---
title: "Nichrome wire for 4.5 kW single-phase oven"
subject: "Industrial Power Distribution and Illumination"
chapter: 5
key: "semester_6/industrial_power_distribution_and_illumination:resistance-oven-nichrome-4-5kw"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/resistance-oven-nichrome-4-5kw
site: BCT Engineering
---

# Nichrome wire for 4.5 kW single-phase oven

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

- Asked in: 2068 Jestha (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/resistance-oven-nichrome-4-5kw
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-5

## Question

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.

## 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.

## Answer

**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

```text
R = V²/P = 200²/4500 = 8.889 Ω
```

#### Step 2: Heat dissipation per m²

```text
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

```text
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

```text
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.**

## Check yourself

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

1. What is the resistance of the element for 4.5 kW at 200 V?
   A. 8.889 Ω
   B. 0.1125 Ω
   C. 44.4 Ω
   D. 22.5 Ω
   Answer: A. R = V^2/P = 40000/4500 = 8.889 Ω. (Key point: Resistance)

2. How is the resistivity 42.5 μΩ-cm used in calculation?
   A. Converted to 42.5 x 10^-8 Ω-m
   B. Used directly as 42.5 Ω-m
   C. Converted to 42.5 x 10^-3 Ω-m
   D. Converted to 42.5 x 10^-6 Ω-cm
   Answer: A. 1 μΩ-cm = 10^-8 Ω-m, so rho = 42.5e-8 Ω-m. (Key point: Unit conversion)

3. What temperatures enter Stefan's law and what is k here?
   A. 1000 and 500 degrees C directly, k = 0.9
   B. Absolute 1273 K and 773 K, with e = 0.9 and k = 1, giving H about 116,812 W/m2
   C. Only 1273 K, k = 0.9
   D. 773 K and 500 K, k = 0
   Answer: B. T1 = 1273 K, T2 = 773 K, e = 0.9, k = 1: H = 5.72e4 x 0.9 x 2.2691 = 116,812 W/m2. (Key point: Heat dissipation)

4. How are d and l found?
   A. d^3 = (l d)/(l/d^2) = 7.465e-10 m^3 so d = 0.907 mm, l = 13.52 m
   B. d = R, l = P
   C. d^2 = l d x l/d^2
   D. d = 9 mm and l = 1.35 m
   Answer: A. Dividing l d = 0.012262 by l/d^2 = 1.6427e7 gives d^3, then d = 0.907 mm and l = 13.52 m. (Key point: Solving)

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