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Simplified calculation for pressure drop for isothermal compressible gas flow 3

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PagoMitch

PagoMitch

Mechanical
Sep 18, 2003
66
Does such a thing exist for O2 at high pressure?
Searched this site, and cannot find anything similar.

I have tried several online calculators for same, with results that vary by an order of magnitude... from 22 psi to 322 psi.
On the TLV website calculator (pretty detailed, and the unit analysis worked out), their calc result (200 psi) is a far cry from their formula that I went through (22 psi)...
Also tried a couple of the Darcy and Iso compressible flow equations in Metric Crane 410 - I could never get the unit analysis to work out.

Conditions:
Client needs 0.7 kg/sec@ 1200 psi. Call it 2520kg/hr.
I have assumed to provide that volume at 1300psi to allow for a ...reasonable...pressure drop.
At 1300 psi = 115 kg/m^3 = 7.2 lb/ft^3 density.
after some conversions = 11.7 ft^3/min (0.0054 m^3/sec) mass flow.

O2 at 1300 psi (89.6 BAR)
Pipeline = Drawn copper at 265LF (81M) at 3/4" ID (19mm) - pipe and O2 tanks are indoors at appx 70-100F (21-38C)
Velocity = 63.6 feet/sec (20 m/sec)
O2 Dynamic viscosity = 0.02125 cP (22.89 x 10^-6 Pa s)
Surface roughness (Epsilon) = 0.0025 mm
Relative Roughness (Epsilon/diameter) = 0.00013
Reynolds #= 2.5Mil, Relative roughness 0.00013, Darcy FF = 0.015
O2 compressibility factor 0.308
O2 molecular weight = 32
Ratio of specific heat "k" = 1.4
Roughness coefficient = 0.0015
Mach # = 0.08; but this was from a site that yielded a pressure drop of 283 psi (???), as well as well as a Reynolds # 2X what I calculated.
A multi-hundred pressure drop for this system just does not feel right...

We have 2-bank O2 Manifold, with H size Cylinders at 2000 psi: so we have lots of P available.

Think I have everything covered above.
I think/thought we fall into the range where Darcy is still applicable; but I cannot seem to make it work.
Can anyone point me to something that has been verified to be reasonably accurate? +/- 10% or so is all I am looking for.

TIA
 
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And your length of tubing is?

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
 
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There is a lot of data there, I know!

265LF, 81M.

 
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Figured I'd provide what I ...think... is the most accurate calcs:

Based upon the calculator at TLV:

DP = mu * L * v^2 * Ro / 24 * d * g

In US units = .015 (Darcy FF) * 265 * 63.6^2 * 7.2 / 24 * 0.0625 * 32.2

= 115,739/48.3 = 2396 lb/ft^2
2396/144 to convert to psi = 16.6 psi
And the unit analysis works!

This feels ...better... than 176 psi, but I cannot find any other equation to duplicate these results.
 
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Here is the classic method of solving pipe flow for both liquids and gas lines
Worked example for both fluids.

attached

The diagram referred to but not copied is the typical
Moody Friction Factor Diagram

f can also be solved for by iteration starting with f=0.017 and it converges in 3 to 4 calculations.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
 
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Required inlet pressure at these conditions is 1530psig.
Error in input compressibility, should be 0.95-0.96. With roughness of 0.03mm for clean service non corrosive piping.
Routine used is isothermal compressible flow.

For roughness of 0.002mm typically used for tubing in clean service, required inlet pressure is 1400psig.
 
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Thanks gents.

1503-44 - I used Moody Diagram to obtain the Darcy FF of 0.015, which was used as "mu" in the TLV calcs.
Your link takes me to the Engineering.com "Free File Sharing" download page, but a click on that link leaves me with a blank screen. Nothing was added to my Downloads file. Not sure what is going on? is it a huge filesize?

georgeverghese - I listed surface roughness in my data at 0.0025. An inlet P of 1530 would mean a dp of 230 psi. That seems...large.

As your "tubing in clean service" value (0.002 mm) is pretty close to what I have, (0.0025), the resulting 200 psi dp is still pretty large; but it is close to the TLV computed number of 175. Still a long way from their formulaic number of 16.6.
Is the "isothermal compressible flow" calc you use computer generated? or was it from Crane 410 equation 6-10? That equation drove me nuts; I can find P(avg) if I know P(1) and P(2). Does not help much if you only know P(1).
 
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I've got an excel sheet for this, and its based on the expression for isothermal compressible flow in Perry Chem Eng Handbook in the chapter on Fluids Transport. In this expression, an average compressibility factor between the inlet and exit is used. In this case, z is about the same going from 1400psig to 1200psig. Since this is compressible flow, the expression self corrects for the actual pressure at inlet and exit.
 
