6. 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 themust 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.
7. Basic Power Calculations
An analogy that
VAR
gy
most can understand.
Mug
i f i
VA
Capacity of equipment
(i.e. xfmr, cable, swgr, etc)
Beer
W
ee
Stuff that you want
Foam (Head)
St ff hi h tStuff which prevents
you from maximizing the
amount of beer that you get
8. Basic Power Calculations
f lV: Reference voltage
IR: Resistive load
IL: Inductive load
I : Capacitive loadIC: Capacitive load
13. Basic Power Calculations
Power Factor [ PF ] = Cos = P / S
S (kVA)
Q (kVAR)
P (W)
Power Triangle
The relationship between S, P, and Q. This figure represents a lagging power
factor. If Q is negative, leading power factor.
g
14. Basic Power Calculations
So how do we know if our
current is lagging or leadingcurrent is lagging or leading
the voltage, and what can we
do to correct it?do to correct it?
Consider this exampleConsider this example,
15. Basic Power Calculations
Incoming service is at
5500 kW
3400 kVAR
g
99.5% capacity
6466 kVA
2000 kW 1000 kW 2500 kW
1000 kVAR 800 kVAR 1600 kVAR
16. Basic Power Calculations
Power Factor [ PF ] = Cos = P / S0.85 lag
S (kVA)
g
Q (kVAR)3400 kVAR
31.8 deg
P (W)
Power Triangle
5500 kW
g
Even though our facility require only 5500 kW to perform Real Work, our
incoming service must be sized for 6466 kVA.
17. Basic Power Calculations
We have a 6500 kVA mug
that is holding 5500 kW
and 3400 kVAR.
VAR
VA
W
With the existing configuration
the facility cannot add any loadsy y
without upgrading the incoming
service.
18. Basic Power Calculations
Incoming service is at
5500 kW
0 kVAR
g
85% capacity
5500 kVA
3400 kVAR
2000 kW 1000 kW 2500 kW
1000 kVAR 800 kVAR 1600 kVAR
19. Basic Power Calculations
Power Factor [ PF ] = Cos = P / S1.0 unity
S (kVA)
y
Q (kVAR)0 kVAR
0 deg
P (W)
Power Triangle
5500 kW
g
The cap bank is providing 3400 kVAR, so our service is now providing only
5500 kVA (reduction from 6466 kVA.
20. Basic Power Calculations
1,500 kVA
COS [ ] = 0.67
COS [ ] = 0.95
Q2 = Q1 + Qc Q1 = 1,118 kVAR
1 000 kW
1,053 kVA
Q2 = 330 kVAR 1,000 kWQ
Qc = 788 kVAR
Required Apparent Power
Before and After
Adding a Power Capacitor Bank
An example of how to calculate the size of a cap bank based on a target power factor
22. Basic Power Calculations
Odds are that the utility is
5500 kW
3400 kVAR
y
charging you for penalties
for a low power factor.
6466 kVA
5500
6466
= 0.85
Typically, penalties are
applied for power factor lessapplied for power factor less
than 0.95%.
2000 kW 1000 kW 2500 kW
1000 kVAR 800 kVAR 1600 kVAR
23. Basic Power Calculations
There are also voltage
5500 kW
3400 kVAR
g
considerations.
Assuming typical values:
6466 kVA
Assuming typical values:
Source impedance =
9% at 6500 kVAVolt Drop = 5.8%
The expected voltage drop at
the main bus will be close to
6%!
2000 kW 1000 kW 2500 kW
1000 kVAR 800 kVAR 1600 kVAR
24. Basic Power Calculations
Capacitor bank reduces the
voltage drop at main bus
5500 kW
0 kVAR
voltage drop at main bus
by 5%!
5500 kVA
Volt Drop = 1.0%
3400 kVAR
2000 kW 1000 kW 2500 kW
1000 kVAR 800 kVAR 1600 kVAR
25. Basic Power Calculations
Note that the capacitor bank
can also raise the bus voltage
5500 kW
-1600 kVAR
g
above the nominal value.
5728 kVA
Volt Drop = -1.5%
5000 kVAR
2000 kW 1000 kW 2500 kW
1000 kVAR 800 kVAR 1600 kVAR
26. Basic Power Calculations
The voltage continues to rise
if the capacitor bank remains
3000 kW
-3200 kVAR
p
connected and the load is
reduced.
