Heat Transfer maximum current a wire can carry

[Rick2015]

Homework Statement

A copper wire has a diameter of 204.3 mil (1 mil = 0.001 in) and a resistance per unit length of 8.31x10^-4 ohm/m. The wire is coated with plastic insulation with thermal conductivity k = 0.20 W/(m K). The upper temperature limit of the plastic is 400 K. The air surrounding the insulation is at 300 K with heat transfer coefficient h = 10 W/(m2 K).
Find the maximum current (A) that the wire can carry. Approximate answer: 200 A.

Homework Equations

r_cr = k/h A (using_r1) = 0.005m^2 k=0.2 w/mK T1 = 400K
q = hA(thetaT) A(using_r_cr) = .02m^2 h=10 w/m^2K Tinf = 300K

The Attempt at a Solution

I have been working on this problem for about an hour. I know that I need to use the critical insulation radius. I have solved for q" and for R using R= (ln(r/r1) / 2pi*r*h) + 1/(2pi*r*h)), but can figure out how to get the answer.
If someone can just give a starting point it would be very helpful.

ps. sorry the bad notation, I m new to the site!

Thanks!

 

[Simon Bridge]

Start with the physics: why would the wire have an upper limit to the current it can carry?

 

[Rick2015]

To protect the circuit, right?!

 

[Simon Bridge]

No... protecting the circuit is not the reason there is a maximum current... its the reason we need to know about it. Think physics, not intent. What happens to the wire when a current goes through it?

 

[Rick2015]

It heats the wire. Friction is involved but to this problem it is not take in consideration. The length of the wire determine the current flow.

 

[Chestermiller]

To solve this problem, you need to know the thickness of the insulation. If you don't know that, you will not be able to determine the maximum current consistent with a plastic insulation temperature less than 400K. Did you forget to mention the thickness of the insulation?

Chet

 

[Simon Bridge]

The wire heats up.
What determines the maximum the wire can heat up?

 

[Rick2015]

The thickness is the critical radius of insulation r_critical = k/h.

Heat = amp x volt

 

[Simon Bridge]

You mean the insulation thickness is r=k/h = 0.02m ?
Your equation for heat is incorrect... amp x volt (you mean current times voltage) because the units don't match. VI is power (Watts) while heat (Q) is energy (Joules).
But what we are interested is in the limiting case - what does the heat do to the wire? Why would there be a limit to that? What role does the insulation play in this process?
If you don't understand the physics you won't be able to use the equations properly.

 

[Chestermiller]

The thickness is the critical radius of insulation r_critical = k/h.

Heat = amp x volt

OK. What is the rate of radial heat flow rate Q per unit length of wire. given that the unknown current is I and the resistance of the wire is 8.31 x 10^-4 ohms/meter? Your answer should be in W/m.

Chet