Demand vs Div2

8/13/2019 Demand vs Div2

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Receptacle Outlet Demand Factor = Demand Interval Factor x Diversity Factor

= (15 minute run time / 15 minutes) x 0.1 = 0.1

Receptacle Outlet Demand Load = 15 x 1500 watts x 0.1 = 2.25 kW

Lathe Demand Factor = Demand Interval Factor x Diversity Factor.

= (5 minute run time / 15 minutes) x 1.0 =0 .33

Lathe Demand Load = 10 hp x .746 x .33 = 2.46 kW

Air Compressor Demand Factor = Demand Interval Factor x Diversity Factor.

= (7.5 minute run time / 15 minutes) x 1.0 = 0.5

Air Compressor Demand Load = 20 hp x .746 x .5 = 7.46 kW

Fire Pump Demand Factor = Demand Interval Factor x Diversity Factor.

= (15 minute run time/ 15 minutes) x 0.0 = 0.0

Fire Pump Demand Load = 15 hp x .746 x 0.0 = 0.0 kW

Summary of Demand Loads :

Equipment kW D.F. Demand KW

Lighting 5 1 5

Receptacle Outlets 22.5 .1 2.25

Lathe 7.5 .33 2.46

Air Compressor 15 0.5 7.46

Fire Pump 11.25 0.0 0.0

TOTAL 61.25 Kw 17.17 Kw

(2) Diversity factor / simultaneity factor (Ks)

Diversity Factor = Sum of Individual Max. Demand. / Max. Demand on Power Station.

Diversity Factor = Installed load. / Running load.

Diversity factor is usually more than one. Since the sum of individual max. demands >Max. Demand)The load is time dependent as well as being dependent upon equipment characteri stics. The diversity

factor recognizes that the whole load does not equal the sum of its parts due to this time

Interdependence (i.e. diverseness).When the maximum demand of a supply is being assessed it is not sufficient to simply add together

the ratings of all electrical equipment that could be connected to that supply. If this is done, a figure

somewhat higher than the true maximum demand will be produced. This is because it is unlikely that

all the electrical equipment on a supply will be used simultaneously.

The concept of being able to De-rate a potential maximum load to an actual maximum demand is

known as the application of a diversity factor.

70% diversity means that the device in question operates at its nominal or maximum load level 70%

of the time that it is connected and turned on.

nd Factor-Diversity Factor-Utilization Factor-Load Factor | EEP http://electrical-engineering-portal.com/demand-factor-diversity-fa

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If total installed full load ampere is twice your running load ampere then the diversity factor is two.

If total installed full load ampere is four times your load a ampere then the diversity factor is four.

If everything (all electrical equipment) was running at full load at the same time the diversity factor

is equal to One

Greater the diversity factor, lesser is the cost of generation of power.

Diversity factor in a distribution network is the ratio of the sum of the peak demands of the

individual customers to the peak demand of the network.

This wil l be determined by the type of service, i.e., residential, commercial, indust rial and

combinations of such.

Example-I:A distribution feeder serves 5 houses, each of which has a peak demand of 5 KW. The

feeder peak turns out to be 20 kw. The diversity is then 20/25 or 0.8. This results from the timing

differences between the individual heating/cooling, appliance usages in the individual customers.

As supply availabili ty decreases, the diversity factor wil l tend to increase toward 1.00. This can be

demonstrated when restoring service after outages (called cold starts) as the system initial surge

can be much greater than the historical peak loads.

Example-II:A sub-station has three outgoing feeders:

feeder 1 has maximum demand 10 MW at 10:00 am,1.

feeder 2 has maximum demand 12 MW at 7:00 pm and2.

feeder 3 has maximum demand 15 MW at 9:00 pm,3.

While the maximum demand of all three feeders is 33 MW at 8:00 pm.4.

Here, the sum of the maximum demand of the individual sub-systems (feeders) is 10 + 12 + 15 =

37 MW, while the system maximum demand is 33 MW. The diversity factor is 37/33 = 1.12. The

diversity factor is usually greater than 1; its value also can be 1 which indicates the maximum demandof the individual sub-system occurs simultaneously.Diversity is the relationship between the rated full loads of the equipment downstream of a

connection point, and the rated load of the connection point. To illustrate:

The building at these co-ordinates is fitted with a 100A main supply fuse.1.

