A
lication of Ca acitors in Distribution Systems
1
Introduction Capacitors provide tremendous benefits to distribution . , reduce losses, free up capacity, and reduce voltage drop: – osses; apac ty: By provi ing the reactive power to motors an othe ot herr lo load adss wi with th lo low w po powe werr fa fact ctor or,, ca capa paci cito tors rs de decr crea ease se th thee li line ne current. Reduced current frees up capacity; the same circuit can serve more load. Reduced current also si ni ni icantl lowers the I 2 R line. – Volt oltage age dro drop: p: Capacitors provide a voltage boost, which cancels part of the drop caused by system loads. Switched capacitors can regulate voltage on a circuit.
2
Introduction Capacitors provide tremendous benefits to distribution . , reduce losses, free up capacity, and reduce voltage drop: – osses; apac ty: By provi ing the reactive power to motors an othe ot herr lo load adss wi with th lo low w po powe werr fa fact ctor or,, ca capa paci cito tors rs de decr crea ease se th thee li line ne current. Reduced current frees up capacity; the same circuit can serve more load. Reduced current also si ni ni icantl lowers the I 2 R line. – Volt oltage age dro drop: p: Capacitors provide a voltage boost, which cancels part of the drop caused by system loads. Switched capacitors can regulate voltage on a circuit.
2
Introduction If applied properly and controlled, capacitors can circuits.
But if not properly applied or controlled, the reactive power from capacitor banks can create losses and high . under light load.
3
Capacitor elements have sheets of polypropylene film, less than one mil thick, aluminum foil sheets.
Capacitor dielectrics must withstand on the order of 78 kV/mm. No other mediumvolta e e ui ment has such high voltage stress.
4
Capacitor units are supplied with an internal discharge resistor. resistor is to provide a path for current to flow in the event that the capacitor is disconnected from the source.
5
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
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 Ratings Capacitors should not be applied when any of the following
• 135% of nameplate kvar. •
o ra e
vo age.
• 135% of nominal RMS current based on rated kvar and . • Capacitors are designed to withstand over-voltages for
9
Capacitor Losses • Capacitor losses are typically on the order of 0.07
to 0.15 W/kvar at nominal fre uenc . • Losses include resistive losses in the foil, , resistor. discharges a capacitor to 50 V or less within 5 min when the ca acitor is char ed to the eak of its rated voltage . This resistor is the major component of losses within a capacitor. 10
Capacitor Connection a) Delta-connection or e a connec on, e s ng e p ase capac or s a wo bushing capacitor unit.
equal to or greater than the nominal line voltage of the system.
11
a) Delta-connection Example-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 .
12
a) Delta-connection Example-1-solution kVAR / phase
3
800 kVAR / phase
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 Connection b) Y-connection , bushing capacitor unit.
medium voltage distribution feeders. 14
Capacitor Connection b) 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-connection Example-2 A 4800 kVAR, 12.47 kV, solidly grounded Y-connected capacitor bank is made of eight 200 kVAR, 7200 V capac or un s per p ase. own use e ec on sc eme 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-connection Example-2-solution Z B Z C j
7200 8 200,000
j 32.4
Z A j
7200 7 200,000
j 37.0
The source volta e references are selected as: V AN 72000,
I A
I C
72000 37 90
7200120 32.4 90
V BN 7200 120,
194.690 A
222.2210 A
I B
V CN 7200120
7200 120 32.4 90
222.2 30 A
I N I A I B I C 27.690 A
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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: =
1
When a shunt capacitor is connected to the load , the new pf is then given by: pf (new) = P/[P2 + (Q1 - QC )2]1/2
19
Power factor corrections values =
20
Example-3 If a 700 kVA load has a 65% ower factor connected to 4160VGrdY/2400V system, it is required to improve the power factor to 92%. Using the following , a) The correction factor required. b) The capacitor size required c) If the capacitor size calculated in (b) is not the standard size, use the list standard of capacitors sizes in the revious Table to calculate the new ossible improved power factor.
21
Power factor corrections values =
22
From the previous Table, the correction factor requ re s . . The real power po wer of the th e 700 kV kVA A load at 0.65 power po wer factor = 700 x 0.65 = 455 kW The capacitor size necessary to improve the power factor from 65% to 92% can de found as apac or s ze = x correc on ac or = 455 (0.74) = 336. 336.7 7 kV kVAR AR
23
higher standard capacitor size is 400 kVAR, therefore the resulting new correction factor can be found to be
=
= .
24
Power factor corrections values =
25
• linear interpolation, the resulting , original power factor of 0.65 and a . New corrected power factor 0.96 (.97 .96) *
(0.879 0.878) (0.918 0.878)
0.96025 0.96
26
, improve the voltage profile for the feeders.
