Simscape Fluids Single-Acting Actuator piston stroke

Hello,

I am developing a fluid system in Simulink with the simscape isothermal liquid library and I am currently struggeling with this Subsystem shown below. I selected parameters are just an example.

I have a single-acting actuator block with 0.458 bar on the pressure side, and an external force on the rod side. The piston stroke is 100mm and the initial piston displacement is 0 mm.

I do not understand why in this example and in steady state I do not have a constand piston displacement of 0 mm. Instead, If I increase the external force in this simulink model, the piston displacement increases (see second picture).

But theoretically and in the steady state, the piston displacement should stay at 0 mm because the external force on the cylinder is higher than the pressure force on the other side. Why is the piston displacement on its max. 100 mm and therefore completely extended, although the piston force on the left side is greater? Is there a problem with the model itself?

[Cris LaPierre]

We don't have enough information to debug. Consider attaching your model.

In the meantime, try simplifying your model until you get the behavior you expect, then add components back in to identify which is causing the unexpected behavior.

Inspect the simulation using Simscape Results Explorer. I suspect what is happenign is your display is showing the final value, but the simulation will show the position is oscillating still rather that at steady state.

[Jonas]

Thank you for your answer. I simplified and attached the model.

I also connected the scope block to the piston stroke signal to see if it is oscillating. But after some testing with different parameters I always get the same steady state without any oscillation.

Basically what i want to do is to push the piston back to the right. But after trying out different external forces the piston stays at it max. stroke of 100mm. I do not understand the logic behind this. Theoretically, the actuator is in balance if the pressure force on the right is the same as the external force on the left. If one of the forces increases, the piston moves to the other side. When I increase the external force, I expect that the piston stays at initial position of 0mm.

Answer 1

[Cris LaPierre]

Sorry, I should have caught this in your screenshots.

In Simscape, you cannot assume direction based on orientation. We have both been making the same mistake - assuming the external force is acting opposite the force pushing the rod up, likely because of the arrows on the block.

However, when solving the physics, all positive forces act in the same direction, and you have entered a force of +100000 N for the external force. If you want it to act opposite the force acting on the rod, you need to give it the opposite sign: -100000 N..

[Cris LaPierre]

You might find this video from our Designing and Modeling Phyiscal Systems course on Coursera helpful.

• Guided Lesson: Modeling Multi-Domain Systems

If you are just getting started, you may find the entire course helpful.

[Jonas]

Got it. Thanks for the explanation.