Power System Distribution (Part 2)

Chapter 5
The Load in
Power System

1
 Course Outcome

Ability to explain and calculate load

characteristics and distribution system
components in power system

2
 CONTENTS
• Load Characteristic
• Load Forecasting
• Load Growth
• Customer Billing
• Power Factor
 LOAD
CHARACTERISTIC
Load Characteristic

• Load Curve
• The load on a power station is not constant. It varies
from time to time.
• The load curve is obtained by plotting the loads against
the time on a graph paper.
• When it plotted from 24 hours of a day it is called daily
load curve and if the time considered is for one year
(8760 hours) then it is called annual load curve.
• The yearly load curve is generally used to determine the
load factor.
• The areas under the load curve represent the energy
generated in the period considered.

5
Load Characteristic

Load Curve
• The area under the curve divided by the total number of
hours gives the average load on the power station.
• The peak curve load indicated by the load curve
represents the maximum demand on the power station.

Example of Typical
Commercial Load Curve

6
Load Characteristic

Load Curve
• The greatest problem for a power supply company is
varying load. The generation should be matched with the
load consistently.

• The base load power is

supplied by power plants
running continuously.
• The Intermediated/Average
load prevails for some part of
the day.
• The peak load prevails only
a few hours of the day

7
Load Characteristic

• Terms and Definitions

a) Demand
 The demand of the installation or system is the load
at the receiving terminals averaged over a specified
time.

b) Demand Interval (t)

 The period over which the load is averaged.
 Selected Δt period may be 15 min, 30 min, 1 hr, or
even longer.

8
Load Characteristic

• Terms and Definitions

c) Maximum Demand
 The maximum demand of an installation or system is
the greatest of all demands which have occurred
during the specified period of time.

d) Demand Factor (DF)

 The ratio of the maximum demand of a system to a
total connected load of the system.
Maximum Demand
DF 
Total Connected Demand
9
Load Characteristic

• Terms and Definitions

e) Diversified Demand / Coincident Demand
 It is the demand of the composite group ( residential,
commercial, industrial, miscellaneous) , as a whole,
of somewhat unrelated loads ( loss in transmission &
distribution) over a specified period of time.
f) Utilization Factor (Fu)
 It is the ratio of the maximum demand of a system to
the rated capacity of the system.
Maximum demand
Fu 
Rated system capacity
10
Load Characteristic

• Terms and Definitions

g) Load Factor (FLD)
 The load factor is defined as the ratio of the average
load to the maximum load during a given period.
 The load factor is always less than one, because the
average load is less than maximum load.
 The greater the load factor, the less will be the cost
per unit.
Average Load xT Unit Served
FLD  
Max/Peak Load x T Max/Peak Load x T
where: T=time, in days, weeks, months or years.
11
Load Characteristic

• Terms and Definitions

h) Plant Capacity Factor
 It is defined the ratio of the total actual energy produced
or served over a designated period of time to the energy
that would have been produced or served if the plant or
unit had operated continously at maximum rating.
E
Plant capacity factor 
where: C xt
E=Energy produced, kWh
C=Plant capacity/rating, kW
t=Total number of hours

12
Load Characteristic

• Terms and Definitions

i) Plant Use Factor
 It is defined as the ratio of the energy produced in given
time to the maximum possible energy that could have
been produced during the actual numbers of hours that
was in operation
E
Plant use factor 
where:
C x t1
E=Energy produced, kWh
C=Plant capacity, kW
t1=The actual number of hours the plant was in
operation
13
Load Characteristic
• Terms and Definitions
j) Diversity Factor , FD
 It is defined as the ratio of the sum of individual
maximum demands to maximum demand on power
station.
Sum of individual max demand
FD 
Coincident max demand
n n

 D  TCD
D1  D2  .......  Dn i 1i i x DFi
   i 1
Dg Dg Dg

 TCD is Total Connected Demand, DF is Demand Factor

 Dg = coincident max demand of group of n loads
14
Load Characteristic

• Terms and Definitions

k) Loss Factor , FLS
 It is defined as the ratio of the average power loss to
the peak load power loss during a specified period of
time. Average power loss
FLS 
Power loss at peak load
 Coincidence factor, Fc =1/FD
l) Load Diversity, LD
 It is the difference between the sum of the peaks of
two or more individual loads and the peak of the
combined load. n
 
LD    Di   Dg
 i 1 
15
Load Characteristic

• Example 1
A generating station had a maximum load of 20,000 kW
and the units generated being 61,500,000 kWh for the
year. Calculate the load factor.

