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Nichrome wire for 4.5 kW single-phase oven

Short answer

Resistance R=200²/4500=8.889 Ω. Heat flux H=116,812 W/m². From R=ρl/(πd²/4) and P=Hπdl, solving gives diameter ≈ 0.907 mm and length ≈ 13.52 m.

A 4.5 kW, 200V, single-phase resistance oven has nichrome wire heating elements. If the wire temperature is to be 1000°C and that of the material is 500°C, estimate the diameter and length of the wire. The resistivity of nichrome alloy is 42.5 μΩ-cm. Assume the radiating efficiency and emissivity of the element is 1 and 0.9 respectively.

Method: the wire must give the right resistance and also radiate 4.5 kW from its surface at the given temperatures. Stefan's law gives heat dissipated per m²:

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

Data: P = 4.5 kW, V = 200 V, ρ = 42.5 μΩ-cm = 42.5 × 10⁻⁸ Ω-m, k = 1, e = 0.9.

Step 1: Resistance

R = V²/P = 200²/4500 = 8.889 Ω

Step 2: Heat dissipation per m²

T₁ = 1000 + 273 = 1273 K,  T₂ = 500 + 273 = 773 K
(1.273)⁴ − (0.773)⁴ = 2.6261 − 0.3570 = 2.2691
H = 5.72 × 10⁴ × 0.9 × 1 × 2.2691 = 116,812 W/m²

Step 3: Equations

R = ρl/(πd²/4) → l/d² = πR/(4ρ)
             = π×8.889/(4×42.5×10⁻⁸) = 1.6427×10⁷  ...(1)
P = H × πdl    → l d = 4500/(π × 116812)
             = 0.012262 m²                ...(2)

Step 4: Solve

d³ = 0.012262 / 1.6427×10⁷ = 7.465 × 10⁻¹⁰ m³
d  = 9.07 × 10⁻⁴ m = 0.907 mm
l  = 0.012262 / 9.071×10⁻⁴ = 13.52 m

Check: ρl/(πd²/4) = 42.5×10⁻⁸ × 13.52 / (6.463×10⁻⁷) = 8.89 Ω ✓

Answer: diameter ≈ 0.907 mm and length ≈ 13.52 m of nichrome wire.

✓ Answer checked by Claude Opus 5.5 (high effort)

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