Power factor improvement using synchronous condenser

1. POWER FACTOR
IMPROVEMENT
IN GRID USING SMALL HYDRO
GENERATING STATION
PRESENTED BY,
AKHIL.A.G TJAKEEE003
ANUSHA.K.BINO TJAKEEE010
VIJISHA.E.V
TJAKEEE051
VISHAG RAVEENDAN TJAKEEE052
Guide: Anil sir
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2. INTRODUCTION
 Power
 Power factor
 Synchronization
 Synchronous machine
 Synchronous condenser
 Current situation
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3. PF IMPROVEMENT EQUIPMENT
Static capacitor
 Synchronous condenser
 Phase advancers
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4. ACT :SYN. GENERATOR
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5. ACT :SYN.CONDENSER
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6. 11
K
V
11/110KV
TRANSFOMER
110KV
SUBSTATION
110/33KV
TRANSFOMER
33KV
SUBSTATION
33/11KV
TRANSFORMER
11KV/415V
TRANSFORMER
LOAD
LOAD LOAD LOAD LOAD
11KV
LINE
PRIMARY
TRANSMISSION
SECONDARY
TRANSMISSION
GENERATING
STATION
PRIMARY
DISTRIBUTI
ON
POWER
SYSTEM
NETWORK
LOAD
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7. 11
K
V
11/110KV
TRANSFOMER
110KV
SUBSTATION
110/33KV
TRANSFOMER
33KV
SUBSTATION
33/11KV
TRANSFORMER
LOAD
11KV/415V
TRANSFORMER
LOAD
LOAD LOAD LOAD LOAD
SYNCHRONO
US
SCOPE
3.3/11KV
TRANSFORM
ER
3.3
KV
11KV
LINE
PRIMARY
TRANSMISSION
SECONDARY
TRANSMISSION
SMALL
GENERATING
STATION
GENERATING
STATION
PRIMARY
DISTRIBUTION
PRIMARY
DISTRIBUTI
ON
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8. 11
K
V
11/110KV
TRANSFOMER
110KV
SUBSTATION
110/33KV
TRANSFOMER
33kV
SUBSTATION
33/11KV
TRANSFORMER
LOAD
11KV/415V
TRANSFORMER
LOAD
LOAD LOAD LOAD LOAD
SYNCHRONO
US
SCOPE
3.3/11KV
TRANSFORM
ER
SYNCH
RONOU
S
CONDE
NSER
11KV
LINE
PRIMARY
TRANSMISSION
SECONDARY
TRANSMISSION
GENERATING
STATION
PRIMARY
DISTRIBUTION
PRIMARY
DISTRIBUTI
ON
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9. SYNCHRONOUS MACHINE
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10. 27-03-2014 10

