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# Calculate Water Volume Of A Gas Cylinder

8 replies to this topic

### #1 tm1274

tm1274

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Posted 22 August 2013 - 03:54 PM

Is it possible to calculate the water volume in gallons if you know the pressure, temperature, and cubic footage of the gas such as air?

I have an application where I know that the volume is 505 cubic feet of air at 6000 psig and temperature would be 70 degrees F., in a cylinder but need to know what the water volume of the cylinder.Can this be calculated without having the diameter and length of the cylinder by using the items I do know? I have found examples online such as 3.14 x (radius* x radius*) x Depth = Cubic Feet.
then Cubic Feet x 7.48 = Total Gallons, but this number is way too high based upon a cylinder that I do know the measurements of. I have also used a different calculation: WP of Cylinder in PSIG divide by 14.7= ATM +1 = ATA (pressure in atmospheres absolute) then Cubic foot rating of cylinder divide by ATA above = actual liquid volume of cylinder in gallons, which is closer to the correct value but still too low based upon the manufacturers data sheet of the cylinder. The cylinder that I do know the dimensions of is a 510.37 cubic foot gas (air) volume at 6000 psig it is 52" long and 9.4" diameter with an actual liquid volume of 1.47 gallons. Using the first method I get 26,979.43 gallons, which is way too high and obviously wrong. Using the second closer calculation I get 1.247 which is close to the manufacturer's specification of 1.47 gallon but still not accurate. Does anyone have an example of how to do this if it is possible? Should I be using a compression factor in the calculation somehow to get an accurate result?

Edited by tm1274, 22 August 2013 - 03:57 PM.

### #2 fseipel

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Posted 22 August 2013 - 10:17 PM

Yes, at 6000 psi air is non-ideal.  Consulting a compressibility chart or spreadsheet gives Z=~1.22.  Bear in mind, your 505 cu. ft figure is STANDARD cubic feet NOT cubic feet at 6000 psig.  So first we need to calculate moles of air at standard conditions (1 atm, 25 deg C).  At standard conditions, Z=1.

1 lb-mole at standard conditions = 359 cubic feet (can confirm this by solving pV=nRT at standard conditions).  So 505 cu. ft /359 cu. ft / lb-mole = 1.406 lb-moles air in cylinder.  We can then proceed to calculate the volume at 6000 psig:

pV=ZnRT, V=ZnRT/p = 1.22 * 1.406 lb-moles x 10.73 ft^3-psia/lb-mole/deg R x 529 R / 6014.7 psia = 1.62 cu. ft x 7.48 gal/cu. ft = 12.11 gallons

Second 510.37 cubic foot cylinder is 9.4" dia x 52" L.  Vol of that cylinder (neglecting heads) is ~3.14 x D^2/4 x Height = 3.14 x 9.4 x 9.4 / 4 x 52 = 3607 cu. in.  Converting to cu. ft by dividing by 12^3 or 1728 gives 2.08 cu. ft.  2.08 cu. ft x 7.48 gal/cu ft gives 15.6 gallons.  Since the first unknown cylinder is 505 cu. ft and the second 510.37, 505/510.37 = 0.989 ratio and so 0.989 x 15.6 gal = 15.4 gallons.  The two figures probably disagree slightly due to my using a flat head calculation whereas obviously a 6000 psi cylinder will have dished heads.  Also your message isn't clear if cylinder diameter is OD or ID; ID should be used in calculations.  Some numbers in your e-mail (such as 1.47 gal figure) don't make sense so check and double check units.  I suspect you meant 1.47 cu. ft.

Typically water capacity is stamped on a cylinder so this is not something one would typically be calculating.

Edited by fseipel, 22 August 2013 - 10:20 PM.