Solution
Unit Served (kWh )
FLD 
M ax load (kW)  T (h)
61,500,000 kWh
 100
20000  365 days  24 hours
 35% 16
Load Characteristic

• Example 2
A power station has two 60 MW units each running for
7500 hours in a year and one 30 MW unit running for
4500 hours in a year. The energy produced per year is 750
x106 kWh.
Determine:
(a) Plant load factor
(b) Plant capacity factor
(c) Plant use factor.
Assume maximum demand to be 80% of the plant
capacity.
17
Load Characteristic

• Solution
a) Plant load factor
Capacity of the power plant  2x60  30  150M W
750 106 kWh
Average load 
8760h
Average load 750 106 kWh
FLD    0.71
M aximum demand 8760h  150 10 W  0.8
6

b) Plant capacity factor

Energy generated 750 106
Plant capacity factor    0.57
Capacity of plant  8760 150 10  8760
3

18
Load Characteristic

• Solution Example 2
c) Plant use factor
Energy that can be generated by two 60MW units and
one 30MW unit
=2 units x 60MW x 7500h + 30MW x 4500h
=103.5x104MWh=1035x106kWh

750 106 kWh

Plant use factor   0.72
1035 10 kWh
6

19
Load Characteristic

• Example 3
Assume the annual peak load of a primary feeder is
2000kW, at which the power loss i.e., total copper, or
I2R, loss is 80kW per three phase. Assuming an annual
loss factor of 0.15.
Determine:
(a) The average annual power loss
(b) The total annual energy loss due to the copper losses
of the feeder circuits.

20
Load Characteristic

• Solution Example 3
a) Average annual power loss
Average power loss = Power loss at peak load x FLS
= 80 kW x 0.15
=12 kW

b) Total annual energy loss

TAELCU = Average power loss x 8760 h
=12 kW x 8760 h
=105,120 kWh

21
 Example of Transformer Sizing
• An application has been made to the utility to connect a commercial
complex to the nearest distribution transformer. The distribution
transformer is rated 310 kVA and the current maximum demand on this
transformer is 250 kVA. The transformer is 20 years old and considering
hot summer, it is suggested not to load it beyond 90% capacity. The load
of the commercial complex are:

A gas station consisting of three unleaded petrol pumps, a diesel pump,

and a LNG pump. Each pump is rated at 5 hp and has an efficiency of
0.9. A general purpose shop with installed load of 5 kW. A Fish and
Chip shop with a connected load of 4 kW. Six street lights each rated at
100 W. The diversity factor of the pumps is 1.8. The demand factor of
the general purpose shop is estimated to be 0.7. The Fish and Chip shop
has a demand factor of 0.8. The diversity factor for the whole load is
estimated to be 1.2. Calculate the the coincident maximum demand of
the whole complex. Is it possible for this load to be connected to the
existing distribution transformer?
22
23
 LOAD
FORECASTING
Load Forecasting

Introduction
• Load forecasting plays an important role in power system
planning, operation and control
• Forecasting means estimating active load at various load
buses ahead of actual load occurrence.
• Planning and operational applications of load forecasting
requires a certain ‘lead time’ also called forecasting
intervals.
• A good forecast reflecting current and future trends,
tempered with good judgment, is the key to all planning,
indeed to financial success.

25
Load Forecasting

• Factors Affecting Load Forecast

26
Load Forecasting

• Nature of Forecast
Nature of Forecast Lead time Application
Very Short Term A few seconds to Generation, distribution
several minutes schedules, contingency analysis
for system security
Short Term Half an hour to a Allocation of spinning reserve;
few hours operational planning and unit
commitment; maintenance
scheduling
Medium Term A few days to a few Planning for seasonal peak-
weeks winter, summer
Long Term A few months to a Planning generation growth
few years

27
Load Forecasting

• Forecasting Methodology
• Forecasting techniques may be divided into two broad
classes. Techniques may be based on extrapolation or on
correlation or on a combination of both.
 Extrapolation
Extrapolation techniques involve fitting trend curves to
basic historical data adjusted to reflect the growth trend
itself.
 Correlation
Correlation techniques of forecasting relate system loads to
various demographic and economic factor.

