Fluid flow past a cylinder

"The flow of an incompressible fluid past a cylinder is one of the first mathematical models that a student of fluid dynamics encounters. This flow is an excellent vehicle for the study of concepts that will be encountered numerous times in mathematical physics, such as vector fields, coordinate transformations, and most important, the physical interpretation of mathematical results."

Mathematical Solution

PD Image
Colors: pressure field. Red is high and blue is low. Velocity vectors.

PD Image
Close-up view of one quadrant of the flow. Colors: pressure field. Red is high and blue is low. Velocity vectors.

Pressure field (colors), streamfunction (black), velocity potential (white).

A cylinder (or disk) of radius Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R} is placed in two-dimensional, incompressible, inviscid flow. The goal is to find the steady velocity vector ${\vec {V}}$ and pressure $p$ in a plane, subject to the condition that far from the cylinder the velocity vector is

${\vec {V}}=U{\widehat {i}}+0{\widehat {j}}$

and at the boundary of the cylinder

${\vec {V}}\cdot {\widehat {n}}=0$

where ${\widehat {n}}$ is vector normal to the cylinder surface. The upstream flow is uniform and has no vorticity. The flow is inviscid, incompressible and has constant mass density $\rho$. The flow therefore remains without vorticity, or is said to be irrotational, with $\nabla \times {\vec {V}}=0$ everywhere. Being irrotational, there must exist a velocity potential $\phi$:

${\vec {V}}=\nabla \phi$

Being incompressible, $\nabla \cdot {\vec {V}}=0$, so $\phi$ must satisify Laplace's equation:

$\nabla ^{2}\phi =0$

The solution for $\phi$ is obtained most easily in polar coordinates <matth>r</math> and $\theta$, related to conventional Cartesian coordinates by $x=r\cos \theta$ and $y=r\sin \theta$. In polar coordinates, Laplace's equation is:

${\frac {\partial ^{2}\phi }{\partial r^{2}}}+{\frac {1}{r}}{\frac {\partial \phi }{\partial r}}+{\frac {1}{r^{2}}}{\frac {\partial ^{2}\phi }{\partial \theta ^{2}}}=0$

The solution that satisfies the boundary conditions is

$\phi (r,\theta )=U\left(r+{\frac {R^{2}}{r}}\right)\cos \theta$

The velocity components in polar coordinates are obtained from the components of $\nabla \phi$ in polar coordinates:

$V_{r}={\frac {\partial \phi }{\partial r}}=U\left(1-{\frac {R^{2}}{r^{2}}}\right)\cos \theta$

and

$V_{\theta }={\frac {1}{r}}{\frac {\partial \phi }{\partial \theta }}=-U\left(1+{\frac {R^{2}}{r^{2}}}\right)\sin \theta$

Being invisicid and irrotational, Bernoulli's equation allows the solution for pressure field to be obtained directly form the velocity field:

$p={\frac {1}{2}}\rho \left(U^{2}-V^{2}\right)+p_{\infty }$

where the constants $U$ and $p_{\infty }$ appear to that $p\rightarrow p_{\infty }$ far from the cylinder, where $V=U$.