Hydraulic motor efficiency conundrum

[Frangelo]

Hello,

I'm trying to compare efficiency of different hydraulic motors by looking at unit of output per unit of input, viz:

$\frac{rpm × torque}{pressure × flow}$

I'm a bit flummoxed by one set of motor specs, for Eaton Duraforce motors, which is attached below.

these motors seem to have variable displacement, but operate at a constant pressure.

From the spec sheet, it looks like these motors operate at constant torque with a given pressure, irrespective of displacement (or rpm). that is, the spec sheet does not suggest that torque varies with respect rpm as I've seen on other spec sheets.

this leads me to the dubious proposition that the motor produces more power (rpm x torque) with constant pressure and lower flow.

For example, on the Duraflow 55, it says minimum displacement is 18.3 cm3 / rev.
and at minimum displacement rpm is 4700. So flow in lpm is (18.3 x 4700)/1000 = 86 lpm.

the chart also indicates that at "continuous pressure" of 250 bar (3625 psi) torque is constant at 218nm (161 ft pounds). It doesn't suggest that it is related to any particular displacement.

At maximum displacement, rpm decreases to 4100 rpm, and flow increase to (54.8 x 4700)/1000 = 258 lpm.

But if torque remains constant and RPM decreases, then power output decreases as well.

In other words decreased flow x constant pressure results in higher power output. So I'm clearly doing something wrong.

Perhaps I am misunderstanding what the top line of chart means?

Does "rated size displacemenet" measured in "cm3/rev" mean the amount of fluid required for each revolution of the motor? Or perhaps it means something else?

Any advice much appreciated.

 

[Baluncore]

Does "rated size displacemenet" measured in "cm3/rev" mean the amount of fluid required for each revolution of the motor?

A specification of "cm3/rev" is probably a maximum possible value at one end of the control range. That specification is important because it dictates the maximum flow and therefore the size of the hoses needed.

Variable displacement motors continuously vary their displacement to control an output parameter. The system designer can usually select the type of control required. The control parameter is used to set the angle of the swash plate and therefore the displacement. The displacement at any time must be adjusted to that required to satisfy conservation of energy.

Efficiency will vary over different parameters. A variable displacement pump permits more efficient control of machine operation than is possible with fixed displacement motors. That is because pressure or volume does not have to be wasted to achieve regulation.

 

[Frangelo]

In this case that's not the issue. Note on the attached spec sheet they supply min and max displacement but they only supply a single value for torque at continuous pressure. If torque is the same with the smaller displacement has or the larger then something isn't adding up -- because torque is quoted at constant pressure,

 

[Baluncore]

but they only supply a single value for torque at continuous pressure.

I believe that value is a maximum torque at maximum displacement.

I think you must interpret all data sheets with caution and a fundamental assumption that the efficiency of hydraulic motors and pumps will be about 80%. If it appears that energy is not being conserved then there is either an omission/error in the data sheet, you are misinterpreting the technical terms used, or there are assumed conditions that have not been specified.

Eaton should have a definition of the terms they use, but I can't find it. You might contact the application engineers at your regional Eaton distributor to clarify the situation. Distributor email addresses are found here;
http://www.eaton.com/Eaton/ProductsServices/Hydraulics/ContactUs/

 

[alphy]

Hi there,

Thank you for sharing your conundrum with us. It's always exciting to see scientists and engineers working to improve efficiency and optimize technology.

From what I understand, you are trying to compare the efficiency of different hydraulic motors by looking at the unit of output per unit of input. However, you have come across some specifications for the Eaton Duraforce motors that are confusing you.

Firstly, it is important to note that the efficiency of a hydraulic motor is not solely determined by its displacement, pressure, and flow. Other factors such as internal friction, fluid viscosity, and design also play a role.

That being said, let's break down the specifications you have shared. The rated size displacement of 18.3 cm3/rev means that for every revolution of the motor, it will displace 18.3 cm3 of fluid. This is a measure of the motor's displacement capacity, not its efficiency.

The chart shows that at a constant pressure of 250 bar, the torque remains constant at 218nm. This means that no matter the displacement or RPM, the motor will produce the same torque. However, as you correctly noted, at maximum displacement, the RPM decreases and the flow increases. This does result in a decrease in power output. However, the motor is still operating at a constant pressure and producing the same torque. This does not necessarily mean that it is more efficient, as other factors may affect its overall efficiency.

It's also worth noting that the spec sheet mentions "continuous pressure" and not "rated pressure." This could mean that the motor is designed to operate at a constant pressure for extended periods of time without overheating or malfunctioning. This could also explain the constant torque at different displacements.

In conclusion, while your calculations may suggest that the motor is more efficient at lower flow and constant pressure, there are other factors at play that may affect its overall efficiency. It's always best to consult with the manufacturer or conduct further testing to determine the true efficiency of the motor.

I hope this helps clarify your confusion. Keep up the great work in your research and experimentation!