Unsteady state heat transfer in an aerosol can

[Roberttt]

Hi,

Since I'm working in an aerosol packing manufacturer, I've been looking to a heat transfer problem. Here is a description of the case:

The filled aerosol cans with propellant LPG need to be tested to make sure that they do not fail when heating the can up. As you can see on aerosol cans, they all have a warning 'do not heat up above 50 degrees'. These cans are tested in this factory, where they will be going through a heated water bath with temperature x using a conveyor belt that is running on speed y (what gives a . My task is to calculate what the temperature x must be while the conveyor is running on speed y what gives a residence time of 97 seconds.

Because I am quite new to unsteady state heat conduction, this is a very hard case for me. Even though this is the situation, I would like to have a look into unsteady state heat transfer. Is there anyone who can give me some information what the best way is to deal with this problem?

Regards,

Rob

 

[chicopee]

Are these cans filled with the propellant LPG and the substance that is labeled on the tank for the buying customer are tested in the water bath? Is the water temperature staying the same while the cans travel thru it? Do you have an idea as the convective heat transfer coefficient on the skin of the cans as water flow around the cans? Do you know the amount of LPG and of the product being sold? What is the product mixed with the LPG? There are few items that we can surmise to be somewhat correct. First there is no temperature gradient in the wall of the aerosol can; secondly the wall temperature is the same as the bulk temperature of the content; thirdly the temperature rise within the aerosol can wall is the same as that of the content; fourthly the convective heat transfer coefficient on the exterior surface does not change to simplify the calculation.

I would start with the basic heat transfer equation: Hc*A*dT= {M*Cp*dT/dt}content+{M*Cv*dT/dt}can; Hc*A*dT: Hc-convective heat transfer coefficient, A-exterior surface area of the aerosol can; dT =(T2-T1) temperature change of the content; M*Cp*dT/dt}content: M-mass of content; Cv specific heat at constant volume of content; The term for content will need to be applied separately for the propellant and the product of interest to the consumer; dT/dt=(Twater-T1)/dt= rate of rise of content temperature; {M*Cp*dT/dt}can: M- mass of can, Cp specific heat of can; DT/Dt ={(Twater-T1)/Dt} rate of temperature rise of can which would be equal to that of the content. Doing a time step on a spread sheet, I would assign the time change dt as a small value such as 0.1 second or even smaller; T1 is the temperature of the content and propellant and I would start it at room temperature, T2 is the final temperature of content and propellant at the end of Dt and would be calculable during the first itineration and would be the value for the next itineration as T1; Hc will be difficult to determine so some further research on your part will be required.

To be noted is the missing heat transfer equation of the product within the can which would have a similar form as(Hc*A*DT). In my opinion, this missing term would be complicated since Hc would again be needed to calculate heat transfer within the can and may be too small of a value, so I disregarded it. At the end of 97 seconds, T2 should be the final value that you are seeking. Let us know about the result if you adopt the above procedure.

 

[racookpe1978]

50 degree temperature CHANGE? Or 50 degree temperature limit?

Degree F or degree C?

 

[Roberttt]

@chicopee:
Are these cans filled with the propellant LPG and the substance that is labeled on the tank for the buying customer are tested in the water bath? Yes, the cans are filled with LPG, those cans are tested in a waterbath. Is the water temperature staying the same while the cans travel thru it? Yes it stays the same. Do you have an idea as the convective heat transfer coefficient on the skin of the cans as water flow around the cans? I discussed with my boss yesterday & he said that it is maybe better to start only with conduction. He said: Assume that the material of the can at 50 degrees is, so I only have to look at the conduction in the LPG (I can assume that it is standing still/not moving). Do you know the amount of LPG and of the product being sold? No, but I can get this information if it is needed. What is the product mixed with the LPG? There are different products mixed with the LPG, but I can assume that it is in 2 layers in the can: 1 layer product and 1 layer LPG.

