application of capacitors in distribution systems

Application of Capacitors in pp cat o o Capac to sDistribution Systems

1

Introduction Capacitors provide tremendous benefits to distribution

system performance Most noticeably capacitorssystem performance. Most noticeably, capacitorsreduce losses, free up capacity, and reduce voltagedrop:

L C i B d h d– Losses; Capacity: By providing the reactive power to motors andother loads with low power factor, capacitors decrease the linecurrent. Reduced current frees up capacity; the same circuit can servemore load. Reduced current also significantly lowers the I2Rg f yline.

– Voltage drop: Capacitors provide a voltage boost, which cancels partof the drop caused by system loads. Switched capacitors can regulatevoltage on a circuit.

2

Introduction If applied properly and controlled, capacitors can

significantly improve the performance of distributionsignificantly improve the performance of distributioncircuits.

But if not properly applied or controlled, the reactivepower from capacitor banks can create losses and highvoltages The greatest danger of over-voltages occursvoltages. The greatest danger of over-voltages occursunder light load.

3

Capacitor ConstructionCapacitor Construction Capacitor elements have

sheets of polypropylene film, less than one mil thick, sandwiched betweensandwiched between aluminum foil sheets.

Capacitor dielectrics must withstand on the order of 78 kV/mm. No other medium-voltage equipment has such g q phigh voltage stress.

4

Capacitor ConstructionCapacitor Construction Capacitor units are supplied

with an internal discharge resistor.

The purpose of the discharge The purpose of the discharge resistor is to provide a path for current to flow in the event that the capacitor is disconnected from the source.

5

Capacitor ConnectionCapacitor Connection

Capacitors are either fixed or switched banks. The fixed capacitors exist all time but the switched

capacitors are switched on based on the system need. A typical switched capacitor bank is shown in the figure

below:below:

6

Capacitor use in the Distribution Network

The application of capacitors in the distribution systems can be summarized as follows:

– 60% of capacitors are applied to feeders.

– 30% of capacitors are applied to substation buses.

– 10% of capacitors are applied to transmission systems.

– Application of capacitors to secondary systems is very rare.

7

Capacitor use in the Distribution Network

8

Capacitor RatingsCapacitors should not be applied when any of the following

limits are exceeded:limits are exceeded:

• 135% of nameplate kvar.

110% f t d RMS lt• 110% of rated RMS voltage.

• 135% of nominal RMS current based on rated kvar and rated voltagerated voltage.

• Capacitors are designed to withstand over-voltages for short periods of time.short periods of time.

9

Capacitor Losses

• Capacitor losses are typically on the order of 0.07 to 0.15 W/kvar at nominal frequency.q y

• Losses include resistive losses in the foil, dielectric losses and losses in the internal dischargedielectric losses, and losses in the internal discharge resistor.

• Capacitors must have an internal resistor that• Capacitors must have an internal resistor that discharges a capacitor to 50 V or less within 5 min when the capacitor is charged to the peak of itswhen the capacitor is charged to the peak of its rated voltage . This resistor is the major component of losses within a capacitor.

10

Capacitor Connectiona) Delta-connection

F d lt ti th i l h it i t For delta connection, the single phase capacitor is a two bushing capacitor unit.

The required voltage rating of the capacitor unit must be

11

The required voltage rating of the capacitor unit must be equal to or greater than the nominal line voltage of the system.

a) Delta-connectionExample-1

Determine the appropriate voltage and kVAR ratings for the capacitor units used to make a 2400 kVAR delta connected capacitor bank to be installed on 13.8 kV feederfeeder.

12

a) Delta-connectionExample-1-solution

2400 phasekVARphasekVAR / 8003

2400/

• The most practical combination would be 2X400 kVAR units per phase or 1X800 kVAR unit per phase.

• The voltage rating of each capacitor is equal to the nominal line-to-line voltage of the system; i.e. 13.8 kV.

13

Capacitor Connectionb) Y-connection

For Y connection the single phase capacitor is a single For Y connection, the single phase capacitor is a single bushing capacitor unit.

