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Heat transfer out of cone

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desktophustler
#1
Feb18-13, 03:48 PM
P: 8
Have a fluid flow in a cylindrical pipe with insulation around it. The insulation is in the shape of a truncated cone. It starts at a thickness with a radius only slightly thicker than the pipe and the radius increases as you move along the pipe. The radius increases at a constant rate.

The goal is to figure out the outer wall temperatures of the cone if we know the bulk fluid temp inside. Things I considered were that a generic 2D model using cylindrical cords could be used and the outer surface area and changing radius could be accounted for, but this would assume that the heat only flows outward radially. This is techinally incorrect as the heat would flow through the path of least resistance (i.e. at an angle away from the pipe to. heat won't travel through 5 inches of insulation when it can go through 3)

So to go around this situation, I figured I'd have to use a 3D model and solve numerically. However, programming isn't my forte. Any ideas? Generally, this appears to be a good MatLab problem. Is this somthing that I could to in VBA? If so, how?

The length of cone is very short compared to the overall process, so cooling of the fluid inside may be able to be neglected. Go ahead and assume that it is at a constant bulk temp
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mfb
#2
Feb18-13, 05:05 PM
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A 2D model is sufficient, if the system has a symmetry axis along the pipe. If the angle between the outer insulation layer and the pipe is small, a 1D simulation (with an analytic solution) might work.

Any ideas?
An easy implementation would be a large grid, with simple equations how temperature gets updated in each time step. Wait until the system reaches its equilibrium.
desktophustler
#3
Feb18-13, 09:44 PM
P: 8
A 2D model couldn't simultaneously take into account the curvature of the cone (e.g. Time marching a flat grid to steady state) and the fact that heat will flow in the path of least resistance (e.g. In the of just using cylindrical cords)

mfb
#4
Feb19-13, 04:22 AM
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Heat transfer out of cone

Sure it can. You have to convert your equations to cylindrical coordinates, so you get a 2D-simulation where the equations are modified a bit. But the temperature does not depend on the angle (and there is no flow along this direction), so you can restrict the analysis to the (z,r)-plane.
AlephZero
#5
Feb19-13, 06:47 AM
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MFB is correct. A "2D" model taking a slice in the radial and axial directions, and including the fact that the "thickness" varies with radius, will give the correct solution.

On the other hand, often in heat transfer problems the thermal constants and especially the boundary conditions are not known very accurately. It's worth considering whether the errors caused heat flowing through the "wrong thickness" of material are sigmificant compared with other sources of error. If the cone has a shallow angle, they may not be.

Writing your own code to solve this is not difficult, but it's probably not an efficient way to use your time. This is a trivial problem to mdoel with a finite element code. We don't know if you have access to a student version of Ansys, etc. If not, Google will find plenty of free open source and shareware FE codes - but I don't have any experience of using them for thermal problems, so I can't recoomend any particular code.
desktophustler
#6
Feb19-13, 08:05 AM
P: 8
The rise is 7"/6".

I think we're on the same page here in understanding that a "2D" model done in slices to account for the changing thickness is truly just a series of 1D models. Or are we thining about different ways to approach this?

See attached for the 2 heat flow lanes. One shows what the model would show and the other something a little more realistic (slightly lol)

I agree with you that a model of this type would work if the angle was shallow, but I'd rather the answer just be as accurate as possible since I couldn't begin to quantify the severity of the errors in any of the assumptions (if I could I would know the correct answer then)

I'm probably going with some software package (tends to make ppl believe your answer a little more). Looks like a free trial of solidworks would get the job done.
Attached Thumbnails
flow.png  
mfb
#7
Feb19-13, 11:34 AM
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No, not slices! The 2D-plane is the same as in your sketch.
desktophustler
#8
Feb19-13, 01:12 PM
P: 8
The 2D plane can account for the fact that the heat won't flow only perpendilcar to the inner pipe. True. But it doesn't account for the curvature of the cone. It assumes that the z-direction can be neglected b/c the the length in the z-direction is very long. See pic attached

Unless there is some way to account for curvature in 2D, the best bet would to model it using software. Mainly b/c ppl like pretty pictures and less assumptions.
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mfb
#9
Feb19-13, 01:37 PM
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Quote Quote by desktophustler View Post
But it doesn't account for the curvature of the cone.
The formulas for the individual elements do that. The problem is similar to this or this one (but with easier boundary conditions).

It assumes that the z-direction can be neglected b/c the the length in the z-direction is very long. See pic attached
No, it does not.
AlephZero
#10
Feb19-13, 02:24 PM
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Quote Quote by desktophustler View Post
Unless there is some way to account for curvature in 2D, the best bet would to model it using software. Mainly b/c ppl like pretty pictures and less assumptions.
You are right that the "2D" model needs to "know" it is axisymmetric about the X axis, but that should be a standard option if your analysis software can deal with rotating machinery for eaxmple.

Therei is no problem including that correctly in a finite element or finite different model.

But if your software doesn't have an option to analyse an axisymmetric model (i.e. it can only handle a "slab" of material of constant thickness as in our drawing) the answers will be wrong, of course.
desktophustler
#11
Feb19-13, 04:11 PM
P: 8
So, in a nutshell, correct me if I'm wrong. There is no way a 2D time-marching model can definitely account for the curvature of the cone.