4386 kVA
Volt Drop = -3.6%
5000 kVAR
Out of service
2000 kW 1000 kW 0 kW
1000 kVAR 800 kVAR 0 kVAR
28. Capacitor Fundamentals
C = eo A / d for a parallel plate capacitor,
where eo is the permittivity of the insulating material (dielectric)o
between plates.
29. Capacitor Fundamentals
We recall that we can add series capacitances to
1 / C = 1 / C1 + 1 / C2
We recall that we can add series capacitances to
obtain an equivalent capacitance.
1 / Ceq 1 / C1 + 1 / C2
31. Capacitor Fundamentals
…but we typically do not have much use for capacitance values.
So we convert capacitance to impedance:So we convert capacitance to impedance:
fCC
XC
2
11
fCC
C
2
39. Capacitor Fundamentals
Note that IEEE Std 18
requires the discharge
Discharge resistor
requires the discharge
resistor to reduce the
terminal voltage to 50 V in
th ti f ifi dthe time frame as specified
in the table below.
41. Capacitor Ratings
Medium voltage capacitors are available in many different stylesMedium-voltage capacitors are available in many different styles.
The main points of differentiation are listed below:
• Voltage rating
• kVAR rating
• Single bushing or dual bushingg g g
• Internally fused, externally fused, or fuseless
42. Capacitor Ratings
IEEE 81 defines the ratings for capacitorsIEEE 81 defines the ratings for capacitors
• Voltage, rms (terminal to terminal)
• Terminal-to-case (or ground) insulation class
• Reactive power
• Number of phasesp
• Frequency
43. Capacitor Ratings
IEEE 18 provides capacitor tolerancesIEEE 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 anywhere fromg y
150 kVAR to 165 kVAR.
44. Capacitor Ratings
IEEE 18 states the capacitor is intended to operate at or belowIEEE 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• 135% of nominal rms current based on rated kVAR and rated
voltage
• 135% of rated kVAR
45. 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
47. Capacitor Application
Power factor correction capacitor banks are typically installed inPower factor correction capacitor banks are typically installed in
the following ways :
• Pole top
• Metal-Enclosed / Pad-Mount
• Open rackOpe ac
• Terminal end at equipment
56. Capacitor Application
Power factor correction capacitor banks can be configured in thePower factor correction capacitor banks can be configured in the
following ways :
• Delta
• Wye - Solidly Grounded
• Wye - Ungroundedy g
A common misconception is that the capacitor bank should be
connected Delta since it is being applied to a delta or highconnected Delta since it is being applied to a delta or high-
impedance grounded system. This is NOT true.
57. Capacitor Application
The driving factor which determine the configuration for the
given application is COST.
Voltage considerationsVoltage considerations
IEEE 1036 suggests that only banks rated 2400 V and below
should be Delta connected This is mainly because standardshould be Delta 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.connected banks for larger bank installations.
58. 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 Therefore unbalancewith capacitors rated phase-to-phase. Therefore, unbalance
detection is not required.
59. Capacitor Application
Wye – Solidly Grounded
• Initial cost of the bank may be lower since the neutral does not
have to be insulated from groundhave to be insulated from ground.
• Capacitor switch recovery voltages are reduced
• High inrush currents may occur in the station ground system
Th d d id l i d f l• The grounded-wye arrangement provides a low-impedance fault
path which may require revision to the existing system ground
protection scheme. Typically not applied to ungrounded systems.p yp y pp g 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 Aprotection relays (consider coordination of 40 A fuse with a 400 A
grounded system).
• Typical for smaller installations (since auxiliary equipment is not
i d)required)
60. 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.
63. Capacitor Protection
Fuseless Capacitors
Constr cted of small capacitor elements hich are arranged inConstructed of small capacitor elements which 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. Themany 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 joint This results in an entireelectrically stable electrical joint. This results in an entire
series section being shorted.
65. Capacitor Protection
Internally Fused Capacitors
Constr cted s ch that each elementConstructed such 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 andto the adjacent elements and
fuses. The current limiting mode
chops the fault current to prevent
the energy stored in the parallelthe energy stored in the parallel
connected elements from being
discharged into the faulted
element.
67. Capacitor Protection
Group Fusing– Considerations for Selecting Fuse
(typical for distribution pole mounted racks)
• Continuous Current• Continuous Current
• Transient Current
• Fault Current
k C C di i• Tank Rupture Curve Coordination
• Voltage on Good Capacitors
68. Capacitor Protection
Continuous Current
• For wye-solidly grounded systems: Fuse > = 135% of rated
capacitor current (includes overvoltage, capacitor tolerances,
and harmonics).