The solidly grounded Y-connection is typically used in

14

The solidly grounded Y-connection is typically used in medium voltage distribution feeders.

Capacitor Connectionb) Y-connection

The voltage rating of the capacitor unit must be equal or more than the nominal line-ground voltage of the feeder.

Additional units may be added in parallel to increase the rating of the bank.

Group fusing is typically provided by fused cutouts. However, individual fusing is provided for larger capacitor banks.

15

b) Y-connectionExample-2

A 4800 kVAR, 12.47 kV, solidly grounded Y-connected capacitor bank is made of eight 200 kVAR, 7200 V

it it h A bl f d t ti hcapacitor units per phase. A blown fuse detection scheme is to be used to determine the presence of a blown fuse. Assume that one fuse of phase A is blown, calculate the current flowing from the neutral of the bank to the ground.

16

b) Y-connectionExample-2-solution

2 2

4.32000,2008

72002

jjZZ CB

0.37000,2007

72002

jjZ A

The source voltage references are selected as:g

1207200 ,1207200 ,07200 CNBNAN VVV

AI A 906.194903707200

AIB 302.222904.321207200

AIC 2102.222904.32

1207200

9027.6 AIIII CBAN

17

a) Power Factor Correction One of the main advantages of the application of

capacitors is the power factor correction.

This reactive power requirement has three adverse effects on distribution system:

– The reactive power increases the generators kVA and consequently all system components sizes and rating have to be increased.

– The reactive current increases the system voltage drop.

– The reactive current increases the system losses.

18

Power Factor Correction Equations

The present power factor (pf) is given by:pf (present) = P/(P2 + Q 2)1/2pf (present) = P/(P2 + Q1

2)1/2

When a shunt capacitor is connected to the load , the new ppf is then given by:

pf (new) = P/[P2 + (Q1 - QC)2]1/2

19

Power factor corrections valuesCorrection factor = Qcap/PloadCorrection factor Qcap/Pload

20

Example-3

If a 700 kVA load has a 65% power factor connected % pto 4160VGrdY/2400V system, it is required to improve the power factor to 92%. Using the following Table determine the following:Table, determine the following:a) The correction factor required.b) The capacitor size requiredc) If the capacitor size calculated in (b) is not the standard size, use the list standard of capacitors sizes in the previous Table to calculate the new possible p pimproved power factor.

21


22

Solution:Solution:

From the previous Table, the correction factor i d i 0 74required is 0.74.

The real power of the 700 kVA load at 0.65 power factor

= 700 x 0.65= 455 kW

The capacitor size necessary to improve the power factor from 65% to 92% can de found as

C it i P ( ti f t )Capacitor size = P x (correction factor)= 455 (0.74)= 336.7 kVAR

23

From the capacitor rating Table the nextFrom the capacitor rating Table the next higher standard capacitor size is 400 kVAR, therefore the resulting new correction factor can be found to be

400/455 0 879= 400/455 = 0.879

24


25

• From power correction Table byFrom power correction Table by linear interpolation, the resulting corrected power factor with ancorrected power factor, with an original power factor of 0.65 and a correction of 0 879 can be found as:correction of 0.879 can be found as:

96.096025.0)878.0918.0()878.0879.0(*)96.97(.96.0

factorpowercorrectedNew)(

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b) Voltage Supportb) Voltage Support

As mentioned earlier, capacitors are used to As mentioned earlier, capacitors are used to improve the voltage profile for the feeders.

The best location for voltage support depends on where the voltage support is needed.g pp

Unlike a regulator, a capacitor changes the U e egu o , c p c o c ges evoltage profile upstream of the bank.

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b) Voltage Supportb) Voltage Support

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Approximate Calculation for Voltage RiRise

“K” Factor:K Factor:The Krise is similar to the Kdrop factor except that the load now is a shunt capacitor. When a leading current flows through an inductive reactance there will be a voltage rise instead of voltage drop.

caprise ZIV

rise ltagePercent voriseK

29

mile .kvar rise

Example-4Example 4

Calculate the K i factor for a feederCalculate the Krise factor for a feeder with an impedance of Z=0.25+j0.6 and a length of 3 mileslength of 3 miles.

