Intermediate6 steps

Restaurant Kitchen Load Calculation

Calculate the electrical load for a commercial restaurant kitchen — cooking equipment demand factors per NEC Table 220.56.

Scenario Overview

Calculate the electrical demand for a full-service restaurant kitchen with commercial cooking equipment, applying NEC Table 220.56 demand factors for commercial kitchen loads.

Given Information

  • Full-service restaurant: 2,800 sq ft
  • 6-burner electric range: 20 kW
  • Electric convection oven: 12 kW
  • Deep fryers (2×): 8 kW each
  • Electric griddle: 10 kW
  • Commercial dishwasher: 6 kW
  • Walk-in cooler: 3 HP (240V, 3Φ)
  • Walk-in freezer: 5 HP (240V, 3Φ)
  • Exhaust hood: 2 HP (240V, 3Φ)
  • HVAC: 15 ton packaged unit (208V, 3Φ)

Calculation Steps

1

List All Kitchen Cooking Equipment

Total connected cooking load:

Range: 20 kW

Convection oven: 12 kW

2× deep fryers: 16 kW

Griddle: 10 kW

Dishwasher: 6 kW (not always counted as cooking equipment — check AHJ)

Total cooking equipment: 58 kW (without dishwasher) or 64 kW (with).

Result: 58-64 kW total cooking equipment

2

Apply NEC Table 220.56 Demand Factor

For 5+ pieces of cooking equipment: NEC Table 220.56 allows 65% demand factor.

58 kW × 0.65 = 37.7 kW demand.

Note: Ventilation required by the equipment is not eligible for this demand factor.

58 kW × 0.65 = 37.7 kW

Result: 37.7 kW kitchen equipment demand

3

Motor Loads (Compressors, Exhaust)

Walk-in cooler 3 HP: NEC Table 430.250 → 9.6A at 230V 3Φ.

Walk-in freezer 5 HP: → 15.2A at 230V 3Φ.

Exhaust hood 2 HP: → 6.8A at 230V 3Φ.

Largest motor (15.2A) at 125%: 19A.

Total motor: 19 + 9.6 + 6.8 = 35.4A.

Motor VA: 35.4 × 208 × √3 = 12,757 VA.

Result: 12,757 VA motor loads

4

General Lighting & Receptacles

Restaurant: NEC Table 220.12 = 2 VA/ft² for restaurants.

2,800 × 2 = 5,600 VA lighting.

Receptacles: 20 outlets × 180 VA = 3,600 VA.

Total: 9,200 VA (under 10 kW, no demand reduction).

Result: 9,200 VA lighting + receptacles

5

HVAC Load

15 ton packaged HVAC: approximately 65A at 208V 3Φ.

VA = 65 × 208 × √3 = 23,423 VA.

100% demand — no demand factor for HVAC.

Result: 23,423 VA HVAC

6

Total Service Demand

Kitchen: 37,700 VA

Motors: 12,757 VA

Lighting/Receptacles: 9,200 VA

HVAC: 23,423 VA

Total: 83,080 VA.

Service amps: 83,080 / (208 × √3) = 230.7A.

Next standard service: 400A (provides growth capacity).

83,080 VA / (208 × 1.732) = 230.7A

Result: 231A demand → 400A service recommended

Final Answer

Total calculated demand: 83,080 VA (231A at 208V 3Φ). Recommended: 400A, 208Y/120V, 3-phase service. Kitchen cooking equipment demand reduced from 58 kW to 37.7 kW using NEC Table 220.56 (65% demand factor for 5+ units).

Key Takeaways

  • NEC Table 220.56 significantly reduces kitchen equipment demand — 65% for 5+ units
  • Motor loads use NEC 430 FLC tables, not nameplate amps, for sizing
  • Restaurant electrical services are typically 400A due to heavy cooking and HVAC loads
  • Exhaust hood motors are NOT eligible for the cooking equipment demand factor

Calculators Used

NEC References

  • NEC Table 220.56 — Kitchen Equipment Demand
  • NEC 220.12 — Lighting by Occupancy
  • NEC 430 — Motor Circuits

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