Air Consumption Rate of Cylinder Actuators on SKG200 Valves

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The discussion focuses on calculating the air consumption rate for cylinder actuators on SKG200 valves. The user has determined the total cylinder volumes for opening and closing strokes, resulting in air consumption values of 42.92 liters for opening and 56.5 liters for closing. They seek clarification on whether the total consumption is 100 liters per cycle or if air is only consumed during one direction of movement. Additionally, they are exploring the sizing of an air receiver for fail-safe purposes, referencing Boyle's law for calculations but expressing uncertainty about the correct application. The conversation highlights the need for accurate compressor sizing based on these air consumption metrics.
WillemBouwer
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Hi all

I am trying to set up a report showing the air consumption rate of my cylinder actuators on different valves. I am stuck at the very end where I should calculate the air consumption rate. All calcs are in metric units.

Okay as an example let's use the following:
Valve: SKG200 with closing thrust 1259kg at 10bar line pressure.
Actuator size: D = 250 mm
Valve stroke: S = 296 mm
Cylinder rod diameter: d = 32 mm
Available air pressure: P2_gauge = 4 bar

Now I calculated the total cylinder volume for the closing and opening strokes:
V_open = ((D-d)/10)^2*∏/4*S = 8584.8 cm^3
V_close = (D/10)^2*∏/4*S = 11290.1 cm^3

So the total air consumption respectively:
Air_open = V_open*(P2_guage+1)/1000 = 42.92 liter
Air_closed = V_close*(P2_guage+1)/1000 = 56.5 liter

And with a maximum operating speed of 25mm/s we calculate that the duration of the stroke is:
t = S/25 = 296/25 = 11.84 sec

So now that I have all this info, how do I calculate the rate of air consumption if I want to know what size compressor to buy... Thanks guys and I hope this is in the correct forum...
 
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Are you saying:

a) it uses 43L to move the piston one way and another 57L to move it back the other way? eg a total of 100L per cycle?

or

b) The cylinder volume changes from 43 to 57 L in one stroke and no air is used for the return stroke? (eg the load or a spring does the return).

Is the duration of the stroke the same in both directions? eg The cycle time is 2 * 11.84 = 23.68 seconds?Assuming you mean a) then..

The max consumption is going to be around...

100L every 23.68 seconds
or
(100/23.68) * 60 = 253 L/min
 
CWatters: Yes as stated in A it is a double acting actuator... Let's say that the duration of the strokes is the same so yes thanks, now you made that seem pretty obvious, haha...
Thanks

Now the other thing I started to have trouble with, say I do not want to place a spring for fail safe purposes (load is to big), I am recommending an air receiver.

I however have to size the receiver accordingly. Let's say for only the closing stroke.
I know I have to use Boyle's law: pV = k or p1V1 = p2V2

I've tried a view solutions by now. I don't know what to use as V1 and V2? Do I use the 57 liters as V1 and say V2 = Vreceiver + V1? One of the engineers said that rule of thumb says multiply it by 4 so that gives 228L! But surely I can get to the answer using Boyle's law.

Thanks for the help
 
I'm afraid that's outside my level of experience.
 
So I know that electrons are fundamental, there's no 'material' that makes them up, it's like talking about a colour itself rather than a car or a flower. Now protons and neutrons and quarks and whatever other stuff is there fundamentally, I want someone to kind of teach me these, I have a lot of questions that books might not give the answer in the way I understand. Thanks
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