Dr.
Amgad Salem
❑ There are many Important terms which must be understood before
performing the load estimation, these terms are:
1- Connected load
It is the sum of all the loads connected to the electrical system, usually
expressed in watts.
2- Demand load
It is the electric load at the receiving terminals averaged over a specified
demand interval of time, usually 15 min., 30 min., or 1 hour based upon the
particular utility’s demand interval. Demand may be expressed in amperes,
kilo-amperes, kilo-watts, kilo-vars, or kilo-volt-amperes.
3- Load factor
The load factor is the ratio of the average load over a designated period of
time, usually 1 year, to the maximum
load occurring in that period.
Load factor = Average load / Maximum load
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EX: Determine the electrical load for a villa (special building) containing:
1. Number of lighting outlets (15000 watts).
2. Electrical sockets: capacity of 2 amps with 30 circuits (each circuit contains 6 socket). Cos Φ=0.85
3. Electrical appliances: (Cos Φ=0.85)
o Garage door powered by 1.2 HP motor
o GYM equipment (1400 watt)
o 7 air conditioners:
2 pcs of capacity (2.6 kW)
3 pcs (1.9 kW)
2 pcs (1.5 kW)
4. Cooking devices:
o Main cooker with a capacity (6000 watts)
o 2 Sub-cookers with a capacity (2000 watts)
o Sub-cooker with a capacity (1200 watts)
5. Pumps
o Water pump (1600 watts)
o Garden irrigation pump (2800 watts)
o Sewage pump in the basement with a capacity of (600 watts)
6. Heaters
o 2 heaters operating continuously, with a capacity of (3000 watts)
o One heater working continuously with a capacity of (2000 watts)
o One intermittent operated heater with a capacity of (6000 watts)
o One Jacuzzi heater (5000 Watt)
o One sauna heater (4000 watts)
Solution
Demand factor calculated by the Egyptian code according to the building type:
1- Lighting= 15000*0.66= 9.9 KW
2-For one circuit (ampere) = 2+(0.4*2*5)= 6 A
For one circuit (KW) =6*220*0.85= 1.122 KW
Total load socket=1.122*30= 33.66 KW.
3-Electrical devices : (KW) devices which greater than 10 Ampere=
0.85*10*220=1.87 KW
Total load of electrical devices
=0.5(2*2.6+3*1.9)+[2*1.5+(1.2*746/1000)+1.4]=10.75 KW
4- Cooking devices (Resistive load), 10*220=2200 W
Total load of electrical devices=1.2+2*2+0.3*6= 7 KW.
5- Total load of electrical motor=2.8+0.5(1.6+0.6)=3.9 KW
6-
(a)Total load of electrical heater (continuously)= 2*3+2=8 KW
(b)Total load of electrical heater (intermittent)= 6+5+4*0.25=12 KW
Total load of electrical=85.21 KW
7- Diversity factor
The diversity factor is the reciprocal of the coincidence factor.
Diversity factor = Sum of individual maximum demands / Maximum
system demand
Notes:
The Diversity Factor is applied to each group of loads (e.g. being
supplied from a distribution or sub-distribution board).
Example#1:
Consider that a feeder supplies 5 consumer with the following load
conditions: On Monday, 1st consumer reaches a maximum demand of
100 amps; on Tuesday, 2nd consumer reaches 95 amps; on Wednesday, 3rd
consumer reaches 85 amps; on Thursday, 4th consumer reaches 75 amps;
on Friday, 5th consumer reaches 65 amps. The feeder’s maximum demand
is 250 amps. Calculate the Diversity Factor for this feeder?
Solution:
The diversity factor can be determined as follows:
Diversity factor = Sum of total demands ÷ Maximum demand on
feeder = 420 ÷ 250 = 1.68
Example#3: Sub DB
Load_1 Load_2
(8 KW) (10 KW)
Note: The Diversity factor equal 1 as the case in the figure shown
Ex : If the measured values of maximum demand loads are
595,160,380,435,504,620 W. But, the maximum demand on
power supply is 900 W. Calculate the diversity factor.
Solution:
Diversity factor= (595+160+380+435+504+620) / (900) =2.99
Ex : According to figure as shown, Calculate the diversity factor
and coincident factor (next slide).
Solution:
Diversity factor= (2000+2000) / (3000) =1.33
Coincident factor= 1/ 1.33 = 0.752
Difference between demand and diversity factor:
Most of the electrical engineers confuse between the
demand and diversity factors, to solve this
confusion, don't forget that:
❑ The classification of non-industrial loads:
1. Lighting loads, which are divided into internal lighting (spaces and
corridors) and external lighting (cosmetic lighting - lighting of the
external space - garage - stairs).
2. Small appliances, including appliances used in offices, and sockets
for general use, such as refrigerators and television.
3. Air conditioning loads
4. Electrical loads for water and sewage equipment such as pumps,
water heaters and firefighting.
