---
title: "Factory hall 55.75 m × 40.25 m LED lighting"
subject: "Industrial Power Distribution and Illumination"
chapter: 7
key: "semester_6/industrial_power_distribution_and_illumination:factory-hall-led-55x40"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/factory-hall-led-55x40
site: BCT Engineering
---

# Factory hall 55.75 m × 40.25 m LED lighting

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

- Asked in: 2078 Chaitra, 12 marks
- 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/factory-hall-led-55x40
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-7

## Question

A factory hall size of 55.75m×40.25m is to be illuminated by LED panel of 30W. Inside hall, an average illumination of 250 lum/m² is to be provided on the working plane. The walls and ceiling are brightly painted. Calculate number of luminaries required to be fitted in the hall. Draw layout diagram showing arrangement of luminaries, switches, and distribution board. Decide the light sub-circuits and the supply is 400V, 3-phase, 4 wire system. Assume value for utilization and maintenance factor are 0.8 and 0.6 respectively, the lamp efficiency is 100 lum/watt.

## Short answer

390 LED panels of 30 W (3000 lm each) in a 26×15 grid, spacing 2.14 m × 2.68 m, achieving 250.3 lux with UF=0.8, MF=0.6, and 100 lm/W. Total lighting load 11.7 kW, supplied by a 25 A TPN MCB.

## Answer

**Given:** hall 55.75 m × 40.25 m, E = 250 lux (lm/m²), 30 W LED panel at 100 lm/W, UF = 0.8, MF = 0.6, supply 400 V, 3-phase, 4-wire.
**Assumptions:** mounting height 3 m above the working plane, maximum spacing-to-height ratio 1.5, LED driver 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        = 55.75 × 40.25 = 2243.938 m²
φ per luminaire = 30 W × 100 lm/W = 3000 lm
Total lumens  = E × A / (UF × MF)
              = 250 × 2243.938 / (0.8 × 0.6)
              = 1168717 lm
N             = 1168717 / 3000 = 389.57
```

Rounding up and fitting a regular grid: **26 × 15 = 390 luminaires** (26 along the 55.75 m side, 15 along the 40.25 m side).

#### Arrangement and spacing

```text
Spacing along length = 55.75/26 = 2.14 m
Spacing along width  = 40.25/15 = 2.68 m
End spacing (to wall) = half of these
Max spacing = SHR × Hm = 1.5 × 3 = 4.5 m
2.14 m and 2.68 m ≤ 4.5 m  → OK
Actual E = 390 × 3000 × 0.8 × 0.6 / 2243.938
         = 250.3 lux (≥ 250 lux required)
```

#### Layout diagram

```text
Plan 55.75 m × 40.25 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 |
| 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 |
| 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 |
|                      :                      |
| 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 |
| 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 |
| 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 |
+---------------------------------------------+
[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 30 W, so up to 10 luminaires per sub-circuit. Number of circuits is made a multiple of 3 for phase balance.

| Phase | Light circuits | Luminaires | Load (W) |
|---|---|---|---|
| R | 13 | 130 | 3900 |
| Y | 13 | 130 | 3900 |
| B | 13 | 130 | 3900 |

Largest light circuit: 300 W → I = 300/(230 × 0.9) = 1.45 A → 6 A MCB, 1.5 mm² Cu.

#### Distribution board design

| Phase | Circuits | Load (W) | Current (A) |
|---|---|---|---|
| R | 13 circuits | 3900 | 18.84 |
| Y | 13 circuits | 3900 | 18.84 |
| B | 13 circuits | 3900 | 18.84 |

```text
Total connected load = 11700 W (lighting)
Most loaded phase    = 3900 W
Iph = 3900 / (230 × 0.9) = 18.84 A
Main MCB ≥ 1.25 × 18.84 = 23.55 A → 25 A TPN MCB
```

**DB specification:** 16-way-per-phase TPN DB (48 ways in all, spares included) with 25 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 2.5 mm² Cu PVC (rating 27 A ≥ 25 A), plus earth continuity conductor.

```text
3-phase, 400/230 V, 4-wire supply
        |
   [Energy meter]
        |
   [25 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. What is the light output of one 30 W LED panel at 100 lm/W?
   A. 30 lm
   B. 300 lm
   C. 3000 lm
   D. 30000 lm
   Answer: C. phi = 30 W x 100 lm/W = 3000 lm. (Key point: Lumen per luminaire)

2. Which calculation gives N for this hall?
   A. N = 250 x 2243.938 x 0.8 x 0.6 / 3000
   B. N = 250 x 2243.938 / (0.8 x 0.6 x 3000) = 389.57
   C. N = 2243.938 / 3000
   D. N = 250 / (0.8 x 0.6)
   Answer: B. UF and MF are in the denominator: N = E x A / (phi x UF x MF) = 389.57, rounded to a 26 x 15 grid of 390. (Key point: Number of luminaires)

3. Why must the actual spacings (2.14 m and 2.68 m) be checked against 4.5 m?
   A. To ensure they do not exceed SHR x Hm, so illumination is uniform
   B. To ensure they exceed it
   C. To count the circuits
   D. To set the MCB rating
   Answer: A. Spacing must stay at or below the maximum SHR x Hm = 1.5 x 3 = 4.5 m for uniform illumination. (Key point: Spacing check)

4. How are the 390 luminaires divided among sub-circuits?
   A. All on the red phase
   B. Equally on R, Y and B (13 circuits and 130 luminaires each, 3900 W)
   C. Only on the neutral
   D. One per phase
   Answer: B. Circuits are a multiple of 3 and loaded equally across phases, 130 luminaires x 30 W = 3900 W per phase, with up to 10 per circuit. (Key point: Phase balance)

## Close variants set in other exams

- [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×
- [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×
- [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×

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