How long to fill a air cylinder? Time vs. Pressure vs. Volume 2

[markdblissmn]

I am trying to figure out how to calculate how long it takes to fill an air storage tank of a certain size to a certain pressure given a compressor HP. What calculation or method should I use?

Imagine a 1000HP central facility compressor connected to a 4" main distribution line. If I have a 1000gallon air receiver and I want to fill it to 5atm pressure, how long would it take? How does the hose/pipe from the main 4in line affect it? How to size this hose? Cv factors like for liquids? Looking for guidance.

Thanks!

Mark Bliss

http://www.2RMengineering.com

 

[racookpe1978]

What class?

Or, if this is a real problem, what is the capacity and efficiency of the compressor? More specifically, what professor believes that a "1000 hp compressor" defines the delivery capacity of a real compressor?

 

[markdblissmn]

It's not for a class. I have been out of college for 6years and am helping my father in law with a design problem. Don't believe me? Check out my website. He works at a plant with a 1000Hp air system that feeds all of the machines. They want to try an experiment in a steel cylinder of 28"ID and length of 30in. He's trying to figure out how to get the vessel up to pressure of 150psi, or 10atm. But, how does he size the hose? 1"? 4"? What impulse concern might there be.

I was trying to make the question generic as to point me out in a direction to solve it. I work on machine design and don't dabble in air systems at all.

Mark Bliss

http://www.2RMengineering.com

 

[zdas04]

You have to ask a couple of questions before you can get to an answer:
[ol 1]
[li]What is the limiting factor in the rate of fill (is it the compressor? the hose? the inlet nozzle?)?[/li]
[li]What mass of air represents the full vessel?[/li]
[/ol]

If your limiting factor is the hose or the nozzle, then the upstream pressure will remain constant. Use something like the Panhandle A equation to calculate a flow rate through the hose with your constant upstream pressure and your starting downstream pressure. Run that for 10 seconds and calculate the pressure of the vessel. Calculate a new rate with the new pressure and run that for 10 seconds and recalculate downstream pressure. Repeat until your downstream pressure reaches the target. You can try it with longer or shorter intervals (I sometimes use a minute, but that overstates the flow rate for most of the time interval and doesn't usually match measured data very well, using 1 second intervals results in a lot of iterations but slightly better results).

If the limiting factor is the compressor, then the problem is a lot more difficult. Compressor discharge pressure will build with vessel pressure and friction, so you have a really hard time calculating the flow rate in the hose with an unknown upstream and downstream pressure.

College Engineering courses would seem to tell you that you can integrate a function to predict the elapsed time without iterating. I've never seen a single time that I was able to make that work with real-world fluid mechanics problems.


David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.
The plural of anecdote is not "data"

 

[hacksaw]

generally receivers and the connecting lines are sized according to the expected demand transients, you probably need a 4" connection to the vessel in your case

as to fill rate that depends on the equipment and how you start it up,

you need a shut-off (gate) valve as a minimum, the fill rate is limited by the load capability of your compressor/drive train.

practically, you open the vent valve on the tank, crack the inlet valve and watch the pressure gauge as the pressure comes up close the vent

 

[EdStainless]

To get the exact answer takes calculus, but an approximation will work.
Do you have a flow vs pressure curve for the compressor? Get one.
Then you can build a model in excel and cut this into time intervals.
I would start with 30sec.
Use starting conditions, pressures and flows for 30 sec, calc the end point, use that as the start for the next 30 sec.
The comp output (flowrate) will drop as the pressure rises.
I would also assume no loss in the lines to start with and see what answer I get.
If the flow velocity sounds high, then either go to a bigger line or add a pressure drop in the line.

The real problem that you will need to account for is the pressure change related to cooling of hte air.
What is the temp out of the compressor?


= = = = = = = = = = = = = = = = = = = =
Plymouth Tube

 

[racookpe1978]

OK, so let's assume it is a real problem - thank you for your reply.

