How long to fill a specific volume to a specific pressure? 1

[borobam182]

Morning!

I've been asked to confirm how long it would take to fill a 445,000 litre volume to a pressure of 1.2kPa using a air source that provides 130 litres a second of air at a pressure of 620kPa.

Firstly, this seems like a very low pressure to set the volume to (1.2kPa). But' I've calculated it would take almost 7 seconds to fill this volume to 1.2kPa using this air supply.

Would that sound about right do we think??

Cheers

Craig

 

[LittleInch]

What pressure are you staying at? Pure vacuum is difficult to achieve....

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[borobam182]

It is being kept at 1.2kPa while a smoke test is performed, to detect any leaks. So not for a huge amount of time....

 

[zdas04]

If I assume by kPa you mean kPag and that you are around sea level (call it 101 kPaa atmospheric pressure) and that the compressor capacity is stated at suction conditions (look at LittleInch's signature).

If those assumptions are correct, then the mass difference between open to atmosphere and 1.2 kPag is 6.44 kg. Your source puts out 0.158 kg/s. That works out to about 41 seconds.

On the other hand if your test pressure is 1.2 kPaa, then this would be an evacuation problem, and you are evacuating 535 kg. Assuming that your compressor capacity is the same at atmospheric pressure and at 1.2 kPaa (an absolutely invalid assumption) then it would take about an hour. With real life equipment figure 12 hours.

Working this problem in volume units is an excellent way to end up outrageously confused.

Do you see what sloppy specifications get you?

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[MortenA]

David, but air weight in at 1.29 kg/m3 at atmospheric pressure/0ºC and 1.2 kPa is just 0.012 bar so i would say that the mass difference is 1.29*0.012*445000=6887 kg?

 

[zdas04]

A L is not an m^3. The last term is 445 m^3, so the mass is 6.8 kg. I did the calcs at 60F instead of 32F and got 6.4 kg.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[BigInch]

Would not the compressor deliver 130L/s (@ STP per normal convention) into the bottle, as long as the pressure in the bottle was less than 620 kPag

Therefore one could convert the 130L/s STP rate into an instantaneous delivery rate into the bottle, then calculate the bottle's instantaneous pressure.

Do that in small time steps, using average delivery pressure of the time step, and arrive at an average injection rate across each timesetp, finally reaching a solution of 17.8 seconds.

(ignoring compressibility and assuming isothermal process and an ideal gas)

 

[zdas04]

BigInch,
That is exactly what you have to do when you get closer to the same pressure on the two systems, but going from over 600 kPag to 1.2 kPag is really a constant flow rate exercise. It is a lot easier to work in mass units. Working in standard volumes is mathematically, correct but an engineer who uses the term "kPa" instead of "kPaa" or "kPag" is going to trip over his feet a couple of times doing that calc in STP (notice that he thinks that it should take 7 seconds to fill).

The convention for air compressors is not STP but actual conditions at the suction plenum at the location of the manufacturer's test. That causes no end of confusion, but it is the convention. I'm so glad that compressors for industrial gases abandoned that stupidity in favor of STP.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[LittleInch]

Borobam,

Your op states "delivers 130 l/sec at 6.2 barg"

If this is correct then your calculation isn't far off IMO.

However if your compressor size is actually like zdas04 states then his calculation makes sense.

Which is it?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[BigInch]

Yes it seems it is a small rate low pressure unit.
If the mgfr's test bench is in Denver he should convert to Sea Level to sell more compressors.

 

[zdas04]

LittleInch,
Are you actually converting 130 L/s at discharge conditions to a mass flow rate (or a volume flow rate at standard conditions)? What temperature are you using? There never has been a compressor whose capacity was rated at max discharge pressure (and the associated discharge temperature) at ACM. If he was filling the system from a tank that is about 10 times the size of his system, then his calculation might be OK, but compressor discharge pressure is absolutely controlled by the pressure downstream of the discharge check valve. If that pressure is 1.19 kPag, then the compressor discharge will be something like 1.21 kPag. The mass flow rate at that point is going to be somewhere pretty close to 130 (std)L/s and the fill time is right around 40 seconds.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[LittleInch]

ZDAS04,

Jeez, I was just trying to point out that the point at which you took the 130litres makes a difference.

If you read the OP carefully, he or she doesn't state they have a compressor, only a "source" of compressed air. You are the one who introduced the compressor into this. Now that "source" could be an air line, a plant air supply or indeed maybe a compressor - who knows, but "source" is what it says. I agree we don't know the temperature, but I think we can assume 15 to 20C for the purposes of the calculation.

