Help Calculating time to recharge air tanks

[JamesCH95]

Originally Posted by Mike Bortfeldt

I think the error you are making is that the CFM compressor performance values you are using are at the inlet to the compressor, not the discharge. The compressor does not put out 0.53 cfm @ 60 psi, it compresses 0.53 cfm of atmospheric air (14 psia) when the discharge is @60 psi.

Mike

Using this tidbit (which makes sense, since it rates the compressor independent of ambient conditions and output temperature) methinks one can do a mass-conservation sort of calculation to take air at STP and a given flow rate and bring it to ST & raised pressure. Integrating the aforementioned flow-rate function (combined with another function or coefficient to account for mass conservation) is still the final way to solve this problem. I think.

[GeeTwo]

1) Assuming temperature stays constant (not a fair assumption maybe): P1 * V1 = P2 * V1 (from the ideal gas law)

Originally Posted by jee7s

This is a bad assumption. The ideal gas law assumes that the energy in the system in constant. The compressor of course puts energy into the system, so that violates PV=PV.

The ideas gas law does not require that the energy in the system remain constant. However, on the other side of that equation, in addition to R (constant) and T (assumed constant) is n - which is definitely changing. You need to add the "new" air, which began at a gauge pressure of zero. This serves as a reminder that you need to add about 15psi to every gauge pressure that you're measuring to use it.

If you did not account for the slowing, the time to fill a 35 in^3 tank from zero pressure would be based on the volume of this gas after it were returned to ambient. Since 120 = 15 * 8, we started with 8 x 35 in^3 = 280 in^3 = .162 ft^3 outside the tank, in the atmosphere. For a 1 cfm pump, this would take about 9.7 seconds to compress. I haven't measured it, but it seems that the sound of an air compressor drops about two octaves as it goes from zero to 120 psi. Assuming the same amount of air is compressed in each cycle, this means that at the end, it is pumping about 1/4 cfm, which would take 39 seconds to fill. Your time of 20 seconds to fill from 60 to 120 sounds about right.

[MrBasse]

Originally Posted by nuclearnerd

Yeah, when it comes down to it, that's what we'll have to do. I would prefer to have a theoretical grasp of it though, so we could make intelligent choices during the design stage about how many tanks to buy, or how often we can actuate our cylinders.

The thing that everyone always seems to overlook is that adding more tanks won't make your compressor run less. It will just run for a longer period of time less times.

Kids always stress over the need to have ten tanks on the robot and then they complain that the compressor runs for the whole match. When we reduce it to three or four tanks, the compressor may run twice a match, but for only 15-20 seconds each time. Unless you need to dump 80 cubic inches of air at a time, you rarely need 80 cubic inches of storage.

[JamesCH95]

Originally Posted by MrBasse

The thing that everyone always seems to overlook is that adding more tanks won't make your compressor run less. It will just run for a longer period of time less times.

Kids always stress over the need to have ten tanks on the robot and then they complain that the compressor runs for the whole match. When we reduce it to three or four tanks, the compressor may run twice a match, but for only 15-20 seconds each time. Unless you need to dump 80 cubic inches of air at a time, you rarely need 80 cubic inches of storage.

To add to this (quite correct) post: the only upside (that I can see) to lots of storage is that one can start the match with a lot of stored energy. This effectively increases the total power that the pneumatic system can expend over the duration of the match.

[Ether]

Originally Posted by JamesCH95

the only upside (that I can see) to lots of storage

You need enough storage so that you can disable the compressor during periods of high current draw and still have enough pressure to operate pneumatics during those (hopefully not too prolonged) blackout periods.

[clicato100]

Originally Posted by JamesCH95 View Post
the only upside (that I can see) to lots of storage

Bit of an aside, I think for many pneumatic shooters last year, the closer your reservoir of air approached "infinite" the higher sustained pressure you maintained at the cylinder throughout the shot.

I believe some students on my team have some interesting graphs we made of "real-time" pressure at the cylinder inlet throughout the shot varying the amount of reservoir tanks both HP and LP side, but besides the point...

[JamesCH95]

Originally Posted by Ether

You need enough storage so that you can disable the compressor during periods of high current draw and still have enough pressure to operate pneumatics during those (hopefully not too prolonged) blackout periods.

I suppose saving that little bit of current could help.

Originally Posted by clicato100

Bit of an aside, I think for many pneumatic shooters last year, the closer your reservoir of air approached "infinite" the higher sustained pressure you maintained at the cylinder throughout the shot.

I believe some students on my team have some interesting graphs we made of "real-time" pressure at the cylinder inlet throughout the shot varying the amount of reservoir tanks both HP and LP side, but besides the point...

Those plots would be most interesting to see.

