- Asked 3/20
- 2068 Bhadra (old course) · 12 marks
Department store 30 × 15 m twin 40 W fluorescent
Short answer
Using lumen method: total lumens required = (200×450)/(0.8×0.8)=140625 lm; luminaire output 4800 lm, N=140625/4800≈29.3, rounded to 32 luminaires in 8×4 grid. Six 6 A SP MCB light circuits balanced across phases, 6 A TPN MCB incomer.
A department store 30m ×15m is illuminated by twin 40W fluorescent luminaries of output 4800 lumens. The lamp being mounted at height of 3m from the work place, the average illumination required is 200lux. Calculate the number of luminaries required to be fitted in the department store, assuming the coefficient of utilization to be 0.8 and maintenance factor to be 0.8. Also design distribution board and layout the lamps, switches and sub-circuit. If the supply system is 3-phase, 400V, 50Hz.
Given: store 30 m × 15 m, twin 40 W fluorescent luminaire (80 W) giving 4800 lm, Hm = 3 m, E = 200 lux, CU = 0.8, MF = 0.8, supply 400 V, 3-phase, 50 Hz. Assumptions: SHR ≤ 1.5 (spacing ≤ 4.5 m); 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 = 30 × 15 = 450 m²
φ per luminaire = 4800 lm
Total lumens = E × A / (UF × MF)
= 200 × 450 / (0.8 × 0.8)
= 140625 lm
N = 140625 / 4800 = 29.3
Rounding up and fitting a regular grid: 8 × 4 = 32 luminaires (8 along the 30 m side, 4 along the 15 m side).
Arrangement and spacing
Spacing along length = 30/8 = 3.75 m
Spacing along width = 15/4 = 3.75 m
End spacing (to wall) = half of these
Max spacing = SHR × Hm = 1.5 × 3 = 4.5 m
3.75 m and 3.75 m ≤ 4.5 m → OK
Actual E = 32 × 4800 × 0.8 × 0.8 / 450
= 218.5 lux (≥ 200 lux required)
A 6 × 5 = 30 grid (5 m × 3 m) or 10 × 3 grid (3 m × 5 m) would put spacing above 4.5 m, so 8 × 4 = 32 is used.
Layout diagram
Plan 30 m × 15 m (letter = phase of lamp circuit)
+-----------------+
| R R R R R R Y Y |
| B B B B Y Y Y Y |
| B B B B B B R R |
| Y Y Y Y Y R R R |
+-----------------+
[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 80 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 | 480 | 2.32 | R | 6 A |
| L2 | 6 | 480 | 2.32 | Y | 6 A |
| L3 | 5 | 400 | 1.93 | B | 6 A |
| L4 | 5 | 400 | 1.93 | B | 6 A |
| L5 | 5 | 400 | 1.93 | R | 6 A |
| L6 | 5 | 400 | 1.93 | Y | 6 A |
Distribution board design
| Phase | Circuits | Load (W) | Current (A) |
|---|---|---|---|
| R | L1, L5 | 880 | 4.25 |
| Y | L2, L6 | 880 | 4.25 |
| B | L3, L4 | 800 | 3.86 |
Total connected load = 2560 W (lighting)
Most loaded phase = 880 W
Iph = 880 / (230 × 0.9) = 4.25 A
Main MCB ≥ 1.25 × 4.25 = 5.31 A → 6 A TPN MCB
DB specification: 4-way-per-phase TPN DB (12 ways in all, spares included) with 6 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 ≥ 6 A), plus earth continuity conductor.
3-phase, 400/230 V, 4-wire supply
|
[Energy meter]
|
[6 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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