Hence if we assume it is a piped supply from some remote air receiver, then if there was a regulator regulating down to 102.2 KPaa, and the supply is being measured in litres at 6.2 barg then it might work. Of course I doubt this regulator exists or whether issues such as temperature drop, sonic velocity, choked flow etc have been considered as filling 455 cubic metres admittedly to a fairly low overpressure in less than 7 seconds just sounds wrong / impractical. I suspect in practice a much slower rate would be required, if indeed the 130l/sec is being measured at 6.2 barg.

I agree it would be much better to use scm and absolute pressures in all sections then it makes life much easier...


Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[zdas04]

Ya got me. To me "source" implies something that takes atmospheric pressure air up to an elevated pressure, but that is just my prejudice. Since he did state a volume flow rate at a pressure, it is fair to assume he's talking about an infinite (i.e., more than about 10 times the required volume) reservoir or a way to supply make-up.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[borobam182]

Everyone,

Thank you for your input with this issue.

I agree that the lack of confirmation over the units would create issues and errors in every engineering problem in all walks of life - the information I included on my origional message was the information that I received from our site team. I have asked if it is kPaa or kPag, and I have not received a response. Sometimes, you just have to find a way to work with what you're given and maybe present a few cases for different scenarios.

But I was asking in the forum to try and get a bit of confidence in the method I was following.

LittleInch,

Yes, the air is delivered at a rate of 130 l/sec at 6.2 barg (I am going to assume that is "g"). It is supplied from a plant air supply, and those are the conditions of the air supply. I agree 7 seconds sounds wrong, and that is why I thought I'd bring it to the forum for a discussion.

.....Every day is a school day, and today, my lesson is, don't ask questions unless you have all your facts straight.

Have a good day everyone.

 

[LittleInch]

That's ok - we just like to tear the odd question apart - if you don't know, say you don't know but assume something.

The 7 seconds is one of those things that looks theoretically possible, but in practice not so.

The 130 litres/second converts to about 850 to 900 litres/sec at the tank pressure. It would be a good check to see what that equates to in velocity terms of the pipe feeding the tank. Anything more than sonic velocity will tend to cap the potential pressurisation speed....

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[itdepends]

If it's a smoke test then the 1.2kPa will be gauge- otherwise you'd have to put the smoke on the outside of the tank.....

If that's gauge pressure than the 620kPa is also probably gauge- and given the time period is so small- it doesn't make much difference (practically speaking) if it's gauge or absolute.

Is the smoke going to be introduced into the air supply or will a smoke bomb be put in the tank followed by pressurising the tank?

In the scheme of things- if the smoke is already inside the tank it'll take <1 minute (as per posters above) regardless of whether the volume quoted is inlet or outlet conditions of a compressor/plant air system.

I've spent longer than that writing this up- don't sweat it. A better use of your time would be to ask how are they going to limit the pressure the tank can achieve during the test in case someone leaves the air supply on longer than they should.

As a chem eng/metallurgist the first part of any answer I give starts with "It Depends"

 

[borobam182]

itdepends - thanks for the message. A smoke bomb will be put in the tank once the tank is up to pressure and then they will look for smoke leaking out.

LittleInch - thanks again for that confirmation.

 

[ione]

Yesterday I’ve read the thread, made my calculations using the ideal gas law and reached approx the same results as others have already presented, that is 40.3 s assuming the volumetric flow rate is referred to standard conditions and 5.7 s assuming it is referred to actual conditions (by the way I consider that working with mass instead of volumes as suggested by zdas04 is the way to go to avoid bothers).

I’d like to add just a note. It has not been mentioned anywhere but I assume the filling of the reservoir will take place through a nozzle. According to the numbers presented in the OP the flow will be choked, so the choked mass flow rate can be calculated with the following equation

M dot = Cd*A*SQRT(k*rho1*P1*0.528)

Being
Mdot = mass flow rate [kg/s]
Cd = coefficient of discharge of the nozzle
A = discharge cross section area [m2]
k = ratio of specific heat (1.4 for air)
rho1 = density at upstream conditions [kg/m3]
P1 = upstream absolute pressure [Pa]

This value will allow you to calculate the min filling time, knowing the upstream conditions and nozzle geometry.

 

[borobam182]

ione - thank you for this confirmation - the team on site were wondering if it would be an overnight / few hours job to reach this pressure - I think we can safely assume from all the comments above that it will be achieved in less than 1 minute!

Thank you all.

 

[LittleInch]

It might take longer if they can't get the 130 litres/second through the nozzle or attachment or hose or regultor. Where did the 130 figure come from? It's still quite a lot of air in a short space of time.

Are you sure they haven't misread a capacity somewhere 130 l/hr maybe?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.