Char Lynn Hydraulic Motors used as Hydraulic Pumps???? Please Help! 2

[Raymond56]

Hi I'm new here. How is everyone. Good I hope.
I hope you Folks can help me with a couple of questions on the Hydraulic Motors I have. I got them for free and they are brand new. I would like to use these motors as pumps. The Motors are Char Lynn S-series Hydraulic motors. The numbers on the Motors read # 103-1620-010 & #103-2006-010 Both Motors Diplacements are 225 The information on the Motors are in the Attached .pdf File Hilighted in yellow. on Page B-3-2 and Page B-3-6
I have a 1.5 HP electric motor that turns at 1725 RPM. According to the charts on the .pdf File, If I am reading it correctly. I would have to gear the electric motor down to about 31 RPM to achieve the required starting Torque (2990 in lbs) to start the Hydraulic motor. Is this correct? If not correct? Could you please provide me the correct information using these Hydraulic motors as Hydraulic pumps with the 1.5 HP Electric motor. They will be used to operate a 2” Hydraulic Cylinder, 24 “ long with a 1” shaft. If you can help me figure this out. That would be great. Also how much push force would that give me per square foot on that Hydraulic cylinder? Again, the Hydraulic Motors are: Char Lynn S-series Hydraulic motors #103-2006-010 & 103-1620-010
Thank you so much for all your help……
Raymond

 

[MikeHalloran]

I wouldn't try it without Char-Lynn's blessing (call and ask for an Application Engineer), mostly because I don't think the bearings would be happy without a nice high pressure supply.

Just call Northern Hydraulics and buy a cheapo pump for your log splitter.

Save the motors for your homemade skid-steer loader.



Mike Halloran
Pembroke Pines, FL, USA

 

[Raymond56]

I tried to contact Eaton the people who sell Char Lynn and this is what they said:

Due to liability reasons we do not offer application support in Tech, I would suggest you contact your local distributor and have their Certified Fluid power agent talk to you.

Thats why I'm here asking you profesionals. Thanks for the reply. It was my understanding that a Hydraulic Motor can be used as a Hydraulic pump with no problem........

 

[MikeHalloran]

So, did you talk to the Certified Fluid Power Agent at your local distributor?


Mike Halloran
Pembroke Pines, FL, USA

 

[Raymond56]

No not yet. I just wanted to run it by you guys first....
Thanks for the help

 

[hydtools]

That is a large displacement pump for a 1.5hp motor.
At 31rpm you will get 4 gpm. Is that enough?
The shaft seal may not seal well on what will be the pump suction side of the porting.
At no load start up there will be no start up torque. That figure is for use as a motor.
Your expected max pressure capacity will be about 640 psi. Is that enough?
Push force at 600psi could be 1884lbs.

Ted

 

[Raymond56]

I am making a vertical standing Can Crusher. The Cylinder Housing will be a little more than a FT. square of 1/4 Inch steel and the crushing plate will be one Ft. square and 5/8 thick steel. The Cylinder is rated for 3,000 PSI

I guess I don't have to worry about the starting Torque, Thanks I didn't think about the no load. What RPM would you recommend for highest effect of the pump? The Max Rpm for the pump is 253 RPM.
Thanks for the info...

 

[Raymond56]

Not Pump I ment Motor.......

 

[Oldhydroman]

It is sort of true that you could use this type of a motor as a pump but it doesn’t make a very good pump.

The job of a pump is to convert the mechanical power input on its shaft to a hydraulic power output from its outlet port. In an open circuit application (which yours will be) the inlet to the pump is connected to tank and the suction gallery in the pump is arranged for a gentle, smooth and low resistance flow path so that the pump can suck in the oil. There will be a slight drop in pressure along the length of the suction line because of fluid friction, and there may be a small drop of pressure because of the change in hydrostatic head (if the pump is above the tank). The suction pipework has to be arranged so that when all the pressure drops are taken into account the pressure at the inlet of the pump is no less than about 0.8 bar absolute, i.e., 0.2 bar below atmospheric pressure. If the pressure ever goes lower than this then you run the risk of cavitation, which will reduce the flow and damage the pump.

