Power Factor : Basics and Benefits of Improvement

Power Factor - What are the basics of Power Factor, how can it be improved and the benefits of improvement.

1. POWER FACTOR
Basics
and
Benefits of Improvement

2. Power Factor Basics
•In all industrial electrical distribution systems, the major loads are resistive and inductive. Resistive loads
are incandescent lighting and resistance heating. In case of resistive loads the voltage (v) , current (I),
resistance relations are linearly related, i.e. V= I x R and Power (kW) = V x I
•Typical Inductive loads are A.C. Motors, Transformers and ballast –type lightings. Inductive loads require
two kinds of power : a) Active power (kW) to perform the work and b) Reactive power ( KVAr) to create
and maintain electro-magnetic field.
•The vector sum of the active power and reactive power make up the total (or apparent) power used
which is measured in KVA

3. •The ratio of KW to KVA is called the Power Factor which is always less than or equal to unity. Theoretically, if the loads have
unity power factor, maximum power can be transferred for the same distribution system capacity.
•However, as the loads are inductive in nature, with the power factor ranging from 0.2 to 0.9, the electrical distribution
network is stressed for the capacity at low power factor.
KVAR
Reactive Power
KW
Active Power
KVA
Total Power
KW, KVArand KVA Vector Diagram
PF= KW / KVA
= Cos ф
Cos ф
•The active power (shaft power required) in KW and the reactive power required in (KVAr) are 900 apart vectorially in a pure
inductive circuit i.e. reactive power lagging active power. The vector sum of two is called the apparent power KVA.

4. A Starch Industry had installed a 1500 KVA transformer. The initial demand of the plant was 1160 KVA with power factor of 0.70 .The %
loading of transformer was about 78% ( 1160/1500 = 77.3 %) .
To improve the power factor and to avoid the penalty, the unit had added about 410 KVArcapacitor in motor load end. This improves
the PF to 0.89 and reduced the required KVA to 913 which is vector sum of KW and KVAr.
KVAR =828
KW=812
Cos ф=0.70
PF =812/1160
= 0.70
KVA=1160
KW=812
KVAR=
828-410=418
KVA=913
PF= 812/913
= 0.89
After improvement, the penalty will be avoided and the 1500 KVA transformer is now
loaded only to 60% capacity. The transformer can now also be loaded more in the future.
Cos ф=0.89
Example

5. Advantages of Power Factor Improvement
•Reactive component of the network reduced
and so also the total current in the system
from the source end.
•I
2
R losses of the system are reduced because
of reduction in current.
•Voltage level at the load end is increased.
•KVA loading on the source Generators,
transformers and lines up to capacitor
reduces giving capacity relief.
•A high power factor can help in utilizing the
full capacity of the electrical system

6. Cost Benefits of PF improvement
•Reduction in KVA demand ( Maximum
demand) charges in utility bill.
•Distribution losses (KWH) reduced within the
plant network.
•Better voltage at motor terminals and
improved performance of motors.
•A high power factor eliminates penalty
charges imposed when operating at low
power factor.
•Reduction in investment on kVA rating of
transformers , Cables, switchgear etc as its
delivering load reduces.

7. KW / KVA = Cos ф1
KVAr/ KVA = Sin ф1
KVAr/ KW = Sin ф1 / Cos ф1
KVAr/ KW = Tan ф1
KVAr= KW X Tan ф1
KVArrequiredtoimprovethepowerfactor
fromCosф1toCosф2willbe
KVAr=KW(tanф1–tanф2)
Example:Theutilitybillshowsanaveragepowerfactor
of0.72withanaveragekwof650.HowmuchKVAris
requiredtoimprovethepowerfactorto0.95.
Cosф1=0.72,tanф1=0.963
Cosф2=0.95,tanф2=0.329
KVArrequired=KW(tanф1–tanф2)
=650(0.963-0.329)=412KVAr
•Theprimarypurposeofcapacitorsistoreducethemaximumdemand.Maximumbenefitsofcapacitorsisderivedby
locatingthemascloseaspossibletotheload.
•Atthislocation,itsKVArareconfinedtothesmallestpossiblesegment,decreasingtheloadcurrent.Thisinturnwill
reducepowerlossesofthesystemsubstantially.
•Powerlossesareproportionaltothesquareofthecurrent.Whenpowerlossesarereduced,thevoltageatthemotor
terminalsincreases,thusmotorperformancealsoincreases.
Selection and location of capacitors

8. Operational Performance of Capacitors
•This can be made by monitoring capacitor charging current via the rated charging current.
•The capacity of fused elements can be replenished as charging current.
•The portable analyzer can be used for measuring KVArdelivered as well to charging current.
•Capacitors consume 0.2 to 6.0 watt per KVAr, which is negligible in comparison to benefits
•Nameplatescanbemisleadingwithrespecttorating.Itisgoodtocheck
bychargingcurrent.
•Capacitorboxesmaycontainonlyinsulatedcompoundandinsulated
terminalswithnocapacitorelementinside.
•Capacitorsforsinglephasemotorstartingandthoseusedforlighting
circuitforvoltageboostarenotpowerfactorcapacitorunitsandthese
cannotwithstandpowersystemconditions.
Some basic checks for the Capacitors :

9. •Inductionmotorsarecharacterizedbypowerfactorlessthanunityi.e.laggingPF.Itleadstoloweroverall
efficiencyandhigheroveralloperatingcost.
•Thecapacitorsareconnectedinparallelwithmotortoimprovepowerfactor.
•TheimpactsofPFcorrectioninclude:
•reducedKVAdemand(andhencereducedutilitydemandcharges),
•reducedI
2
Rlossesincablesupstreamofthecapacitors(andhencereducedenergycharges),
•reducedvoltagedropinthecables(leadingtoimprovedvoltageRegulation).
•Thisincreasestheoverallefficiencyoftheplant.
•Thesizeofcapacitorsrequiredforaparticularmotorshouldnotexceed90%ofthenoloadKVArofthe
motor.
Power factor correction of Motors

10. The capacitor rating for power connection by direct connection to induction motors
is shown below.
Motor
Rating ( HP)
Capacitors Rating ( KVAr) for Motor Speed
3000 1500 1000 750 600 500
5 2 2 2 3 3 3
7.5 2 2 3 3 4 4
10 3 3 4 5 5 6
15 3 4 5 7 7 7
20 5 6 7 8 9 10
25 6 7 8 9 9 12
30 7 8 9 10 10 15
40 9 10 12 15 16 20
50 10 12 15 18 20 22
60 12 14 15 20 22 25
75 15 16 20 22 25 30
100 20 22 25 26 32 35
125 25 26 30 32 35 40
150 30 32 35 40 45 50
200 40 45 45 50 55 60
250 45 50 50 60 65 70