Hydraulic Motor Displacement question

[woodlandfarms]

I keep running into this and keep getting confused.

What is the point of cubic inches in a hydraulic motor? I look at a 4.9 CU Inch motor that spins at the speed I want, and accepts the pressures and GPM I want. But then I look at a 10.5 CU Inch motor and it gives the same results.

So, who wants to educate me on hydrualic motor theory or point me to a site for morons like me.

 

[jgrreed]

I keep running into this and keep getting confused.

What is the point of cubic inches in a hydraulic motor? I look at a 4.9 CU Inch motor that spins at the speed I want, and accepts the pressures and GPM I want. But then I look at a 10.5 CU Inch motor and it gives the same results.

So, who wants to educate me on hydrualic motor theory or point me to a site for morons like me.

I don't know a whole lot more than you, but try this as a start. It helped me a bit.....

Burden Sales Surplus Center - Tech Help

-Jer.

 

[dfkrug]

I keep running into this and keep getting confused.

What is the point of cubic inches in a hydraulic motor? I look at a 4.9 CU Inch motor that spins at the speed I want, and accepts the pressures and GPM I want. But then I look at a 10.5 CU Inch motor and it gives the same results.

The spec is displacement per revolution. In your system design, you will
be operating in a range of hydraulic flows and a max pressure from your
hydraulic pump. Whatever you want to rotate with the motor needs to
turn at a certain RPM and deliver a certain force or torque. Since we
are probably talking about a tractor's implement pump, you are working
with about 5GPM and 2500psi max pressure. The motor may drive the
load directly or thru a gearbox. A gearbox will change the torque and
speed, so you will have to allow or design for that. For simplification,
assume direct drive. Divide your pump volume by your design speed to
get your motor displacement. e.g. 5GPM = 1155 ci/min ==> for 200 RPM
motor speed, I need 1155/200 or 5.8 ci/rev. Torque is another concern.
Different motor designs yield different torque outputs. You will have to
look at the tables for the motor in the manufacturer catalog to see if
the one you want delivers the torque you need.

What are you designing?

 

[woodlandfarms]

I am designing a stump grinder. Direct Drive...

My tractor is all hydraulic, think skid steer. 18GPM at 3100 PSI...

JJ recommended a motor that was 4.9 CU In... I was looking at a wheel motor that was 8.9 CU In...

I have to say I still don't get it.. I guess the CU Inch is least important? Focus on RPM listed with the ability to handle my capacities and as much torque? Man, there are lists of motors....

 

[Mith]

If you have a fixed flow, then by changing the displacement of the motor you will change the speed that it rotates at. The displacement also has an effect on the torque.
With fixed flow, a larger displacement roatates slower but has a higher torque.
Seeming as you have a fixed flow then you will need to size your motor so that you get the right output speed at that flow.
Flow / displacement = rpm
The motor specs will indicate the torque. The rpm values and flows they list will be the max values, they all depend on the application, so dont think that just because the motor specs list it as a 1000rpm motor it will do that with your flow. You need to calculate what the motor will do when connected to your system.
Ignore the rpm listed (as long as it is above what you want to run it at), displacement is EVERYTHING. Ignore the flow listed, use the flow that you will be giving it to calculate the motor speed.
Torque will also be effected by the pressure, but seeming as I expect you'll just run it at the max pressure your tractor will allow you'll just have to size the motor to run at the correct speed with the flow you have available, and hope the torque produced is enough. If you want to increase torque you will have to either increase the system pressure, or get a larger pump on your tractor and then get a higher displacement motor.


OK, so you have a fixed 18gpm flow.

18 gal = 4158ci (I work in metric, so you might have to check my conversion)

So you have 4158ci of fluid coming out of your tractor per minute.

A motor is 4.9ci displacement, so every time 4.9ci goes through it, the output shaft turns once. So if you put 4158ci of fluid through it in a minute, the motor shaft will turn 849 turns in 1 minute, or 849rpm (4158/4.9=848.6)

A 8.9ci motor will turn at 467rpm.

The larger displacement motor will have a higher torque, but you will need a higher flow from your tractor to get the same speed out of it.

Presumably you know the speed you want the motor to turn at. If so you can calculate the size motor you need.
You have 4158ci of flow. You know the rpm you want (I'll say 800rpm)
So 4158/800 = 5.20ci displacement for the motor.
Clearly, you can sub in the numbers you want to find the right answer.


just as another though, I dont think that a wheelmotor will take the abuse that you'll give it if you connect a stump grinder disc to it directly, side loads on a grinder are massive.
You really want to use a belt drive and have the disc on a seperate shaft.