The distribution board has 2no. 6A breakers, 1no. 20A breaker and 5no. 32A breakers, a total,

potentially, of 192A.

2.

Not all these rated loads are turned on at once. If they were, then the 100A supply fuse would

rupture, as it cannot pass 192A. So the diversity factor of the distribution board can be said to be

192A/100A, or 1.92, or 52%.Many designers prefer to use unity as the diversity factor in calculations for planning conservatism

because of plant load growth uncertainties. Local experience can justify using a diversity factor larger

than unity, and smaller service entrance conductors and transformer requirements chosen accordingly.

The diversity factor for all other instal lations will be different, and would be base d upon a local

evaluation of the loads to be applied at different moments in time. Assuming it to be 1.0 may, on

some occasions, result in a supply feeder and equipment rating that is rather larger than the local

installation warrants, and an over-investment in cable and equipment to handle the rated load current.

It is better to evaluate the pattern of usage of the loads and calculate an acceptable diversity factor


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for each particular case.

In the case of the example given above, achieving a diversity of 1.0 or 100% would require well over

twice the cross-sectional area of copper cable to be installed in a deep trench underneath a field, the

rebuild of a feeder cabinet to larger dimensions, more substantial overhead supply cables for a

distance exceeding 2km northwards and a different tariff, where one pays rather more for a kWh

than at present. The investment required to achieve 1.0 simply isnt justifiable in this particular case.

Diversity factor is mostly used for distribution feeder size and transformer as well as to determine

the maximum peak load and diversity factor is always based on knowing the process. You have to

understand what will be on or off at a given time for different buildings and this will size the feeder.

Note for typical buildings diversity factor is always one. You have to estimate or have a data records

to create 24 hours load graph and you can determine the maximum demand load for node then you

can easily determine the feeder and transformer size.

The diversity factor of a feeder would be the sum of the maximum demands of the individual

consumers divided by the maximum demand of the feeder. In the same manner, it is possible to

compute the diversity factor on a substation, a transmission line or a whole utility system.

The residential load has the highest diversity factor. Industrial loads have low diversity factors usually

of 1.4, street light practically unity and other loads vary between these limits.

Diversity Factor in distribution Network

Elements of System

Diversity Factors

Residentia l Commercia lGeneral

Power

Large

Industrial

Between individual users 2.00 1.46 1.45

Between transformers 1.30 1.30 1.35 1.05

Between feeders 1.15 1.15 1.15 1.05

Between substations 1.10 1.10 1.10 1.10

From users to transformers 2.00 1.46 1.44

From users to feeder 2.60 1.90 1.95 1.15

From users to substation 3.00 2.18 2.24 1.32

From users to generating station 3.29 2.40 2.46 1.45

Diversity Factor for distribution switchboards


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Number of circuits Diversity Factor (ks)

Assemblies entirely tested 2 and 3 0.9

4 and 5 0.8

6 to 9 0.7

10 and more 0.6

Assemblies partially tested in every case choose 1

Diversity Factor for according to circuit function (IEC 60439)

Circuits Function Diversity Factor (ks)

Lighting 0.9

Heating and air conditioning 0.8

Socket-outlets 0.7

Lifts and catering hoist

For the most powerful motor 1

For the second most powerful motor 0.75

For all motors 0.8

Diversity Factor for an apartment block

Apartment Diversity Factor (ks)

2 To 4 1

5To 19 0.78

10To 14 0.63

15To 19 0.53

20To 24 0.4925To 29 0.46

30 To 34 0.44

35 To 39 0.42

40To 40 0.41

50 To Above 0.40


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the sum of the demand loads is 6 kW but the maximum load presented by the system at any

time is only 1.5 kW.

Diversity factor =Sum of Individual Max. Demand / Max. Demand = 6 Kw / 1.5 Kw =4.

