14.2 Continuous Loads and Equipment Demand

The 125% Continuous Load Rule

The NEC defines a continuous load as one where the maximum current is expected to continue for 3 hours or more. Commercial lighting, refrigeration, and some process loads are typically continuous. Two parallel rules govern conductor sizing:

• 215.2(A)(1) — Feeders: Feeder conductors shall have an ampacity not less than 125% of the continuous load plus 100% of the noncontinuous load.
• 210.19(A)(1) — Branch circuits: Branch-circuit conductors shall have an ampacity not less than 125% of the continuous load plus 100% of the noncontinuous load.

The formula is straightforward:

Conductor ampacity ≥ (1.25 × continuous load) + (1.00 × noncontinuous load)

Important interaction with Table 220.42(A): The general lighting unit values in Table 220.42(A) already include the 125% continuous load factor. When you use those table values for the load calculation, you do not apply 125% again. However, for other continuous loads not calculated from Table 220.42(A) — such as show window lighting (220.14(B)), sign lighting, or process loads — you must apply the 125% factor per 215.2(A)(1) or 210.19(A)(1).

Fixed Electric Space Heating — 220.51

NEC 220.51 requires fixed electric space-heating loads to be calculated at 100% of the total connected load (nameplate rating). No demand factor is applied to the heating load itself. The feeder or service load current rating cannot be less than the rating of the largest branch circuit supplied.

Exception: Where reduced loading results from duty cycling, intermittent operation, or units not operating simultaneously, the AHJ may permit smaller conductors — but only if the conductors have adequate ampacity for the actual load.

Noncoincident Loads — 220.60

Where two or more loads are unlikely to be operating simultaneously, NEC 220.60 permits using only the larger load in the calculation. The classic exam application is electric heating versus air conditioning — a building will not run both at full capacity at the same time.

Special rule: If a motor or air-conditioning load is part of the noncoincident pair but is not the largest, you must include 125% of the motor or AC load (per 220.60, referencing 220.50) in the total, not just the larger load alone.

Motors and Air-Conditioning — 220.50

NEC 220.50 governs how motor and air-conditioning equipment loads are included in feeder and service calculations. Motor loads are generally taken at nameplate full-load current, with the largest motor added at 125% per 430.62 for feeder OCPD sizing. Motor conductor sizing (430.22 for a single motor, 430.24 for multiple motors) is covered in Chapter 11 — this chapter references it but does not re-teach it.

Table 220.56 — Commercial Kitchen Equipment Demand Factors

For commercial electric cooking equipment, dishwasher boosters, water heaters, and other kitchen equipment (other than dwelling units), NEC 220.56 permits demand factors from Table 220.56:

Floor rule: In no case shall the feeder or service calculated load be less than the sum of the largest two kitchen equipment loads.

These demand factors apply only to equipment with thermostatic control or intermittent use as kitchen equipment. They do not apply to space heating, ventilating, or air-conditioning equipment in the kitchen.

Example: 4 kitchen units

• Range: 8,000 VA, Oven: 6,000 VA, Fryer: 4,000 VA, Griddle: 3,000 VA
• Total connected = 21,000 VA
• 4 units → 80% demand factor
• Demand = 21,000 × 0.80 = 16,800 VA
• Floor check: 8,000 + 6,000 = 14,000 VA
• 16,800 ≥ 14,000 → use 16,800 VA

If the calculated demand had been less than 14,000 VA, you would use 14,000 VA instead.

Worked Example: Commercial Feeder with Continuous and Noncoincident Loads

Given: A 240 V single-phase feeder supplies:

• Continuous lighting load: 10,000 VA (not from Table 220.42(A))
• Noncontinuous receptacle load: 5,000 VA
• Electric heating: 15,000 VA
• Air conditioning: 12,000 VA

Step 1 — Noncoincident Loads (220.60)

• Electric heat (15,000 VA) vs. AC (12,000 VA) → use larger = 15,000 VA
• The 12,000 VA AC load is dropped from the calculation.

Step 2 — Apply Continuous Load Factor (215.2(A)(1))

• Continuous: 10,000 × 1.25 = 12,500 VA
• Noncontinuous: 5,000 + 15,000 = 20,000 VA
• Total = 12,500 + 20,000 = 32,500 VA

Step 3 — Feeder Current

• I = 32,500 VA ÷ 240 V = 135.4 A

Step 4 — Conductor Selection (Table 310.16, 75°C column per 110.14(C))

• #1/0 Cu = 150 A ≥ 135.4 A ✓
• Select: 1/0 AWg Cu THWN

Step 5 — OCPD Selection (240.6(A))

• Next standard rating ≥ 135.4 A = 150 A
• 150 A (conductor) ≥ 150 A (OCPD) ✓
• Select: 150 A breaker

This example shows three rules working together: 220.60 eliminates the smaller noncoincident load, 215.2(A)(1) applies the 125% continuous factor to the continuous lighting only, and the conductor and OCPD are selected from the verified tables.