Hydraulic Radial Piston Motor 3

[goodguy1405]

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

 

[hydtools]

To generate 6000 lb-ft torque with 1000psi you need a displacement of 452.4 in^3.
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

 

[goodguy1405]

Thank you Hydtools. I did get the same thing as you. I went back and saw the errors of my way. Now I have to come up with a motor that will displace 452 in^3 at a differential pressure of 1000 psi turning at 200 RPM to generate 6000 ft-lbs of torque. Piece of cake lol. Limited by O.D. and I.D. and even then with the inlet and outlet ports acting just right. Fun FUN!!!! Thanks for all your help everyone. Off to make some unobtanium.