Assuming a load of 7000 kVA and power factor of 0 9 lagging and apower factor of 0.9 lagging and a nominal line to line voltage = 11 kV determine the rating of a three phasedetermine the rating of a three phase capacitor bank to limit the voltage drop to 1 5%

30

to 1.5%.

Example-4 SolutionExample 4 Solution

900525.0903

1cap kV

kVARI3 LLkV

V 034.0900525.0)6.025(. jIZV caprise

milekVARriseKrise ./ %000537.03/11000

034.0

8.25367)9.0(cos113

7000 1

loadI

V 6.238).( IZVdrop

31

Example-4 SolutionExample 4 Solution

%76.33/11000

6.238% dropV3/11000p

However, it is required to limit the voltage drop to 1.5%, so:

%26.25.176.3 riseV

kVARmileK

VkVARrise

rise 14033000537.0

26.2

32

c) Reducing Line Lossesc) Reducing Line Losses

• One of the main benefits of applying capacitors• One of the main benefits of applying capacitors is that they can reduce distribution line losses.

L f t th h th• Losses come from current through the resistance of conductors.

• Some of that current transmits real power, but some flows to supply reactive power.

• It is desirable to determine the size and location of capacitors to maximize reduction in li l

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line losses.


• The magnitude of the line current can be• The magnitude of the line current can be expressed as follows:

2/122 III qpL III

• Where:

-Ip = magnitude of in-phase component of line current

- Iq = magnitude of quadrature component of line current

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• The current absorbed by a capacitor bank will• The current absorbed by a capacitor bank will subtract from the quadrature component of the line current resulting in the following:g g

2/122cqpL IIII

• Where:

I i d f h i-Ic = magnitude of the capacitor current

35


Example-5Example-5

If the load (700 kVA) in example 3 was t d t th i f d ith thconnected to the source via a feeder with the

following impedance: Z = 0.5+j1.3, find the line losses before and after power factor correctionlosses before and after power factor correction. Also, find the optimum location of the capacitor for maximum line loss reduction.

36


Example-5-solution:Example-5-solution:kW 455)65(.700 P

455 kW1614L

RILosses 23

A9765.16.43

4551

LI kW16.14Losses

455 kW192LA 2.8396.16.43

4552

LI kW19.2Losses

37

Where is the best place for this capacitor?


Practical considerations:

• Determining the size and location of a capacitor for a uniformly distributed load is more

c c co s de o s:

for a uniformly distributed load is more complicated, why?

Th i i f h l d ill l b• The time-varying nature of the loads will also be a significant factor in determining capacitor requirements why?requirements, why?

38


Example-6Example-6

For the reactive load shown below for a 4.16 f d d t i th fi d d it h dfeeder, determine the fixed and switched capacitor to be added to correct the power factor?factor?

39


Example-6-solutionExample-6-solution

• Solution (b) is better as it delivers better compensation for the reactive current.

40

• However, solution (b) requires the switching of two capacitors instead of one for solution (a) which is not desirable in power system.

c) Reducing Line Lossesc) Reducing Line LossesOptimum capacitor size and location:

• Consider the following radial system with uniform reactive current.

111 .)( Ix

LIkIxi

L• The active power loss per phase due to reactive component of load current is:

dxRIxIkIPL

L

2

111

41

dxRIxL

PLoss ...0

1


RKKILPLoss .1..3

221 R is the resistance per unit length

If the load has only lumped load, so K = 1 and:

RILPLoss .. 21

If the load has only distributedIf the load has only distributed load, so K = 0 and:

RILPL .. 21

42

RIPLoss ..3 1


If a single capacitor bank is added to the circuit, the reactive load profile is modified as shown below:reactive load profile is modified as shown below:

'1

11 0for .)( xxIIxL

IkIxi C

L

LxxIxL

IkIxi

'1

11 for .)(L

dxRIIxL

IkIPx

CLoss ...2'

01

11

So:

43

dxRIxL

IkIL

x

...2

'1

11


RKKILIIIxKIILxP CCCLoss .)1(

3)2(')1((' 22

112

1

2

F i l d fil li l th• For a given load profile, line length, and resistance, the quantities K, I1, R and L are constant.• The only two variables are I and x’• The only two variables are IC and x .• To determine the optimum capacitor size and location to minimize losses, the partial derivatives are taken for thesepartial derivatives are taken for these two variables, IC and x’.