### #3 tm1274

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Posted 22 August 2013 - 11:24 PM

Hi Fseipel,

Thanks for responding to my question. I am very much a novice when it comes to gas compression or gas laws in general. It appears that the information that I had is slightly different than the manufacturer now has listed on their site: http://www.norriscyl...ifications.php#

Attached is the data sheet for the specific example i mentioned as being 510.37 cf. Their website shows their part number 8HP586C and has a volume listed as 2641 cubic inches, which would only be 11.43 gallons. This cylinder at one time was listed as 1.47 I thought gallons but it must have been ACF?? I cannot seem to grasp the reason the numbers are not adding up. I suppose the manufacturer could be wrong but that seems strange as well. I am not sure how to calculate the number in the first example if the second does not match up with the factory. I would assume their engineers or at least during testing someone would have determined the correct volume for accuracy. Would you mind taking a look at the link to their website and the attached drawing and giving me your thoughts as to whether you believe their number is accurate based upon their actual documentation? Also, you mentioned a spreadsheet for the compressibility, do you have this or a sheet that can show me the basics on the calculation with the factor or do you know what I should look for in the forums to get the information? Or wher I can find the equation you used broken down so I can study each needed item to appropriately understand how to do the calculation. Any assistance wold be greatly appreciated..

#### Attached Files

Edited by tm1274, 23 August 2013 - 06:45 AM.

### #4 PaoloPemi

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Posted 23 August 2013 - 03:06 AM

your question is not clear,

you know the vessel volume (for the first item in your second post link 40 liters),

that is the volume of water contained in vessel (40 liters),

if you wish to calculate the volume of gas at atmospheric conditions (for example 15 C and 1 Bar.a)

you need only to calculate the density at rated conditions (15 C, 200 Bar.a)

and at atmospheric conditions (15 C, 1 Bar.a)

for Nitrogen density at 15 C 200 Bar.a is 227.85 Kg/M3

at 15 C 1 Bar.a 1.1699 Kg/M3

(values calculated in Excel with PRODE PROPERTIES and Peng Robinson)

then the volume of Nitrogen at atmospheric conditions is

0.04 M3  * 227.85 Kg/M3 / 1.1699 Kg/M3

or 7.79 M3 which is the aproximate value given by manufatcurer

supposing you know the rated pressure (and temperature) and

you can measure the volume of gas at atmospheric condition

you can estimate the vessel's volume

7.79 M3 * 1.1699 Kg/M3 / 227.85 Kg/M3

or 0.04 M3

you can expect 1-5% errors in gas density (depending from method and condtions) and estimated volumes with this procedure,

by the way you can apply the same procedure for different fluids

however I would prefer to calculate volume from sketch or

weight the water required to fill the vessel,

Edited by PaoloPemi, 23 August 2013 - 04:11 AM.

### #5 tm1274

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Posted 23 August 2013 - 12:29 PM

I apologize for not being clear on my statement. Honestly, it is the lack of knowledge and understanding of how to do the calculation and my trying to explain is difficult to speak laymen terms on a subject that I am trying to grasp. Please accept my apologiies. I am very thankful to have found a forum that is willing to take time to help with my thirst for knowledge on how to do this calculation accurately. I think I may understand but am still tying to understand each part of the equation.

One note though in case you were looking at a different number on the cylinder list, the cylinder that I was using as an example was HP586C and it shows 43.3 liters not 40, but I don't think that matters based upon your equation.

Did you create your own spreadsheet that you mentioned (values calculated in Excel with PRODE PROPERTIES and Peng Robinson) or was it one from the forums?

### #6 fseipel

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Posted 23 August 2013 - 10:33 PM

You should be able to look up the compressibility factor in a table such as http://www.enggcyclo...y-factor-table/ and simply interpolate -- or alternately you can use an online calculator for it e.g. http://www.peacesoft...rte/luft_e.html

PREOS.XLS will use Peng-Robinson and has youtube tutorials & will calculate compressibility & other parameters also.

