc making with the fiber lining of the fuse tube. The link is cooled and stretched as it is forced out the tube. The fuse continues to conduct until a natural current zero occurs. The current zero is causedd bby th the power system t fault f lt currentt crossing i zero. If other th capacitors it are connected in parallel with the failed unit, all the stored energy in these capacitors will be absorbed in either the fuse operation or the failed capacitor unit. Most of gy is absorbed in the failed capacitor. p the energy
Capacitor Fundamentals Current Limiting Fuses: Uses a long uniform cross section element. This configuration makes the fuse a current chopping fuse. The fuse develops a back voltage per inch of element across the entire length of the element. When this voltage exceeds the available voltage across the fuse, the fuse forces the arc to extinguish. The result is that a trapped voltage may and probably will remain on the other capacitors in the series group. The fuse by its design avoids absorbing all of the available energy on the series group. This fuse is used for capacitor banks with a large number of parallel ll l capacitors. it It can be b usedd on applications li ti with ith essentially ti ll infinite i fi it parallel ll l stored energy, as long as sufficient back voltage can be developed to force the current to extinguish. This is the fuse we apply to series, large shunt, and DC banks. Because of the high back voltage that is developed, this fuse must be used with several capacitors in parallel to limit the voltage build up or a flashover may occur elsewhere in the capacitor rack. rack
Capacitor Fundamentals Current Limiting Fuses vs Expulsion Fuses:
Expulsion Fuse
C Current Limiting Li i i Fuse F
Capacitor Fundamentals Current Limiting Fuses vs Expulsion Fuses: Expulsion Fuses Operate mechanically and provide a visual indication Require additional space for operation Typically applied for outdoor application due to ionized gas release. Combination C bi ti expulsion l i with ith currentt limiting li iti characteristic h t i ti fuses f can be b usedd in i indoor metal-enclosed equipment. p Less expensive
Capacitor Fundamentals
Capacitor Fundamentals Current Limiting Fuses Do not emit ionized gases during operation. operation Ionized gases are undesirable because they can cause bushing and insulator flashovers that result in additional damage. Do not require ventilation. Fast current-limiting operation High interrupting capacity, noiseless operation Can be specified for indoor and outdoor applications . No pressure build-up, therefore, no vents or special reinforced compartments are required. i d More expensive
Capacitor Fundamentals Note no pigtail and blown fuse indication
Capacitor Fundamentals Current Limiting with Expulsion
Capacitor Fundamentals What about arresters? How and where should they be applied? Depending on application, environment, exposure to switching etc, switching, etc arresters may be necessary. necessary We recall that when a travelling wave meets a high impedance, the wave can double in size. For this reason, arresters (if used) should be installed as close to the capacitor bank as ppossible. Installation of arresters at the breaker feeding the capacitor bank will not do much for protection of the capacitor bank.
Capacitor Protection A basic three (3) arrester method is shown below. This is typical for solidly grounded systems and wye wye-grounded grounded capacitor banks.
Capacitor Protection Depending on type of installation, system parameters and level of protection required, a six(6) arrestor method may be applied.
Capacitor Protection For an ungrounded system or a high-impedance grounded system, a four (4) arrestor grounding method might be considered an wye ungrounded bank.
VLL
Ground Fault VNG = VLN VAG = VBG = VLL = 1.732 p.u.
VLL
Phase to Neutral Fault VNG = VLN VAN = VBN = VLL = 1.732 p.u.
ArrPH
If faults can be maintained,
ArrN ArrPH must be rated VLL ArrN must be rated VLN The effective arrester MCOV is VLL + VLN
Capacitor Protection Note that if a basic three (3) arrester method is applied to an ungrounded bank, the arresters must be rated high enough to sustain a temporary overvoltage condition during a phase-toground fault on the system. This may not provide an adequate level of protection for the capacitors. Phase to Neutral Fault VNG = VLN VAN = VBN = VLL = 1.732 1 732 p.u.
Arresters do not provide protection across the capacitor bushings. Note that the BIL applies to bushing-tocase insulation. i l ti
VLL
If faults can be maintained, Arresters must be rated VLL
VLL
Ground Fault VNG = VLN VAG = VBG = VLL = 1.732 1 732 p.u.
Capacitor Protection Good Presentations on Capacitor – Arrester Applications “Guidelines “G id li ffor S Selection l ti off S Surge A Arresters t ffor Sh Shuntt Capacitor Banks” – ABB Technical Information “Surge Arrester Application of MV-Capacitor Banks to Mitigate Problems of Switching Restrike” – CIRED 19th International Conference on Electricity Distribution, Distribution Vienna, Vienna 21-24 May 2007. Both B th off these th papers also l address dd phase-to-phase h t h arrester t connections.
Harmonics
Harmonics Recall that the impedance of a capacitor is inversely proportional to the system frequency. frequency
1 1 XC C 2fC Harmonics flow to the point of lowest impedance. The higher the harmonic, harmonic the lower the impedance of the capacitor. capacitor As capacitors absorb harmonics, the capacitor heats up and th life the lif expectancy t is i reduced. d d The Th voltage lt harmonics h i stress t the capacitor dielectric and reduce the life expectancy of the capacitor.
Harmonics
Harmonics Where do harmonics come from? • Power Electronics (drives, rectifiers, computer power supplies, etc) • Arcingg Devices ((welders,, arc furnaces,, florescent lights, g , etc)) • Iron Saturating Devices (transformers) • Rotating Machines (Generators) • Parallel Resonance (between cap bank and power equipment) IEEE Std 519 provides recommended limits of harmonic distortion at the point-of-common-coupling (PCC) with the y utility.
Harmonics
Harmonics
Harmonics Resonance When a number of harmonic current sources are injecting currents into the supply and the frequency of one of the harmonics coincides with the resonant frequency of the supply transformer and Power Factor Correction capacitor combination, the system resonates and a large circulating harmonic current is excited between these components components. The result of this is that a large current at this harmonic flows in the supply transformer, this resulting in a large harmonic voltage distortion being imposed upon the load voltage.
Harmonics A studyy should be pperformed to determine levels of harmonics on a system to determine if any filters are necessary when installing a capacitor bank. Care should be taken to determine if the filtered capacitor bank will introduce any resonance problems. If resonance problems exist, the fil design filter d i must be b adjusted. dj d
Harmonics
An example of a 13 13.8 8 kV harmonic filter
Capacitor Bank Design Considerations
Design Considerations So how do we size a capacitor bank? Determine D t i your primary i goall • Voltage • Lower utility bill (avoid penalties) • Increase capacity of system It can be all three, or any combination of the above. Note that correcting to unity power factor at maximum load is costly tl andd may nott be b necessary.
