Pressure Vessel Heat Loss

[jshepard]

Am after a little help with some rough calcs I am doing.

What I have is a gas bottle (i.e. diving bottle) with hot air within. I am trying to calculate the time taken for it to cool.

By simplifying to a cylinder, a figure for conduction heat transfer can be calculated (I realise it's not this simple, but bear with me).

My problem is, I have the heat transfer firgure (in Watts), but I have been unable to work out how this corresponds to a reduction in temperature in the diving bottle??

Anyone got any ideas? PV/T = constant doesn't apply as the system is losing energy.

Any help appreciated. Sorry if it's obvious, this is not an area of expertise!

Cheers

Jeff

 

[Latexman]

Can you get your hands on Process Heat Transfer by Donald Q. Kern? See Chapter 18 Batch and Unsteady State Processes. Or, Perry's Chemical Engineers' Handbook? In my 6th Edition, see Chapter 10, page 38 for Batch Operations: Heating and Cooling of Vessels.

Good luck,
Latexman

 

[IRstuff]

There are also a lot of resources on the web:

http://web.mit.edu/16.unified/

http://www/FALL/thermodynamics/notes/node129.html

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize

 

[fegenbush]

Maybe I'm looking at this one too simply, but the base formula is Q = m * Cp * deltaT

Since you have a know the heat transfer, Q, in Watts, you can assume an average Cp for your material (air) from the reference charts.

 

[balagast]

"Maybe I'm looking at this one too simply, but the base formula is Q = m * Cp * deltaT

Since you have a know the heat transfer, Q, in Watts, you can assume an average Cp for your material (air) from the reference charts. "

I would have to agree with this methodology, assuming the temperature change is not too great the difference in Cp shouldn't be large enough to create a significant amount of error.

 

[Latexman]

fegenbush,

Thats for steady-state heat transfer. "I am trying to calculate the time taken for it to cool" means unsteady-state heat transfer since deltaT will be decreasing with time. Therefore the base differential equation is:

dQ/d[Θ] = -MCpdT/d[Θ] = UA[Δ]T

Good luck,
Latexman

 

[sailoday28]

Don't forget the cylinder metal?

Regards

 

[racookpe1978]

In open air (tropic or arctic) or in the water?

How big a tank (what weight of compressed gas vs weight of steel tank?

Seems like a foolish (or academic-level) question to me, since the heat of compression is going to be added in the compressor, lost in the piping to the HP air tank, then lost (very quickly!) to the atmosphere and ocean water before it will matter.
\
\You can't "overpressurize" a HP tank just to squish more air in so that you can get more air after it cools - not enough to matter on a dive time limit in any case. And if you are that close to a air limit (with safety factor) in your dive calc's to require thinking about cooling the HP air to get another two minutes dive time, you're acting unsafe in the first place.