[Jennifer0430]

Hi,
I am doing simulation of a scroll compressor in CFX.I have built my mesh in Twinmesh. A mesh is generated at intervals of one degree. I'm using transient calculation.

When the revolution speed is 1000rpm, CFX can calculate correctly. But when I increase the revolution speed to 2000rpm, en error will always occur at about 120 step.

The warning message is like：
ERROR #004100018 has occurred in subroutine FINMES. | | Message: | | Fatal overflow in linear solver.

I have tried to use the result of the 1000rpm calculation as the initial value as the 2000rpm calculation, but the error would still occur.

I wonder what's the reason of the problem and how can I solve it.

Thanks in advance!

[Gert-Jan]

In this forum, look for "004100018" or "FINMES" and you will read what to do.

[Opaque]

You have not provided enough information, but assuming everything remained the same. Here is a hint for you.

If you set a time step to integrate the solution for a given RPM, and you now increase the RPM by a factor of 2, what do you think is happening with the physics?

In the first setup, for the given timestep the scroll will move a certain angle as a function of RPM and timestep, correct? At twice the speed, the traversed angle will now be?

If you move the scroll too fast you will be skipping the physics interaction between the stationary and rotating components.

Hope the above helps you.

[Jennifer0430]

Originally Posted by Gert-Jan

In this forum, look for "004100018" or "FINMES" and you will read what to do.

I have searched for relative information and made some attempts, but I still can't solve the problem above. I will search for more information. Thank you!

[Jennifer0430]

Originally Posted by Opaque

You have not provided enough information, but assuming everything remained the same. Here is a hint for you.

If you set a time step to integrate the solution for a given RPM, and you now increase the RPM by a factor of 2, what do you think is happening with the physics?

In the first setup, for the given timestep the scroll will move a certain angle as a function of RPM and timestep, correct? At twice the speed, the traversed angle will now be?

If you move the scroll too fast you will be skipping the physics interaction between the stationary and rotating components.

Hope the above helps you.

Hi,

Thank you for your time. I really appreciate your answer.

At twice the speed, the traversed angle will still be 1 degree. But the Time Steps, which has been set as a function of RPM, will be a half as before.

I'm not quite clear about "If you move the scroll too fast you will be skipping the physics interaction between the stationary and rotating components."What can I do to avoid this?

Thanks in advance!

[Gert-Jan]

Originally Posted by Jennifer0430

I have searched for relative information and made some attempts, but I still can't solve the problem above. I will search for more information. Thank you!

What has been said on the forum many time is that an error message like this does not help us. E.g. FINMES means Final message. So, we need the context, meaning the last part of your output file. In that way it is easier to understand the problem and to help you.

But
- remember, you are hitting the capabilities of CFX using Twin mesh with probably a double scew compressor. I never wanted to burn my fingers on it.
- You mentioned you created a grid for 1 degree rotation at 1000rpm. Are you sure you can use this for 2000 rpm? Does that match?

[Opaque]

Originally Posted by Jennifer0430

Hi,

Thank you for your time. I really appreciate your answer.

At twice the speed, the traversed angle will still be 1 degree. But the Time Steps, which has been set as a function of RPM, will be a half as before.

I'm not quite clear about "If you move the scroll too fast you will be skipping the physics interaction between the stationary and rotating components."What can I do to avoid this?

Thanks in advance!

I would admit I did not set my context appropriately; however, if I am an observer on the stationary side, and I look at a tooth/blade of the rotating part passing, there is no way that at twice the speed the traversed angle is the same in the same period of time; otherwise, it is not moving at twice the angular speed.

You did not mention your timestep is directly computed from the RPM for the specific case, and that is a great setting already.

Since you adjusted your timestep as a function of the RPM, how many timesteps are within a passing between two teeth/blades of the rotating component, i.e how many timesteps/pitch passing?

[Jennifer0430]

Originally Posted by Gert-Jan

What has been said on the forum many time is that an error message like this does not help us. E.g. FINMES means Final message. So, we need the context, meaning the last part of your output file. In that way it is easier to understand the problem and to help you.