28
LOAD GROWTH
Load Growth

Introduction
• In planning to accommodate future electric energy needs,
it is necessary that an estimate of the rate at which those
needs will grow.

• Examples of a typical
energy requirement
projection for a certain
country

30
Load Growth

Growth Rates
• Suppose a certain quantity M grows at a rate that is
proportional to the amount of M that is present.
• It gives dM
 aM
dt
• Where a is the constant of proportionality, known as the
per-unit growth rate. The above equation may be written
as
M  M 0e at

• Where M0 is the value of M at t=0.

31
Load Growth

Growth Rates
• At any two values of time, t1 and t2, the inverse ratio of
the corresponding quantities M1 and M2 is

M2
 e a (t2 t1 )
M1
• Based on above equation, the doubling time td may be
obtain such that M2 = 2M1 and t2 - t1= td

ln 2 0.693
td  
a a
32
Load Growth

Growth Rates
• Power system planners also need to know how much
power will be demanded.

• The approximation equation

in this curve
P  P0e bt

• Where P0 is the peak power at

t =0, b is the per unit growth

Examples of the peak power demand

for a certain country over several years.
33
Load Growth

Growth Rates
• The area under this curve over a given period is a
measure of the energy Q consumed during that period.

• If the per unit growth rate has not

changed, then the energy
consumed in one doubling period
equals the energy consumed for
the entire time prior to that
doubling.

34
Load Growth

• Growth Rates
• Evaluating the energy Q1 consumed up to t1 and the
energy Q2 consumed during the doubling time td = t2 - t1
t1
P0 bt1
Q1   0 dt 
bt
P e e

b
t2
P0 btd
Q2   P0 e dt  ( e  1 )e bt1
bt

t1
b
• Td = (ln 2/b), therefore Q2 becomes
P0 P0 bt1
Q2  ( 2  1 )e  e  Q1
bt1

b b
35
Load Growth

• Example 1
• Suppose that the consumption of the energy in a certain
country has a growth rate of 4 percent per year.
In how many years will the energy consumption be
tripled?
• Solution Example 1
Q2
3 then 3  e 0.04t or ln 3  0.04t
Q1
ln 3
t  27.47 years
0.04

36
Load Growth

• Example 2
• In certain country the energy consumption is expected to
be double in 10 years.
Calculate the growth rate in percent.
• Solution Example 2
ln 2
a
td
0.693
a x100  6.93%
10

37
Load Growth

• Example 3
• Assume that one of the distribution transformer of the
Riverside substation supplies three primary feeders. The
30-min annual max demands per feeder as listed as
following table, together with the power factor (PF) at the
time of annual peak load. Assume a diversity factor of
1.15 among the three feeder for both real power (P) and
reactive power (Q).
Feeder Demand (kW) PF
1 1800 0.95
2 2000 0.85
3 2200 0.90

38
Load Growth

• Example 3
a) Calculate the 30-min annual max demand on the substation
transformer in kW and in kVar
b) Find the load diversity in KW
c) Select a suitable substation transformer size if zero load growth is
expected and if the company policy permits as much as 25 % short
time overload on the distribution substation transformer. Use the
suitable standard 3-phase as follow
2500/3125 KVA self-cooled/forced air cooled
3750/4687 KVA self-cooled/forced air cooled
5000/6250 KVA self-cooled/forced air cooled
7500/9375 KVA self-cooled/forced air cooled

39
Load Growth

• Example 3
d) Now assume that the substation load will increase at a
constant percentage rate per year and will double in 10
years. If the 7500/9375 KVA rated transformer is
installed, in how many years will be loaded to its fans-
on rating. Assume the load growth equation is
Pn=P0(1+g)n
where: Pn=Load at the end of the nth year
P0=Initial Load
g=Annual growth rate
n=Numbers of years

