---
title: "Process room 30 × 15 × 5 m fluorescent lighting"
subject: "Industrial Power Distribution and Illumination"
chapter: 7
key: "semester_6/industrial_power_distribution_and_illumination:process-room-30x15-fluorescent"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/process-room-30x15-fluorescent
site: BCT Engineering
---

# Process room 30 × 15 × 5 m fluorescent lighting

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

- Asked in: 2071 Magh, 8 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-30x15-fluorescent
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-7

## Question

A process room measuring 30 m × 15 m × 5 m is to be provided with illumination of 200 lux. Assuming the coefficient of utilization and maintenance factors are 0.8 and 0.7 respectively. Calculate the number of luminaries for the process room in an industry, illuminated by twin 40 watt fluorescent lamps. The efficiency of lamp is 60 lum/watt and distance between working plane and mounting height of luminaries is 3 m. Draw a neat sketch showing the arrangement of lamps and switches with light sub-circuits.

## Short answer

Using the lumen method N = E×A/(φ×CU×MF) with φ=80 W×60 lm/W=4800 lm, A=450 m², CU=0.8, MF=0.7 gives 33.5, rounded to 36 luminaires. Grid 9×4, spacing 3.33 m×3.75 m, actual illuminance 215 lux.

## Answer

**Given:** room 30 m × 15 m × 5 m, E = 200 lux, CU = 0.8, MF = 0.7, twin 40 W fluorescent (80 W per luminaire) at 60 lm/W, Hm = 3 m.
**Assumptions:** SHR ≤ 1.5 (spacing ≤ 4.5 m); electronic ballast, pf 0.9, ballast loss neglected; 3-phase 4-wire supply for the DB.

#### Number of luminaires

Lumen method: N = (E × A) / (φ × UF × MF), where φ is the output of one luminaire.

```text
Area A        = 30 × 15 = 450 m²
φ per luminaire = 80 W × 60 lm/W = 4800 lm
Total lumens  = E × A / (UF × MF)
              = 200 × 450 / (0.8 × 0.7)
              = 160714 lm
N             = 160714 / 4800 = 33.48
```

Rounding up and fitting a regular grid: **9 × 4 = 36 luminaires** (9 along the 30 m side, 4 along the 15 m side).

#### Arrangement and spacing

```text
Spacing along length = 30/9 = 3.33 m
Spacing along width  = 15/4 = 3.75 m
End spacing (to wall) = half of these
Max spacing = SHR × Hm = 1.5 × 3 = 4.5 m
3.33 m and 3.75 m ≤ 4.5 m  → OK
Actual E = 36 × 4800 × 0.8 × 0.7 / 450
         = 215 lux (≥ 200 lux required)
```

#### Layout diagram

```text
Plan 30 m × 15 m (letter = phase of lamp circuit)
+-------------------+
| R R R R R R Y Y Y |
| B B B B B B Y Y Y |
| R R R R R R Y Y Y |
| B B B B B B 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 80 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 | 6 | 480 | 2.32 | R | 6 A |
| L2 | 6 | 480 | 2.32 | Y | 6 A |
| L3 | 6 | 480 | 2.32 | B | 6 A |
| L4 | 6 | 480 | 2.32 | R | 6 A |
| L5 | 6 | 480 | 2.32 | Y | 6 A |
| L6 | 6 | 480 | 2.32 | B | 6 A |

#### Distribution board design

| Phase | Circuits | Load (W) | Current (A) |
|---|---|---|---|
| R | L1, L4 | 960 | 4.64 |
| Y | L2, L5 | 960 | 4.64 |
| B | L3, L6 | 960 | 4.64 |

```text
Total connected load = 2880 W (lighting)
Most loaded phase    = 960 W
Iph = 960 / (230 × 0.9) = 4.64 A
Main MCB ≥ 1.25 × 4.64 = 5.8 A → 6 A TPN MCB
```

**DB specification:** 4-way-per-phase TPN DB (12 ways in all, spares included) with 6 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 ≥ 6 A), plus earth continuity conductor.

```text
3-phase, 400/230 V, 4-wire supply
        |
   [Energy meter]
        |
   [6 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. A twin 40 W fluorescent luminaire at 60 lm/W gives what flux?
   A. 40 x 60 = 2400 lm
   B. 80 x 60 = 4800 lm
   C. 80 / 60 lm
   D. 60 x 2 lm
   Answer: B. The luminaire has 2 x 40 = 80 W, so flux is 80 x 60 = 4800 lm. (Key point: Output per twin luminaire)

2. Total lumens = 200 x 450 / (0.8 x 0.7) = 160714. What is fitted?
   A. 33
   B. 36 luminaires in a 9 x 4 grid, giving 215 lux
   C. 40 in 10 x 4
   D. 30 in 6 x 5
   Answer: B. 160714 / 4800 = 33.48, rounded to the regular grid 9 x 4 = 36; spacings 3.33 m and 3.75 m are below 4.5 m. (Key point: Number of luminaires)

3. 36 luminaires of 80 W are split into 6 circuits of 6. What is the load per phase and incomer?
   A. 960 W per phase, 4.64 A at pf 0.9, so a 6 A TPN MCB
   B. 2880 W per phase, so 16 A
   C. 480 W per phase, so 2 A
   D. 960 W per phase, so 32 A
   Answer: A. Two 480 W circuits per phase give 960 W; 960/(230 x 0.9) = 4.64 A; 1.25 x 4.64 = 5.8 A, so 6 A. (Key point: Sub-circuits and phase load)

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