Velocity and Pressure Distribution for Flow Over a Cylinder

For starters, let's just look at the velocity and pressure fields relating to flow past a circular cylinder to gain a qualitative understanding.  Rest assured we will delve into the matter sufficiently to determine which aspects of the solution are valid and which are not.

The above figure shows the velocity of flow past the cylinder; vectors are shown and color represents the magnitude of the velocity  (warmer colors, faster velocity).  Green in this case corresponds to the velocity far away from the cylinder, the free stream velocity which you can think of as the wind, all moving at the same velocity from left to right.  As they get close enough , fluid elements approaching the cylinder straight on (at the equator) begin to slow (shifting to blue color).  In fact, the fluid element on the surface of the upstream side of the cylinder is stopped -- velocity = 0.  This is called a stagnation point.  Fluid elements passing either above or below the cylinder increase their velocity magnitude (shifting to red color).  It will turn out that velocity at both the top and bottom of the cylinder (the poles) is twice the free stream velocity.  Velocity along the surface of the cylinder is in a tangential direction, i.e. parallel to the surface of the cylinder.

Fluid elements continuing around the downstream side of the cylinder begin to slow down, reaching a second stagnation point at the downstream equator.  Note that this part of the solution is quite unrealistic and we'll soon consider the reason for this.  Elements continuing downstream begin to increase and velocity, gradually returning to the free stream value sufficiently far downstream.

The velocity magnitude exhibits 2 lines (planes) of symmetry.  A line drawn vertically through the cylinder divides the velocity magnitude into right and left sides that are geometric mirror images.  Note that the velocity itself is not symmetric about this line, i.e. velocity is directed to the right on both sides of the image.  Velocity, and velocity magnitude, are symmetric about a horizontal line through the center of the cylinder.

The next image shows the magnitude of the velocity in color and streamlines suggesting the paths of fluid elements as they flow past the cylinder.  Streamlines do indeed coincide with element paths when the flow is steady (no variation with time as in pulsatile blood flow).  The horizontal streamlines at the cylinder equator clearly suggest the stagnation points at the upstream and downstream sides of the cylinder.

You know from the Bernoulli equation that the value of  $$p + \rho |u|^2/2$$ remains a constant on a streamline (assuming negligible friction and steady flow, both of which are exactly true for the simplified model).  Note that all the streamlines extend out into the free stream, far away from the cylinder where velocity is all the same and so is pressure (an arbitrary value of 0 pressure is assumed far away from the cylinder).  Consequently the value of $$p + \rho |u|^2/2$$   is the same for all the streamlines and is the same throughout the entire solution.  Since pressure is 0 in the free stream and the magnitude of the velocity is a specific value, call it $$U$$   , the value of  $$p + \rho |u|^2/2\equiv \rho U^2/2$$ everywhere in the solution.

In the next image, the pressure distribution around the cylinder is shown using color and contour lines; warmer colors correspond to higher pressure.  In addition, vectors are shown that represent the magnitude and direction of the (negative) pressure gradient -- the pressure force acting on the fluid elements.  This is the identical flow problem as depicted above so that the pressure field here is the one corresponding to the previous velocities.

Obviously enough, pressure increases as velocity decreases in accordance with the Bernoulli equation.  The relationship between pressure and velocity is nonlinear, i.e. pressure is proportional to the square of the velocity.  This particular mathematical relationship may not be clearly evident by simple observation of the colors, but it is clear that the pressure hotspots coincide with the stagnation points where velocity is zero.  At these two locations, fluid elements have given up all of their kinetic energy and pressure is at its highest.  Similarly, velocity at the poles (top and bottom) of the cylinder is at a maximum and pressure at a minimum.

Note that the pressure force vectors, due to the pressure gradient ($$-\nabla p$$ ),  are aligned perpendicularly to the pressure contours throughout the solution and are directed from the high pressure side of each contour to the low-pressure side.  Pressure is the only force that is operative in the solution; friction due to viscosity is omitted for the ideal fluid.  (See previous discussion of gradient of a scalar field.)

