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Re: Hydraulics Question - Low pressure High-Flow source to drive Hydraulic Motor - Feasible?
Re: Hydraulics Question - Low pressure High-Flow source to drive Hydraulic Motor - Feasible? -- matsoda	Post Reply	Top of thread	Forum

Posted by: randykimball ® 10/08/2006, 16:27:36 Author Profile eMail author Edit

By incompressible fluid you mean a liquid. No liquid is compressible. Only the gasses traped in a fluid are compressible. If you move a liquid through a line from a pump towards a hydraulic motor and the liquid has no place to escape, the hydraulic motor will have to either operate at a velocity based on volumn displaced by the pump and the motor, leak by bass, exceed the ability of the pump to pump, or something will break. These are the simplistic physical laws..nailing down the rates, when, friction losses, and how, become more complex. Basically, if a liquid moves in a closed circuit and has nowhere to go something is going to move, the containment or the load. Seeing that a hydraulic motor is in a circuit between the pump and the ambient pressure in the supply tank, the operating pressure at the hydraulic motor is going to be determined by the load acting against the hydraulic motor. The operating pressure at the unloader or relief valve needs to exceed the max caliculated operating pressure required to move the load at the hydraulic motor at the rate required (volumn) while maintaining the required pressure while not exceeding maximum temperature. The size of the supply tank and its heat sink method determin the temperature of the fluid in the supply tank while doing work for a given length of time. The worst suggestion of your lifetime may be the catalyst to the grandest idea of the century, never let suggestions go unsaid nor fail to listen to them. Modified by randykimball at Sun, Oct 08, 2006, 16:45:05

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Re: Re: Hydraulics Question - Low pressure High-Flow source to drive Hydraulic Motor - Feasible?
Re: Re: Hydraulics Question - Low pressure High-Flow source to drive Hydraulic Motor - Feasible? -- randykimball	Post Reply	Top of thread	Forum

Posted by: matsoda ® 10/09/2006, 13:32:40 Author Profile eMail author Edit

Thankyou kindly for you comments, very much appreciated ;-) In this case, it is a closed system so i understand the operating pressure on the motor is ultimately determined by the operating load. In more detail, assume a 98kN force is available that acts linearly on some fluid contained in a piston/actuator (of 2.5m in diameter). The force available acts over a relatively long distance of approx 10 meters. Due to the pistons large diameter (a long story), the pressure is relative low, but the volume of fluid that the piston can move will be relatively high. The force is naturally limited to a speed of around 1-1.5 m/s. I am looking to convert work done by the force into Hydraulic pressure to use in a hydraulic motor. Am i correct in assuming that a 98kN force acting on a cylinder of 2.5m in diameter (with total resistance) will create a maximum static pressure of: Pmax = Force / Area Pmax = 98,000(N) / (pi x 1.25 x 1.25) Pmax = 98,000(N) / 4.9(sqm) Pmax = 20,000 (N/sqm) ---------------- Pmax = 2.9 (psi) ---------------- If the force is naturally limited to travelling at maximum of 1-1.5 m/s, then the maximum flow rate (during the 10m of travel) might be calculated as: Pflow = Volume(m3) / Time(secs) PFlow = (Area x Length) / (Length/Speed) Pflow = (4.9 x 10m) / (10/1) ---------------- Pflow = 4.9 m3/s (assuming 1m/s) ---------------- Question: With a force of 98kN pushing on a piston in a cylinder of 2.5m in diameter, creating that kind of pressure and flow.. Is it possible to use that fluid to do significant work??? I have thought about the possibility of feeding the fluid directly into a second stage of the cylinder where the diameter is larger say 3.5m. That would produce a much larger force that travels a much smaller distance out of the second stage. The movement of the second piston out of the cylinders second stage could then be used to drive a standard hydraulic ram. Here are some of my own calculations (Assuming a diameter of 2.5m in stage-1 and a dia of 3.5m in stage-2): Volume(stage1) = Volume(stage2) (as all the fluid from stage-1 must fill exactly into stage-2 of the cylinder) Volume1 = Volume2 Area1 x length1 = Area2 x length2 (pi x 1.25 x 1.25 x 10) = (pi x 1.75 x 1.75) x length2 ----------------- Length2 = 3.38(m) ----------------- Pressure(stage1) = Pressure(stage2) Force(stage1)/Area(stage1) = Force(stage2)/Area(stage2) Force(stage2) = Force(stage1) x (Area(stage2)/Area(stage1) Force(stage2) = 98,000(N) x (9.62(sqm)/4.9(sqm)) -------------------------- Force(stage2) = 192.42(kN) -------------------------- So my question is.. can the fluid flow out of the cylinder be used directly to drive a hydraulic motor at that low pressure and flow rate? Or.. could that force of 192kN travelling a distance 3.38m every 8secs be used to generate a signifiant amount of power?? Many thanks for you assistance so far ;-)

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Re: Hydraulics Question - Low pressure High-Flow source to drive Hydraulic Motor - Feasible?
Re: Hydraulics Question - Low pressure High-Flow source to drive Hydraulic Motor - Feasible? -- matsoda	Post Reply	Top of thread	Forum

Posted by: Zip ® 10/05/2006, 23:53:06 Author Profile eMail author Edit

How are you going to handle approx. .8 million litres per minute ? I assume you mean 12 cubic metres a second? Zip

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Re: Re: Hydraulics Question - Low pressure High-Flow source to drive Hydraulic Motor - Feasible?
Re: Re: Hydraulics Question - Low pressure High-Flow source to drive Hydraulic Motor - Feasible? -- Zip	Post Reply	Top of thread	Forum

Posted by: matsoda ® 10/09/2006, 13:39:27 Author Profile eMail author Edit

Yes, i did mean 12 cubic meters / second. I was hoping i could convert the relatively large flow into a higher pressure/ and or moderate movement that could then be used to generate soem useful power. Any ideas? i did think of feeding it into second stage hydraulics (please see my reply). Many thanks ;-)

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