Design Considerations For a 20 MVA load at 0.88 0 88 power factor (17.6 (17 6 MW, MW 9.5 9 5 MVAR) To achieve 95% power factor, a 3.72 MVAR bank is required To achieve 98% power factor, a 5.93 MVAR bank is required To achieve unity power factor, a 9.50 MVAR bank is required
Design Considerations Determine if current limiting reactors or tuning reactors are required. Harmonics and resonance may dictate tuning reactors Back to back switching may require current limiting reactors Back-to-back (unless another method is used to mitigate the switching surges, i.e. pre-insertion closing resistors/reactors, zero-crossing breakers etc) breakers,
Design Considerations Determine the proper voltage. Capacitors are very susceptible to voltage transients and harmonics. Increasing the rated voltage increases the protective margin on the insulation. The voltage at the capacitor terminals will be higher than bus voltage lt if reactors t are utilized. tili d It is i important i t t to t t for f this thi voltage difference. Determine the voltage swing of the system. Will the capacitors remain on-line while the facility is lightly loaded.
Design Considerations We listed some reasons for specifying higher than bus nominal rating of capacitors. However, care must be taken to ensure that the kVAR rating is properly adjusted as a result. Three (6) 150 kVAR, 7960 V wye-connected capacitors provide a nominal 901 kVAR when connected to a 13.8 kV bus. Three (6) 150 kVAR, 8320 V wye-connected capacitors provide a nominal 825 kVAR when connected to a 13.8 13 8 kV bus
Design Considerations Determine optimal size and number of stages. Depending D di on swing i in i plant l t load, l d a single i l bank b k sized i d for f full f ll plant capacity may not be the answer. IEEE 1036 recommends limiting the voltage change to 2-3%. The delta voltage can be estimated by: V
MVAR 100% MVASC
Switch of a capacitor applies high stresses to the insulation. insulation Limiting the number of stages and limiting the frequency of switching will increase the life. Ideally, a capacitor is switched on and left on.
Design Considerations Determine best location for the installation installation. The most effective placement for power factor correction capacitor banks is at the load. However, this is not always practical or cost effective. Typically, a capacitor bank is installed on each bus of a main-tiemain switchgear. If capacitors are installed at the motor pecker head (running capacitors) ensure that the capacitor VAR rating does not exceed capacitors), 90% of the motor no-load VAR. Otherwise, it is possible to damage the motor by overexcitation.
Design Considerations Use caution when sizing motor running capacitors. Logic would suggest that installation of a power factor correction capacitor at the motor terminals sized to provide y ppower factor makes sense. unity THIS IS NOT THE CASE. Do not exceed 90% of the motor no-load no load kVAR demand. demand Exceeding this value can result in damage to the motor insulation as a result of overexcitation.
Design Considerations As an example for a 4000 V, 4000 hp motor: 100% load current = 495 A @ 89.7% pf 100% load kVAR = 1516 kVAR No load current = 117 A @ 6.3 % pf No load kVAR = 809 kVAR Max size of running capacitor is 0.90 0 90 x 809 kVAR = 728 kVAR
Design Considerations M: Motor Magnetizing Curve C1: Capacitor size at 100% motor mag current C2: Capacitor sized > 100% motor mag current C4: Capacitor sized < 100% motor mag current
If the capacitive reactance of the capacitor is less less than that of the motor reactance (this occurs when to large of a capacitor is chosen). This combination of reactance will result in a resonant frequency below 60 hertz hertz. Therefore, Therefore as the motor slows in speed, speed the frequency of the motor terminal voltage will decrease from a value of near 60 hertz toward zero. When the motor's terminal voltage frequency es through the resonant frequency setup between the capacitor reactance and the motor reactance, the terminal voltage will become very high, only limited by the properties of the iron Depending on the inertia of the motor, iron. motor this resonance (or high voltage) may be present for a considerable period of time.
Design Considerations Determine the most optimal type of installation. installation Will the capacitor bank be installed within a fenced substation? Metal-enclosed, pad mount, or open rack may be good choices Will the capacitor bank be installed in a process area? Metal-enclosed or pad mount may be good choices Will the capacitor bank be pole mounted on a distribution line?
Bank Failures
Design Considerations Consider the impact to personnel safety adjacent equipment when deciding between a metal-enclosed and open-rack system. Porcelain can resemble shrapnel when a capacitor bushing fails.
Design Considerations Determine the most optimal configuration. Higher reliability costs more. y g (1) ( ) 800 kVAR pper pphase 2400 kVAR,, 13800 V,, wye-grounded bank will be a smaller footprint and cost less than 2400 kVAR, kVAR 13800 V V, wye-ungrounded wye ungrounded (8) 100 kVAR per phase bank. However, the H h reliability li bili off the h wye-ungrounded d d bbank k iis significantly higher
Design Considerations Determine the switching equipment When breakers are used for switching capacitors (single bank or back-to-back switching), the breakers must be rated for p switching. g capacitor IEEE C37.99 provides the equations for calculating the inrush current and frequency. frequency
Design Considerations
Design Considerations Consider a single 4800 kVAR wye-ungrounded bank switched (with nominal inductance from equipment): 3253A pk @ 600 Hz, the product is 0.20 x 107 Switching a second similar bank on the same bus without current limiting reactor: 24,058 A pk @ 7.66 kHz, the product is 18.4 x 107 B adding By ddi a 100 mH H current limiting li i i reactor, the h inrush i h is: i p is 1.7 x 107 7254 A ppk @ 2.31 kHz,, the product
Design Considerations
Design Considerations
Design Considerations Energization of a single capacitor bank.
Design Considerations
B k b k switching Back-to-back i hi off the h same unit. i
Design Considerations
E Example l off breaker b k with i h pre-insertion i i resistor. i
Design Considerations
Another application.
Design Considerations
A Pre-Insertion Manufacture’s Perspective. p
Design Considerations Another concern is voltage amplification as a result of switching a second capacitor bank.