Walk over to the compressor and read the spec sheet (label) on the compressor: We need to know how many cubic feet per minute the compressor is rated at. The net power (1000 HP) available does not necessarily mean anything in this context.
What is the pressure coming out of the compressor? What is the capacity of the on-line storage tank?
What is the current line diameter between the tank and compressor and dryer?
What is the diameter of the closest existing air line near the tank loading point?
Why do you care how fast it will take to load the tank to 1500 psig? If this time is not extremely critical to either the testing, the cycle time of the product, or to your bottom line profits, the smaller existing line or smallest new line will be cheaper and faster to build.

A 1 inch dia or even 3/8 diameter will pressurize that small a tank up the pressure of the storage tank very rapidly, but how important is it to shave off 1 or 2 minutes? 10 minutes?

 

[markdblissmn]

The entire goal of my question is to minimize the fill time of a pressure vessel. If I can fill the vessel in 5seconds, great. 30seconds is too long for the application. 5-10 is ideal.

Here is an easier question. Imagine I have two tanks of equal size connected with a one way valve with internal diameter X. One is at 100atm pressure. The other is at 1 atm. I open the valve the connects the two, how long will it take to fill the one vessel up to 10atm?

Mark Bliss

http://www.2RMengineering.com

 

[sailoday28]

With regard to your last question. If you submerge your tanks in water you will have Joule's experiment. Use those equations on internal energy to answer the quasi static flow process coupled with the dynamics of the compressor and flow resistance.
Regards

 

[racookpe1978]

Come on now; Make up your mind:

From your first post:

Imagine a 1000HP central facility compressor connected to a 4" main distribution line. If I have a 1000gallon air receiver and I want to fill it to 5atm pressure, how long would it take? How does the hose/pipe from the main 4in line affect it?

Then, you claim

"He works at a plant with a 1000Hp air system that feeds all of the machines. They want to try an experiment in a steel cylinder of 28"ID and length of 30in. He's trying to figure out how to get the vessel up to pressure of 150psi, or 10atm. But, how does he size the hose? 1"? 4"?

Now, you want to fill an equal sized tank through a hose, but the source tank is only 5 ATM (approximately 75 psig), but you want to load that tank in 5 seconds.

The entire goal of my question is to minimize the fill time of a pressure vessel. If I can fill the vessel in 5seconds, great. 30seconds is too long for the application. 5-10 is ideal.

Through a hose? Or a quick-acting valve and a permanent pipe?

Again, please tell us
(1) the actual nameplate delivery capacity of the compressor. We don't care what the size of its engine/motor is. That 1000 HP is not telling us how fast it can fill a tank or pipe. And, in truth, ONLY the size of the receiver tank at the compressor matter if you want to fill a tank in 5 seconds from that receiver tank!
(2) the output pressure of the compressor, measured at its storage tank.
(2a)The actual desired pressure of the test tank once it is filled.
(3) The actual size of the receiver (current storage tank) now on-line.
(4) Confirm the size of the desired test tank: double headed domed tank 28 inch ID and 30 (top to bottom? or seam to seam - seems unlikely though.)
(5) the approximate linear distance between the compressor and the proposed test tank.

 

[crshears]

I won't venture to comment on the math...but Mark asks:

Here is an easier question. Imagine I have two tanks of equal size connected with a one way valve with internal diameter X. One is at 100atm pressure. The other is at 1 atm. I open the valve the connects the two, how long will it take to fill the one vessel up to 10atm?

And sailoday28 responds:

With regard to your last question. If you submerge your tanks in water you will have Joule's experiment. Use those equations on internal energy to answer the quasi static flow process coupled with the dynamics of the compressor and flow resistance.

To which I add:

You want to fill a cylinder of that volume to THAT pressure THAT quickly? You are definitely going to have to take cylinder cooling into consideration! Have a look at your adiabatic expansion/compression tables to see what I mean...

SCUBA divers are well acquainted with this issue; cylinder fill rates are limited due to heat build-up [this is a MAJOR safety consideration].

