3. RESISTIVE CIRCUIT
For a purely resistive circuit (such as that in
Fig. 19.2), v and i are in phase, and θ = 0°,
as appearing in Fig. 19.3.
FIG. 19.2 Determining the power delivered to a purely
resistive load.
8. Most plant loads are Inductive and require a magnetic field to
operate:
Motors
Transformers
Florescent lighting
The magnetic field is necessary, but produces no useful work
The utility must supply the power to produce the magnetic field and
the power to produce the useful work: You pay for all of it!
These two types of current are the ACTIVE and REACTIVE
components
ACTIVE & REACTIVE POWERS
9. Advantages of Phasor Analysis
0
2 2
Resistor ( ) ( )
( )
Inductor ( )
1 1
Capacitor ( ) (0)
C
Z = Impedance
R = Resistance
X = Reactance
X
Z = =arctan( )
t
v t Ri t V RI
di t
v t L V j LI
dt
i t dt v V I
j C
R jX Z
R X
R
Device Time Analysis Phasor
(Note: Z is a
complex number but
not a phasor)
10. INDUCTIVE CIRCUIT AND REACTIVE
POWER
FIG. 19.8 Defining the power level for a purely inductive
load.
11. INDUCTIVE CIRCUIT AND REACTIVE POWER
FIG. 19.9 The power curve for a purely inductive load.
Q
S
VARind
P
12. CAPACITIVE CIRCUIT
For a purely capacitive circuit (such as that
in Fig. 19.12), i leads v by 90°, as shown in
Fig. 19.13.
FIG. 19.12 Defining the power level for a purely capacitive
load.
14. Power Factor Fundamental
Definitions:
Working /Active Power: Normally measured in
kilowatts (kW). It does the "work" for the system--
providing the motion, torque, heat, or whatever else
is required.
Reactive Power: Normally measured in kilovolt-
amperes-reactive (kVAR), doesn't do useful "work."
It simply sustains the electromagnetic field.
Apparent Power: Normally measured in kilovolt-
amperes (kVA). Working Power and Reactive
Power together make up apparent power.
15. Complex Power
*
cos( ) sin( )
P = Real Power (W, kW, MW)
Q = Reactive Power (var, kvar, Mvar)
S = Complex power (VA, kVA, MVA)
Power Factor (pf) = cos
If current leads voltage then pf is leading
If current
V I V I
V I
S V I j
P jQ
lags voltage then pf is lagging
(Note: S is a complex number but not a phasor)
17. Power Factor:The Beer Analogy
Mug Capacity = Apparent Power (KVA)
Foam = Reactive Power (KVAR)
Beer = Real Power (kW)
Power Factor =
Beer (kW)
Mug Capacity (KVA)
Capacitors provide the Foam (KVAR),
freeing up Mug Capacity so you don’t have
to buy a bigger mug and/or so you can pay
less for your beer !
kVAR
Reactive
Power
kW
Active
Power
kVA
Apparent
Power
18. 18
WHAT IS POWER FACTOR?
Power Factor is the ratio of ACTIVE
POWER to the TOTAL POWER (apparent
power):
= Active Power = P
Total Power S
S = Total power of Generator (or used)
P = Power consumed in the load (active power)
Q = Reactive power stored in magnetic field. Or
wasted power
Power Factor
19. 19
WHAT IS POWER FACTOR?
Vectorial Representation:
S P.Q Φ
P
S
Q
j=90o
Generator
Load
Total power = S = VI = (units = KVA)
Active power = P = VI CosΦ = (units = KW)
Reactive power= Q = VI SinΦ = (units = KVAR)
Φ
V
I
V = Voltage : Volts
I = Current : Ampere
Φ = Physical displacement of V&I
Power Factor = CosΦ
20. 20
HOW TO IMPROVE THE
POWER FACTOR ?
