Economics of Generation,
Transmission and Operation
Lectures 4: Load Factors
HOSSEIN SEIFI AND
MOHAMMAD SADEGH
SEPASIAN,
“ELECTRIC POWER
SYSTEM
PLANNING
ISSUES, ALGORITHMS
AND SOLUTIONS”
REFERENCE BOOK
Load Characteristics and
Forecasting
• The design of a power distribution
system must take into account the
future increase in load.
• Consumers (domestic), commercial,
industrial, municipal, agricultural etc.
3
Load Characteristics
• The connected load
• the sum of continuous ratings of all the devices and outlets
installed on his distribution circuit.
• The maximum demand
• maximum power that the consumer’s circuit is likely to
draw at any time.
• Consumers do not use all the devices at full load
simultaneously
Demand Factor=Maximum Demand (MD)/Connected Load
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Type of Load Description Demand
Factor
Residential 0.5 KW and less 0.7
0.5 to 1 KW 0.6
1KW and above 0.5
Commercial Restaurants, Shops and Offices 0.7
Educational Institutions 0.3-0.5
Theatre, Cinema and Hotels 0.5
Industrial Cottage (less than 5kW) 0.8
Small Scale (5kW to 25kW) 0.7
Medium (25kW to 100kW) 0.6
Large (More than 100kW) 0.5 5
• Example1: A residential consumer has the following
connected load: 8 bulbs of 100 W each, 2 fans of 60 W
each and 2 light plug points of 100 W each. His use of
electricity during a day as under:
• 12 midnight to 5 am one fan
• 5 am to 7 am 2 fans and one light point
• 7 am to 9 am Nil
• 9 am to 6 pm 2 fans
• 6 pm to midnight 2 fans and 4 bulbs
• Find
(a) connected load (b) maximum demand (c) demand factor
(d) energy consumed during 24 hours
(e) energy consumed in 24 hours if all devices are used all
the day. 6
Solution.
(a) Connected load = 8Lx100 +2Fx60 + 2Px100 = 1120 W
(b) Total wattage at different times is:
12 midnight to 5 am: one fan x 60 = 60 W
5 am to 7 am : 2 fans x 60 + 1 light point x 100 =220
W
7 am to 9 am: Nil
9 am to 6 pm: 2 fans x 60 = 120 W
6 pm to midnight: 2 fans x 60 + 4 bulbs x 100 = 520 W
Then the maximum demand is 520 W
(c) Demand factor = MD/Connected Load
= 520 / 1120 = 0.464
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(d) Energy consumed kWh:
12 midnight to 5 am: is 5 hours period, then
e1 = 60 W x 5 h = 300 Wh
5 am to 7 am is 2 hours period,
e2 = 220 W x 2 h = 440 Wh
7 am to 9 am Nil
9 am to 6 pm is 9 hours period, then
e3 = 120 W x 9 h = 1080 Wh
6 pm to 12 midnight is 6 hours period then
e4 = 520 W x 6 h = 3120 Wh
Total energy consumed during 24 hours is
E=300+440+1080+3120Wh=4940Wh = 4.94kWh
(e) If all devices are used throughout the day,
the energy consumed is =
Connected Load x 24 hours =
1120 x 24 = 26880 Wh = 26.88 kWh
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PV Sizing
• 1 kW considering real demand
consumption
• 5 kW or more considering connected load
consumption
Group Diversity Factor
• Each one of the consumers will have a certain
maximum demand.
• It is likely that the maximum demands of different
consumers occurs at different times. This is due to
different habits and requirements.
• the maximum demands of the group will be less
than the sum of their maximum demands.
Sum of individual Maximum Demands
Group Diversity Factor equals
Maximum Demand of the group
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What is diversity factor of this
load
Example 2
• Flat 1:
• Lighting: 6 kW
• HVAC: 4 kW
• Appliances: 3 kW
• Total Connected Load for Flat 1: 6+4+3=13, Maximum Demand for Flat 1: 10 kW
• Flat 2:
• Lighting: 5 kW
• HVAC: 3 kW
• Appliances: 4 kW
• Total Connected Load for Flat 2: 5+3+4=12, Maximum Demand for Flat 2: 9 kW
• Flat 3:
• Lighting: 7 kW
• HVAC: 5 kW
• Appliances: 2 kW
• Total Connected Load for Flat 3: 7+5+2=14, Maximum Demand for Flat 3: 11 kW
Solution
Knowing that Maximum Demand of the System=25 kW, find
diversity factor and demand factor of the building.
