Potential flow around a circular cylinder

Potential flow with zero circulation

In mathematics, potential flow around a circular cylinder is a classical solution for the flow of an inviscid, incompressible fluid around a cylinder that is transverse to the flow. Far from the cylinder, the flow is unidirectional and uniform. The flow has no vorticity and thus the velocity field is irrotational and can be modeled as a potential flow. Unlike a real fluid, this solution indicates a net zero drag on the body, a result known as d'Alembert's paradox.

Mathematical solution

Colors: pressure field. Template:Color is high and Template:Color is low. Velocity vectors.

Close-up view of one quadrant of the flow. Colors: pressure field. Template:Color is high and Template:Color is low. Velocity vectors.

Pressure field (colors), stream function (black) with contour interval of Template:Math from bottom to top, velocity potential (white) with contour interval Template:Math from left to right.

A cylinder (or disk) of radius Template:Mvar is placed in a two-dimensional, incompressible, inviscid flow. The goal is to find the steady velocity vector Template:Math and pressure Template:Mvar in a plane, subject to the condition that far from the cylinder the velocity vector (relative to unit vectors Template:Math and Template:Math) is:

$𝐕=U𝐢+0𝐣,$

where Template:Mvar is a constant, and at the boundary of the cylinder

$𝐕\cdot \widehat{𝐧}=0,$

where Template:Math is the 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 Template:Mvar. The flow therefore remains without vorticity, or is said to be irrotational, with Template:Math everywhere. Being irrotational, there must exist a velocity potential Template:Mvar:

$𝐕=\mathrm{\nabla }\varphi .$

Being incompressible, Template:Math, so Template:Mvar must satisfy Laplace's equation:

${\mathrm{\nabla }}^2\varphi =0.$

The solution for Template:Mvar is obtained most easily in polar coordinates Template:Mvar and Template:Mvar, related to conventional Cartesian coordinates by Template:Math and Template:Math. In polar coordinates, Laplace's equation is (see Del in cylindrical and spherical coordinates):

$\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial \varphi }{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^2\varphi }{\partial {\theta }^2}=0.$

The solution that satisfies the boundary conditions is

$\varphi (r,\theta )=Ur(1+\frac{R^2}{r^2})\cos \theta .$

The velocity components in polar coordinates are obtained from the components of Template:Math in polar coordinates:

$V_r=\frac{\partial \varphi }{\partial r}=U(1-\frac{R^2}{r^2})\cos \theta$

and

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

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

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

where the constants Template:Mvar and Template:Math appear so that Template:Math far from the cylinder, where Template:Math. Using Template:Math,

$p=\frac{1}{2}\rho U^2(2\frac{R^2}{r^2}\cos (2\theta )-\frac{R^4}{r^4})+p_{\mathrm{\infty }}.$

In the figures, the colorized field referred to as "pressure" is a plot of

$2\frac{p-p_{\mathrm{\infty }}}{\rho U^2}=2\frac{R^2}{r^2}\cos (2\theta )-\frac{R^4}{r^4}.$

On the surface of the cylinder, or Template:Math, pressure varies from a maximum of 1 (shown in the diagram in Template:Color) at the stagnation points at Template:Math and Template:Math to a minimum of −3 (shown in Template:Color) on the sides of the cylinder, at Template:Math and Template:Math. Likewise, Template:Mvar varies from Template:Math at the stagnation points to Template:Math on the sides, in the low pressure.

Stream function

The flow being incompressible, a stream function can be found such that

$𝐕=\mathrm{\nabla }\psi \times 𝐤.$

It follows from this definition, using vector identities,

$𝐕\cdot \mathrm{\nabla }\psi =0.$

Therefore, a contour of a constant value of Template:Mvar will also be a streamline, a line tangent to Template:Math. For the flow past a cylinder, we find:

$\psi =U(r-\frac{R^2}{r})\sin \theta .$

Physical interpretation

Laplace's equation is linear, and is one of the most elementary partial differential equations. This simple equation yields the entire solution for both Template:Math and Template:Mvar because of the constraint of irrotationality and incompressibility. Having obtained the solution for Template:Math and Template:Mvar, the consistency of the pressure gradient with the accelerations can be noted.

