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circuit understanding and calculation 2
- Thread starter Pravin086
- Start date
The pump max displacement is 250 cc and at 1500 rpm that produces a max flow of 375 L/min or 6.25 L/sec.
If your cylinder numbers are...
Bore 140mm
Rod 110 mm
and
Stroke 600 mm
Bore displacement = 9.236 L
Capped side displacement = 3.534 L
(Rod volume = 5.702 L)
The theoretically extend stroke time will be 9.236 / 6.25 = 1.478 sec and the extend speed will be 0.600 / 1.478 = 0.406 m/sec or 406 mm/sec.
The theoretically retract stroke time will be 3.534 / 6.25 = 0.565 sec and accordingly, the theoretical retract speed will be 0.600 / 0.565 = 1.061 m/sec or 1061 mm/sec
The answer is NO, the pump can't even theoretically extend the piston @ 900 mm/sec
You would need a much higher pump rpm, (3225+ rpm) or a "554+" cc pump
If your cylinder numbers are...
Bore 140mm
Rod 110 mm
and
Stroke 600 mm
Bore displacement = 9.236 L
Capped side displacement = 3.534 L
(Rod volume = 5.702 L)
The theoretically extend stroke time will be 9.236 / 6.25 = 1.478 sec and the extend speed will be 0.600 / 1.478 = 0.406 m/sec or 406 mm/sec.
The theoretically retract stroke time will be 3.534 / 6.25 = 0.565 sec and accordingly, the theoretical retract speed will be 0.600 / 0.565 = 1.061 m/sec or 1061 mm/sec
The answer is NO, the pump can't even theoretically extend the piston @ 900 mm/sec
You would need a much higher pump rpm, (3225+ rpm) or a "554+" cc pump
- Thread starter
- #4
Pravin086 said:
"I can explain to you but I can't understand it for you"
OK, Regeneration. regen, means basically that instead of pushing the entire piston area we only push on the rod area.
Here is a little math trick now.
The piston area relates to the rod area like this. 1402:1102 = 19600:12100 (=1.620)
This means that the regen speed will be 1.620 times higher. The piston extend speed was 0.406 m/sec and now the piston-rod regen speed 1.620 x 0.406 = 0.658 m/sec or 658 mm/sec.
The conventional math would be to calculate the regen extend-time by taking the rod displacement volume divided with the pump flow ie 5.702 / 6.25 = 0.912 sec, and the regen extend velocity by dividing the stroke with the time of regen stroke ie 600 / 0.912 = 658 mm/sec
NO, the regeneration option won't make it either. With a "94" mm rod diameter you can theoretically make a regen speed of 900mm/sec.
Regeneration is a tough nut to crack with your capped side area vs rod side area ratio. In your case you will have extreme flow intensification when retracting the piston, your meter-in flow on the rod side is 375 L/min and the capped side meter-out flow will be 980 L/min. This number must match the flow rating on your valves and your hydraulic lines. Sun Hydraulics has a great info document about how to design regen systems.
Another flow makes it go thread.
Peter Nachtwey
Delta Computer Systems
-
1
- #7
You guys aren't thinking.
I have done a lot of application support. If the system is designed right it will run right.
The first questions I ask is
what is the application?
How much mass?
How far must it move?
In how much time?
Is the rod point up, down or horizontal?
Is there an external force to be applied. The mass and external force are two different things.
The OP didn't provide any information to the forum so I don't think he deserves much of an answer.
I would ask on more very important question. What are the cycle times and dwell time at the end of strokes? This is extremely important.
Apparently everyone ignore my previous thread about hydraulic sizing. You can see what I think is important.
So the pump isn't big enough. There is a cure. It is called an accumulator.
The pump only needs to supply the average flow, plus a little more, for the whole cycle if you size the accumulator correctly.
If there is time between the strokes then there is time to re-charge the accumulator.
However, the OP didn't give us this information so when I gave my smart ass response that is all that was deserved.
Look at my thread about hydraulic sizing where I look at the motion profile and the resulting oil demands and how the HPU can be much smaller than the peak oil flow.
The average HPU flow required is a little more than 14 L/min but the peak flow is is 70.7 L/min. You can see the pressure doesn't vary by more than 10%.
One of the purposes of my hydraulic sizing program is to optimize hydraulic design and make it as efficient as possible. It is too bad people ignore it.
I have been doing hydraulic servo control for over 35 years now. I have seen too many projects gone wrong due to poor design or projects not attempted due to lack of a good design.
This thread is more about energy. The pump can supply so much energy over a period of time. The accumulators store energy. As long as the energy required per cycle is not required more than supplied per cycle then the actuator should be able to do what is required. The cycles just need to slow down so the energy of the pump is not exceeded.
This forum is titled "Hydraulic Power Engineering"
I just showed you. Happy now?
Peter Nachtwey
Delta Computer Systems
I have done a lot of application support. If the system is designed right it will run right.
The first questions I ask is
what is the application?
How much mass?
How far must it move?
In how much time?
Is the rod point up, down or horizontal?
Is there an external force to be applied. The mass and external force are two different things.
The OP didn't provide any information to the forum so I don't think he deserves much of an answer.
I would ask on more very important question. What are the cycle times and dwell time at the end of strokes? This is extremely important.
Apparently everyone ignore my previous thread about hydraulic sizing. You can see what I think is important.
So the pump isn't big enough. There is a cure. It is called an accumulator.
The pump only needs to supply the average flow, plus a little more, for the whole cycle if you size the accumulator correctly.
If there is time between the strokes then there is time to re-charge the accumulator.
However, the OP didn't give us this information so when I gave my smart ass response that is all that was deserved.
Look at my thread about hydraulic sizing where I look at the motion profile and the resulting oil demands and how the HPU can be much smaller than the peak oil flow.
The average HPU flow required is a little more than 14 L/min but the peak flow is is 70.7 L/min. You can see the pressure doesn't vary by more than 10%.
One of the purposes of my hydraulic sizing program is to optimize hydraulic design and make it as efficient as possible. It is too bad people ignore it.
I have been doing hydraulic servo control for over 35 years now. I have seen too many projects gone wrong due to poor design or projects not attempted due to lack of a good design.
This thread is more about energy. The pump can supply so much energy over a period of time. The accumulators store energy. As long as the energy required per cycle is not required more than supplied per cycle then the actuator should be able to do what is required. The cycles just need to slow down so the energy of the pump is not exceeded.
This forum is titled "Hydraulic Power Engineering"
I just showed you. Happy now?
Peter Nachtwey
Delta Computer Systems
- Thread starter
- #10
in worst condition what pressure drop we should consider for logic cartriage valve?
Can you help understanding the values written on the catirage valve & how to calculate it as it shows different on both side.
Selecting 2 pumps of 250cc will do?
Can you help understanding the values written on the catirage valve & how to calculate it as it shows different on both side.
Selecting 2 pumps of 250cc will do?
Where? I don't see it! Pravin086 seems obsessed with the pump and not answering any questions.
BTW, die casting machines store oil under pressure to travel at speeds up to 10m/s. 5 m/s - 8 m/s is more common. Obviously the HPUs cannot supply that much oil or do it instantaneously. Die casting machines use servo piloted cartridge valves for control. Even then the cartridge valves are not close to linear. Even the fastest servo piloted cartridge valves have response times of about 20 ms although 30 ms is more typical.
Peter Nachtwey
Delta Computer Systems
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1
- #13
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