PSA off gas line sizing

[sav455]

Hi,
I am trying to size a low pressure gas line (Hydrogen PSA off gas,MW = 27.89,actual vol flow = 8.749e4 m3/h,inlet pressure = 131 kPa, 32 deg C) for feeding to CO2 recovery unit.The 24" NPS Sch 40 line is giving me a total pressure drop of 6.5 kpa(total equivlent length is 250 mtrs),and the velocity of 107 ft/sec.
The problem is this velocity is 3 times higher than the given in guidelines.My question is: Is it possible that even at three times higher velocity the pressure drop is so low?? Am I making any mistake? Is it unusual?
Any suggestions??
Regards,

 

[Montemayor]

Give us the following:

1) The gas density at the entry to the pipe;
2) The gas absolute viscosity;
3) The allowable pressure drop in the 250 mt length pipe.

We’ll assume the pipe’s roughness as 0.046 mm.

Question: What are you feeding a 27.89 molecular weight gas to a CO2 recovery unit for?

 

[sav455]

Hi Montemayor,

Gas density at the inlet of the pipe = 1.444 kg/m3
Gas viscosity at 32 degC = 1.525e-2cp
Total maximum available pressure drop in the system = 15.9kPa

The H2 PSA purge gas contains H2,CO,CO2,CH4 and very small amount of water vapor.We have to recover CO2 from this stream and use it in froth treatment unit. The technology selection report suggests that PSA technology is the most economical for this.

Regards,
Sav455

 

[Montemayor]

SAV455:

When I plug your conditions into my compressible flow spreadsheet, I get:

P2 = 16.73102 psig
Delta P = 2.268984 psi = 15.6441 kPa

This is within your allowable pressure drop. In fact, for practical purposes, it is there. So the 24" line size is the correct one. I don't see anything bad about your velocity of 107 ft/sec for a gas flow.

 

[sav455]

Thanks Montemayor.

 

[25362]

Professor Lydersen (the University of Trondheim, Norway) proposes a simple formula for the friction drop in gas lines, with Re from 100,000 to 500,000, which I believe can be used as a rough approximation in this case, although Re ~ 1.5[×]10⁶:

[Δ]P_f ~ (L/50) (1/D) [ρ] V²/2​

[Δ]P_f = (250/50)[1/(18.81[×]0.0254)] (1.444) (107[×]0.3048)²/2 = 8036 Pa ~ 8 kPa

 

[sav455]

Thanks 25362.