The dynamic pressure at both the upstream and the downstream stagnation point has a value of Template:Math. This value is needed to decelerate the free stream flow of speed Template:Mvar to zero speed at both these points. This symmetry arises only because the flow is completely frictionless.

The low pressure on the lateral sides of the cylinder is needed to provide the centripetal acceleration of the flow:

$\frac{\partial p}{\partial r}=\frac{\rho V^2}{L},$

where Template:Mvar is the radius of curvature of the flow. But Template:Math, and Template:Math. The integral of the equation for centripetal acceleration over a distance Template:Math will thus yield

$p-p_{\mathrm{\infty }}\approx -\rho U^2.$

The exact solution has, for the lowest pressure,

$p-p_{\mathrm{\infty }}=-\frac{3}{2}\rho U^2.$

The low pressure, which must be present to provide the centripetal acceleration, will also increase the flow speed as the fluid travels from higher to lower values of pressure. Thus we find the maximum speed in the flow, Template:Math, in the low pressure on the sides of the cylinder.

A value of Template:Math is consistent with conservation of the volume of fluid. With the cylinder blocking some of the flow, Template:Mvar must be greater than Template:Mvar somewhere in the plane through the center of the cylinder and transverse to the flow.

Comparison with flow of a real fluid past a cylinder

The symmetry of this ideal solution has a stagnation point on the rear side of the cylinder, as well as on the front side. The pressure distribution over the front and rear sides are identical, leading to the peculiar property of having zero drag on the cylinder, a property known as d'Alembert's paradox. Unlike an ideal inviscid fluid, a viscous flow past a cylinder, no matter how small the viscosity, will acquire a thin boundary layer adjacent to the surface of the cylinder. Boundary layer separation will occur, and a trailing wake will exist in the flow behind the cylinder. The pressure at each point on the wake side of the cylinder will be lower than on the upstream side, resulting in a drag force in the downstream direction.

Janzen–Rayleigh expansion

Template:Main The problem of potential compressible flow over circular cylinder was first studied by O. Janzen in 1913 and by Lord Rayleigh in 1916 with small compressibility effects. Here, the small parameter is the square of the Mach number ${\mathrm{M}}^2=U^2/c^2\ll 1$, where Template:Mvar is the speed of sound. Then the solution to first-order approximation in terms of the velocity potential is

$\varphi (r,\theta )=Ur(1-\frac{a^2}{r^2})\cos \theta -{\mathrm{M}}^2\frac{Ur}{12}[(\frac{13a^2}{r^2}-\frac{6a^4}{r^4}+\frac{a^6}{r^6})\cos \theta +(\frac{a^4}{r^4}-\frac{3a^2}{r^2})\cos 3\theta ]+\mathrm{O}\left({\mathrm{M}}^4\right)$

where $a$ is the radius of the cylinder.

Potential flow over a circular cylinder with slight variations

Regular perturbation analysis for a flow around a cylinder with slight perturbation in the configurations can be found in Milton Van Dyke (1975). In the following, Template:Mvar will represent a small positive parameter and Template:Mvar is the radius of the cylinder. For more detailed analyses and discussions, readers are referred to Milton Van Dyke's 1975 book Perturbation Methods in Fluid Mechanics.

Slightly distorted cylinder

Here the radius of the cylinder is not Template:Math, but a slightly distorted form Template:Math. Then the solution to first-order approximation is

$\psi (r,\theta )=Ur(1-\frac{a^2}{r^2})\sin \theta +\epsilon \frac{Ur}{2}(\frac{3a^2}{r^2}\sin \theta -\frac{a^4}{r^4}\sin 3\theta )+\mathrm{O}\left({\epsilon }^2\right)$

Slightly pulsating circle

Here the radius of the cylinder varies with time slightly so Template:Math. Then the solution to first-order approximation is

$\psi (r,\theta ,t)=Ur(1-\frac{a^2}{r^2})\sin \theta +\epsilon Ur(\frac{a^2}{Ur}\theta f^{\prime }(t)-\frac{2a^2}{r^2}f(t)\sin \theta )+\mathrm{O}\left({\epsilon }^2\right)$

Flow with slight vorticity

In general, the free-stream velocity Template:Mvar is uniform, in other words Template:Math, but here a small vorticity is imposed in the outer flow.

