Calculate number of turns for an electric heater

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    Electric Heater
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Homework Help Overview

The original poster is attempting to calculate the number of turns of nichrome wire needed to create an electric heater with a specified resistance of 40 Ω. The problem involves understanding the relationship between resistance, wire length, and cross-sectional area, with specific parameters provided such as wire diameter and cylinder radius.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for resistance and how to derive the length of wire needed from the resistance value. There are questions about the calculations leading to the expected number of turns and the discrepancies in results. Some participants explore the implications of using different values for specific resistance.

Discussion Status

The discussion is ongoing, with participants sharing their calculations and questioning the assumptions made regarding the specific resistance of the wire. There is a recognition of potential errors in the calculations, but no consensus has been reached on the correct approach or resolution.

Contextual Notes

Participants note that the specific resistance value may vary, which could affect the calculations. There is also mention of the need to clarify the relationship between wire length and the number of turns, as well as the implications of the diameter of the wire on the overall resistance.

MisterP
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Homework Statement


Need to calculate, how many turns of nichrome wire do I need on a cylinder to make a heater with resistance of 40 Ω.
Wire diameter = 1mm
Cylinder radius = 2.5cm
R = 40 Ω
Wire diameter = 1mm
Correct answer - 200 turns

Homework Equations


R = ρ*l/S
R = resistance
ρ = specific resistance (I guess) and it is 1.1 Ω⋅mm2/m
l = length of wire
S = area of wire "cross section"

The Attempt at a Solution


Area of wire = Π*r2 = 3.14 * 0.52 = 0.785mm2
So I used this formula, but get nowhere close to 200 turns. 40 = 1,1l/0,785...
 
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MisterP said:
40 = 1,1l/0,785...
Solving that will give you a length. What is it in turns?
 
haruspex said:
Solving that will give you a length. What is it in turns?
Yes, I know, but it does not add up to near "200" or 0,2 or something like that..
To calculate turns I did this: 3.14 * 5cm (2*r) = 15.7cm is for one turn.
So, 200 turns would be 3140cm.. where did I go wrong?

If I continue with "40 = 1,1l/0,785..." I get 28.57.. let`s just change it to 2857cm.. too short for answer..
 
Last edited:
MisterP said:
Yes, I know, but it does not add up to near "200" or 0,2 or something like that..
To calculate turns I did this: 3.14 * 5cm (2*r) = 15.7cm is for one turn.
So, 200 turns would be 3140cm.. where did I go wrong?

If I continue with "40 = 1,1l/0,785..." I get 28.57.. let`s just change it to 2857cm.. too short for answer..
The difference is about 10%, right?
Note that if you solve it algebraically first the πs cancel and you end up with 200/1.1. So to get 200, you have to take ρ as 1mm2/m.
Looking it up, I see it quoted as 1.1 to 1.5.
 
Last edited:

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