So what is the exact case:
I can assume that the can is 50 degrees & that there will be no convection (only conduction!). My task is to calculate how much time it takes to heat up the LPG in the can if the can temperature is 50 degrees.

 

[chicopee]

I would forget conduction thru the aerosol can as, in my opinion, is an insignificant heat transfer condition thru the thin wall of the can. If we had a wall thickness of let's say 1/2" then yes transient heat transfer by conduction would need to be considered but for the aerosol can wall thickness is probably 1/16" which in a water bath would not have much transient effect on heating the LPG and Product. I would tackle this problem by considering an increase in internal energy of the LPG and of the other product and that of the aerosol as I proposed in my first reply. Convection heat transfer in your case is real and should be included.

 

[racookpe1978]

I would expect not much more wall thickness than 0.020 to 0.010 inch

Putting a pressurized can of air (PC keyboard cleaner) in 120 degree F water for 15 seconds heats up the outside of the can from <60 F to 85 F. Leaving it for 20-25 seconds heats the internal air sufficiently while agitating the can in the water is enough to make a substantial increase in blown force of the air inside.

 

[Roberttt]

The thickness of the can wall is 0.20mm (0.00787 inch).

 

[chicopee]

Not much of a wall thickness to think about transient heat transfer. I would not expect much of a temperature gradient within the first milliseconds.

 

[LittleInch]

I think you have too many unknowns and transient events to be able to calculate your water bath temperature with any accuracy.

Trial and error and measurement of the can temperature as it comes out of the bath would seem to be the best solution here and start with the water bath temp at 51C and then work up from there so you don't accidentally overheat your can. How much margin / accuracy do you need on your 50C can temperature??

Aerosol cans are not big volumes of liquid and a minute and a half immersed in a moving bath of water kept at a constant 50 C swirling around it is intuitively enough to get the can contents pretty close to the water bath temp. If I had to make an estimate I would go for somewhere between 55 and 60 C for the bath temperature.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Roberttt]

@LittleInch I already did it that way and got an end temperature of 55 C, but because I want to get to know more about unsteady state heat transfer I'd prefer to calculate it on the theoretical way as well.

 

[EdStainless]

If the cans are not being shaken could you use the properties of the can contents and go to a Heisler chart?
The catch is that you have two fluids inside, but how much mass of each and what are the thermal properties of each?
It may be that you can ignore one of them (lower mass and lower HC).

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube

 

[LittleInch]

The missing factor is the Hc as chicopee says. That can only be really gathered from experimental data. You can back calculate that but not really sure why you want to?

I wasn't far out with my estimate though

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Roberttt]

@EdStainless The cans are not being shaken. I have looked at that already but the Fourier Number is too low to use the Heisler Chart. I got a heat transfer book that says that there is another way to calculate it, so that is what I did and I found a time of around 35 seconds to heat up the LPG from 20 C to 50 C. Is this an acceptable time or is it too low?

 

[Roberttt]

@LittleInch Can't I calculate the hc factor by calculating the Nusselt number (with Prandtl number and Grashof Number), and then h=(Nu*k)/L ? When I do that I find a h value from 312 W/m2K.

And I am a bit stuck of what the exact meaning is of the characteristic time. I need this value to calculate the Fourier Number.

 

[LittleInch]

Roberttt - I don't know - I'm not a heat calculation guru so it's out of my knowledge envelope.

I was just trying to look at this from a practical point of view, not a theoretical calculation view.

LI

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[MisterE2]

gah... that's an awful method of "testing" even if you heat the can up to 150' C the gas isn't going to heat up and go to high pressure to cause a failure. It'd probably be more effective to just randomly pull cans and throw them in a stove for batch testing

 

[racookpe1978]

The original poster is very reluctant to do actual testing (put can in hot and cold water, measure can temperature over time) to get the actual results he needs.

Since that test could have been done many times in many different levels of agitated water (or still water) in the weeks this thread has been active, I can think of no valid theoretical calc that can help: Everything requires approximations plated over approximations covered up by more theoretical assumed properties of theoretical spaces, shapes, and weights.