- Asked 4/20
- 2069 Bhadra · 8 marks
Production room 27 × 45 m twin 36 W tubes
Short answer
Lumen method N = 150×1215/(5040×0.7×0.8)=64.6, rounded to 66 twin 36 W luminaires, grid 11×6. Spacing 4.09 m and 4.5 m within 4.5 m limit, actual illuminance 153.3 lux.
A production room of a factory size of 27 m × 45 m is to be illuminated by twin fluorescent tube of 36W. Inside the room an average illumination of 150 lum/m² is to be provided on the working plane. The walls and ceiling are brightly painted. Calculate the no. of luminaries required to be fitted in the room. Draw single line diagram showing arrangement of luminaries, switches and distribution board. Decide the light sub-circuits and the supply is 3-phase 4 wire system. Assume suitable value for utilization and maintenance factor and lamp efficiency is 70lum/watt.
Given: room 45 m × 27 m, twin 36 W fluorescent (72 W per luminaire) at 70 lm/W, E = 150 lm/m², 3-phase 4-wire supply. Assumptions: brightly painted walls and ceiling → UF = 0.7; MF = 0.8; Hm = 3 m, SHR ≤ 1.5; electronic ballasts, pf 0.9.
Number of luminaires
Lumen method: N = (E × A) / (φ × UF × MF), where φ is the output of one luminaire.
Area A = 45 × 27 = 1215 m²
φ per luminaire = 72 W × 70 lm/W = 5040 lm
Total lumens = E × A / (UF × MF)
= 150 × 1215 / (0.7 × 0.8)
= 325446 lm
N = 325446 / 5040 = 64.57
Rounding up and fitting a regular grid: 11 × 6 = 66 luminaires (11 along the 45 m side, 6 along the 27 m side).
Arrangement and spacing
Spacing along length = 45/11 = 4.09 m
Spacing along width = 27/6 = 4.5 m
End spacing (to wall) = half of these
Max spacing = SHR × Hm = 1.5 × 3 = 4.5 m
4.09 m and 4.5 m ≤ 4.5 m → OK
Actual E = 66 × 5040 × 0.7 × 0.8 / 1215
= 153.3 lux (≥ 150 lux required)
Layout diagram
Plan 45 m × 27 m (letter = phase of lamp circuit)
+-----------------------+
| R R R R R R R R Y Y Y |
| B B B B B B Y Y Y Y Y |
| B B R R R R R R R Y Y |
| B B B B B B Y Y Y Y Y |
| B R R R R R R R Y Y Y |
| 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.
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 |
| L7 | 7 | 504 | 2.43 | R | 6 A |
| L8 | 7 | 504 | 2.43 | Y | 6 A |
| L9 | 7 | 504 | 2.43 | B | 6 A |
Distribution board design
| Phase | Circuits | Load (W) | Current (A) |
|---|---|---|---|
| R | L1, L4, L7 | 1584 | 7.65 |
| Y | L2, L5, L8 | 1584 | 7.65 |
| B | L3, L6, L9 | 1584 | 7.65 |
Total connected load = 4752 W (lighting)
Most loaded phase = 1584 W
Iph = 1584 / (230 × 0.9) = 7.65 A
Main MCB ≥ 1.25 × 7.65 = 9.57 A → 10 A TPN MCB
DB specification: 4-way-per-phase TPN DB (12 ways in all, spares included) with 10 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 ≥ 10 A), plus earth continuity conductor.
3-phase, 400/230 V, 4-wire supply
|
[Energy meter]
|
[10 A TPN MCB]
|
[25 A 4P RCCB]
|
==R====Y====B====N== busbars
| | | | |
L1 L2 L3 L4 L5 ...
✓ Answer checked by Claude Opus 5.5 (high effort)
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