Design Considerations
Design Considerations Consider other accessories: • Disconnect switch • Grounding switch • Kirk-keyy interlock • Ventilation requirements • Control power
Design Considerations
Design Considerations
Design Considerations Be aware that larger medium voltage motors may include surge g ppacks. The surge pack will decrease the crest voltage and rate of rise of the impending surge. surge High rates of rise damage end turns while high crest voltage damage winding to core insulation. Typically the capacitance of the is small enough that it can g but this should be verified. be neglected,
Design Considerations
Typical Surge Pack Application
Design Considerations Do not confuse Harmonic Filter Banks with Power Factor Correction Banks. Banks The voltage ratings of harmonic filter banks are substantially higher because they are connected on the back end of a tuning reactor where the voltage is substantially higher. As a result of thee higher g e voltage, vo ge, thee installed s ed kVAR V ccan be anywhere yw e e from o 25% to 40% higher than nominal design. The capacitor cans must be capable of this output. output
Grounding of Wye Banks If multiple wye-grounded banks are in close proximity, use ppeninsula ggrounding g or single-point g p ggrounding. g Single-Point Grounding The neutrals of all banks of a given voltage are connected together with insulated cable/bus and tied to the ground grid only at one point. This prevents high-frequency currents (due (d to back-to-back b k b k switching) i hi ) from f flowing fl i into i the ground grid.
Grounding of Wye Banks Peninsula Grounding One or more isolated gground ggrid conductors are carried underneath the capacitor rack of each phase and tied to the station ground grid at one point at the edge of the capacitor area All capacitor bank neutral connections are made to area. this isolated peninsula ground grid conductor.
Grounding of Wye Banks
Grounding of Wye Banks
References • IEEE Std. 18 • IEEE C37.99 • NEMA -1 • IEEE Std 1036 •IEEE IEEE Std 399 (Brown Book)
Outline Day 1 • Basic Power Calculations • Capacitor Fundamentals • Capacitor Ratings • Capacitor Application • Capacitor p Protection Day 2 • Harmonics • Capacitor Bank Design Considerations
Shunt Capacitors • Medium Voltage pp • Substation Applications • Power Factor Correction
Basic Power Calculations
Basic Power Calculations C Common Q Questions: ti What are VARs? Wh do Why d we care about b t VARs? VAR ?
Basic Power Calculations Most plant loads (motors, transformers, etc) are INDUCTIVE and require a magnetic field to operate. The magnetic field is necessary, but produces NO USEFUL WORK. The utility must supply the power to produce the magnetic field and the power to produce USEFUL work. The ACTIVE component produces the USEFUL work, the REACTIVE component produces the magnetic field.
VAR
Basic Power Calculations An analogy gy that most can understand.
W
VA
Mug Capacity i off equipment i (i.e. xfmr, cable, swgr, etc) ee Beer Stuff that you want Foam (Head) St ff which Stuff hi h prevents t you from maximizing the amount of beer that you get
Basic Power Calculations
V: Reference f voltage l IR: Resistive load IL: Inductive load IC: Capacitive load
Basic Power Calculations
Vref Ires
Power = Re(V I*)
Basic Power Calculations
Vref Ilag
Power = Re(V I*) Note that due to phase shift (30 degrees) only 86.66% of current is applied to calculate work.
Basic Power Calculations
Vref Ilead
Power = Re(V I*) Regardless whether leading or lagging, power calculation yields similar results
Basic Power Calculations
Vref I90
Power = Re(V I*) In the case where current leads or lags the voltage by 90 degrees, P = 0
Basic Power Calculations Power Factor [ PF ] = Cos
=P/S
A) V k S(
P (W) Power Triangle g The relationship between S, P, and Q. This figure represents a lagging power factor. If Q is negative, leading power factor.
Q (kVAR)
Basic Power Calculations So how do we know if our current is lagging or leading the voltage, and what can we do to correct it? Consider this example, example
Basic Power Calculations Incomingg service is at 99.5% capacity 5500 kW 3400 kVAR 6466 kVA
2000 kW 1000 kVAR
1000 kW 800 kVAR
2500 kW 1600 kVAR
Basic Power Calculations Power Factor [ PF ] = Cos =gP / S 0.85 lag A) V k S(
31.8 deg P (W) 5500 kW Power Triangle g Even though our facility require only 5500 kW to perform Real Work, our incoming service must be sized for 6466 kVA.
Q (kVAR) 3400 kVAR
We have a 6500 kVA mug that is holding 5500 kW and 3400 kVAR.
W
VA
VAR
Basic Power Calculations
With the existing configuration the facilityy cannot add anyy loads without upgrading the incoming service.
Basic Power Calculations Incomingg service is at 85% capacity 5500 kW 0 kVAR 5500 kVA
3400 kVAR
2000 kW 1000 kVAR
1000 kW 800 kVAR
2500 kW 1600 kVAR
Basic Power Calculations Power Factor [ PF ] = Cos = Py / S 1.0 unity A) V k S(
0 deg P (W) 5500 kW Power Triangle g The cap bank is providing 3400 kVAR, so our service is now providing only 5500 kVA (reduction from 6466 kVA.
Q (kVAR) 0 kVAR
Basic Power Calculations COS [ ] = 0.67 COS [ ] = 0.95 Q2 = Q1 + Qc
VA k 00 1,5
Q1 = 1,118 kVAR
VA 1,053 k
Q2 = 330 kVAR Q
1 000 kW 1,000 Required Apparent Power Before and After Adding a Power Capacitor Bank
Qc = 788 kVAR
An example of how to calculate the size of a cap bank based on a target power factor
Basic Power Calculations
But wait, there’s more….
Basic Power Calculations Odds are that the utilityy is charging you for penalties for a low power factor.
5500 kW 3400 kVAR 6466 kVA
5500 = 0.85 6466 Typically, penalties are applied for power factor less than 0.95%. 2000 kW 1000 kVAR
1000 kW 800 kVAR
2500 kW 1600 kVAR
Basic Power Calculations There are also voltage g considerations. 5500 kW 3400 kVAR 6466 kVA Volt Drop = 5.8%
Assuming typical values: Source impedance = 9% at 6500 kVA The expected voltage drop at the main bus will be close to 6%!
2000 kW 1000 kVAR
1000 kW 800 kVAR
2500 kW 1600 kVAR
Basic Power Calculations Capacitor bank reduces the voltage drop at main bus by 5%! 5500 kW 0 kVAR 5500 kVA Volt Drop = 1.0% 3400 kVAR
2000 kW 1000 kVAR
1000 kW 800 kVAR
2500 kW 1600 kVAR
Basic Power Calculations Note that the capacitor bank can also raise the bus voltage g above the nominal value. 5500 kW -1600 kVAR 5728 kVA Volt Drop = -1.5% 5000 kVAR
2000 kW 1000 kVAR
1000 kW 800 kVAR
2500 kW 1600 kVAR
Basic Power Calculations The voltage continues to rise if the capacitor p bank remains connected and the load is reduced.
3000 kW -3200 kVAR 4386 kVA Volt Drop = -3.6%
5000 kVAR
Out of service
2000 kW 1000 kVAR
1000 kW 800 kVAR
0 kW 0 kVAR
Capacitor Fundamentals
Capacitor Fundamentals
C = eo A / d for a parallel plate capacitor, where eo is the permittivity of the insulating material (dielectric) between plates.
