16 July 2011

Electricity: Calculation of Loads




PAUL HAY Capital Projects

  

Topic:            Calculating Electrical Load
Author:          Paul Hay
e-mail:             paul.hay@phcjam.com
profile:            www.linkedin.com/in/phcjam


Electricity - Calculating Service Amps

Example 8.1:             Calculate (A) the service current required for a 200 m2 residence, with appliance loads stated below; (B) main breaker size, and (C) size of service entrance conductor.

BRANCH CIRCUIT                                     REQUIREMENT/SPEC       POWER (W)
1.         Lighting and General Use Circuits:          Area of house x 30W =                     6,000

2.                  Small Appliances Circuits:
a.                   Kitchen                                               2 x 1,500W =                                         3,000
b.                  Laundry                                              1 x 1,500W =                                         1,500
c.                   Additional circuits                              0 x 1,500W =                                                0
                                                                                                                                                 ---------
Subtotal =                                                                                                          4,500

3.                  Dedicated Circuits:
a.                   Cooking Range                                   1 x 8,000W =                                         8,000
b.                  Water Heater                                       1 x 5,000W =                                         5,000
c.                   Microwave Oven                                1 x 1,200W =                                         1,200
d.                  Refrigerator                                         1 x   600W =                                             600
                                                                                                                                                ----------
Subtotal =                                                                                                        14,800

4.                  General Load:                                                 Sum of items 1 thro= 3                        25,300

5.                  Derated Load:             10,000 + [0.4 x (General Load - 10,000)] =                            16,120
6.                  Heating and Cooling:                                      A/C @ 10,000W                                  10,000
                                                                                                                                              -----------
7.                  Total Electrical Load:                                     Derated Load + Cooling =                  26,120

8.                  Service Current:          Total Electrical Load ) Service Voltage (220V) =                118.73A

9.                  Min. Allowable Conductor Current:               1.5 x Service Current =                       178.1A
10.              Breaker Amperage:                                         0.8 x Conductor Rating =                   142.5A


NB.     The nearest breaker available is 150A. Therefore, minimum copper service conductor should be #1 AWG.

  

FURTHER READING

            JLC Field Guide to Residential Construction: A Manual of Best Practices, Vol. 2, Manley Wood LLC, USA, 2006


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