(1) )2()1('20' 1

21 CCC

Loss IIIKIILxP

44

' Lx

(2) 22)1('0 11 IIKILx

IP

CC

Loss

c) Reducing Line Lossesc) Reducing Line LossesOptimum capacitor size and location:Solving equation no.1 will result in:

2)1()2(' 1

2 LKIIIIIx CC

2)1(1 KII C

It is convenient to express the capacitor current IC as a function of the reactive current I1

(3)2'

LxII (3)

121

KxIIC

Substituting equations (3) in (2) will results in:

45

results in:

321

230

c) Reducing Line Lossesc) Reducing Line LossesOptimum capacitor size and location:So the size of the capacitor is 2/3 of the total reactive current entering the feeder. If this value is substituted in equation 3, then:

1

132'

K

Lx

So it can be seen from this equation that the 2/3 capacitor size is only true for K value is up to 1/3. If K is more than 1/3 th ’ ill b th L hi h i tthen x’ will be more than L which is not logic. If K exceeds 1/3, the optimum location is x’=L and the capacitor size will be:

46

is x =L and the capacitor size will be:

2

1

K


Capacitor size and placement:C p c o s e d p ce e :If K = 0 (only uniformly distributed load), then x’ = 2/3L

47


Capacitor size and placement:• A generalization of the 2/3 rule for applying n capacitors to a circuit is to size each one to 2/(2n+1)

C p c o s e d p ce e :

capacitors to a circuit is to size each one to 2/(2n+1) of the circuit var requirements.

• Apply them equally spaced starting at a distance of• Apply them equally spaced, starting at a distance of 2/(2n+1) of the total line length from the substation and adding the rest of the units at intervals of g2/(2n+1) of the total line length.

48


Capacitor size and placement:• The total vars supplied by the capacitors is 2n/(2n+1) of the circuit’s var requirements.

C p c o s e d p ce e :

2n/(2n+1) of the circuit s var requirements.

• So to apply three capacitors, size each to 2/7 of the total vars needed and locate them at per unittotal vars needed, and locate them at per unit distances of 2/7, 4/7, and 6/7 of the line length from the substation.

49

Example 7:

A section of a 12.47 kV distribution line has a length of 3 miles. The reactive power loading was measured as 2000 kVAR at the distribution substation line exit. The reactive power loading at the end of the line section was estimated as 600 kVAR. Determine the optimum capacitor rating and p p glocation to minimize line loss of this section.

50

Example 7-solution:

The ratio of reactive power at the end of the line section to the reactive power at the beginning of the line is:

3.02000600

K

Since K is less than 1/3, the optimum capacitor rating is two-thirds time the reactive loading at the beginning of the line section i e kVAR = (2/3)*2000=1333 3 kVARsection, i.e. kVARCAP = (2/3)*2000=1333.3 kVAR

The optimum capacitor location is given by:p p g y

miles 86.2301

1)3(32'

x

51

3.013

d) Released Capacity• In addition to reducing losses and improving voltage, capacitors release capacity.

• Improving the power factor increases the amount of real-power load the circuit can supply.

52

d) Released Capacity

53

Example 8:I th f ll i Fi i li ith if l di t ib t d l dIn the following Figure a primary line with uniformly distributed load. The voltage at the distribution substation low-voltage bus is held at 1.03 pu V with bus voltage regulation. When there is no capacitor bank installed on the feeder the per unit voltage at the end of the line at annualinstalled on the feeder, the per unit voltage at the end of the line at annual peak load is 0.97. Use the nominal operating voltage of 13.8 KV of the three-phase as the base voltage. Assume that the off peak load of the system is about 25% of the on peak load. Also, assume that the linesystem is about 25% of the on peak load. Also, assume that the line reactance is 0.80 Ω /(phase.mi) but the line resistance is neglected and determine the following:

a When the shunt capacitor bank is not used find the Vx voltages at thea- When the shunt capacitor bank is not used, find the Vx, voltages at the times of peak load and off-peak load.

b- Apply an un-switched capacitor bank and locate it at the point of X = 4 mi on the line, and size the capacitor bank to yield a voltage of 1.05 per unit at point X=0 at the time of zero load. Find the size of the capacitor in three phase kilovars.

