Single Family Dwelling Calculation, Optional Method.
Published on July 18, 2016 by Dennis Bordeaux
If you are preparing to take an electrical licensing exam, you will need to know how to calculate the feeder or service load for a single family dwelling. Refer to section 220.82(A)&(B) in the 2014 National Electrical Code (NEC). Here is a sample problem. Use the optional calculation to find the load on the following single family dwelling:
2800 sq. ft.
14 kW range
3 kW water heater
5 kW clothes dryer
1.5 kW dishwasher
15 kW central heat
29 amp, 240 volt air conditioning
Step 1:
Multiply the sq. ft. area by 3 VA per sq. ft. 2800 sq. ft. x 3 VA =8,400 VA. VA = volt-amperes.
220.82(B)(1).
Step 2:
Add in 1500 VA for each 2-wire, 20-amp small-appliance branch circuit and the laundry circuit. 1500 VA x 3 = 4500 VA.
220.82(B)(2).
Step 3:
Add in the appliances at nameplate value.
220.82(B)(3).
Range 14,000 VA
Water heater 3,000 VA
Clothes dryer 5,000 VA
Dishwasher 1,500 VA
Step 4:
Add all appliance loads together.
220.82(B).
Total = 36,400 VA
Step 5:
Take the first 10kW at 100%. 10,000 VA
Take the remainder (26,400 VA) at 40%. 10,560 VA
220.82(B).
Step 6:
Add the two values from Step 5 together to find the general load: 10,000 VA + 10,560 VA = 20,560 VA.
220.82(B).
Step 7:
Compare the heating load to the AC load and take the larger of the two loads. 220.82(C).
AC load at 100%. 29 amps x 240 volts = 6,960 VA
Heat at 65%. 15,000 VA x .65 = 9,750 VA (largest load).
220.82(C)(4).
Step 8:
Add the general load to the largest of the AC or heating load.
General load = 20,560 VA
Heating load = 9,750 VA
Total = 30,310 VA
220.82(A)
Step 9:
Divide the load in VA by the voltage. 30,310 VA ÷ 240 volts = 126 amps.
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