Question regarding heat transfer through a solid bar 1

[AaronH]

Hi all,

It's been a while since Thermodynamics so my memory is a bit faded. I've got Mark's Hanbook cracked, I'm sure all of the info is in there, but I'm a bit slow in getting started. I'm hoping someone can give me a gentle shove in the right direction. I have a titanium shaft partially inside a furnace. An external pneumatic cylinder is connected via threads to the titanium shaft. I need to make sure the cylinder will not be damaged by the heat conducted through the shaft.

An alloy steel block 2.5"x6"x10" is inside an oven at 1400 F. The block is attached to (4) 3/4" dia. titanium solid bars 9.375" long. Each bar is connected independently to a cylinder. When the cylinder is extended 5.5" of the bar is exposed inside the oven, then it passes through a 2" thick heater element and enters into 3" of insulation. Inside the insulation the rod of the pneumatic cylinder is threaded onto the end of the titanium shaft. The cylinder rod continues through the insulation, followed by a 2.75" (I think) airgap and finally into the cylinder body.

Can someone provide some insight on how to handle this calculation? I appreciate any input you guys have.

Thanks in advance,
Aaron

 

[prex]

You would first estimate the amount of heat going into each bar, safely assuming that at the inside face of insulation is at the furnace temperature and at the outside face it is at the temperature the cylinder can withstand.
Assuming the latter is, as an example, 100°C, this would be:
Q=kA[Δ]T/L and should give a figure not far from 50 W, if I'm not grossly mistaken.
That amount of heat would have to be evacuated into the surrounding air, using the outer surface of the cylinder. To evacuate 50 W under a temperature differential of 50°C in air a surface of 1000 cm² is sufficient. It is a fairly high surface for a small cylinder, however much of the heat would also go into the cylinder support.

prex

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[DRWeig]

Prex has the method -- but I always say never calculate anything if it's possible to measure it instead. Is there a chance that you could experiment and see how hot the end of the shaft gets? Solving overheat would simply amount to adding surface area (with either fins or distance).

Let us know how it ends!

Good on ya,

Goober Dave

 

[chicopee]

My suggestion:
1)let ends of titanium bar (TB) be at 1400 dF
2)Estimate radiation heat transfer at the exposed end of TB's with equation Q=e*s*SA*T^4 where e,s are the emissivity and plank's constant,SA=surface areas exposed to radiation,T=oven temperature in absolute units. Also discount convective HT.
3) Use PREX conduction equation to determine temperature at threaded end to cylinder rod; you already know Q and T on one side of TB. Also discount HT thru insulation.
4) Knowing T at threaded connection, determine temperature gradient throughout length of cylinder rod using the steady state relaxation method. It's a little bit complicated to explain here, so get a book on heat transfer for reference. In essence you establish nodes along the length of the threaded road and develop two dimensional heat transfer equations at each node.
You'll have conduction and convection at each node.
Conductive HT in - conductive HT out - conductive HT out thru the air gap= Residual HT value.
To do this method you will need to estimate initial temperatures at each node and establish a deleta length for node.
If you establish 10 nodes of equal lengths, you will have 10 algebraic equations which can be solved by the relaxation method. If you can program a spread sheet or a use FORTRAN or BASICs, you can solve these equations a lot faster.
As I said get a book on HT to relearn this relaxation method.
Good Luck!

 

[AaronH]

All, thanks for the replies. I am considering everything you have contributed. I apologize for taking so long to get back to this thread. I had an old Email in my profile and wasn't recieving the reply notifications. I had assumed that my thread was ignored.

That being said, I found my old themo book and have started reading about Fourier's. Q=(k * dT)/dX Am I correct that Q is not a constant based soley on material performace, rather it is also based on geometry? As such, if geometry changes so does Q?

I have to correct a bit on my first message. The rid design is actually two piece. The first piece is 3/4" dia. and is 316 stainless. Then the previous designer used a 1" dia. X 2" piece of titanium to act as an insluator between the stainless rod and the pneumatic cylinder rod. So, I need to figure out how to calculate the temperature at the joint between the stainless rod and the titanium insulator as well as the between the insulator and the pneumatic cylinder rod.

As I said, I am considering all of your input. If you have any other thoughts to throw at me, by all means fire away.

Thanks again,
Aaron

 

[IRstuff]

You gave the equation, Q=(k * dT)/dX, but apparently ignored it. Q is proportional to the conduction coefficient and temperature difference, and inversely proportional to length.

TTFN

FAQ731-376

 

[prex]

Mmmm...what about Q=kAdT/dx?
AaronH, there is not much more to say with your data, besides that fundamental equation and the other one for convection to ambient air Q=hS(T-T_a).
If the SS section is also within the insulation, you solve it just by putting it in series with the titanium section (the same Q flowing through both of them). However you'll need a better understanding of heat transfer to manipulate the formulae and get to a meaningful result.