54

Example 8, solutionExample 8, solutionThe current flowing through any segment along a feeder with uniformly distributed load can be calculated from theuniformly distributed load can be calculated from the following equation (no installed capacitors exist):

xII 1

lII Sx 1

The voltage drop across this segment can be calculated from h f ll i ithe following equation:

dxzIdVD xx xx

55

Example 8, solutionExample 8, solutionThe total voltage drop from the source point to point x along the feeder is given by:the feeder is given by:

xx

dxzIdVDVD xxx dxzIdVDVD00

x

x

Sx dxzlxIVD

0

1

lxxzI

lxxzIVD SSx 2

12

2

56

Example 8, solutionExample 8, solutionThe total voltage drop from the source point to the feeder end-point is given by:end-point is given by:

2 lzIllzIVD

22

zIl

lzIVD SSl

lx

lx

ll

xx

VDVD

VDVD xx 2

21

%%

lllVDVD ll

2%

57

Example 8, solutionExample 8, solutionThe total voltage drop from the source point to the feeder end-point (no installed capacitors exist) at the peak load isend-point (no installed capacitors exist) at the peak load is given by:

%606.097.003.1, puVD pul , p

lx

lx

VDVDx 888.0

98

322

322

puxVDllVD

x

l

0533.006.0888.0933

puVDVV xox 9767.00533.003.1

58

kVxVx 47846.138.139767.0

Example 8, solutionExample 8, solutionThe total voltage drop from the source point to the feeder end-point (no installed capacitors exist) at the no loadingend-point (no installed capacitors exist) at the no loading condition is given by:

41

%% ,

peak

off

peakl

offl

DD

VDVD

, peakpeakl

Therefore, at off-peak conditions:1 %5.1015.006.041% , puxVD offl

lx

lx

VDVD

ffl

offx 888.098

322

322,

puxVD

llVD

offx

offl

0133.0015.0888.0

933

,

,

puVDVV 0167101330031

59

puVDVV xox 0167.10133.003.1

kVxVx 03.148.130167.1

Example 8, solutionExample 8, solutionThe voltage at point X with no capacitor is 1.03 pu (because there is no load and the voltage at the bus will equal thethere is no load and the voltage at the bus will equal the voltage at X), after installing the capacitor bank the voltage at point X becomes 1.05 pu. Therefore, the per unit voltage rise t i t X i 0 02 2 %at point X is 0.02 pu or 2 %.

90042.0908.13*3

1

kVARIcap

VIzV caprise 0336.0]*[

milerise/kvar%00042200336.0K milerise/kvar.%000422.03/8.13riseK

milekVARKV ii **% 711852KVAR

60

milekVARKV riserise% 7.11854*000422.0KVAR

Capacitor banks switching controlCapacitor banks switching control

Several options for controls are available for pcapacitor banks. They can be classified to:

a) Simple control: these techniques does not i l i lrequire any electrical measurements.

– Time clock: The simplest scheme: the controller switches capacitors on and off based on the time ofswitches capacitors on and off based on the time of day. This control is the cheapest but also the most susceptible to energizing the capacitor at the wrong time.time.

– Temperature: Another simple control; the controller switches the capacitor bank on or off depending on temperature

61

depending on temperature.

Capacitor banksCapacitor banks

b) More complicated control: these techniquesb) More complicated control: these techniques require different electrical measurements like: – Voltage:

• The capacitor switches on and off, based on voltage magnitude.

• Voltage control is most appropriate when the primary role Vo tage co t o s ost app op ate w e t e p a y o eof a capacitor is voltage support and regulation.

