---
title: "Drawing hall 40 × 25 × 6 m, 90 lm/m² LED"
subject: "Industrial Power Distribution and Illumination"
chapter: 7
key: "semester_6/industrial_power_distribution_and_illumination:drawing-hall-40x25-90lux-led"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/drawing-hall-40x25-90lux-led
site: BCT Engineering
---

# Drawing hall 40 × 25 × 6 m, 90 lm/m² LED

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

- Asked in: 2072 Magh, 16 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/drawing-hall-40x25-90lux-led
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-7

## Question

A drawing hall 40 m × 25 m × 6 m high is to be illuminated LED lamps to an average lux is 90 lm/m² on a working plane 1 m above the floor. Estimate suitable number of lamps. Assume coefficient of utilization of 0.8 depreciation factor of 1.2 and spacing / height ratio of 1.2.

| Size of lamps | 45 W |
|---|---|
| Luminous efficiency (in lm/W) | 90 |

Draw layout diagram showing arrangement of luminaries, switches, power socket and design the distribution board, if the supply system is 3-φ, 4-wire system.

## Short answer

Needed 45 W LED luminaires: 35, arranged 7×5. Lumen method: MF=1/1.2=0.833; φ=4050 lm; required lumens = 90×1000×1.2/0.8 = 135000 lm; N = 33.33, rounded up for spacing limit. Six light circuits (6 A MCB) and six power circuits (20 A MCB); DB: 32 A TPN MCB.

## Answer

**Given:** hall 40 m × 25 m × 6 m, E = 90 lm/m² on a plane 1 m above floor, CU = 0.8, depreciation factor 1.2, SHR = 1.2, 45 W LED at 90 lm/W, 3-phase 4-wire supply.
**Assumptions:** luminaires mounted on the ceiling, so Hm = 6 − 1 = 5 m; 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
              = 90 × 1000 × 1.2 / 0.8
              = 135000 lm
N             = 135000 / 4050 = 33.33
```

Rounding up and fitting a regular grid: **7 × 5 = 35 luminaires** (7 along the 40 m side, 5 along the 25 m side).

#### Arrangement and spacing

```text
Spacing along length = 40/7 = 5.71 m
Spacing along width  = 25/5 = 5 m
End spacing (to wall) = half of these
Max spacing = SHR × Hm = 1.2 × 5 = 6 m
5.71 m and 5 m ≤ 6 m  → OK
Actual E = 35 × 4050 × 0.8 × 0.833 / 1000
         = 94.5 lux (≥ 90 lux required)
```

Here the spacing limit (SHR 1.2) decides the grid: 34 lamps are needed for lux, and 7 × 5 = 35 just satisfies spacing ≤ 6 m. If the lamps were suspended 1 m (Hm = 4 m), spacing ≤ 4.8 m would need a 9 × 6 = 54 grid, so ceiling mounting is chosen.

#### Layout diagram

```text
Plan 40 m × 25 m (letter = phase of lamp circuit)
+---------------+
| R R R R R R Y |
| B B Y Y Y Y Y |
| B B B B R R R |
| Y Y Y Y R R R |
| Y Y B B B B B |
+---------------+
[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 | 6 | 270 | 1.3 | R | 6 A |
| L2 | 6 | 270 | 1.3 | Y | 6 A |
| L3 | 6 | 270 | 1.3 | B | 6 A |
| L4 | 6 | 270 | 1.3 | R | 6 A |
| L5 | 6 | 270 | 1.3 | Y | 6 A |
| L6 | 5 | 225 | 1.09 | 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, P1, P4 | 4540 | 21.93 |
| Y | L2, L5, P2, P5 | 4540 | 21.93 |
| B | L3, L6, P3, P6 | 4495 | 21.71 |

```text
Connected load = 1575 W light + 12000 W power = 13575 W
Most loaded phase    = 4540 W
Iph = 4540 / (230 × 0.9) = 21.93 A
Main MCB ≥ 1.25 × 21.93 = 27.42 A → 32 A TPN MCB
```

**DB specification:** 6-way-per-phase TPN DB (18 ways in all, spares included) with 32 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 ≥ 32 A), plus earth continuity conductor.

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

## Check yourself

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

1. Total lumens = 90 x 1000 x 1.2 / 0.8 = 135000 with phi = 4050 lm. What does the lumen method alone give?
   A. 33.33, so 34 lamps
   B. 135000 / 45 = 3000 lamps
   C. 20 lamps
   D. 50 lamps
   Answer: A. 135000 / 4050 = 33.33, so 34 lamps are needed for the illuminance. (Key point: Number from lumens)

2. Why is 35 lamps (7 x 5) chosen when 34 meet the lux?
   A. 34 does not fit a regular grid that keeps spacing within SHR x Hm = 1.2 x 5 = 6 m; 7 x 5 gives 5.71 m and 5 m
   B. 35 is a standard pack size
   C. Spacing must exceed 6 m
   D. Because 34 is prime
   Answer: A. A regular grid is needed with spacing at most 6 m; 7 x 5 = 35 satisfies this. (Key point: Spacing limit decides grid)

3. Why is ceiling mounting (Hm = 5 m) preferred to suspending lamps 1 m (Hm = 4 m)?
   A. With Hm = 4 m the spacing limit is 4.8 m, which needs a 9 x 6 = 54 grid
   B. Lower mounting gives less light
   C. Suspension is not allowed in drawing halls
   D. Hm does not change spacing
   Answer: A. Smaller Hm reduces maximum spacing, so more lamps are needed; ceiling mounting needs only 35. (Key point: Effect of mounting height)

4. Light circuits are 6,6,6,6,6,5 luminaires. What results for phase loads?
   A. R and Y carry 4540 W, B carries 4495 W, so the design is nearly balanced and the incomer is 32 A
   B. All phases exactly equal
   C. Only R phase is loaded
   D. B phase carries 9000 W
   Answer: A. The 5-luminaire circuit makes B slightly lower; the worst phase 21.93 A gives 1.25 x 21.93 = 27.4 A, so 32 A. (Key point: Unbalanced phase loads)

## Close variants set in other exams

- [Drawing hall 40 × 25 × 6 m, 250 lux LED](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Adrawing-hall-40x25-250lux) — Industrial Power Distribution and Illumination, chapter 7, asked 3×

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