hydraulics question (tie'ing together)

bomrat · Oct 5, 2010

hi, i have a toro lawn mower. it uses 3 hydraulic motors to run 3 separate reels. each real is .43cu. in/rev on 2000psi.

i want to run a single motor. larger of course. to run one pulley.

the setup has 3 separate circuits to drive the motors.

can i just tie the outputs of the circuits together and run them into one larger motor?

the output circuits are already tied together and run back to the reservoir

and what motor size would your run?

thanks

J_J · Oct 6, 2010

As the displacement gets larger, the rpm decreases. You will have to step up the rpm via pulleys. You should be able to feed all three into one hose for the larger motor. A manifold with three in, and one out. One hyd motor with a displacement of around 1.29 cu in. at 2000 psi.

I believe someone on here has already done something similar.

bomrat · Dec 10, 2010

hi, ok, i've been trying to find a motor that is .43cu. in rev on 2000psi

everything i can find is listed in different specs.

most motors i find are in gallons per minute.

how can i convert or find what i need.

i am hoping i can find something that will run at

1.29cu in at 2000psi and around 700 rpm.

thanks

J_J · Dec 10, 2010

4 GPM flowing through your selected motor will run it at 710 rpm.

Close but not exact.

https://www.surpluscenter.com/item.asp?item=9-1156&catname=hydraulic

oldnslo · Dec 10, 2010

there are 231 cubic inches per gallon.

so if you have a motor with 1.29 CIR (cubic inches per revolution) operating at 700 RPM the formula is:

1.29 CIR x 700 = 903 cubic inches per minute divide by 231 = 3.91 GPM

If the motor data gives flow in GPM Vs RPM

multiply the flow by 231 and then divide by the stated RPM will give you the CIR.

NOTE: NONE of these calculations take ineffiencies into account.

Roy

bomrat · Dec 10, 2010

hi i found the pump specs. and please verify my math on this.

.99 in/rev 2850 rpm max pump (this is a 2000psi pump)

so:
( .99 x 2850 = 2,821.5 ) / 231 = 12.21 gpm.

am i right?

thanks