14.4 Commercial Building Load Capstone

Capstone Problem: 10,000 sq ft Office Building

This capstone integrates every rule from Sections 14.1 through 14.3 into a single end-to-end calculation.

Given:

• 10,000 sq ft office building
• 40 receptacle outlets (one per strap/yoke)
• Occupancy classification: office building, which triggers the 220.14(K) receptacle comparison
• Continuous display lighting: 2,000 VA (separate from Table 220.42(A) general lighting)
• Electric heating: 14,000 VA
• Air conditioning: 11,000 VA
• Commercial kitchen: range 4,000 VA + oven 3,000 VA (2 units)
• System: 208 V, three-phase

Step 1 — General Lighting (Table 220.42(A))

The 125% continuous load factor is already included in the Table 220.42(A) unit value — do not apply it again.

Step 2 — Receptacles (220.14(I), Table 220.47, and the 220.14(K) floor)

Two things are happening here. First, 7,200 VA is under the 10 kVA threshold, so no 50% reduction applies under Table 220.47. Second — and this is the step most candidates drop — 220.14(K) requires banks and office buildings to use the larger of the 220.14(I) result or 1 VA per square foot. At 10,000 ft² the floor is 10,000 VA, which beats the 7,200 VA strap count, so 10,000 VA is what moves forward. The two figures are never added.

Step 3 — Continuous Display Lighting (215.2(A)(1))

This display lighting is a separate continuous load — it is not part of the Table 220.42(A) general lighting, so the 125% factor must be applied explicitly.

Step 4 — Noncoincident Loads (220.60)

Per 220.60, only the larger load is used. Per 220.51, fixed heating is calculated at 100% of connected load.

• Heating demand = 14,000 VA

Step 5 — Commercial Kitchen (Table 220.56)

Step 6 — Total Computed Load

Step 7 — Service Current (208 V Three-Phase)

Three-phase current formula:

I = VA ÷ (V × √3)

Substitution:

• I = 46,500 ÷ (208 × 1.732)
• I = 46,500 ÷ 360.26
• I = 129.1 A

Step 8 — Service Conductor Selection

Per 110.14(C)(1)(b), equipment rated over 100 A uses the 75°C column of Table 310.16.

Select: 1 AWG Cu THWN-2 (130 A at 75°C)

Step 9 — OCPD Selection (240.6(A) and 240.4(B))

• Load current = 129.1 A
• 130 A is not a standard rating, so the next standard rating from 240.6(A) at or above the load is 150 A
• 240.4(B) permits the next higher standard rating above the conductor ampacity where the ampacity does not correspond to a standard rating, the conductors do not supply cord-and-plug-connected portable loads on a multioutlet branch circuit, and the rating is 800 A or less. 230.90(A), Exception No. 2 carries that allowance to service conductors.
• Verify: load 129.1 A ≤ conductor 130 A ✓, and 150 A is the next standard rating above 130 A ✓

Select: 150 A device on 1 AWG Cu. Upsizing to 1/0 Cu (150 A at 75°C) is the conservative alternative and removes the need to invoke 240.4(B) at all.

Verification Summary

Cross-Reference

Chapter 4 covers the service-disconnect requirements (230.70–230.79) and service-entrance conductor rules (230.42) that complete the service design. Chapter 5 covers multi-family demand calculations (Table 220.84) and separately derived systems, which are distinct from this single-tenant commercial building calculation.

What Could Change the Answer?

• If the building were 240 V single-phase instead of 208 V three-phase: I = 46,500 ÷ 240 = 193.8 A. In the 75°C column, 2/0 Cu is only 175 A, so the conductor moves to 3/0 Cu at 200 A, protected by a 200 A device — 240.4(B) is not needed here because 200 A is both a standard rating and the exact conductor ampacity.
• If the building included a hospital wing: the lighting demand factor from Table 220.45 would change from 100% (All Others) to 40%/20% (hospitals), potentially reducing the total load.
• If a voltage-drop check were required for a long service run: the conductor might need to be upsized beyond the ampacity-based selection, as demonstrated in Section 14.3.