compressed air flow calculation 5

[Ductmate]

I'm looking for an equation to figure what the flow rate of compressed air (in CFM) would be through a seamless tube of diameter 0.180", 24" long, at 100psi. One end is open to the atmosphere.

I looked through my old engineering books, but it's been a long time since I've done flow stuff!

Thanks
Vince

 

[zdas04]

Ductmate,
That is a truly ugly set of conditions. Tubing that small and that long will have a significant friction drop, probably enough to result in the exhaust not being in choked flow. Every time I've tried to calculate flow rate to atmosphere at less than choked flow I've been wrong by at least an order of magnitude.

If you didn't have the tube (i.e., if the line was open ended from a reservoir through a 0.180" orifice) then it would be a simple relief-valve flow calc. With your configuration, the way I'd do it is measure the pressure 5 ft downstream of the reservoir and use the AGA equations to get a standard flow rate (SCFM) that you could use as a surogate for mass flow rate to get to the pressure at the end of the tube. If the pressure at the end of the tube results in choked flow, then use the relief-valve calc to get to a new SCFM and use that to calculate a dP down the entire tube (using AGA). If the new dP results in choked flow, you are probably as close as you're going to get. If not then itterate again.

The problem is that none of the equations deals properly with the transition to/from sonic velocity. The relief valve equations assume you are always sonic. AGA assumes you are never sonic. Neither assumption is appropriate for your conditions.

David Simpson, PE
MuleShoe Engineering

http://www.muleshoe-eng.com

 

[Ductmate]

David,
Thanks for the quick response.

The tube is 24 inches long, not 24 feet. Can't measure 5' downstream from the reservoir, it doesn't exist!

where can i find the AGA equations?

I don't need an exact number, just something within +/- 2cfm.

Vince

 

[Montemayor]

Ductmate:

zdas04 has stated the case in an excellent manner. The basic dilemma is that we don't know if you have established Choke Flow conditions or not. We lack the basic data (flow rate, temp., pressures) so we can't make the determination. But you can. Normally, if the pressure drop is 1.8 to 2.0, you expect and obtain Choke Flow. However, we can't tell is the 2 ft of tubing consumes the pressure down to where the outlet pressure drop is less than 1.8. Therefore, what zdas04 recommends is the practical step to take. I suspect Choke Flow, but that has to be proven first.

Art Montemayor
Spring, TX

 

[israelkk]

In my opinion you can use the choked flow equations. However because in the 24" tube it will function as an orifice with a reduced discharge coefficent (Cd). To my knowledge a tube with 0.18" inner diameter is a "large" tube for air therefore, the resistance probably be almost the same as a short orifice.

Calulations for a short (L/D~1) 0.18" diameter orifice gives 39.8 SCFM. To be on the safe side assume 30 SCFM. An actual test is recommended.

 

[zdas04]

My eyes aren't what they used to be. Now I'm mistaking " for ', the old folks home can't be far behind.

At 2 ft, I'd roll the dice and use the choked-flow equations. They give you a result in lbm/hr that you can covert to CFM:

m(dot) = (pi/4)d^2 * C(m) * P * K(total) * [MW/(T*Z)]^0.5

where:
C(m) = 520 * {k * (2/(k+1))^[(k+1)/(k-1)]}^0.5
k = ratio of specific heats
d = orifice diameter (ft)
P = upstream pressure (psia)
K(total) is a stream of constants that is close to 1.0 for your application
MW = molecular weight
T = temp in Rankine
Z = compressibility (1.0 for air)

You ain't going to get within 2 CFM with this equation. With israelkk's evaluation of around 40 cfm you can expect to get to +/- 6 cfm.

David Simpson, PE
MuleShoe Engineering

http://www.muleshoe-eng.com

 

[Montemayor]

zdas04:

Don't feel too bad. There's plenty of company here at the old engineers' convalescent home. Your eyes can't be as bad as my attempt at speed typing w/o proofing. Look at how I left out that the 1.8-2.0 is a pressure ratio, not a pressure drop! I hope everyone forgives me and understands that it is the downstream-to-upstream absolute pressure ratio.

Now, I've got to get back to oiling this squeaky rocking chair. The noise is driving me crazy.

Regards

Art Montemayor
Spring, TX

 

[katmar]

For this type of calculation I would assume isothermal flow, although in reality it will be more nearly adiabatic. But isothermal calcs are so much easier and the error, even in an extreme situation like this, is unlikely to be more than 20% (flow too high).

By my calc the flow is definitely choked. The good thing about choked flow is that as the assumed outlet pressure decreases towards the limit, the flowrate varies very little because the outlet density increases so rapidly. So you should get a reasonable number.

I assumed the air reservoir was at 60 F and 114.7 psia. This gives a flowrate of 160 lb/h or 35 scfm. I used a roughness of 5.0E-6 ft (drawn tubing). It is interesting to note that because the friction factor is a function of the roughness divided by the pipe diameter (e/d), and the diameter is small, the flow is very sensitive to the assumed roughness. If you used the usual value for commercial piping (1.6E-4 ft)you would get a flow of only 120 lb/h or 26 scfm.

Hope this helps
Katmar

 

[zdas04]

I was just working on something unrelated to this and saw that the "d" term in the equation above should have been in inches instead of feet.

Sorry about that, sometimes us old guys just work from memory and that is always a mistake.

David

 

[Ductmate]

Thanks to everyone for their help on this. This has given me the info I need.

Vince