Capacitor Fundamentals
We recall that we can add series capacitances to obtain an equivalent capacitance.
1 / Ceq = 1 / C1 + 1 / C2
Capacitor Fundamentals
Similarly, we can add parallel capacitances to obtain an equivalent capacitance.
Ceq = C1 + C2
Capacitor Fundamentals …but we typically do not have much use for capacitance values. So we convert capacitance to impedance:
1 1 XC C 2fC
Capacitor Fundamentals Z R jX c Z R jX c Z XC 2
V Z S
2
V S XC
Capacitor Fundamentals 2
V [kV ] S [kV kVAR] *1000 000 Z [] Example: The capacitance of a capacitor is 6.22 F and the nameplate voltage is 8000 V. Calculate the power rating.
XC
1 426.7[] 6 2(3.14)(60)(6.22 x10 ) (8) 2 S 1,000 150[kVAR] 426.7
Capacitor Fundamentals
Capacitor Fundamentals
Capacitor Fundamentals
Capacitor Fundamentals
Capacitor Fundamentals
Capacitor Fundamentals Note that IEEE Std 18 requires the discharge resistor to reduce the terminal voltage to 50 V in th time the ti frame f as specified ifi d in the table below.
Discharge resistor
Capacitor Ratings
Capacitor Ratings Medium-voltage Medium voltage capacitors are available in many different styles. styles The main points of differentiation are listed below: • Voltage rating • kVAR rating • Single g bushingg or dual bushingg • Internally fused, externally fused, or fuseless
Capacitor Ratings IEEE 81 defines the ratings for capacitors • Voltage, rms (terminal to terminal) • Terminal-to-case (or ground) insulation class • Reactive power • Number of phases p • Frequency
Capacitor Ratings IEEE 18 provides capacitor tolerances • The capacitance shall not vary more than -0% to +10% of nominal value based on rated kVAR, voltage, and frequency measured at 25 deg C. This means that a new 150 kVAR unit can range g anywhere y from 150 kVAR to 165 kVAR.
Capacitor Ratings IEEE 18 states the capacitor is intended to operate at or below rated voltage. Capacitors shall be capable of continuous operation given that none of the following limitations are exceeded: • 110% of rated rms voltage (temporary overvoltage parameters will be discussed later)) • 120% of peak voltage, including harmonics but excluding transients • 135% of nominal rms current based on rated kVAR and rated voltage • 135% of rated kVAR
Capacitor Ratings
* Impulse tests shall be applied between terminals and case, with the terminals connected together. For capacitors having bushings with two different BIL ratings, this test shall be based on the bushing with the lower BIL. The nameplate shall show both BIL ratings, e.g. 150/95 kV BIL. ** Not applicable to indoor ratings
Capacitor Application
Capacitor Application Power factor correction capacitor banks are typically installed in the following ways : • Pole top • Metal-Enclosed / Pad-Mount • Ope Open rack ac • Terminal end at equipment
Capacitor Application
P l Top Pole T Installation I t ll ti
Capacitor Application
Transient inrush reactors
P l Top Pole T Installation I t ll ti
Capacitor Application
Pad Mounted Installation Pad-Mounted
Capacitor Application
Metal-Enclosed Metal Enclosed Substation Installation
Capacitor Application
Three-phase iron core harmonic filter reactor Metal-Enclosed Metal Enclosed Substation Installation
Capacitor Application
Open Rack, Medium-Voltage Substation Installation
Capacitor Application
Open Rack, High-Voltage Substation Installation
Capacitor Application
Installation in Equipment
Capacitor Application Power factor correction capacitor banks can be configured in the following ways : • Delta • Wye - Solidly Grounded • Wye y - Ungrounded g A common misconception is that the capacitor bank should be connected Delta since it is being applied to a delta or high highimpedance grounded system. This is NOT true.
Capacitor Application The driving factor which determine the configuration for the given application is COST. Voltage considerations IEEE 1036 suggests that only banks rated 2400 V and below should be Delta connected. connected This is mainly because standard voltage ratings for wye connected banks may not be available. Cost of phase-to-phase vs phase-to-neutral rated capacitors at higher voltages tends to point installations towards wye connected banks for larger bank installations.
Capacitor Application Delta Lower voltages (<= 2400 V) • Standard capacitors are typically not available at 1380 V Distribution systems (pole top) • Units are configured with a single series group of capacitors with capacitors rated phase-to-phase. phase to phase Therefore Therefore, unbalance detection is not required.
Capacitor Application Wye – Solidly Grounded • Initial cost of the bank may be lower since the neutral does not have to be insulated from ground. ground • Capacitor switch recovery voltages are reduced • High inrush currents may occur in the station ground system • The Th grounded-wye d d arrangement provides id a low-impedance l i d fault f l path which may require revision to the existing system ground pprotection scheme. Typically yp y not applied pp to ungrounded g systems. y • When applied to resistance-grounded systems, difficulty in coordination between capacitor fuses and upstream ground protection relays (consider coordination of 40 A fuse with a 400 A grounded system). • Typical for smaller installations (since auxiliary equipment is not required) i d)
Capacitor Application The most common capacitor bank configurations for larger substation applications are Wye-Ungrounded
Three of the most common unbalance protection schemes are shown. Discussion of the protection schemes will be presented later.
Capacitor Protection Fusing • Fuseless • Internally I t ll Fused F d • Externally Fused
Capacitor Protection
B k Protection Bank P t ti Summary S
Capacitor Protection Fuseless Capacitors Constr cted of small capacitor elements which Constructed hich are arranged in series and parallel. The elements are constructed of aluminum foil with a dielectric of electrical grade polypropylene. This design provides a safe failure mode. In the event that the dielectric fails, the energy in the resulting small arc punctures many layers of the thin film and foil within the element. The arc causes the film layer to receded allowing many layers of the aluminum foil electrodes to touch and weld together forming an electrically stable electrical t. t This results in an entire series section being shorted.
Capacitor Protection Example of Fuseless Installation
Capacitor Protection Internally Fused Capacitors Constr cted such Constructed s ch that each element is protected with a series connected current limiting fuse. The design is such that isolated fusing prevents potential damage to the adjacent elements and fuses. The current limiting mode chops the fault current to prevent the energy stored in the parallel connected elements from being discharged into the faulted element.
