Heat transfer out of coneby desktophustler Tags: cone, heat transfer, heat transfer pipe, temperature, truncated cone 

#1
Feb1813, 03:48 PM

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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 



#2
Feb1813, 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.




#3
Feb1813, 09:44 PM

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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)




#4
Feb1913, 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 2Dsimulation 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.




#5
Feb1913, 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. 



#6
Feb1913, 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. 



#7
Feb1913, 11:34 AM

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No, not slices! The 2Dplane is the same as in your sketch.




#8
Feb1913, 01:12 PM

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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 zdirection can be neglected b/c the the length in the zdirection 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. 



#9
Feb1913, 01:37 PM

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#10
Feb1913, 02:24 PM

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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. 



#11
Feb1913, 04:11 PM

P: 8

So, in a nutshell, correct me if I'm wrong. There is no way a 2D timemarching model can definitely account for the curvature of the cone.
See attached If incorrect explain why. 



#12
Feb1913, 04:54 PM

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#13
Feb1913, 05:08 PM

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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 zdirection and round.
If not, you sir are wrong. 



#14
Feb1913, 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. 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): 



#15
Feb1913, 09:15 PM

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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. 



#16
Feb2013, 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)(T1Tn)/Δx + 2*pi*k*(T2Tn)/ln((Rn+Δy)/Rn) + k(Δy)(T3Tn)/Δx + 2*pi*k*(T4Tn)/ln(Rn/(RnΔy)) = 0 referencing the nodes below, moving from 2 to 4 would be equivalent to moving outward from the cone 



#17
Feb2013, 09:26 AM

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What boundary conditions are you using for the exterior surface?




#18
Feb2013, 10:10 AM

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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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