March 6, 2007
Ok, I changed the title for part 2 of this series. The title “planning” was too general and would have made this post too long.
Now that you have completed your research, it’s time to get started on your house. But, what size electrical service do you need? Well, there is a calculation to determine this. Open up your National Electrical Code book to article 220. Here you will find everything you need to calculate the minimum sized electrical service for your new house.
Let’s get started with the general lighting and general use receptacle load. To determine this you need to know the square footage of your new house measured from the outside dimensions of the house. This measurement does not include open porches, garages, or an unfinished basement not adaptable for future use. However, if you do have such a space that is suitable for conversion to family rooms, bedrooms, home offices, or other habitable areas, these spaces need to be included in the general lighting load calculation.
Let’s assume your house is 2,000 square feet with a gas furnace and water heater. So, according to NEC table 220.12, you calculate your general lighting and general use receptacle load at 3 volt-amperes (VA) per square foot. This calculation equals (3VA x 2,000 square feet) 6,000 VA. So what does this mean? Well, 6,000 VA ÷ 120 volts = 50 amps. This means you need a minimum of 3 – 20 amp circuits or 4 – 15 amp circuits or a combination of the 2 totaling 50 amps or more for general lighting and general use receptacles.
Now, according to NEC section 210.11(C)(1) you are required 2 small appliance circuits. Each small appliance circuit shall be calculated at 1,500 VA [NEC section 220.52(A)]. You are also required 1 laundry circuit [NEC section 210.11(C)(2)] calculated at 1,500 VA [NEC section 220.52(B)].
So far we have:
General lighting and receptacles = 6,000 VA
Small appliance circuits = 3,000 VA
Laundry circuit = 1,500 VA
Total = 10,500 VA
Now we need to refer to NEC table 220.42 for lighting load demand factors. These demand factors shall apply to the portion of the total branch circuit load calculated for general illumination. They shall not be applied in determining the number of branch circuits for general illumination. That said, the demand factor percentage for the first 3,000 VA or less is 100%. The demand factor percentage from 3,001 VA – 120,000 VA is 35%. Now we can use these numbers to determine our net load.
Total load from above = 10,500 VA
3,000 VA at 100% = 3,000 VA
(10,500 VA – 3,000 VA) 7,500 VA at 35% = 2,625 VA
Net load = 4,875 VA
Now we need to add in the range and dryer loads to determine the net calculated load.
Net load from above = 4,875 VA
Range = 8,000 VA
Dryer = 5,500 VA
Net load = 18,375 VA
18,375 VA ÷ 240 volts = 77 amps
So, according to the calculation above you need a 77 amp service minimum. However, NEC sections 230.42(B) and 230.79 require service conductors and disconnecting means rated not less than 100 amperes. So, you need a 100 amp service minimum. However, I recommend installing a 200 amp service. This will allow for future growth, like a hot tub, air conditioner or whatever you desire, at a later date without needing to upgrade your electrical service.
If you need further clarification or have questions please submit them in the comment section of this post.
Tune in for part 3 tomorrow.
January 1, 2006
Q: I have a barn on my property that I would like to get power to. This barn is 377 feet from the nearest power. I would like to have 50 amps @240 volts and I’m going to run my power in conduit. What size wire do I need to run?
A: The formula for voltage drop is: Vd = 2K x L x I / Cm
Vd = Voltage Drop
I = Current in Conductor (Amperes)
L = One way Length of Circuit
Cm = Cross Section Area of Conductor (Circular Mils)
K = Resistance in ohms of one circular mil foot of conductor
K = 12.9 for Copper Conductors @ 75 degrees C
K = 21.2 for Aluminum Conductors @ 75 degree C
/ = Divided by
I will assume you are going to use copper conductors and your temperature is @ 75 degrees C.
Reasonable operating efficiency is achieved if the voltage drop of a feeder or a branch circuit is limited to 3 percent. However, the total voltage drop of a branch circuit plus a feeder can reach 5% and still achieve reasonable operating efficieny (210.19(A)(1)FPN No. 4 or 215.2(A)(4)FPN No. 2).
8 AWG = 50 amps @ 75 degrees C = 16510 Cm
Vd = 2 x 12.9 x 377 x 50 / 16510 = 30 volts
30 volts / 240 volts = 0.125 = 12.5% = Not Acceptable
6 AWG = 65 amps @ 75 degrees C = 26240 Cm
Vd = 2 x 12.9 x 377 x 50 / 26240 = 19 volts
19 volts / 240 volts = 0.079 = 7.9% = Not Acceptable
4 AWG = 85 amps @ 75 degrees C = 41740 Cm
Vd = 2 x 12.9 x 377 x 50 / 41740 = 12 volts
12 volts / 240 volts = 0.05 = 5% = Not Acceptable – This is not acceptable because the 5% voltage drop is at your sub panel. If you were to run any wire beyond the sub panel, your voltage drop would exceed 5%. I’m assuming you are going to install a light and some receptacles in your barn.
3 AWG = 100 amps @ 75 degrees C = 52620 Cm
Vd = 2 x 12.9 x 377 x 50 / 52620 = 9 volts
9 volts / 240 volts = 0.038 = 3.8% = Acceptable
Your equipment grounding conductor (ground wire) is sized off of table 250.122. You need to run a 10 AWG copper ground wire for this circuit.
To summarize; you need to run 3 – 3 AWG (2-hots and 1-neutral) branch circuit conductors and 1 – 10 AWG equipment grounding conductor (ground wire).
Tip: The rule of thumb is to plan for voltage drop at 100 feet and increase one wire size for every 100 feet thereafter.
Happy New Year!