40
Load Growth

• Solution Example 3
a) Given the diversity factor FD = 1.15
Sum of individual max demand
FD 
Coincident max demand
D1  D2  .......  Dn 1800kW  2000kW  2200kW
   1.15
Dg Dg
Therefore, Annual maximum demand Dg,kW
(1800kW  2000kW  2200kW) 6000kW
Dg ,kW    5217.39kW
1.15 1.15

41
Load Growth

• Solution Example 3
a) Dg =5217.39kW ( in KW), in KVA, then find the power
factor angle PF Angle
0.95 18.2
0.85 31.79
0.90 25.84
The total Reactive Power (Q)
3
Q   Pi  tan  i  (1800)(tan18.2)  (2000)(tan 31.79)
i 1

 (2200)(tan 25.84)  2896.79kVar

2896.79kVar
Dg ,kVar   2518.95kVar
1.15 42
Load Growth

• Solution Example 3
a) Therefore Dg

Dg  Dg ,kW  Dg ,kVar  5217 2  2518.822  5793.60 kVA

2 2

b) The load diversity (LD) is

 n 
LD    Di   Dg ,kW  6000 kW  5217 kW  783 kW
 i 1 

43
Load Growth

• Solution Example 3
c) The transformer size capacity if permits of 25% short
time overload
The maximum demand is 5793.60 KVA
Tx Size 25% overload Remarks
2500/3125 KVA 3125x1.25=3906.25 Under size
3750/4687 KVA 4687x1.25=5858.75 Nearest
5000/6250 KVA 6250x1.25=7812.5 Over size
7500/9375 KVA 9375x1.25=11718.75 Over size

The most suitable distribution transformer is 3750/4687

KVA self-cooled/ forced air cooled
44
Load Growth

• Solution Example 3
d) The term fans-on means the forced air cooled rating.
The increase annual growth rate (g) per year,
given Pn=P0(1+g)n
hence (1+g)10=2 , 1+ g =1.07177
g = 0.07177%/year
Therefore, (1.07177)n x 5793.60 KVA = 9375 KVA
Or ln 1.6182
n  6.944 or 7 Years
ln 1.07177
If the 7500/9375 KVA rated transformer is installed, it
will be loaded to its fan-on rating in about 7 years.
45
CUSTOMER
BILLING
Costumer Billing

• Introduction
• Customer billing is done by taking the difference in
readings of the meter at two successive times, usually at
an interval of 1 month.
• The difference in readings indicates the amount of
electricity, in kilowatt hours, consumed by the customer
in that period.
• The amount is multiplied by the appropriate rate or the
series of rates and the adjustment factor, and the bill is
sent to the customer.

47
Costumer Billing

Sample of Electricity Bill

1. Tariff
2. Bill & Payment
3. Current Bill
4. Other Charges
5. Bill Amount
6. Meter Reading
7. Billing Date
8. Payment Due Date

48
Costumer Billing

• Tariff
• ‘Electricity Tariff’ can be define as a list of fixed rate
electricity prices which has been approved by a government.
• In Malaysia the tariff for electricity are divided into 7
categories as follow:
1) Domestic
2) Commercial
3) Industrial
4) Mining
5) Street Lighting
6) Specific Agriculture
7) Top up & Stand By

49
 Costumer Billing

• Tariff
TARIFF CATEGORY UNIT RATES
1. Tariff A - Domestic Tariff
For the first 200 kWh (1 - 200 kWh) per month sen/kWh 21.8

For the next 100 kWh (201 - 300 kWh) per month sen/kWh 33.4

For the next 100 kWh (301 - 400 kWh) per month sen/kWh 40.0

For the first 100kWh (401 - 500 kWh) per month sen/kWh 40.2

For the next 100 kWh (501 - 600 kWh) per month sen/kWh 41.6

For the next 100 kWh (601 - 700 kWh) per month sen/kWh 42.6

For the next 100 kWh (701 - 800 kWh) per month sen/kWh 43.7

For the next 100 kWh (801 - 900 kWh) per month sen/kWh 45.3

For the next kWh (901 kWh onwards) per month sen/kWh 45.4

The minimum monthly charge is RM3.00 50

 Costumer Billing
TARIFF CATEGORY UNIT RATES
1. Tariff B - Low Voltage Commercial Tariff
For Overall Monthly Consumption Between 0-200 kWh/month
For all kWh sen/kWh 39.3
The minimum monthly charge is RM7.20

