- 2079 Chaitra · 16 marks
Workshop hall lighting and power distribution
Short answer
423 twin-tube 40 W LED luminaires (4000 lm each) in a 47×9 grid, spacing 2.55 m × 3.33 m, providing 300.8 lux. Total lighting load 16.92 kW. Power demand 48.19 kVA, transformer size 83.74 kVA with 25% growth margin.
A hall 120 m×30 m is to be illuminated by twin tube 40W LED fitting. Required illumination for the space is 300 lux, maintenance factor 0.8, and utilization factor 0.8, luminous efficiency of 2×20W LED Tube is 100 lumens/watt. Calculate no. of luminaries required to be fitted. Draw the layout for the Lighting and Power sub-circuit. This hall space will house a Workshop having the following loads:
| Types of Loads | Workshop |
|---|---|
| Lathe Machine (5kVA) | 3 |
| Pedestal Drill (2 kVA) | 2 |
| Ventilation (2.5kVA) | 2 |
| Compressor (15kVA) | 1 |
| Oven (15 kVA) | 1 |
| 7.5 hp pump with η = 80%, p.f = 0.8 | 2 |
Draw the Single Line Diagram and DBs for the power distribution of above loads. Use separate DB for power loads and Lights load. Assuming suitable ku and ks, and with the load calculation, determine the size of transformer and CBs required this workshop.
Given: hall 120 m × 30 m, E = 300 lux, UF = 0.8, MF = 0.8, twin-tube LED fitting 2 × 20 W = 40 W at 100 lm/W. Assumptions: mounting height 3 m above work plane, SHR ≤ 1.5, pf 0.9 for LED, 3-phase 400/230 V supply, all motors and machines 3-phase.
Number of luminaires
Lumen method: N = (E × A) / (φ × UF × MF), where φ is the output of one luminaire.
Area A = 120 × 30 = 3600 m²
φ per luminaire = 40 W × 100 lm/W = 4000 lm
Total lumens = E × A / (UF × MF)
= 300 × 3600 / (0.8 × 0.8)
= 1687500 lm
N = 1687500 / 4000 = 421.88
Rounding up and fitting a regular grid: 47 × 9 = 423 luminaires (47 along the 120 m side, 9 along the 30 m side).
Arrangement and spacing
Spacing along length = 120/47 = 2.55 m
Spacing along width = 30/9 = 3.33 m
End spacing (to wall) = half of these
Max spacing = SHR × Hm = 1.5 × 3 = 4.5 m
2.55 m and 3.33 m ≤ 4.5 m → OK
Actual E = 423 × 4000 × 0.8 × 0.8 / 3600
= 300.8 lux (≥ 300 lux required)
Layout diagram
Plan 120 m × 30 m (o = luminaire)
+---------------------------------------------+
| o o o o o o o o o o ... o o o o o o o o o o |
| o o o o o o o o o o ... o o o o o o o o o o |
| o o o o o o o o o o ... o o o o o o o o o o |
| o o o o o o o o o o ... o o o o o o o o o o |
| o o o o o o o o o o ... o o o o o o o o o o |
| o o o o o o o o o o ... o o o o o o o o o o |
| o o o o o o o o o o ... o o o o o o o o o o |
| o o o o o o o o o o ... o o o o o o o o o o |
| o o o o o o o o o o ... o o o o o o o o o o |
+---------------------------------------------+
[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 40 W, so up to 10 luminaires per sub-circuit. Number of circuits is made a multiple of 3 for phase balance.
| Phase | Light circuits | Luminaires | Load (W) |
|---|---|---|---|
| R | 15 | 141 | 5640 |
| Y | 15 | 141 | 5640 |
| B | 15 | 141 | 5640 |
Largest light circuit: 400 W → I = 400/(230 × 0.9) = 1.93 A → 6 A MCB, 1.5 mm² Cu.
Power load calculation
Pump rating: 7.5 hp × 746 = 5595 W output; input = 5595/0.8 = 6993.75 W; kVA = 6993.75/0.8 = 8.742 kVA.
Assumed utilisation factor ku: 0.75 for machine tools, 0.8 for compressor and pumps, 1.0 for oven and ventilation. CB rating ≥ 1.25 × full-load current, D-curve for motors (high starting current), C-curve for the oven.
| Load | kVA each | No. | Connected kVA | ku | Demand kVA | I each (A) | CB (each) |
|---|---|---|---|---|---|---|---|
| Lathe machine | 5 | 3 | 15 | 0.75 | 11.25 | 7.22 | 10 A TP MCB (D) |
| Pedestal drill | 2 | 2 | 4 | 0.75 | 3 | 2.89 | 6 A TP MCB (D) |
| Ventilation fan | 2.5 | 2 | 5 | 1 | 5 | 3.61 | 6 A TP MCB (D) |
| Compressor | 15 | 1 | 15 | 0.8 | 12 | 21.65 | 32 A TP MCB (D) |
| Oven | 15 | 1 | 15 | 1 | 15 | 21.65 | 32 A TP MCB (C) |
| 7.5 hp pump | 8.742 | 2 | 17.48 | 0.8 | 13.99 | 12.62 | 16 A TP MCB (D) |
| Total | 71.48 | 60.24 |
PDB demand = ks × Σ(ku × kVA) = 0.8 × 60.24 = 48.19 kVA
Lighting load = 16920 W / 0.9 pf = 18.8 kVA
Total max demand = 48.19 + 18.8 = 66.99 kVA
With 25 % future growth: 1.25 × 66.99 = 83.74 kVA
Transformer: standard 100 kVA, 11/0.4 kV, Dyn11 (full-load LV current = 100000/(√3 × 400) = 144.3 A).
Breakers and DBs
PDB current = 48.19 kVA/(√3×400 V) = 69.56 A
→ 1.25I = 86.95 A
→ 100 A TP MCCB for PDB feeder
LDB current = 27.25 A per phase → 1.25I = 34.06 A
→ 40 A TPN MCB + 40 A 4P RCCB (100 mA)
Main LT breaker (transformer secondary 144.3 A)
→ 160 A TPN MCCB
- Power DB (PDB): 12-way TP DB with 100 A TP MCCB incomer and outgoing TP MCBs as in the table (3 × lathe, 2 × drill, 2 × ventilation, compressor, oven, 2 × pump = 11 ways + 1 spare).
- Lighting DB (LDB): TPN DB, 16 ways per phase, 45 light circuits of 9–10 luminaires (6 A SP MCB each), 15 per phase, with 40 A TPN MCB + RCCB incomer.
Single line diagram
11 kV supply
|
[HT fuse/VCB]
|
(( 100 kVA, 11/0.4 kV ))
|
[160 A TPN MCCB] Main LT panel
|
====+==========+==== LT busbar
| |
[100 A MCCB] [40 A TPN MCB+RCCB]
| |
PDB LDB
| | | | | | ...
L1 L2 L3 ... L1..L45 light
(lathes, circuits
drills, fans, (6 A each)
comp, oven,
pumps)
Layout of power and light sub-circuits
+-------------------- 120 m --------------------+
| o o o o o o o o ... 47 luminaires per row |
| (9 rows; 45 circuits of 9–10 lamps, R-Y-B) |
| |
| [Lathe][Lathe][Lathe] [Drill][Drill] [Oven] | 30 m
| |
| [Fan] [Compressor] [Pump][Pump] [Fan]|
+--[LDB]-[PDB]--- entrance ---------------------+
Power circuits: radial TP cables PDB → each machine
✓ Answer checked by Claude Opus 5.5 (high effort)
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