Linear shear

Here a linear shear in the velocity is introduced.

$\begin{array}{cc}\psi & =U(y+\frac{1}{2}\epsilon \frac{y^2}{a}),\\ \omega & =-{\mathrm{\nabla }}^2\psi =-\epsilon \frac{U}{a}\text{as }x\to -\mathrm{\infty },\end{array}$

where Template:Mvar is the small parameter. The governing equation is

${\mathrm{\nabla }}^2\psi =-\omega (\psi ).$

Then the solution to first-order approximation is

$\psi (r,\theta )=Ur(1-\frac{a^2}{r^2})\sin \theta +\epsilon \frac{Ur}{4}(\frac{r}{a}(1-\cos 2\theta )+\frac{a^3}{r^3}\cos 2\theta -\frac{a}{r})+\mathrm{O}\left({\epsilon }^2\right).$

Parabolic shear

Here a parabolic shear in the outer velocity is introduced.

$\begin{array}{cc}\psi & =U(y+\frac{1}{6}\epsilon \frac{y^3}{a^2}),\\ \omega & =-{\mathrm{\nabla }}^2\psi =-\epsilon U\frac{y}{a^2}\text{as }x\to -\mathrm{\infty }.\end{array}$

Then the solution to the first-order approximation is

$\psi (r,\theta )=Ur(1-\frac{a^2}{r^2})\sin \theta +\epsilon \frac{Ur}{6}(\frac{r^2}{a^2}{\sin }^2\theta -3r\ln r\sin \theta +\chi )+\mathrm{O}\left({\epsilon }^2\right),$

where Template:Mvar is the homogeneous solution to the Laplace equation which restores the boundary conditions.

Slightly porous cylinder

Let Template:Math represent the surface pressure coefficient for an impermeable cylinder:

$C_{\mathrm{p}\mathrm{s}}=\frac{p_s-p_{\mathrm{\infty }}}{\frac{1}{2}\rho U^2}=1-4{\sin }^2\theta =2\cos 2\theta -1,$

where Template:Math is the surface pressure of the impermeable cylinder. Now let Template:Math be the internal pressure coefficient inside the cylinder, then a slight normal velocity due to the slight porousness is given by

$\frac{1}{r}\frac{\partial \psi }{\partial \theta }=\epsilon U(C_{\mathrm{p}\mathrm{i}}-C_{\mathrm{p}\mathrm{s}})=\epsilon U(C_{\mathrm{p}\mathrm{i}}+1-2\cos 2\theta )\text{at }r=a,$

but the zero net flux condition

$\underset{0}{\overset{2\pi }{\int }}\frac{1}{r}\frac{\partial \psi }{\partial \theta }\mathrm{d}\theta =0$

requires that Template:Math. Therefore,

$\frac{\partial \psi }{\partial \theta }=-2\epsilon rU\cos 2\theta \text{at }r=a.$

Then the solution to the first-order approximation is

$\psi (r,\theta )=Ur(1-\frac{a^2}{r^2})\sin \theta -\epsilon U\frac{a^3}{r^2}\sin 2\theta +\mathrm{O}\left({\epsilon }^2\right).$

Corrugated quasi-cylinder

If the cylinder has variable radius in the axial direction, the Template:Mvar-axis, Template:Math, then the solution to the first-order approximation in terms of the three-dimensional velocity potential is

$\varphi (r,\theta ,z)=Ur(1+\frac{a^2}{r^2})\cos \theta -2\epsilon Ub\frac{K_1\left(\frac{r}{b}\right)}{{K_1}^{\prime }\left(\frac{r}{b}\right)}\cos \theta \sin \frac{z}{b}+\mathrm{O}\left({\epsilon }^2\right),$

where Template:Math is the modified Bessel function of the first kind of order one.

See also

• Joukowsky transform
• Kutta condition
• Magnus effect