--- title: "Process room 40 × 20 m twin 40 W fluorescent" subject: "Industrial Power Distribution and Illumination" chapter: 7 key: "semester_6/industrial_power_distribution_and_illumination:process-room-40x20-fluorescent" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/process-room-40x20-fluorescent site: BCT Engineering --- # Process room 40 × 20 m twin 40 W fluorescent Industrial Power Distribution and Illumination · Chapter 7: Illumination Design Principles - Asked in: 2068 Magh (old course), 12 marks - Video: [Lumen method of Interior lighting](https://www.youtube.com/watch?v=I5jayDiwEsk) — Electrical Learning Aids - 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/process-room-40x20-fluorescent - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-7 ## Question Calculate the number of luminaries for the process room in an industry. The size of room is 40m × 20m illuminated by twin 40 Watt fluorescent lamp. The efficiency of lamp is 60 lumen/watt and distance between working plane and mounting height of luminaries is 3m and coefficient of utilization and depreciation factor are 0.75 and 1.2 respectively. Draw neat sketch showing the arrangement of lamps, sub-circuit and switches. Design the distribution board to supply lighting loads, if the supply system is 3 phase, 400V and 50Hz. ## Short answer Assuming 200 lux, lumen method: luminaire output=80 W×60 lm/W=4800 lm, MF=1/1.2=0.833, total lumens=(200×800×1.2)/0.75=256000 lm; N=256000/4800≈53.33, rounded to 54 luminaires in 9×6 grid. Six sub-circuits, balanced, 10 A TPN MCB incomer. ## Answer **Given:** room 40 m × 20 m, twin 40 W fluorescent (80 W) at 60 lm/W, Hm = 3 m, CU = 0.75, depreciation factor 1.2, supply 400 V, 3-phase, 50 Hz. **Assumption:** the required illumination is not given, so **E = 200 lux** (general process/industrial work) is taken; SHR ≤ 1.5; electronic ballasts, pf 0.9. #### Number of luminaires Lumen method: N = (E × A) / (φ × UF × MF), where φ is the output of one luminaire. ```text Area A = 40 × 20 = 800 m² φ per luminaire = 80 W × 60 lm/W = 4800 lm MF = 1/DF = 1/1.2 = 0.833 Total lumens = E × A × DF / UF = 200 × 800 × 1.2 / 0.75 = 256000 lm N = 256000 / 4800 = 53.33 ``` Rounding up and fitting a regular grid: **9 × 6 = 54 luminaires** (9 along the 40 m side, 6 along the 20 m side). #### Arrangement and spacing ```text Spacing along length = 40/9 = 4.44 m Spacing along width = 20/6 = 3.33 m End spacing (to wall) = half of these Max spacing = SHR × Hm = 1.5 × 3 = 4.5 m 4.44 m and 3.33 m ≤ 4.5 m → OK Actual E = 54 × 4800 × 0.75 × 0.833 / 800 = 202.5 lux (≥ 200 lux required) ``` #### Layout diagram ```text Plan 40 m × 20 m (letter = phase of lamp circuit) +-------------------+ | R R R R R R R R R | <- L1 | Y Y Y Y Y Y Y Y Y | <- L2 | B B B B B B B B B | <- L3 | R R R R R R R R R | <- L4 | Y Y Y Y Y Y Y Y Y | <- L5 | B B B B B B B B B | <- L6 +-------------------+ [DB] [SB] <- entrance: DB and switch board S1 S2 S3 S4 S5 S6 (one switch per sub-circuit) ``` Each row of 9 luminaires forms one sub-circuit, controlled by one 6 A one-way switch. All switches are grouped on a switch board (SB) next to the entrance, and the DB is mounted beside it. #### Sub-circuits Rule (IS 732 / NEA practice): a lighting sub-circuit carries at most 10 points and 800 W. One luminaire here is 80 W, so up to 10 luminaires can go on one sub-circuit. The number of circuits is made a multiple of 3 for phase balance. | Circuit | Luminaires | Load (W) | Current (A) | Phase | MCB | |---|---|---|---|---|---| | L1 | 9 | 720 | 3.48 | R | 6 A | | L2 | 9 | 720 | 3.48 | Y | 6 A | | L3 | 9 | 720 | 3.48 | B | 6 A | | L4 | 9 | 720 | 3.48 | R | 6 A | | L5 | 9 | 720 | 3.48 | Y | 6 A | | L6 | 9 | 720 | 3.48 | B | 6 A | #### Distribution board design | Phase | Circuits | Load (W) | Current (A) | |---|---|---|---| | R | L1, L4 | 1440 | 6.96 | | Y | L2, L5 | 1440 | 6.96 | | B | L3, L6 | 1440 | 6.96 | ```text Total connected load = 4320 W (lighting) Most loaded phase = 1440 W Iph = 1440 / (230 × 0.9) = 6.96 A Main MCB ≥ 1.25 × 6.96 = 8.7 A → 10 A TPN MCB ``` **DB specification:** 4-way-per-phase TPN DB (12 ways in all, spares included) with a 10 A TPN MCB as incomer, a 25 A 4-pole RCCB (100 mA) for earth-leakage protection, and 6 A SP MCBs (B-curve) for the light circuits. Incoming cable: 4-core 1.5 mm² Cu PVC (rating 21 A ≥ 10 A), plus an earth continuity conductor. ```text 3-phase, 400/230 V, 4-wire supply | [Energy meter] | [10 A TPN MCB] | [25 A 4P RCCB] | ==R========Y========B========N== busbars | | | | | | L1 L4 L2 L5 L3 L6 (6 A SP MCBs) ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The question gives no illuminance. What is done? A. The problem cannot be solved B. A suitable value (200 lux for process work) is assumed C. Zero lux is used D. The lux is taken as the lamp efficiency Answer: B. A reasonable standard level of 200 lux for general industrial work is stated as an assumption. (Key point: Assumption of lux) 2. With DF = 1.2 and CU = 0.75, which gives total lumens? A. 200 x 800 x 1.2 / 0.75 = 256000 lm B. 200 x 800 x 0.75 / 1.2 = 100000 lm C. 200 x 800 / 1.2 = 133333 lm D. 200 x 800 x 1.2 x 0.75 = 144000 lm Answer: A. MF = 1/DF, so lumens = E x A x DF / CU = 256000 lm; divided by 4800 lm gives 53.3 and 54 luminaires. (Key point: Depreciation factor use) 3. Why is the 9 x 6 grid acceptable? A. It has exactly 53.33 luminaires B. Spacings 4.44 m and 3.33 m are within the 4.5 m limit C. Spacing is greater than 4.5 m D. Because 9 is a multiple of 3 phases only Answer: B. Maximum spacing = 1.5 x 3 = 4.5 m, and both spacings are below it; actual E = 202.5 lux. (Key point: Grid 9 x 6 check) 4. Most loaded phase is 6.96 A. What incomer is chosen? A. 6 A TPN MCB B. 10 A TPN MCB (1.25 x 6.96 = 8.7 A) C. 25 A TPN MCB D. 32 A SP MCB Answer: B. A 1.25 margin gives 8.7 A, so the next standard rating, 10 A, is used; 6 A would be too small. (Key point: Incomer rating) ## Close variants set in other exams - [Industry room 30 × 20 m LED lighting and DB](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aindustry-room-30x20-led) — Industrial Power Distribution and Illumination, chapter 7, asked 2× ## Related in Industrial Power Distribution and Illumination - [Process room 30 × 15 × 5 m fluorescent lighting](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aprocess-room-30x15-fluorescent) — Industrial Power Distribution and Illumination, chapter 7 - [Production room 27 × 45 m twin 36 W tubes](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aproduction-room-27x45) — Industrial Power Distribution and Illumination, chapter 7, asked 4× - [Department store 30 × 15 m twin 40 W fluorescent](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Adepartment-store-twin-40w-fluorescent) — Industrial Power Distribution and Illumination, chapter 7, asked 3× - [Drawing hall 40 × 20 × 6 m lighting design](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Adrawing-hall-40x20-led) — Industrial Power Distribution and Illumination, chapter 7, asked 2× - [Room 40 × 15 m, 75 lux FTL design](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aroom-40x15-ftl-75lux) — Industrial Power Distribution and Illumination, chapter 7 - [Drawing hall 40 × 25 × 6 m, 2×36 W fluorescent](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Adrawing-hall-2x36w-fluorescent) — Industrial Power Distribution and Illumination, chapter 7