Capacitor Protection
Group Fusing
Individual d v du Fusing us g
Capacitor Protection Group Fusing– Considerations for Selecting Fuse (typical for distribution pole mounted racks)
• Continuous Current • Transient Current • Fault Current • Tankk Rupture Curve C C di i Coordination • Voltage on Good Capacitors
Capacitor Protection Continuous Current • For wye-solidly grounded systems: Fuse > = 135% of rated capacitor current (includes overvoltage, capacitor tolerances, and harmonics). • For wye-ungrounded y g systems: y Fuse > = 125% of rated capacitor current (includes overvoltage, capacitor tolerances, and harmonics). Care should be taken when using NEMA Type T and K tin links which are rated y 1.50. 150%. In this case, the divide the fuse ratingg by
Capacitor Protection Transient Current • Capacitor switching (specifically back-to-back switching) • Lightning surges Back-to-back is typically yp y not a factor for ppole mounted capacitors p banks. High frequency lightning surges: Use NEMA T tin links for ampere ratings up to 25 A. Use NEMA K tin links for ampere ratings above 25 A.
Capacitor Protection Fault Current Ensure that the fuse can interrupt the available fault current Tank Rupture Coordination Ensure that the fuse maximum clearingg TCC curve for the fuse link is plotted below the capacitor tank rupture curve. In cases of high fault currents, the tank rupture curve should be compensated for asymmetry. asymmetry Voltage on Good Capacitors For an ungrounded d d system, a fault f l on one phase h results l in i a 1.73 times overvoltage on the un-faulted phases. Ensure that the fault p failure. is cleared before the second capacitor
Capacitor Protection Problems with Fusing of Small Ungrounded Banks Consider a 12.47 kV, 1500 kVAR cap bank made of three (3) 500 kVAR single-phase i l h units. i 1500[kVAR] 69.44[ A] 69.44 1.5 104[ A] 100[ A]Fuse 3 12.47[kV ]
If a capacitor fails, we will expect approximately 3x line current. pp y 500 seconds to clear this It will take a 100 A fuse approximately fault (3 x 69.44 A = 208.32 A). The capacitor case will rupture long before the fuse clears the fault. The solution is using smaller units (explanation to follow).
Capacitor Protection Individual Fusing– Considerations for Selecting Fuse (typical for substation capacitor banks)
• Continuous Current • Transient Current • Fault Current • Tank Rupture Curve Coordination • Voltage on Good Capacitors • Energy Discharge into Faulted Unit • Outrush Current • Coordination with Unbalance Detection System
Capacitor Protection Continuous Current • Fuse > = 135% of rated capacitor current (includes overvoltage, capacitor tolerances, and harmonics) Care should be taken when using NEMA Type T and K tin links which are rated 150% In this case, 150%. case the divide the fuse rating by 1.50. 1 50
Capacitor Protection Transient Current • Lightning surges • Capacitor switching (specifically back-to-back switching) High g magnitude, g , high g frequency q y lightning g g surges g are typically yp y not a concern for substation installations. Back-to-back Back to back switching is typically controlled with pre-insertion pre insertion closing resistors or current limiting reactors. By the h nature off installation, i ll i the h parallel ll l fuses f will ill share h the h transient current and will not be a factor.
Capacitor Protection Fault Current Ensure that the fuse can interrupt the available fault current. In substation banks with multiple series groups, fault current will not flow through g a failed capacitor p unit unless other units experience a simultaneous failure. For this reason expulsion fuses are commonly used rather than current limiting fuses Tank Rupture Coordination Ensure that the fuse maximum clearing TCC curve for the fuse li k is link i plotted l d below b l the h capacitor i tankk rupture curve. In cases off high fault currents, the tank rupture curve should be compensated y y for asymmetry.
Capacitor Protection Example of a Definite Tank Rupture Curve. The time Th i between b the h rupture curve and the fuse maximum clear curve is the coordination margin.
Capacitor Protection Example of a 10% and 50% Rupture Curve for a 100 kVAR Capacitor. Probability y based tank rupture p curves are developed when there is too much variance in rupture test data. Based on the 10% and 50% curves, one can extrapolate the curves for f any probability. b bili
Capacitor Protection Voltage on Good Capacitors When a short-circuit on one unit occurs, an overvoltage results on the un-faulted phases. Ensure that the fault is cleared before the second capacitor failure. A table summarizes this voltage rise on the un-faulted units
Per Unit Voltage on Un-failed Capacitors
Capacitor Protection Energy Discharge Into a Failed Unit When a capacitor failure occurs, the stored energy in the parallel connected capacitors can discharge thro through gh the failed capacitor and its fuse. The total calculated parallel stored energy should not exceed the energy capability (Joule rating) of the capacitor and fuse. If the energy capabilities are exceeded, a failure of the fuse and/or rupture of the capacitor tank can result. Typical rating of film capacitors is 15,000 Joules (4650 kVAR in parallel) and 10,000 Joules (3100 kVAR in parallel) for paperfilm capacitors. capacitors Expulsion fuses are typically rated 30 30,000 000 Joules. Current limiting fuses are required if ratings are exceeded. (1 Joule = 1 W x sec sec, use 00.2 2 cycle clearing time for calculation)
Capacitor Protection Outrush Current When a capacitor failure occurs, the parallel connected capacitors can discharge high frequency current into the failed capacitor. The fuses of the un-failed capacitors should be able to withstand the high g frequency q y discharge g currents. These calculations and measurements are complex and are determined by the manufacturer. Coordination with Unbalance Detection Scheme The individual fuse must clear the fault before the unbalance protection i scheme h trips i the h entire i capacitor i bank. b k
Capacitor Protection Fusing Recommendations by McGraw Edison
Capacitor Protection Recall Problem with Fusing of Small Ungrounded Banks 12.47 kV, 1500 kVAR cap bank made of three (3) 500 kVAR units 1500[kVAR] 69.44[ A] 69.44 1.5 104[ A] 100[ A]Fuse 3 12.47[kV ]
It will take a 100 A fuse approximately 500 seconds to clear this fault (3 x 69.44 A = 208.32 A). The capacitor case will rupture long before the fuse clears the fault. The solution is using smaller units with individual fusing. Consider five ((5)) 100 kVAR capacitors p pper phase, p , each with a 25 A fuse. The clear time for a 25 A fuse @ 208.32 A is below the published capacitor rupture curve.
Capacitor Protection Why is the Current 3 x Nominal Line Current for a Phase-toNeutral Fault on a Wye-Ungrounded Capacitor Bank? B
A N
A N
VNG Since V = I*Z, where Z is constant t t (assuming ( i f = 60 hz) h ) If voltage across capacitor is increased by 1.732, the current also increases byy factor of 1.732
C
IA = 3.0 3 0 p.u.
