Nickel-chrome strip for 30 kW furnace

Assumptions: the strip radiates from both faces (edges neglected); power per phase is supplied at phase voltage of the star connection; Stefan's law for heat dissipation:

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

where e = emissivity, k = radiating efficiency, T₁ and T₂ are absolute temperatures of strip and charge.

Electrical data per phase

Power per phase P   = 30/3       = 10 kW
Phase voltage Vph   = 400/√3     = 230.94 V
Resistance per phase R = Vph²/P  = 230.94²/10000 = 5.333 Ω

Heat dissipated per m² of strip surface

T₁ = 1100 + 273 = 1373 K,  T₂ = 700 + 273 = 973 K
(1.373)⁴ − (0.973)⁴ = 3.5537 − 0.8963 = 2.6574
H = 5.72 × 10⁴ × 0.9 × 0.6 × 2.6574
  = 82,082 W/m²

Equations for width w and length l (thickness t = 0.25 cm)

Resistance: R = ρl/(w t) → l/w = R t/ρ

l/w = 5.333 × 0.0025 / 1.03×10⁻⁶ = 12,945        ...(1)

Heat balance (two faces): P = H × 2 w l

w × l = 10000 / (2 × 82082) = 0.06091 m²           ...(2)

Multiplying and dividing (1) and (2):

w² = 0.06091 / 12945  → w = 2.17 × 10⁻³ m = 2.17 mm
l  = 12945 × w        → l = 28.08 m

Answer: width ≈ 2.17 mm and length ≈ 28.08 m of strip per phase (with t = 0.25 cm as given).

Note: a width smaller than the thickness is not a practical strip, which suggests the intended thickness is 0.25 mm (0.025 cm). With t = 0.25 mm: l/w = 1294.5, w = √(0.06091/1294.5) = 6.86 mm and l = 8.88 m per phase.