Tank volume calculations..what was the temp used?

[MikeFerrara]

Mike -

I don't usually fill with a low volume compressor, so I was sort of calculating on the fly. When we fill for the school, we've got 4 banks (only one of which is open at a time) and 4 whips. We try to keep the fill rate between 100 and 200 psi per minute for the AL80s, and that seems to keep things pretty stable in the long run. YEs, the tanks do get warm, but nothing exorbent enough to noticably affect the fill.

When I'm mixing trimix I fill extremely slow in an attempt to keep things cool so my metering is accurate. I fill slow enough to keep the outside of the tank at room temp (by feel). Using a digital gauge I'll still see it drop by 50+ psi easy after cooling. Just because you don't feel it on the outside doesn't mean that you haven't heated up inside.

 

[Big-t-2538]

When I'm mixing trimix I fill extremely slow in an attempt to keep things cool so my metering is accurate. I fill slow enough to keep the outside of the tank at room temp (by feel). Using a digital gauge I'll still see it drop by 50+ psi easy after cooling. Just because you don't feel it on the outside doesn't mean that you haven't heated up inside.

Unfortunatyely, I haven't had the need to mix something other than air....or the training. In that case, I can see a 50 psi difference meaning something...as well as a 5 foot change in depth, etc., etc.

 

[MikeFerrara]

Unfortunatyely, I haven't had the need to mix something other than air....or the training. In that case, I can see a 50 psi difference meaning something...as well as a 5 foot change in depth, etc., etc.

My point was though that even with a really slow fill there is a significant temperature change in the tank.

I don't use a water bath because it's too much trouble but it make getting good fills and mixing gas easier.

When I did use one I never had a problem getting water in tanks.

Now it doesn't matter because I can mix gas over multiple days and I just let things cool and recheck the pressure.

unless you have all day you just won't be able to fill slow enough to keep things cool enough to get a good fill without overshooting the mark or doing something to keep the tank cool.

 

[will_tekkie]

When the tank is immersed in air, dissipation of heat is by convection which is inefficient. In a water bath, heat is carried away by conduction which is very efficient given the physical characteristics of water.


this statement have no sense.....heat transfer between solid and fluid surfaces is due to convection...

in water heat is transfer basically by free convection (a thermal and hydrodinamic boundary layer is a good caracteristic due to this transfer mechanics)..there are two important ways....in laminar flow, conduction between layers (in the thermal boundary layer is the main thing to consider)..in turbulent flow , mass transport is the main responsible)...

Heat is transfer by conduction in the tank walls from the air inside and the water or air outside...

 

[pescador775]

Tekkie, your critique is interesting. However, convection implies pick up and physical transfer of energy contained within the flow and deposited elsewhere. The temp of a tank immersed in water does not get hot enough to cause the convection currents, IMO. Nor is there any particular destination for this energy to be deposited or transferred. It stays within the water and, I believe, enters the water by direct contact and conduction into and through the water. OTOH, when the tank is exposed to air there is enough temp change at the surface of the metal to cause welling of hot air and subsequent recirculation of cold air, through gravity, to cause a return circuit (to the tank). Thus, I see this as a convection current. It is not a closed loop as the hot air presumably goes elsewhere with its higher energy and buoyancy.

 

[will_tekkie]

Tekkie, your critique is interesting. However, convection implies pick up and physical transfer of energy contained within the flow and deposited elsewhere. The temp of a tank immersed in water does not get hot enough to cause the convection currents, IMO. Nor is there any particular destination for this energy to be deposited or transferred. It stays within the water and, I believe, enters the water by direct contact and conduction into and through the water. OTOH, when the tank is exposed to air there is enough temp change at the surface of the metal to cause welling of hot air and subsequent recirculation of cold air, through gravity, to cause a return circuit (to the tank). Thus, I see this as a convection current. It is not a closed loop as the hot air presumably goes elsewhere with its higher energy and buoyancy.

dear friend...i recomend you to make some review about free convection, boundary layers, thermal mechanics related to this and interlayer conduction (laminar flow). For a low flow regimen ( which is supposed to be shown in the tank filling case), the heat transfer is due mostly to the temp gradient (by conduction) between layers within boundary layer but this mechanics is known as convection not conduction (like do happen at the tank walls as i said)..i don´t whish to begin a trolley for this (non important for divers, actually) so i have nothing more to say...

 

[will_tekkie]

Dear friends...do you remember the question about the temp used in volume calculations??..here the answer i received from lufex about the method they used for determined it...

this follows:

Dear Sir,

Thank you for your recent enquiry via our website, I apologise for the
delay
in replying to you.

I belive that you have been looking at the values given for our US
cylinder
range and so I have been in touch with a colleuague in the USA and
asked for
his comments on your query. His reply was as follows:

"We have traditionally used data published by the Department of
Explosives
(now Department of Weights and Measures, I believe). This data was
published for the compressibility of air at standard room temperature
of 70F
(21C). This data was originally printed in the early 70's and so could
be
outdated. The way the calculation works is as follows:

Take the internal volume of the cylinder divide by a residual gas
factor for
the pressure in question which gives the standard cubic foot capacity
of the
cylinder.

The data we use may be slightly different to other published values by
other
organizations. It should all be relatively close however. Also please
bear
in mind that we do some rounding of the numbers to come up with the
capacity. So although an S80 will state 80 cu ft of air at 3000 psi,
the
real number could be anywhere from 75 to 84.9.

The residual gas factor I have for air at 3000 psi is 8.763 cu in water
capacity/cu ft of air. The internal volume of an S80 is 678 cu in.
Therefore, in this case, the capacity in cu ft is 678/8.763 = 77.37 cu
ft of
air."