The Bernoulli equation is an expression of energy conservation.  Now we will take a look at the solution as it relates to motion of the fluid elements -- conservation of momentum.  This will be done qualitatively so as to allow a glimpse of the interrelationships.  In the next image the previous pressure solution is represented by color, but the streamlines (fluid path lines) have been retained.  The pressure force vectors are also retained.

Consider a fluid element approaching from the left at position 1.  We know that the particle is slowing down as it approaches the equator of the cylinder.  Pressure force vectors are directed exactly  opposite to the direction of motion -- the force acts entirely to decrease the fluid speed (velocity magnitude). From the above image showing pressure contours, the pressure retarding force actually increases in magnitude as the element approaches the cylinder (2) and brings the element to a halt at the surface of the cylinder.  Fluid elements moving around the top side of the cylinder are subjected to a dramatically varying pressure force.  Initially it appears that the forces are directed primarily along the direction of motion, adding to the velocity of the elements there (3).  As the elements approach the top of the cylinder, forces are directed less and less in the direction of motion and gradually become perpendicular to the motion (4).  Forces aligned with the direction of motion change the speed of an object;  forces directed perpendicular to the direction of motion change the direction of the object. BOTH constitute acceleration. Velocity is a vector and a change in EITHER magnitude or direction constitutes an acceleration.  Engineers typically apply the term acceleration whether an object is speeding up or slowing down.  It is only the direction of the acceleration (force) vector that determines whether speed is added or subtracted to the object and whether the direction of motion is changing.

As fluid elements continue around the downstream side of the cylinder, they continue to be subjected to a large pressure force which acts to keep the elements moving tangentially to the cylinder surface (direction of velocity) and also to slow them down (decelerate, magnitude of velocity).  Principally, this is where the solution loses its relationship to realism for reasons to be discussed.

We turn now to mathematical issues that will be useful in understanding certain aspects of the flow model.  The solutions depicted above are simply graphical displays of functions where velocity and pressure are expressed in terms of physical location.  In the following, $$u$$ and $$v$$ are velocities in the $$x$$ and $$y$$ directions respectively.  They are expressed here initially as functions of $$r$$ and $$\theta$$ which often is more convenient than $$x$$ and $$y$$ when dealing with a cylindrical geometry.

$$\Large u(r,\theta)= U \left[1-\Large\frac{a^2\cos(2\theta)}{r^2}\right]$$

$$\Large v(r,\theta)=-U\Large\frac{a^2\sin(2\theta)}{r^2}$$

$$\Large p(r,\theta)=\Large\frac{\rho}{2}[U^2-(u^2+v^2)]=\Large\frac{\rho U^2}{2}\Large\frac{a^2[2r^2 \cos(2\theta)-a^2]}{r^4}$$

$$U$$ in the above is the free stream velocity that is directed from left to right. $$a$$ is the radius of the cylinder.  Values of $$r < a$$ are not part of the physical solution; they are outside of the domain of the problem.

While the determination of these functions is somewhat more difficult, we can examine them for certain important characteristics.  This is an approach that should be considered whenever examining equations that relate to the physical universe.

