Rotordynamic coefficients of a plain seal

[Michael1968]

Hello,

I'm working in the moment on the rotordynamic coefficients of plain seals in pumps and I'm a little bit confused about the positive or negative sign of the crosscoupled stiffness and damping coefficient. In detail:

I have a coordinate system: x is to the right and y is to the top (so z is estimated with the right hand..). z is the rotation axis. If you look on the x-y system, the rotation direction is from +x to +y-axis (positive around z, counter clockwise in x-y). I have now a stiffness matrix
cxx=10 cxy=20
cyx=-20 cyy=10

First the rotor is in the center, that means x=y=0 -->no force on the rotor.
Then the rotor moves 1mm to the right (new position: x=1mm, y=0mm).

The forces are
Fx=10 * 1mm + 20 * 0mm =10 N
Fy=-20* 1mm + 10 * 0mm =-20 N
Are this the forces acting from the stator on the rotor or from the rotor on the stator ?

Sorry for the confusing question
Michael

 

[electricpete]

This will probably not be much help, but here's the simplified way I would look at it.

Assuming you had normal positive damping... the direction of the damping force is opposite the velocity of the machine.

If the machine were in a circular orbit, this force is tangentential in direction opposite rotation. If yout multiply by eccentricity, you have torque. If you then multiply by 2*pi*speed, you have friction losses associated with this positive damping.

An important point, that damping force is changing direction over time. To make sense of which direction the force should be, you need to know where the rotor is.

If the rotor is at (x=1,y=0), rotating CCW, then the positive damping force would be straight down (have no x component and a negative y component).

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