The power factor can be improved by
supplying KVAR to the loads (inductive type)
“Capacitor is source of KVARs”
Therefore the power factor of connected load
can be improved by installing power factor
improvement capacitors/capacitor banks
21. 59.7 kV
17.6 MW
28.8 MVR
40.0 kV
16.0 MW
16.0 MVR
17.6 MW 16.0 MW
-16.0 MVR28.8 MVR
Power System Notation
Power system components are usually shown as
“one-line diagrams.” Previous circuit redrawn
Arrows are
used to
show loads
Generators are
shown as circles
Transmission lines
are shown as a
single line
22. Reactive Compensation
44.94 kV
16.8 MW
6.4 MVR
40.0 kV
16.0 MW
16.0 MVR
16.8 MW 16.0 MW
0.0 MVR6.4 MVR
16.0 MVR
Key idea of reactive compensation is to supply reactive
power locally. In the previous example this can
be done by adding a 16 Mvar capacitor at the load
Compensated circuit is identical to first example with
just real power load
23. Reactive Compensation, cont’d
Reactive compensation decreased the line
flow from 564 Amps to 400 Amps. This has
advantages
Lines losses, which are equal to I2 R decrease
Lower current allows utility to use small wires, or alternatively, supply
more load over the same wires
Voltage drop on the line is less
Reactive compensation is used extensively by
utilities
Capacitors can be used to “correct” a load’s
power factor to an arbitrary value.
24. 24
HOW TO IMPROVE THE
POWER FACTOR ?
LOAD
LOW POWER FACTOR
LOAD
IMPROVED POWER FACTOR
CAPACITOR
Fig.I
25. Calculation
If the original inductive load has apparent
power S1, then
P = S1 cos 1 and Q1 = S1 sin 1 =
P tan 1
If we desired to increased the power
factor from cos1 to cos2 without
altering the real power, then the new
reactive power is
Q2 = P tan 2
The reduction in the reactive power is caused by the shunt
capacitor is given by
QC = Q1 – Q2 = P (tan 1 - tan 2)
26. 26
Calculation
rms
2
21
rms
2
C
ωV
)tanθP(tanθ
ωV
Q
C
The value of the required shunt capacitance is
determined by the formula
Notice that the real power, P dissipated by the
load is not affected by the power factor correction
because the average power due to the capacitor
is zero
27. 27
Example 11.15
When connected to a 120V (rms), 60Hz power line, a
load absorbs 4 kW at a lagging power factor of 0.8.
Find the value of capacitance necessary to raise the
pf to 0.95.
28. 28
Solution
VA5000
8.0
4000
cos 1
1
P
S
If the pf = 0.8 then,
cos1 = 0.8 1 = 36.87o
where 1 is the phase difference between the voltage and current.
We obtained the apparent power from the real power and the pf
as shown below.
The reactive power is
VAR300087.36sin5000sin 111 SQ
29. 29
Solution
VA5.4210
95.0
4000
cos 2
2
P
S
When the pf raised to 0.95,
cos2 = 0.95 2 = 18.19o
The real power P has not changed. But the apparent power has
changed. The new value is
The new reactive power is
VAR4.1314sin 222 SQ
30. 30
Solution
VAR6.16854.1314300021 QQQC
The difference between the new and the old reactive power is
due to the parallel addition of the capacitor to the load.
The reactive power due to the capacitor is
The value of capacitance
added is
μF5.310
)120)(60(2
6.1685
22
rms
C
V
Q
C
31. In this example, demand
was reduced to 8250 kVA
from 10000 kVA.
1750KVA Transformer
Capacity Release.
The power factor was
improved from 80% to
97%
Before After
Why do we install Capacitors?
32. 32
DISADVATAGES OF LOW
POWER FACTOR
1. For a given power to be supplied, the current
is increased.
2. The current thus increased in-return causes
increase in copper losses (PL=I2R) and
decrease in the efficiency of both apparatus
and the supply system, which results in
overloading and hence burning of the
associated equipment.
33. 33
DISADVATAGES OF LOW
POWER FACTOR
3. Copper losses in transformers also increases.
4. Generators, transformers, switches, transmission
lines and other associated switchgear becomes
over-loaded.
5. Voltage regulation of generators, transformers
and transmission lines increases.
6. Hence, cost of generation, transmission and
distribution increases.
