This document provides an example calculation of the electrical load of a building with the following specifications: it has 5 floors with 3 flats on each floor, each flat has a 240V single phase supply and 20 wiring points, there is one 15HP lift motor and two 5HP machines, the building receives a 415V 3-phase supply. The summary calculates: 1) The total electrical load is 52.137KW 2) The main switch size is 100A 3) The main cable is 25mm sq copper cable 4) Each flat has 5 sub-circuits

1. Calculation of Electrical load
of Building

2. Example
A newly constructed building having 5 floors, each floor is
having 3 flats; Each flat supplied with 240 volt; single phase-
supply through a main switch of 16 amp/250V rating .and each
flat is to be carried out 20 points wiring. This building has one
lift; operated by 15 H.P., 3 phase motor; and two 3 phase
electric motors of 5 H.P. capacity. This building is supplied with
3 phase, 4 wire, 415 volt supply (Assume additional data)
calculate following. (1) Total electrical load on installation. (2)
Size of main switch. (3) Size and type of main cable. (4) No. of
sub circuits of individual flats.

3. First We assume load of one flat.
In the example given that 20 points,
So we take
4 points of fans, 8 points of lights and 7 points of 5Amp
sockets and 1 point of 15Amp socket.
Wattage of fan point is 60watts/ 80watts
Wattage of light point is 40watts
Wattage of 5amp point socket is 100watts
Wattage of 15amp point socket is 10000watt
Wattage of lamp point is 60watts

4. So total wattage(load) of one flat = no. of lamp point × 60
watt + no. of fan points × 60 watts + no. of 5amp socket points
× 100 watts + no. of light point × 40 watts + no. 15amp
socket(power point) × 1000 watts [Note: it is formula]
Now for our example
Total load of one flat = no. of fan points × 60 watts + no. of
5amp socket points × 100 watts + no. of light point × 40 watts
+ no. 15amp socket(power point) × 1000 watts
Total load of one flat = 4 × 60+7 × 100+8 × 40 + 1 × 1000 = 2260
watts

5. Number of sub circuits of one flat
As per IE rules,
one light and fan sub circuit have 8 to 10 points or 800 watts (
in this light points , fan points and 5 amp sockets point)
one power sub circuit have 2 points of 3000 watts (in this 15
amp socket points)
But in general we take no. of points for calculate number of
sub circuits.
so total no. of light sub circuit = (fan points + light point +
lamp points + 5amp socket points) / 8

6. Our example
No. of light sub circuit == (fan points + light point + 5amp
socket points) / 8 = (4 + 8 + 7)/8 = 2.375
So no. of light sub circuit =3
No of power sub circuit = ( no. power points)/2
= ½ = 0.5 so no/ of power sub circuit = 1
Total no. of sub circuits = 3+1+1(extra for future)
=5

7. Load current of electrical motor =
𝐻𝑃 ×735.5
√3×𝑉×cos ɸ×ɳ
Load of electrical motor = HP × 735.5 watts
Our example
Load of lift = 15 hp × 735.5 = 11032 watts
Load of other 2 motors = 2 × 5 h.p × 735.5 = 7355 watt
So total motor load = 11032+7355 = 18387 watts = 18.387 KW

8. Total electrical load of all flats = No. of floors × no. of flat on
each floor × electrical load of one flat
Our example
Total electrical load of all flats = 5 × 3 × 2250
= 33750 watts = 33.750 KW
So,
Total Electrical Load of building = electrical load of all flats +
electrical load of lifts + electrical load of other motors and
machines

9. Total electrical load of building = 33750 + 11032 +7355 =
52137 watts = 52.137 KW
Total Load current of building
=
𝑡𝑜𝑡𝑎𝑙 𝑙𝑎𝑜𝑑 𝑜𝑓 𝑏𝑢𝑖𝑙𝑑𝑖𝑛𝑔
√3×𝑉𝑜𝑙𝑡𝑎𝑔𝑒
So,
Load current of building =
52137
3 ×415
= 72.5 Amp

10. Now selection of cable and main switch

11. From given data sheet
Our load current is 73 amp, so we select 25 sq. mm, 4 core
copper conductor PVC insulated cable with current capacity of
85 amp for main supply.
Our load current is 73 amp, so we select 100 amp. 4 pole 600
volt grade MCCB for main switch.

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