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AC POWER CALCULATIONAC POWER CALCULATIONPower Factor CorrectionPower Factor Correction

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Power FactorPower Factor

As we have seen before (for sinusoidal voltage and current),

p.f. = cos (vi)

The REAL power (or average power) is transformed into useful energy

e.g. heat, mechanical, light, sound, etc

For a given Vrms and P, loads with high power factor draw LESS current compared with loads with low power factor

P = Vrms Irms cos (vi)

Load+Vrms

Irms Vrms = Vrmsv

Irms = Irmsi

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Power FactorPower Factor

For a given Vrms and P, loads with high power factor draw LESS current compared with loads with low power factor

Pave

p.f. = cos 1

Vrms

I1, rms

Pave

p.f. = cos 2

Vrms

I2, rms

cos 1 > cos 2

1 < 2

PavePave

1 2

rmsrms,1 V

I 1S

rmsrms,2 V

I 2S

1SQ1

Q2

rms,2rms,1 II

2S

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Power FactorPower Factor

For a given Vrms and P, loads with high power factor draw LESS current compared with loads with low power factor

Less current results in LESS losses during transmission

Utility company (TNB) charge more to loads with LOW power factor

Therefore, it is desirable to increase the power factor

LoadSource+

I2R

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Power Factor CorrectionPower Factor Correction

Process of increasing the power factor without altering the voltage or current to the original load

PL

QL1

2

QC

QT

(voltage and current to original load retained)

Before C added, S = PL + jQL p.f. = cos 1

After C added, S = PL + j(QL – QC) p.f. = cos 2 i.e. increased

+

VL

IL

+Vs (rms)

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Power Factor CorrectionPower Factor Correction

PL

1

QL

QC

2

QT

How do we calculate C?How do we calculate C?

Voltage across C = Vs (rms)

Impedance of C = C1j

221

2

2)tan(tan

1s

L

s

csc

PQCC

QVV

V

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LV Static CapacitorLV Static Capacitor3 phase connection cct3 phase connection cct

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LV Capacitor Bank c/w P.F RegulatorLV Capacitor Bank c/w P.F Regulator

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LV Capacitor Bank c/w P.F RegulatorLV Capacitor Bank c/w P.F Regulator

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Prob 1#

A mud pump has an output power of 2 kW, an efficiency of 70% and a power factor of 0.5 lagging when operated from a 230V, 50Hz supply. It is required to improve the power factor to 0.95 lagging by connecting a capacitor in parallel with the motor. Determine:

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a)The current taken by the motor and its phase angle.

01

3

3

6050.0cos

45.2550.0110104.1

cos

4.1)102)(70.0(

Axx

VPI

kWxPP

iM

oi

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b) The supply current after power factor correction

AIIoa

AIoaM

40.1319.18cos73.12

73.1260cos45.2560cos

19.1895.0cos'6050.0cos

0

0

0

0101

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c) The current taken by the capacitor.

AAacabIA

IacA

Iab

C

M

86.17)18.404.22(18.419.18sin)40.13(

19.18sin04.2260sin45.25

60sin19.1895.0cos'6050.0cos

0

0

0

0

0101

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d) Draw phasor diagram by showing phase angle and magnitude.

IIC C = 17.86A= 17.86A

O

IImm =25.45 =25.45

a

c

b

18.1918.1900

606000 I =13.4AI =13.4A

IIC C = 17.86A= 17.86A

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e) The capacitance of the capacitor.

FfVIC

fCV

fC

VXVI

c

cC

17.247)230)(50(2

86.172

2

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