[nuclearnerd]

Originally Posted by Mike Bortfeldt

I think the error you are making is that the CFM compressor performance values you are using are at the inlet to the compressor, not the discharge. The compressor does not put out 0.53 cfm @ 60 psi, it compresses 0.53 cfm of atmospheric air (14 psia) when the discharge is @60 psi.

Oh crap, so it's SCFM then, not cfm? Ugh, that changes things significantly. Now I get closer to 13 seconds to recharge one tank from 60 to 120 psi. I'll compare that with the robot performance tonight, but it sounds more like what we were seeing. I wish that were clearer on the compressor spec.

- that moment when you realize you need to make a big change, and it's almost week 5

[Mr V]

If you go to the link Mr Forbes provided you'll find another tab labeled refill rates. Of course those are only valid with the listed sizes and pressure ranges. Of course the listed pressure ranges are not exactly what we see in FRC and it is unlikely that you are using a .5 gallon tank. However you could probably use that data along with the CFM at the different pressures to derive how it would work with your tank size and pressure range.

The thing to keep in mind is that it is a piston type compressor. That piston has a set volume however you don't get to use all of that volume with each stroke. At zero psi the entire volume of the cylinder is discharged. As the pressure rises the initial portion of the stroke is used to reach the current tank pressure and no air flows out of the cylinder until the current pressure is met. The other thing is that at zero psi there is minimal load on the compressor's motor. As the pressure rises the load on the motor increases so it's rpm decreases. Not that the chart lists different current consumption at different pressures to reflect that change in load and rpm of the motor.

The listed CFM in the chart is what flows out of the compressor when working against a given pressure but the air does cool and contract when it reaches the tank further complicating the calculations.

[Jon Stratis]

Originally Posted by MrBasse

Kids always stress over the need to have ten tanks on the robot and then they complain that the compressor runs for the whole match. When we reduce it to three or four tanks, the compressor may run twice a match, but for only 15-20 seconds each time. Unless you need to dump 80 cubic inches of air at a time, you rarely need 80 cubic inches of storage.

This.

When designing our pneumatic system this year, our first focus was on actual air usage - how often would our "ideal" system be able to actuate. From there, we worked backwards to determine how much air was required at working pressure. This helped to directly inform us of the required storage volume - how many times do you want to be able to activate your mechanism before you run out of storage? How much airflow do you need to support the system over the entire match? Can the compressor handle it?

In the end, our math showed that we could handle the load at our theoretical top actuation speed (a speed we'll probably never actually hit, but it's good to have a bit of a cushion in these calculations). It also showed us the affect we would have if we had 2 tanks for storage, or 4, or 8. Doing the math to figure out how much the storage pressure drops with each activation is critical!

[MrBasse]

Originally Posted by clicato100

Bit of an aside, I think for many pneumatic shooters last year, the closer your reservoir of air approached "infinite" the higher sustained pressure you maintained at the cylinder throughout the shot.

This is where smart design takes over and eliminates the need for "infinite" air storage. Our shooter used a 2" bore 12" stroke cylinder and we were never wanting for air. We precharged our cylinder and let springs throw the ball.

[jee7s]

Originally Posted by GeeTwo

The ideas gas law does not require that the energy in the system remain constant. However, on the other side of that equation, in addition to R (constant) and T (assumed constant) is n - which is definitely changing.

I stand corrected. Boyle's Law (PV=PV) assumes that mass (n) and temperature (T) -- hence energy -- are constant.

[GeeTwo]

Originally Posted by jee7s

I stand corrected. Boyle's Law (PV=PV) assumes that mass (n) and temperature (T) -- hence energy -- are constant.

Even in isothermal processes (T remains constant), the gas does work as it expands and work must be done on it to compress it.

[Michael Hill]

I'm not sure I trust Viair's CFM data...has anyone actually measured it. It doesn't jive mathematically if you look at the 0.5 gallon and 1.0 gallon performance...(assuming my calculation is right). If there's a math mistake please late me know.

[GeeTwo]

Originally Posted by Michael Hill

I'm not sure I trust Viair's CFM data...has anyone actually measured it. It doesn't jive mathematically if you look at the 0.5 gallon and 1.0 gallon performance...(assuming my calculation is right). If there's a math mistake please late me know.

I'm not sure exactly what you're doing in your spreadsheet, but I certainly do find it suspicious that you wind up with a temperature of over 1000R (550F). I believe that this temperature would cause plastic tubing (and maybe tanks) to soften and rupture. This may be correct for an adiabatic compression of air, but there is certainly some temperature equalization taking place.
I did the calculations based on constant temperature (not right, but probably closer), and came up with 2 minutes and 9 seconds for 0-120 on a 1 gallon tank. This is still shorter than Viair's 3 minutes, but a good bit closer. There does seem to be a disagreement between the "performance" and "fill rate" data.