For Overall Monthly Consumption More Than 200 kWh/month

For all kWh (From 1kWh onwards) sen/kWh 43.0

The minimum monthly charge is RM7.20

2. Tariff C1 - Medium Voltage General Commercial Tariff
For each kilowatt of maximum demand per month RM/kW 25.9

For all kWh sen/kWh 31.2

The minimum monthly charge is RM600.00
3. Tariff C2 - Medium Voltage Peak/Off-Peak Commercial Tariff
For each kilowatt of maximum demand per month during the peak period RM/kW 38.60

For all kWh during the peak period sen/kWh 31.2

For all kWh during the off-peak period sen/kWh 19.2

The minimum monthly charge is RM600.00

51
 Costumer Billing
TARIFF CATEGORY UNIT RATES
1. Tariff D - Low Voltage Industrial Tariff
For Overall Monthly Consumption Between 0-200 kWh/month

For all kWh sen/kWh 34.5

The minimum monthly charge is RM7.20
For Overall Monthly Consumption More Than 200 kWh/month

For all kWh (From 1kWh onwards) sen/kWh 37.7

The minimum monthly charge is RM7.20

Tariff Ds – Special Industrial Tariff (for consumers who qualify only)

For all kWh sen/kWh 35.9

The minimum monthly charge is RM7.20
2. Tariff E1 - Medium Voltage General Industrial Tariff

For each kilowatt of maximum demand per month RM/kW 25.3

For all kWh sen/kWh 28.8

The minimum monthly charge is RM600.00
Tariff E1s – Special Industrial Tariff (for consumers who qualify only)

For each kilowatt of maximum demand per month RM/kW 19.9

For all kWh sen/kWh 28.3

The minimum monthly charge is RM600.00 52
Costumer Billing

• Example 1
• A domestic costumer consume 460 kWh for 1 month.
Calculate the monthly bill.

• Solution:

1) First 200 kWh (1-200 kWh) per month: (200 x 21.8)/100 = RM 43.6
2) Next 100 kWh (201-300 kWh) per month: (100 x 33.4)/100 = RM 33.4
3) Next 100 kWh (301-400 kWh) per month: (100 x 40.0)/100 = RM 40
4) Next 100 kWh (401-500 kWh) per month: (60 x 40.2)/100 = RM 24.12

Monthly Bill = RM(43.6 + 33.4 + 40 + 24.12) = RM 141.12

53
Costumer Billing

• Example 1
• Assume a domestic customer use the following domestic
tariff rate schedule.

54
Costumer Billing

• Example 1(cont)
a) Assume that an average month is 730h and find the
monthly load factor
b) Find the reasonable size of continuous KVA rating of the
distribution transformer
c) Calculate the monthly bill
d) What size of capacitor (in kVar) would rise the PF of the
customer to 0.9

55
Costumer Billing

• Solution Example 1
a) Customer A , FLD
Unit served 1200kWh
FLD    0.205
Peak load x T 8kW x 730h

b) Continuous KVA rating

PA 8 kW
SA    9.41 kVA
Cos  0.85

The continuous suitable rating for the distribution

transformer is 10 KVA

56
Costumer Billing

• Solution Example 1
c) The monthly bill
First 200kWh = 200kWh x 21.8 sen/kWh = RM43.6
Next800kWh = 800 kWh x 28.9 sen/kWh =RM 231.2
Over 1000kWh = 200 kWh x 31.2 sen/kWh = RM 62.4
The total monthly Bill= RM 337.20

d) Size of capacitor
Current PF=0.85, the kVarh value is
1200kWh
xSin( Cos 1 0.85 )  743.69kVarh
0.85

57
Costumer Billing

• Solution Example 1

d) Size of capacitor (cont.)