• For wye-ungrounded systems: Fuse > = 125% of ratedy g y
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, the divide the fuse rating by 1.50.g y
69. Capacitor Protection
Transient Current
• Capacitor switching (specifically back-to-back switching)
• Lightning surges
Back-to-back is typically not a factor for pole mounted capacitorsyp y p p
banks.
High frequency lightning surges: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.
70. Capacitor Protection
Fault Current
Ensure that the fuse can interrupt the available fault current
Tank Rupture Coordination
Ensure that the fuse maximum clearing TCC curve for the fuseg
link is plotted below the capacitor tank rupture curve. In cases of
high fault currents, the tank rupture curve should be compensated
for asymmetryfor asymmetry.
Voltage on Good Capacitors
d d f l h l iFor an ungrounded system, a fault on one phase results in a 1.73
times overvoltage on the un-faulted phases. Ensure that the fault
is cleared before the second capacitor failure.p
71. Capacitor Protection
Problems with Fusing of Small Ungrounded Banks
Consider a 12.47 kV, 1500 kVAR cap bank made of three (3)
500 kVAR i l h i500 kVAR single-phase units.
FuseAAA
kV
kVAR
][100][1045.144.69][44.69
][47.123
][1500
kV ][47.123
If a capacitor fails, we will expect approximately 3x line current.
It will take a 100 A fuse approximately 500 seconds to clear thispp y
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).
72. Capacitor Protection
Individual Fusing– Considerations for Selecting Fuse
(typical for substation capacitor banks)
• Continuous Current• Continuous Current
• Transient Current
• Fault Current
• Tank Rupture Curve Coordination
• Voltage on Good Capacitors
• Energy Discharge into Faulted UnitEnergy Discharge into Faulted Unit
• Outrush Current
• Coordination with Unbalance Detection System
73. 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 the divide the fuse rating by 1 50150%. In this case, the divide the fuse rating by 1.50.
74. Capacitor Protection
Transient Current
• Lightning surges
• Capacitor switching (specifically back-to-back switching)
High magnitude, high frequency lightning surges are typically notg g , g q y g g g yp y
a concern for substation installations.
Back to back switching is typically controlled with pre insertionBack-to-back switching is typically controlled with pre-insertion
closing resistors or current limiting reactors.
h f i ll i h ll l f ill h hBy the nature of installation, the parallel fuses will share the
transient current and will not be a factor.
75. 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 a failed capacitor unit unless other unitsg p
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 i l d b l h i k flink is plotted below the capacitor tank rupture curve. In cases of
high fault currents, the tank rupture curve should be compensated
for asymmetry.y y
76. Capacitor Protection
Example of a DefiniteExample of a Definite
Tank Rupture Curve.
Th i b hThe time between the
rupture curve and the
fuse maximum clear
curve is the
coordination margin.
77. Capacitor Protection
Example of a 10% and 50%
Rupture Curve for a 100 kVARRupture Curve for a 100 kVAR
Capacitor.
Probability based tank rupturey 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
f b bilicurves for any probability.
78. 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
79. Capacitor Protection
Energy Discharge Into a Failed Unit
When a capacitor failure occurs, the stored energy in the parallel
connected capacitors can discharge thro gh the failed capacitorconnected capacitors can discharge through 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 paper-
film capacitors Expulsion fuses are typically rated 30 000film capacitors. Expulsion fuses are typically rated 30,000
Joules. Current limiting fuses are required if ratings are exceeded.
(1 Joule = 1 W x sec use 0 2 cycle clearing time for calculation)(1 Joule = 1 W x sec, use 0.2 cycle clearing time for calculation)
80. 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 frequency discharge currents. These calculations andg q y g
measurements are complex and are determined by the
manufacturer.
Coordination with Unbalance Detection Scheme
The individual fuse must clear the fault before the unbalance
i h i h i i b kprotection scheme trips the entire capacitor bank.
82. Capacitor Protection
Recall Problem with Fusing of Small Ungrounded Banks
12.47 kV, 1500 kVAR cap bank made of three (3) 500 kVAR units
FuseAAA
kV
kVAR
][100][1045.144.69][44.69
][47.123
][1500
It will take a 100 A fuse approximately 500 seconds to clear thisIt 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 per phase, each with a( ) p p p ,
25 A fuse. The clear time for a 25 A fuse @ 208.32 A is below
the published capacitor rupture curve.
