# Stuck on a conductivity question - Conductor in a metal crucuible

1. Jun 20, 2012

### Hazzattack

Hey guys,

I'm hoping someone can help me out, as i have an exam tomorrow and have no began panicing as i've found a question i can't seem to get to the end of...

Here is the question itself;

A large crucible contains a metal of thermal conductivity, κ = 12.5 Watts. m-1.K-1, electrical resistivity, ρ = 5 × 10-7 Ω.m, and melting point 70°C. A vertical electrical conductor ends in a hemisphere of radius 10 mm which is just embedded in the surface of the metal. If the crucible is maintained at 50 °C, what is the greatest current that can flow from the contact to the crucible without melting taking place? You may ignore heat losses through the conductor and the atmosphere and that all dimensions of the crucible are large compared to the radius of the conductor.

My thinking was;

As the current flows out it heats the metal by ohmic resistance.
The heat generated flows out to infinity (the crucible is very
large) by conduction along a temperature gradient.
To solve the problem you i assume you need to set up the equations for heat
generation and flow in a radial element (a thin hemispherical
shell).

Hence,

I started by expressing my heat flow as followed (Q) (k = conductivity of material);

dQ/dt = k A dT/dx

I then evaluated both dQ/dt and dT/dr and the points r and r+dr

The left hand side for dQ/dt gives you the heat flow dQ(dr)/dt - this i reconginised was the power.. which can be expressed as P = I^2 * R and R can be changed for (rho * L)/A
This meant i could take out several constants.. including I and would now be evaluated at points r/A(r)

I continued to try and evaluate points to get to a point where i can actually solve for I.. Struggling to say the least. And some advice and guidance would be so appreciated!

Thanks guys... hope you can follow my thinking.

2. Jun 20, 2012

### Staff: Mentor

A sketch of your system would make things easier. Can you assume a perfect spherical system?

I would begin with the electric flow, as this is independent of your temperature here (otherwise κ should include that dependence).

The electric current I is the same at every radius, but the differential resistance at radius r drops with 1/A: dR/dr=ρ/A=ρ/(4 pi r^2).
Heating power P is now related to the current via dP/dr=I^2 dR/dr. A simple integration from the contact outwards gives you P(r), which is the total heating power which has to be conducted at radius r.
This is equivalent to your dQ/dt (r). Using this, you can evaluate dT/dr, and an integration over the dimensions of the crucible should give you a total temperature difference, which then depends on I.

3. Jun 20, 2012

### Hazzattack

Ok, so i've uploaded the picture provided with the question. Then added R as the radius onto it(given in the question). The conductor is assumed to be perfect, thus no heat loss to the environment or the conductor. Ok, the first part makes perfect sense, as i'm pretty sure thats what i've ended up doing, i came to the conclusion that the electrical current is the same.

I^2 * rho/a * dr - is this the same? (as i used the resistivity equation R = rho l/a but said that L in this case is dr, therefore [1/r^2] integrated between two points (which i thought would be r and r+dr - is this incorrect?)

I then thought this was the same as dQ/dt which then meant i just needed to evaluate the RHS - this is the part i really struggle with i was also evaluating this at the same points as dQ/dt ( r and r+dr).

Thanks for the help! really confused by this now and been doing it for about 4 hours... (thats just today)

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4. Jun 20, 2012

### Hazzattack

Is it at all possible to evaluate the temperature in the same way you would evaluate the potential in EM?

5. Jun 20, 2012

### LawrenceC

"If the crucible is maintained at 50 °C, what is the greatest current that can flow from the contact to the crucible without melting taking place?"

If it is maintained at 50 C and has a melting point of 70 C, how can it melt? Where is it maintained at 50 C?

6. Jun 20, 2012

### Staff: Mentor

@LawrenceC: I think at the surface, far away (with the limit r->infinity).

@Hazzattack: You cannot use the same equations for current and heat flow, as you have additional heat sources in your material.

This is not an equation at all. However, d/dr (dQ/dt) = I^2 * rho/a

7. Jun 20, 2012

### Hazzattack

Ok - and d/dr (dQ/dt) is equal to dP/dr - which i get to be I^2 * rho/a . I thought this was separation of variables so i was able to say that dP = I^2 * rho/a * dr.

What i ended up with was;

dQ/dt = -k A dT/dr

=> dQ/dt = dE/dt = P = I^2 * R. However as you pointed out the Resistance varies with r, therefore,

dP/dr = I^2 * dR/dr = I^2 * rho/a -> a = 2*pi*r^2

therefore; dP/dr = I^2 * rho/(2*pi*r^2) -> dP = I^2 * rho/2*pi * [1/r^2] dr

RHS -> k * A * dT/dr

should this 'A' be dA ?

Thanks again!

Lawrence: As current is passed through the crucible there is a variation in the power as it travels through the metal. Thus there is a temperature gradient (although its constant i believe), So if you evaluate it at the radius and at infinity this will be the maintained heat 50 (at infinity) and melting at the point of contact.

8. Jun 20, 2012

### LawrenceC

What about taking the Poisson equation in spherical coordinates and solving it? The Poisson equation differs from the LaPlace equation by an nonhomogenous term. That term should the power as a function of radius and contain the current as an unknown quantity. The equation is:

1/r^2 * d/dr ((r^2)*dt/dr) + q'''(r)/k = 0

where q'''(r) = i * f(r). k is thermal conductivity.

Integrate twice and eliminate constants of integration by boundary conditions of t = 70 at r = R1 (10 mm) and t = 50 at r = infinity. The function you get for the temperature distribution will contain the current which is the only unknown.

9. Feb 10, 2014

### RichardK

I have a rather similar problem and have been attempting to form an equation using the 3D spherical poisson equation in the case where the RHS is only dependent on r, however I have been having difficulty setting up the RHS. I've been trying to remember my math notes, and if I recall correctly t on the LHS will be something akin to a temperature potential, while on the RHS I was thinking I should have something like the heat density as a function of r? Or something like that? I'm really confused and I've been trying to do this for ages :/