How long will it take for heat to flow through this sample?

[examorph]

I have a rubber sample here, basically it is a slice with a diameter of 45mm and a depth of 3mm. If this sample was at 25 degrees Celsius and I applied 180 degrees temperature all around it (top, bottom faces and outside diameter) how long would it take for the centre point (last point to get heated) to reach the 180 degrees Celsius temperature?

The materials thermal properties are:

Thermal Conductivity: 0.36 W/m*K
Specific Heat Capacity: 2 kJ/Kg K

I am really stuck with this and would appreciate any guidance at all. Thank you.

 

[marty007]

Is this a homework question? If so, just know that those are not allowed on this forum.

If not, given those dimensions I would just ignore the diameter and consider it a linear transient conductive heat transfer problem. Any introductory heat transfer textbook should provide adequate formulas. (Checked my Holman, 9th edition, and very first example in unsteady-state conduction chapter was this exact problem).

 

[examorph]

Thank you for your help, this is not homework, it is for part of a work related simulation that has been conducted. I will look into this book when I get back, do you have any more information that may aid me with this? Thanks.

 

[marty007]

The text I mentioned above has a whole derivation for the solution, which dives into differential equations, series solutions and Fourier sine expansions...

In short however, the final unsteady heat equation can be written as per the attached file. The text also goes into cases where the entire surface is exposed to convection rather than an instantaneous external surface temperature. For that it provides charts for solutions.

If this is critical to design, PLEASE obtain your own reference material. This is for illustration only, I cannot guarantee that I've typed this correctly.

Cheers,

 

[examorph]

Thank you very much for your help it is greatly appreciated. I thought that this would be a simple equation which I could directly plug the values into and get an average time however this goes way beyond my knowledge within this subject. I therefore, used a simple heat transfer simulation to receive a value and got about 25 seconds. Also, this is not critical to the design, it was just something I wanted to know. Again, thanks for your help.

 

[Compositepro]

You question displays a a fundamental lack of understanding some basic principles. To get to 179C = some finite answer. To get to 181C = never. To get to 180 C = indeterminate.

Other things you don't know:
heat transfer coefficient
heating medium properties (flow rates)
Is the rubber curing and releasing heat? (Your question type often applies to a molding operation).

 

[marty007]

Compositepro - very true, I hadn't noticed that problem...

 

[examorph]

Thank you for the replies, I did some more research and found that modifying the energy equation so that it applies to 1D transient flow I get:

αt/b^2=1.2

I plugged the values in and solved for t and found that it comes to 15 seconds which was close to my simulation results as when I re-checked the temperature result logs and they actually showed that the center temperature was very close to the wall temperature at about 15 seconds.

However, this was based on a standard material with the rubbers thermal properties and heat loss during the curing was not considered. Would the heat loss's be large? I ask because the entire surface of the rubber has an applied temperature of 180 degrees therefore, I cannot understand why heat would leave.

 

[boyze]

here's a recent blog posting that describes and provides an Excel spreadsheet simulation of rubber state of cure for various heating scenarios. Part of the calculation process is to determine the temperature versus time for given mold temperatures. This particular spreadsheet is limited to 2D.

I plugged your parameters in and got ~ 17.5 seconds for the sides to reach 179C and 30 seconds for the center. This obviously is an upper bound on time by the fact that it's limited to 2D.



Have Fun!

James A. Pike

http://www.xl4sim.com

http://www.erieztechnologies.com

 

[boyze]

Correction on my earlier post. The 179 C in ~ 17.5 sec was at .1 mm from the heated sides. The heated sides would, of course, be at the 180.

Have Fun!

James A. Pike

http://www.xl4sim.com

http://www.erieztechnologies.com

 

[examorph]

Wow this is amazing, Thank You!

Could you please tell me how you calculated how the rubber is curing? The method I had used is I ran a heat transfer simulation on the MDR testing geometry and once the temperature of the geometry was the same everywhere I subtracted that time from the T90 shown on the MDR cure graph and used this time in my simulations so that when each point reached a certain initial temperature the time would start at that node.

 

[racookpe1978]

22 Feb 14 20:01
Correction on my earlier post. The 179 C in ~ 17.5 sec was at .1 mm from the heated sides. The heated sides would, of course, be at the 180.

If and only if this thin rubber washer were firmly clamped between two perfectly flat disks of heated metal kept perfectly regulated to a permanent temperature of 180 F !!

You were assuming no heat loss across the disk-to-rubber interface,
that neither heating "disk" cooled off as it transferred heat to the rubber,
that both sides had no losses across the gap to the 0.1 depth.

