---
title: "Nichrome wire size for 4.5 kW resistance oven"
subject: "Utilization of Electrical Energy"
chapter: 5
key: "semester_7/utilization_of_electrical_energy:nichrome-oven-4-5kw-wire-size"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/nichrome-oven-4-5kw-wire-size
site: BCT Engineering
---

# Nichrome wire size for 4.5 kW resistance oven

Utilization of Electrical Energy · Chapter 5: Electric Heating

- Asked in: 2081 Bhadra, 5 marks
- Video: [ELECTRICAL HEATING  - PART - 03 - THREE PROBLEMS IN DESIGN OF HEATING ELEMENT](https://www.youtube.com/watch?v=4tFLfZXrAeM) — zenmurali Sachithanandam
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/nichrome-oven-4-5kw-wire-size
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-5

## Question

A 4.5 kW, 200 V, and 1-ϕ resistance oven is to have nichrome wire heating elements. If the wire temperature is to be 1,000°C and that of the charge 500°C. Estimate the diameter and length of the wire. The resistivity of the nichrome alloy is 42.5 μΩ m. Assume the radiating efficiency and the emissivity of the element as 1.0 and 0.9, respectively.

## Short answer

Use R = V²/P = 8.889 ohm and Stefan-Boltzmann heat loss H = 1.168×10⁵ W/m². Solving R = 4ρl/(πd²) and P = πdlH gives diameter about 4.21 mm and length about 2.91 m.

## Answer

The wire must have the resistance that gives 4.5 kW at 200 V, and enough surface area to radiate that power at the given temperatures.

**Given:** $P = 4500$ W, $V = 200$ V, $T_1 = 1000 + 273 = 1273$ K, $T_2 = 500 + 273 = 773$ K, $\rho = 42.5 \times 10^{-6}\ \Omega$m (as given), $k = 1.0$, $e = 0.9$.

#### Heat dissipated per unit surface area (Stefan's law)
$$
\begin{aligned}
H &= 5.72\,e\,k\left[\left(\frac{T_1}{100}\right)^4 - \left(\frac{T_2}{100}\right)^4\right]\ \text{W/m}^2 \\
&= 5.72 \times 0.9 \times 1.0\,[12.73^4 - 7.73^4] \\
&= 5.148 \times (26\,261 - 3\,570) = 1.168 \times 10^5\ \text{W/m}^2
\end{aligned}
$$

#### Resistance condition
$$
R = \frac{V^2}{P} = \frac{200^2}{4500} = 8.889\ \Omega, \qquad R = \frac{4\rho l}{\pi d^2}
$$
$$
\frac{l}{d^2} = \frac{\pi R}{4\rho} = \frac{\pi \times 8.889}{4 \times 42.5\times10^{-6}} = 1.6427 \times 10^5 \quad (1)
$$

#### Heat-dissipation condition
$$
P = \pi d l H \Rightarrow d\,l = \frac{4500}{\pi \times 1.168 \times 10^5} = 0.012262 \quad (2)
$$

#### Solving
Dividing (2) by (1): $d^3 = \dfrac{0.012262}{1.6427 \times 10^5} = 7.465 \times 10^{-8}$
$$
\begin{aligned}
d &= 4.21 \times 10^{-3}\ \text{m} = 4.21\ \text{mm} \\
l &= \frac{0.012262}{4.21 \times 10^{-3}} = 2.91\ \text{m}
\end{aligned}
$$

**Answer:** Diameter $\approx 4.21$ mm, length $\approx 2.91$ m.

Note: the given resistivity is used as printed; real nichrome is about $1.1\ \mu\Omega$m, which would give a thinner, longer wire.

## Check yourself

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

1. Which two conditions fix the wire diameter and length?
   A. Only melting point and cost
   B. Resistivity and emissivity alone
   C. Resistance (from V^2/P) and radiated heat equal to the power
   D. Only voltage and length
   Answer: C. R = V^2/P = 8.889 Ω with R = 4ρl/(πd^2), and P = π d l H from Stefan's law. (Key point: Two design conditions)

2. What temperatures must be used in the radiation formula?
   A. Celsius values directly
   B. Kelvin: 1273 K and 773 K
   C. Wire temperature only
   D. Charge temperature only
   Answer: B. Stefan's law needs absolute temperature: T1 = 1273 K and T2 = 773 K, in H = 5.72 e k [(T1/100)^4 - (T2/100)^4]. (Key point: Temperatures in Stefan's law)

3. Dividing dl (from heat) by l/d^2 (from resistance) gives:
   A. d^2
   B. d^3
   C. d
   D. d^4
   Answer: B. (dl)/(l/d^2) = d^3, so d = 4.21 mm and l = 2.91 m. (Key point: Solving for d)

## Close variants set in other exams

- [Nichrome wire size for 5.5 kW resistance oven](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Anichrome-oven-5-5kw-wire-size) — Utilization of Electrical Energy, chapter 5, asked 2×
- [Heating element characteristics; 25 kW nichrome heater](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aheating-element-characteristics-nichrome-25kw) — Utilization of Electrical Energy, chapter 5, asked 2×

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