duration of flow at 100psi from bulk storage tank 1

[stanfea]

I am interested in calculating the duration of flowing gas (nitrogen, argon, or air) from storage tanks (cylinders, receivers, etc). The purpose is to determine if the demand can be met without changing the tank every day.

For example, I have a tank of compressed nitrogen at 2000psi, that has a compressed volume of 8ft^3. If I need 10CFH at 100psi, how long will the 100psi be able to be maintained flowing?

If I calculate the volume of gas released that doesn't really do me any good. I need sustained 100 psi flowing. Any guidance on this subject is much appreciated.

Stano

 

[quark]

I will leave the precise calculation to chemical engineers, yet I will show you a simple and rough calculation to have a feeling.

When the pressure in the main receiver comes to 100 psi, you can't draw any gas out of it. So the total quantity of useful gas in the receiver (expanded to atmospheric pressure) is (1900+14.7) x 8 = 15317.6 cu.ft

The consumption of gas (again expanded to atmospheric pressure) is (100+14.7) x 10 = 1147 cu.ft/hr

So, the receiver can deliver for a period of 15317.6/1147 = 13.35 hours.

 

[stanfea]

Quark

I appreciate the feedback, however something seems off with the units.

If I multiply the following I get:

(100psi + 14.7psi) x 10ft^3/HR = 165,168 lb-ft/hr

Not 1147ft^3/hr? As you suggested.

Can you have another look at this? Am I missing something?

Everyone else is also welcome to chime in. I started out thinking this should be a well documented and simple process, however I have not managed to nail it down yet. I have several books on compressors, mechanical handbooks, etc to no avail... All help appreciated.

Thanks
Stan

 

[Zapster]

Find the density of the fluids of interest at both 2000 psi and 100 psi.

http://webbook.nist.gov/chemistry/fluid/

can be a helpful site.

Then convert volume of fluid to mass for the 2000 psi supply
Then convert the mass to a volume at 100 psi.
Subtract the volume of the cylinder from above.
This should result in the volume of gas you have available at 100 psi.

You will need to know the temperature at states, 100 psi and 2000 psi, to determine the density. So what is the process you will use to drop the pressure? Critical flow through an orifice or nozzle. Isothermal?

 

[stanfea]

One of the processes consists of a gas (ie Argon) flowing thru a nozzle at pressure. The purpose is shielding/piercing gas for a CNC machine. The gas comes from a bulk storage tank, which I am sizing. The tank starts at 2200psi and slowly declines to 100psi. Here is what I get?

I assumed a constant temp of 60deg F (isothermal)throughout the process of emptying the tank. It is not clear to me that this is the case, comments welcome.

Tank volume = 8ft^3
High pressure (full tank) = 2200psi
Desity@2200psi = 17 lb-m/ft^3 (NIST)
Mass@2200psi = 17 x 8 = 136 lb-m

Low pressure (empty tank)= 100psi
Density@100psi = 0.75 lb-m/ft^3 (NIST)
Volume@100psi = 136lb-m / 0.75lb-m/ft^3 = 181ft^3

Therefore total gas @100psi available = 181-8=173ft^3

Is this correct? Comments appreciated.

Thanks
Stan

 

[quark]

You have taken the literal meaning of the calculation. What I meant is isothermal expansion (P1V1 = P2V2). To keep out ambiguity, let us take mass instead of volume.

Going to the NIST database, assuming 60F (BTW, what temperature you considered and what process you choose to get the data?)and isothermal process, density of nitrogen at 2214.7 psia (or 2200 psig) is 10.925 lbm/cu.ft and at 114.7 psia (or 100 psig) it is 0.57736 lbm/cu.ft.

Total mass of gas in the receiver is 10.925 lbm/cu.ft x 8 cu.ft = 87.4 lbm

As I already said, when the receiver pressure reaches 100 psig, the nitrogen inside is of no use to you. So, unused mass of nitrogen is 8 cu.ft x 0.57736 lbm/cu.ft = 4.618 lbm. Available mass of nitrogen is 87.4lbm - 4.618lbm = 82.78 lbm

Actual volumetric consumption rate of nitrogen is 10 cfh (at 100psi), so mass rate is 10 cf/hr x 0.57736 = 5.7736 lbm/hr.

So, your receiver can support you for 82.78 lbm/5.7736 (lbm/hr) = 14.33 hours (if the chemical engineers excuse me, I would take it as close to my initial value of 13.35 hours)

 

[quark]

Now I came to know how stupid I was. As I always work with SI units, I missed out the important factor. You are right.

The corrected calculation is as follows.

Volume of gas at 2000psig is (2000+14.7)x8/14.7 = 1096.43 cu.ft.

At 100 psig it is 114.7x8/14.7 = 62.42 cu.ft

Available gas is 1096.43 - 62.42 = 1034 cu.ft

10 cfh at 100psi is 10 x (114.7/14.7) = 78.02 cfh at 14.7 psia

So the time for which you can use the receiver is 1034/78.02 = 13.25 hours (actually, my initial answer was not wrong as I had to divide both numerator and denominator with 14.7 and this only to clarify you about the irregularity of the units)

Alternately, on mass basis, if you consider 2000psig (instead of 2200 psig), the density is 10.005lbm/cu.ft. So, the time is (10.005 * 8 - 0.57736 * 8)/(10 * 0.57736) = 13.06 hours.

 

[BigInch]

Due to the Joule-Thomson effect, you will not be getting gas out of the tank at the tank's ambient temperature. Tip: It will not be an isothermal calculation.

Going the Big Inch!

http://virtualpipeline.spaces.msn.com

 

[25362]

Don't forget that the continuously reduced mass of gas, left behind in the supplying tank, cools progressively by expansion (ie, a polytropic pressure reduction). The tank itself will also cool down. If one adds the JT cooling effect -mentioned by BigInch- the expansion should, no doubt, result in cooling of the exit gas, and the whole process wouldn't be isothermic.
See the discussion in thread391-78680.

 

[BigInch]

I was thinking the expansion rate wouldn't be enough inside the tank to cool it, but it is a possibility. Good you brought that part to attention.

Going the Big Inch!

http://virtualpipeline.spaces.msn.com

 

[25362]

I pressume that, if the tank were thermally insulated, the [Δ]T (on cooling) by depressurizing could be even greater than the one produced by the JT effect.