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Heat transfer through a hollow cone 4
- Thread starter mprice
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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
kAd[sup]2[/sup][θ]/dx[sup]2[/sup]=hP[θ]
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
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For a fin of cross section A and perimeter P (only the perimeter where exchange occurs is to be counted) the equation is
kAd[sup]2[/sup][θ]/dx[sup]2[/sup]=hP[θ]
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
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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
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
Oops! Blindly assumed this was like a fin of triangular shape, but in fact:
A=[π]D(x)t (t=thickness)
P=[π]D(x) (assuming exchange at one face only, as you said).
Hence the term [π]D(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]=C[sub]1[/sub]e[sup]nx[/sup]+C[sub]2[/sub]e[sup]-nx[/sup]
Now assuming the end conditions are
[θ]=[θ][sub]o[/sub] at the small end
and
[θ]=0 at the large end
the temperature distribution is simply
[θ]=[θ][sub]o[/sub]e[sup]-nx[/sup]
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
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A=[π]D(x)t (t=thickness)
P=[π]D(x) (assuming exchange at one face only, as you said).
Hence the term [π]D(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]=C[sub]1[/sub]e[sup]nx[/sup]+C[sub]2[/sub]e[sup]-nx[/sup]
Now assuming the end conditions are
[θ]=[θ][sub]o[/sub] at the small end
and
[θ]=0 at the large end
the temperature distribution is simply
[θ]=[θ][sub]o[/sub]e[sup]-nx[/sup]
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
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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.
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.
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1
- #6
sailoday28
Mechanical
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
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
- Thread starter
- #7
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
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
So you assume no heat losses along the tube.
In this case
kAd[θ]/dx=const.=your thermal input=Q
Now
A=A(x)=[π]D(x)t=[π][D[sub]0[/sub]+(D[sub]1[/sub]-D[sub]0[/sub])x/L]t
Rearranging
d[θ]/dx=Q/(a+bx)
with a and b easily derived.
A quick tour on 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
[θ][sub]0[/sub]=(Q/b)log(a)+C
Assuming [θ][sub]0[/sub] is known, this allows to calculate the constant of integration C.
prex
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In this case
kAd[θ]/dx=const.=your thermal input=Q
Now
A=A(x)=[π]D(x)t=[π][D[sub]0[/sub]+(D[sub]1[/sub]-D[sub]0[/sub])x/L]t
Rearranging
d[θ]/dx=Q/(a+bx)
with a and b easily derived.
A quick tour on 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
[θ][sub]0[/sub]=(Q/b)log(a)+C
Assuming [θ][sub]0[/sub] is known, this allows to calculate the constant of integration C.
prex
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sailoday28
Mechanical
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
Also interesting is that there is "0" heat flux at the appex. BUT-What is the temperature at the apex?
Regards
sailoday28
Mechanical
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
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
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
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
sailoday28
Mechanical
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.
With insulated surface boundaries and a constant heat input, the temperature will rise.
- Thread starter
- #13
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
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
Developing my math above a little more (and correctly accounting for the sign of Q ![[blush]](/data/assets/smilies/blush.gif "[blush] [blush]")
), I get to the following:
[θ][sub]1[/sub]-[θ][sub]0[/sub]=QL/([π]kt(D[sub]1[/sub]-D[sub]0[/sub])) [·] log(D[sub]1[/sub]/D[sub]0[/sub])=50x15.5E-3/(3.14x390x0.8E-3x(37.5E-3-11.5E-3))log(37.5/11.5)=36°C
prex
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), I get to the following:[θ][sub]1[/sub]-[θ][sub]0[/sub]=QL/([π]kt(D[sub]1[/sub]-D[sub]0[/sub])) [·] log(D[sub]1[/sub]/D[sub]0[/sub])=50x15.5E-3/(3.14x390x0.8E-3x(37.5E-3-11.5E-3))log(37.5/11.5)=36°C
prex
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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
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
- Thread starter
- #16
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
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
sailoday28
Mechanical
What would happen to a sodering iron if the heat iput (constant wattage) could not be dissapated from it surface?
Regards
Regards
sailoday28
Mechanical
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.
If temperature distribution is required and heat is dissipated form one end of a truncated cone, then a boundary condition must be stated.
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
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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
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sailoday28
Mechanical
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
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
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