Dwelling Unit Feeder/Service Conductor Calculations 
4/28/2003 
How do you size
conductors for residential services and feeders?
By Mike Holt,
Extracted from my Electrical Calculations textbook.
Most electrical
licensing exams (e.g., journeyman, master electrician, contractor) require you
to calculate residential loads and feeders. To do this, you might use the
standard method contained in Article 220, Part II. Usually, you can use the
optional method in Article 220, Part III. In either case, you are free to exceed
the NEC requirementsthese are minimum requirements, not design
specifications.
Though the standard method involves more steps, many
people use it exclusively to avoid using the wrong method. However, most
residential construction qualifies for the optional method. So, it's prudent to
understand both methods. The standard method is where we'll startit requires
six sets of calculations:
General lighting, receptacles, smallappliance,
and laundry Air conditioning versus heat Appliances Clothes dryer
Cooking equipment Conductor size General lighting and receptacles,
smallappliance, and laundry
The NEC recognizes these circuits will not
all be on (loaded) simultaneously. Thus, you may apply a demand factor to the
total connected load [220.16]. To determine the feeder demand load, refer to
Table 220.11 and follow these steps:
Total Connected Load. Determine the
total connected load for general lighting, receptacles (3 VA per sq ft), two
smallappliance circuits each at 1,500 VA, and one laundry circuit at 1,500 VA.
Demand Factor: Apply Table 220.11 demand factors to the total connected
load. First 3,000 VA at 100% demand. Remainder VA at 35% demand Example:
What is the general lighting, smallappliance and laundry demand load for a
2,700 sq ft dwelling unit? (See Figure 913).
Graphics are not included
in this Newsletter.
(a) 8,100 VA (b) 12,600 VA (c) 2,700 VA (d) 6,360
VA
Answer: (d) 6,360 VA General lighting/receptacles: 2,700 sq ft x 3
VA = 8,100 VA Smallappliance circuits: 1,500 VA x 2 = 3,000 VA Laundry
circuit: 1,500 VA x 1 = 1,500 VA Total Connected Load: 2,600 VA First
3,000 VA at 100% = 3,000 VA x 1.00 = 3,000 VA Remainder at 35% = 9,600 VA x
0.35 = 3,360 VA Total = 3,000 VA + 3,360 VA = 6,360 VA Airconditioning
versus heat
Because airconditioning and heating loads are not on
simultaneously, you may omit the smaller of the two loads [220.21]. Calculate
each of these at 100% [220.15].
Example: What is the service demand load
for 5hp, 230V A/C versus three 3 kW baseboard heaters? (see Figure
914).
(a) 6,400 VA (b) 3,000W (c) 8,050 VA (d) 9,000W
Answer: (d)
9,000W A/C = 230V x 28A = 6,440 VA. Omit this, per 220.21. Heat = 3,000W
x 3 = 9,000W Appliances
Per 220.17, you can use a 75% demand factor
when four or more "fastened in place" appliances (e.g., dishwasher, waste
disposal) are on the same feeder.
Example: What is the demand load for a
waste disposal (940 VA), dishwasher (1,250 VA), and water heater (4,500
VA)?
(a) 5,018 VA (b) 6,690 VA (c) 8,363 VA (d) 6,272 VA
Answer:
(b) 6,690 VA Waste disposal: 940 VA Dishwasher: 1,250 VA Water
heater: 4,500 VA Total: 6,690 VA Clothes dryer
Per 220.18, the
feeder or service demand load for electric clothes dryers in a dwelling unit
shall not be less than 5,000W. Exception: if the nameplate rating exceeds
5,000W, use that rating as the load. You can omit this calculation if the unit
has no electric dryer provisionhowever, it's common to provide both gas and
electric sources. If you see gas on the plans, verify electric will be
omitteddo not just assume.
Example: What is the service and feeder
demand load for a 4 kW dryer? (see Figure 916).
(a) 4,000W (b) 3,000W
(c) 5,000W (d) 5,500W
Answer: (c) 5,000W. The dryer load must not be less
than 5,000 VA. Cooking equipment
For household cooking appliances
rated over 13/4 kW, you can use the demand factors in 220.19, Table and Notes 1,
2, and 3.
Example: What is the service and feeder demand load for a 13.6
kW range in a dwelling unit?
(a) 8.8 kW (b) 8 kW (c) 9.2 kW (d) 6
kW
Answer: (a) 8.8 kW 13.6 kW exceeds 12 kW by one kW and one major
fraction of a kW. Column C value (8 kW) must be increased 5% for each kW or
major fraction of a kW (0.5 kW or larger) over 12 kW. 8 kW x 1.1 = 8.8 kW
Feeder and Service Conductor Size.