The job of a motor is to convert the hydraulic power delivered to its ports into a mechanical power output. The torque required to turn the shaft is reflected as the pressure required to push the oil through the motor. Most motors are bi-directional so both ports are compact, high pressure galleries. When you try to use the motor as a pump you have no suitable suction port available (low resistance flow path) and the pump will cavitate unless you boost its inlet port with a small pressure (which we’ll call “boost pressure”). In a nutshell: the motor, when used as a pump, has to be force fed with oil because it can’t suck for itself.

These motors do actually work as pumps in certain hydraulic circuits when they are used with an overrunning load, e.g., when a winch drive is lowering the load, but special facilities are provided in the circuit to ensure that the motor inlet has sufficient pressure to avoid the cavitation.

How much boost pressure is needed? It is unfortunate that Char-Lynn have been so uncooperative. Maybe they’re thinking there will be no sales involved in helping you so they don’t need to bother. I’m sure they won’t like you telling all the participants of this forum that they wouldn’t help you and they’ll be lots of people sufficiently turned off by that approach that they might think twice about relying on Char-Lynn for their next project. Never mind, you know the old saying: “As ye sow so shall ye reap”.

So how much boost pressure is needed? You can find this out by experimentation: connect your motor to a hydraulic supply and run it with no load on the shaft and with the outlet port running straight back to tank with minimum back pressure, then measure the pressure needed to drive the motor at the speed you want. This inlet pressure is that which is needed to overcome the friction in the motor and the pressure drop in the inlet gallery plus the pressure drop in the outlet gallery. For a conservative estimate of the pressure drops let’s assume that the motor has no friction at all – so all the pressure drops are fluid friction. Then assume that both galleries are identical (that’s fair) which means that half the pressure needed to drive the bare shaft motor will be enough to overcome the flow resistance of one gallery. This is the boost pressure which you need to apply to run the motor as a pump.

Is there an easier way? Yes. Sauer-Danfoss make a very similar motor and they publish a graph of the pressure needed to run the bare shaft motor (see attached file)

How fast can you run the motor (pump)? You need to start at the other end of your circuit to work this out. What is the maximum force you want to get out of your cylinder? Divide this force by the area of the cylinder to work out the pressure at which you want to run. Pressure multiplied by flow equals power and you only have 1.5 HP to work with. Make some broad brush allowances for losses and you will find that you only have about 1 HP available as hydraulic power. So from your pressure requirements you can work out the flow – and this will enable you to work out the boost pressure requirement. The flow will also enable you to work out the pump speed and then you can work out your reduction ratio. But remember you will need another pump to provide the boost pressure, or (I’ve seen it done but wouldn’t recommend it) pressurise your tank with compressed air, i.e., use a certified air receiver as the tank, or position the tank high enough above the pump for the static head to give you the required inlet pressure.

Or, and this would be my preferred option, sell the motors on ebay and use the money to buy a cheap gear pump from your local supplier.

DOL

 

[hydtools]

Raymond56, I have to ask the question the other way. How much flow and pressure do you need to do what you want to do? Your input available is 1.5hp. Effective output may be 1.1hp, 75%. psi X gpm / 1714 must equal 1.1, no matter the combination of flow and pressure.
That is what you have. What do you need is what you have to answer? Even if you sell the CharLynns and buy something else you still must answer the same question.

Ted

 

[Oldhydroman]

Hi Raymond

It’s a very convoluted and error prone calculation, but if you really do want to try to work out the performance of your proposed hydraulic system then it goes like this:

Image you had a motor of 225 cc/rev, when you give this motor a flow of 45.4 L/min [45,400 cc/min] you would expect a [theoretical] speed of 45,400/225 = 201.7 rpm. You will find, however that the speed is less than that because of internal leakage. The ratio of the actual speed to the theoretical speed is called the “volumetric efficiency” and you will see from your datasheet that this is a function of torque (a higher pressure causes a greater internal leakage) and also a function of speed (flow).