See attached

If incorrect explain why.
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mfb
#12
Feb19-13, 04:54 PM
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So, in a nutshell, correct me if I'm wrong. There is no way a 2D time-marching model can definitely account for the curvature of the cone.
Fine: You are wrong. See all previous posts for explanations.
desktophustler
#13
Feb19-13, 05:08 PM
P: 8
In 200 words or less explain how the the grid in the last attached picture can both simulate a object that is long in the z-direction and round.

If not, you sir are wrong.
mfb
#14
Feb19-13, 05:39 PM
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The formulas how grid cells are influenced by their surrounding cells depend on the setup you want to simulate.

111 characters.

If not, you sir are wrong.
That does not depend on the text length.

With more than 200 words, but in all details, as quote from another forum (the bottom and that green thing were fixed in the setup):
Hmm... quick&dirty in Excel:



The idea: Use a cylindrical coordinate system with the obvious orientation. The system is invariant under phi-rotation, therefore it is sufficient to model the two-dimensional (rho,z)-plane.
Determine the potential phi at each point of this plane with some finite resolution. The sample and the electrode get a fixed phi of 0 and 1, respectively. For all other points, the Maxwell equations give [itex]0 = \nabla^2 \phi = \frac{1}{\rho} \partial_\rho \left( \rho\; \partial_\rho \phi \right) + \partial^2_z \phi[/itex]. The electric field is then given by the gradient of the potential.

To apply this to a grid, the derivatives have to be transformed to differences of grid cells i: [itex]\partial f(i+1/2) = f(i+1) - f(i)[/itex]

This allows to calculate the second derivatives as
[tex]\partial^2 f(i) = \partial f(i+1/2)-\partial f(i-1/2) = f(i+1) + f(i-1) - 2f(i)[/tex]
and
[tex]\frac{1}{i} \partial \left(i\; \partial f(i)\right) = \frac{1}{i} \left[ \left(i+\frac{1}{2}\right) \partial f(i+1/2) - \left(i-\frac{1}{2}\right) \partial f(i-1/2) \right] = \frac{1}{i} \left[ \left(i+\frac{1}{2}\right) \left(f(i+1) - f(i)\right) - \left(i-\frac{1}{2}\right) \left(f(i) - f(i-1)\right) \right][/tex]

Putting everything together, solving for a specific cell and using the index i for the rho-coordinate and j (columns) for z, I get
[tex]\phi(i,j)==\frac{1}{2}\left((\phi(i,j-1)+\phi(i,j+1))/2+\frac{1}{2i}\left((i+1/2)*\phi(i+1,j)+(i-1/2)\phi(i-1,j)\right)\right)[/tex]
I hope that the prefactors are right.

As the simulated area is finite, there are 4 borders to consider:
- the sample border is simple, as it has a fixed potential.
- rho=0 is a coordinate singularity and cannot be calculated using the formula given above. But there, it is easy to get the appropriate formula from cartesian coordinates:
[tex]\phi(0,j)=\frac{1}{6}\left(\phi(0,j+1)+\phi(0,j-1)+4\phi(1,j)\right)[/tex]
- far away from the sample. This needs a bit more thought. Here, I just assumed that the field is nearly constant with z, and therefore [itex]\phi(i,0)=\phi(i,1)[/itex]. This is a good approximation for the region close to the electrode, but not so good far away from both sample and electrode.
- far away from the electrode. This is the most tricky case. Depending on the precision you need, it might be sufficient to just assume a nearly constant potential here or a constant first derivative in rho-direction or something similar.


The component of the electric field in one direction is then given by [tex]E = \partial \phi(i) \approx \frac{1}{2} \left(\partial \phi(i+1/2) + \partial \phi(i-1/2)\right) = \frac{1}{2}\left(\phi(i+1)-\phi(i-1)\right)[/tex]

Here is the magnitude of the E-field as graph.

Every reasonable programming language which supports arrays can calculate this, and as all calculations are quite basic it should be easy to implement it.
There are certainly better ways to calculate the electric potential and field, and it might be useful to decrease the grid size close to the electrode (especially the corner of it, as it has the highest field strength). Smaller grid cells everywhere can do the same job, they just need more computing power.
AlephZero
#15
Feb19-13, 09:15 PM
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Quote Quote by desktophustler View Post
In 200 words or less explain how the the grid in the last attached picture can both simulate a object that is long in the z-direction and round.
The only change between 2D plane and 2D axysymmetric heat transfer is inserting "r" in a few places in the differential equation. The change to the finite element or finite different formulation is equally trivial. For example see http://books.google.co.uk/books?id=9...ations&f=false - but you will have to buy the book to get the full story.

If you think this is "wrong", I recommend you never fly with a cmmercial airline, because every jet engine manufacturer has been doing heat transfer calculations on turbine disks this way for the past 40 or 50 years.
desktophustler
#16
Feb20-13, 09:12 AM
P: 8
Nice. I like being wrong. Honestly I don't really care about the transient problem.

As a final check what would be wrong with using that grid we were discussing. An energy balance in and out of any node would yield:

k(Δy)(T1-Tn)/Δx + 2*pi*k*(T2-Tn)/ln((Rn+Δy)/Rn) + k(Δy)(T3-Tn)/Δx + 2*pi*k*(T4-Tn)/ln(Rn/(Rn-Δy)) = 0

referencing the nodes below, moving from 2 to 4 would be equivalent to moving outward from the cone
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nodes.png  
Integral
#17
Feb20-13, 09:26 AM
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What boundary conditions are you using for the exterior surface?
desktophustler
#18
Feb20-13, 10:10 AM
P: 8
Convective eventually.

If any wants to critique anything though, they can use constant wall temps just to dumb down the problem.


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