How to calculate room illuminance per SP 52.13330 and IS 3646

Why calculate illuminance

Lighting is not "hang a chandelier and done". Insufficient illuminance leads to eye strain, reduced productivity, and even accidents. Excessive — energy waste and discomfort. The electrical engineer must be able to calculate illuminance precisely.

In Russia, calculations follow SP 52.13330.2016. In India — IS 3646 (Parts 1-3). Both use the utilization factor method. Let's work through a concrete example.

Initial data

Room: kitchen 12 m² (4 × 3 m), ceiling height 2.7 m. Finishes: light wallpaper (walls 50%), white ceiling (70%), porcelain floor tiles (30%). Purpose: cooking, dining — required illuminance 200 lux (SP 52.13330 / IS 3646 for residential kitchens).

Kitchen illuminance standards comparison:

• SP 52.13330: 200 lx (general), 300 lx (countertop work zone)
• IS 3646 Part 1: 200 lx (general), 300 lx (work zone)
• EN 12464-1: 300-500 lx (kitchen overall)

  Utilization Factor Method

  Formula: Φ = (E × S × Ks × Z) / (N × η)

  Where:

• Φ — required luminous flux per luminaire (lm)
• E — required illuminance (lx) = 200 lx
• S — room area (m²) = 12 m²
• Ks — maintenance factor (accounts for dust and lamp aging) = 1.3 for LED
• Z — non-uniformity factor = 1.1 for spot lights
• N — number of luminaires
• η — utilization factor (determined by room index and reflectance)

  Step 1: Room Index

  i = (A × B) / (h × (A + B))

  Where:

• A = 4 m, B = 3 m — room dimensions
• h = H - hw - hs = 2.7 - 0.8 - 0.15 = 1.75 m — calculation height

  i = (4 × 3) / (1.75 × (4 + 3)) = 12 / 12.25 ≈ 0.98

  Step 2: Utilization factor η

  From SP 52.13330 tables for cosine-distribution luminaire (Type D) at i ≈ 1.0 and reflectance 70/50/30:

  η ≈ 0.48 (48% of luminous flux reaches the working plane)

  Step 3: Luminous flux

  Φtotal = (E × S × Ks × Z) / η = (200 × 12 × 1.3 × 1.1) / 0.48 = 3432 / 0.48 = 7150 lm

  Step 4: Luminaire selection

  Option 1 — 4 spot lights (N = 4): Φ1 = 1788 lm → LED 1800-2000 lm (≈ 15-18 W each)

Option 2 — 2 linear luminaires (N = 2): Φ1 = 3575 lm → LED panels 3600-4000 lm (≈ 30-36 W each) Option 3 — central chandelier (N = 1): Φ1 = 7150 lm → LED 7000-8000 lm (≈ 60-70 W)

Step 5: Power density check

Power density for kitchen: 10-15 W/m². All LED options ≈ 6 W/m² — economical, 2.5-3× better than fluorescent.

Work zone: local lighting

Countertop requires 300 lx. For 2.5 m² work zone and η = 0.5: Φ = (300 × 2.5 × 1.3 × 1.1) / 0.5 = 2145 lm → LED strip 2200 lm (18-20 W) under cabinets.

What GorkyCAD automates

- Calculates room index from floor plan geometry

• Selects η from SP/IS/IEC tables
• Proposes luminaire placement variants
• Calculates total lighting group load
• Checks compliance with standards and power density