Hydraulic powered mowing deck

[Truckerdad]

Hello all..any know what the minimum reguirements/specs would be for a hydraulic motor that would be capable of powering a 38"(2 blade) or 48" (3 blade) mowing deck? Thanks

 

[SuperDuty335]

More details.

 

[Truckerdad]

I have an extra deck or 2 and am wanting to use it as a tow behind in an offset position. Want to power it with hydra motor. Have no real idea on what size(gpm,rpm displacement etc.) would be required/sufficient to power it.
Thanks

 

[Iplayfarmer]

That's a good question.

I know that these little 10 hp mowers would power a 38" deck plus have some power availble to power the wheels. I think 10 Hp would be a good place to start. I don't know how to translate that into GPM and pressure, though. RPM is going to be subject to the size of pulley you have on it. If you match the pulley to the same diameter that's under most riding mowers, you'll need a motor capable of turning about 3600 RPM.

 

[Truckerdad]

Iplayfarmer,thanks. I have called some suppliers and they all want me tell them what I need...I dont know. It's all the GPM,Disp,RPM that they want to know. I was thinking less than 10 hp...but thats why I'm here. I could change the RPM at the deck with diff pulley sizes?? Thank again

 

[Iplayfarmer]

I did a little more looking. Surplus Center rates their high speed hydraulic motors in torque. Horsepower = (torque X 2 pi X RPM)/33000. In this formula, torque has to be measured in foot-pounds. Some hydraulic motors are rated in inch-pounds. 1 foot-pound equals 12 inch-pounds.

Therefore, a 10 hp engine at 3600 rpm will put out a theoretical torque of about 14.5 ft. lbs. (175 inch-pounds). Most of those engines are rated for a few more horsepower than they actually produce. My SWAG would be that you need to look for a motor that puts out more that 10 foot-pounds of torque. This one is a potential candidate.

Now comes the fun part. Your selection of motor is going to depend also on what you have powering it. You'll need to know what pressure and what flow (GPM) you have feeding the motor in order to know what your resultant speed and torque are going to be. More pressure equals more torque and more flow equals more speed. Notice that the potential candidate that I linked to requires 15 gpm at 2000 psi to achieve the maximum torque and speed.

 

[bobodu]

And some way to cool all that oil..........

 

[J_J]

If you had a 10 HP engine, you can use a hyd pump of .3 cu in, pressure of 3000 psi, turning at 3600 rpm, and this will produce 5 GPM.

If you power a 1 cu in hyd motor with this setup, with an efficiency rating of 85%, the hyd motor will developing 8.5 HP, and turn at 1155 rpm, but produces only 478 in lbs. Not good enough.

That is why you don't see small hyd powered decks on smaller lawn tractors.

The Power-Tracs run a 48 in, three spindle deck, but it has a 25 HP Kohler engine, and pumps 8 GPM, so it does take force to get force.

 

[quicksandfarmer]

Google "hydraulic horsepower calculator." The horsepower equivalent of a hydraulic flow can be calculated from the PSI, the gallons per minute, and the efficiency of the motor.

Decide how much HP you think you need. I'm going to assume that your using open-center tractor hydraulics to power this. Open center systems deliver a constant gpm and vary the PSI to vary power. Your tractor's owner's manual should tell you the gpm at operating speed (it varies with engine speed) and maximum PSI. Using the calculator, make sure your hydraulic system can deliver enough fluid at enough pressure to deliver that HP. Then figure out what the PSI will be for the amount of HP you want. If you don't know the efficiency of the motor you'll be using, use 85% as a guess. (Example: 10 gpm at 2000 psi and 85% efficiency gives 13.7 HP).


Then decide what speed you want the motor going at, and use that to figure out what size displacement you need. Motors are rated in cubic inch displacements, which is how much fluid it takes to do one rotation of the motor. There are 231 cubic inches in a gallon, so multiplying the GPM by 231 and dividing by the RPM gives displacement. (Example: At 10 gpm 3600 RPM requires a displacement of 2310/3600 = 0.641 cu in).

Calculate how much torque is required to provide that much HP at that RPM. Torque= HP*33000/( 2 pi X RPM). (Example: 13.7 HP at 3600 RPM= 20 ft-lbs.)

Now, look in the specifications for a motor with the appropriate displacement. Make sure its rated PSI, RPM and torque are greater than the numbers you calculate.

Example: Surplus Center has a motor with the following specifications:
Burden Sales Surplus Center Item Detail

Disp. 0.70 cu. in/rev
Pressure:
1500 PSI Cont.
2000 PSI Int.
Motor Torque:
166 in-lbs. cont.
222 in-lbs. int.
Speed:
5000 RPM (motor)
3000 RPM (pump)

Displacement is good, speed is good. But 1500 PSI is below the calculated pressure, and 166 in-lbs (13.8 ft-lbs) is below the calculated torque, so this motor would not be suitable.

 

[Iplayfarmer]

Truckerdad,

Did you get some specs on what you'd be powering the motor with? If you'll post some flow and pressure specifications, we'll all be able to help you a lot better.