83. Capacitor Protection
Why is the Current 3 x Nominal Line Current for a Phase-to-
Neutral Fault on a Wye-Ungrounded Capacitor Bank?
A
B
A
B
V
A
C
N
A
N
VNG C C
I 3 0
Since V = I*Z, where Z is
t t ( i f 60 h ) IA = 3.0 p.u.constant (assuming f = 60 hz)
If voltage across capacitor is
increased by 1.732, the current
also increases by factor of 1.732y
84. 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
ratingrating.
85. Capacitor Protection
Unbalance Protection
As single-phase units in a multiple unit/phase installation fail and
are removed from service the remaining units are experience anare removed from 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 c cles 2 0015 cycles 2.00
1 s 1.70
15 sec 1.40
1 min 1.301 min 1.30
An unbalance protection scheme must by implemented to prevent
the failure of the overvoltaged units.g
89. Capacitor Protection
Ungrounded or Impedance
Grounded System
VA
V
V
VN
VG
VNG
Normal Conditions
VN = VG
VAN = VBN = VCN = 1.0 p.u.
VC VB
VAN VBN VCN .0 p.u.
92. Capacitor Protection
Ungrounded or Impedance
Grounded System
VA
V
V
VN
VG
VNG
Normal Conditions
VN = VG
VAN = VBN = VCN = 1.0 p.u.
VC VB
VAN VBN VCN .0 p.u.
93. Capacitor Protection
Ungrounded or Impedance
Grounded System
VA
V
Gnd
VN
VNG
VC
VB
Ground Fault
VNG = VLN
V = V = V = 1 732 p u
VC =VG
Ground Fault at Cap Bank or
A h h SVAG = VBG = VLL = 1.732 p.u. Anywhere on the System
100. Capacitor Protection
A phase B phase C phase
P: Number of
units in groupunits in group
(P=6)
S: Number ofS: Number of
series groups
(S=4)
Reference Figure for Calculations to Follow
101. Capacitor Protection
# of Series Groups Grounded Y or Delta Ungrounded Y
Split Ungrounded Y
(equal sections)
1 - 4 21 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
104. 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 7321.732.
Does this mean that capacitors must be rated phase-to-phase?
Certainly nothing wrong with this, but cost will be significantly
higher.g
105. Capacitor Fundamentals
Recall:
X
V
S
2
CX
This means that a 150 kVAR 12470 V unit applied at 7200 VThis means that a 150 kVAR, 12470 V unit applied at 7200 V
will provide only 50 kVAR.
][50
][150
][12470
][7200
2
2
kVARS
kVAR
V
V
SNEW
][50 kVARSNEW
106. Capacitor Fundamentals
Using 12470 V capacitors on a 12470 V Ungrounded or
Resistance-Grounded System will require 3x more cans.y q
It should be noted that the 12470 V cans will also be larger than
the 7200 V cansthe 7200 V cans.
Results in a much larger and more costly installation.
This solution would be required if a ground fault could be
maintained for extended periods of time.p
107. Capacitor Fundamentals
Perhaps a 150 kVAR or 200 kVAR, 7620 V or 7960 V units
applied at 7200 V would be a better solution.
][150
][7620
][7200
2
2
kVAR
V
V
SNEW ][200
][7620
][7200
2
2
kVAR
V
V
SNEW
][134
OR
kVARSNEW ][178
2
OR
kVARSNEW
][123
][150
][7960
][7200
2
2
kVARS
kVAR
V
V
SNEW
][163
][
][7960
][7200
2
2
kVARS
kVAR
V
V
SNEW
][123 kVARSNEW
Note that the 7620 V unit provides an additional 6%
The 7960 V unit provides an additional 11%
][163 kVARSNEW
The 7960 V unit provides an additional 11%
108. 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. Thereactance and the other capacitor units in series with the failed capacitor unit. The
pressure is generated by the hot arc making contact 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
d b th t f lt t i If th itcaused by the power system fault current crossing zero. If other capacitors 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
the energy is absorbed in the failed capacitor.gy p
109. 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 avoltage 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
ll l it It b d li ti ith ti ll i fi it ll lparallel capacitors. It can be used on applications with essentially infinite parallel
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 rackoccur elsewhere in the capacitor rack.
111. Capacitor Fundamentals
Current Limiting Fuses vs Expulsion Fuses:
Expulsion Fuses
Operate mechanically and provide a visual indication
Require additional space for operationRequire additional space for operation
Typically applied for outdoor application due to ionized gas release.
