×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Heat transfer through a hollow cone
4

Heat transfer through a hollow cone

Heat transfer through a hollow cone

(OP)
Does anyone know how to caclulate temperature gradient along a hollow cone. Heat flow is from small end of cone towards larger end. I am only interested in heat flow inside the walls of the cone.

RE: Heat transfer through a hollow cone

You can treat it as a fin exposed to a fluid.
For a fin of cross section A and perimeter P (only the perimeter where exchange occurs is to be counted) the equation is
kAd2θ/dx2=hPθ
where θ is the difference of metal temperature to that of the surrounding fluid.
In your case A and P are linear functions of x, and I'm afraid that an analytical solution to that equation requires the use of Bessel functions.
However we should always remember to be engineers: I guess that a solution obtained for a tube with the average diameter of the cone (with a much simpler exponential solution of the equation) would give results within a few percent to that of the cone, with an approximation certainly better than the approximation for the heat exchange coefficient.

prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads

RE: Heat transfer through a hollow cone

2
Difficult without a diagram, but here goes...

Consider a cone, length L, base diameter D, thickness Z
Assume no heat losses other than heat flowing out of the base. Assume Z is constant.

Assume that the hot boundary is a conic circle at a distance Xh from the apex, temperature Th. You have to do this rather than define the apex temperature, as a point source of heat with zero flow area gives zero heat flow!

Assume that the cold boundary is the base, temperature Tc
Assume thermal conductivity is k
General distance from the apex is x
Denote pi() as p
Heat transfer area is A
Temperature difference is dT
ln() is natural logarithm of ()

Thermal conduction equation is Q=kA.dT/dx

At any point on the cone, distance x from the apex, where Xh<x<L, the diameter is Dx/L. Hence the heat transfer area is pZDx/L

So Q=kpZDx/L.dT/dx

First, consider the whole cone and evaluate Q. It's difficult to do this without math symbols, so I hope this is understandable (or just skip to the end!)

integral between x=L and x=Xh of(dx/x) = integral between T=Tc and T=Th of (kpZD/LQ. dT)
From that you can work out
Q = kpZD(Th-Tl)/[L.ln(Xh/L)]

In a similar way you can work out the temperature at point x and get

Tx = Th - LQ/kpZD. ln(Xh/x)

Substituting for Q we get the much simpler result

Tx = Th - (Th-Tc).ln(x/Xh)/ln(Xh/L)

If I bung this into Excel using example figures and graph Tx vs x, I get a nice gentle curve, steeper at the hot end than the cold, as you'd expect from the varying flow area.

I hope this helps!

Stuart

RE: Heat transfer through a hollow cone

Oops! Blindly assumed this was like a fin of triangular shape, but in fact:
AD(x)t (t=thickness)
PD(x) (assuming exchange at one face only, as you said).
Hence the term πD(x) cancels out in the equation that becomes the same as for a fin of constant cross section.
Then if you put n=√(h/kt), the general solution is:
θ=C1enx+C2e-nx
Now assuming the end conditions are
θ=θo at the small end
and
θ=0 at the large end
the temperature distribution is simply
θ=θoe-nx
I see now the answer from c2sco: this one is of course for pure conduction (no convection), is valid only for a full cone (including the apex), if my understanding is correct, but is singular at the apex.
Please mprice, better specify your conditions if you need more help.

prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads

RE: Heat transfer through a hollow cone

I'd assumed there is no heat loss from his comment "I am only interested in heat flow inside the walls of the cone."

Convection could be built into the calculus treatment but it would get awfully complicated. My current knowledge of Bessel functions is close to that before I went to university many moons ago.

RE: Heat transfer through a hollow cone

What is the orientation, vertical, horizontal,etc?
What will the inside surfaces see (thermal radiation)?
What is the medium inside the cone?
What is thickness distribution along the length of the cone?
What are the boundary conditions on the outside of the cone?
BETTER YET-What is the application?

regards

RE: Heat transfer through a hollow cone

(OP)
Many thanks to all for your contributions,
Stuart (c2sco) - your reply has been particularly helpful.
I gave very few specifics and I can see that there are several approaches.


What I am trying to do is estimate the dT through a tube body.  The cone is part of the body. I am ignoring convection and radiation.
The truncated cone has a Small Diameter 11.5mm, Large Diameter 37.5mm length 15.5mm, wall thickness 0.8mm,K = 390Wm-1K-1
Medium in and around cone is High Vacuum.
Thermal input varies from 50 to 100 W.

Stuart - I rearranged your formula to give dT(Th-Tc) =[Q/(L.ln(Xh/L)]/kpZD. Is this correct? - I got very large negative dT values.

