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

 

[BillPSU]

At what RPM does this motor have to turn?

Bill

 

[goodguy1405]

300 max rpm

 

[tygerdawg]

Sorry I don't have an answer. The lack of response is probably due to our shock discovering there is machinery that can bore & mill inch dimensions to 10 decimal places. That's impressive.

TygerDawg
Blue Technik LLC
Virtuoso Robotics Engineering

http://www.bluetechnik.com

 

[Pud]

tygerdawg: Impressive (the sarcasm and/or irony)!

www.tynevalleyplastics.co.uk

It's ok to soar like an eagle, but weasels don't get sucked into jet engines.

 

[hydtools]

Torque is proportional to the product of pressure times displacement. Google hydraulic motor torque.

Ted

 

[racookpe1978]

600 gpm fluid flow?

Nonsense. (If any of the other dimensions given are correct.)

 

[tbuelna]

I would agree with racookpe1978. A fluid flow velocity of 60.67 ft/sec thru the motor intake orifice is extremely high, and would result in significant flow losses.

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.

 

[btrueblood]

I'll third that, unless some dozens of identical motors are linked in banks.

Oh, and "Torque is proportional to the product of pressure times displacement" : minus friction losses, which will be significant.

 

[goodguy1405]

lol The 10 decimal places are only for calculations. Don't like rounding errors in something this tolerance tight. The pistons are in series 3 stacks of 6 piston which total 18 pistons in all. The 600 gpm is flow into the system, the inlet ports are only grabbing a portion of that total flow in. How much total flow the inlet gets is one of my questions? How to calculate the pressure difference? The torque? There will be some back flow pressure as well. This is not your typical everyday hydraulic piston motor. Any help would be appreciated. Thanks

 

[hydtools]

Well, if you don't know flow into the motor cylinders you don't know speed of rotation.
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/engine...alculations/hydraulic-motor-calculations.aspx

Ted

 

[BrianPetersen]

What you have not given us (or perhaps I have missed it) is the pressure. You have given us the flow rate, but not the pressure. Need both, in order to get anywhere.

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.

 

[goodguy1405]

The change in pressure is 1000 psi. Hydtools: I am unable to divulge certain information due to proprietary reasons. I do understand what your saying. Thanks for the input.

 

[hydtools]

I don't know, psi x gpm / 1714 = hp seems pretty easy.

Ted

 

[BrianPetersen]

1000 psi = 7 MPa (close enough for what we are talking about)
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.

 

[hydtools]

1000 x 600/1714 = 350hp

But, not all 600gpm flow goes through the motor cylinders.

Ted

 

[goodguy1405]

Brian And Hydtools thank you both for your replies. This design process gets more complicated by the minute. I have to design a motor with Q max= 600 into the system and then cheat some of that flow to turn a cam that will produce 6000 ft-lb of torque. It will have from 1000-1500 psi differential pressure.It will need to rotate at 200-300 rpm max. I am limited by size of pistons. I have a blank slate on how to accomplish this task. The tool has to be somewhere between 5-6 ft long with a max o.d. of 6.63 and min i.d. of 1.38. Good to be me today. Haha!

 

[goodguy1405]

This is what I came up with so far. I figured about 229 hp needed to get the 6000 ft-lbs of torque required at 200 rpm ( I divded by 5252 for ft-lbs). The displaqcment I got is .884 in^3 from 1.50 diameter piston with a .50 stroke. The gpm needed to push those pistons came out to be .018 with 16 pistons [gpm= cylinder volume x # of cylinders x (60/200)/231]. This is my source link to equations

http://www.milwaukeecylinder.com/pdfs/mc_design_engineers_guide.pdf

page 190. How is that so far? Thanks for any input.

 

[goodguy1405]

Correction: GPM needed is 1.102 [GPM=cylinder volume x # of cylinders x (60/3.333333333)/231] Had to use seconds instead of minutes for RPM.

 

[hydtools]

You do not have enough pressure available to deliver torque required with your motor displacement.


Ted