C bi ti l i ith t li iti h t i ti f b d iCombination expulsion with current limiting characteristic fuses can be used in
indoor metal-enclosed equipment.
Less expensivep
113. Capacitor Fundamentals
Current Limiting Fuses
Do not emit ionized gases during operation Ionized gases are undesirableDo not emit ionized gases during 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
i drequired.
More expensive
116. Capacitor Fundamentals
What about arresters? How and where should they be
applied?applied?
Depending on application, environment, exposure to
switching etc arresters may be necessaryswitching, etc, arresters may be 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 possible. Installation of arresters at the breakerp
feeding the capacitor bank will not do much for protection of
the capacitor bank.
117. Capacitor Protection
A basic three (3) arrester method is shown below. This is
typical for solidly grounded systems and wye-groundedtypical for solidly grounded systems and wye grounded
capacitor banks.
118. Capacitor Protection
Depending on type of installation, system parameters and
level of protection required, a six(6) arrestor method may belevel of protection required, a six(6) arrestor method may be
applied.
119. Capacitor Protection
For an ungrounded system or a high-impedance grounded
system, a four (4) arrestor grounding method might besystem, a four (4) arrestor grounding method might be
considered an wye ungrounded bank.
Phase to Neutral Fault
VNG = VLNNG LN
VAN = VBN = VLL = 1.732 p.u.
Ground Fault
VNG = VLN
VLL
VLL
ArrNG LN
VAG = VBG = VLL = 1.732 p.u.
If faults can be maintained,
ArrN
V
ArrPH
ArrPH must be rated VLL
ArrN must be rated VLN
The effective arrester MCOV
N
is VLL + VLN
120. 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-to-
ground fault on the system. This may not provide an adequate level of protection for the
capacitors.
Phase to Neutral Fault
VNG = VLN
V V V 1 732VAN = VBN = VLL = 1.732 p.u.
Ground Fault
VNG = VLN
V V V 1 732
VLL
VLL
VAG = VBG = VLL = 1.732 p.u.
If faults can be maintained, Arresters
must be rated VLL
V
Arresters do not provide protection
across the capacitor bushings. Note
that the BIL applies to bushing-to-
i l ticase insulation.
121. Capacitor Protection
Good Presentations on Capacitor – Arrester Applications
“G id li f S l ti f S A t f Sh t“Guidelines for Selection of Surge Arresters for Shunt
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 ViennaInternational Conference on Electricity Distribution, Vienna,
21-24 May 2007.
B th f th l dd h t h tBoth of these papers also address phase-to-phase arrester
connections.
123. Harmonics
Recall that the impedance of a capacitor is inversely
proportional to the system frequencyproportional to the system frequency.
fCC
XC
2
11
fCC 2
Harmonics flow to the point of lowest impedance. The higher
the harmonic the lower the impedance of the capacitorthe harmonic, the lower the impedance of the capacitor.
As capacitors absorb harmonics, the capacitor heats up and
th lif t i d d Th lt h i tthe life expectancy is reduced. The voltage harmonics stress
the capacitor dielectric and reduce the life expectancy of the
capacitor.
125. Harmonics
Where do harmonics come from?
• Power Electronics (drives, rectifiers, computer power
supplies, etc)
• Arcing Devices (welders, arc furnaces, florescent lights, etc)g ( , , g , )
• Iron Saturating Devices (transformers)
• Rotating Machines (Generators)
• Parallel Resonance (between cap bank and power• 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
utility.y
128. 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 andwith 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 The result of this is that a largebetween these 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.
129. Harmonics
A study should be performed to determine levels of harmonics on ay p
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 d i b dj dfilter design must be adjusted.
132. Design Considerations
So how do we size a capacitor bank?
D t i i lDetermine your primary goal
• Voltage support
• Lower utility bill (avoid penalties)
• Increase capacity of system
It can be all three, or any combination of the above.It can be all three, or any combination of the above.
Note that correcting to unity power factor at maximum load is
tl d t bcostly and may not be necessary.
133. Design Considerations
For a 20 MVA load at 0 88 power factor (17 6 MW 9 5 MVAR)For a 20 MVA load at 0.88 power factor (17.6 MW, 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
134. Design Considerations
Determine if current limiting reactors or tuning reactors areDetermine if current limiting reactors or tuning reactors are
required.
