How does the rise in temperature of fuse wire depend upon its radius?

[Darshit Sharma]

Homework Statement

The question could be understood as if we have two fuse wires one of current rating 1 A and one of current rating 8A then what should be the ratio of their radius?

Relevant Equations

Q = mc delta t
Joule's law of heating

I tried the following:

We know, Q = mc (delta t)

but by Joule's law of heating, we also know that H = I^2 R t

Assuming that no heat is lost in the surroundings,
we get Q = H
=> mc (delta t) = I^2 R t
=> (delta t) = (I^2 R t)/(mc)

But m = V d = AL d
and R = (rho L) / A

=> (delta t) = (I^2 rho t) / (A^2 d)

Here we come across a surprising reveal that the rise in temperature doesn't depend on the length of the fuse wires.

Moving ahead, assuming a wire of circular cross-section ( as they generally are),

A = pi * r^2

=>(delta t) = (I^2 rho t) / (pi^2 * r^4 d)

The temperature rise should be inversely proportional to the tesseract of radius, i.e. r to the fourth power, which essentially implies that the ratio of the radius of two fuse wires one of current rating 1 A and one of current rating 8A should be *1:2*.

However, my science textbook not only does say that it should be "1:4" but also that the temperature rise should be inversely proportional to the cube of radius, i.e. r to the third power.

Who is correct, why and how?

 

[haruspex]

You have assumed the current is the same for both. I would have thought they were to be subjected to the same voltage.
Edit: you are right to do so.

Also, if you assume no heat is lost to the surroundings then they will both get arbitrarily hot over a long enough time. You need to consider how the radius allows heat to be dissipated.

 

[Darshit Sharma]

You have assumed the current is the same for both. I would have thought they were to be subjected to the same voltage.
Also, if you assume no heat is lost to the surroundings then they will both get arbitrarily hot over a long enough time. You need to consider how the radius allows heat to be dissipated.

How have I assumed the current same for both sir? Could you navigate me to the point where I committed this flaw?

I am not sure but while checking my equation I plugged in different current ratings for them.
And then solved further keeping the rise in temperature same for both the wires so as to get a same parameter to compare values.

"Also, if you assume no heat is lost to the surroundings then they will both get arbitrarily hot over a long enough time. You need to consider how the radius allows heat to be dissipated."
That's a textbook question so I could not think of a better way to progress further.


(Why do I feel that I must have not assumed the same rise in temperature for both the wires as they are different so they must melt at different temperature?)

What about a better derivation by you instead of handling this whole mess, sir?

 

[Darshit Sharma]

You have assumed the current is the same for both. I would have thought they were to be subjected to the same voltage.
Also, if you assume no heat is lost to the surroundings then they will both get arbitrarily hot over a long enough time. You need to consider how the radius allows heat to be dissipated.

Lemme try with voltages; just a second. well, then how will I come to another claim by my textbook that says that the rise in temperature is proportional to the square of the current rating?

 

[Lnewqban]

(Why do I feel that I must have not assumed the same rise in temperature for both the wires as they are different so they must melt at different temperature?)

The type of metal should be the same for both, in order to make a proper comparison.
In that case, the temperature of fusion should be the same for both.

One fuse will melt if the current reaches 1 amp, and the other will if the current reaches 8 amps.

What related subject are you studying at this time?
Heat transfer?

 

[haruspex]

How have I assumed the current same for both sir? Could you navigate me to the point where I committed this flaw?

You arrived at:

>(delta t) = (I^2 rho t) / (pi^2 * r^4 d)

and deduced that the temperature rise goes like $r^-4$. That is only true if no other variables in that expression depend on r, in particular, if I does not depend on r.
If instead you take V as fixed then you use $V^2/R$, meaning a higher resistance reduces the power.
But, having said all that, I withdraw my original comment. You are right to take I as fixed, and I apologise. I have edited my comment.

"Also, if you assume no heat is lost to the surroundings then they will both get arbitrarily hot over a long enough time. You need to consider how the radius allows heat to be dissipated."
That's a textbook question so I could not think of a better way to progress further.

Is the title of this thread an exact copy of the question as given to you? If not, please provide the exact text. Without that we cannot be sure whether you are to find the initial rate of rise of temperature (for which ignoring heat loss from the wire is fine) or how the radius affects the steady state temperature (which is what you really want to know for a fuse wire).
Taking heat loss into account is not difficult. Since the wire itself is a very good conductor, you can assume the whole wire is at the same temperature. The rate of loss will be proportional to the surface area and to the rise in temperature.

Edit: Seems to me this approach gives the book answer.

What about a better derivation by you instead of handling this whole mess, sir?

Don't get petulant. We are trying to help you.

 

[DaveE]

What about a better derivation by you instead of handling this whole mess, sir?

Because this is a homework forum, where we provide feedback and suggestions, not answers. It's your mess, not ours.

Perhaps you should review the forum rules:
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

 

[Darshit Sharma]

Is the title of this thread an exact copy of the question as given to you? If not, please provide the exact text. Without that, we cannot be sure whether you are to find the initial rate of rise of temperature (for which ignoring heat loss from the wire is fine) or how the radius affects the steady state temperature (which is what you want to know for a fuse wire).

Actually, in my book, it is just written that (delta t) is proportional to I^2/r^3
They just provided that I A example and that is it. However, out of curiosity, I did try to derive the relation and was struck at that point.
I must add my textbook images.

 

[Darshit Sharma]

Because this is a homework forum, where we provide feedback and suggestions, not answers. It's your mess, not ours.

Perhaps you should review the forum rules:
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

. Sorry sir. I do understand it is my mess.

 

[Darshit Sharma]

Taking heat loss into account is not difficult. Since the wire itself is a very good conductor, you can assume the whole wire is at the same temperature. The rate of loss will be proportional to the surface area and to the rise in temperature.

How can we do that? We are not taught that as yet.

I think that is the thing which I am missing out.

 

[haruspex]

How can we do that? We are not taught that as yet.

Since you have not been asked to show the inverse cube relationship, maybe you are not expected to be able to do this.
Suppose it reaches a steady state rise in temperature of $\Delta T$. As I wrote, the rate of loss of heat is proportional to that and to the surface area. This means that, again, the length does not matter. You only have to consider heat loss per unit length, which will be proportional to $r\Delta T$. Set that as being proportional to the rate of generation of heat per unit length that you found.

 

[Darshit Sharma]

Yes, I understood it now. Thanks, sir. So one r in the numerator cancels one r in the denominator and brings us to the book's supposed relation with inverse r cube.

 

[haruspex]

Yes, I understood it now. Thanks, sir. So one r in the numerator cancels one r in the denominator and brings us to the book's supposed relation with inverse r cube.

Not exactly. It would not cancel an r in your $r^4$, but increase it to $r^5$.

You had

(delta t) = (I^2 rho t) / (pi^2 * r^4 d)

I did not understand your derivation of that because I did not know what all your variables meant, but it seems to be based on a fatter fuse having greater thermal capacity so being slower to heat up. But with the view that what matters is the steady state temperature that becomes irrelevant. Your $A^2$ term now becomes just $A$, and we have that the rate of heat generation per unit length $\propto \frac{I^2}{r^2}$, and the rate of heat dissipation per unit length $\propto \Delta\theta r$, whence $\Delta\theta \propto \frac{I^2}{r^3}$.