---
title: "Nichrome wire for 15 kW furnace"
subject: "Industrial Power Distribution and Illumination"
chapter: 5
key: "semester_6/industrial_power_distribution_and_illumination:furnace-nichrome-wire-15kw"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/furnace-nichrome-wire-15kw
site: BCT Engineering
---

# Nichrome wire for 15 kW furnace

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

- Asked in: 2068 Magh (old course), 8 marks
- 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-wire-15kw
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-5

## Question

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.

## 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.

## Answer

**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

```text
R = V²/P = 220²/15000 = 3.2267 Ω
```

#### Step 2: Heat dissipation per m²

```text
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

```text
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)

```text
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.**

## Check yourself

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

1. What two conditions must the heating wire satisfy?
   A. Right resistance to draw 15 kW at 220 V, and enough surface area to radiate 15 kW
   B. Low resistance and low surface area
   C. High resistivity only
   D. Large diameter and short length only
   Answer: A. R = V^2/P sets the electrical need, and the radiating surface must dissipate the power at the given temperatures. (Key point: Two conditions)

2. What is the wire resistance for 15 kW at 220 V?
   A. 14.7 Ω
   B. 3.2267 Ω
   C. 0.068 Ω
   D. 48.4 Ω
   Answer: B. R = V^2/P = 220^2/15000 = 3.2267 Ω. (Key point: Resistance value)

3. Which equation expresses the radiation condition for a round wire?
   A. P = H x (pi d^2/4)
   B. P = H x pi d l
   C. P = H x l/d
   D. P = H x 2 d l
   Answer: B. The lateral surface of a wire is pi x d x l, so P = H pi d l, giving l x d = 0.0756 m^2. (Key point: Surface equation)

4. How are d and l obtained from the two equations?
   A. Dividing (l d) by (l/d^2) gives d^3 = 3.03e-8 m^3, so d = 3.12 mm and l = 24.24 m
   B. Adding the equations, giving d = 15 mm
   C. Setting d equal to l
   D. Multiplying them to give l = 3 m
   Answer: A. (l d)/(l/d^2) = d^3, so d = 3.118 mm and l = 0.0756/0.003118 = 24.24 m. (Key point: Solving for d)

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