# Heat conduction problem

1. Apr 27, 2014

### joshmccraney

1. The problem statement, all variables and given/known data
A cylindrical wire with radius $r_i$ carries a current $I \frac{A}{cm^2}$ with a resistance of $\Omega \: ohms\times cm$. It is insulated with a material with radius $r_o$ whose thermal conductivity is $k \frac{W}{cm \times K}$. The insulator is exposed to air that convects heat at $h \frac{W}{cm^2 \times K}$. Find the temperature at the wire\insulator interface.

2. Relevant equations
Heat equation with fourier's law using constant thermal conductivity and steady state conditions: $$k\nabla ^2 T = \dot{g}$$
where $\dot{g}$ is the heat generation term.

3. The attempt at a solution
I will use cylindrical coordinates. i first want to conduct a relationship with the temperature passing through the insulator. Thus, i think the first thing I can do is state that: $$k \nabla ^2 T_1 = AI^2$$ where my units are $\frac{W}{cm^3}$. I do need boundary conditions, however. I should state, before proceeding, that I assume heat only transfers radially. now, to work on those boundary conditions, isn't the following relationship among heat fluxes true: $$\underbrace{AI^2 \times r_i}_{\text{heat flux from generation term at r=r_i}} = \underbrace{-k \nabla T_1}_{\text{heat flux from conductive term at } r=r_i}$$ and $$\underbrace{-k\nabla T_2}_{\text{heat flux from conductive term at } r=r_o} = \underbrace{h(T_{r_o}-T_{air})}_{\text{heat flux of convective term at }r=r_o}$$
although I do not know $T_{air}$. $T_1$ and $T_2$ are the heat profiles throughout the copper wire and then the insulator respectively.

Now, solving $k \nabla^2 T_1 = AI^2$ to solve yields: $$T_1(r) = \frac{kAI^2 r^2}{4} + C_1 \ln r + C_2$$
and $$\nabla T_1 = \frac{kAI^2r}{2} + \frac{C_1}{r}$$
It seems now we can solve for $C_1$ using the first boundary condition.
Assuming we find $C_1$ (this is easy enough) i really don't know how to proceed? can someone please help?

thanks!!

2. Apr 27, 2014

### Staff: Mentor

Let Q represent the rate of heat generated in the wire per unit length of the wire. The same heat flow rate has to pass through each and every radial cross section (since it has no where else to go). So, the heat flux at any radial location in the insulation is given by:

$$\frac{Q}{2πr}=-k\frac{dT}{dr}$$

You can integrate this over the insulator from r = ri to r = ro, where the temperatures are Ti and To respectively. Also, at r = ro you have

$$\frac{Q}{2πr_o}=-k(\frac{dT}{dr})_{r_o}=h(T_o-T_{air})$$

This should give you enough to solve for Ti.

Chet

3. Apr 28, 2014

### joshmccraney

thanks!

i'm not sure we are given a generation per unit length, Q? i think the generation, if we take $\Omega I^2$, gives us watts per cubic centimeter. shouldn't we multiply this by $r$, which would give us the heat flux passing through the $z \times \theta$ surface?

if not, can you help me understand geometrically why we divide Q by $2 \pi r$ (and how we get Q such that its units are Watts per unit length)?

since we don't know $T_{air}$ do we even need to worry about the convective term?

thanks a ton for your help. please let me know what you think!

4. Apr 28, 2014

### Staff: Mentor

Well, to me, the units they give for I and $\Omega$ are problematic. But, in any case, if $\Omega I^2$ represents the energy generation per unit volume of wire, if we multiply by πri2, we get the rate of heat generated per unit length of wire. That's Q: Q = $π\Omega I^2r_i^2$ So the radial heat flux at the surface r = r0 is Q/(2πr0). At any other radial location, the radial heat flux is Q/(2πr). That is, the heat flow per unit length Q has to pass through every cylindrical section of wire.
Yes. you can't get the temperature at r0 unless you know the air temperature. They expect you to specify the air temperature symbolically as Tair.

Chet

5. Apr 28, 2014

### joshmccraney

Thanks! Also, why are we multiplying by $2 \pi r_i ^2$ instead of a general $2 \pi r^2$?

6. Apr 28, 2014

### Staff: Mentor

To obtain the equation Q = $π\Omega I^2r_i^2$, I've multiplied the uniform heat generation rate per unit volume within the wire by $πr_i^2$ (the cross sectional area of the wire); this gives the generation rate per unit length of the wire Q. Another way of getting this is as follows: Over the entire wire, the heat generation rate is $π\Omega I^2r_i^2L$ , where πri2L is the volume of the wire. If we divide this by the length of the wire, we get Q = $π\Omega I^2r_i^2$, which is the heat generation rate per unit length.

The radial heat flux (radial heat transfer rate per unit area) at inside surface of the insulation r = ri is Q/(2πri); the radial heat flux at the outside surface of the insulation is Q/(2πro); the radial heat flux at any arbitrary radial location r within the insulation is Q/(2πr).

Chet

7. Apr 28, 2014

### joshmccraney

Excellent! thanks for the thorough explanation of the physical situation! it seems odd we did not even need to use the heat equation i initially proposed. in fact, using it only offers the unknown constant $C_1$ (in the first derivative), so it seems not only do we not need it, but it can't even help with anything, right?

also, would the temperature throughout the wire be constant (since it generates heat constantly) or would it be hottest in the middle? to really find this out, we would have to solve the heat equation so we would have the temperature profile, right (which we can NOW do because we have $T(r_i) = T_i$ and $T(r_o) = T_o$ which are both known)???

8. Apr 28, 2014

### Staff: Mentor

For the insulation, your equation $k\nabla ^2 T = \dot{g}$ becomes $k\nabla ^2 T = \frac{k}{r}\frac{d}{dr}(r\frac{dT}{dr})=0$. If we integrate this equation, we obtain:
$$-k\frac{dT}{dr}=\frac{C}{r}$$

From the boundary conditions, we get C = Q/2π. So, your original equation does apply.

No. It would be highest in the middle.

Yes.

9. Apr 28, 2014

### joshmccraney

Thanks! You're the man (or woman) :)