---
title: "Heating element characteristics; 25 kW nichrome heater"
subject: "Utilization of Electrical Energy"
chapter: 5
key: "semester_7/utilization_of_electrical_energy:heating-element-characteristics-nichrome-25kw"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/heating-element-characteristics-nichrome-25kw
site: BCT Engineering
---

# Heating element characteristics; 25 kW nichrome heater

Utilization of Electrical Energy · Chapter 5: Electric Heating

- Asked in: 2079 Bhadra, 3+5 marks
- 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/heating-element-characteristics-nichrome-25kw
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-5

## Question

With example, write about the characteristics of heating element? A 25 kW, 240 V, 50 Hz, single phase resistance heater is to have nichrome wire as a heating element. If the wire temperature is to be 1400°C and that of the charge is 500°C. Estimate the diameter and length of the wire. The resistivity of nichrome is 42.5 μΩ-m. Assume the radiating efficiency and the emissivity of the element as 0.9 and 0.85 respectively.

## Short answer

Heating elements should have high resistivity, high melting point, low temperature coefficient, oxidation resistance, strength, and low cost. For the 25 kW, 240 V heater, H = 3.272×10⁵ W/m²; solving R = 2.304 ohm with P = πdlH gives d ≈ 8.30 mm and l ≈ 2.93 m.

## Answer

#### Characteristics of heating elements
A heating element produces heat by $I^2R$ loss, so its material should have:
1. **High resistivity** — a short, compact element gives the needed resistance. Example: nichrome ($\approx 1.1\ \mu\Omega$m) is about 60 times copper.
2. **High melting point** — to work at high temperature. Example: tungsten (3400°C) for very high-temperature furnaces.
3. **Low temperature coefficient of resistance** — power stays steady as the element heats. Example: nichrome changes very little from cold to hot.
4. **Resistance to oxidation** — no scaling at working temperature. Example: nichrome forms a protective chromium-oxide layer; molybdenum must be used in inert gas.
5. **Mechanical strength and ductility** — can be drawn into wire or strip and survives thermal cycling.
6. **Low cost.**

Common elements: nichrome (up to 1150°C), Kanthal (up to 1300°C), silicon carbide rods (up to 1450°C).

#### Numerical: 25 kW, 240 V nichrome heater
**Given:** $T_1 = 1400 + 273 = 1673$ K, $T_2 = 500 + 273 = 773$ K, $\rho = 42.5 \times 10^{-6}\ \Omega$m (as given), $k = 0.9$, $e = 0.85$.

Heat dissipated per m²:
$$
\begin{aligned}
H &= 5.72\,e\,k\left[\left(\frac{T_1}{100}\right)^4 - \left(\frac{T_2}{100}\right)^4\right] \\
&= 5.72 \times 0.85 \times 0.9 \times (16.73^4 - 7.73^4) \\
&= 4.3758 \times (78\,340 - 3\,570) = 3.272 \times 10^5\ \text{W/m}^2
\end{aligned}
$$

Resistance condition:
$$
R = \frac{240^2}{25\,000} = 2.304\ \Omega, \qquad \frac{l}{d^2} = \frac{\pi R}{4\rho} = \frac{\pi \times 2.304}{4 \times 42.5\times10^{-6}} = 42\,578 \quad (1)
$$

Surface condition:
$$
P = \pi d l H \Rightarrow d\,l = \frac{25\,000}{\pi \times 3.272\times10^5} = 0.024322 \quad (2)
$$

Solving:
$$
\begin{aligned}
d^3 &= \frac{0.024322}{42\,578} = 5.712 \times 10^{-7} \\
d &= 8.30 \times 10^{-3}\ \text{m} = 8.30\ \text{mm} \\
l &= \frac{0.024322}{8.30 \times 10^{-3}} = 2.93\ \text{m}
\end{aligned}
$$

**Answer:** Diameter $\approx 8.30$ mm, length $\approx 2.93$ m (using the resistivity as printed).

## Check yourself

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

1. Why does nichrome resist oxidation at working temperature?
   A. It is used only in vacuum
   B. It has a very low melting point
   C. It has zero resistivity
   D. It has a protective chromium-oxide layer
   Answer: D. The chromium-oxide layer protects against scaling, whereas molybdenum needs inert gas. (Key point: Oxidation resistance example)

2. In the 25 kW problem, the radiating efficiency k and emissivity e are:
   A. k = 0.85, e = 0.9
   B. k = 1, e = 1
   C. k = 0.9, e = 0.85
   D. k = 0.5, e = 0.5
   Answer: C. Given radiating efficiency 0.9 and emissivity 0.85; H = 5.72 x 0.85 x 0.9 x [...]. (Key point: Stefan's law inputs)

3. Required resistance for 25 kW at 240 V is:
   A. 23.04 Ω
   B. 2.304 Ω
   C. 0.1 Ω
   D. 9.6 Ω
   Answer: B. R = V^2/P = 240^2/25,000 = 2.304 Ω. (Key point: Resistance requirement)

4. The result of the numerical is:
   A. d about 8.30 mm, l about 2.93 m
   B. d about 0.83 mm, l about 29 m
   C. d about 30 mm, l about 1 m
   D. d about 2 mm, l about 5 m
   Answer: A. d^3 = 5.712x10^-7 gives d = 8.30 mm, and l = 0.024322/d = 2.93 m. (Key point: Final size)

## Close variants set in other exams

- [Nichrome wire size for 4.5 kW resistance oven](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Anichrome-oven-4-5kw-wire-size) — Utilization of Electrical Energy, chapter 5, asked 3×

## Related in Utilization of Electrical Energy

- [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×
- [Nichrome strip width for 40 kW 3-phase oven](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Anichrome-strip-40kw-oven-width) — Utilization of Electrical Energy, chapter 5, asked 2×
- [Nichrome strip width for 30 kW 3-phase oven](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Anichrome-strip-30kw-oven-width) — Utilization of Electrical Energy, chapter 5, asked 2×

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