---
title: "Drawing hall 40 × 25 × 6 m, 250 lux LED"
subject: "Industrial Power Distribution and Illumination"
chapter: 7
key: "semester_6/industrial_power_distribution_and_illumination:drawing-hall-40x25-250lux"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/drawing-hall-40x25-250lux
site: BCT Engineering
---

# Drawing hall 40 × 25 × 6 m, 250 lux LED

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

- Asked in: 2073 Bhadra, 12 marks
- Video: [How to solve illumination problem's !! question from  VTU QP !! By Sharan Sir!!](https://www.youtube.com/watch?v=-pIazs-fjYs) — Knowledge center
- 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/drawing-hall-40x25-250lux
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-7

## Question

A drawing hall 40m×25m×6m high is to be illuminated with LED lamps to an average illumination of 250 lux on a working plane 1 m above the floor. Estimate suitable number, size and mounting height of lamps. Assume coefficient of utilization of 0.8 depreciation factor of 1.2 size of lamp is 45 watt and luminous efficiency of lamp is 90 lm/Watt.
Draw layout diagram showing arrangement of luminaries, switches, power socket and distribution board. Design 3-phase distribution system, if supply system is 400 V, 150 Hz.

## Short answer

96 LED luminaires (12×8 grid) at 4m mounting height provide 259 lux. Twelve 6A lighting sub-circuits (8 luminaires each) and six 20A power sub-circuits (2 sockets each) balanced across three phases. DB: 40A TPN MCB incomer, 40A 4P RCCB (100mA), 6A/20A SP MCBs, 4-core 6mm² Cu incoming cable.

## Answer

**Given:** hall 40 m × 25 m × 6 m, E = 250 lux on a plane 1 m above floor, CU = 0.8, depreciation factor 1.2, 45 W LED at 90 lm/W. The supply "400 V, 150 Hz" is read as 400 V, **50 Hz** (standard frequency in Nepal).
**Assumptions:** lamps suspended 1 m below ceiling → 5 m above floor, so **Hm = 4 m**; SHR ≤ 1.5; pf 0.9; 12 power sockets of 16 A (1 kW each).

#### Number of luminaires

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

```text
Area A        = 40 × 25 = 1000 m²
φ per luminaire = 45 W × 90 lm/W = 4050 lm
MF            = 1/DF = 1/1.2 = 0.833
Total lumens  = E × A × DF / UF
              = 250 × 1000 × 1.2 / 0.8
              = 375000 lm
N             = 375000 / 4050 = 92.59
```

Rounding up and fitting a regular grid: **12 × 8 = 96 luminaires** (12 along the 40 m side, 8 along the 25 m side).

#### Arrangement and spacing

```text
Spacing along length = 40/12 = 3.33 m
Spacing along width  = 25/8 = 3.13 m
End spacing (to wall) = half of these
Max spacing = SHR × Hm = 1.5 × 4 = 6 m
3.33 m and 3.13 m ≤ 6 m  → OK
Actual E = 96 × 4050 × 0.8 × 0.833 / 1000
         = 259.2 lux (≥ 250 lux required)
```

#### Layout diagram

```text
Plan 40 m × 25 m (letter = phase of lamp circuit)
+-------------------------+
| R R R R R R R R Y Y Y Y |
| B B B B B B B B Y Y Y Y |
| R R R R R R R R Y Y Y Y |
| B B B B B B B B Y Y Y Y |
| R R R R R R R R Y Y Y Y |
| B B B B B B B B Y Y Y Y |
| R R R R R R R R Y Y Y Y |
| B 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. Power sockets (16 A, 3-pin, with earth) are placed along the walls at about 1 m height, spaced roughly evenly.

#### Sub-circuits

Rule (IS 732 / NEA practice): a light sub-circuit carries at most 10 points and 800 W. Here one luminaire is 45 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 | 360 | 1.74 | R | 6 A |
| L2 | 8 | 360 | 1.74 | Y | 6 A |
| L3 | 8 | 360 | 1.74 | B | 6 A |
| L4 | 8 | 360 | 1.74 | R | 6 A |
| L5 | 8 | 360 | 1.74 | Y | 6 A |
| L6 | 8 | 360 | 1.74 | B | 6 A |
| L7 | 8 | 360 | 1.74 | R | 6 A |
| L8 | 8 | 360 | 1.74 | Y | 6 A |
| L9 | 8 | 360 | 1.74 | B | 6 A |
| L10 | 8 | 360 | 1.74 | R | 6 A |
| L11 | 8 | 360 | 1.74 | Y | 6 A |
| L12 | 8 | 360 | 1.74 | B | 6 A |

Power sub-circuits (assumed): 12 sockets of 16 A, each taken as 1 kW, at most 2 sockets per power sub-circuit (IS 732), 2.5 mm² Cu wire, 20 A MCB.

| Circuit | Sockets | Load (W) | Phase | MCB |
|---|---|---|---|---|
| P1 | 2 | 2000 | R | 20 A |
| P2 | 2 | 2000 | Y | 20 A |
| P3 | 2 | 2000 | B | 20 A |
| P4 | 2 | 2000 | R | 20 A |
| P5 | 2 | 2000 | Y | 20 A |
| P6 | 2 | 2000 | B | 20 A |

#### Distribution board design

| Phase | Circuits | Load (W) | Current (A) |
|---|---|---|---|
| R | L1, L4, L7, L10, P1, P4 | 5440 | 26.28 |
| Y | L2, L5, L8, L11, P2, P5 | 5440 | 26.28 |
| B | L3, L6, L9, L12, P3, P6 | 5440 | 26.28 |