The best location for voltage support depends on where the volta e su
ort is needed.
Unlike a re ulator a ca acitor chan es the voltage profile upstream of the bank.
27
28
Approximate Calculation for Voltage se “ ” The K rise 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.
V rise ZI cap K r se
Percent voltage rise kvar . mile 29
r se
with an impedance of Z=0.25+j0.6 and a .
Assuming a load of 7000 kVA and . nominal line to line voltage = 11 kV capacitor bank to limit the voltage drop . . 30
I cap
1kVAR
90 0.052590
L L
V rise Z I cap (.25 j 0.6) 0.052590 0.034 V K rise
I load
0.034 11000 / 3 7000 3 11
0.000537% rise / kVAR.mile
cos 1 (0.9) 367 25.8
V drop ( Z .I ) 238.6 V
31
%V drop
238.6
3.76%
However, it is required to limit the voltage drop to 1.5%, so: V rise 3.76 1.5 2.26% kVAR
V rise K rise mile
2.26 0.000537 3
1403 kVAR
32
• is that they can reduce distribution line losses. • osses come rom curren resistance of conductors.
roug
e
• 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 ne osses. 33
• expressed as follows: L
2 p
2
1/ 2 q
• Where: -I p = magnitude of in-phase component of line current - Iq = magnitude of quadrature component of line current
34
• subtract from the quadrature component of the line current resultin in the followin :
I L I
2 p
I q I c
2 1/ 2
• Where: - c = magn tu e o t e capac tor current
35
If the load (700 kVA) in example 3 was connec e o e source v a a ee er w e following impedance: Z = 0.5+j1.3, find the line . Also, find the optimum location of the capacitor for maximum line loss reduction.
36
P 700(.65) 455 kW Losses 3 I 2 R I L1 I L 2
3 4.16 .65
3 4.16 .96
97 A 38.2 A
osses
.
osses .
Where is the best place for this capacitor? 37
Practical considerations:
• Determining the size and location of a capacitor complicated, why? • e t me-vary ng nature o t e oa s w a so e a significant factor in determining capacitor ,
38
For the reactive load shown below for a 4.16 ee er, e erm ne e xe an sw c e capacitor to be added to correct the power
39
Solution (b) is better as it delivers better compensation for the reactive current. However, solution (b) requires the switching of two capacitors instead of one for solution (a) which is not desirable in power system. •
•
40
Optimum capacitor size and location:
• Consider the following radial system with uniform reactive current. i ( x )
kI 1 I 1
. x I 1
• The active power loss per phase due to reactive component of load current is:
kI 1 I 1 . L 0 L
Loss
2
. . 41
Optimum capacitor size and location: P Loss
L
I . 12 . K 2 K 1 . R
R is the resistance per unit length
3 If the load has only lumped load, so K = 1 and: P Loss L I . 12 R .
load, so K = 0 and: L 2 oss
3 42
Optimum capacitor size and location: If a single capacitor bank is added to the circuit, the reactive load rofile is modified as shown below: kI 1 I 1 . x I 1 I C
i ( x)
kI 1 I 1 . x I 1
i ( x )
x '
So:
kI 1 I 1
P Loss 0
L
for 0 x x '
for x ' x L
2
. x I 1 I C . R.dx 2
kI I 1 1 . x I 1 . R.dx L x ' L
43
Optimum capacitor size and location: P Loss
x'2 L ( I 1 I C (1 K ) x' ( I C 2 2 I 1 I C ) I 12 ( K 2 K 1).R 3 L
or a g ven oa pro e, ne eng , and resistance, the quantities K, I 1, R and L are constant. ’. C To determine the optimum capacitor size and location to minimize losses, the artial derivatives are taken for these two variables, IC and x’. •
•
P Loss x
0
2 x'
I 1 I C (1 K ) ( I C 2 2 I 1 I C )
P Loss x ' 0 I 1 (1 K ) 2 I C 2 I 1 I C L
(1)
(2) 44
Optimum capacitor size and location: Solving equation no.1 will result in:
( I C 2 2 I 1 I C ) L x ' I 1 I C (1 K ) 2 It is convenient to express the capacitor current IC as a function of the reactive current I1
C
1
'
L 2
2 1 K
Substituting equations (3) in (2) will resu s n:
0
3 2
1
2 3
45
Optimum 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:
2 1 x ' L 3 1 K 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 en x w e more an w c s no logic. If K exceeds 1/3, the optimum location ’
K 1
2
46
Ca acitor size and lacement: If K = 0 (only uniformly distributed load), then x’ = 2/3L
47
Ca acitor size and lacement:
• A generalization of the 2/3 rule for applying n + of the circuit var requirements. , 2/(2n+1) of the total line length from the substation and addin the rest of the units at intervals of 2/(2n+1) of the total line length.
