Load Calculations - Part 2, based on the 2020 NEC^®

Figure 01

For EC&M Magazine
By Mike Holt, NEC^® Consultant

To catch up or re-read Load Calculations - Part 1, click here.

Note: This article is based on the 2020 NEC.

Do understand feeder and service load calculations?

Before performing feeder and service load calculations, characterize the loads and determine where demand factors apply. For example, not all luminaires are expected to be on at the same time. So you can apply demand factors to the general lighting load per Table 220.42. These demand factors do not apply to the calculated load of feeders or services supplying areas where the entire lighting is likely to be used at once, such as ballrooms or dining rooms.

For dwelling units, the demand factors of Table 220.42 apply to the two small-appliance circuits of 1,500 VA [220.52(A)] and a laundry circuit of 1,500 VA [220.52(B)]. Include these as part of the general lighting load calculation, along with the required lighting and general-use receptacle load of 3 VA per sq ft [220.14(J)].

Remember to subtract 3,000 VA from 120,000 VA when using Table 220.42, since the 35% applies only to the 3,001 VA to 120,000 VA range.

Non-dwelling unit receptacles
To determine the load for general-purpose receptacle outlets [220.14(I)] and fixed multioutlet assemblies [220.14(H)]:

• Add the receptacle and fixed multioutlet assembly VA load to the general lighting load [Table 220.12] and adjust this value by the demand factors in Table 220.42, or
• Apply a 50 percent demand factor to that portion of the receptacle and fixed multioutlet receptacle loads that exceed 10 kVA [220.44].
  

Example: What is the demand load for 150 general-purpose receptacles and 100 ft of fixed multioutlet assembly in a commercial occupancy? The appliances powered by the multioutlet assembly are not used simultaneously. Figure 01
(a) 18,200 VA   (b) 19,400 VA   (c) 20,100 VA   (d) 20,300 VA
Answer: (d) 20,300 VA
Solution:
Step 1:‚Determine the total connected load.
Receptacle Load = 150 receptacles × 180 VA
Receptacle Load = 27,000 VA [220.14(I)]
Multioutlet Load = 100 ft/5 ft
Multioutlet Load = 20 sections × 180 VA
Multioutlet Load = 3,600 VA [220.14(H)]
Step 2:‚Apply the Table 220.44 demand factors.
Total Connected Load = 30,600 VA
First 10,000 VA at 100% = 10,000 VA × 100% = 10,000 VA
Remainder at 50% = 20,600 VA × 50% = +10,300 VA
Receptacle Calculated Load = 20,300 VA

Motors and appliances in dwelling units
Calculate the feeder conductor for motor loads per 430.24. Conductors supplying motors and other load(s) must have an ampacity of at least the sum of each of the following:
(1) 125 percent of the full-load current rating of the highest rated motor.
(2) Sum of the full-load current ratings of all the other motors in the group.
(3) 100 percent of the noncontinuous nonmotor load(s).
(4) 125 percent of the continuous nonmotor load(s).

Small-appliance and laundry circuits in dwelling units
The load for each 20A circuit:

• Small-appliance circuit covered by 210.11(C)(1) is 1,500 VA; you can apply the general lighting demand factors in Table 220.42 [220.52(A)].
• Laundry circuit covered by 210.11(C)(2) is 1,500 VA; you can apply the general lighting demand factors in Table 220.42 [220.52(B)].
  

Appliance loads in dwelling units
You can apply a demand factor of 75 percent to the total connected load of four or more appliances rated ¼ hp or greater, or 500 watts or greater, that are fastened in place [220.53].

Do not apply this demand factor to:
(1) Household electric cooking equipment fastened in place [220.55].
(2) Clothes dryers [220.54].
(3) Space-heating equipment [220.51].
(4) Air-conditioning equipment [220.50].

The service/feeder load for electric clothes dryers must be at least 5,000W (5,000 VA), or the nameplate rating of the equipment if more than 5,000W (5,000 VA). For load calculations in this section, kVA is the same a kW and VA is the same as W [220.54].

Household electric dryers in dwelling units
When a building contains five or more dryers, you can apply the demand factors in Table 220.54 to the total connected dryer load.

Household electric appliances in dwelling units
For household cooking appliances rated over 1.75 kW, you can calculate the feeder/service demand load per the demand factors of Table 220.55.

The four notes to Table 220.55 may affect your calculations. For example, Note 1 describes one set of criteria for which you must raise the maximum demand in Column C by 5 percent for each additional kilowatt rating and Note 2 describes another. Note 4 provides both a permissible and a must.

Commercial kitchen equipment
You can use Table 220.56 to calculate the demand load for thermostat-controlled or intermittently used commercial kitchen equipment [220.56]. The kitchen equipment feeder/service calculated load must be at least the sum of the two largest kitchen equipment loads.

