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Street lighting a 2 km road with sodium lamps

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%.

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.

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

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

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:

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).

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):
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.
 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 %).

✓ Answer checked by Claude Opus 5.5 (high effort)

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