Help me choose a correct hyd motor.

Jd44A_1

Member
I'm working om a IH 595 manure spreader.
The angle drive box for the rear beater
is shot past fixing. No gear boxes are
available that we can find soooooo my
idea is to replace the 90? gear box with
a hyd motor. Im sure itll work fine.
Going to be ran with a JD 4440 or 4840.
2250 psi amd 23 gpm flow. Im not sure
other which motor to use other then using
a Low Speed High Torque (LSHT) gear hyd
motor. Need some hyd masters to help me
figure this out. Thanks !!!
 
You first need to determine what speed you want the motor to turn at. After knowing this you can use the equation DISP=GPM/RPM*231 to calculate the needed displacement of the motor where DISP is in units of cubic inches. You don't want to overestimate the needed speed since the faster the motor is capable of running (in other words, the smaller it is) the less peak torque it will deliver.

Using your figures of 2250 psi and 23 gpm you can get a theoretical maximum of 30 hp out of the system but by the time you factor in efficiency you will get much less. To push this much flow through remote valves will result in a lot of pressure drop which means a lot of heat generation.

Also keep in mind that the pressure and flow values you list are absolute maximum numbers - there is no "reserve capacity". In order to deal with big slugs of manure and other transient loads you probably don't want the system running at much over 50% max capacity in normal use. If the beater takes more than about 10 hp to operate in normal conditions you may not be happy with the performance of a hydraulic drive.
 
There are a few new spreaders using hydraulic motors to drive the floor chains, Bunnings is one, i think Delgelman does too, might be worth finding a dealer and pricing a motor... I suspect they may be too much motor for your 595, but it will give you an idea.
 
It will work very well, sorry I do not have the formulas to tell you a size but I have seen may things converted, one thing you want to know is gas or electric or hyd power do not follow the same hp paths,, been 35 years since I had the books with all the info you need they listed what each was equal to. say you had a 4 hp gas engine, you could get the same "power" from a electric motor of 2 hp, the hyd ratio was even better than half,, seems like a 1 to 1 1/2 hp hp hyd would do the same as 4 hp gas but that might be a bit off,, if i was doing this I would aslo install a flow control valve in the system so you can dial-in and "set" the speed ,just my thoughts
cnt
 
The amount of power being transmitted by a given oil flow rate and pressure is determined by physics with the equation HP = GPM * PSI / 1714. This equation assumes 100% efficiency so in the real world a correction factor needs to be used either to lessen the power extracted from a motor or increase the power needed to drive a pump, depending on which end of the system you're looking at.

If the motor size is known you can calculate power a different way by first finding motor torque with the equation TORQUE = DISP * PSI / 6.28 where torque is in in-lbs and displacement is in cubic inches. Then calculate the speed of the motor using RPM = 231 * GPM / DISP. Combine the torque and speed and you'll get the same answer for power as you did using the equation above.

It is correct that there is a "rule of thumb" relating the power required to drive a hydraulic pump depending on if it is powered by a gas engine or electric motor but that has to do with the power characteristics of the prime mover - it has nothing to do with the amount of power that can be extracted from the system at the other end of the hose. In the case at hand we know a maximum flow rate and maximum pressure but how that power got into the oil doesn't make any difference when looking at the motor side of the system.
 
Thanks Brendon !!! Still dont quite understand how to figure your figures but will get it figured out.
 

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