At PF=0.9, the kVarh value is

1200kWh
 sin(cos 1 0.9)  581.19kVArh
0.9
Therefore, the capacitor size required is

(743.69  581.19)kVarh
 0.22kVAr
730h

58
Power Factor Surcharge

Percent of Condition
surcharge from
the current bill
For every 0.01 less than
1.5% 0.85 power factor
For every 0.01 less than
3% 0.75 power factor

59
60
POWER FACTOR
Power Factor

• Definition of Power Factor

• Power factor is the ratio between actual (true) load power
(kW) and the apparent load power (kVA)
Actual load power(kW)
p. f . 
Apparentload power (kVA)

• It is a measure of how effectively the current is being

converted into useful work output and more particularly
is a good indicator of the effect of the load current on
the efficiency of the supply.

62
Power Factor

• Fundamental of Basic Electricity - The Power Triangle

P - kW
Q - kVar
S - kVA

S  P2  Q2
kW
Power Factor 
kVA

63
Power Factor

• Equipment Causing Poor Power Factor

• Lightly loaded induction motor. Examples of this type of
equipment and their approximate power factor are:
 70% power factor or better: Air conditioners, pumps,
center less grinders, cold header, up setter, fans or
blower
 60% to 70% power factor: Induction furnaces,
standard stamping machines and weaving machines
 60% power factor and below: Single-stroke presses,
automated machine tools, finish grinders, welders

64
Power Factor

• Reactive Power Problem (Motor)

• Example that a motor is rated at 10,000W at 0.8 power
factor. The resistance is 5ohm. At 415V, the motor will
require the following amount of current:
I=10000/(√3x0.8x415)=17.39A
Losses when pf =0.8 : I2R=(17.39)2(5)=1,512W
The same motor rated at 0.65 power factor will require:
I=10000/(√3x0.65x415)=21.403A
Losses when pf=0.65 : I2R=(21.403)2(5)=2290.4W

65
Power Factor

• Reactive Power Problem (Transformer)

• Example that 11/0.433 kV, 1000 kVA transformer has
maximum loading of 800 kW and power factor of 0.45
What is the % loading of the transformer?
P
PF   0.45
S
P 800
S (load )    1777 kVA
PF 0.45
S (load ) 1777
%TX load  100  100  177%
TX Capacity 1000

66
Power Factor

• Reactive Power Problem (Transformer)

• Example that 11/0.433 kV, 1000 kVA transformer has
maximum loading of 800 kW and power factor of 0.9
What is the % loading of the transformer?

P
PF   0.9
S
P 800
S (load )    888.88 kVA
PF 0.9
S (load ) 888.88
%TX load  100  100  88.88%
TX Capacity 1000

67
Power Factor

• Reactive Power Problem (Transformer)

• Condition 1
 PF=0.45
 TX Size=1000kVA
 Load KVA=1777 KVA
 %TX Load=177%
• Condition 2
 PF=0.9
 TX Size=1000kVA
 Load KVA=888 KVA
 %TX Load=88%
68
Power Factor

• Minimum Power Factor

• Customers are advise to maintain power factor at
minimum of 0.85

69
 Power Factor

How to Improve Power Factor

• Customers are advised to follow these steps:-
• Install capacitors (KVAR Generators)
– Capacitor
– Corrector
– Synchronous generators
– Synchronous motors
• Minimise operations of idling or lightly loaded motors.
• Avoid operating equipment above its rated voltage.
• Replace standard motors as they burn out with energy efficient
motors.
70
Power Factor

• Power Factor Improvement

Example Of Power Flow Diagram Of Industrial Plant

71
Power Factor

Benefits of Improving Power Factor

Benefit 1: Reducing KW billing demand
• Low Power Factor requires high reactive power (KVAR) and apparent
power (KVA), which is the power that electric utilities supplies. Therefore,
a facility’s low power factor forces electric utilities to increase its
generation and transmission capacity in order to handle this extra demand.
• By increasing power factor, customers use less KVAR. This results in less
KW, which equates to cost savings for electric utilities .

Benefit 2: Eliminating power factor surcharge

• Utility companies all around the world charge customers an additional
surcharge when their power factor is less than 0.95. In fact, some utilities
are not obliged to deliver electricity to their customers at any time the
customer’s power factor falls below 0.85.
Power Factor

• Thus, customer can avoid this additional surcharge by increasing power

factor.

