*Originally published at Electrical Notes – Articles*

## (1) Demand factor

**Demand Factor = Maximum demand of a system / Total connected load on the system***Demand factor is always less than one.*- Example: if a residence having 6000W equipment connected has a maximum demand of 300W,Than demand factor = 6000W / 3300W = 55%.
*The lower the demand factor, the less system capacity required to serve the connected load.***Feeder-circuit**conductors should have an ampere sufficient to carry the load; the ampere of the feeder-circuit need not always be equal to the total of all loads on all branch-circuits connected to it.

Remember that the demand factor permits a feeder-circuit ampere to be less than 100% of the sum of all branch-circuit loads connected to the feeder.

**Example:**One Machine Shop has

**Fluorescent fixtures**=1 No, 5kw each, Receptacle outlets =1 No, 1500w each.

Lathe=1No, 10 Hp, Air Compressor=1 No, 20 Hp, Fire Pump=1 No, 15 Hp.

- After questioning the customer about the various loads, the information is further deciphered as follows:
- The shop lights are on only during the hours of 8 a.m. to 5 p.m.
- The receptacle outlets are in the office only, and will have computers and other small loads plugged into them.
- The lathe is fully loaded for 5 minutes periods. The rest of the time is setup time. This procedure repeats every 15 minutes.
- The air compressor supplies air to air tools and cycles off and on about half the time.
- The fire pump only runs for 30 minutes when tested which is once a month after hours.

## Calculation:

**Lighting**Demand Factor = Demand Interval Factor x Diversity Factor.- = (15 minute run time/ 15 minutes) x 1.0 = 1.0
**Lighting Demand Load = 5 kW x 1.0 = 5 kW**- 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 characteristics. 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.
- 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 will be determined by the type of service, i.e., residential, commercial, industrial 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 availability decreases, the diversity factor will 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,
- feeder 2 has maximum demand 12 MW at 7:00 pm and
- feeder 3 has maximum demand 15 MW at 9:00 pm,
- While the maximum demand of all three feeders is 33 MW at 8:00 pm.

- 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 demand of 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.
- The distribution board has 2no. 6A breakers, 1no. 20A breaker and 5no. 32A breakers, a total, potentially, of 192A.

- 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 installations will be different, and would be based 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 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 isn’t 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 |
| |||

Residential | Commercial | GeneralPower | LargeIndustrial | |

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

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 | 0.49 |

25To 29 | 0.46 |

30 To 34 | 0.44 |

35 To 39 | 0.42 |

40To 40 | 0.41 |

50 To Above | 0.40 |

**Example**: 5 storey apartment building with 25 consumers, each having 6 kVA of installed load.

The total installed load for the building is: 36 + 24 + 30 + 36 + 24 = 150 kVA

The apparent-power supply required for the building is: 150 x 0.46 = 69 kVA- It is a matter of common experience that the simultaneous operation of all installed loads of a given installation never occurs in practice, i.e. there is always some degree of diversity and this fact is taken into account for estimating purposes by the use of a simultaneity factor / Diversity Factor (ks).
- The Diversity factor ks is applied to each group of loads (e.g. being supplied from a distribution or sub-distribution board). The determination of these factors is the responsibility of the designer, since it requires a detailed knowledge of the installation and the conditions in which the individual circuits are to be exploited. For this reason, it is not possible to give precise values for general application.

#### Designing Size of Electrical Switchgear by use of Demand Factor and Diversity Factor:

- Diversity factors are used by utilities for distribution transformer sizing and load predictions.
- Demand factors are more conservative and are used by NEC for service and feeder sizing.
- Demand factors and diversity factors are used in design.
- For example, the sum of the connected loads supplied by a feeder is multiplied by the demand factor to determine the load for which the feeder must be sized. This load is termed the maximum demand of the feeder. The sum of the maximum demand loads for a number of sub feeders divided by the diversity factor for the sub feeders will give the maximum demand load to be supplied by the feeder from which the sub feeders are derived.
**Example-1:**Suppose We have four individual feeder-circuits with connected loads of 250 kVA, 200 kVA, 150 kVA and 400 kVA and demand factors of 90%, 80%, 75% and 85% respectively.Use a diversity factor of 1.5.- Calculating demand for feeder-circuits
- 250 kVA x 90% = 225 kVA
- 200 kVA x 80% = 160 kVA
- 150 kVA x 75% = 112.5 kVA
- 400 kVA x 85% = 340 kVA
- 837.5 kVA
- The sum of the individual demands is equal to
**837.5 kVA.** - If the main feeder-circuit were sized at unity diversity: kVA = 837.5 kVA ÷ 1.00 = 837.5 kVA.
- The main feeder-circuit would have to be supplied by an
**850 kVA**transformer. - However, using the diversity factor of 1.5, the kVA = 837.5 kVA ÷ 1.5 = 558 kVA for the main feeder.
- For diversity factor of 1.5, a
**600 kVA**transformer could be used. **Example-2:**A conveyor belt made up of six sections, each driven by a 2 kW motor. As material is transported along this belt, it is first carried by section 1, and then each section in succession until the final section is reached. In this simple example only one section of conveyor is carrying material at any point in time. Therefore five motors are only handling no-load mechanical losses (say .1 kW) keeping the belts moving whilst one motor is handling the load (say 1 kW). The demand presented by each motor when it is carrying its load is 1 kW, 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 the cost 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 that the motor is in use.