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1503... I don't do this type of work, not since university, but that appears to be a great link...

-----*****-----
So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik
 
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If it is really isothermal flow then you should use the isothemal gas flow equation.
 
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georgeverghese - Thanks for running that. So based upon the 1200 psi needed at the point of use (not the 1300 I also listed), I have about a 200 psi dp. Can you confirm if you used 1200 or 1300?

1503-44/dik - I still cannot access that link. Although I can access the link provided by pierreick fine.

pierrick - thanks. Good article - That reminded me (I graduated about 40 years ago!) this is an iterative solution; have not used that much in designing mechanical systems for hospitals!
Found a ...reputable... site (Chemical Engineering Calcs, "Che Calc") with an Isothermal flow calculator, verified my variables, and after half a dozen runs got to a number.
I have attached my final run.

I was initially getting a calculated value for density of 352 kg/m^3; which seemed odd, as my calculated value was about 115 kg/m^3 at 1300 psi. See upper right of attachment. Seems I had a value of Z (compressibility) of 0.308, which I found on a website. A couple other websites defined Z (at 1400 psi) at 0.95-0.97; including one from the National Bureau of Standards, circa 1947! That seems to have solved the problem.

This meets my requirement to provide a nominal 0.7 kg/sec (2500 kg/hr), with a resulting dp of 221 psi; which seems (not coincidentally, I hope) to closely match the value provided by georgeverghese.

Thanks again all.
 
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What's up with that 221 psi pressure drop? How many hundreds of meters of tubing you got there?
I asked before how long is the tubing? Still don't know.

I tried your che calculator using closely matching data as I could .z=0.9

I used 10m length.

As you can see i used a pressure drop 0.1B
And I got a flow 5X what you need.

I could have 200m length and still get 2500 m3/h flow. See 2nd run.

It's been awhile since I did a gas run, but I think I could get a few 100K std cubic feet per day through a short 3/4" gathering line at a nominal pressure drop at 1200 psig.

Screenshot_20230916-204535_Brave_xu9f71.jpg


Screenshot_20230916-205511_Brave_bc5orm.jpg


BTW, D'Arcy is an isothermal equation. No temperature is lost or gained.
I didn't run D'Arcy, but I do believe we are talking about a relatively few psi,or 1/10th of bar pressure drop no matter what equation you want to use.

The particular link isn't important, you just need a Moody diagram. Your choice which one, they are all the same. Or just Iterate f until you've got equality. Usually only takes 3 or 4 iterations no matter what your first choosen number is.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
 
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I did a calculation on my spreadsheet using the Darcy Equation attached. However I used a compressibility factor of 0.96 based on the attached chart. Looks like 1" ID tubing is required as there is a big difference to 3/4" pressure drop.

The calculation is attached to this post. The O2 chart is attached to my next post.

Note I used a value of flow pressure 1250 psig which is average of inlet and outlet pressiures and this gives close enough approximate results. For the Darcy equation to get more accurate results you would do a step by step calculation where only 5% of pressure drop versus intial pressure is lost in each step since the pressure drop per length is a function of the original pressure. However between 1300 and 1200 psig there is not that big a difference.
 
 https://files.engineering.com/getfile.aspx?folder=26ce6010-3ba6-4cc3-8eae-15f9fee97743&file=O2_Pressure_Drop.pdf
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OK I now see the 265ft length. Pretty far for tubing at 1300psig
Is this in a protected environment?

Isothermal needs 1.5 pipe D to get a 6 bar dP.

Screenshot_20230916-235410_Brave_xpp2dj.jpg


--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
 
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Based upon the calculator at TLV:


DP = mu * L * v^2 * Ro / 24 * d * g


This looks like the friction loss equation of a liquid

dH = f L/D (V)2/2g Then multiply by density in #/ft3 divided by 12? Some kind of conversion from feet head to pressure?

Don't believe everything you see on the internet.

Also you made a mistake in your calculation above with this equation - you show d as 0.0625 but it is 0.75
 
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Yeah, I need a 12". That calculator is messed up.



--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
 
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That is the liquid head loss equation as follows

Friction loss for liquid

dH = (fL/(D/12))(V2/2g) in feet of head. Where D is in inches - needs to be converted to feet by dividing by 12

To convert to PSI multiply by density and divide by 144

dP = {(fL/(D/12))(V2/2g)} (Rho/144)

rearrange and you get same equation which is a liquid flow equation.
 
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