5. Light current loads (alarms and telephones)
6. Elevators and escalators are also called dynamic loads.
❑Industrial loads are classified into heavy, medium
and light loads. Of course, you need detailed
information to estimate them, not tables.
Load Estimation for Industrial Load
✓ The Primary Stage of Load Estimation
First Method: Loads are
estimated according to residential
areas and social level.
Second Method: The loads are
estimated according to the nature of
the building’s use (pre-knowledge of
the loads of similar buildings)
Third Method: (estimated
calculation for the load most usage)
such as lighting load, sockets air
condition and general loads.
1. Primary load estimation for
lighting load:
The lighting load represents about 20%:
50% from the electrical load. The table
(1,2) illustrate the standard loads for
lighting system according to NEC
(National Electrical Code)
2. Primary load estimation for Socket:
Because socket loads are not fixed, there are many ways to
estimate their loads:
Estimate for one socket = 180 W
Estimate for one socket = 1.5 ampere
According to (NEC), estimated for each apartment= 3000W, in
addition to the heavy loads such as (laundry , dryer and
electric oven) for easy to calculations, the rated for each
socket = 100 watts, in addition to the boiler and air
conditioning (put in its value).
3.Air condition estimation
The responsibility for estimating the loads of air conditioning to the
mechanical engineer, not the electrical engineer, but he must be
familiar with the approximate calculations of the air conditioning
loads .
Note: Calculations are made on cooling because it is generally largest
heating load.
For quick calculations, the air conditioning load can be considered as 1
ton / 10 m 2 but 1 ton / 20 m 2 for central air conditioning.
There are many variables when estimating air-conditioning loads that
are taken into account, such as wall insulators, trees around the
building and sit of the sun.
The elevator load can be
estimated at (15 - 25 KW)
depending on the height
of the building and for
each pump has 5 KW.
The AC loads were estimated
by the table
4. General service loads :Such as elevators - pumps - (water -
sewage) are estimated by mechanical engineering and then
added to the loads of lighting, sockets and air conditioning.
Example: Calculate the electrical load for a residential
building consisting of 14 apartments and the actual area
of each apartment is 130 m2
Solution
Lighting load for one apartment =130*15=4500 W
Air condition for one apartment =130*65=8450 W
Air condition for one apartment =130*50=6500 W
Total load of one apartment = 6500+4500+8450=19450 W
Total load of residential building = 14*19450=272300 KW
✓ Final Stage of Load Estimation
The project engineer must prepare the following
information before sitting with the engineer of the
distribution company to agree on the contractual power
capacity:
Total connected load.
Determining the power factor of the loads and designing the
power factor correction panel.
Estimation of load reduction coefficients (demand factor and
diversity factor ).
Determine the required reliability.
Determining the largest load and its impact on the network at
the moment of switching especially projects with large loads.
EX.: Suppose that the general loads of a building are
1128 KVA and the service loads 95 KVA. Calculate the
KVA of the transformer needed to feed these loads.
Solution
Total loads=1128+95=1223 KVA
The transformer capacity=1223/0.8=1528 KVA
So, the transformer capacity=2000 KVA
If we take account the demand factor with 0.7 and applied on the
general load only. Because the DF do not apply on the service loads
Total loads=1128*0.7+95=885 KVA
The transformer capacity=885/0.8=1106 KVA
So, the transformer capacity=1500 KVA
Classification and Types of Electrical Loads
The electrical loads can be classified into various categories
according to various factors as follows:
1- According to Load Natural Components
Resistive Electrical Loads.
Capacitive Electrical Loads.
Inductive Electrical Loads.
Combination Electrical Loads.
2- According to Load Properties
Linear Electrical Load.
None-Linear Electrical Load.
3- According to Load Function
Lighting Load.
General / Small Appliances Load.
Power Loads.
4- According to Load Consumer Category
Residential Electrical Loads (Dwelling Loads).
Commercial Electrical Loads.
Industrial Electrical Loads.
Municipal / Governmental Electrical Loads (Street Lighting, Power
Required For Water Supply and Drainage Purposes, Irrigation Loads
And Traction Loads).
5- According to Load Grouping
Individual Loads (Single Load).
Load Centers (Area Loads).
6- According to Load Planning
Existing Electrical Loads.
Future Electrical Loads (Electrical Loads Growth).
New Electrical Loads (Additional Electrical Loads).
7- According to Load Operation Time
Continuous Electrical Loads.
Non-Continuous Electrical Loads.
Duty, Intermittent Electrical Loads.
Duty, Periodic Electrical Loads.
Duty, Short-Time Electrical Loads.
Duty, Varying Electrical Loads.
8- According to Load Importance
Vital Electrical Loads (Life Safety Electrical Loads).
Essential Electrical Loads (Emergency Electrical Loads).
Non-Essential Electrical Loads (Normal Electrical Loads).
9- According to number of Electrical Loads phases
Single phase Electrical Loads.
Three phase Electrical Loads.
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