Flow in and out of hydraulic motor

In summary, it is possible to create a simple model of the total forces acting on the piston rod if you know how the flow is affected.
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Assuming no losses, is the flow in and out of a hydraulic motor - connected to an electric motor with an electrical load - the same?
Hey all,
It is possible I am over complicating this, but is the flow in and out of a hydraulic motor with an electrical load the same? Assuming no losses.
Scenario is a hydraulic piston (moving by some external force), driving a bi-directional hydraulic motor which itself is driving an electric motor with an electrical load presenting some damping on the system. Would the damping from R_load affect Q2 out of the hydraulic motor since the necessary current would produce some opposing torque on the rotor shaft?
flow.png
 
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  • #2
Think about Conservation of Mass.
 
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T-osu said:
It is possible I am over complicating this, but is the flow in and out of a hydraulic motor with an electrical load the same? Assuming no losses.
You should be aware that the hydraulic cylinder is asymmetrical. The area of the rod reduces the volume of fluid flowing at the rod end port. There is an imbalance of fluid volumes, and no reservoir for your closed system.

You will need to change your hydraulic circuit so that the cylinder cyclically pumps fluid from a reservoir through the motor, to return to the reservoir for filtering and cooling. To do that, you will need some one-way check valves. Does your electrical motor need to reciprocate, or can it keep turning in the same direction.

If you need equal flow volumes from the cylinder, you might use a double-ended cylinder that has a through-rod, passing through both ends of the cylinder, or maybe you can connect two identical cylinders in parallel with one reversed, so the total of the piston areas are equal.

What are you trying to achieve with this system?
 
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T-osu said:
Would the damping from R_load affect Q2 out of the hydraulic motor since the necessary current would produce some opposing torque on the rotor shaft?
The mechanical power into the electric generator is equal to the electric power out of the electric generator divided by the efficiency of the generator. The mechanical power out of the hydraulic motor is equal to the mechanical power into the electric generator. Torque on the hydraulic motor shaft affects the pressure difference between the motor inlet and outlet. Whether that affects the flow through the motor depends on the source of the hydraulic oil.

T-osu said:
Assuming no losses.
You can assume no losses. Unfortunately, in the real world, the best hydraulic systems are about 90% efficient at their optimal operating point. A realistic rule of thumb estimate is that a typical hydraulic system is about 50% efficient. Your system would likely have even lower efficiency. An assumption of no losses is a really bad assumption in this system.

If your external force has a fixed displacement, a crankshaft would be a far more efficient method of spinning a generator. If not, you can search wave power for some ideas for converting random reciprocating motion into electric power.
 
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jrmichler said:
You can assume no losses. Unfortunately, in the real world, the best hydraulic systems are about 90% efficient at their optimal operating point.
Energy losses in a hydraulic system are high.
Fluid losses are usually low.
Some hydraulic fluid will escape past the seals in the motor. The motor will probably have a small third port. That port returns the escaped fluid to the reservoir tank.
 
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  • #6
Hi folks,
Thanks for all the quick replies. Goal is to get the total force that is acting on the hydraulic rod. Once I know how flow is affected, I can get pressure to get force. This post is related to my other post about an air spring Air Spring post. However the overall goal: I'm trying to make a simpler model of the total forces acting on the piston rod b/c the current very complex simscape model takes too long to simulate.
I should mention that this is an already built system that has been tested and used for about a decade so there's no changing its design (adding reservoirs or check valves). As I stated in my other post, I'm an electrical engineer, so all the hydraulics/thermodynamics is new to me therefore I can't justify the developer's reasons of not including a reservoir or making it unidirectional or the type of oil they chose, but it is a successful working system.

To inform all regarding including losses, I do have tabulated data of efficiencies from the manufacturer which I am intending on using for the hydraulic motor at least, I just wanted to present the question in its simplest form.

@Baluncore it is actually a through rod system, the other side goes to the air spring; I was just trying to simplify my post and not make it too detailed, but I see that detail was an important one.

@jrmichler Could you elaborate more on this please?
jrmichler said:
Whether that affects the flow through the motor depends on the source of the hydraulic oil.

After staring at it for a day, I'm thinking that when the electric motor is providing power to a load, this would act against the system and equally slow down Q1 and Q2:
##Q_1 = A_1\,v_{rod} - Q_{change}##, ##Q_2 = A_2\,v_{rod} - Q_{change}##
##Q_{change} = displacement_{hydmotor} \, w_{elec}##, where ##w_{elec}## can be calculated from the torque produced by the hydraulic motor ## \tau_{hydmotor} = displacement_{hydmotor}\, \Delta p ## and ##w_{elec} = \frac{ \tau_{hydmotor}}{b_{elec}}## where ##b_{elec}## is damping on the electric motor (load, friction)

maybe?? Spitting out thoughts...
 
  • #7
A lot depends on whether your external force is really an external force or external displacement. If it is truly an external force, then changing the load on the hydraulic motor will change the motor speed, which changes the flow rates Q1 and Q2. If it is really an external displacement, then changing the load on the hydraulic motor will change the motor inlet and outlet pressures, but not the flow rates Q1 and Q2.

Or is the external force somewhere in between? For instance, a system generating power from ocean waves by means of a float moving up and down with the waves would be in between. With no load on the motor, the flow would follow the waves and would be a displacement input. Pressures 1 and 2 would not change, but Q1 and Q2 would. With a very high load on the hydraulic motor, the float would not move and would be a force input. Pressures 1 and 2 would change, but Q1 and Q2 would be zero. The maximum electrical power out would be at a load in between, where both Pressures 1 and 2, and Q1 and Q2 would all be changing.

In the physics world, we have pure forces or pure displacements. In the engineering world, we have systems that are in between pure forces and pure displacements. These can be challenging to understand when the output affects the input.

T-osu said:
After staring at it for a day
Wrapping your mind around this will take at least another day, but it will be well worth the time spent.
 

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