---
title: "Production room 27 × 45 m twin 36 W tubes"
subject: "Industrial Power Distribution and Illumination"
chapter: 7
key: "semester_6/industrial_power_distribution_and_illumination:production-room-27x45"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/production-room-27x45
site: BCT Engineering
---

# Production room 27 × 45 m twin 36 W tubes

Industrial Power Distribution and Illumination · Chapter 7: Illumination Design Principles

- Asked in: 2069 Bhadra, 8 marks
- Video: [ILLUMINATION ENGINEERING - PART - 11 - LUMEN OR FLUX METHOD OF  CALCULATION AND THREE PROBLEMS](https://www.youtube.com/watch?v=mB573i9S39A) — zenmurali Sachithanandam
- 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/production-room-27x45
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-7

## Question

A production room of a factory size of 27 m × 45 m is to be illuminated by twin fluorescent tube of 36W. Inside the room an average illumination of 150 lum/m² is to be provided on the working plane. The walls and ceiling are brightly painted. Calculate the no. of luminaries required to be fitted in the room. Draw single line diagram showing arrangement of luminaries, switches and distribution board. Decide the light sub-circuits and the supply is 3-phase 4 wire system. Assume suitable value for utilization and maintenance factor and lamp efficiency is 70lum/watt.

## Short answer

Lumen method N = 150×1215/(5040×0.7×0.8)=64.6, rounded to 66 twin 36 W luminaires, grid 11×6. Spacing 4.09 m and 4.5 m within 4.5 m limit, actual illuminance 153.3 lux.

## Answer

**Given:** room 45 m × 27 m, twin 36 W fluorescent (72 W per luminaire) at 70 lm/W, E = 150 lm/m², 3-phase 4-wire supply.
**Assumptions:** brightly painted walls and ceiling → **UF = 0.7**; **MF = 0.8**; Hm = 3 m, 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        = 45 × 27 = 1215 m²
φ per luminaire = 72 W × 70 lm/W = 5040 lm
Total lumens  = E × A / (UF × MF)
              = 150 × 1215 / (0.7 × 0.8)
              = 325446 lm
N             = 325446 / 5040 = 64.57
```

Rounding up and fitting a regular grid: **11 × 6 = 66 luminaires** (11 along the 45 m side, 6 along the 27 m side).

#### Arrangement and spacing

```text
Spacing along length = 45/11 = 4.09 m
Spacing along width  = 27/6 = 4.5 m
End spacing (to wall) = half of these
Max spacing = SHR × Hm = 1.5 × 3 = 4.5 m
4.09 m and 4.5 m ≤ 4.5 m  → OK
Actual E = 66 × 5040 × 0.7 × 0.8 / 1215
         = 153.3 lux (≥ 150 lux required)
```

#### Layout diagram

```text
Plan 45 m × 27 m (letter = phase of lamp circuit)
+-----------------------+
| R R R R R R R R Y Y Y |
| B B B B B B Y Y Y Y Y |
| B B R R R R R R R Y Y |
| B B B B B B Y Y Y Y Y |
| B R R R R R R R Y Y Y |
| B B B B B B B Y Y Y Y |
+-----------------------+
[DB]  [SB]  <- entrance: DB and switch board
       S S S S ... one switch per sub-circuit
```

Each light sub-circuit is 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 light sub-circuit carries at most 10 points and 800 W. Here one luminaire is 72 W, so up to 10 luminaires per sub-circuit. Number of circuits is made a multiple of 3 for phase balance.

| Circuit | Luminaires | Load (W) | Current (A) | Phase | MCB |
|---|---|---|---|---|---|
| L1 | 8 | 576 | 2.78 | R | 6 A |
| L2 | 8 | 576 | 2.78 | Y | 6 A |
| L3 | 8 | 576 | 2.78 | B | 6 A |
| L4 | 7 | 504 | 2.43 | R | 6 A |
| L5 | 7 | 504 | 2.43 | Y | 6 A |
| L6 | 7 | 504 | 2.43 | B | 6 A |
| L7 | 7 | 504 | 2.43 | R | 6 A |
| L8 | 7 | 504 | 2.43 | Y | 6 A |
| L9 | 7 | 504 | 2.43 | B | 6 A |

#### Distribution board design

| Phase | Circuits | Load (W) | Current (A) |
|---|---|---|---|
| R | L1, L4, L7 | 1584 | 7.65 |
| Y | L2, L5, L8 | 1584 | 7.65 |
| B | L3, L6, L9 | 1584 | 7.65 |

```text
Total connected load = 4752 W (lighting)
Most loaded phase    = 1584 W
Iph = 1584 / (230 × 0.9) = 7.65 A
Main MCB ≥ 1.25 × 7.65 = 9.57 A → 10 A TPN MCB
```

**DB specification:** 4-way-per-phase TPN DB (12 ways in all, spares included) with 10 A TPN MCB as incomer, 25 A 4-pole RCCB (100 mA) for earth-leakage protection, 6 A SP MCBs (B-curve) for light circuits. Incoming cable: 4-core 1.5 mm² Cu PVC (rating 21 A ≥ 10 A), plus 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   L2   L3   L4   L5  ...
```

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is a utilization factor of 0.7 and maintenance factor of 0.8 assumed here?
   A. They were given in the question
   B. Light walls/ceiling give reasonable utilization, and 0.8 allows for dirt and lamp ageing
   C. They are fixed values for all rooms
   D. Because the supply is 3-phase 4-wire
   Answer: B. The question omits UF and MF, so suitable values are assumed from the bright surface finish and normal upkeep. (Key point: Assumed factors)

2. With 150 lux, 1215 m^2, UF 0.7, MF 0.8 and 5040 lm per luminaire, what is N before rounding?
   A. About 32
   B. About 64.6
   C. About 129
   D. About 36
   Answer: B. 150 x 1215/(0.7 x 0.8) = 325446 lm, divided by 5040 lm gives about 64.57, rounded to a 11 x 6 = 66 grid. (Key point: Number of luminaires)

3. What limits the number of luminaires on one light sub-circuit?
   A. At most 10 points and 800 W per circuit
   B. At most 30 points per circuit
   C. Only one luminaire per circuit
   D. Limit set by the energy meter
   Answer: A. IS 732/NEA practice allows at most 10 points and 800 W per light sub-circuit; here 8 or 7 luminaires of 72 W are used. (Key point: Sub-circuit loading rule)

4. How is the load shared in the DB design?
   A. All 9 circuits on the R phase
   B. Circuits in multiples of 3 spread over R, Y, B, giving 1584 W per phase
   C. Neutral carries all the load
   D. Circuits are paired with one phase carrying double
   Answer: B. Nine circuits are spread three per phase so each phase carries 1584 W and 7.65 A, leading to a 10 A TPN incomer. (Key point: Phase balancing)

## Close variants set in other exams

- [Factory hall 55.75 m × 40.25 m LED lighting](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Afactory-hall-led-55x40) — Industrial Power Distribution and Illumination, chapter 7, asked 4×
- [Factory hall 15 × 12 m metal halide lighting](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Afactory-hall-metal-halide) — Industrial Power Distribution and Illumination, chapter 7, asked 4×
- [Hall 16 × 12 m LED lighting](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Ahall-16x12-led) — Industrial Power Distribution and Illumination, chapter 7, asked 4×

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