Download - Chapter 4 (power factor correction)
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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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