Let’s imagine that the motor was driving a load such that the torque was 158 Nm. The theoretical [differential] pressure induced at the ports of the motor would be 225/(20*pi) Nm per bar, i.e., 3.58 Nm/bar. Theoretically, you would need 44.1 bar to drive that motor, but when you look it up on the datasheet, you would find that you actually needed 55 bar. The ratio of the actual torque to the theoretical torque is called the “hydro-mechanical efficiency” and you will see that this is also a function of torque and speed (pressure and flow).

So this is our situation: it takes a pressure of 55 bar to push 45.4 L/min of oil into the motor when the shaft has to overcome a torque of 158 Nm and, when the motor is loaded like this, our flow only gives us a speed of 191 rpm. Our input power (hydraulic) is 45.4*55/600 = 4.16 kW and our output power (mechanical) is 191*2*pi/60*158/1000 = 3.16 kW. So it looks like the “overall efficiency” is 3.16/4.16 = 76% (which is just like Hydtools Ted said it would be).

When driving this Char-Lynn unit like a pump we have no option other than to assume that the efficiencies as a pump are the same as the efficiencies as a motor (as long as the operating conditions are the same). Our volumetric efficiency was 191/201.7 = 94.7% at 191 rpm and our hydro-mechanical efficiency was 44.1/55 = 80.2% at 55 bar. When running as a motor the internal losses in the unit meant that the speed was less than we expected for the flow input and the pressure differential was more than we expected for the torque output. If we run it as a pump then it’s all the other way round: the losses now mean the flow output will be less than we expect for the speed and the torque input will be more than we expect for the pressure differential.

Driving the “pump” at 191 rpm against a circuit resistance which causes the outlet pressure to be 55 bar would give us a flow of 94.7% of the theoretical flow figure: 191*225/1000*0.947 = 40.7 L/min and would require an input torque of 55*3.58/0.802 = 245.5 Nm. To get a hydraulic power output of 40.7*55/600 = 3.73 kW we had to put in a mechanical power input of 191*2*pi/60*245.5/1000 = 4.91 kW. The overall efficiency is 3.73/4.91 = 76%. It’s not surprising that this overall efficiency figure appears to the same for the pump duty and the motor duty: I used the same efficiency figures for both equations.

[Note that the big assumptions I’ve made here are that both the volumetric efficiency and the hydro-mechanical efficiency remain constant if shaft speed stays at 191 rpm and the motor input pressure and the pump outlet pressure are both at 55 bar – ignoring the fact that the motor inlet flow is slightly higher than the pump outlet flow.]

This is something like what your application works out to be: your power input is much more limited (sorry – got to work in SI units): 1.5 HP = 1.118 kW so we will have to work at significantly lower pressures and speeds than used in the example above. We have yet to learn what sort of outlet pressure your pump is going to have to work against, so imagine you put enough cans in the crusher to cause the pump outlet pressure to be 55 bar – what will be the pump’s hydro-mechanical efficiency? Let’s guess that the speed is around 23 rpm (because that’s a number on the table in the datasheet), the motor torque output at 23 rpm and 55 bar is 169 Nm but it should have been 55*3.58 = 196.9 Nm so your hydro-mechanical efficiency is 85.8%. If you want to run this as a pump you will have to put in 55*3.58/0.858 = 229.5 Nm.

Your available input power is 1.118 kW (1118 Watts) and your electric motor shaft speed is 1725 rpm (180.6 rad/s) so your available electric motor output torque is 1118/180.6 = 6.19 Nm. Let’s say you lose 3% of this in your gearing/pulley/chain arrangement so you only have 6.00 Nm left to driv ethe pump and your gear ratio will have to be 229.5/6.00 = 38.25:1. This will give you a resultant pump shaft speed of 45.1 rpm.