 

[MortenA]

I hope he pays attention to that last post! Assuming it's adiabatic and that the constant is 1.4 and initial air temp is say 300k you get a final temp close to 600 K or 620F or 325C

 

[tbuelna]

MortenA,

Great point. The airflow discharged from the compressor will have an elevated temperature, due to efficiency losses in the compression system. The compressed air mass accumulated in the receiver will have a drop in pressure as it cools down. So if you want to ensure a pressure of 5atm in the receiver after the compressed air charge has had a chance to cool down to ambient conditions, then you'll need to create a somewhat higher initial pressure in the receiver to compensate for this effect.

 

[MortenA]

tbuelna, i think you are doing the same mistake that i fear the priginator might do: You dont consider that the air INSIDE the vessel that is being filled will also "heat up" to to the compression - and so will the "first air" that you fill in, and so forth untill the cyling is full - and the final temperature will be much higher than your inlet temperature.

 

[markdblissmn]

Thanks for all of the information everyone. All I'm really after is information on moving a set amount of compressed air from a higher pressure source into a vessel initially at atmospheric pressure. I can use a facility compressor, gas bottle, set of gas bottles, etc.

If I had a 2000psi cylinder of compressed air (source) and a pressure regulator set to 10atm, how long would it take to fill a 10ft3 vessel to 10atm. The second vessel has a relief valve set to something slightly over 10atm, 160psi lets say. Yes, I assume adiabatic process. I don't care about temperature rise of the 10ft3 vessel - I know it will increase. And I know the source bottle over time will have to make up for the reduction in pressure as the vessel cools and pressure drops.

If I were to use a bottle of nitrogen or argon gas, how would this change things? I won't have to worry about ignition based on these pressures correct?



Mark Bliss

http://www.2RMengineering.com

 

[prex]

You should care about the temperature rise, as this is likely to be the limiting factor for determining the time required.
People have killed themselves in doing this kind of operation.
What's the goal of your request, if this is not a real situation?

prex
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[sailoday28]

To do your last problem
You need internal volume of each cylinderInitial temp and press of each
To get fastest time Provide minimum flow area connecting vessels.
I would then assume the gas in each cyliner to be adiabatic. Therefore the source cyinder will follow an isentropic process. The decrease in internal energy of the high pressure cylinder will be the increase in internal energy of the low pressure cylinder.For ideal perfect gas, HP cylinder press vs mass will follow P/m^gamma = constant. Then use choked relation of flow from HP to LP

 

[sailoday28]

Final temp in low press cylinder is approx 246F based on assumptions of
perfect gas,
adiabatic ----no heat trans to gas in either cylinder
equal volume cylinders
constant specific heats
hp cylinder initial condition 100 atm, 70F
lp cylinder initail condition 1 atm, 70F
lp cylinder final condition 10 atm

Flow is from hp cylinder to lp cylinder. Therefore hp cylinder is isentropic
P/m^gamma is constant where P is pressure and m, mass

Subscript 1, HP cylinder, 2 LP cylinder
subscript i represents initial conditions
Consv of mass and energy yields P1-P1i +P2-P2i=0
P1=91atm
m2i/m1i=0.01
mi/m1i=0.9349 isentropic relation
m2/mi1=0.0751
m2/m2i=7.51
P2/p2i=7.51(T2/T2i)
T2/T2i= 10/7.51 T2/T2i=1.332 T2=705.7R or 246F

 

[FredRosse]

Looks like you are getting answers to questions other than yours. Starting and ending points are describes, but nothing about the time.

The time it takes to charge a cylinder from another depends on the flow capacity of the regulator or valve between the two cylinders, the differential pressure (or just the upstream pressure IFF the flow path is choked flow), and the gas being transferred, and its properties. Fairly simple calculations, and easy to evaluate if it is done in small steps appropriate to the boundary conditions. Name the regulator flow capacity (or the equivalent orifice size), and conditions upon which the regulator capacity is based, and the answer is easily calculated.

 

[sailoday280]

FredRosse(Mech)
A closed form solution of upstream mass vs time is obtainable under the following assumptions/conditions.
Source tank--adiabatic
Perfect gas with constant specific heats.
Connecting piping is adiabatic and frictionless.
Flow is choked.

Regards
Sailoday28