---
title: "Nickel-chrome strip for 30 kW furnace"
subject: "Industrial Power Distribution and Illumination"
chapter: 5
key: "semester_6/industrial_power_distribution_and_illumination:furnace-nichrome-strip-30kw"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/furnace-nichrome-strip-30kw
site: BCT Engineering
---

# Nickel-chrome strip for 30 kW furnace

Industrial Power Distribution and Illumination · Chapter 5: Electrification of Industrial Building

- Asked in: 2068 Bhadra (old course), 8 marks
- Video: [Module1- Problems on Design of Heating Elements](https://www.youtube.com/watch?v=GwhtwHPxPwY) — Sai Vidya Institute of Technology - SVIT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/furnace-nichrome-strip-30kw
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-5

## Question

A 30KW, 3 phase, 400V resistance furnace is to employ nickel-chrome strip 0.25cm thick for the 3 phase star connected heating elements. If the strip temperature is 1100°C and that of charge is to 700°C, calculate length and width of the strip per phase. The radiating efficiency is 0.6 and emissivity is 0.9. The specific resistance of the micron alloy is 1.03×10⁻⁶ Ωm. State any assumption made.

## Short answer

Per phase: 10 kW, 230.94 V, R=5.333 Ω. Radiated heat flux H=82,082 W/m². Solving R=ρl/(wt) and P=H×2wl gives width ≈ 2.17 mm and length ≈ 28.08 m per phase (with strip thickness 0.25 cm).

## Answer

**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

```text
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

```text
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

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

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

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

Multiplying and dividing (1) and (2):

```text
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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For the 30 kW, 400 V star-connected furnace, what are the per-phase power and resistance?
   A. 30 kW and 5.33 Ω
   B. 10 kW and 5.333 Ω (using Vph = 230.94 V)
   C. 10 kW and 16 Ω (using 400 V)
   D. 30 kW and 16 Ω
   Answer: B. P/phase = 30/3 = 10 kW; star phase voltage = 400/sqrt3 = 230.94 V, so R = Vph^2/P = 5.333 Ω. (Key point: Phase quantities)

2. In the heat dissipation formula, what temperatures are used?
   A. Celsius temperatures of strip and charge
   B. Absolute temperatures (K) of strip and charge, in H = 5.72e4 x e x k x [(T1/1000)^4 - (T2/1000)^4]
   C. Only the strip temperature
   D. Difference of Celsius temperatures
   Answer: B. Radiation uses absolute temperatures: T1 = 1373 K and T2 = 973 K, and e and k scale the heat. (Key point: Stefan's law)

3. Which pair of relations is solved for strip length l and width w?
   A. R = rho l/(w t) giving l/w = R t/rho, and P = H x 2wl
   B. R = rho w/(l t) and P = H x wl
   C. P = V I and R = V/I only
   D. R = rho l t/w and P = H x t l
   Answer: A. Resistance fixes l/w (12,945); radiating from two faces fixes w x l = P/(2H) = 0.0609 m^2. (Key point: Two equations)

4. What strip size results with t = 0.25 cm, and what does the answer suggest?
   A. w about 2.17 mm and l about 28.08 m; width below thickness suggests the intended t was 0.25 mm
   B. w about 28 mm and l about 2 m
   C. w about 6.86 mm and l about 8.88 m for t = 0.25 cm
   D. w about 2.17 m and l about 28 mm
   Answer: A. Solving gives w = 2.17 mm, l = 28.08 m. Since w < t is impractical, t = 0.25 mm is likely intended, giving w = 6.86 mm, l = 8.88 m. (Key point: Result and sanity)

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