Velocity Potential of a Moving Cylinder

[dimasmr21]

Hi, i need help in deriving the velocity potential of a moving 2D cylinder (circle) in potential fluid.

The cylinder is moving in negative x direction U(t).

I can derive the velocity potential of fluid past a cylinder (cylinder is stationary, in which it is the scalar summation of uniform flow, and a doublet). But i don't know how to derive the potential for a moving cylinder.

My simple thoughts were:

a.) There is no need to use a uniform flow potential.
b.) The velocity potential shall be in term of U(t), or -dx/dt.

Please kindly help me. Thank you so much.

 

[NFuller]

I can derive the velocity potential of fluid past a cylinder (cylinder is stationary, in which it is the scalar summation of uniform flow, and a doublet). But i don't know how to derive the potential for a moving cylinder.

The flow of a fluid moving past a cylinder and the flow around a cylinder moving through a stationary fluid are identical. They just use different reference frames. If the velocity of the cylinder is $-U_{x}\hat{x}$, then in the frame of the cylinder, you can assume the cylinder to be at rest and the far-field velocity of the fluid to be $U_{x}\hat{x}$.

 

[dimasmr21]

The flow of a fluid moving past a cylinder and the flow around a cylinder moving through a stationary fluid are identical. They just use different reference frames. If the velocity of the cylinder is $-U_{x}\hat{x}$, then in the frame of the cylinder, you can assume the cylinder to be at rest and the far-field velocity of the fluid to be $U_{x}\hat{x}$.

Thank you so much for your reply. That is my objective. I mean i have to prove that the "pressure distribution" shall be the same.

During the process, i managed to understand that the velocity potential must satisfy two boundary condition (this is what i wrote on my report).

1. Fluid at the far region from the cylinder shall remain stationary (no flow, or one can simply say that the fluid velocity at the far region is zero)
2. If one “moves” along with the cylinder, he shall have the same velocity as the cylinder. One can simply interpret that the fluid velocity at every point on cylinder surface is equal to the cylinder velocity.

My problem right now is, i can't seem to derive it in a scientific manner (as i have already found the velocity potential by trial and error (lol) which satisfied the boundary condition).

Thank you, i hope you can help me! :)

 

[NFuller]

My problem right now is, i can't seem to derive it in a scientific manner (as i have already found the velocity potential by trial and error (lol) which satisfied the boundary condition).

The proof that the two solutions are similar is enforced by Galilean invariance which states that the laws of motion (including fluid motion) are identical in all reference frames. If the fluid velocity in the frame of the moving cylinder is given by $U'(\mathbf{x}',t)$ then the fluid velocity $U(\mathbf{x},t)$ in the rest frame of the fluid is
$$U(\mathbf{x},t)=U'(\mathbf{x}',t)+\mathbf{v}$$
where $\mathbf{v}$ is the velocity of the moving cylinder and
$$\mathbf{x}'=\mathbf{x}-\mathbf{x}_{0}-\mathbf{v}t$$
where $\mathbf{x}_{0}$ is the initial location of the cylinder in the stationary frame.
The potential $\phi(\mathbf{x},t)$ likewise undergoes the transformation
$$\phi(\mathbf{x},t)=\phi'(\mathbf{x}',t)$$