- This factor must be applied to each individual load, with particular attention to electric motors, which are very rarely operated at full load. In an industrial 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 example, 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,
- the power factor is usually less than unity so current is greater, and
- 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.

- So long as the power factor of a discharge lighting circuit is not less than 0.85, the current demand for 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.

## (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

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 Sheller’s | |||

a) Working Season | 0.70 | 0.80 | 0.56 |

b) Non-Working Season | 0.05 | 0.30 | 0.01 |

Ice Candy Units | |||

a) Working Season | 0.50 | 0.65 | 0.32 |

b) Non-Working Season | 0.50 | 0.10 | 0.05 |

Ice Factories | |||

a) Working Season | 0.80 | 0.65 | 0.52 |

b) Non-Working Season | 0.80 | 0.10 | 0.08 |

Cotton Ginning | |||

a) Working Season | 0.70 | 0.25 | 0.17 |

b) Non-Working Season | 0.10 | 0.10 | 0.01 |

Spinning Mills | 0.60 | 0.80 | 0.48 |

Textile Industry | 0.50 | 0.80 | 0.40 |

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 | 0.10 |

Bottling 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:**

Individual Facilities | Demand Factor | Load Factor |

Communications – buildings | 60-65 | 70-75 |

Telephone exchange building | 55-70 | 20-25 |

Air passenger terminal building | 65-80 | 28-32 |

Aircraft fire and rescue station | 25-35 | 13-17 |

Aircraft line operations building | 65-80 | 24-28 |

Academic instruction building | 40-60 | 22-26 |

Applied instruction building | 35-65 | 24-28 |

Chemistry and Toxicology Laboratory | 70-80 | 22-28 |

Materials Laboratory | 30-35 | 27-32 |

Physics Laboratory | 70-80 | 22-28 |

Electrical and electronics systems laboratory | 20-30 | 3-7 |

Cold storage warehouse | 70-75 | 20-25 |

General warehouse | 75-80 | 23-28 |

Controlled humidity warehouse | 60-65 | 33-38 |

Hazardous/flammable storehouse | 75-80 | 20-25 |

Disposal, salvage, scrap building | 35-40 | 25-20 |

Hospital | 38-42 | 45-50 |

Laboratory | 32-37 | 20-25 |

Dental Clinic | 35-40 | 18-23 |

Medical Clinic | 45-50 | 20-23 |

Administrative Office | 50-65 | 20-35 |

Single-family residential housing | 60-70 | 10-15 |

Detached garages | 40-50 | 2-4 |

Apartments | 35-40 | 38-42 |

Fire station | 25-35 | 13-17 |

Police station | 48-53 | 20-25 |

Bakery | 30-35 | 45-60 |

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 | 15-20 |

Museum | 75-80 | 30-35 |

Great explanation sir! Thanks!

very clearly explained.

This is a great job. Your contributions to engineering profession have been immense. They shall speak after you.

Muchas gracias por compartir esta información; mucho trabajo y horas de estudio. Saludos desde México.

thanks

I am very much happy to your highly appreciateable work.

Please let me know, what will be the diversity factors needs to be considered for mechanical workshop equipment electrical load for sizing transformer? (This workshop basically for oil and gas industry, and most of the equipment will be very rarely used like once in month)

pls share your no i want to ask something.how to find 15mints peak load and 1mints peak load for a steel melt shop having a 100mva and a 60mva furnace

Very good inputs on critical design factors to be considered for load estimations.

I need to re-estimate a typical energy requirement for a steel rolling mill, based on connectd load and assumd average loading of motors. Could you suggest the international standadrs that I must quote and refer in my report ofcalculations?

I think there is a wrong in 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.

is the answer should be 25/20 = 1.25 . I wish you reply

Dear Sir,

I want know that motor solo run shall be 4hours even small motor?

I try to find in ASME, but cannot..

please advice to shorten 2hours for solo run test..

Please check following in above article

Demand factor

Example: if a residence having 6000W equipment connected has a maximum demand of 300W,Than demand factor = 6000W / 3300W = 55%.

I Think, the below is correct

Example: if a residence having 6000W equipment connected has a maximum demand of 3300W,Than demand factor = 3300W / 6000W = 55%

If Not so please clarify??..

Your calculation of DF is the right one.

Sir, I have a question related to distribution please give me a best answer…

Show how to share the supply of the available electric power of 150KVA between 2 electrical resistive load each of 15 KVA rating using individual switches…

nice article

Dear author,

I want to know how to calculate daily loss of load factor when demand and time is given