Hydraulic Radial Piston Motor
Hydraulic Radial Piston Motor
(OP)
Good morning, I am currently designing a hydraulic radial piston motor that will require 6000 ft-lbs of torque, as the end result. What I have so far is: I have 18 pistons (6 per cam) in series. I have a fluid flow of 600 gpm min. I have an inlet port of hydraulic diameter of .2832211511 in. An outlet port hydraulic diameter of .2868575741 in. Velocity into the system is 60.67 ft/s. There is a thru hole in the center. I am trying to calculate How much displacement is needed for 6000 ft-lbs of torque. I am trying to cheat some of the thru holes 600 gpm to drive the motor thru the inlet and outlet port. The pistons volume per chamber is .371836943 in^3 x18 pistons =6.69306497 in^3 of displacement. Any thoughts, suggestions, comments, help would greatly be appreciated. Please ask questions as to that I may have made unclear above. Thanks
I have attached a file showing some of what was discussed above. Thanks
I have attached a file showing some of what was discussed above. Thanks





RE: Hydraulic Radial Piston Motor
Bill
RE: Hydraulic Radial Piston Motor
RE: Hydraulic Radial Piston Motor
TygerDawg
Blue Technik LLC
Virtuoso Robotics Engineering
www.bluetechnik.com
RE: Hydraulic Radial Piston Motor
www.tynevalleyplastics.co.uk
It's ok to soar like an eagle, but weasels don't get sucked into jet engines.
RE: Hydraulic Radial Piston Motor
Ted
RE: Hydraulic Radial Piston Motor
Nonsense. (If any of the other dimensions given are correct.)
RE: Hydraulic Radial Piston Motor
If your motor displaced volume is 6.69 cu. in. per rev, and the flow is 138,600 cu. in. per minute (600 x 231), then I believe the motor speed should be 20,717 rpm. And 6000 ft-lb torque at 20,717 rpm works out to over 23K shp, which is one big honkin' hydraulic motor.
RE: Hydraulic Radial Piston Motor
Oh, and "Torque is proportional to the product of pressure times displacement" : minus friction losses, which will be significant.
RE: Hydraulic Radial Piston Motor
RE: Hydraulic Radial Piston Motor
If you don't know pressure pushing on the pistons, you don't know torque capacity.
Do you know the anticipated load demand to be placed on the motor output? If the anticipated load exceeds the motor capacity, which you don't know, then the motor will not turn.
Do you know the pressure differential across the motor, inlet to outlet, at zero motor speed? With that delta P you may calculate, based on the motor displacement, a starting torque capacity. Will that exceed the load placed on the motor output drive? If not, then the motor will not start and there will be no flow through the motor.
What do you know? 600gpm through the system. Require 6000 lb-ft torque. 6.7 cu. in. motor displacement. Yet you ask how to calculate how much displacement is required to generated how much torque by how much pressure. And you ask how much flow will go through the motor at some unknown speed.
See some formulas here: http://www.womackmachine.com/engineering-toolbox/f...
Ted
RE: Hydraulic Radial Piston Motor
If you do your calculations in SI units, life becomes a whole lot easier, in my opinion.
The pressure (in pascals) x the volume flow rate (in cubic metres per second) = the power (in watts).
At the shaft, the torque (in N.m) x the rotation speed (in radians per second, which is revolutions per second multiplied by 2 x pi) = the power (in watts).
Obviously assuming that the output shaft power = the input hydraulic power makes an implicit assumption of 100% efficiency, but start with that and see where it puts you.
The volume flow rate is obviously the piston displacement per revolution x revolutions per second.
RE: Hydraulic Radial Piston Motor
RE: Hydraulic Radial Piston Motor
Ted
RE: Hydraulic Radial Piston Motor
600 USgpm = 10 US gallons per second = 37.85 L/second = 0.03785 m3/s
So the power we are talking about is 265 kW
300 rpm and 6000 ft.lbs is 342 hp which is 255 kW so plausibility checks out (it's ominously near 100% efficiency, though)
So now we know that 300 rpm = 5 revolutions per second and ^ that flow rate will get it near what you are looking for.
So each revolution has to displace 2 US gallons or 7.57 litres or 462 cubic inches of fluid. Each of your 18 cylinders has to displace 25.66 cubic inches if it goes through one stroke per revolution.
Right now this does not consider friction, fluid losses, etc which are going to considerably affect the real outcome.
RE: Hydraulic Radial Piston Motor
But, not all 600gpm flow goes through the motor cylinders.
Ted
RE: Hydraulic Radial Piston Motor
RE: Hydraulic Radial Piston Motor
RE: Hydraulic Radial Piston Motor
RE: Hydraulic Radial Piston Motor
Ted
RE: Hydraulic Radial Piston Motor
T = psi x in^3/(2*pi) lb-in or in^3 = 6000*12*2*pi/1000
To turn 200rpm with 452.4in^3 displacement requires 391.7gpm
rpm = 231 * gpm / in^3 or gpm = 200 * 452.4 / 231
1000psi * 452.4gpm / 1714 = 228.2 hydraulic horsepower. Rounding errors account for the difference between your 229hp and my 228.2 hydraulic hp. Close enough.
Ted
RE: Hydraulic Radial Piston Motor