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Nichrome wire for 15 kW furnace

Short answer

Resistance R=220²/15000=3.227 Ω. Heat flux H=63,174 W/m². From R=ρl/(πd²/4) and P=Hπdl, solving yields diameter ≈ 3.12 mm and length ≈ 24.24 m.

Workout the size and length of nichrome wire required for a furnace of 15kW capacity operating at 220V. Assume working temperature of wire 1000°C, temperature of charge 600°C, radiating efficiency as 0.6 and emissivity as 0.9 and resistivity of nichrome as 1.016 × 10⁻⁶ Ω-m.

Method: the wire must (a) have the resistance that draws 15 kW at 220 V and (b) have enough surface to radiate 15 kW at the given temperature. Heat dissipated per unit surface (Stefan's law):

H = 5.72 × 10⁴ × e × k × [(T₁/1000)⁴ − (T₂/1000)⁴] W/m²

Assumption: single-phase supply, one continuous wire element.

Step 1: Resistance

R = V²/P = 220²/15000 = 3.2267 Ω

Step 2: Heat dissipation per m²

T₁ = 1000 + 273 = 1273 K,  T₂ = 600 + 273 = 873 K
(1.273)⁴ − (0.873)⁴ = 2.6261 − 0.5808 = 2.0453
H = 5.72 × 10⁴ × 0.9 × 0.6 × 2.0453 = 63,174 W/m²

Step 3: Two equations in d and l

R = ρl/(πd²/4)  → l/d² = πR/(4ρ)
              = π×3.2267/(4×1.016×10⁻⁶) = 2.4943×10⁶  ...(1)
P = H × πdl     → l × d = 15000/(π × 63174)
              = 0.075579 m²                 ...(2)

Step 4: Solve

Dividing (2) by (1): d³ = 0.075579 / (2.4943 × 10⁶)

d³ = 3.030 × 10⁻⁸ m³
d  = 3.118 × 10⁻³ m = 3.12 mm
l  = 0.075579 / 0.003118 = 24.24 m

Check: ρl/(πd²/4) = 1.016×10⁻⁶ × 24.24/(7.634×10⁻⁶) = 3.227 Ω ✓

Answer: diameter d ≈ 3.12 mm, length l ≈ 24.24 m of nichrome wire.

✓ Answer checked by Claude Opus 5.5 (high effort)

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