Harmonics and resonance may dictate tuning reactors
Back-to-back switching may require current limiting reactorsBack to back switching may require current limiting reactors
(unless another method is used to mitigate the switching surges,
i.e. pre-insertion closing resistors/reactors, zero-crossing
breakers etc)breakers, etc)
135. 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.margin on the insulation.
The voltage at the capacitor terminals will be higher than bus
lt if t tili d It i i t t t t f thivoltage if reactors are utilized. It is important to account for this
voltage difference.
Determine the voltage swing of the system. Will the capacitors
remain on-line while the facility is lightly loaded.
136. Design Considerations
We listed some reasons for specifying higher than bus nominalWe 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 kV busnominal 825 kVAR when connected to a 13.8 kV bus
137. Design Considerations
Determine optimal size and number of stages.
D di i i l t l d i l b k i d f f llDepending on swing in plant load, a single bank sized for full
plant capacity may not be the answer.
IEEE 1036 recommends limiting the voltage change to 2-3%. The
delta voltage can be estimated by:
MVAR
Switch of a capacitor applies high stresses to the insulation
%100
SCMVA
MVAR
V
Switch of a capacitor applies high stresses to the 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.
138. Design Considerations
Determine best location for the installation The most effectiveDetermine best location for the 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-tie-
main switchgear.main switchgear.
If capacitors are installed at the motor pecker head (running
capacitors) ensure that the capacitor VAR rating does not exceedcapacitors), ensure that the capacitor VAR rating does not exceed
90% of the motor no-load VAR. Otherwise, it is possible to
damage the motor by overexcitation.
139. 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
unity power factor makes sense.y p
THIS IS NOT THE CASE. Do not exceed 90% of the
motor no load kVAR demand Exceeding this value canmotor no-load kVAR demand. Exceeding this value can
result in damage to the motor insulation as a result of
overexcitation.
140. 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 x 809 kVAR = 728 kVARMax size of running capacitor is 0.90 x 809 kVAR = 728 kVAR
141. Design Considerations
M: Motor Magnetizing CurveM: 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 Therefore as the motor slows in speed the frequency of the motorfrequency below 60 hertz. Therefore, as the motor slows in 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 passes 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 this resonance (or high voltage) may be present for airon. Depending on the inertia of the motor, this resonance (or high voltage) may be present for a
considerable period of time.
142. Design Considerations
Determine the most optimal type of installationDetermine the most optimal type of 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?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?Will the capacitor bank be pole mounted on a distribution line?
144. 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.
145. Design Considerations
Determine the most optimal configuration.
Higher reliability costs more.
2400 kVAR, 13800 V, wye-grounded (1) 800 kVAR per phase, , y g ( ) p p
bank will be a smaller footprint and cost less than
2400 kVAR 13800 V wye ungrounded (8) 100 kVAR per2400 kVAR, 13800 V, wye-ungrounded (8) 100 kVAR per
phase bank.
H h li bili f h d d b k iHowever, the reliability of the wye-ungrounded bank is
significantly higher
146. 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
capacitor switching.p g
IEEE C37.99 provides the equations for calculating the inrush
current and frequencycurrent and frequency.
148. 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 ddi 100 H li i i h i h iBy adding a 100 mH current limiting reactor, the inrush is:
7254 A pk @ 2.31 kHz, the product is 1.7 x 107p @ , p
161. Design Considerations
Be aware that larger medium voltage motors may include
surge packs.g p
The surge pack will decrease the crest voltage and rate of
rise of the impending surge High rates of rise damage endrise of the impending 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
be neglected, but this should be verified.g
163. Design Considerations
Do not confuse Harmonic Filter Banks with Power Factor
Correction BanksCorrection 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
the higher voltage, the installed kVAR can be anywhere frome g e vo ge, e s ed V c be yw e e o
25% to 40% higher than nominal design.
The capacitor cans must be capable of this outputThe capacitor cans must be capable of this output.
164. Grounding of Wye Banks
If multiple wye-grounded banks are in close proximity, use
peninsula grounding or single-point grounding.p g g g p g g
Single-Point Grounding
The neutrals of all banks of a given voltage are connectedThe 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
(d b k b k i hi ) f fl i icurrents (due to back-to-back switching) from flowing into
the ground grid.
165. Grounding of Wye Banks
Peninsula Grounding
One or more isolated ground grid conductors are carriedg g
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 toarea. All capacitor bank neutral connections are made to
this isolated peninsula ground grid conductor.
