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10-30-2017 10:49 AM

Hi,

When simulating an oscillating hydraulic system consisting of a rotary actuator (motor), an inertia and 2 volumes, the vibrations I get are slightly damped whereas I would like to simulate a no-loss system: no friction, no pressure loss, no leakage, ideal motor,...

When doing a similar case but with a linear actuator (cylinder) and a mass, the simulation results show no damping.

Where does the damping come from in the rotary case? Why is there a difference in terms of damping between rotary and linear cases?

Many thanks for any advise

Solved! Go to Solution.

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6 REPLIES

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10-30-2017 12:09 PM

Hi Huitante,

The damping you see is purely numerical and it is due to the hyperbolic tangent used to obtain the *fact *coefficient. This coefficient is used to compute the reference pressure needed to calculate the density ratio, which is then applied to compute the output flow rate. You can refer to the MO001 submodel documentation to see the equations I am referring to.

To get rid of this effect, you just have to increase the parameter "typical speed of motor" (wtyp) to a very large value (1E+9 for example).

I hope it helps.

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10-30-2017 12:53 PM

Thanks a lot @FedericoC, increasing the *typical speed of motor* to 1E+09rev/min does solve my problem!

However there is a last bit I'd like to understand, related to my second question. Why is the hyperbolic tangent needed for any rotary motion and not for linear motions (see for instance HJ021)?

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10-31-2017 04:31 AM

Hi Huitante,

in rotary motion, the hyperbolic tangent is a numerical means used to switch between pressures (inlet and outlet) avoiding discontinuities.

In order to compute the density ratio used to convert the nominal flow at current pressure to the flow at reference pressure, an average pressure between the motor inlet and outlet is calculated. However, as the machine is rotating, the definition of inlet and outlet depends on the sign of the rotational speed. The hyperbolic tangent allows to switch from one to the other.

Regarding the linear motion, there is no need to compute an average pressure of the piston's chambers to apply the density ratio mentioned above. The ratio is simply between the density evaluated at the chamber's pressure divided by the density at the reference pressure. For this reason, there is no hyperbolic tangent in linear motion.

Regards,

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10-31-2017 07:28 AM

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10-31-2017 09:47 AM

Now, for the linear motion, this ratio is straightforward. You just take the density evaluated in the piston chamber and the density at the reference pressure. Then, you apply this ratio to the flow trough the piston. For pumps and motors, Amesim consider the average value between inlet and outlet pressure to compute the numerator of this ratio. Then, the ratio is applied to the pump nominal flow and the corrected flow is applied to inlet and outlet (with inverted signs). Therefore, if you compare a linear system with a rotational, the ratio applied is not the same and you may see some differences. This is accentuated especially if there is a huge pressure difference between inlet and outlet (I believe that in your case this difference is 2000bar, right?).

If you apply the same pressure in inlet and outlet to the two systems, you will get identical flow rates.

May I ask you what is the purpose of this comparison you are doing?

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10-31-2017 12:47 PM

Why I came to this is because Amesim introduces a numerical damping in a very basic oscillating system where it should be no loss (see my first post). This leads to misinterpretation of results especially when you care about energy efficiency of a system. While trying to understand this incorrect behavior, I came across the fact that this behavior is indeed correct for linear motion. I may have missed something, but I would have also expected pump and motor submodels to correct the flow at each port by considering the actual pressure of each port, exactly like it is done for cylinders. One would end up with a motor having different volumetric flow rates at inlets and outlets (again exactly like for cylinders, see my last post) but I would consider it to be more accurate than the way it is computed today.

Now that I am aware of this "inaccuracy", I am OK. I will increase the typical speed up to 1E+09rpm any time I need energy to be conserved!

Thanks for this interesting exchange of views!

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