Length of a rod cooled by free convection 1

[samoht]

Let's say that a small a small cylindrical rod with no insulation is welded to a large insulated vessel with temperature x. What is the solution for finding the length of the rod based on a maximum allowed temperature y at the free end of the cylinder? I have decided not to consider radiation of heat in this case. The rod is cooled by free convection.



Thomas A

 

[sailoday28]

If the radial temperature gradient in the rod is small, then use the standard fin equations. See any good heat transfer text.

Regards

 

[trashcanman]

You won't get the right answer by ignoring radiation.

 

[IRstuff]

Is it really a rod, or is it a pipe? This will also radically affect your answer.

TTFN

FAQ731-376

 

[samoht]

To Trashcanman
I am aware that I will not get the correct answer by ignoring radiation. Since I am trying to solve this in a general manner, very little is known on the surroundings, exept that it is cooler. Let say just say that the heat radiated from the "rod" will be a safety margin.

To IRstuff
OK I admit that I tried to simplify this case by calling it rod. But since my knowledge on Thermodynamics is currently limited the idea whas that I first need to be able to solve this case by using a massive rod. It is actually a piece of instrument tubing (O.D. is 10mm, I.D. is 7mm) connected to a pressure gauge. The temperature of the vessel is 250deg C and the temperature at the pressure gauge should not exceed 70deg C.



Thomas A

 

[prex]

If you are just checking whether the instrument is too hot, just take h=10 w/m²°C as the heat transfer coefficient. This value includes the effect of radiation for low temperatures (up to some 100°C) in calm air, and, as the contribution of radiation raises with temperature, you'll be on the safe side above.
Temperature goes down along the bar with the law (it doesn't make much difference whether it's solid or a pipe):
T(x)=T_we^-nx
with n=[√](hP/kA) and:
h=heat transfer coefficient
P=perimeter exposed to air (outer perimeter only)
k=conductivity of material
A=cross section area of bar

prex

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[sailoday28]

prex (Structural)What are your boundary conditions at the end of the fin?

 

[sailoday28]

prex (Structural)What are your boundary conditions at the end of the fin?
I believe your solution is for a fin of infinite length.
Solution for fin of finite length should be
(T(x)-Tambient)=(Tw-Tambient-C)e^nx + Ce^-nx
with n=?(hP/kA
Where C is determined from conditions at x=L

Regards

 

[prex]

That's right, sailoday28, but for checking the maximum temperature at a certain point of the bar the method I proposed is safe: any equipment at a short distance along the bar will be a heat sink and the temperature will tend to go down more quickly.

prex

http://www.xcalcs.com

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[sailoday28]

Max temp at a certain point would be obtained by putting in boundary condition of dt/dx=0 at that point.
Regards

 

[prex]

Again: yes, you are right, but what is the purpose of a pipe terminating with a plug and connected to nothing?

prex

http://www.xcalcs.com

: Online tools for structural design

http://www.megamag.it

: Magnetic brakes for fun rides

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: Air bearing pads

 

[sailoday28]

Prex-I thought the original question was to determine the max temp. Clearly if another boundary condition can is reasonably known at length, L, then it should be applied.

Refering to your solution, T(x)=Twe-nx, I believe
T(x) should be T(x)-Tambient and
Tw should be Tw- Tambient.

Further, using dt/dx=0 as a boundary condtion, the closed form solution will involve cosh(nx) and will probably be safer to use.

Regards

 

[samoht]

First thanks to Sailoday28 and Prex for helping me solve this problem. As I try to follow the discussion I fail to fully understand a couple of the variables used above. Therefore I would appreciate if you could explain the following variables.

Tw
dt
dx

To Sailoday28
Ref your answer 29 oct 07 13:32 you write n=?(hP/kA)should the question mark be a sqrt?

Thomas A

 

[sailoday28]

dt/dx refers to temperature gradient in axial direction.
Tw refers to temperature at location where x=0
Not understanding this terminology indicates you are weak or rusty in conduction heat transfer.

With no heat transfer at length x=L and letting
THETA = T - Tambient
THETAO = Tw -Tambient
THETAL = TL - Tambieet
n=sqrt(hP/kA)
THETAL/THETAO=1/COSH(nL) WHERE cosh is the hyperbolic cosine function.
Please work with an engineer having a heat transfer background.

Regards