--- title: "Workshop hall lighting and power distribution" subject: "Industrial Power Distribution and Illumination" chapter: 7 key: "semester_6/industrial_power_distribution_and_illumination:workshop-hall-lighting-power" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/workshop-hall-lighting-power site: BCT Engineering --- # Workshop hall lighting and power distribution Industrial Power Distribution and Illumination · Chapter 7: Illumination Design Principles - Asked in: 2079 Chaitra, 16 marks - Video: [Lecture#5, ET-233, Design a Given Lighting Scheme](https://www.youtube.com/watch?v=2YJJ2l7_rCk) — Kahuta Institute of Technology - 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/workshop-hall-lighting-power - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-7 ## Question 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. ## 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. ## Answer **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. ```text 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 ```text 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 ```text 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** | | | ```text 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 ```text 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 ```text 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 ```text +-------------------- 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 ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the number of luminaires 47 x 9 = 423 rather than 421.88? A. Because the lumen method gives exactly 423 B. Because the result is rounded up to a regular grid, and actual E is then 300.8 lux C. Because SHR forces 423 D. Because 423 is the number of circuits Answer: B. N = 421.88 is rounded up to a regular grid that fits the hall, and the check E = 300.8 lux meets the 300 lux requirement. (Key point: Number of luminaires) 2. How is the kVA of a 7.5 hp pump (eta 0.8, pf 0.8) found? A. 5595 W / 0.8 = 6993.75 W input, then / 0.8 = 8.742 kVA B. 7.5 x 746 x 0.8 x 0.8 C. 5595 x 0.8 = 4476 kVA D. 7.5 / 0.8 = 9.4 kVA Answer: A. Output is 5595 W; input = output / efficiency = 6993.75 W; kVA = input / pf = 8.742 kVA. (Key point: Pump kVA) 3. How is the power DB demand obtained from the loads? A. Sum of connected kVA without any factor B. ks x sum of (ku x kVA) = 0.8 x 60.24 = 48.19 kVA C. Sum of kVA divided by 2 D. Only the largest load Answer: B. Each load's demand is ku x kVA; the sum (60.24) is multiplied by the simultaneity factor ks = 0.8. 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