If you placed the rubber disk in a "hot air" environment at 180 F, then you need to again "reheat" that air back to 180 F, and you need to agitate the 180 degree air to keep it uniform, AND you need to account for the finite and measurable temperature loss across the air-film-rubber resistance "forced convection ?? " (film loss). With that thin a disk, the reheat air time is going to be a substantial part of the total time to get the middle of the rubber hot. (That is, the air in the chamber needs to be replaced or reheated at the same time as the air left in the chamber itself heats up the rubber disk.)

On the plus side, while you can NEVER get the outside nor the inside of the rubber to get hotter than 180 degrees (can't burn the rubber or over-vulcanize it or damage it by being too hot for too long), you will never be able to get the rubber to 180 degrees either.

 

[boyze]

racookpe1978, all good points. The simulation I offered was based on a rubber molding scenario where the uncured rubber is held under pressure in a mold that is temperature controlled. The original simulation was done awhile back for a simple curing study to understand if we were potentially heat aging a molded part by running an excessive cure cycle.

examorph, the simulation is a transient thermal finite difference model of the rubber volume that numerically integrates the time temperature to obtain the state-of-cure at each node (position) and time using a user input cure curve defined by ts2 and tc90 and the cure sample cure T. The transient finite difference thermal modeling technique is covered in many thermal textbooks. The cure calc is less readily available but a web search by your favorite search engine will get you some usable techniques.

Some of the assumptions in the model:

- no heat gain from curing kinetics
- intimate contact between the pressurized rubber and the mold cavity
- the entire uncured rubber volume starts at the initial temperature
- there is sufficient heat in the mold and heating platens to maintain the mold T at the controlled level
- the model is only 1D (2D if you use the radial option), i.e., it does not account for heat conduction from the ends
- constant thermal properties (conductivity, cp) of the rubber through the entire cure cycle
- ....

This model has worked pretty well for predicting the cure cycle of several high aspect ratio molded parts we've recently developed.

Have Fun!

James A. Pike

http://www.xl4sim.com

http://www.erieztechnologies.com

 

[examorph]

Thank you so much, this is very helpful!

Could you please explain to me what the Cure Temperature Coefficient is and how do I find it based on MDR test results?

Also, could you please guide me on what I should follow to understand what calculations you used for calculating the state of cure used in your analysis? I ask because when I searched online I found many different methods of determining this which seemed to be far more complicated that what you had used as your spreadsheet only requires a small amount of variables easily available from rehometer tests.

If this 1D analysis was ran using explicit methods rather than implicit would this have a large effect on the overall accuracy of the simulation in such a simple case?

Thanks.

 

[examorph]

Hi Boyze,

Forget about my last questions, I will be need to read up further to understand these as I have just realized how large this topic really is and it could not possibly be explained in a single post.

For my purposes the simulation that you have created will do the job. Could you please confirm for me if I am correct in presuming that the state of cure works as a percentage therefore, the value will be >=1 for 100% or over cure? Does this also mean that I should be taking the component out of the moulding die when SOC is 0.9 as only 90% cure is required?

Thank you.

 

[chicopee]

My submission is definitely outside the recommended time element for submitting an entry into this post. Refer to the PDF attachment for my example of a numerical analysis on the rubber pad. For clarity purposes, I exaggerated the thickness of the sample. The analysis also shows one nodal point to which the energy conservation applies and that method of that energy conservation applies to all other nodes. It is an easy method which is discussed in heat transfer books. Since we have isothermal boundaries on all exposed surfaces, the analysis only need to be carried up to the center of the circular pad and to the thickness midpoint.

 

[boyze]

examorph,

sorry for the delayed reply - I've been traveling. Your assumption for SOC is correct for the simulation.

Have Fun!

James A. Pike

http://www.xl4sim.com

http://www.erieztechnologies.com

 

[examorph]

Thank you chicopee, I really appreciate the time you have spent helping me with this.

Thank you for confirming boyze, your simulation gave me good results when tested against real cases. Could you please explain to me how I would obtain the cure temperature coefficient from the MDR results. Thanks.

 

[sailoday280]

For those inclined to do a some applied math. Used product solution for a disc and superposition. Result will be in terms of Bessel and Fourier series. This will give a closed form solution for temp at any point (for example, center) vs. time.

Regards
Sailoday 28

 

[boyze]

examorph,

one way to get the T coef is to run the cure curve for a few different cure temperatures on the subject material.

Have Fun!

James A. Pike

http://www.xl4sim.com

http://www.erieztechnologies.com