Solution
Total Connected Load=Flat 1+Flat 2+Flat 3=13 kW+12 kW+14 k
W=39 kW
Demand Factor=Maximum demand of system / Connected load
of system = 0.64
Total Individual Maximum Demand=10 kW+9 kW+11 kW=30 k
W
Diversity Factor= 1.20
Comments on the example
• Connected Load vs. Maximum Demand:
• The total connected load of 39 kW represents the total capacity
of all electrical devices in the flats if they were all running
simultaneously.
• The maximum demand of 25 kW reflects the peak load that the
building actually experiences, taking into account that not all
devices are used at the same time.
• Demand Factor (0.641): This indicates that only 64.1% of the
total connected load is utilized during peak demand,
suggesting potential for optimizing usage.
• Diversity Factor (1.20): This shows that the maximum
demands of the individual flats do not occur simultaneously,
allowing for a more efficient distribution of electrical load
across the building.
District example
• A district composed of 10 house holds has
• Total Connected Load for all households in the area: 12 MW
• Maximum Demand observed during peak usage (e.g., a winter
evening): 8 MW
• Average Demand over a 24-hour period: 6 MW
• Household 1 has a maximum demand of 1.0 MW, while
Household 2 uses 0.8 MW. Household 3 has a slightly higher
demand at 1.2 MW, and Household 4 requires 0.9 MW.
Household 5's demand is 1.1 MW, whereas Household 6 has a
lower maximum demand of 0.7 MW. Household 7 shows the
highest individual demand at 1.3 MW, while Household 8 has a
demand of 0.6 MW. Lastly, Household 9 also reaches 1.0 MW, and
Household 10 tops the list with a demand of 1.4 MW.
Request and solution
• Please calculate demand factor and diversity factor.
Solution
Demand Factor: 0.67
Diversity Factor: 1.275
A demand factor of 0.67 indicates that, at peak demand, about
67% of the connected load is being utilized, suggesting unused
capacity.
A diversity factor of 1.275 suggests that the loads are generally
synchronized, with peak demands occurring close to each other.
This can happen in densely populated areas or during specific
events (e.g., cold weather).
Assignment
A group of two consumers has the following electricity
demand pattern on a typical winter day:
• Consumer A: Connected load 2.5 kW
• Load from 12 midnight to 5 am: 100 W
from 5 am to 6 am : 1.1 kW
from 6 am to 8 am : 200 W
from 8 am to 5 pm : NIL
from 5 am to 12 midnight: 500 W
• Consumer B: Connected load 3 kW
Load from 1 pm to 7 am : NIL
from 7 am to 8 am: 300 W
from 8 am to 10 am; 1 kW
from 10 am to 6 pm: 200 W
from 6 pm to 11 pm : 600 W 17
(a) Calculate demand factors of both consumers.
(b) Plot the variation in demand versus time of the day for
each consumer and the group.
(c) Find group diversity factor.
(d) Find energy consumed by each consumer in 24 hours.
(e) Find the maximum energy which each consumer would
consume in 24 hours if his load were constant and equal to
his maximum demand.
(f) Find the ratio of actual energy to maximum energy for each.
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• From the solution note that:
• The ratio of actual energy to maximum energy of a consumer is
known as load factor of the consumer.
• The ratio of actual energy to maximum energy of a group of
consumers is the load factor of the group.
• The ratio of the actual energy to maximum energy of a system is
the load factor of the system.
• The diversity between the two consumers has reduced the
maximum demand of this group of two consumers.
• If their pattern of using electricity were similar, the total maximum
demand would have been 2.1 kW.
• Diversity has reduced the combined maximum demand to 1100
W.
• That has an important effect on power system economics
because the power supply authorities will have to make lesser
investment in generation, transmission and distribution facilities. 19
Load Factor
Average Load
Load Factor
Peak Load
Energy consumed during a time of t hours
Load Factor
Peak Load x t
• The load factor depicts the variation of load
during a certain period but it does not give any
indication of the shape of the load duration
curve.
• load duration curve might be for a day, a week,
or a year.
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Example 3: A consumer has the following
connected load:
10 lamps each of 60 W
2 heater each of 1000 W
Maximum demand 1500 W
On the average uses 8 lamps for 5 hours per day,
each heater 3 hours per day.
Find
(a) average load,
(b) monthly energy consumption,
(c) load factor.
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Solution.
Daily average load
= Actual Daily Energy Consumption / day Time
=(8Lx60Wx5h+2Hx1000Wx3h) / 24 = 350W
Monthly energy consumption
= Daily average load x 24 h x 30 day =252kWh
Daily Load Factor
= Daily average load / Day Maximum Demand
= 350 / 1500 = 0.233
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