Eng-Tips is the largest engineering community on the Internet
Intelligent Work Forums for Engineering Professionals
Method for finding stroke speed in an air cylinder?
- Thread starter PEinOHIO
- Start date
Hi PEinOHIO
This might be of help:-
desertfox
This might be of help:-
desertfox
MiketheEngineer
Structural
Maybe I missed the point - but how high does it need to go??
Theoretically at 100 psi you could lift about 15,400 lbs if you you were using say water or oil. Not sure what the compressibility will do but if you can maintain 100 psi (BIG COMPRESSOR) it should not affect the overall performance. I think...
Theoretically at 100 psi you could lift about 15,400 lbs if you you were using say water or oil. Not sure what the compressibility will do but if you can maintain 100 psi (BIG COMPRESSOR) it should not affect the overall performance. I think...
If you can maintain 100psi throughout the stroke, you can use F=ma
to calculate constant acceleration. Then use some form of kinematics, v = sqrt(2*a*s). v, velocity; a, acceleration; s, distance traveled. This neglects forces opposing the movement of the load, friction, flow loss of air exhausting from the opposite end, return spring(if single acting).
Ted
to calculate constant acceleration. Then use some form of kinematics, v = sqrt(2*a*s). v, velocity; a, acceleration; s, distance traveled. This neglects forces opposing the movement of the load, friction, flow loss of air exhausting from the opposite end, return spring(if single acting).
Ted
The real question is what speed can you obtain in what distance? It takes a lot of air to increase the pressure enough to accelerate the load quickly. Basically it is hard to change the acceleration rate of change quickly with air. You are jerk ( the derivative of acceleration ) limited.
I can do the calculations but it involves solving a system of differential equations. What makes this calculation trickier than doing hydraulic calculations is that the bulk modulus of air changes as a function of pressure.
So what is the motion profile? The general formulas I use are
v=1.5*?x/?t
a=4.5*?x/?t^2
so if ?x=14 inches and ?t=3 seconds then
v=7 which isn't to high
a=7 which I think is easy to do.
I think you should be OK IF you have a big enough valve. The problem is that the induced pressure will be relatively high, 45 psi. That means the air flow from 100 psi to 45 psi will not be as great as 100 psi to 0 psi. The flow will not be high. Sizing the valve will be critical.
Peter Nachtwey
Delta Computer Systems
I can do the calculations but it involves solving a system of differential equations. What makes this calculation trickier than doing hydraulic calculations is that the bulk modulus of air changes as a function of pressure.
So what is the motion profile? The general formulas I use are
v=1.5*?x/?t
a=4.5*?x/?t^2
so if ?x=14 inches and ?t=3 seconds then
v=7 which isn't to high
a=7 which I think is easy to do.
I think you should be OK IF you have a big enough valve. The problem is that the induced pressure will be relatively high, 45 psi. That means the air flow from 100 psi to 45 psi will not be as great as 100 psi to 0 psi. The flow will not be high. Sizing the valve will be critical.
Peter Nachtwey
Delta Computer Systems
Here is how I do this problem.
Let's say you want to lift the 7000 lb 21 feet in 3 seconds.
Then the average velocity is ;
21/3= 7'/sec
Then I choose a valve whose area is obtained from
d^2/14^2=7./1000= .007
d=1.17" dia valve
where the 1000 is the approximate velocity (acoustic) at the valve 114.7*.53=60psia or 45 psig, 0.53 critical ratio for air.Next we check the time for the cylinder to reach 7'/sec
t=v/a
a=F/w*g=45*14^2*pi/4/700*32=6923/7000*32=31.6'/sec^@
t=7/31.6=0.22 sec
So it takes only .2 sec to come to speed, which is negligible.
The acceleration force is
45*14^2*pi/4=
Let's say you want to lift the 7000 lb 21 feet in 3 seconds.
Then the average velocity is ;
21/3= 7'/sec
Then I choose a valve whose area is obtained from
d^2/14^2=7./1000= .007
d=1.17" dia valve
where the 1000 is the approximate velocity (acoustic) at the valve 114.7*.53=60psia or 45 psig, 0.53 critical ratio for air.Next we check the time for the cylinder to reach 7'/sec
t=v/a
a=F/w*g=45*14^2*pi/4/700*32=6923/7000*32=31.6'/sec^@
t=7/31.6=0.22 sec
So it takes only .2 sec to come to speed, which is negligible.
The acceleration force is
45*14^2*pi/4=
Hydtools, yes, 45 psi just hold position and that is IF there is no pressure on the top of the piston.
I miss read the original post and assumed the length/travel was 14 inches. 21*12=252 inches in 3 seconds. Wow!!
Even so I missed the obvious and so has everyone else.
Peter Nachtwey
Delta Computer Systems
I miss read the original post and assumed the length/travel was 14 inches. 21*12=252 inches in 3 seconds. Wow!!
Even so I missed the obvious and so has everyone else.
Peter Nachtwey
Delta Computer Systems
It would be good to hear from PEinOhio about now.
The project sounds like a pneumatic pile driver hammer.
Over the length of the stroke the load must be accelerated, decelerated, stopped, returned. Unless the load is impacting something at the end of the stroke, peak velocity will be somewhere mid-stroke.
Ted
The project sounds like a pneumatic pile driver hammer.
Over the length of the stroke the load must be accelerated, decelerated, stopped, returned. Unless the load is impacting something at the end of the stroke, peak velocity will be somewhere mid-stroke.
Ted
"Won't this require pressure greater than 45psi to get the load moving up? 45psi just balances the load if the load is being lifted and there is no accelerating force until pressure is >45psi."
No, because initially the pressure as the mass lifts will be closer to 100 psig which will quickly accelerate the mass upward (close to 1 g) to some velocity which will remain constant under the 45 psig at choked flow ; I choose the valve size to keep the average speed 7'/sec.
As I mentioned, I originally did the problem for horizontal motion, but since fortuitously the 45psig is the choked flow back pressure, it is close to the vertical solution.
No, because initially the pressure as the mass lifts will be closer to 100 psig which will quickly accelerate the mass upward (close to 1 g) to some velocity which will remain constant under the 45 psig at choked flow ; I choose the valve size to keep the average speed 7'/sec.
As I mentioned, I originally did the problem for horizontal motion, but since fortuitously the 45psig is the choked flow back pressure, it is close to the vertical solution.
So far the OP hasn't declared how far he wants to move this object as far as I can see. Zekeman - your 21' in 3s might be a bit ambitious?
I'm finding the analysis interesting - thank you to all who are contributing.
----------------------------------
If we learn from our mistakes I'm getting a great education!
I'm finding the analysis interesting - thank you to all who are contributing.
----------------------------------
If we learn from our mistakes I'm getting a great education!
"zekeman, what is that velocity? I believe that is the OP's question. "
Depends on the valve size; I assumed an average velocity of 7'/sec and came up with a valve of about 1.17 inch diameter.Obviously, if you increase the valve size the flow rate increase and with it, the velocity. Implicit is that the valve diameter is much smaller than the tank dimensions and the pipe length from compressor tank to cylinder is short.
Depends on the valve size; I assumed an average velocity of 7'/sec and came up with a valve of about 1.17 inch diameter.Obviously, if you increase the valve size the flow rate increase and with it, the velocity. Implicit is that the valve diameter is much smaller than the tank dimensions and the pipe length from compressor tank to cylinder is short.
- Status
Similar threads
- Locked
- Question
- Locked
- Question
- Locked
- Question