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[AaronH]

On my way home yesterday I realized I had written the formula wrong. It should be Qa=kdT/dX or as you have written it, Prex. The part I have been struggling with is Q. This all seems rather elementary, but it's been so long since I have had to do any thermal it's not even funny. Q, if I understand it, is the amount of energy that is conducted per unit time. In the formula I want to calculate dT. However, I haven't been able to figure out what I am supposed to put in for Q. Then I had a thought. I have four heating elements that are rated at 3000 Watts each. Is 12 KW my missing Q value? Will that now allow me to calculate dT? The only other option I could think of is DRWeig's experimental method to determine dT (as well as Q).

 

[AaronH]

Well, I don't think Q can be 12KW. Unless I have my units wrong it produces a dT of 1.13 million. Back to the drawing board.

 

[prex]

No it isn't 12 kW.
You normally wouldn't calculate dT (or [Δ]T), as this is known to a close extent. It is in fact close to the difference between the inside temperature and the outside one. What you don't know and need to calculate is Q.

prex

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[AaronH]

The issue I have come up against is that I don't know dT. I am trying to predict that based on material properties and temperature input. What is the temperature at the cold end of the shaft? I can't find a formula that will let me calculate dT without knowing Q first. I could assume the cold temp is ambient when the oven is first fired up. That will allow me to calculate a Q value, but I'm not sure that accomplishes anything. Unfortunately, we don't have the thermal part of our FEA package or I would just run a study. I'd really like to make a calculated estimate as the components in question are burried deep inside the machine we are assembling. It's a hard job to tear it down to replace them. As DRWeig pointed out, this may require some experimenting if I can't get my brain wrapped around the theoretical side of it.

 

[vpl]

Aaron

Sometimes you just have to iterate until you arrive at a reasonable answer.

Patricia Lougheed

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.

 

[IRstuff]

Actually, you can do this:

> Figure out the entire serial path for the heat. If there are parallel paths, you'll need to add the thermal conductivities
> Take each individual segment of different material or cross-section, and determine the °C/W, the thermal resistance of each segment
> Sum all the thermal resistances to get net_thermal_resistance
> Determine the deltaT across the entire stack
> DeltaT/net_thermal_resistance = Q
> Use that Q with the individual segments to determine each segment's deltaT


TTFN

FAQ731-376

 

[AaronH]

Did a bit of an experiment this morning. We fired up an oven and got it to approx 1400 F inside. Our ovens have a two piece door. Half open up, half down. So, we assembled a 3/4" dia. x 7.375" long SS shaft to a 1" dia. x 1.75" long titanium insulator and a pneumatic cylinder. Then we took a piece of left over insulation and drilled a 3/4" hole in the center. Next we slid the SS shaft through the hole in the insulation up to the titanium insulator. Then we cracked the oven doors and slid the 3/4" SS shaft between them. The doors could no longer seal on the face of the oven, there was about a 1" gap. We let the shaft/cylinder soak for about 15 minutes and removed it. Finally, we removed the parts from the oven and I measured the temperatures at various locations with a Fluke IR thermometer. The hot end of the shaft was about 300 F, the cold end about 120. The titanium insulator was about 110 on the hot end and 100 on the cold. The cylinder rod was initially about 90 F. Various items near the furnace checked at about 85 F, so I would say that is the ambient around the oven. Then we did the same experiment using a ceramic insulator of same dimensions as the TI insulator. The SS shaft had the same temps as previous, the ceramic part was 100 on the hot end, 85 on the cold end. The cylinder rod was 85 as well. As the two test pieces sat for a half hour, heat continued to be drawn into the cylinder rod with the TI insulator, whereas the ceramic insulator did not allow heat to be passed to the cylinder rod.

I believe the tests clearly show that the TI insulator will not provide enough thermal resistance @ 1400 F to protect the seals in the pneumatic cylinder. However, I don't know for certain if the ceramic insulator will. Moreover, the ceramic may not mechanically hold up to the abuse of this application. We're running out of time and may end up trying the ceramic insulators while at the same time search for a material with equal (or better) thermal properties, but improved strength properties.

 

[HDS]

Thank you for sharing the end results and closing the loop.

 

[prex]

Titanium is for sure not a thermal insulator. However if you use a tube instead of a solid bar, the thickness being the minimum required for strength, the heat flow to the outside will be reduced more or less in proportion to the reduction of the cross section.

prex

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: Online engineering calculations

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: Magnetic brakes for fun rides

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: Air bearing pads

 

[GuyFromDenver]

One thing you also must figure in it your thermal balance: heat in plus generated equals heat stored plus heat out. the input heat is from the furnace. There should be little if and heat generated unles there is some reaction going on. The heat stored is calculated from lumped capacitance and will be a good amoutn at titanium has a fairly good specific heat, and heat out will be through conduction along the length of the rod and convection to ambient. i didn't see a lot about the convective cooling, but that's the easiest way to maintain a constant temp.