- Asked 2/20
- 2068 Magh (old course) · 12 marks
Process room 40 × 20 m twin 40 W fluorescent
Short answer
Assuming 200 lux, lumen method: luminaire output=80 W×60 lm/W=4800 lm, MF=1/1.2=0.833, total lumens=(200×800×1.2)/0.75=256000 lm; N=256000/4800≈53.33, rounded to 54 luminaires in 9×6 grid. Six sub-circuits, balanced, 10 A TPN MCB incomer.
Calculate the number of luminaries for the process room in an industry. The size of room is 40m × 20m illuminated by twin 40 Watt fluorescent lamp. The efficiency of lamp is 60 lumen/watt and distance between working plane and mounting height of luminaries is 3m and coefficient of utilization and depreciation factor are 0.75 and 1.2 respectively. Draw neat sketch showing the arrangement of lamps, sub-circuit and switches. Design the distribution board to supply lighting loads, if the supply system is 3 phase, 400V and 50Hz.
Given: room 40 m × 20 m, twin 40 W fluorescent (80 W) at 60 lm/W, Hm = 3 m, CU = 0.75, depreciation factor 1.2, supply 400 V, 3-phase, 50 Hz. Assumption: the required illumination is not given, so E = 200 lux (general process/industrial work) is taken; 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 = 40 × 20 = 800 m²
φ per luminaire = 80 W × 60 lm/W = 4800 lm
MF = 1/DF = 1/1.2 = 0.833
Total lumens = E × A × DF / UF
= 200 × 800 × 1.2 / 0.75
= 256000 lm
N = 256000 / 4800 = 53.33
Rounding up and fitting a regular grid: 9 × 6 = 54 luminaires (9 along the 40 m side, 6 along the 20 m side).
Arrangement and spacing
Spacing along length = 40/9 = 4.44 m
Spacing along width = 20/6 = 3.33 m
End spacing (to wall) = half of these
Max spacing = SHR × Hm = 1.5 × 3 = 4.5 m
4.44 m and 3.33 m ≤ 4.5 m → OK
Actual E = 54 × 4800 × 0.75 × 0.833 / 800
= 202.5 lux (≥ 200 lux required)
Layout diagram
Plan 40 m × 20 m (letter = phase of lamp circuit)
+-------------------+
| R R R R R R R R R | <- L1
| Y Y Y Y Y Y Y Y Y | <- L2
| B B B B B B B B B | <- L3
| R R R R R R R R R | <- L4
| Y Y Y Y Y Y Y Y Y | <- L5
| B B B B B B B B B | <- L6
+-------------------+
[DB] [SB] <- entrance: DB and switch board
S1 S2 S3 S4 S5 S6 (one switch per sub-circuit)
Each row of 9 luminaires forms one sub-circuit, 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 lighting sub-circuit carries at most 10 points and 800 W. One luminaire here is 80 W, so up to 10 luminaires can go on one sub-circuit. The number of circuits is made a multiple of 3 for phase balance.
| Circuit | Luminaires | Load (W) | Current (A) | Phase | MCB |
|---|---|---|---|---|---|
| L1 | 9 | 720 | 3.48 | R | 6 A |
| L2 | 9 | 720 | 3.48 | Y | 6 A |
| L3 | 9 | 720 | 3.48 | B | 6 A |
| L4 | 9 | 720 | 3.48 | R | 6 A |
| L5 | 9 | 720 | 3.48 | Y | 6 A |
| L6 | 9 | 720 | 3.48 | B | 6 A |
Distribution board design
| Phase | Circuits | Load (W) | Current (A) |
|---|---|---|---|
| R | L1, L4 | 1440 | 6.96 |
| Y | L2, L5 | 1440 | 6.96 |
| B | L3, L6 | 1440 | 6.96 |
Total connected load = 4320 W (lighting)
Most loaded phase = 1440 W
Iph = 1440 / (230 × 0.9) = 6.96 A
Main MCB ≥ 1.25 × 6.96 = 8.7 A → 10 A TPN MCB
DB specification: 4-way-per-phase TPN DB (12 ways in all, spares included) with a 10 A TPN MCB as incomer, a 25 A 4-pole RCCB (100 mA) for earth-leakage protection, and 6 A SP MCBs (B-curve) for the light circuits. Incoming cable: 4-core 1.5 mm² Cu PVC (rating 21 A ≥ 10 A), plus an 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 L4 L2 L5 L3 L6 (6 A SP MCBs)
✓ Answer checked by Claude Opus 5.5 (high effort)
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