Demand Factor = Maximum demand / Total connected load = 1.5 Kw / 12 Kw = 0.125.

(3) Load factor

Load Factor = Average load. /Maximum load during a given period.

It can be calculated for a single day, for a month or for a year.

Its value is always less than one. Because maximum demand is always more than avg. demand.

It is used for determining the overall cost per unit generated. Higher the load factor, lesser will be thecost per unit.Load Factor = Load that a piece of equipment actually draws / Load it could draw (full load).

Example: Motor of 20 hp drives a constant 15 hp load whenever it is on.

The motor load factor is then 15/20 = 75%.

Load factor is term that does not appear on your utility bill, but does affect electricity costs. Load

factor indicates how efficiently the customer is using peak demand.

Load Factor = ( energy (kWh per month) ) / ( peak demand (kW) x hours/month )

A high load factor means power usage is relatively constant. Low load factor shows that occasionally

a high demand is set. To service that peak, capacity is sitting idle for long periods, thereby imposing

higher costs on the system. Electrical rates are designed so that customers with high load factor are

charged less overall per kWh.

For Example

Customer A High Load Factor

82% load factor = (3000 kWh per month x 100%) / 5 kW x 730 hours/month.

Customer B Low Load Factor

41% load factor = (3000 kWh per month x 100%) / 10kW x 730 hours/month.

To encourage the efficient use of installed capacity, electricity rates are structured so the price per

kWh above a certain load factor is lower. The actual structure of the price blocks varies by rate.

(4) Utilization factor (Ku)

In normal operating conditions the power consumption of a load is sometimes less than that

indicated as its nominal power rating, a fairly common occurrence that justifies the application of an

utilization factor (ku) in the estimation of realistic values.

Utilization Factor = The time that a equipment is in use./ The total time that it could be in

use.

Example:The motor may only be used for eight hours a day, 50 weeks a year. The hours of

operation would then be 2000 hours, and the motor Utilization factor for a base of 8760 hours per

year would be 2000/8760 = 22.83%. With a base of 2000 hours per year, the motor Utilization factor

would be 100%. The bottom line is that the use factor is applied to get the correct number of hours


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that the motor is in use.

This factor must be applied to each individual load, with particular attention to ele ctric motors,

which are very rarely operated at full load. In an industria l installation this factor may be estimated

on an average at 0.75 for motors.

For incandescent-lighting loads, the factor always equals 1.

For socket-outlet circuits, the factors depend entirely on the type of appliances being supplied from

the sockets concerned.

Maximum demand

Maximum demand (often referred to as MD) is the largest current normally carried by circuits,

switches and protective devices. It does not include the levels of current flowing under overload or

short circuit conditions.

Assessment of maximum demand is sometimes straightforward. For example, the maximum demand

of a 240 V single-phase 8 kW shower heater can be calculated by dividing the power (8 kW) by the

voltage (240 V) to give a current of 33.3 A. This calculation assumes a power factor of unity, which

is a reasonable assumption for such a purely resistive load.

There are times, however, when assessment of maximum demand is less obvious. For exam ple, if a

ring circuit feeds fifteen 13 A sockets, the maximum demand clearly should not be 15 x 13 = 195 A,

if only because the circuit protection will not be rated at more than 32 A. Some 13 A sockets may

feed table lamps with 60 W lamps fitted, whilst others may feed 3 kW washing machines; others

again may not be loaded at all.

Lighting circuits pose a special problem when determining MD. Each lamp-holder must be assumed

to carry the current required by the connected load, subject to a minimum loading of 100 W per

lamp holder (a demand of 0.42 A per lamp holder at 240 V). Discharge lamps are particularly

difficult to assess, and current cannot be calculated simply by dividing lamp power by supply voltage.

The reasons for this are:

Control gear losses result in additional current,1.

the power factor is usually less than unity so current is greater, and2.

Chokes and other control gear usually distort the waveform of the current so that it contains

harmonics which are additional to the fundamental supply current.

3.