B
C
Capacitor Protection
Minimum Conductor Size It was noted that capacitors are rated 135% of rating. This requires the conductor to be sized 135% of the nominal capacitor rating. rating
Capacitor Protection Unbalance Protection As single-phase units in a multiple unit/phase installation fail and are removed from service, service the remaining units are experience an overvoltage condition. IEEE Standard 1036 provides overvoltage limitations. Duration
Max Voltage (x rated RMS)
6 cycles
2.20
15 ccycles cles
2 00 2.00
1s
1.70
15 sec
1.40
1 min
1.30
An unbalance protection scheme must by implemented to prevent g units. the failure of the overvoltaged
Capacitor Protection
Capacitor Protection
Neutral Voltage Unbalance with Unbalance Compensation
Capacitor Protection
VA
VNG
Normal Conditions VN = VG VAN = VBN = VCN
VG VN VC
VB
Capacitor Protection Ungrounded or Impedance Grounded System
VA
VNG
Normal Conditions VN = VG VAN = VBN = VCN = 1.0 .0 p.u.
VG VN VC
VB
Capacitor Protection
VNG Phase to Neutral Fault VNG = VLN VAN = VBN = VLL = 1.732 p.u.
Capacitor Protection
VA
VNG VG One Can Removed VNG = 0.2 p.u. VCN = 1.2 1 2 p.u. pu
VC
VN VB
Capacitor Protection Ungrounded or Impedance Grounded System
VA
VNG
Normal Conditions VN = VG VAN = VBN = VCN = 1.0 .0 p.u.
VG VN VC
VB
Capacitor Protection Ungrounded or Impedance Grounded System
VA Gnd VNG
Ground Fault VNG = VLN VAG = VBG = VLL = 1.732 1 732 p.u. pu
VN VGC VC =V Ground Fault at Cap Bank or A Anywhere h on the h S System
VB
Capacitor Protection
Wye-Ungrounded: Voltage Between Capacitor Bank Neutral and Ground vs. Percentage of Capacitor Units Removed from Series Group
Capacitor Protection
Wye-Ungrounded: Voltage on Remaining Capacitor Units in Series Group vs. Percentage of Capacitor Units Removed from Series Group
Capacitor Protection
Wye-Grounded: Neutral Current vs Percentage of Capacitor Units Removed from Series Group
Capacitor Protection
Wye-Grounded or Delta: Voltage on Remaining Units in Series Group vs. Percentage of Capacitor Units Removed from Series Group
Capacitor Protection
Double Wye-Ungrounded, Neutrals Tied Together: Neutral Current vs. Percentage of Capacitor Units Removed from Series Group
Capacitor Protection
Double Wye-Ungrounded, Neutrals Tied Together: Voltage on Remaining Capacitor Units in Series vs. Percentage of Capacitor Units Removed from Series Group
Capacitor Protection A phase
B phase
C phase P: Number of units in group (P=6)
S: Number of series groups (S=4)
Reference Figure for Calculations to Follow
Capacitor Protection # of Series Groups
Grounded Y or Delta
Ungrounded Y
Split Ungrounded Y (equal sections)
1
-
4
2
2
6
8
7
3
8
9
8
4
9
10
9
5
9
10
10
6
10
10
10
7
10
10
10
8
10
11
10
9
10
11
10
10
10
11
11
11
10
11
11
12 and over
11
11
11
Minimum recommended number of units in parallel per series Group to limit voltage on remaining units to 110% with one unit out
Capacitor Protection
Many more configurations and calculations shown in IEEE C37.99
Day 2
Capacitor Fundamentals Further discussion on capacitor voltage ratings: On a ungrounded or impedance grounded system, a ground fault on one phase will cause the other two phases will be elevated by 1 732 1.732. Does this mean that capacitors must be rated phase-to-phase? Certainly nothing wrong with this, but cost will be significantly higher. g
Capacitor Fundamentals Recall:
V2 S XC
This means that a 150 kVAR, kVAR 12470 V unit applied at 7200 V will provide only 50 kVAR. S NEW S NEW
7200[V ]2 150[kVAR] 2 12470[V ] 50[kVAR]
Capacitor Fundamentals Using 12470 V capacitors on a 12470 V Ungrounded or Resistance-Grounded System y will require q 3x more cans. It should be noted that the 12470 V cans will also be larger than the 7200 V cans. cans Results in a much larger and more costly installation. This solution would be required if a ground fault could be maintained for extended pperiods of time.
Capacitor Fundamentals Perhaps a 150 kVAR or 200 kVAR, 7620 V or 7960 V units applied at 7200 V would be a better solution. 7200[V ] 150[kVAR ] 2 7620[V ] 134[kVAR ]
S NEW S NEW
2
S NEW OR
OR 2
7200[V ] 150[kVAR ] 2 7960[V ] 123[kVAR ]
S NEW S NEW
S NEW
7200[V ]2 200[kVAR ] 2 7620[V ] 178[kVAR ]
S NEW S NEW
7200[V ]2 [kVAR] 2 7960[V ] 163[kVAR ]
Note that the 7620 V unit provides an additional 6% The 7960 V unit provides an additional 11%
Capacitor Fundamentals Explusion Fuses: Provides a means of disconnecting a failed capacitor from the circuit by melting a tin-lead low current link. The shorted capacitor unit causes a large increase in the current through the fuse. The current is limited only by the power system reactance and the other capacitor units in series with the failed capacitor unit. The pressure is generated by the hot arc making with the fiber lining of the fuse tube. The link is cooled and stretched as it is forced out the tube. The fuse continues to conduct until a natural current zero occurs. The current zero is causedd bby th the power system t fault f lt currentt crossing i zero. If other th capacitors it are connected in parallel with the failed unit, all the stored energy in these capacitors will be absorbed in either the fuse operation or the failed capacitor unit. Most of gy is absorbed in the failed capacitor. p the energy
Capacitor Fundamentals Current Limiting Fuses: Uses a long uniform cross section element. This configuration makes the fuse a current chopping fuse. The fuse develops a back voltage per inch of element across the entire length of the element. When this voltage exceeds the available voltage across the fuse, the fuse forces the arc to extinguish. The result is that a trapped voltage may and probably will remain on the other capacitors in the series group. The fuse by its design avoids absorbing all of the available energy on the series group. This fuse is used for capacitor banks with a large number of parallel ll l capacitors. it It can be b usedd on applications li ti with ith essentially ti ll infinite i fi it parallel ll l stored energy, as long as sufficient back voltage can be developed to force the current to extinguish. This is the fuse we apply to series, large shunt, and DC banks. Because of the high back voltage that is developed, this fuse must be used with several capacitors in parallel to limit the voltage build up or a flashover may occur elsewhere in the capacitor rack. rack
Capacitor Fundamentals Current Limiting Fuses vs Expulsion Fuses:
Expulsion Fuse
C Current Limiting Li i i Fuse F
Capacitor Fundamentals Current Limiting Fuses vs Expulsion Fuses: Expulsion Fuses Operate mechanically and provide a visual indication Require additional space for operation Typically applied for outdoor application due to ionized gas release. Combination C bi ti expulsion l i with ith currentt limiting li iti characteristic h t i ti fuses f can be b usedd in i indoor metal-enclosed equipment. p Less expensive
Capacitor Fundamentals
Capacitor Fundamentals Current Limiting Fuses Do not emit ionized gases during operation. operation Ionized gases are undesirable because they can cause bushing and insulator flashovers that result in additional damage. Do not require ventilation. Fast current-limiting operation High interrupting capacity, noiseless operation Can be specified for indoor and outdoor applications . No pressure build-up, therefore, no vents or special reinforced compartments are required. i d More expensive
Capacitor Fundamentals Note no pigtail and blown fuse indication
Capacitor Fundamentals Current Limiting with Expulsion
Capacitor Fundamentals What about arresters? How and where should they be applied? Depending on application, environment, exposure to switching etc, switching, etc arresters may be necessary. necessary We recall that when a travelling wave meets a high impedance, the wave can double in size. For this reason, arresters (if used) should be installed as close to the capacitor bank as ppossible. Installation of arresters at the breaker feeding the capacitor bank will not do much for protection of the capacitor bank.