Benefit 3: Increased system capacity and reduced system losses in

electrical system
• Low power factor causes power system losses in the customer’s electrical
system. By improving power factor, these losses can be reduced. With the
current rise in the cost of energy, increased facility efficiency is
important. Moreover, with lower system losses, customers are able to add
additional load in their electrical system.
Power Factor

Benefit 4: Increased voltage level in electrical system, resulting in more

efficient motors
• As mentioned before, low power factor causes power system losses in
customer’s electrical system. As power losses increase, customer may
experience a voltage drop. Excessive voltage drops can cause overheating
and premature failure of motors and other inductive equipment.
• Therefore, by raising the power factor, customers can minimize these
voltage drops along feeder cables and avoid related problems. Motors will
run more efficiently, with a slight increase in capacity and starting torque.
Power Factor

• Power Factor Improvement

Power Triangle To Illustrate Power Factor Correction

75
Power Factor

• Power Factor Improvement

• The amount of reactive compensation supplied by the
capacitor bank is
QCAP  Q1  Q2
• Apparent power consumed by the load before adding
capacitors PLOAD PLOAD
S1  
PF1 Cos 1
• Apparent power supplied by the source after adding
capacitors
PLOAD PLOAD
S2  
PF2 Cos 2
76
Power Factor

• Power Factor Improvement

• Where PF1, and PF2 are the actual load power factor and
desired system power factors, respectively.
2
P 
Q1  S1  PLOAD   LOAD   PLOAD
2 2 2

 PF1 
2
 PLOAD 
Q2  S 2  PLOAD     PLOAD 2
2 2

 PF2 
  1     1  
QCAP  PLOAD x    1    1 
 PF     PF  
 2 2
 1     2  

77
Power Factor

• Example 1
• An industrial plant has an active power demand of
500kW at a power factor of 0.76 lagging. Determine the
reactive power rating of the capacitor bank required to
improve the power factor to the following:
a) 0.8 lagging
b) 0.9 lagging
c) Unity
Assume the capacitor steps are available in 50 kVar
increments

78
Power Factor

• Solution Example 1
  1     1  
QCAP  PLOAD x      
 PF 2   1     PF 2   1 
  1     2  

  1     1  
a ) QCAP  500kW x   2 
1   2 
 1   52.6kVar  50kVar
  0.76 
    0.80  
  1     1  
b ) QCAP  500kW x   2 
1   2 
 1   185.4kVar  200kVar
  0.76 
    0.90  

  1     1  
c ) QCAP  500kW x   2 
 1     2   1   427.6kVar  400kVar
  0.76     1  


79
Power Factor

• Example 2
Assume that a 700 kVA load has a 65% power factor. It is
desired to improve the power factor to 92%.Using the
power factor correction table (Table 1).
Determine the following:
a) The correction factor required
b) The capacitor size required
c) What would be the resulting power factor if the next
higher standard capacitor size is used. Assume the
capacitor steps are available in 50 kVar increments

80
Power Factor

• Solution Example 2
a) From correction factor table (Table 1), the correction
factor required can be found as 0.743
b) The 700 KVA load at 65% power factor is
PL=SL x cos  W
=700k x 0.65 W
=455 kW
The capacitor size necessary to improve the power factor
from 65 to 92% can be found as
Capacitor size = PL x (Correction Factor)
=(455)(0.743)
=338.065 kVar

81
Power Factor

• Solution Example 2
c) Assume that the next higher standard capacitor size (or
rating) is selected to be 350 kVar. Therefore the resulting
new correction factor can be found from
Standard Capacitor Rating
New Correction Factor 
PL
350kVar
  0.769
455kW
From the table by using interpolation method, based on
original power factor (65%) and new correction factor
(0.769), refer to Table 1.

82
Power Factor

• Solution Example 2
the close value of desired correction factor 0.769 is 0.774 at
93% power factor and;
The difference between the correction factors at 92% (0.743)
and 93% (0.774) is 0.031.
Thus, the new corrected % PF is obtained by an interpolation
method:
0.774 - 0.769 0.769 - 0.743
  92%   93%
0.031 0.031
 92.8387%
 0.928

83
Table 1: Power Factor Correction Table

84