--- title: "Flood lighting procedure and 50 × 15 m frontage" subject: "Industrial Power Distribution and Illumination" chapter: 8 key: "semester_6/industrial_power_distribution_and_illumination:flood-procedure-building-50x15" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/flood-procedure-building-50x15 site: BCT Engineering --- # Flood lighting procedure and 50 × 15 m frontage Industrial Power Distribution and Illumination · Chapter 8: Outdoor Lighting Design - Asked in: 2075 Baisakh, 4+4 marks - Video: [flood lighting in illumination | flood ilghting in hindi | flood light outdoor | what is flood light](https://www.youtube.com/watch?v=btEEfvxKcTU) — Mind of Engineer - 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/flood-procedure-building-50x15 - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-8 ## Question Explain briefly the design procedure of flood lighting system. A building frontage 50m × 15m is to be illuminated by floodlighting projectors situated 25m away. If the illumination is 100 lux, coefficient of utilization is 0.5, depreciation factor is 1.5, and waste light factor 1.2, estimate the number of projectors of each of size 200W with luminous efficiency of 90 lumen/watt. ## Short answer 15 projectors of 200 W each (90 lm/W). Total lumens: 100×750×1.5×1.2/0.5 = 270,000 lm. Beam spread required ≈ 33.4°. Use medium-beam projectors. ## Answer #### Design procedure of floodlighting 1. **Survey the object:** size, shape, colour and reflection factor of the surface and the brightness of surroundings. 2. **Choose the illumination level (E)** from tables (higher for dark surfaces and bright surroundings). 3. **Find the area** to be illuminated: A = length × height. 4. **Fix projector location** – distance about 0.7–1 times the building height, out of the viewers' sight line. 5. **Select beam spread** to suit the throw: θ = 2 tan⁻¹(half-height / distance). 6. **Choose factors:** coefficient of utilisation (CU), depreciation factor (DF), waste light factor (WLF). 7. **Total lumens:** F = E × A × DF × WLF / CU. 8. **Lamp lumens** = wattage × efficacy; **number** N = F / lamp lumens (round up). 9. **Arrange and aim** projectors for uniform, glare-free lighting; design cabling and control. #### Numerical **Given:** 50 m × 15 m, distance 25 m, E = 100 lux, CU = 0.5, DF = 1.5, WLF = 1.2, 200 W lamps of 90 lm/W. ```text Area A = 50 × 15 = 750 m² Total lm = E·A·DF·WLF / CU = 100 × 750 × 1.5 × 1.2 / 0.5 = 270 000 lm Lamp lm = 200 × 90 = 18 000 lm N = 270 000 / 18 000 = 15 ``` Vertical beam spread needed (aimed at mid-height): 2 tan⁻¹(7.5/25) = 33.4°, so medium-beam projectors are suitable. **Answer: 15 projectors of 200 W each.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which formula gives the total lumens needed in floodlighting? A. F = E x A x CU / (DF x WLF) B. F = E x A / (DF x WLF x CU) C. F = E x A x DF x WLF / CU D. F = E x A x CU x DF x WLF Answer: C. Depreciation and waste light increase the lumens required, while CU (useful fraction) divides. (Key point: Total lumens formula) 2. For 100 lux, 750 m2, CU 0.5, DF 1.5, WLF 1.2, what are the total lumens? A. 270 000 lm B. 135 000 lm C. 75 000 lm D. 180 000 lm Answer: A. 100 x 750 x 1.5 x 1.2 / 0.5 = 270 000 lm. (Key point: Numerical result) 3. With 18 000 lm per 200 W, 90 lm/W lamp, how many projectors are needed? A. 12 B. 15 C. 20 D. 30 Answer: B. N = 270 000 / 18 000 = 15. (Key point: Number of projectors) 4. How far from the building are projectors usually placed, and how is beam spread found? A. About 3 times the height; theta = tan^-1(height/distance) B. Right against the wall; theta = 90 degrees C. About 0.7-1 times the width; theta = height x distance D. About 0.7-1 times building height; theta = 2 tan^-1(half-height/distance) Answer: D. The procedure puts projectors at 0.7-1 x building height and uses theta = 2 tan^-1(half-height/distance). 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