|Feb27-13, 06:03 AM||#1|
Air consumption Rate
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 lets 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_guage = 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...
|Feb28-13, 02:58 AM||#2|
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?
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
(100/23.68) * 60 = 253 L/min
|Feb28-13, 03:14 AM||#3|
CWatters: Yes as stated in A it is a double acting actuator... Lets say that the duration of the strokes is the same so yes thanks, now you made that seem pretty obvious, haha...
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. Lets 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 dont 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
|Feb28-13, 03:27 AM||#4|
Air consumption Rate
I'm afraid that's outside my level of experience.
|Similar Threads for: Air consumption Rate|
|Burnup Vs. Consumption Rate||Nuclear Engineering||5|
|rate of oxygen consumption while sleeping||Biology, Chemistry & Other Homework||6|
|Steam consumption rate||Materials & Chemical Engineering||0|
|Finding rate of oxygen consumption||Biology, Chemistry & Other Homework||2|
|Burnup rate and Consumption rate||Nuclear Engineering||1|