• Voltage-controlled capacitor banks have bandwidths which should be at least 3 or 4 V (on a 120-V scale). ( )

62

Capacitor banksCapacitor banks

– Vars:Vars:• The capacitor uses var measurements to

determine switching. • This is the most accurate method of ensuring

that the capacitor is on at the appropriate times for maximum reduction of lossesfor maximum reduction of losses.

• Like the voltage control technique, there is a bandwidth for switching of each capacitor bank to prevent excessive switching operations in most cases.

63

Control Methods used for Switched Capacitors

Type of Control Pole Mounted Banks on FeedersPercent

Distribution Substation BanksPercentPercent Percent

Voltage 16.6 30.8

Current 4.9 2.4

Time 59.8 16.3

Voltage-Current 7.2 12.6

Voltage-Time 5.1 6.3

Manual* 6.2 28.4

Others 0.2 3.2

Total 100.0 100.0

* Manual includes any switching directly or indirectly caused by the dispatcher

64

* Manual includes any switching directly or indirectly caused by the dispatcher

Capacitor Switching ConsiderationCapacitor Switching Consideration

In many cases it is desirable to installIn many cases it is desirable to install several steps of switched capacitor units.

This is particularly true if the load reactiveThis is particularly true if the load reactive power requirements fluctuates during the dayday.

When a de-energized capacitor is energized, h i b h h i ithe capacitor behaves as a short circuit.

The inductance of the source/line will limit

65

the current.

Capacitor Switching ConsiderationCapacitor Switching Consideration

The calculation of currents during capacitor The calculation of currents during capacitor switching is extremely important in capacitor applications.

Both contactors and circuit breakers used in capacitor switching are limited in the amount of momentary current the contacts can safely withstand.

This current will be also at high frequency compared to system frequency which will produce high frequency voltage spikes in the system

66

high frequency voltage spikes in the system.

Switching Single Capacitor BankSwitching Single Capacitor Bank

Exact calculations of capacitor switching currents Exact calculations of capacitor switching currents are extremely difficult manually, so the following assumptions will be made:

a) The system will be analyzed on a single phase basis.

b) The source will be modeled as a DC voltage source.

c) The DC voltage will have a magnitude equal to the peak line to neutral system voltage.

67

d) Resistances will be ignored.


The equivalent circuit is shown below The equivalent circuit is shown below

3.2 LL

oVV 3

The capacitance per phase of the capacitor bank is:

2_ ).(..2 ratedLLrated

rated

kVfMVARC

Th i i hi iThe capacitor switching current is:

/)( sVI

68

)/1(/)(

sCsLsVsI

s

o


)/1(/)( 2 CLs

LVsI so

Re-arranging the equation:)/1( CLs s

2/1C

Where:

22)(

o

oo sL

CVsI

Where:

CL so

1

)sin()(2/1

tLCVti oo

2/1

max

LCVI o

69

)()(L oo

max Lo

Example 9:

A 1200-kVAR, 4.16-kV capacitor bank is installed on a plant bus. The plant bus is supplied from a p p pp5000-kVA, 69kV-4.16/2.4 kV transformer having an impedance of 7%. Neglecting the impedance of the source and resistance determine the maximum instantaneous value and the frequency of the inr sh c rrent Also determine the ind ctance ofinrush current. Also, determine the inductance of the inductors that must be added to reduce the inrush currentinrush current.

70

Example 9-solution:

The transformer inductive reactance is:

2 242.0

516.4.07.0

2

MVAkVX

The transformer inductance is: HL 41043.6242.0 The transformer inductance is: HL 1043.660.2

The capacitance per phase is equal to:F

kVMVARC 4

2 1084.1)16.4.(60.2

2.1

41602The peak source voltage is: VVo 3396

34160.2

108412/14

71

AI 18171043.61084.13396 4max

Example 9-solution:

The frequency of the transient inrush current is1

Hzrado 46329091043.6.1084.1

12/144

The total amount of inductance to limit theThe total amount of inductance to limit the maximum current to 1000 A is:

V 32

HCIVL o

s3

2max

1012.2.

The inductance to be added will beThe inductance to be added will be equal to:

72

mHLadded 5.11043.61012.2 43

application of capacitors in distribution systems

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