What is the temperature at the wire/insulator interface?

[member 428835]

Homework Statement

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.

Homework 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.

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!

 

[Chestermiller]

You've done some very nice thinking about this problem. But the problem is simpler than that.

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 = r_i to r = r_o, where the temperatures are T_i and T_o respectively. Also, at r = r_o 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 T_i.

Chet

 

[member 428835]

You've done some very nice thinking about this problem.

thanks!

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:

$$\frac{Q}{2πr}=-k\frac{dT}{dr}$$

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)?

You can integrate this over the insulator from r = r_i to r = r_o, where the temperatures are T_i and T_o respectively. Also, at r = r_o 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 T_i.

Chet

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!

 

[Chestermiller]

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)?

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 πr_i², 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.

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

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 T_air.

Chet

 

[member 428835]

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 πr_i², 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.

Chet

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

 

[Chestermiller]

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

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 πr_i²L 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 = r_i is Q/(2πr_i); the radial heat flux at the outside surface of the insulation is Q/(2πr_o); the radial heat flux at any arbitrary radial location r within the insulation is Q/(2πr).

Chet

 

[member 428835]

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)?

 

[Chestermiller]

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?

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.

also, would the temperature throughout the wire be constant (since it generates heat constantly) or would it be hottest in the middle?

No. It would be highest 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)?

Yes.

 

[member 428835]

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