Lighting design and a replacement heater element for a motor repair workshop
A 12 m × 8 m × 3.5 m repair workshop for electric motors is to be lit for fine winding work. The client asks for 300 lux on the bench top, 0.8 m above the floor. The proposal is to use 100 W LED high-bay luminaires rated 10,000 lm each, suspended 0.3 m below the ceiling. The room has light-coloured walls, and a utilisation factor of 0.5 and a maintenance factor of 0.8 have been agreed. Lighting is single-phase 230 V. Separately, the workshop's 1,500 W, 230 V soldering bench heater has failed and the owner has a spare nichrome coil whose resistance at working temperature is 35 Ω. He wants to fit that coil. In the evening the supply in that feeder falls to 205 V.
1. Calculate the number of high-bay luminaires needed and propose a layout. Check the spacing-to-height ratio and the connected lighting current.
Read the worked answer
Area: 12 × 8 = 96 m².
Mounting height above the work plane: 3.5 − 0.3 (suspension) − 0.8 (bench) = 2.4 m.
Lumen method: N = (E × A) / (Φ × UF × MF) = (300 × 96) / (10,000 × 0.5 × 0.8) = 28,800 / 4,000 = 7.2, so 8 luminaires.
Layout: 8 fittings as 4 rows of 2 across the 12 m × 8 m plan gives about 3 m spacing along the 12 m side and 4 m across the 8 m side. Spacing-to-height ratio = 4 / 2.4 ≈ 1.67, which is within the 1.5–2.0 normally accepted for high-bay industrial fittings. If the client wants better uniformity for fine work, 9 fittings in a 3 × 3 grid at 4 m × 2.7 m spacing is the safer choice, because 8 fittings leave the corners darker.
Load: 8 × 100 W = 800 W, i.e. 800 / 230 ≈ 3.5 A - fine on a 6 A lighting MCB in 1.5 mm² cable, provided the lighting is not shared with socket outlets. Leave a spare way in the DB for future fittings.
2. The owner fits the 35 Ω nichrome coil in place of the failed 1,500 W element. What power will it actually deliver at 230 V and at 205 V, what current will it draw, and what do you check before energising?
Read the worked answer
At 230 V: P = V² / R = 230² / 35 = 52,900 / 35 ≈ 1,512 W, current = 230 / 35 ≈ 6.6 A.
At 205 V: P = 205² / 35 = 42,025 / 35 ≈ 1,201 W, current = 205 / 35 ≈ 5.9 A.
So the coil is a fair match for a 1.5 kW heater, but at evening voltage it delivers about 20 % less heat and will feel weak. That is a supply problem, not an element problem - do not 'fix' it by fitting a lower-resistance coil, which would overload the circuit.
Checks before energising:
- Element terminals clean and tight; the coil must not touch the metal case or the reflector.
- Insulation resistance of the heater to earth, at 500 V, at least 1 MΩ - a new element that has absorbed moisture from storage often fails this and needs drying out.
- The supply cable and switch rated for at least 6.6 A: 1.5 mm² cable with a 10 A MCB is right here; a 6 A MCB at 6.6 A is marginal and may nuisance-trip.
- The heater is a fixed appliance in a workshop; a 30 mA RCCB on the circuit and a proper earth connection to the metal frame are mandatory, and the frame earth continuity must be proved with the meter.
- Confirm the coil is actually rated for the working temperature - a coil from a different appliance can be dimensionally right and thermally wrong.
3. After installation, a light meter shows an average of 255 lux on the benches instead of 300, with a minimum of 150 lux. What do you check, and what do you recommend?
Read the worked answer
First, confirm the measurement: meter calibrated, cosine correction used, readings taken at bench height with the workshop in its normal working state, and the supply voltage at the lighting DB noted - a 205 V evening supply will pull LED output down several percent.
Then check the assumptions used in the design:
- Maintenance factor: if the fittings and lenses are already dusty from motor work, the real MF is below 0.8. Cleaning may recover 10–15 %.
- Utilisation factor: 0.5 assumed light-coloured walls; if the walls are dark or the benches are dark, the real UF is lower.
- Lumen output: cheap '10,000 lm' high-bays often deliver less, and ageing or poor-quality drivers reduce it further.
Uniformity: minimum/ average = 150 / 255 ≈ 0.59, below the 0.7 normally wanted for fine work - the layout, not just the quantity, is wrong. Corners are under-lit.
Recommendation: clean and re-measure first; if it is still short, add one row of fittings (from 8 to 12) to raise both the average and the uniformity, or move to a 3 × 3 layout with the fittings brought in from the walls. Record the measured values and hand them to the client with the completion report, so the basis of the design is in writing.