Correct me if I'm wrong, but..

[20HP2N]

Hey, all. I've been pondering hydraulics for some time. In particular, I am trying to determine the maximum size pump I can use on my LT 133 to drive a wood splitter.
I'm taking a stab at the problem from a point of view not typical to the calculators and equations that seem to dominate the field.

The engine is 13 hp and I want to run a pump that is properly sized for the engine, but the twist is, I want to run the pump at the lowest recommended rpm.

To elaborate... one can realize that pump flow is a variable governed by engine speed. So, for example, I am looking at a pump with a flow of 17 gpm@ 2600rpm, but with a minimum rpm of 500, effectively reducing the flow to about 3.4 gpm. Obviously, I could not calculate my required HP using that flow rate, because the engine would stall.

So correct me if I'm wrong, but.. it seems that most HP/Flow@PSI calculations must assume maximum pump flow to determine required HP?

And for my question.. is there a formula/chart that sizes engine hp to pump displacement@PSI, rather than HP/Flow@PSI?

I used to play left field, and maybe I'm still out there? :laughing:

 

[J_J]

Go to the Tech Help section of Surplus Center - Hydraulic Cylinders Pumps Motors

Select Pump displacement & HP

Enter the figures and it will compute the results.

 

[20HP2N]

Go to the Tech Help section of Surplus Center - Hydraulic Cylinders Pumps Motors

Select Pump displacement & HP

Enter the figures and it will compute the results.

Thanks for that.. I've been doing more research and I think what I'm looking for is torque. (I'm trying my best)

I found this formula Torque=Pressure X Pump displacement (cipr) divided by 6.28.

I have this flywheel thing from a small punch press I want to throw into the equation so I can complete a full stroke with one step on the pedal, the flywheel storing energy for the inital moment of splitting if the pump speed drops to 500 rpm.

The engine can keep the flow rate up on the pump to a certain degree, but the flywheel is there for backup, basically.

I'm trying to figure on how fast I need to spin the flywheel to store the necessary energy.:confused2:

I'm not greatly smart, but I like tinkering and find that learning becomes a necessity if I want to avoid trial and error......

 

[20HP2N]

Thanks for that.. I've been doing more research and I think what I'm looking for is torque. (I'm trying my best)

I found this formula Torque=Pressure X Pump displacement (cipr) divided by 6.28.

I have this flywheel thing from a small punch press I want to throw into the equation so I can complete a full stroke with one step on the pedal, the flywheel storing energy for the inital moment of splitting if the pump speed drops to 500 rpm.

The engine can keep the flow rate up on the pump to a certain degree, but the flywheel is there for backup, basically.

I'm trying to figure on how fast I need to spin the flywheel to store the necessary energy.:confused2:

I'm not greatly smart, but I like tinkering and find that learning becomes a necessity if I want to avoid trial and error......

....so I found this other formula for determining torque from horsepower.

Horsepower x 5252/ RPM

If I calculate 13 x5252 divided by 1500 rpm, I get about 45 ftlb. torque.

if I use the torque formula for the pump using 250 psi...

250psi x 1.22(cipr) divided by 6.28, I get 49 ftlb. torque

If I multiply 250 psi x 24 in.sq (cylinder piston area) I get around 6000lbf.

Is three tons enough for most splitting?

 

[J_J]

Depends on the type wood.

Why are your limiting yourself if you have a 13 HP engine and a large pump.

You haven't mentioned a cyl, and it is the cyl and pressure that determines the force.

So if using the cyl for force, where does the flywheel come in to play?

Some splitters use only the flywheel with rack and pinion for the splitting force.

http://www.supersplit.com/principle.htm

 

[20HP2N]

Depends on the type wood.

Why are your limiting yourself if you have a 13 HP engine and a large pump.

You haven't mentioned a cyl, and it is the cyl and pressure that determines the force.

So if using the cyl for force, where does the flywheel come in to play?

Some splitters use only the flywheel with rack and pinion for the splitting force.

Principle behind the log splitting power of Super Split(R)

Yeah, the cylinder is 5.5" bore....and I think I miscalculated the engine torque. I should have divided by 3600 rpm. Substantially lower torque value. I don't think it could squeeze a lemon with a pump that big.

Maybe I'm missing something.

The flywheel and pump as well is to be driven by the engine, but the flywheel would transfer added torque to the pump through a sprague clutch when the pump speed fell to around 500 rpm, at the point where the flywheel would start to overrun the pump.
This would help the engine through the high initial moment of splitting....

mmmm...maybe I better go out tonight and chase skirts..

What I'm trying to achieve in the end is not running the engine at WOT, but having enough torque on hand to do the work.
There doesn't seem to be any formula or chart that matches a pump to an engine without running the engine at full speed. ie cipr to engine torque@*rpm... I should just try to buy an old stationary engine. :licking:

 

[J_J]

If you use your engine, at 13 HP, it can support a pump, pumping 6.5 GPM at 3000 psi using a single stage hyd pump.

You can pump more GPM with a two stage pump at a lower pressure.

To do that, you need a .4 cu in hyd pump running at 3600 rpm.

If you ran that pump and motor at 500 rpm, you would pump .8 GPM.

To pump .8 GPM at 3000 psi, will use 2 HP.

So knowing all this, what is your plan with the flywheel?

 

[FredH]

Way over my pay grade , but if you ran the engine slower as I can do on my splitter , it simply slows down the " Cycle " or split / travel time of the cylinder . ( not sure if it builds same pressure , seems to , just takes longer ) .
Were as my brothers splitter runs at a constant rate with no throttle adjustment .

Still trying to understand this flywheel thing ???

Fred H.

 

[glastron23]

I think what he's after is he's trying to use the kinetic energy from the flywheel to keep the pump turning with the engine at a slow speed, when the pressure rises,the torque requirements demanded by the pump rises and it slows the engine, the flywheel (through the sprague clutch) kicks in and uses the flywheels stored energy to keep turning the pump for the short time thats required. Your into a REALLY BIG CAN OF WORMS here..... yess... something to keep me up at night.... try this , it's a start Flywheel Effect or Polar Moment of Inertia - Engineers Edge if you float around in here you will find ALL KINDS of stuff. Jim

 

[FredH]

After watching the super split video , I kind of understand the flywheel aspect . But also in watching that video , while some people like the super split , myself anyway , would have a guard or cage between me and the wood . Go to the thread in safety about the near miss accident were the wood " Exploded " apart .

I guess I am still missing something though , As I mentioned above , by slowing engine , thus slowing pump , does it not just slow cycle time with still the same pressure ? Why not just put in a gear box that rather than reduces rpm's to the pump , increases rpm's to the pump while reducing rpm's required from the motor ?

All I really know for sure is JJ is the man in Hydraulics !!!:thumbsup:

Fred H.