pf compensation

1. 1
AC POWER CALCULATIONAC POWER CALCULATION
Power Factor CorrectionPower Factor Correction

2. 2
Power FactorPower Factor
As we have seen before (for sinusoidal voltage and current),
p.f. = cos (θv−θi)
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 (θv−θi)
Load
+
Vrms
−
Irms
Vrms = Vrms∠θv
Irms = Irms∠θi

3. 3
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 θ1
Vrms
I1, rms
Pave
p.f. = cos θ2
Vrms
I2, rms
cos θ1 > cos θ2
θ1 < θ2
Pave
Pave
θ1
θ2
rms
rms,1
V
I 1S
=
rms
rms,2
V
I 2S
=
1S
Q1
Q2
rms,2rms,1 II <
2S

4. 4
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
+
−
I2
R

5. 5
Power Factor CorrectionPower Factor Correction
Process of increasing the power factor without altering the
voltage or current to the original load
PL
QL
θ1
θ2
QC
QT
(voltage and current to original load retained)
Before C added, S = PL + jQL
p.f. = cos θ1
After C added, S = PL + j(QL – QC) p.f. = cos θ2 i.e. increased
+
VL
−
IL
+
Vs (rms)
−

6. 6
Power Factor CorrectionPower Factor Correction
PL
θ1
QL
QC
θ2
QT
How do we calculate C?How do we calculate C?
Voltage across C = Vs (rms)
Impedance of C =
C
1
j
ω
−
2
21
2
2
)tan(tan
1 s
L
s
cs
c
PQ
C
C
Q
VV
V
ω
θθ
ω
ω
−
==⇒=

7. 7
LV Static CapacitorLV Static Capacitor
3 phase connection cct3 phase connection cct

8. 8
LV Capacitor Bank c/w P.F RegulatorLV Capacitor Bank c/w P.F Regulator

9. 9
LV Capacitor Bank c/w P.F RegulatorLV Capacitor Bank c/w P.F Regulator

10. 10

11. 11

12. 12

13. 13
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:

14. 14
a)The current taken by the motor and its phase angle.
01
3
3
6050.0cos
45.25
50.0110
104.1
cos
4.1)102)(70.0(
==
===
===
−
θ
θ
η
A
x
x
V
P
I
kWxPP
i
M
oi

15. 15
b) The supply current after power factor correction
AI
Ioa
A
Ioa M
40.13
19.18cos73.12
73.1260cos45.25
60cos
19.1895.0cos'6050.0cos
0
0
0
0101
=∴
==
==
=
==⇒== −−
θθ

16. 16
c) The current taken by the capacitor.
AAacabI
A
Iac
A
Iab
C
M
86.17)18.404.22(
18.419.18sin)40.13(
19.18sin
04.2260sin45.25
60sin
19.1895.0cos'6050.0cos
0
0
0
0
0101
=−=−=
==
=
==
=
==⇒== −−
θθ

17. 17
d) Draw phasor diagram by showing
phase angle and magnitude.
IICC = 17.86A= 17.86A
O
IImm =25.45=25.45
a
c
b
18.1918.1900
606000 I =13.4AI =13.4A
IICC = 17.86A= 17.86A

18. 18
e) The capacitance of the capacitor.
F
fV
I
C
fCV
fC
V
X
V
I
c
c
C
µ
ππ
π
π
17.247
)230)(50(2
86.17
2
2
2
1
===
===