Thanks again for any further input

Matt

RE: Heat transfer through a hollow cone

So you assume no heat losses along the tube.
In this case
kAdθ/dx=const.=your thermal input=Q
Now
A=A(x)=πD(x)t=π[D0+(D1-D0)x/L]t
Rearranging
dθ/dx=Q/(a+bx)
with a and b easily derived.
A quick tour on www.integrals.com gives
θ=(Q/b)log(a+bx)+C
(was easy to calculate directly of course, but for lazy people like me...)
Now for x=0
θ0=(Q/b)log(a)+C
Assuming θ0 is known, this allows to calculate the constant of integration C.

prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads

RE: Heat transfer through a hollow cone

c2sco (Chemical)Good derivation. ButI have to digest your rleation of area vs. distance, x.  You should however define what is a positive x as Q=minus kAdT/dx.
Also interesting is that there is "0" heat flux at the appex. BUT-What is the temperature at the apex?
Regards

RE: Heat transfer through a hollow cone

c2sco (Chemical)As I further digest your approach, I don't believe it is applicable to the stated problem AND I don't believe that the stated problem makes sense.
For no heat flow from the surface boundaries all the input heat must exit and the point of zero area. Since that imposed boundary condition cannot occur, a transient will result.
Your derivation would be similar for the following:
Consider a solid cylinder (insulated at its ends) or a solid sphere initially at a cold temperature with an applied hot temperature at its surface. The body in question would heat up until Thot is reached throuh out.

mprice (Mechanical)should specify some other boundary conditions.

Regards



RE: Heat transfer through a hollow cone

mprice, yes, you re-arranged the formula correctly. As sailoday28 pointed out I've omitted the minus sign - pure laziness! You can ignore convection, but beware ignoring radiation unless your temperature is similar to the surrounding temperature.

If I substitute into this equation, I get for 50W input

dT = [50 x 0.0155 x ln(4.75/15.5)]/[390 x 3.142 x 0.0008]
   = 24.92 degC

Lack of a diagram makes it difficult to check we're talking the same things here. I'm assuming that if you were to sit the cone on its base, its apex would be 15.5mm high (0.0155m). Since the small diameter (hot boundary)=11.5mm, large diameter = 37.5mm, then the hot boundary distance Xh is 11.5/37.5 x 15.5mm = 4.75mm from the apex.


sailoday28, if we extrapolate backwards to the apex, we need an infinite temperature at the point to get a heat flow. That is why we need to specify the hot boundary as something apart from the point source! The problem is fine, just unusual.

Stuart

RE: Heat transfer through a hollow cone

c2sco (Chemical)Its still not clear to me how the heat is dissipated if all the surface boundaries are insulated.
With insulated surface boundaries and a constant heat input, the temperature will rise.


RE: Heat transfer through a hollow cone

(OP)
C2sco,
Thanks for your response.
Without a diagram, as you say, it is difficult to explain geometry. Actually if the Truncated cone were sitting on its 37.5mm diameter base then its apex would be 22.35mm high. The Small diameter is 11.5mm. The hot boundary distance Xh is 11.5/37.5 x 22.35 = 6.85mm from the apex. The distance from the Large Diameter to the Small Diameter 22.35-6.85=15.5mm was what I wrongly called the height of the truncated cone.

So I calculate:
dT =[Q/(L.ln(Xh/L)]/kpZD
dT=50/(0.02235 x ln(6.85/22.35)/(390 x 3.142 x 0.0008 x 0.0375)
dT=-51460

My result is very different from yours but I guess its probably my maths - any clues as to what I am doing wrong here?

Matt

RE: Heat transfer through a hollow cone

Developing my math above a little more (and correctly accounting for the sign of Q blush), I get to the following:
θ1-θ0=QL/(πkt(D1-D0)) · log(D1/D0)=50x15.5E-3/(3.14x390x0.8E-3x(37.5E-3-11.5E-3))log(37.5/11.5)=36°C

prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads

RE: Heat transfer through a hollow cone

Mprice,

I've been reading these posts on this subject for several days, and looked over all the fancy mathematical derivations of the problem with total amazement.

I think that SailOday28 came closest to putting his finger on the whole thing.  OK.  So there is a frustum of a hollow cone.  Since the cone is hollow, it implies that it's like a paper or metal cone (maybe a pointed ice cream cone)with the tip cut off.  So far so good.  

Then some form of heat or energy is applied at the smaller end of the cone by some unknown source.  However, none of the energy is transmitted from the cone to its surrounding be either convection or radiation.

