Thermal Resistance of a Hollow Circular Cone

[alex_amvdor]

Homework Statement

This is actually not an assigned homework question, rather it is for my research. I need to determine the thermal resistance of a set of stainless steel edge-welded vacuum bellows, but it seems like a problem that is very similar to something that would be assigned to an undergrad.

The edge-welded bellows can be treated as a set of 40 or so stacked hollow truncated circular cones. Each cone has a small end with radius $R_1$ and a large end with radius $R_2$. They have a wall thickness of $t$, and $t<<R_1, R_2$ so that the cross sectional area can be treated as $$A = 2 \pi Rt$$.

The cones' main axes are concentric with the z-axis.

The total length of the bellows as a whole is L, so the height of each cone is L/40, and the resistance of the complete bellows will be 40 times that of one cone.

Relevant Equations

$$R = \frac{l}{\lambda A}$$
Where l is the length traversed by the heat, i.e. $\sqrt{h^2 + (R_2-R_1)^2}$ from the bottom of a cone to the top.

$\lambda = 14.4 \frac{W}{m \cdot K}$ for stainless steel
$R_1 = 9.525 mm$
$R_2 = 14.288 mm$
$t = 1.245 mm$
$L = 31.75 mm$

We can write our radius as a function of the height, z, of our cone: $$R(z) = \frac{R_2 - R_1}{h} z + R_1$$
Where h is the height of our cone, $h = \frac{L}{40}$.

Our cross sectional area, $$A = 2 \pi R t$$ can then be written as $$A = 2 \pi t [\frac{R_2 - R_1}{h} z + R_1]$$

This I am all relatively sure about.Here is where I am not so sure:

For a small change in z, $\Delta z$, we will have a corresponding change in the cross sectional area: $$\Delta A = 2 \pi t \frac{R_2-R_1}{h} \Delta z$$ which, using our equation for thermal resistance, would give us: $$\Delta R = \frac{\sqrt{(\Delta z)^2 + (\frac{R_2-R_1}{h} \Delta z)^2}}{2 \lambda \pi t \frac{R_2-R_1}{h} \Delta z}$$

We can then pull out the $\Delta z$'s from the radical and cancel them, then clean the equation up a bit, giving $$\Delta R = \frac{\sqrt{1+(\frac{h}{R_2-R_1})^2}}{2 \lambda \pi t}$$

This result confuses me, as there is no $\Delta z$ dependence, and I am not entirely sure how to integrate it. My hope is to find an expression which I can integrate from $z = 0$ to $z = h$. If anyone can point out a problem with my math, or give me some clues as to where to go from here, I would greatly appreciate it.

I realize I may be missing the use of the definition of a derivative, i.e. $\frac{dR}{dz} = \lim_{\Delta z \to 0} \frac{R(z + \Delta z) - R(z)}{\Delta z}$, but I'm not really sure how to work this into the problem.

 

[mfb]

which, using our equation for thermal resistance, would give us

That step doesn't work. You cannot replace the whole length with a differential length, and you cannot use the difference in area in the denominator, you need the full area. You'll need to integrate over the z coordinate or over the length.

I'm puzzled by the given dimensions. h=1/40 L is smaller than 1mm. Over this length of 1 mm the radius of the bellow increases from 9.5 mm to 14.3 mm? That is more like a disk! It would also make the definition of the thickness and the corresponding area questionable as (R2-R1)>>h is not satisfied.

How exactly are subsequent bellows connected to each other (sketch?)? As R2-R1>>t this sounds relevant.

 

[alex_amvdor]

The bellows are similar to this, but much taller. The drawing in the question statement is an exaggeration in terms of height of the cone, it is simply to show the sort of shape that I mean.

You are right that it is practically flat enough to be a disk (h approx. 1mm), but it is still a cone. It may be more practical, and accurate enough, to solve the problem as though each cone were a disk, but I am not entirely convinced of that, and I would still be unsure of the next step to take.

The wall thickness is constant, and the heat flow is always going to be exactly perpendicular to this thickness, as no heat is able to flow from the inside or outside faces of the cones. This means that the cross-sectional area through which the heat flows is going to be approx. $A = 2 \pi R t$. If the cross-section were taken horizontally, you would be correct in assuming that the area approximation would be bad given the problem setup, but it is instead taken perpendicular to the surface of the cone.