48
Ca acitor size and lacement:
• The total vars supplied by the capacitors is + ’ . • So to apply three capacitors, size each to 2/7 of the , 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 o timum ca acitor ratin and location 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: K
600
0.3
2000 Since K is less than 1/3, the optimum capacitor rating is twothirds time the reactive loading at the beginning of the line , . . . CAP
The o timum ca acitor location is iven b :
1 x' (3) 2.86 miles . 2
51
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: n t e o ow ng gure a pr mary ne w t un orm y str ute oa . 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 , 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 . , reactance is 0.80 Ω /(phase.mi) but the line resistance is neglected and determine the following: 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
The current flowing through any segment along a feeder with following equation (no installed capacitors exist): x
x S l
The voltage drop across this segment can be calculated from t e o ow ng equat on:
dVD I z dx
55
The total voltage drop from the source point to point x along
x
x
x
x
0
x
0
x
VD x I S 1 0
x
z dx l
x 2 x VD x I S z x I S z x 1 2 l 2 l 56
The total voltage drop from the source point to the feeder -
l
S
2 l
2 l
VD x l
% VD x
x 1
S
l
2
x
2 l x x 2
l
2
57
The total voltage drop from the source point to the feeder given by:
VDl ,
VD x
u
1.03 0.97 0.06 pu 6 %
x x 2 2 8 2 2 0.888
l
VD x 0.888 x 0.06 0.0533 pu
V x V o VD x 1.03 0.0533 0.9767 pu
V x 0.9767 x 13.8 13.47846 kV 58
The total voltage drop from the source point to the feeder condition is given by: % VDl , off Doff 1 % VD ,
4
D
Therefore, at off-peak conditions: % VDl , off x 0.06 0.015 pu 1.5 %
4
VD x , off
x x 2 2 8 2 2 0.888
,o
VD x , off 0.888 x 0.015 0.0133 pu x
o
x
.
.
V x 1.0167 x 13.8 14.03 kV
.
pu 59
The voltage at point X with no capacitor is 1.03 pu (because voltage at X), after installing the capacitor bank the voltage at point X becomes 1.05 pu. Therefore, the per unit voltage rise at po nt s . pu or . 1kVAR I cap 90 0.04290
3 *13.8
V rise [ z * I cap ] 0.0336V rise
r se
0.0336 13.8 / 3
r se *
. *
. 2 0.000422 * 4
.
60
Several o tions for controls are available for capacitor banks. They can be classified to: a) Simple control: these techniques does not require any electrical measurements. – Time clock: The simplest scheme: the controller switches ca acitors 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 – Temperature: Another simple control; the controller switches the capacitor bank on or off . 61
require different electrical measurements like: – Voltage: • The capacitor switches on and off, based on voltage magnitude. • Volta e control is most a ro riate when the rimar role of 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
– • 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 . • Like the voltage control technique, there is a bandwidth for switching of each capacitor bank to prevent excessive switching operations in most cases.
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Control Methods used for Switched Capacitors Type of Control
Pole Mounted Banks on Feeders Percent
Distribution Substation Banks 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
64
several steps of switched capacitor units. u y u v power requirements fluctuates during the .
When a de-energized capacitor is energized, t e capac tor e aves as a s ort c rcu t.
The inductance of the source/line will limit the current. 65
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 . 66
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. d) Resistances will be ignored. 67
2 .V LL
V o
The capacitance per phase of the capacitor bank is: C
MVAR
rated
2 . . f rated .( kV LL _ rated ) 2
e capac tor sw tc ng current s: s)
V o / s sL
s
(1 / sC ) 68
V o / L s
I ( s )
2 s
Re-arranging the equation: 1/ 2
I ( s ) V o
L
2 o 2 s o
Where:
o
i ( t ) V
1 L s C
C
1/ 2
sin(
t )
I
V
C
1/ 2
69
Example 9:
A 1200-kVAR, 4.16-kV capacitor bank is installed on a lant bus. The lant bus is su lied from a 5000-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 nrus curren . so, e erm ne e n uc ance o the inductors that must be added to reduce the .
70
Example 9-solution:
The transformer inductive reactance is: X 0.07.
4.16kV 5 MVA
0.242
The transformer inductance is:
0.242 2 .60
.
4
The capacitance per phase is equal to: C
1.2 MVAR 2 .60.(4.16kV ) 2
1.84 10 4 F V o
The peak source voltage is: I max 3396
.
4
3
3396V
1/ 2
6.43 10 4
.
1817 A 71