I hope this helps, and I hope I've done my sums right. A couple months ago I was in your position with regards to motors, but with a little help I ended up building a tractor like yours which works, so I must have done something right

 

[Farmerford]

Mith: That's a great explanation. You did get your sums right. I agree with everything you say with one qualification.

When you said "If you want to increase torque you will have to either increase the system pressure, or get a larger pump on your tractor and then get a higher displacement motor", you assumed that Woodlandfarms' machine has more horsepower than the 18gpm/3100 psi pump requires. 18gpm/3100psi of pump output is about 32horsepower at the pump outlet. If the pump is 85% efficient, an engine of 36-37 horsepower is needed to drive the pump.

So, if Woodlandfarms' machine has more than, say, 35 horsepower available to drive the hydraulic pump, he could increase torque as you describe with either a larger displacement pump or a higher pressure pump. But if 35 horsepower is his limit, then, as you say, his torque will be determined by the wheel rpm he chooses.

 

[woodlandfarms]

Wow guys, now I am starting to get it..

The engine in my tractor is rated at 65HP. For that amount of HP I have always felt as it is a full hydraulic tractor that the fluid flow is a bit under its potential, but I am not the designer so...

As for a belt drive. The design I intend to rip off was the S-18

..:: MTBMFG ~ Stump Grinders ::..

It appears to be direct drive or at least some sort of coupler...

Mith, you MUST post pictures... No way around this one.

 

[Mith]

FF, thanks. I was aware that Woodlands tractor had alot of HP, but its definately a valid point that fiddling with flow and pressure is going to alter the HP requirements.

Woodland,
I expect PT have designed your tractor so that all the hydraulic systems can run at nearly full power at the same time, which is why your hydraulic PTO is relatively low flow. If you had a PTO that required the full engine HP then every time you tried to drive forward then the engine would stall.

Looking at the pictures of the S-18 it looks like the disc is running on a shaft, then the motor is coupled to the end of that shaft.
If you do a similar thing then you dont need to worry about side loads on the shafts, so no need for a wheel motor. This seems to meet the specs.

If I were you, I would figure out what size motor you need based on your flow 18gpm and the speed that you want the disc to spin at, and get a motor with that displacement that meets the spec for max speed etc.
I wouldnt worry about the torque too much, you'd be better off concentrating on getting a heavy disc for the flywheel effect.
Unless you fancy adding another pump to your tractor to get higher flow (and a bigger motor) then there isnt much point worrying about the torque, because there is nothing you can do to change it (and clearly those ones on the link work OK, and they are using the same flow as you have got).


I dont want to detract from your thread, but you can read the story on my build here
It isnt done yet, but it works....

 

[dfkrug]

My tractor is all hydraulic, think skid steer. 18GPM at 3100 PSI...

WOOD, I would say that first you need to find a motor series that can make
full use of the pressure you have (3100). MITH suggests the Charlynn 104
series, but that only handles 3000psi continuous pressure. You DO need
to protect from side loads as he points out, and that means you prob want
belt-drive or a gearbox, or another way to protect it.

For a stump grinder, you are turning a flywheel (as mentioned) and that
requires high speed but lower torque, so a speed-increaser final drive can
allow you to use fairly high displacement motors, or a 1:1 final drive ratio
and a low displacement motor. The S-18 grinder you linked must be a beast
that makes use of the massive hyd power available to skid steers.

 

[Farmerford]

Mith:

That is quite a project! I am going to join the site so I can open all the pictures.

One point on the 4.9 cu in Charr-Lynn motor you identified. From personal experience I can tell you that Charr-Lynn mills a 5/16" key slot on the 1 1/4" shaft rather than the more common 1/4". The SAE specs do allow either keyway, and I suppose the 5/16" key is stronger, but the problem is that virtually all 1 1/4" bore hubs (couplings, gears, pulleys, etc.) have a 1/4" keyway. And after quite a bit of hunting for a 1/4"-5/16" stepped key (unsuccessfully) I wound up making my own.

In contrast, the same 4.9 cu in motor with a 1" shaft has the common 1/4" keyway, since the SAE specs do not provide a choice. The smaller shaft is rated for a slightly smaller radial load, but that should not be a problem in most applications.

I am looking forward to learning more about your project.