Electrical

Diversity Factor with Formulas and Examples

Diversity Factor

Diversity Factor with Formulas and Examples- We know that the entire 100% load capacity installed in a building is never used simultaneously or at a moment (i.e., the entire load installed in a residential building is not used at a time), rather some of the load remains ON whereas some other remains OFF. Even during the peak load hours, some of the outlet points are to be such that are not brought into use. There are such electrical points in the building as well, which are fixed keeping in view the future requirements, whereas some points are used only during special occasions (e.g., marriage, function, or birthday, etc.)

For example, if three light points, two fan points, two 5-ampere socket outlets, and a 15–ampere socket outlet are installed in a drawing room, rarely such an occasion may come when all these electrical points are brought into use at a time. Suppose, that even if all these points of the drawing room are used simultaneously on a special occasion, even then it is possible that several points in bedrooms, store, kitchen, and veranda, etc., may not have been connected. Thus, the total load installed in a building is never used at a time. However, this point is practicable only for residential buildings. 




As the total load of any residential building is not used simultaneously, therefore at the time of selecting the most suitable size wiring cable or wire for such a building, it is kept in mind that its size should be according to the load being used under normal circumstances so that a low size cable can be used for sub-main circuit and the main – circuit. For the achievement of this objective, the diversity factor is being resorted to for assistance. The diversity factor is expressed as a percentage, and its definition is as below;

During normal conditions, the ratio of the total connected load full current to the full load rated current is known as the diversity factor. In other words, the ratio of total connected load and the actual maximum is known as the diversity factor, i.e., 

Diversity factor = Total connected load / Actual maximum load x 100

It ought to be remembered that in the event of a final subcircuit, a diversity factor is not normally being used, because at the time of selecting a cable for the final subcircuit, it is always kept into mind that the current passing capacity of a cable or wire to the corresponding appliance, should be equal to the maximum capacity of that particular appliance (i.e., the capacity of a wire supplying electricity to a 15 – ampere socket must necessarily be equal to 15 – amperes). 

Apart from knowing the size and rating of sub-main cable and main–cable of any residential building, diversity factor is always kept in mind while asking questions related to the rating of switch–gear (circuit breaker). The detail of the practicable percentage value of the diversity factor for different types of residential loads is as below;



Lighting and Fans

In the presence of lighting and fans in the residential installations, keeping in view the diversity factor, it is always assumed that only 66 % of the total load remains ON at a time. In the case of hotels and hostels, the diversity factor can increase up to 75%, whereas its value for commercial buildings increases to 90%. 

Cooking Appliances

The diversity factor value for a 10A rating main cable being used for cooking appliances in a residential building tends to be 100 % and for the value of all other loads above this load, the diversity factor value is adopted as 50%. 

When the main cable is desired to be selected for a large block of residential flats, the following different values of the diversity factor are kept in mind with respect to the cooking appliances. 

(a). Full load should be taken for the largest cooker

(b). Plus 50% for the Second largest cooker

©. Plus 33% for the Third largest

(d). Plus 25% for the Fourth largest

(e). Plus 20% for the remainder

However, this maximum diversity value is not acceptable for hotels, hostels, and other places of public interest. In such a situation, keeping a full load for the largest appliance, 80% load for the second-largest, and 60% load for the rest of the appliances is considered better, and safe.



Socket Outlets

In a situation of socket outlets, the permissible diversity value on a circuit wiring is not taken too high (approx. 66%), however, its value is taken on a vast basis while calculating the load of cables supplying electricity to the sub-distribution boards. For residential buildings, a full load rating (i.e., 100%) for the largest socket outlet, while 40% diversity for the reminders is taken. In a situation of socket outlets installed in hotels, hostels, and other public places, the value of the diversity factor of the largest point is taken as 100%, whereas the diversity of socket outlets points installed in the main rooms is considered 75% and 40% for the remainder of points. In the situation of a water heater, the diversity of each water heater is also taken as 100%. The different allowances of diversity have been shown in the data tables given at the end of this book.

Connected load Amp. Maximum To Be Expected Amp.
Lighting points 10 66% of load 7
Four 15 Amp. Socket outlets 60 100% of the largest (15ampere)

Plus 40% of remainder

33
Six 5 – Ampere socket outlets 30 100% of largest (5amp.)

Plus 40% of remainder

15
Water heater 12 100% 12
Cooker 9 100% 9
TOTAL Load 121 Amps. 76 Amps.

Keeping in view the future extension or emergency requirements, if only 10% extra load is also be included, (normally 20% additional load is taken into account for drawing out the size of cable), then in such a situation, total expected maximum current demand will turn out to be approx.76 + 7.6 = 84 amperes. Thus, the rating of sub – main cable for this particular installation will only be 84 instead of 121.



