Calculation of compressed air to valve actuators

Hi ! , not sure if anyone here knows how to calculate this.. but just in case ;

I will instal a cluster of 20 valves and have to calculate the compressed air consumption.

The valves consumes 0,25Liter/cycle.

I asume I will have 4 actuations per hour..

but sometimes i may have 4 actuations in 10 minutes during that hour , and some times 1 actuation per 15 mins during the same hour..

So doing 0,25*4 doesnt sound good to me to calculate the air consumption..

because although in both cases i consume 1litre per hour, is not the same to conume it in 10 minutes (0,15 liter per 2,5min) or consume it in 1 hour (0,15 liter per 15 min) I mean the pipe to reach the cluster shouldnt be the same in both cases right ?

Any one with compressed air experience ?

thanks!

what is the speed of 1 cycle, probably around 2 or 3 seconds, so thats your airspeed in the tubes.

Hi hooter, could you specify a bit more..

thanks

Are you trying to calculate the amount of air that the compressor must generate per hour or the instantaneous air flow through the plumbing??
Either way I think you will have to calculate the worst case.

I have to calculate both..

I think i will do the following:

For the pipe calculation, i will use the instantaneouse flow, lets say if the valve consumes 0,25 NL per cycle and the valve actuates in 3 secs, then to design the pipe i will asume the flow is 0,25/3 NL/s

But to dimension the compressor i wil use the actuations per hour,

If i have 4 actuations per hour, then the capacitiy of the compressor must be 0,25*4 NL/hour

This is for the case of 1 valve, of course.

mendonsy said:

Are you trying to calculate the amount of air that the compressor must generate per hour or the instantaneous air flow through the plumbing??
Either way I think you will have to calculate the worst case.

Click to expand...

I concur.

This should be then: max flow rate = (number of actuators opening at once) x (volume per actuation) / (smallest actuation time) = (20)(0.25)/3 = 1.667 L/sec

You need to make sure that your regulator supports the max flow rate (for the worst case when all fire in 3 seconds), but the compressor itself doesn't have to support that flow rate though because you're not going to be maintaining that high rate, i.e. 20 actuators open in 3 seconds, but that only happens every 10 minutes or so.

thanks all !

You need to standardize the amount of air required. Pneumatics is tricky. The basic unit of air is SCFM
https://en.wikipedia.org/wiki/Standard_cubic_feet_per_minute
That mean we need to know the pressure required to move .25 liters per cycle. The higher the required pressure the higher the SCFM.

If you have 0.25L x 4 per hour, then this is 1L/h, however you need to know what the regulator pressure is, generally circa 600~700kPa. Compressors are rated in standard (normalised) volumentric rates (S-Ft3/min) or NM3/min.
If you have a 6bar (600kPa) regulated air supply, by boyles law:
P1 * V1 = P2 * V2. Given that the pressure for the standard (normalised) pressure is atmospheric pressure (101.3kPa ~ 1 bar), then the normalised volume will be 6 x greater. So if the cylinder consumption at 6bar = 1L, then at atmospheric pressure there is 6 times the volume = 6L.
So consumption is 6L/h. Generally the compressed air system will have a large air receiver, capable of delivering the air volume 6L_normalised without losing much pressure. The 6L/h is how much the compressor has to deliver in the hour to replenish, but the pressure may be reduced below the required operating pressure if the 6L is consumed in a one minute period from a small air receiver.
So the guts of it is that the compressor may need to be larger to recharge the air receiver.
Think of it as a tank with a pump. If the tank is only 10L and you take 6L from it in 1minute, then the tank will only have 4L for 59minutes, which may not provide enough pressure for what the tank feeds.
6L/h of air is tiny, but I guess everything is relative. Maybe you're running from a compressed air bottle?