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Electrical designing for construction

 • Example :


G+7 floor : Residential+ commercial complex

        ( Commercial for super market)

-Under ground parking

-Ground floor→super market→1500 sqft

-1 to 7 floor→Resident→1500 sqft



1 sqft for commercial 3.5 watt only lighting and In gulf, 4.5 watt only lighting.


For commercial,

        10000 sqft × 3.5 

        = 35000 w

        = 35 kw


HAVC → 120 sqft - 1 ton (1 ton- 1400w)

                 10000 - ?

               = 10000 × 1400/120

               = 14000000/120

               = 116666.6 w

               = 116666.6/1000 = 116.66 KW


Therefore, take 120 kw for AC load.


Total connected load for commercial - 

         35+ 120 = 155 kw

  Extra 25% = 155 × 1.25 = 193.75

Therefore, take 195 kw 


    Maximum demand = 0.8

    Peak demand = max. demand× T.C.l

                              = 0.8 × 195 

                              = 156 kw

Therefore, transformer selection for commercial 

              = KVA →kw÷P.F

                             156÷ 0.8 = 195kw

      Loading factor = 195/0.7 = 278.75 KVA


Therefore, 315 transformer is recommended for commercial


Main circuit breaker for L.T panel for super market

        A = KVA × 1.4

            = 315 × 1.4

            = 441 A

# 630 A MCCB is recommended


Electrical load calculation for residential complex.


Common load :

  Left = 7.5 HP × 2 = 11.19 kw

  Borewell →5HP = 3.73 kw

  Pump → 5HP = 3.73 kw

  Parking = 5kw

  Loby = 5kw


Flat : 

 3BHK - 7.5kw

 4BHK - 10 kw


Total flat in each floor - 6

6(flat) × 7 (floor) = 42 flats

= 42 × 7.5 = 315 kw


T.C.L = 315 + 11.19 + 3.73 + 3.73 + 5 + 5 

         = 343.65 kw

         

Demand factor = 0.6

   Peak demand = T.C.L × 0.6(D.F)

                             = 345 × 0.6

                             = 207 kw


KVA = KW ÷ P.F = 207 ÷ 0.8 = 258.75


Loading factor = 258.75/0.7 = 369.6


Therefore, 400 KVA transformer is recommended for Residential complex.


Circuit Breaker for Residential for L.T panel

               A = KVA × 1.4

                   = 400 × 1.4 = 560 A

Therefore, 630 A MCCB is recommend.


When load is connected in three-phase and distribution in 3 different phases. Then, we have to take C.B double the Amper.


Each flat per kw - 7.5 kw

    7.5 × 2 = 15A 

    15 × 2 = 30 A 


P.F = 0.8 @ 6A

P.F = 0.6 @ 8A


Example :


  A - block

  B - block 

  C - block


Light - 60 W

Fan - 100 W

AC - 1400 W

5A - 200 W

15A - 1000 W


A- block = 30 class room

          Each class room load :

            4 - light, 4 - fan, 1 - AC, 5A - 4 socket

, 15A - 1 socket.


B- block = 25 class room

         Each class room load same as A-block

Office : 20- light, 10- fan, 5A - 16, 15A - 6.


C- block = over all canteen -20kw


Conference hall : 40 - light, 20- fan, 5A - 10, 15A - 6.


Common load : washroom, outdoor lighting, water pump etc - 15kw


Calculate :

A- block




B- block




C- block



 

T.C.L = 20+ 12.4

          = 32.4 kw 

          = 32.4× 1.9 = 61.58

          = 61.58× 1.25 = 76.95


Common load : 15 kw

         

        15 kw × 1.9 = 28.5

          28.5 × 1.25 = 35.6


T.C.L = A- block+ B - block+ C - block+ common load

        = 115.2 + 114.4 + 32.4+ 15

        = 277 kw


Demand factor = 0.6

    Peak demand = 277 × 0.6 = 166.2 kw


KVA = KW/ P.F = 166.2/0.8 = 207.75


Loading factor = 207.75/0.7 = 296.78KVA


     315 transformer is recommended.


Figure : SLD






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