Examples:

Example 1; Find the rating of sub–main cable keeping allowance for diversity to supply a building of one flat of a large block flats in which the following load is connected.

(i). Lightpoint 10 with a total load of 10A.

(ii). Four 15A socket outlets

(iii). Six 5A socket outlets

(iv). Water heater with a load of 12A

(v). Cooker with a load of 9A

Solution;

From table allowance for diversity, 2nd column will be selected because one flat of a large block is given.

Connected Load Total load Maximum to be Expected Amp.
(i). Light points 10A 66% of load = 66 / 100 x 10 =6.6 7

 

(ii). Four 15A socket outlets 60A 100% of largest + 40% of remainder

15A + 40 / 100 x (15 x 3)

15 + 18

33

 

 

 

(iii). Six 5 Amp socket outlets

 

30A

 

100% of largest + 40% of remainder

5A + 40 / 100 x (5 x 5)

5 + 10

 

15

 

 

(iv). Water heater 12A 100% 12

 

(v). Cooker 9A 100% 9
Total 121A 76A

Adding say 10% for future addition, it makes a total expected maximum current demand of 76 + 7.6 = 84A approximately. Thus, the sub–main cable need only have a rating of 84A for this particular installation instead of 121A.

Note;

(1). Radial and ring final sub-circuit for socket outlets are used for industrial installation.

(2). Socket outlets and stationary appliances other than those listed above are used for residential buildings, flats, hotels, shops, offices, etc.




Example 2; Find the rating of sub–main cable keeping allowance for diversity to supply a boarding house in which the following load is connected. 

(1). 20 light points each taking 1A

(ii). Ten 5A socket outlets in rooms

(iii). Water heater with a load of 15A

(iv). Cooker with a load of 13A

(v). Two motors taking current 8A and 6A respectively.

Solution;

From table allowance for diversity, 4th column will be selected because boarding house is given;

Connected Load Total Load Maximum to be Expected Amp.
`(i). 20 light points 20A 75% of load = 75 / 100 x 20 15

 

(ii). Ten 5A socket outlets 50A 100% of largest + 40% of remainder

5A + 40 / 100 x (5 x 9)

5 + 18

 

23

(iii). Water heater 15A 100% 15

 

(iv). Cooker 13A 100% 13

 

(v). Two motors (8A & 6A) 14A 100% largest + 50% of 2nd

8A + 50/100 x 6

8 + 3

 

 

11

Total 112A 77A

Adding say 10% for future addition, = 10 / 100 x 77 = 7.7A

Total expected maximum current demand = 77 + 7.7 = 84.7A

Thus, the sub–main cable need only have a rating of 85A for this particular installation instead of 112 Amp.



Example 3; Find the rating of the sub–main cable keeping allowance of diversity to supply a building of one flat of a large block flats in which the following load is connected.

(i). Light points 12 with total load of 12A 

(ii). Five 15A socket outlets

(iii). Seven 5A socket outlets

(iv). Electric heater with a load of 10A

(v). Electric cooker with a load of 9A

Solution;

From the table allowance for diversity, 2nd column will be selected because one flat of a large block is given.

Connected Load Total Load Maximum to be Expected Amp.
(i). Light points 12A 66% of load = 66 / 100 x 12 = 7.92 ~ 8

 

(ii). Five 1 5A socket outlets 75A 100% of largest + 40 % of remainder

15 A + 40 / 100 x (15 x 4)

15 + 24

 

 

 

39

 

(iii). Seven 5 Amp. Socket outlets 35A 100% of largest + 40 % of remainder

5A + 40 / 100 x (5 x 6)

5 + 12

 

 

17

 

 

(iv). Electric heater 10A 100% 10

 

(v). Electric cooker 9A 100% 9
Total 141A 81A

Adding say 10% for future addition, it makes a total expected maximum current demand of 81 +8.1= 89.1 A approximately. Thus, the sub–main cable need only have a rating of 89.1 A for this particular installation instead of 141 A.



Next Topic: Class of Excess Current Protection

Previous Topic: Depreciation Factor with Formulas and Examples

For electronics and programming-related projects visit my YouTube channel.

My YouTube channel Link

Related Articles

Leave a Reply

Your email address will not be published. Required fields are marked *

Back to top button