So long as the power factor of a discharge lighting circuit is not less than 0.85, the current demandfor the circuit can be calculated from:

current (A) = (lamp power (W) x 1.8) / supply voltage (V)

For example, the steady state current demand of a 240 V circuit supplying ten 65 W fluorescent

lamps would be: I = 10X65X1.8A / 240 = 4.88A

Switches for circuits feeding discharge lamps must be rated at twice the current they are required to

carry, unless they have been specially constructed to withstand the severe arcing resulting from the

switching of such inductive and capacitive loads.


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(5) Coincidence factor

The coincidence factor =Max. demand of a system / sum of the individual maximum

demands

The coincidence factor is the reciprocal of the diversity factor

Demand Factor & Load Factor according to Type of Industries


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Type of Industry Demand

Factor

Load

Factor

Utilization

Factor (DF x

LF)

Arc Furnace 0.55 0.80 0.44

Induction Furnace 0.90 0.80 0.72

Steel Rolling mills 0.80 0.25 0.20

Mechanical/ Electrical

a) Single Shift 0.45 0.25 0.11

b) Double Shift 0.45 0.50 0.22

Cycle Industry 0.40 0.40 0.16

Wire products 0.35 0.40 0.14

Auto Parts 0.40 0.50 0.20

Forgings 0.50 0.35 0.17

Cold Storage

a) Working Season 0.60 0.65 0.39

b) Non-Working Season 0.25 0.15 0.04

Rice Shellers



Ice Candy Units



Ice Factories


b) Non-Working Season 0.80 0.10 0.08Cotton Ginning



Spinning Mills 0.60 0.80 0.48

Textile Industry 0.50 0.80 0.40


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Dyeing and Printing 0.40 0.50 0.20

Ghee Mills 0.50 0.50 0.25

Oil Mills 0.70 0.50 0.35

Solvent Extraction Mills 0.45 0.50 0.22

Plastic 0.60 0.25 0.11

Soap 0.50 0.25 0.12

Rubber (Foot Wear) 0.45 0.35 0.16

Distilleries 0.35 0.50 0.17

Chemical Industry 0.40 0.50 0.20

Gas Plant Industry 0.70 0.50 0.35

Pain and Colour Factory 0.50 0.40 0.20

Sugar 0.30 0.45 0.13

Paper 0.50 0.80 0.40

Flour Mills(Single Shift) 0.80 0.25 0.20

Atta Chakies 0.50 0.25 0.12

Milk Plants 0.40 0.80 0.32

Printing Presses 0.35 0.30 0.10

Repair Workshops 0.40 0.25 0.10Bottling Plants 0.40 0.35 0.14

Radio Stations 0.55 .0.45 0.25

Telephone exchange 0.50 0.90 0.45

Public Water Works 0.75 0.40 0.30

Medical Colleges 0.60 0.25 0.15

Hospitals 0.25 0.90 0.22

Nursing Homes 0.50 0.50 0.25

Colleges and Schools 0.50 0.20 0.10

Hotels and Restaurants 0.75 0.40 0.30

Marriage Palaces 1.00 0.25 0.25

Demand Factor & Load Factor according to Type of Buildings:


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Laundry/dry cleaning plant 30-35 20-25

K-6 schools 75-80 10-15

7-12 schools 65-70 12-17

Churches 65-70 5-25

Post Office 75-80 20-25

Retail store 65-70 25-32

Bank 75-80 20-25

Supermarket 55-60 25-30

Restaurant 45-75 15-25

Auto repair shop 40-60 15-20

Hobby shop, art/crafts 30-40 25-30

Bowling alley 70-75 10-15

Gymnasium 70-75 20-45

Skating rink 70-75 10-15

Indoor swimming pool 55-60 25-50

Theater 45-55 8-13

Library 75-80 30-35

Golf clubhouse 75-80 15-20Museum 75-80 30-35

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jiguparmar - Jignesh Parmar has completed his B.E(Electrical) from Gujarat University. He is member of

Institution of Engineers (MIE),India. Membership No:M-1473586.He has more than 12 years experience in

Transmission -Distribution-Electrical Energy theft detection-Electrical Maintenance-Electrical Projects

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