---
title: "Conference hall 2×36 W CFL lighting"
subject: "Industrial Power Distribution and Illumination"
chapter: 7
key: "semester_6/industrial_power_distribution_and_illumination:conference-hall-cfl"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/conference-hall-cfl
site: BCT Engineering
---

# Conference hall 2×36 W CFL lighting

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

- Asked in: 2072 Asoj, 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/conference-hall-cfl
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-7

## Question

In a conference hall 20 m× 40 m × 5.5 m is to be illuminated using 2×36 W CFL lamps. Inside hall an average illumination of 108 lumen/m² is to be provided on the working plane. Calculate the number of lamp set required to be fitted in the hall. Decide light and power sub-circuit. Draw layout diagram showing arrangement of lamp set, switches and power socket. Design the distribution board to supply lighting load and power load, the supply system is 3 phase 4 wire system and efficiency of lamp is 60 lm/W.

## Short answer

Lumen method: needed lumens = 108×800/(0.6×0.8) = 180000 lm. With 4320 lm per 2×36W CFL set, N = 41.67 → 45 sets arranged 9×5. Six light circuits (6 A MCB) and six power circuits (20 A MCB) on a TPN DB with 32 A incomer.

## Answer

**Given:** hall 40 m × 20 m × 5.5 m, 2 × 36 W CFL set (72 W per set) at 60 lm/W, E = 108 lm/m², 3-phase 4-wire supply.
**Assumptions:** UF and MF not given, so **UF = 0.6** (medium-light room surfaces) and **MF = 0.8** (clean room); working plane 0.8 m and sets suspended 0.5 m below ceiling → Hm = 5.5 − 0.5 − 0.8 = 4.2 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 × 20 = 800 m²
φ per luminaire = 72 W × 60 lm/W = 4320 lm
Total lumens  = E × A / (UF × MF)
              = 108 × 800 / (0.6 × 0.8)
              = 180000 lm
N             = 180000 / 4320 = 41.67
```

Rounding up and fitting a regular grid: **9 × 5 = 45 luminaires** (9 along the 40 m side, 5 along the 20 m side).

#### Arrangement and spacing

```text
Spacing along length = 40/9 = 4.44 m
Spacing along width  = 20/5 = 4 m
End spacing (to wall) = half of these
Max spacing = SHR × Hm = 1.5 × 4.2 = 6.3 m
4.44 m and 4 m ≤ 6.3 m  → OK
Actual E = 45 × 4320 × 0.6 × 0.8 / 800
         = 116.6 lux (≥ 108 lux required)
```

#### Layout diagram

```text
Plan 40 m × 20 m (letter = phase of lamp circuit)
+-------------------+
| R R R R R R R R Y |
| B B Y Y Y Y Y Y Y |
| B B B B B B R R R |
| Y Y Y Y Y R R R R |
| Y Y B B 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 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 |

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 | 5080 | 24.54 |
| Y | L2, L5, P2, P5 | 5080 | 24.54 |
| B | L3, L6, P3, P6 | 5080 | 24.54 |

```text
Connected load = 3240 W light + 12000 W power = 15240 W
Most loaded phase    = 5080 W
Iph = 5080 / (230 × 0.9) = 24.54 A
Main MCB ≥ 1.25 × 24.54 = 30.68 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. A 2 x 36 W CFL set at 60 lm/W gives what output per set?
   A. 36 x 60 = 2160 lm
   B. 72 x 60 = 4320 lm
   C. 72 / 60 = 1.2 lm
   D. 36 x 2 / 60 = 1.2 lm
   Answer: B. A set has 72 W, so lumens = 72 x 60 = 4320 lm. (Key point: Output of one lamp set)

2. UF and MF are not given. What does the answer do?
   A. Assumes UF = 0.6 and MF = 0.8 and states them
   B. Takes both as 1
   C. Leaves the answer in terms of UF and MF
   D. Uses UF = 0.8 and MF = 0.6 without stating
   Answer: A. Missing factors are assumed from typical values for the room type and stated. (Key point: Assumed UF and MF)

3. Hall is 5.5 m high, sets hang 0.5 m below ceiling, plane is 0.8 m. What is Hm?
   A. 5.5 m
   B. 4.2 m
   C. 5.0 m
   D. 4.7 m
   Answer: B. Hm = 5.5 - 0.5 - 0.8 = 4.2 m, which gives a maximum spacing of 1.5 x 4.2 = 6.3 m. (Key point: Mounting height)

4. Why are 6 light circuits used (8 and 7 sets per circuit) rather than 5?
   A. Each set is 72 W and sets must be a multiple of 3 circuits for phase balance, within 10 points per circuit
   B. Because 45 sets must be split in 6 equal parts of 8
   C. Because 800 W / 72 W = 6
   D. Because there are 6 sockets
   Answer: A. Circuits are a multiple of 3 for balance; 45 sets over 6 circuits gives 8 or 7 per circuit, under the 10-point limit. (Key point: Sub-circuit loading)

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