• Functions involved in physical problems should have consistent physical units. $$u$$ and $$v$$ are both velocities in the above.  Each is depicted as   $$U$$  multiplying a fraction. Consequently the fraction should be a pure number with no physical units.  Since $$a$$ and $$r$$ both have units of length, the condition is satisfied ($$1$$ , $$\cos(2\theta)$$ and $$\sin(2\theta)$$ are all pure numbers.)  We know also from the Bernoulli equation that pressure and $$\rho u^2/2$$ must have the same physical units.  Once again, the fraction multiplied by $$\rho U^2/2$$ must be nondimensional in the above (condition satisfied!).
• Examine the functions for agreement with expectations for limiting values of the independent variables ($$r$$ and $$\theta$$ in this case).  We know that sufficiently far from the cylinder we should have $$u\rightarrow U$$ and $$v\rightarrow 0$$ .  This is apparent from the fractions which approach zero in proportion to $$1/r^2$$ as the distance (radius) from the center of the cylinder increases.  This is true for the pressure function also.  On the surface of the cylinder ($$r=a$$ ), we have the following velocities: $$u(r,\theta)= U \left[1-\cos(2\theta)\right]$$ and $$v(r,\theta)=-U\sin(2\theta)$$ .  Careful consideration of these equations will convince you that velocities are directed perpendicular to the cylinder radius, i.e. tangential to the cylinder surface.  The velocity magnitude exhibits a minimum value ($$0$$ ) where $$\theta = 0$$ and $$\theta = \pi$$ , i.e. at the upstream and downstream points of the cylinder, and a maximum value of $$2U$$ at $$\theta = \pi/2$$ and $$\theta = -\pi/2$$ , the poles at the top and bottom of the cylinder.
• Are the equations linear? The equations representing the velocities, both $$u$$ and  $$v$$ , are multiplied by the free stream velocity $$U$$ .  Increasing the free stream velocity, simply increases the values of $$u$$ and  $$v$$ proportionately.  The shape of the solution doesn't change at all with increasing $$U$$ .  Linearity is an extremely important aspect of an equation and, in this case, not entirely justified as will soon be discussed.  Fluid dynamics solutions are notoriously nonlinear and it is only by taking a simplified approach to this problem that we have been able to develop a solution that seems applicable.  Beware -- the simplified Bernoulli equation is not always applicable in cardiology nor is the concept of vascular resistance or impedance.  The pressure solution is nonlinear with respect to $$U$$ as you are fully aware of from the Bernoulli equation.  Even so, we are able to readily derive the pressure solution since we have the velocity solution available.  This again is a consequence of taking a simplified approach.

The equations for velocity above include 1 or more terms in the solution.  The simplest term in the equation for $$u$$ is simply $$U$$ , a constant that represents the free stream velocity.  Plotting vectors for this component of the velocity yields the following image (click the "Uniform" checkbox in the vector group box).

(Pressure is represented by color in each of these two images).  By itself, the uniform component describes the velocity sufficiently far away from the cylinder, but obviously takes no account of the presence of the cylinder.  The uniform velocity component does not fulfill the boundary condition of 0 velocity in the direction of the cylinder surface.
The more complicated term, depending on both $$r$$ and $$\theta$$ , is a dipole.  You have likely encountered dipoles in discussions of electrocardiography.  Clicking on the "Dipole" checkbox in the vector group box (and unchecking the "Uniform" checkbox) displays the dipole component of the velocity solution only.

Notice that the dipole vectors diminish in magnitude with distance from the cylinder surface.    A little bit of imagination will help you to visualize how the sum of these two components fulfills the velocity boundary condition requirements, both at the cylinder surface and in the free stream, far away from the cylinder.

Some would say that the equations above for $$u(r,\theta)$$ and $$v(r,\theta)$$ have their metaphors mixed; $$u$$ is velocity in the $$x$$ direction and $$v$$ is velocity in the $$y$$ direction so these velocity should be expressed as functions of the Cartesian coordinates, $$u(x,y)$$ and $$v(x,y)$$ .  These expressions will come in handy in the future and so are given here.  Note that $$r = \sqrt{x^2+y^2}$$ , $$\cos(\theta)=x/r$$ , $$\sin(\theta)=y/r$$ , $$\sin(2\theta)=2\cos(\theta)\sin(\theta)$$ and  $$\cos(2\theta)=\cos(\theta)^2-\sin(\theta)^2$$ .  Algebraic substitution of these facts into the previous equations allows solution for the velocities strictly in terms of the Cartesian coordinates.

$$\Large u(x,y)=U\Large\frac{a^2(y^2-x^2) +\left(x^2+y^2\right)^2}{\left(x^2+y^2\right)^2}$$

$$\Large v(x,y)=-U\Large\frac{2a^2xy}{\left(x^2+y^2\right)^2}$$

$$\Large p(x,y)=\Large\frac{\rho U^2}{2}\Large\frac{a^2(2x^2-2y^2-a^2)}{\left(x^2+y^2\right)^2}$$