--- title: "Drawing hall 40 × 25 × 6 m, 2×36 W fluorescent" subject: "Industrial Power Distribution and Illumination" chapter: 7 key: "semester_6/industrial_power_distribution_and_illumination:drawing-hall-2x36w-fluorescent" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/drawing-hall-2x36w-fluorescent site: BCT Engineering --- # Drawing hall 40 × 25 × 6 m, 2×36 W fluorescent Industrial Power Distribution and Illumination · Chapter 7: Illumination Design Principles - Asked in: 2070 Magh, 8 marks - Video: [How to solve illumination problem's !! question from VTU QP !! By Sharan Sir!!](https://www.youtube.com/watch?v=-pIazs-fjYs) — Knowledge center - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/drawing-hall-2x36w-fluorescent - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-7 ## Question A drawing hall 40m×25m×6m height is to be illuminated with 2×36W fluorescent lamp to an average illumination of 90 Lm/m² on a working plane 1 m above the floor. Estimate suitable numbers, sizes of lamps. Mounting height 1m below the ceiling. Sketch the spacing layout. Assume: coefficient of utilization 0.8 depreciation factor 1.2, space height ratios 1.4 and the lamp efficiency is 60lumen/watt. ## Short answer With mounting height 4 m above working plane, lumen method yields 31.3 but spacing limit 5.6 m requires 8×5=40 luminaires of 2×36 W, spaced 5 m both ways, giving 115.2 lux average illuminance. ## Answer **Given:** hall 40 m × 25 m × 6 m, E = 90 lm/m² on a plane 1 m above floor, 2 × 36 W fluorescent (72 W per luminaire) at 60 lm/W, CU = 0.8, depreciation factor 1.2, SHR = 1.4, luminaires 1 m below ceiling. Mounting height above working plane: Hm = (6 − 1) − 1 = **4 m**. #### Number of luminaires Lumen method: N = (E × A) / (φ × UF × MF), where φ is the output of one luminaire. ```text Area A = 40 × 25 = 1000 m² φ per luminaire = 72 W × 60 lm/W = 4320 lm MF = 1/DF = 1/1.2 = 0.833 Total lumens = E × A × DF / UF = 90 × 1000 × 1.2 / 0.8 = 135000 lm N = 135000 / 4320 = 31.25 ``` Rounding up and fitting a regular grid: **8 × 5 = 40 luminaires** (8 along the 40 m side, 5 along the 25 m side). #### Arrangement and spacing ```text Spacing along length = 40/8 = 5 m Spacing along width = 25/5 = 5 m End spacing (to wall) = half of these Max spacing = SHR × Hm = 1.4 × 4 = 5.6 m 5 m and 5 m ≤ 5.6 m → OK Actual E = 40 × 4320 × 0.8 × 0.833 / 1000 = 115.2 lux (≥ 90 lux required) ``` The lumen calculation needs only 32 luminaires, but the 5.6 m spacing limit governs: 40/5.6 = 7.1 → 8 luminaires along the length and 25/5.6 = 4.5 → 5 across the width. A 7 × 5 grid gives 5.71 m spacing (> 5.6 m) and an 8 × 4 grid gives 6.25 m, so both are too wide. So **8 × 5 = 40 luminaires of 2 × 36 W** are used at 5 m × 5 m spacing (2.5 m from the walls). Total load = 40 × 72 = 2880 W. #### Layout diagram ```text Plan 40 m × 25 m (o = luminaire) +-----------------+ | o o o o o o o o | | o o o o o o o o | | o o o o o o o o | | o o o o o o o o | | o o o o o o o o | +-----------------+ Spacing between luminaires = 5 m both ways End spacing to each wall = 2.5 m Mounting height: 5 m above floor (Hm = 4 m) ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a 6 m high hall with working plane 1 m above floor and luminaires 1 m below ceiling, what is Hm? A. 6 m B. 5 m C. 4 m D. 3 m Answer: C. Hm is measured from working plane to luminaire: (6 - 1) - 1 = 4 m. (Key point: Mounting height Hm) 2. Which total lumen requirement is used for E = 90 lux, A = 1000 m^2, CU = 0.8, DF = 1.2? A. 90 x 1000 x 0.8 / 1.2 = 60000 lm B. 90 x 1000 / (0.8 x 1.2) = 93750 lm C. 90 x 1000 x 1.2 / 0.8 = 135000 lm D. 90 x 1000 x 1.2 x 0.8 = 86400 lm Answer: C. Depreciation factor multiplies the requirement (MF = 1/DF) and CU divides it: 135000 lm. (Key point: Total lumens with depreciation factor) 3. The lumen calculation needs about 32 luminaires, yet 40 are installed. Why? A. The spacing limit 5.6 m forces an 8 x 5 grid B. Extra luminaires are added as spares C. Each luminaire has only 36 W D. The room has no walls to reflect light Answer: A. A 7 x 5 grid gives 5.71 m and 8 x 4 gives 6.25 m, both above the 5.6 m maximum, so 8 x 5 = 40 is used. (Key point: Why 40 luminaires not 32) 4. How much light does one 2 x 36 W luminaire give at 60 lm/W? A. 2160 lm B. 4320 lm C. 72 lm D. 8640 lm Answer: B. 72 W x 60 lm/W = 4320 lm per luminaire. 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