```text
Connected load = 4320 W light + 12000 W power = 16320 W
Most loaded phase    = 5440 W
Iph = 5440 / (230 × 0.9) = 26.28 A
Main MCB ≥ 1.25 × 26.28 = 32.85 A → 40 A TPN MCB
```

**DB specification:** 8-way-per-phase TPN DB (24 ways in all, spares included) with 40 A TPN MCB as incomer, 40 A 4-pole RCCB (100 mA) for earth-leakage protection, 6 A SP MCBs (B-curve) for light circuits and 20 A SP MCBs for power circuits. Incoming cable: 4-core 6 mm² Cu PVC (rating 45 A ≥ 40 A), plus earth continuity conductor.

```text
3-phase, 400/230 V, 4-wire supply
        |
   [Energy meter]
        |
   [40 A TPN MCB]
        |
   [40 A 4P RCCB]
        |
==R====Y====B====N== busbars
  |    |    |    |    |
  L1   L2   L3   P1   P2  ...
```

**Result:** 96 LED lamps of **45 W**, at **5 m above floor (Hm = 4 m)**, 12 × 8 grid.

## Check yourself

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

1. The question states a 400 V, 150 Hz supply. How does the answer treat this?
   A. Uses 150 Hz in all current calculations
   B. Reads it as 50 Hz, the standard frequency in Nepal
   C. Designs a special 150 Hz distribution board
   D. Divides lumens by 3
   Answer: B. 150 Hz is taken as a misprint of the standard 50 Hz; frequency does not enter the lighting calculation. (Key point: Reading the 150 Hz typo)

2. Lamps hang 1 m below the 6 m ceiling and the plane is 1 m above the floor. What is Hm?
   A. 6 m
   B. 5 m
   C. 4 m
   D. 1 m
   Answer: C. Lamps are at 5 m above floor; minus 1 m working plane gives Hm = 4 m, so max spacing is 1.5 x 4 = 6 m. (Key point: Mounting height Hm)

3. Total lumens = 250 x 1000 x 1.2 / 0.8 = 375000, phi = 45 x 90 = 4050 lm. What is chosen?
   A. 93 lamps
   B. 96 lamps in a 12 x 8 grid, giving 259 lux
   C. 100 lamps
   D. 375000 / 45 = 8333 lamps
   Answer: B. 375000 / 4050 = 92.59, rounded to the regular grid 12 x 8 = 96. (Key point: Number of lamps)

4. Most loaded phase is 5440 W, 26.28 A at pf 0.9. Which incomer is chosen?
   A. 32 A, since 26.28 is below 32
   B. 40 A TPN MCB, since 1.25 x 26.28 = 32.85 A needs the next size up
   C. 25 A
   D. 63 A
   Answer: B. The incomer must be at least 32.85 A, and 32 A is too small, so 40 A is selected. (Key point: DB incomer rating)

## Close variants set in other exams

- [Drawing hall 40 × 25 × 6 m, 90 lm/m² LED](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Adrawing-hall-40x25-90lux-led) — Industrial Power Distribution and Illumination, chapter 7, asked 2×
- [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×

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