400A and Less: For 3wire,
120/240V, 1Ø systems, size the feeder or service conductors to Table
310.15(B)(6). For all others, use Table 310.16. Size the grounded (neutral)
conductor to the maximum unbalanced load [220.22] per Table 310.16.
Over
400A: Size the ungrounded and grounded (neutral) conductors per Table
310.16.
What size THHN feeder or service conductor (120/240V, 1Ø) does
the NEC require for a 225A service demand load? (a) 1/0 AWG (b) 2/0 AWG (c)
3/0 AWG (d) 4/0 AWG
Answer: (c) 3/0 AWG
Optional Method
You
can use the easier optional method ( in 220.30) only when the total connected
load is served by a single 3wire, 120/240V or 208Y/120V set of service or
feeder conductors with an ampacity of 100A or greater. Because this condition
describes the typical residential service, the optional method is likely to
apply. Using it can simplify the design process and save you time because you
have so many fewer sets of calculations.
General loads.
The
calculated load shall not be less than 100% for the first 10 kW, plus 40% of the
remainder of the following loads:
Smallappliance and laundry branch
circuits: 1,500 VA for each 20A circuit. General lighting and receptacles: 3
VA per sq ft Appliances: The nameplate VA rating of all appliances and
motors fastened in place (permanently connected) or on a specific circuit. Be
sure to use the range and dryer at nameplate rating! HVAC
Include the
largest of the following:
100% of the nameplate rating of the
airconditioning equipment. 100% of the heatpump compressors and
supplemental heating, unless the controller prevents simultaneous operation of
the compressor and supplemental heating. 100% of the nameplate ratings of
electric thermal storage and other heating systems where you expect the usual
load to be continuous at the full nameplate value. Do not configure such systems
under any other selection in this table. 65% of the nameplate rating(s) of
the central electric space heating, including integral supplemental heating in
heat pumps where the controller prevents simultaneous operation of the
compressor and supplemental heating. 65% of the nameplate rating(s) of
electric space heating, if there are less than four separately controlled units.
40% of the nameplate rating(s) of electric space heating of four or more
separately controlled units. Sizing Service/Feeder Conductors
Now
that we've seen how to determine residential loads, let's size the
service/feeder conductors.
400A and Less. Size the ungrounded conductors
per Table 310.15(B)(6) for 120/240V, 1Ø systems up to 400A. You must size the
grounded (neutral) conductor to carry the unbalanced load per Table
310.16.
Over 400A. Size the ungrounded and grounded (neutral) conductors
per Table 310.16.
An example will help illustrate how to do these. What
size service conductor does a 1,500 sq ft dwelling unit need, if it contains the
following loads?
Dishwasher (1,200 VA) Water Heater (4,500 VA)
Disposal (900 VA) Dryer (4,000 VA) Cooktop (6,000 VA) Two Ovens
(each 3,000 VA) A/C (5hp) Electric Heating (10 kW) (a) 6 AWG (b) 4
AWG (c) 3 AWG (d) 2 AWG
Answer: (c) 3 AWG
Step 1: Calculate the
general loads [220.30(B)]. Smallappliances = 1,500 VA x 2 circuits = 3,000
VA General lighting = 1,500 sq ft x 3 VA= 4,500 VA Laundry = 1,500 VA
Appliances (nameplate) = All the appliance loads summed = 1,200 VA + 4,500
VA + 900 VA + 4,000 VA + 6,000 VA + (3,000 VA x 2 units) + 6,000 VA. Total
Connected Load = 31,600 VA First 10,000 VA at 100% = 10,000 VA x 1.00 =
10,000 VA
Remainder VA at 40% = 21,600 VA x 0.40 = 8,640 VA
Demand
load = 18,640 VA
Step 2: AirConditioning Versus Heat [220.30(C)]. AC
= 230V x 28A = 6,440 VA; Electric heat: 10,000 VA x 0.65 = 6,500 VA. Thus, omit
the AC in the calculations. Step 3: Service/Feeder Conductors
[310.15(C)(6)]. General Loads = 18,640 VA Heat = 6,500 VA Total
Demand Load = 25,140 VA I = VA/V I = 25,140 VA / 240V = 105A Note:
The feeder/service conductor is sized to 110A, 3 AWG.
Now that we've
walked through the process of calculating residential feeders, you can see that
doing so is fairly easy. You need to calculate the loads, then the feeder size.
The NEC provides the requirements in Articles 220 and 230. Doing these
calculations correctly can save you money during design and construction, while
providing safe homes for the families that occupy them.