34. 34
NATURAL POWER FACTORS
o CEILING FAN 0.5 TO 0.7
o CABIN FAN 0.5 TO 0.6
o EXAUST FAN 0.6 TO 0.7
o SEWING MACHINE 0.6 TO 0.7
o WASHING MACHINE 0.6 TO 0.7
o RADIO 0.9
o VACUUM CLEANER 0.6 TO 0.7
o TUBE LIGHT 0.5 TO 0.9
o CLOCK 0.9
o ELECTRONIC EQUIPMENT 0.4 TO 0.95
35. 35
NATURAL POWER FACTORS
o NEON SIGN 0.5 TO 0.55
o WINDOW TYPE AIR CONDITIONER 0.62 TO 0.85
o HAIR DRYERS 0.7 TO 0.8
o LIQUIDISER 0.8
o MIXER 0.8
o COFFEE GRINDER 0.75
o REFRIGERATOR 0.65
o FREEZER 0.7
o SHAVER 0.6
o TABLE FAN 0.5 TO 0.6
36. 36
NATURAL POWER FACTORS
o MERCURY VAPOUR LAMP O.4 TO 0.6
o INDUSTRIAL INDUCTION MOTOR:
◘ NO LOAD O.18
◘ 25% FULL LOAD 0.56
◘ 75% FULL LOAD 0.81
◘ 100% FULL LOAD 0.85
◘ 125% FULL LOAD 0.86
o COLD STORAGE O.76 TO 0.80
o CINEMAS 0.78 TO 0.80
o METAL PRESSING O.57 TO 0.72
38. 38
NATURAL POWER FACTORS
o OIL MILLS O.51 TO 0.59
o WOOLEN MILLS O.70
o POTTERIES 0.61
o CIGARETTE MANUFACTURING 0.80
o FOUNDRIES 0.59
o STRUCTURAL ENGINEERING 0.53 TO 0.68
o CHEMICALS 0.72 TO 0.87
o MUNICIPAL PUMPING STATIONS 0.65 TO 0.75
o OIL TERMINALS 0.64 TO 0.83
o ROLLING MILLS 0.60 TO 0.72
39. 39
NATURAL POWER FACTORS
o PLASTIC MOLDING 0.57 TO 0.73
o FILM STUDIOS O.65 TO 0.74
o HEAVY ENGINEERING WORK 0.48 TO 0.75
o RUBBER EXTRUSION AND MOLDING 0.48
o PHARMACEUTICALS 0.75 TO 0.86
o OIL AND PAINT MANUFACTURING 0.51 TO 0.69
o BISCUIT FACTORY 0.60
o LAUNDRIES 0.92
o FLOUR MILLS 0.61
o GLASS WORKS 0.87
40. 40
NATURAL POWER FACTORS
o IRRIGATIONS PUMPS O.62 TO 0.80
o REPAIR SHOP, AUTOMATIC LATHE, 0.6
WORKSHOP, SPINNING MILLS,
WEAVING MILL
o WELDING SHOP 0.5 TO 0.6
o HEAT TREATMENT SHOP, STEEL 0.65 TO 0.8
WORKS, ROLLING MILLS
o TEXTILE 0.65 TO 0.75
o CEMENT 0.8 TO 0.85
o OFFICE BUILDING O.8 TO 0.85
41. Three Options for Applying Power Factor
Capacitors:
A) Fixed capacitors @ individual motors or @ MCC
B) Automatic Banks at Main Switch Board
C) De-tuned Automatic Capacitor Bank at Main Switch Board
M M M M M
A B C A
Harmonic
Source
e.g. Variable
Speed Drive
Capacitor Locations
42. 42
ADVANTAGES OF POWER
FACTOR IMPROVEMENT
i. Adding capacitor, releases circuit capacity for
more load or relieves the overloaded circuit. The
capacitor KVAR per KVA of load increase is of
particular interest as this establishes the average
cost of supplying each additional KVA of load.
This cost can be compared with the cost per KVA
of increasing the transformer or supply circuit
rating and would justify the application of
capacitors.
PFI Capacitor’s addition, thus can be viewed in two lights.
43. 43
ADVANTAGES OF POWER
FACTOR IMPROVEMENT
ii. Capacitors applied to given load reduce the I2R
losses in the supply circuit. For a 70 percent power
factor load with 40-KVAR of capacitors added for
each 100 KVA of circuit capacity, the I2R loss will
be 59% of its former value. The losses are not only
reduced in the circuit in which the capacitors are
applied but in all the circuit back to and including
the source generator.
44. 44
ADVANTAGES OF POWER
FACTOR IMPROVEMENT
Automatic Power Factor improvement capacitors or
capacitor banks applied on the load end of circuit,
with lagging power factor (more than 95% loads),
have particular effects, one or more of which may be
the reason for the application.
1. Improves the power factor at the source.
2. Reduces system losses as current in
conductors decreases.
45. 45
ADVANTAGES OF POWER
FACTOR IMPROVEMENT
3. Improves voltage level at the load.
4. Decreases KVA loading on the source.
5. Reduces investment in system facilities per
KW of load supplied.