Capacitor Protection A basic three (3) arrester method is shown below. This is typical for solidly grounded systems and wye wye-grounded grounded capacitor banks.
Capacitor Protection Depending on type of installation, system parameters and level of protection required, a six(6) arrestor method may be applied.
Capacitor Protection For an ungrounded system or a high-impedance grounded system, a four (4) arrestor grounding method might be considered an wye ungrounded bank.
VLL
Ground Fault VNG = VLN VAG = VBG = VLL = 1.732 p.u.
VLL
Phase to Neutral Fault VNG = VLN VAN = VBN = VLL = 1.732 p.u.
ArrPH
If faults can be maintained,
ArrN ArrPH must be rated VLL ArrN must be rated VLN The effective arrester MCOV is VLL + VLN
Capacitor Protection Note that if a basic three (3) arrester method is applied to an ungrounded bank, the arresters must be rated high enough to sustain a temporary overvoltage condition during a phase-toground fault on the system. This may not provide an adequate level of protection for the capacitors. Phase to Neutral Fault VNG = VLN VAN = VBN = VLL = 1.732 1 732 p.u.
Arresters do not provide protection across the capacitor bushings. Note that the BIL applies to bushing-tocase insulation. i l ti
VLL
If faults can be maintained, Arresters must be rated VLL
VLL
Ground Fault VNG = VLN VAG = VBG = VLL = 1.732 1 732 p.u.
Capacitor Protection Good Presentations on Capacitor – Arrester Applications “Guidelines “G id li ffor S Selection l ti off S Surge A Arresters t ffor Sh Shuntt Capacitor Banks” – ABB Technical Information “Surge Arrester Application of MV-Capacitor Banks to Mitigate Problems of Switching Restrike” – CIRED 19th International Conference on Electricity Distribution, Distribution Vienna, Vienna 21-24 May 2007. Both B th off these th papers also l address dd phase-to-phase h t h arrester t connections.
Harmonics
Harmonics Recall that the impedance of a capacitor is inversely proportional to the system frequency. frequency
1 1 XC C 2fC Harmonics flow to the point of lowest impedance. The higher the harmonic, harmonic the lower the impedance of the capacitor. capacitor As capacitors absorb harmonics, the capacitor heats up and th life the lif expectancy t is i reduced. d d The Th voltage lt harmonics h i stress t the capacitor dielectric and reduce the life expectancy of the capacitor.
Harmonics
Harmonics Where do harmonics come from? • Power Electronics (drives, rectifiers, computer power supplies, etc) • Arcingg Devices ((welders,, arc furnaces,, florescent lights, g , etc)) • Iron Saturating Devices (transformers) • Rotating Machines (Generators) • Parallel Resonance (between cap bank and power equipment) IEEE Std 519 provides recommended limits of harmonic distortion at the point-of-common-coupling (PCC) with the y utility.
Harmonics
Harmonics
Harmonics Resonance When a number of harmonic current sources are injecting currents into the supply and the frequency of one of the harmonics coincides with the resonant frequency of the supply transformer and Power Factor Correction capacitor combination, the system resonates and a large circulating harmonic current is excited between these components components. The result of this is that a large current at this harmonic flows in the supply transformer, this resulting in a large harmonic voltage distortion being imposed upon the load voltage.
Harmonics A studyy should be pperformed to determine levels of harmonics on a system to determine if any filters are necessary when installing a capacitor bank. Care should be taken to determine if the filtered capacitor bank will introduce any resonance problems. If resonance problems exist, the fil design filter d i must be b adjusted. dj d
Harmonics
An example of a 13 13.8 8 kV harmonic filter
Capacitor Bank Design Considerations
Design Considerations So how do we size a capacitor bank? Determine D t i your primary i goall • Voltage • Lower utility bill (avoid penalties) • Increase capacity of system It can be all three, or any combination of the above. Note that correcting to unity power factor at maximum load is costly tl andd may nott be b necessary.
Design Considerations For a 20 MVA load at 0.88 0 88 power factor (17.6 (17 6 MW, MW 9.5 9 5 MVAR) To achieve 95% power factor, a 3.72 MVAR bank is required To achieve 98% power factor, a 5.93 MVAR bank is required To achieve unity power factor, a 9.50 MVAR bank is required
Design Considerations Determine if current limiting reactors or tuning reactors are required. Harmonics and resonance may dictate tuning reactors Back to back switching may require current limiting reactors Back-to-back (unless another method is used to mitigate the switching surges, i.e. pre-insertion closing resistors/reactors, zero-crossing breakers etc) breakers,
Design Considerations Determine the proper voltage. Capacitors are very susceptible to voltage transients and harmonics. Increasing the rated voltage increases the protective margin on the insulation. The voltage at the capacitor terminals will be higher than bus voltage lt if reactors t are utilized. tili d It is i important i t t to t t for f this thi voltage difference. Determine the voltage swing of the system. Will the capacitors remain on-line while the facility is lightly loaded.
