--- title: "Street lighting a 2 km road with sodium lamps" subject: "Industrial Power Distribution and Illumination" chapter: 8 key: "semester_6/industrial_power_distribution_and_illumination:street-lighting-sodium-2km" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/street-lighting-sodium-2km site: BCT Engineering --- # Street lighting a 2 km road with sodium lamps Industrial Power Distribution and Illumination · Chapter 8: Outdoor Lighting Design - Asked in: 2076 Bhadra, 8 marks - Video: [Calculate Street Light Pole’s Distance, Fixture Watt, And Lighting Area](https://www.youtube.com/watch?v=-IZ_YOqcgfQ) — Engineering Infinity - 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/street-lighting-sodium-2km - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-8 ## Question A main road 2 km long and 7m wide is required to be illuminated by 125 watt sodium vapour lamps. The lamps are mounted on poles 9m high, so that the minimum level of illumination is 1.5 lux. Design a suitable street lighting scheme using 3.5 core underground cable feeder, if coefficient of utilization is 0.5 and efficiency of sodium vapour lamp is 85 lumen per watt. Assume length of span between lamp posts is 30m. ## Short answer 68 lamps of 125 W on 9 m poles at ~30 m spacing, single-side arrangement. Average illumination 25.3 lux, minimum midway 2.84 lux (>1.5 lux). Feed with 3.5-core 16 mm² Al cable from centre; voltage drop 3.2%. ## Answer **Given:** road length 2 km, width W = 7 m, 125 W SOX/HPSV lamps, efficacy 85 lm/W, pole height h = 9 m, span S = 30 m, CU = 0.5, minimum illumination 1.5 lux, feeder: 3.5-core underground cable. #### Step 1: Number of lamps ```text Spans = 2000 / 30 = 66.7 → 67 spans Actual spacing = 2000 / 67 = 29.85 m Poles (one side) = 67 + 1 = 68 lamps ``` For a 7 m road a **single-side** arrangement is enough. #### Step 2: Check average illumination ```text Lamp lumens F = 125 × 85 = 10 625 lm E_av = F × CU / (S × W) = 10 625 × 0.5 / (30 × 7) = 25.3 lux ``` #### Step 3: Check minimum illumination (midway between poles) Assume the lamp radiates uniformly: I = F / 4π = 10 625 / 4π = 845.5 cd. The midway point is 15 m from each pole: ```text d² = 9² + 15² = 306 E (one lamp) = I h / d³ = 845.5 × 9 / 306^1.5 = 1.42 lux E_min (two lamps) = 2 × 1.42 = 2.84 lux > 1.5 lux ✓ ``` So the scheme meets the minimum of 1.5 lux. (Directly below a lamp: E = 845.5/81 = 10.4 lux.) #### Step 4: Electrical load and cable Assume 15 W ballast loss per lamp, supply 400/230 V, 3-phase, pf 0.9 (with capacitor). ```text Total load P = 68 × (125 + 15) = 9520 W Line current I = 9520 / (√3 × 400 × 0.9) = 15.3 A ``` - Lamps are connected **phase-to-neutral**, taken in turn on the R, Y and B phases (≈ 23 lamps per phase), so the load is balanced and the half-size (0.5) core serves as the neutral. - Feed from the **middle** of the road, so each side carries about 7.6 A over 1 km. - For a uniformly distributed load, voltage drop = I × R × L/2. For 16 mm² Al (R ≈ 1.91 Ω/km): ```text ΔV = 7.6 × 1.91 × 0.5 = 7.3 V = 7.3 / 230 × 100 = 3.2 % (< 5 % allowed) ``` - Choose a **3.5-core, 16 mm² aluminium, PVC/XLPE armoured cable** laid underground, with a feeder pillar containing MCBs/fuses, a contactor and a **photocell/timer** for automatic switching. ```text Feeder pillar (middle, 3φ+N) ◄───── 1 km ─────┼───── 1 km ─────► o o o o o o │ o o o o o o R Y B R Y B R Y B R Y B ``` **Answer: 68 lamps of 125 W on 9 m poles at ~30 m spacing on one side; E_av ≈ 25.3 lux, E_min ≈ 2.84 lux (> 1.5 lux); 3.5-core 16 mm² Al cable fed from the centre (voltage drop ≈ 3.2 %).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a 2 km road with 30 m span (about 29.85 m actual), how many lamps are needed on one side? A. 67 lamps B. 68 lamps (67 spans + 1) C. 66 lamps D. 133 lamps Answer: B. Spans = 2000/30 rounded up to 67, and poles = spans + 1 = 68. (Key point: Number of lamps) 2. Which formula gives the average illumination used in the answer? A. E_av = F / (S x h) B. E_av = F x CU x W / S C. E_av = F x CU / (S x W) D. E_av = I h / d^3 Answer: C. Average lux = lamp lumens x CU divided by the area per lamp, S x W. (Key point: Lumen method for average illumination) 3. Where is the minimum illumination checked, and how? A. Midway between poles, adding the contribution of the two nearest lamps B. Directly below a lamp, using one lamp C. At the road edge below the pole, using the lumen method D. At the feeder pillar, using the voltage drop Answer: A. The darkest point is midway between poles (15 m from each), so two lamps' E = I h / d^3 are added: 2 x 1.42 = 2.84 lux > 1.5 lux. (Key point: Minimum illumination check) 4. Why are lamps taken on R, Y and B phases in turn and fed from the middle of the road? A. To increase the voltage drop on each side B. To make the lamps run at line voltage 400 V C. To avoid needing a photocell D. To balance the load so the half-size neutral core is enough, and to halve the length each side carries Answer: D. Balanced phases keep neutral current small, and feeding from the centre means each side carries about half the current over 1 km, cutting voltage drop. 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