When the motor is running at 55 bar the tabulated figure we have for its speed is 23 rpm for an input flow of 7.6 L/min. But the theoretical input speed is 7.6*1000/225 = 33.8 rpm so the volumetric efficiency is 23/33.8 = 68%. If we assume this volumetric efficiency figure doesn’t change with speed (that’s not really true but it will actually improve at higher speeds) then we can say that a pump shaft speed of 45.1 rpm will give an actual outlet flow of 225/1000*45.1*0.68 = 6.9 L/min.

Here we are then; if your can crusher presented a resistance of 55 bar then you would get a flow of 6.9 L/min from your “pump” which is running at 45.1 rpm. Your mechanical input power would be 1.118 kW, your hydraulic output power would be 6.9*55/600 = 0.632 kW and your overall efficiency would be 56%. Your 2” hydraulic cylinder would extend at 56.7 mm/sec and the force would be 11.1 kN (assuming no losses in the cylinder, pipework, back pressures etc.). Then converting that back to US units: ~2¼ in/sec cylinder extension speed and ~2490 lbf compression, ~800 psi system pressure, ~1.84 (US)GPM pump flow and motor input power equals 1.5 HP.

Of course if your load were more you would overload your electric motor, if you wanted to run at a higher pressure you would need a greater reduction ratio and you would get less flow. As the pressure goes up the volumetric efficiency (at low speeds) gets even worse and eventually will reach zero (the pump shaft turns and there is no outlet flow – it all goes into leakage before you get up to the required pressure).

Actually now I’ve done some sums it doesn’t look too bad – except that’s one rather large input torque you need to put on your motor shaft and the effort involved in gearing down by nearly 40:1 isn’t a small task either. And your pump still needs a 1 bar boost pressure on its inlet (that’s a header tank 40 ft above the pump inlet and a fat enough, 40 ft long suction line so that your ~ 2 gpm doesn’t cause the pump inlet pressure to drop below 1 bar). Are you sure you don’t want to buy a simple gear pump?

DOL

 

[Raymond56]

Hi hydtools (Ted)
Thanks for the feedback. You’re very knowledgeable on this subject. I think your rite. I am going to sell the Motors and buy a Two stage Hydraulic pump. 11 GPM and a 5 HP Electric Motor.
But just to let you know I was planning to mount the Hydraulic oil reservoir about a 1 foot above the Motor to have good head pressure. I was hoping to get at least 5 GPM from the motor. The Motors MAX RPM is 253 for that Char Lynn Motor. My cylinder is rated for 3000 psi so I hoped for around 2500 or best I could that that 1.5 HP Electric Motor. Just ain’t gone to happen….. 
Yes maybe just easier to get the rite equipment to get it all done
As for the Eaton Char Lynn people…. Yes would have been nice If they could of helped me. But being optimistic as I am…. I’m thinking they were looking after my safety. Thinking I would have the whole thing Blow up on me or something.
But thank all of you for all the feedback. I really appreciate it.
When I’m done with the Project. I will post some pic’s
Thanks Again To All Of You……..
Thanks hydtools ( Ted )

 

[Raymond56]

Wow! You guys are smart! I mean that………
Thanks for all the info. I had to read the post a few times to sink in and it all makes a fogy kind of sense…
But I do understand. Yes oldhydroman the Torque I was first trying to achieve was high but Ted pointed to me it was unnecessary because of the no load at startup.
Both of you guys made a lot of sense and I’m all the wiser for it.
Thank you so much again….. oldhydroman …..and hydtools…….
As I said I will post pics later when the project is done and working………  Thanks!

 

[Raymond56]

Just to let you all know .... Scraping the 2" Cylinder and geting a 4" X 24" Tierod Cylinder.....
Taks care!