As PaoloPemi indicates, there will be variations in different estimate methods.  Also note one column in mfr table indicates minimum water capacity -- cylinder itself will have manufacturing tolerances per dwg (+- ~1/4").  Due to these variations, manufacturer states in their table (in link you provided), APPROXIMATE capacity at rated service pressure.

Since pV=ZnRT, or p=ZnRT/V, so for Z>1 (which it typically is for higher pressures), the gas exerts higher pressure than an ideal gas.  Similarly, solving for V you have V=ZnRT/p -- the real gas takes up more space (higher volume) than an ideal gas at a fixed pressure.  The reason is at elevated pressures of ~410 times that of atmospheric pressure, the repulsive intermolecular forces play a larger role and thus require this correction factor -- when molecules are closer together the repulsive force matters so it takes more pressure to force the gas into a given volume vs an ideal gas -- so Z=1.22 means gas deviates 22% from an ideal gas and thus takes up that much more volume than an ideal gas at same pressure.

My initial post as stated didn't account for dished heads; also wall thickness is 3/8" on these high pressure cylinders -- both factors reducing volume vs my initial calc improperly using OD rather than ID since I didn't have it.  The bottom head is dished IN to make a foot upon which cylinder can rest vertically.  Overall length also includes threaded coupling at top.  So those factors reduce actual volume vs a cylinder with flat heads & thus explain the calculated vs actual vol difference.  It is also possible the manufacturer capacity figure is based upon filling on a cold day, but allowing pressure to rise if stored outside where cylinder may heat to some maximum temperature.

Edited by fseipel, 23 August 2013 - 10:46 PM.

### #7 PaoloPemi

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Posted 24 August 2013 - 01:52 AM

tm1274,

you can utilize the same equation (or the inverse in case you prefer to calculate gas volumes instead of densities) for different fluids and conditions,

for example from your data sheet HP586C has a volume of 43.3 liters (0.0433 M3) and a working pressure of 414 Bar,

supposing you wish to store oxygen at 414 Bar.a 15C the density is 513.48 Kg/M3

at atmospheric conditions (1.013 Bar.a, 15 C) oxygen has a density of 1.353 Kg/M3

so the total volume *at atmospheric conditions) will be

=0.0433 M3 * 513.48 Kg/M3 / 1.353 Kg/M3

16.42 M3

values calculated with Soave Redlich Kwong Extended in PRODE PROPERTIES,

as suggested by fseipel you can calculate densities (or volumes) with different methods and software,

in my opinion REFPROP (NIST) and PRODE PROPERTIES are two good tools for these purposes,

a free version of PRODE is available  at http://www.prode.com

and there is online access to REFPROP from NIST site.

As said in my previous post at high pressures different methods can predict quite different values,

for example for oxygen at 414 Bar.a 15 C

PRODE (with SRK EXT) calculates a density of 513.48 Kg/M3

REFPROP calculates a density of 520.91 Kg/M3

so there is about 1.4% difference in calculated values

in my opinion a more accurate way to measure the internal volume is to weigh the vessel (before and after filing with water)

### #8 Art Montemayor

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Posted 24 August 2013 - 04:18 PM

Paolo:

I started my career in the Industrial Gas business, operating, managing, and later designing and installing these plants.  I filled a LOT of industrial gas cylinders during 8 years.  I think you'll be interested in knowing that what you stated in your last sentence is exactly the truth.  The most accurate way to measure compressed gases is to weigh the cylinders before and after compression.  Some cylinders have their tare weight stamped on their top crown.  All compressed liquid cylinders (CO2, ammonia, chlorine, propane, butane, CO, nitrous oxide, etc.) are filled and sold in this manner.

Cylinders of Oxygen and Nitrogen and other gaseous substances are measure by their filling pressure (and not by weight) because of the convenience factor of filling them much faster.

### #9 PaoloPemi

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Posted 25 August 2013 - 02:55 AM

thanks Art

Edited by PaoloPemi, 25 August 2013 - 03:10 AM.