Each cone is welded to the next, alternating their orientation from big end to small end, then small to big, then big to small, and so on.

I understand now why I cannot replace the area with a differential area, but why can't I replace the length with the differential length? In a small volume, is not the length traversed only that differential length?

 

[alex_amvdor]

Thinking about it, it seems reasonable to replace the cone with a disk that has a hole in the center. That changes our cross-sectional area through which heat flows from approximately $A = 2 \pi r t$ to exactly $A = 2 \pi r t$. We can then integrate over the radius itself rather than the z-coordinate. Using the formula I used before, but including the total area plus the differential area, we find: $$\Delta R = \frac{\Delta r}{2\pi \lambda t(r + \Delta r)}$$ This then gives $$\frac{\Delta R}{\Delta r} = \frac{1}{2 \pi \lambda t(r+ \Delta r)}$$ and taking $lim_{\Delta r \to 0}$, we find $$\frac{dR}{dr} = \frac{1}{2 \pi \lambda t r}$$ Integrating this from $r_1$ to $r_2$, we find $$R = \frac{1}{2 \pi \lambda t} \ln{\frac{r_2}{r_1}}$$
I still used the differential length which I think is correct to use given that the differential radial length covered at some radius is not dependent on that radius, but I may be wrong.

 

[Chestermiller]

I get a different answer. I agree that the cross section area for heat flow is $\2\pi rt$ if t is measured perpendicular to the shell. The heat flux is along the shell, and if dR is differential position along the shell, the heat flux is $$q=-\lambda\frac{dT}{dR}$$ So the total rate of heat flow along the shell is given by $$Q=qA=-2\pi r t \lambda\frac{dT}{dR}$$where Q is constant. The differential position along the shell dR is related geometrically to dr by $$dR=\frac{dr}{\sin{\theta}}$$where $\theta$ is the half-angle of the cone. So, we have $$Q=-2\pi r t \lambda \sin{\theta}\frac{dT}{dr}$$

 

[alex_amvdor]

That actually ends up giving very nearly the same answer as my disk answer, except that it has the $\sin{\theta}$ in it and the argument of $\ln{\frac{r_1}{r_2}}$ is flipped. $$\frac{dT}{dr} = -\frac{\dot{Q}}{2 \pi rt \lambda \sin{\theta}} \frac{1}{r}$$ Integrating this gives $$\Delta T = \frac{\dot{Q}}{2 \pi t \lambda \sin{\theta}} \ln{\frac{r_1}{r_2}}$$ and the thermal resistance R is defined as $R \dot{Q} = \Delta T$, so this gives $$R = \frac{1}{2 \pi \lambda t \sin{\theta}} \ln{\frac{r_1}{r_2}}$$ Thank you for the help! I think that the argument of the logarithm needs to be flipped, so somehow that minus sign is messing things up, but the $\sin{\theta}$ makes perfect sense.

 

[Chestermiller]

That actually ends up giving very nearly the same answer as my disk answer, except that it has the $\sin{\theta}$ in it and the argument of $\ln{\frac{r_1}{r_2}}$ is flipped. $$\frac{dT}{dr} = -\frac{\dot{Q}}{2 \pi rt \lambda \sin{\theta}} \frac{1}{r}$$ Integrating this gives $$\Delta T = \frac{\dot{Q}}{2 \pi t \lambda \sin{\theta}} \ln{\frac{r_1}{r_2}}$$ and the thermal resistance R is defined as $R \dot{Q} = \Delta T$, so this gives $$R = \frac{1}{2 \pi \lambda t \sin{\theta}} \ln{\frac{r_1}{r_2}}$$ Thank you for the help! I think that the argument of the logarithm needs to be flipped, so somehow that minus sign is messing things up, but the $\sin{\theta}$ makes perfect sense.

It depends on how you express $\Delta T$.

 

[mfb]

Looking at the geometry: Are you sure radiative heat transfer will be negligible? Most surfaces mainly see the adjacent cone which has a somewhat similar temperature but the interior has wider "views" as well.

 

[alex_amvdor]

Yes, the heat loss due to radiation will be negligible. This bellows is going to be used in a cryogenic setting (around 15K) in a vacuum of $10^{-6}$ Torr, so the radiation will be miniscule.

 

[mfb]

Ah, cryogenic, okay.

Is the thermal conductivity for 15 K? It is quite close to room temperature values.