Powering a hydraulic motor w/ the tractor remotes

[Darren]

I'm looking at an auger to move lime. The utility auger I found requires a 3/4 hp electric motor. It can also be purchased with a hydraulic head that will take a 4 bolt hydraulic motor.

I know many tractors are not designed to run heavy hp loads via the remotes because of heat buildup, etc. What I don't know is how to determine what the tractor could run. Flow for the Kubota L3710 is 7.8 gpm. I don't have the specs for the auger motor, but I figure I can ask the auger maker for the specs on the one they sell.

Even after I have the specs that still won't tell me if the tractor can handle it. Any ideas?

 

[jasonf]

HP = f * p / 1714
where-
HP = hydraulic horsepower
f = flow, in GPM
p = pressure, in psi

So assuming that you have 7.8 GPM at the hose and the Relief valve is set to 2000 PSI you would have HP=7.8*2000 / 1714 or HP = 9.1 Now If you knew what the brand of the motor is, you could look it up on the manufacturer's site and get the performance data. For example Prince Hyd makes Hyd motor model called the ADM that I used for a chute rotator I recently built for my front mounted snowblower. If you go to PRINCEHYD.COM, click on PRODUCTS, then on CATALOG.
Click on products, then motors and select ADM. This will take a while to load the catalog . click on ADM and then GO TO THIS PRODUCT. This take awhile even on a T1 conection. Hang in there it is worth it. Click the displacement size you want I used 100 (10 Cu in) then click 4 for 4 bolt pattern, then click R for standard 1" keyed shaft, then P for 1/2 inch port size for the hose connection. Then NO for Case drain port option. Unfortunately I can't copy and paste the perfomance data into this post so I will just pick some values that you might be interested in:

At 7 GPM flow and 2400PSI this pump will put out 2016 in/LBS of torque and 176 RPM. At 7 GPM and 1600 PSI it puts out 1334 In/LBS of Torque and 214 RPM. Since they don't have 2000 PSI on the chart lets call it 1500 In/LBS of Torque and maybe 190 RPM as a guess. If you take the formula that is on the performance page HP = RPM * Torque (in In/Lbs) / 63025, We get HP=190*1500/63025 and that gives us 285000/63025=HP or HP = 4.5. Sooo running this motor at these pressure and flow values provide 4.5 HP and you say that the electric motor supplied is 3/4 HP ? Sounds like this motor would be more than enough. If we back off a little and say that the GPM is only 5 and the pressure is only 1600 PSI then the chart says that the torque is 1344 In/LBS and the RPM is 142. Put that into the HP formula and we get 142*1344/63025=HP, that translates to 190848/63025=3.0 HP; still 4 times the value that the electric motor can supply.

The motor that I am using as an exaple is available in at NORTHERNTOOL.COM and is $157. I have used this very motor and it is teriffic. Very powerful and heavy. Not a toy motor. They also make larger ones that you can see in the catalog and best of all the motor is reverseable when you switching the hoses.

I hope that I haven't gone too far with this, I'm no math wiz, but I did do all the research for my project, so I finally have something to say....... Hope this helps..

 

[rbargeron]

Clear, accurate, documented and very helpful. Great job - going to cut and paste it for future reference. /w3tcompact/icons/clever.gif

 

[Darren]

Jason, youy post is a huge help. After following your calculations, there's no question that the tractor remote can handle it. I'm confused about the information in the Northern Catalog which suggests that the operating pressure should be kept below 1000 psi by selecting the next larger size motor.

Is that from the performance curves? I'm using a 486 with Windows 3.1 (spent the money on the tractor) so I can't get to the curves on the Prince site. If the curves do give you that info, I'll have to get copies from Prince or Northern.

Thanks for the response. I know that took some time and effort.

 

[jasonf]

Attached is the performance chart from the the Prince site, (I finally figured out how to capture the file). If you look at it you will see numbers that are shaded in Gray or Blue, indicating the areas of greatest efficiency. Note that the PSI values are the difference between the Input pressure and the Output pressure. I would think that You would use the pressure value at the hose because there is little pressure that I can see on the output side of the motor, as it is just returning into the Hydraulics tank for reuse. There may be some small back pressure however created by the diameter of the hose that returns the oil from the motor to the Tank. I have used 3/8 diameter hoses in my project for both the supply and the return sides and it worked very well. Please let me know if there are any other questions of concerns.....

 

[Darren]

I talked to someone at Prince. They recommended a relief valve before the hydraulic motor to maintain the inlet pressure to 1000 psi. Which means that because of the possibility of either connecting the hoses backwards or operating the remote lever incorrectly means that the motor's inlet and outlet will need relief valves.

That brings up the question whether it makes a difference which way the fluid flows through a relief valve. With two relief valves even when the auger is turning correctly, flow will be reversed through the relief valve on the outlet side of the motor.

Thanks for the curve. It's a help.