---
title: "Nichrome strip width for 30 kW 3-phase oven"
subject: "Utilization of Electrical Energy"
chapter: 5
key: "semester_7/utilization_of_electrical_energy:nichrome-strip-30kw-oven-width"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/nichrome-strip-30kw-oven-width
site: BCT Engineering
---

# Nichrome strip width for 30 kW 3-phase oven

Utilization of Electrical Energy · Chapter 5: Electric Heating

- Asked in: 2070 Asar
- 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-strip-30kw-oven-width
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-5

## Question

A 30 KW, 3 phase, 400 V resistance oven is to employ nickel-chrome strip 0.025 cm thick for a 3 phase star connected heating elements. If the wire temperature is to be 1100°C and that of charge is to be 700°C, estimate a suitable width for the strip. Assume radiating efficiency as 0.6 and emissivity as 0.9. The specific resistance of the nichrome-alloy is 10.3×10⁻⁶ Ωm. State any assumption made.

## Short answer

Assume radiation from both faces, star connection, and black-body charge at 700C. Radiative heat dissipation is about 82,082 W/m2; per-phase resistance is 5.333 ohm. Suitable strip width is about 21.7 mm with length about 2.81 m per phase.

## Answer

**Assumptions:** the strip radiates from both faces (edges neglected, surface $= 2wl$); the elements are star connected so each phase gets $V_L/\sqrt3$; the charge absorbs heat as a black body at 700°C; values are used as given ($t = 0.025$ cm, $\rho = 10.3\times10^{-6}\ \Omega$m).

**Given:** $P = 30$ kW, 400 V, 3-phase; $T_1 = 1373$ K, $T_2 = 973$ K; $k = 0.6$, $e = 0.9$; $t = 2.5\times10^{-4}$ m.

#### Per-phase values
$$
\begin{aligned}
P_{ph} &= \frac{30\,000}{3} = 10\,000\ \text{W}, \quad V_{ph} = \frac{400}{\sqrt3} = 230.94\ \text{V} \\
R &= \frac{230.94^2}{10\,000} = 5.333\ \Omega
\end{aligned}
$$

#### Heat dissipated per m²
$$
\begin{aligned}
H &= 5.72\,e\,k\left[\left(\frac{T_1}{100}\right)^4 - \left(\frac{T_2}{100}\right)^4\right] \\
&= 5.72 \times 0.9 \times 0.6 \times (13.73^4 - 9.73^4) \\
&= 3.0888 \times 26\,574 = 82\,082\ \text{W/m}^2
\end{aligned}
$$

#### Strip equations
$$
\begin{aligned}
R = \frac{\rho l}{w t} &\Rightarrow \frac{l}{w} = \frac{R t}{\rho} = \frac{5.333 \times 2.5\times10^{-4}}{10.3\times10^{-6}} = 129.45 \quad (1)\\
P_{ph} = 2 w l H &\Rightarrow w l = \frac{10\,000}{2 \times 82\,082} = 0.060915\ \text{m}^2 \quad (2)
\end{aligned}
$$

#### Solving
$$
\begin{aligned}
w^2 &= \frac{0.060915}{129.45} = 4.706\times10^{-4} \\
w &= 0.02169\ \text{m} = 21.7\ \text{mm} \\
l &= 129.45 \times 0.02169 = 2.81\ \text{m per phase}
\end{aligned}
$$

**Answer:** Suitable strip width $\approx 21.7$ mm (about 2.2 cm), length $\approx 2.81$ m per phase.

## Check yourself

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

1. For 30 kW, 400 V star, the resistance per phase is:
   A. 16 Ω
   B. 5.333 Ω
   C. 1.78 Ω
   D. 12 Ω
   Answer: B. P_ph = 10 kW, V_ph = 230.94 V, R = 230.94^2/10,000 = 5.333 Ω. (Key point: Per-phase resistance)

2. Why is H the same (82,082 W/m^2) as in the 40 kW problem?
   A. Temperatures, e and k are the same
   B. Power is the same
   C. Thickness is the same
   D. Resistivity is the same
   Answer: A. H depends only on T1, T2, e and k: 1373 K, 973 K, 0.9 and 0.6. (Key point: Heat per m^2)

3. Which assumption is used for the strip?
   A. Only one face radiates
   B. Delta connection
   C. Charge is at the wire temperature
   D. Both faces radiate (2wl), edges neglected
   Answer: D. Surface = 2wl and star connection gives V_ph = V_L/sqrt3. (Key point: Assumptions)

4. Width of the strip is about:
   A. 2.17 mm
   B. 217 mm
   C. 21.7 mm
   D. 12 mm
   Answer: C. l/w = 129.45 and wl = 0.060915 give w^2 = 4.706x10^-4, w = 21.7 mm. (Key point: Width)

## Close variants set in other exams

- [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×

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