11. DESIGN
Generating capacity of Peechi generating station is
1.25MW. The. We know that ,
If act as generator
Active power = 3 𝑉𝐼 𝐶𝑜𝑠∅ = 1.250MW.
PF=0.9 lagging; ∅ = 25.840
Apparent power = 3 𝑉𝐼 = 1.388𝑀𝑉𝐴.
Reactive power = 3 𝑉𝐼 𝑆𝑖𝑛 ∅ = 0.604𝑀𝑉𝐴𝑅
Full load current I = 1.388MVA = 243 A
3 ∗ 3.3 KV
Full load active current = 218.70A.
Full load reactive current = 105.91A.
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12. If act as synchronous condenser
An over excited synchronous motor operated at no load
called synchronous condenser. This MVA we are going to
run as synchronous condenser with 0.1pf leading.
Apparent power = 1.388 MVA.
cos−1
(.1) = 84.260
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13. This 1.388 MVA can be supplied to 33KV Substation in
kannara where Load =3Mw, pf = .6 that means it draws 5
MVA from feeding Ollur 33kV Substation
Pf =0.6
∅ = 53.130
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14. Power factor = cos⁡
〖(39.72)= 0.769.〗
So Power factor improved from 0.6 to 0.769. With 3 MW
= 0.169.
5MVA – 4.08MVA = .92 MVA.
I.e. The working of synchronous motor produces .92
MVA as profit.
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15. Consider now a machine that:
1 Is operated at successively smaller and
smaller torque angle
2. Greater and greater field excitation
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16. FULL LOAD
We know that,
Output of AC Generator = 1250000 MW
3 𝑉𝐼 𝑐𝑜𝑠∅ = 1250000.
Full load current IA =243A
Input of AC Generator = 1314434MW
𝟑 ∗ 𝑬 ∗ 𝑰 ∗ 𝒄𝒐𝒔ɸ =1306200
E = 3.44 kV
Voltage drop across the generator = .147kV
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17. 75% LOAD
Assume ɸ= 20º
Output of AC Generator = 9375000 MW
√3 VI cos∅=9375000
IA = 174.54A
Excitation current = 165.5A
Voltage drop across the generator = jIaX = 0.082 kV
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18. 50% LOAD
Assume ɸ =150
Output of AC Generator = 625000 MW
3 𝑉𝐼 𝑐𝑜𝑠∅ = 625000
3 ∗ 3.3 ∗ 𝐼A*.96 = 625000
IA = 113A
Excitation current = 147A
Voltage drop across the generator = jIaX = 29.24*1.38
= 0.043 kV
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19. 25% LOAD
Assume ɸ=100
Output of AC Generator = 312500 MW
3 𝑉𝐼 𝑐𝑜𝑠∅ = 312500
3 ∗ 3.3 ∗ 𝐼A*.98 = 312500
IA = 55.51A
Excitation current = 128.50A
Voltage drop across the generator = jIaX = 9.63*1.38
= 0.013 kV
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20. NO LOAD
Assume ɸ= 20
Output of AC Generator = 17136.85 MW
3 ∗ 3.3 ∗ 𝐼A*.99 = 17136.85
IA = 3A
Excitation current = 110A
Voltage drop across the generator = jIaX = 0.10*1.38
= 0.000138 kV
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21. SYNCHRONOUS CONDENSER OPERATION
Synchronous generator has the capability of operating
as a synchronous condenser. Increasing the field
excitation causes E to increase.
We know that
Vt = E + jIaX
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22. SUCCESSIVELY GREATER FIELD EXCITATION
Case1
Apparent Power = 317320.81MW
Active Power = 31736.65 MVA
Reactive Power = 315729.76 MVAR
Let we assume that 0.1 power factor ie, ɸ = 84.260
Ia, act =5.50 A Ia, react =54.72 A
Resultant current I =55A
Voltage drop across the generator, jIaX = 1.38 * 54.72 =75.51 V
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23. Case 2
Apparent Power = 647047.61MW
Active Power = 64714.08 MVA
Reactive Power = 643803.30MVAR
Assume ɸ = 84.260
Ia, act =11.30 A
Ia, react =112.43 A
Resultant current I =113A
Voltage drop across the generator, jIaX = 1.38 * 113 =155.15 V
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24. Case 3
Apparent Power = 997666.66 MW
Active Power = 99781.03 MVA
Reactive Power = 992664.34MVAR
Assume ɸ = 84.260
Ia, act =17.45 A
Ia, react=173.66 A
Resultant current I =174.54A
Voltage drop across the generator, jIaX = 1.38 * 173.66
=239.65V
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25. Case 4
Apparent Power = 1388866.19MW
Active Power = 138906.62 MVA
Reactive Power = 1381902.4MVAR
Assume ɸ = 84.260
Ia , act =24.30A Ia, react =241.76 A
Resultant current I =242.98A
Voltage drop across the generator, jIaX =155.15 V
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26. STARTING & EXCITATION CONTROL
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27. SYNCHRONOUS MACHINE & REGULATOR
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28. SIMULATION RESULT
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29. 27-03-2014 29

30. CONCLUSION
 Low power factor problems in the power system
network can be analysed.
 Power factor improved from 0.6 to 0.769. With
3MW
 The working of synchronous motor produces .92
MVA as profit.
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31. REFERENCE
 C.JAYARAMAN – Concept of reactive
power compensation
 Juan Dixon and Luis Moran -“ Reactive
Power Compensation Technologies”.
 T.J Millen- “ Reactive Power Control in
Electrical Systems.”
 RICARDO DOMKE -Power factor
improvement
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32. THANKS
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