But
- remember, you are hitting the capabilities of CFX using Twin mesh with probably a double scew compressor. I never wanted to burn my fingers on it.
- You mentioned you created a grid for 1 degree rotation at 1000rpm. Are you sure you can use this for 2000 rpm? Does that match?

Hi,

Thank you very much for your time.

I set the timestep as a function of RPM, so the grid and setting should match with the 2000rpm case.

There were only two pieces of message in the last part of Out File:
ERROR #004100018 has occurred in subroutine FINMES. Message: Fatal overflow in linear solver.
An error has occurred in cfx5solve: The ANSYS CFX solver exited with return code 1. No results file has been created.

There was no warning or other message.

Besides, the maximal temperature and maximal pressure suddenly went up at about the 90th step, and they kept on going up to a very big value until the error occurred. But this wouldn't happen when the speed was 1000rpm. There were at most some small fluctuations at 1000rpm.

[Jennifer0430]

Originally Posted by Opaque

I would admit I did not set my context appropriately; however, if I am an observer on the stationary side, and I look at a tooth/blade of the rotating part passing, there is no way that at twice the speed the traversed angle is the same in the same period of time; otherwise, it is not moving at twice the angular speed.

You did not mention your timestep is directly computed from the RPM for the specific case, and that is a great setting already.

Since you adjusted your timestep as a function of the RPM, how many timesteps are within a passing between two teeth/blades of the rotating component, i.e how many timesteps/pitch passing?

Hi,

Thank you very much.

In Time Duration, I set Number of Timesteps per Run as 360, which corresponds to 360 grids in one turn. In Time Steps, I set Timeteps as a expression, whose value is equal to the time it takes the rotor to turn 1 degree. For example, the Timesteps will be 1.67e-04s at 1000rpm， and it will be 8.33e-05s at 2000rpm. The initial time is 0.0s. In solver control, I set the Max. Coeff,Loops as 10.

[Opaque]

Let me illustrate the idea with a simple problem. Say we have a differential equation like
Example 1:

dy_dt = cos(2*pi*w*t)

and you want to integrate it between t = 0, and t = 1. How many timesteps will you need to obtain a reasonable solution?

Example 2:
Now the interpretation for a bladed device. Let's say the device has 12 blades, and the observer in the stationary frame tries to integrate the signal of the 12 passing blades during one revolution.

How many peaks of the signal are in 1 rev?
How many timesteps are between each period of the signal?
Are those timesteps similar to the ones for example 1?

My point is that saying the resolution of the timestep is 1[deg] is meaningless w/o taking into account the number of blades, speed of the machine, and flow conditions. The timestep serves at least two purposes: accuracy, and stability. Below a certain value, the solution will not get any more accurate (timestep independent: converged in the L2-norm), and above a certain value the solution CANNOT be computed at all.

[Jennifer0430]

Originally Posted by Opaque

Let me illustrate the idea with a simple problem. Say we have a differential equation like
Example 1:

dy_dt = cos(2*pi*w*t)

and you want to integrate it between t = 0, and t = 1. How many timesteps will you need to obtain a reasonable solution?

Example 2:
Now the interpretation for a bladed device. Let's say the device has 12 blades, and the observer in the stationary frame tries to integrate the signal of the 12 passing blades during one revolution.

How many peaks of the signal are in 1 rev?
How many timesteps are between each period of the signal?
Are those timesteps similar to the ones for example 1?

My point is that saying the resolution of the timestep is 1[deg] is meaningless w/o taking into account the number of blades, speed of the machine, and flow conditions. The timestep serves at least two purposes: accuracy, and stability. Below a certain value, the solution will not get any more accurate (timestep independent: converged in the L2-norm), and above a certain value the solution CANNOT be computed at all.

Thank you for taking the time to explain to me. I understand what you mean. I'll give it a try. Thanks again.

[Jennifer0430]

Hi，
I have solved the above problem and I want to share what I have done here in case that anyone meet the same problem in the future.
First, improve the models, grids and CFX settings. I adjusted my meshing generating method and refined my mesh.
Second, shorten the time step. In simulation of compressor, the time step reflects the rotation speed, so we can't use adaptive time step. With the speed increasing, Courant Number increase. We need to shorten the time step to decrease Courant Number and ensure the stability of the culculation.

Jennifer