Design Considerations We listed some reasons for specifying higher than bus nominal rating of capacitors. However, care must be taken to ensure that the kVAR rating is properly adjusted as a result. Three (6) 150 kVAR, 7960 V wye-connected capacitors provide a nominal 901 kVAR when connected to a 13.8 kV bus. Three (6) 150 kVAR, 8320 V wye-connected capacitors provide a nominal 825 kVAR when connected to a 13.8 13 8 kV bus
Design Considerations Determine optimal size and number of stages. Depending D di on swing i in i plant l t load, l d a single i l bank b k sized i d for f full f ll plant capacity may not be the answer. IEEE 1036 recommends limiting the voltage change to 2-3%. The delta voltage can be estimated by: V
MVAR 100% MVASC
Switch of a capacitor applies high stresses to the insulation. insulation Limiting the number of stages and limiting the frequency of switching will increase the life. Ideally, a capacitor is switched on and left on.
Design Considerations Determine best location for the installation installation. The most effective placement for power factor correction capacitor banks is at the load. However, this is not always practical or cost effective. Typically, a capacitor bank is installed on each bus of a main-tiemain switchgear. If capacitors are installed at the motor pecker head (running capacitors) ensure that the capacitor VAR rating does not exceed capacitors), 90% of the motor no-load VAR. Otherwise, it is possible to damage the motor by overexcitation.
Design Considerations Use caution when sizing motor running capacitors. Logic would suggest that installation of a power factor correction capacitor at the motor terminals sized to provide y ppower factor makes sense. unity THIS IS NOT THE CASE. Do not exceed 90% of the motor no-load no load kVAR demand. demand Exceeding this value can result in damage to the motor insulation as a result of overexcitation.
Design Considerations As an example for a 4000 V, 4000 hp motor: 100% load current = 495 A @ 89.7% pf 100% load kVAR = 1516 kVAR No load current = 117 A @ 6.3 % pf No load kVAR = 809 kVAR Max size of running capacitor is 0.90 0 90 x 809 kVAR = 728 kVAR
Design Considerations M: Motor Magnetizing Curve C1: Capacitor size at 100% motor mag current C2: Capacitor sized > 100% motor mag current C4: Capacitor sized < 100% motor mag current
If the capacitive reactance of the capacitor is less less than that of the motor reactance (this occurs when to large of a capacitor is chosen). This combination of reactance will result in a resonant frequency below 60 hertz hertz. Therefore, Therefore as the motor slows in speed, speed the frequency of the motor terminal voltage will decrease from a value of near 60 hertz toward zero. When the motor's terminal voltage frequency es through the resonant frequency setup between the capacitor reactance and the motor reactance, the terminal voltage will become very high, only limited by the properties of the iron Depending on the inertia of the motor, iron. motor this resonance (or high voltage) may be present for a considerable period of time.
Design Considerations Determine the most optimal type of installation. installation Will the capacitor bank be installed within a fenced substation? Metal-enclosed, pad mount, or open rack may be good choices Will the capacitor bank be installed in a process area? Metal-enclosed or pad mount may be good choices Will the capacitor bank be pole mounted on a distribution line?
Bank Failures
Design Considerations Consider the impact to personnel safety adjacent equipment when deciding between a metal-enclosed and open-rack system. Porcelain can resemble shrapnel when a capacitor bushing fails.
Design Considerations Determine the most optimal configuration. Higher reliability costs more. y g (1) ( ) 800 kVAR pper pphase 2400 kVAR,, 13800 V,, wye-grounded bank will be a smaller footprint and cost less than 2400 kVAR, kVAR 13800 V V, wye-ungrounded wye ungrounded (8) 100 kVAR per phase bank. However, the H h reliability li bili off the h wye-ungrounded d d bbank k iis significantly higher
Design Considerations Determine the switching equipment When breakers are used for switching capacitors (single bank or back-to-back switching), the breakers must be rated for p switching. g capacitor IEEE C37.99 provides the equations for calculating the inrush current and frequency. frequency
Design Considerations
Design Considerations Consider a single 4800 kVAR wye-ungrounded bank switched (with nominal inductance from equipment): 3253A pk @ 600 Hz, the product is 0.20 x 107 Switching a second similar bank on the same bus without current limiting reactor: 24,058 A pk @ 7.66 kHz, the product is 18.4 x 107 B adding By ddi a 100 mH H current limiting li i i reactor, the h inrush i h is: i p is 1.7 x 107 7254 A ppk @ 2.31 kHz,, the product
Design Considerations
Design Considerations
Design Considerations Energization of a single capacitor bank.
Design Considerations
B k b k switching Back-to-back i hi off the h same unit. i
Design Considerations
E Example l off breaker b k with i h pre-insertion i i resistor. i
Design Considerations
Another application.
Design Considerations
A Pre-Insertion Manufacture’s Perspective. p
Design Considerations Another concern is voltage amplification as a result of switching a second capacitor bank.
Design Considerations
Design Considerations Consider other accessories: • Disconnect switch • Grounding switch • Kirk-keyy interlock • Ventilation requirements • Control power
Design Considerations
Design Considerations
Design Considerations Be aware that larger medium voltage motors may include surge g ppacks. The surge pack will decrease the crest voltage and rate of rise of the impending surge. surge High rates of rise damage end turns while high crest voltage damage winding to core insulation. Typically the capacitance of the is small enough that it can g but this should be verified. be neglected,
Design Considerations
Typical Surge Pack Application
Design Considerations Do not confuse Harmonic Filter Banks with Power Factor Correction Banks. Banks The voltage ratings of harmonic filter banks are substantially higher because they are connected on the back end of a tuning reactor where the voltage is substantially higher. As a result of thee higher g e voltage, vo ge, thee installed s ed kVAR V ccan be anywhere yw e e from o 25% to 40% higher than nominal design. The capacitor cans must be capable of this output. output
Grounding of Wye Banks If multiple wye-grounded banks are in close proximity, use ppeninsula ggrounding g or single-point g p ggrounding. g Single-Point Grounding The neutrals of all banks of a given voltage are connected together with insulated cable/bus and tied to the ground grid only at one point. This prevents high-frequency currents (due (d to back-to-back b k b k switching) i hi ) from f flowing fl i into i the ground grid.
Grounding of Wye Banks Peninsula Grounding One or more isolated gground ggrid conductors are carried underneath the capacitor rack of each phase and tied to the station ground grid at one point at the edge of the capacitor area All capacitor bank neutral connections are made to area. this isolated peninsula ground grid conductor.
Grounding of Wye Banks
Grounding of Wye Banks
References • IEEE Std. 18 • IEEE C37.99 • NEMA -1 • IEEE Std 1036 •IEEE IEEE Std 399 (Brown Book)
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