So what happens?  In a steady state, absolutely nothing.  The entire cone remains at one temperature.  If it's viewed as transitional , then, as long as the temperature of the soure is above that to the cone, the temperature of the entire cone continues to increase to be eventually at the same level as the source.  If it's a constant source of ENERGY (from where?) then the temperature of the cone simply increases to infinity or it melts, catches fire, or something.

Regards,

speco

RE: Heat transfer through a hollow cone

(OP)
speco,

At the small diameter of the truncated cone is a target heat source of about 50W. I would like to estimate how this heat energy will be conducted away from the target area by the conical wall which supports it. The cone is just a part of a larger tube body part of which is eventually cooled by forced air convection.
The cone itself is surrounded by high vacuum (inside and out)
My question relates to calculating the theoretical temperature gradient along the cone because this was the bit that I was not sure about.
 c2Sco and prex have both been kindly helping me with this.

Regards

Matt

RE: Heat transfer through a hollow cone

What would happen to a sodering iron if the heat iput (constant wattage) could not be dissapated from it surface?

Regards

RE: Heat transfer through a hollow cone

Sorry to keep harping about insulated surfaces.
 If temperature distribution is required and heat is dissipated form one end of a truncated cone, then a boundary condition must be stated.

RE: Heat transfer through a hollow cone

Well sailoday28, no supplementary boundary condition is required besides a given source temperature (that's even not necessary if only the temperature drop is sought): this problem assumes constant thermal flow, that's why the boundary condition at the other end is not required.
Imagine, as mprice states, that this is a section of a longer pipe: somewhere upstream heat is input at a known rate and somewhere downstream heat is extracted, of course at the same rate. How (boundary conditions) heat is input and taken off influences the local temperature distributions, but far from the source and the sink the temperature distribution is known: it will be linear if the section area is constant along the axis, logarithmic as shown above if the area varies linearly, and so on.

prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads

RE: Heat transfer through a hollow cone

prex (Structural)
1--I'm not clear on what you mean by constant isothermal flow. Doesn't heat flow require a temperature gradient?
q=-kAdt/dx ??
2 "that this is a section of a longer pipe: somewhere upstream heat is input at a known rate and somewhere downstream heat is extracted, of course at the same rate."
One boundary condition is a known heat flux   Btu/sq ft/time. What is the other boundary condition? If at the same rate, then the other boundary condition is known.
The flux at both boundaries should then be specified in the governing differnetial equation. That is, dt/dx is specified at both boundaries and not temperature.

Regards


RE: Heat transfer through a hollow cone

prex (Structural)
Please disregard my last post. I believe by constant isothermal flow you mean constant q (BTU/hr, etc).
And since you specify a frustum of a cone, there is no need for x=0.   X can be a reference, ie at boundaries x=x1 and x=x2 etc.
In that case, yes, integration of the equations will be in terms of a temperature difference.

RE: Heat transfer through a hollow cone

mprice,
Sorry, I gave a false "OK" to your re-arrangement which explains your problem with substituting numbers since. The correct re-arrangement of my equation is
dT = [Q.L.ln(Xh/L)]/kpZD
If you look at my worked example you'll see that's what I used to get the sensible answer of 24.92 degC. You substituted a divide between the Q and the L which is an error, and leads to the answer you correctly identify to be wrong.

Might I respectfully suggest that other respondents actually read the initial problem and contributions which make the problem clear. Heat flow is constant, entering at the small diameter and leaving by convection only at the cone's base. There is a temperature profile (which is the whole point of the question), assumed to be constant due to the initial unsteady state having died away over time.

Regards,
Stuart

RE: Heat transfer through a hollow cone

c2sco,
you assumed in your calculations that the length of the cone from the base to the apex is 15.5 mm, whilst from mprice's data it is clear that this the height of the truncated cone (from major base to minor base). That's why our results are different.
To check our results we may simply replace the cone with a cylinder of the same length or height, that I'll call L as above.
In this case the result is straightforwardly calculated as:
ΔT=QL/(πktD)
If we now take for D the major diameter of the cone, we obviously will get a ΔT lower than the actual one, a bigger value with the minor diameter, and an hopefully close approximation with the average diameter.
So:
-with the major diameter ΔT=21°C
-with the minor diameter ΔT=69°C
-with the avg. diameter ΔT=32°C
The last value is not so close, but is anyway the closest one to my formerly calculated value of 36°C.

prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads

RE: Heat transfer through a hollow cone

(OP)
c2sco, thanks for sorting that out - all is clear.
prex , thanks for the further anaylsis. Using the corrected height of the truncated cone I now get the same result with your formula and the formula developed by c2sco. That is DT=36°C.

Many thanks for your inputs

Matt

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources