Suppose that the fluid is flowing into the U-shaped tube at velocity V and the tube is vibrating at angular velocity . Consider a small section of the fluid that is on the inlet side away from the point of flexture at distance r.
The Coriolis force on the small fluid section m is
During the down cycle, the tube applys an upward resisting force to the fluid or the fluid pushes the tube down. On the outlet side, the Coriolis force has the opposite direction.
To simply the problem, we assume that the tube has a perfect U shape with a cross section area of A. The length and width are l, d, respectively. The opposite directions of Coriolis forces on inlet and outlet sides result in a twisting moment Tc
A K factor can be introduced to compensate for the more generalized U-shape.
where Qm = AV is the mass flow rate.
The governing equation of twisting is
where Iu is the inertia of the U-shaped tube, Cu is the damping coefficient, Ku is the stiffness, is the twist angle, and t is time.
Recall that the Coriolis flowmeters are vibrating the U-shaped tube to generate the rotation, the real angular velocity is function of vibrating frequency :
Assuming that the damping term Cu is negligible, the equation of twisting becomes
The particular solution (steady-state solution) of the twist angle is
Furthermore, the velocity of the turning corners of the U-shaped tube are and the displacement difference between these two corners is d/2. Therefore, the time lag between these two corners is
By measuring the time lag , the mass flow rate can be obtained
In vibration analysis, it is custom to use the natural frequency as a basis and normalize frequency terms against it. The natural frequency of the U-shaped tube system is (note that Iu includes the mass of the fluid in the tube)
The mass flow rate then becomes