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Economics When Appling
Shunt Capacitors
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Raised Voltage Levels |
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Released Generation Capacity |
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Released system capacity |
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Reductions of System Loses |
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Regulations of Voltage Levels |
The utility can witness benefits on their generation, EHV transmission, sub-transmission and distribution systems with power capacitors.
Voltage Drop
Simply, the voltage drop is basic and is due to the impedance in the line. The impedance consists of resistance, which creates IR voltage drop and reactance, which creates IXL voltage drop. The combination of these two drops is known as the impedance drop, or IZ drop.
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Figure 1 |
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Table 1 |
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Table 2 |
%Voltage Drop =
D =Line length
KV = phase to phase voltage
LET
KW = 1000
Pf = 85%
R = 0.699/mile
XL = 0.712 Ohms
D = 10 Miles
THEN
KV = 12.47
ARCCOS 0.85 = 31.80SIN
= 0.527
With a 7.33% voltage drop the total voltage drop will be 914.57 volts. The line voltage at this point will be 11,555.43 volts. If the secondary voltage is 480 volts normally this voltage drop will be 444.8 volts.
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Figure 2 |
D =Line lengthFrom above
KV = phase to phase voltageKV = 12.47
KW = 1000
Pf = 85%
XL = 0.712 Ohms
D = 10 Miles
To correct the power factor to approximately 95%, we will use 300 KVAR. This will correct this system to approximately 97%
THEN
With a 1.374% voltage rise the voltage at the load VL will be increased by 159 volts. The line voltage at this point will be 11,714.2 volts. With the secondary voltage of 444.8 volts, voltage will raise to 461.44 volts.
Percentage Voltage Drop In a Transformer
The above discussion shows the voltage drop and rise on the power line and drop in the transformers. Every transformer will also experience a voltage rise from generating source to the capacitors. This rise is independent of load or power factor and may be determined as follows:
Using the 300 Kvar bank given above and assuming a 1200 KVA transformer with 5.75% reactance we would have:
Kvar =Applied Kilo-vars
Kva = Kva of the transformer
Xt = Transformer Reactance in %
New Total Voltage Improved Drop
New net Voltage Drop =Total Drop – Total RiseUsing the values from above, the new improved voltage drop will be:
| Voltage Drop in Line = | 7.3300% |
| Voltage Drop in the Transformer = | 0.7480% |
| Total Voltage Drop = | 8.0780% |
| Voltage Rise in Line = | 1.3740% |
| Voltage Rise in the Transformer = | 1.4375% |
| Total Voltage Rise = | 2.8115% |
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Net Voltage Drop = |
8.078% - 2.8115% = 5.2665% |
A quick approximation of KVA increase can be as follows:
To determine this capacity released the formula used for the Substation Capacity Release can also be used. We would replace the substation transformer (KVAS) with the generating capacity (KVAG). Assuming the KVAG is 20MVA we have the following:
Or, by the simplified formula
Feeder capacity is limited basically by permissible voltage drop rather than thermal conditions. The application of capacitor, as shown above, reduces the voltage drop so more KVA of load may be freed up so more load can be added without over taxing the transformer. This may be calculated by the following formula, which does not incorporate the generating or substation capacity release.
The possible increase in KW due to the released feeder KVA can be obtained by multiplying ΔKVA by the corrected power factor.
With the decrease conductor losses due to the addition of capacitors, less kilowatt-hours of electrical energy are dissipated annually. The quantity of energy saved and the resulting economy can be computed as:
where:
All these benefits will have financial benefits to the utility, or whoever provides the capacitor bank on their system.KVAR = Three phase kilovars applied.
KVA = Uncorrected three-phase load (24 hour rms value…for 1200 KVA assume 720 KVA)
KV = Phase to phase voltage in KVThe value of this energy savings can be obtained by multiplying the energy saved by the cost per kw-hr.
1. Power Factor Relation
Where
Kw = kilowatt load
Kva = kilovolt-amperes
Pf = power factor
2. Ratio of I2R Losses
3. Voltage Drop in A Line
Where
Kva = three phase kva
L = Line length in miles (1 wire only)
R = Ohms resistance per mile (table 2)
X = Ohms reactance per mile (table 2)
= Uncorrected power factor
= Sine of power factor angle (table 1)
kv= Phase to phase kilovolts
4. Voltage Rise in A Line
WhereKvar = three-phase kilovars applied
See Formula 3 for other units
5. Voltage Rise in A Transformer
Where
Kvar = Three-phase kilovars applied
KvaT = kva of transformer
XT = Transformer Reactance in percent.
6. Increase in feeder Capacity
Where
= Increase in kva capacity
See Formula 3 for other units
7. Reduced Energy Losses in A Line
Where
Ea = Annual conserved energy in kw-hrs.
R = Resistance to load center in Ohms.
Kva = Uncorrected 3-phase kva (24-hour rms value)
Kvar = here phase kilovars applied.
See Formula 3 for other units
8. Increased Revenue Due to Voltage Improvement.
Where
V2 = Average voltage after adding capacitors.
V1 = Average voltage before adding capacitors.
9. Reduced Substation Capacity
When kvar is small compared to kvaS10. Reduced Substation CapacityWhen kvar is more than 10% of kvaS
Where
= Released kva of the substation capacity at original power factor.
KvaS = Substation kva capacity
Kvar = Three phase kilovars applied.
See Formula 3 for other units
When kvar is small compared to kvaG
else
Where
= Released kva beyond maximum generating capacity at original power factor.
kvaG = Generating station kva capacity
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