Table 220.56 demand factors do not apply to space-heating, ventilating, or air-conditioning equipment.

Noncoincident loads
If it is unlikely that two or more loads will be used simultaneously, use only the largest load for load calculations [220.60]. Where a motor is part of the noncoincident load, and is not the largest noncoincident load, use 125 percent of the motor load if it is the largest motor.

Neutral load
The neutral load for feeders or services is based on the maximum calculated load between the neutral conductor and any one phase conductor [220.61(A)]. Line-to-line loads are not considered in the calculation.

Permitted reductions:

• Cooking loads and dryers. The feeder/service neutral load for household electric ranges, wall-mounted ovens, or counter-mounted cooking units can be calculated at seventy percent of the cooking equipment demand load as determined per Table 220.55 [220.61(B)(1)]. You can apply the same reduction to dryers, per Table 220.54 [220.61(B)(1)].
• Over 200A neutral. The feeder/service calculated neutral load for a 3-wire, single-phase, or 4-wire, three-phase system can be calculated at seventy percent for that portion of the unbalanced load over 200A [220.61(B)(2)]. See Annex D, Example D4(a) Multifamily Dwelling.
  

Prohibited reductions:

• 3-wire circuits from 4-wire, wye-connected systems [220.61(C)(1)].
• Nonlinear loads [220.61(C)(2)].
  

Optional load calculation, dwellings
If the load for a dwelling unit is at least 100A, you can calculate the service load by adding the calculated general load [220.82(B)] to the calculated HVAC load [220.82(C)].

Some rules for calculating the general load:

• Determine the neutral load per 220.61.
• The demand load must be at least 100 percent of the first 10 kVA, plus 40 percent of the remainder kVA for the following loads:
• Base the general load on 3 VA per sq ft for general lighting and general-use receptacles.
• Add 1,500 VA for each 20A small-appliance circuit as required by 210.11(C)(1)(a) with at least two circuits per dwelling unit, and 1,500 VA for each 20A laundry circuit as required by 210.11(C)(2).
  

Example: Using the optional calculation method, what size service is required for a 1,500 sq ft dwelling unit containing the following loads?
Dishwasher 1,200 VA
Garbage Disposal 900 VA
Cooktop 6,000 VA
Oven 3,000 VA
Dryer 4,000 VA
Water heater 4,500 VA
Heat-pump compressor rated 28A at 240V, with supplemental electric heat having a rating of 7 kW.
(a) 100A   (b) 110A   (c) 125A   (d) 150A
Answer: (c) 125A
Solution:
Step 1:‚Determine the total feeder/service calculated load.
(1) Lighting, receptacles, and appliance calculated load
Small-Appliance 1,500 VA × 2 = 3,000 VA
Laundry 1,500 VA × 1 = 1,500 VA
General Lighting 1,500 sq ft × 3 VA/sq ft = 4,500 VA
Dishwasher 1,200 VA × 1 = 1,200 VA
Garbage Disposal 900 VA × 1 = 900 VA
Cooktop 6,000 VA × 1 = 6,000 VA
Oven 3,000 VA × 1 = 3,000 VA
Dryer 4,000 VA × 1 = 4,000 VA
Water Heater 4,500 VA × 1 = + 4,500 VA
Calculated Load = 28,600 VA
First 10,000 VA at 100% 10,000 VA× 100% = 10,000 VA
Remainder at 40% 18,600 VA × 40% = + 7,440 VA
220.82(B) Demand Load 17,440 VA
(2) Largest of Air-Conditioning or Heat [220.82(C)]
Heat Pump Compressor at 100% 240V × 28A = 6,720 VA
Supplemental Heat at 65% = 7,000 VA × 65% = +4,550 VA
220.82(C) Demand Load 11,270 VA
Total Demand Load [220.82(B) and (C)] = 17,440 VA + 11,270 VA
Total Demand Load [220.82(B) and (C)] = 28,710 VA
Step 2:‚Determine the service size in amperes.
I = Volt-Ampere/Volts
I = 28,710 VA/240V
I = 120A, 125A Service [240.6(A)]

You'd do something similar for a multifamily building, per 220.84. What if you have only two units supplied by a single feeder? Where the standard calculated load per 220 Part III exceeds that for three identical units calculated per 220.84, you can use the lesser of the two calculated loads.

Take a load off
Improper application of demand factors can result in a costly overbuild on the one hand or a dangerous underbuild on the other. A tip for avoiding this error is to annotate your calculations with the Table reference at each place where you use a Table. Then when you review your calculations, double-check that you used the correct Table and the correct value from it.