do you have examples for actuator speed that is pushing/pulling a load? or do these calculations apply regardless since the fluid is assumed incompressible?
For simple scenarios in which a load is supported (ie: not overrunning or acted on by any outside source) these calculations remain valid for hydraulic (ie: non compressible) systems. Consider a cylinder pushing a weight across a table on extension. Actuator speed would be dependent upon flow rate into the cap end volume. This same cylinder would pull on retraction. Actuator speed would be dependent upon flow rate into the rod end volume. All bets are off in pneumatic (ie: compressible) and overrunning scenarios!
@@bigbadtech Thanks for the clarification and for all the videos you’ve made, they’re a big help. Do you have any videos for speed calculations for pneumatic actuators?
Unfortunately I haven't yet produced a lecture discussing actuator speed calculation for pneumatic systems principally because air is a compressible substance and at different actuation pressures the volume of a given quantity of air changes thus there's a complicated interdependency between pressure and flow. This being said I do have a lecture discussing general flow control methods in pneumatic systems at: ruclips.net/video/Gj7AeZDmbLQ/видео.htmlsi=gIo7JGTuXtnW7NbH
Hey just curious as someone who would be interested in going through all of your playlists, what would be a logical and coherent order for an interested learner? Much appreciated for all of this!!
Check out this lecture here: ruclips.net/video/iqu2gJ2OmaI/видео.htmlsi=qCCiTSvTGQcMuEVL It lays out how I've organized the lectures in playlists and in which order to watch them.
Can I just confirm that with a constant displacement pump, the flow rate is the same going into the rod end as coming out of the bore end, presumably due to the principle of the conservation of mass?
Not exactly. A hydraulic motor might have same input and output flow rate because flow is traveling through the motor, however a double acting cylinder won't because there is no connection from input to output because the piston acts as a seal. Consider a double acting cylinder with a cap end volume of 2 gallons and a rod end volume of 1 gallon. An input flow rate of 1 gpm means the cylinder takes 2 min to extend. Emptying the rod end also takes 2 mins so exhaust flow rate is .5gpm. During retraction an input flow rate of 1gpm means the cylinder retracts in 1 min and exhaust flow from the cap end is 2 gpm. This is often why tank hoses are oversized to accommodate differential flow rates.
You are a gentleman and a scholar.
do you have examples for actuator speed that is pushing/pulling a load? or do these calculations apply regardless since the fluid is assumed incompressible?
For simple scenarios in which a load is supported (ie: not overrunning or acted on by any outside source) these calculations remain valid for hydraulic (ie: non compressible) systems. Consider a cylinder pushing a weight across a table on extension. Actuator speed would be dependent upon flow rate into the cap end volume. This same cylinder would pull on retraction. Actuator speed would be dependent upon flow rate into the rod end volume. All bets are off in pneumatic (ie: compressible) and overrunning scenarios!
@@bigbadtech Thanks for the clarification and for all the videos you’ve made, they’re a big help. Do you have any videos for speed calculations for pneumatic actuators?
Unfortunately I haven't yet produced a lecture discussing actuator speed calculation for pneumatic systems principally because air is a compressible substance and at different actuation pressures the volume of a given quantity of air changes thus there's a complicated interdependency between pressure and flow. This being said I do have a lecture discussing general flow control methods in pneumatic systems at: ruclips.net/video/Gj7AeZDmbLQ/видео.htmlsi=gIo7JGTuXtnW7NbH
Hey just curious as someone who would be interested in going through all of your playlists, what would be a logical and coherent order for an interested learner? Much appreciated for all of this!!
Check out this lecture here: ruclips.net/video/iqu2gJ2OmaI/видео.htmlsi=qCCiTSvTGQcMuEVL
It lays out how I've organized the lectures in playlists and in which order to watch them.
Can I just confirm that with a constant displacement pump, the flow rate is the same going into the rod end as coming out of the bore end, presumably due to the principle of the conservation of mass?
Not exactly. A hydraulic motor might have same input and output flow rate because flow is traveling through the motor, however a double acting cylinder won't because there is no connection from input to output because the piston acts as a seal. Consider a double acting cylinder with a cap end volume of 2 gallons and a rod end volume of 1 gallon. An input flow rate of 1 gpm means the cylinder takes 2 min to extend. Emptying the rod end also takes 2 mins so exhaust flow rate is .5gpm. During retraction an input flow rate of 1gpm means the cylinder retracts in 1 min and exhaust flow from the cap end is 2 gpm. This is often why tank hoses are oversized to accommodate differential flow rates.