next up previous
Next: References Up: Steady flows of incompressible Previous: Aplications of Bernoulli

Drag and D'Alembert's "paradox"

Let's now consider one particularly simple nonviscous flow, the irrotational flow of a fluid around a cylinder. This problem will demonstrate how some fluid mechanics problems are solved, and brings out an important paradox in the theory of nonviscous flow.

We consider a cylinder of radius R, with an imposed flow of velocity tex2html_wrap_inline1199 far from the cylinder--see Fig. 2.7.

Figure 2.7: Flow past a cylinder

We know that the fluid cannot penetrate the surface of the cylinder, so that the normal component of the velocity at the surface of the cylinder must be zero. In radial coordinates this is expressed as


What is the equation for tex2html_wrap_inline1019 ? Since the flow is irrotational, tex2html_wrap_inline1088 , and from vector calculus we know that this implies that tex2html_wrap_inline1019 may be written as the gradient of some scalar function: tex2html_wrap_inline1207 . If the fluid is also incompressible, then tex2html_wrap_inline1090 , so that


The potential function  tex2html_wrap_inline1211 is a solution of Laplace's equation, which should be familiar to those of you who have studied advanced electrostatics. In radial coordinates Laplace's equation is


In terms of tex2html_wrap_inline1211 the boundary condition at infinity is tex2html_wrap_inline1215 ; in radial coordinates tex2html_wrap_inline1217 , so this becomes


At the surface of the cylinder we have


The strategy is that we solve Laplace's equation for tex2html_wrap_inline1211 , and then calculate the velocity field from the potential.gif1 To do this we first need to realize that in order to satisfy the boundary condition at infinity, tex2html_wrap_inline1225 ; therefore, we set


Substituting this into Laplace's equation, Eq. (2.29), we obtain a differential equation for f:


To solve this equation, we'll guess the solution tex2html_wrap_inline1229 , and see if we can find an appropriate value of tex2html_wrap_inline1231 . Substituting, we find tex2html_wrap_inline1233 , so that our solution is


Using the boundary condition at infinity, we see that A=-U; applying the boundary condition at the surface of the cylinder, we find tex2html_wrap_inline1237 , so our final solution is


The components of the velocity are



The streamlines in this case are symmetric on the upstream and downstream sides of the cylinder.

How does the pressure vary around the cylinder? First, the square of the velocity on the surface of the cylinder is


We can now use Bernoulli's equation; since the flow is irrotational, the constant is the same everywhere in the fluid. In particular, at infinity the constant is tex2html_wrap_inline1239 , where tex2html_wrap_inline1241 is the ambient pressure (which we can choose to be zero without any loss of generality). The pressure on the surface of the cylinder is then


From the pressure we can calculate the net force acting on the cylinder. First, recall that the pressure at the surface of the cylinder is the force per unit area acting on the cylinder; therefore the force tex2html_wrap_inline1243 acting upon an element of the cylinder of surface area dS is tex2html_wrap_inline1247 , where tex2html_wrap_inline1249 is a unit vector normal to the surface, and the minus sign indicates that this force acts inward. Next, we integrate this over the entire surface of the cylinder to find the net force:


For the cylinder, tex2html_wrap_inline1251 , with z along the axis of the cylinder. The unit vector depends upon the angle tex2html_wrap_inline1255 as tex2html_wrap_inline1257 . Then the x and y components of the force are



where tex2html_wrap_inline1263 is the length of the cylinder in the z-direction. If we now substitute our expression for the pressure distribution on the surface of the cylinder, Eq. 2.39, into our expressions for the drag and the lift, Eqs. 2.41 and 2.42, and perform the tex2html_wrap_inline1255 integration, we find that tex2html_wrap_inline1269 ! Although we have performed an explicit calculation for a cylinder, the result is more general: irrotational flow of a nonviscous fluid about an object produces no drag on the object. This peculiar result is known as d'Alembert's paradox.

Physically, the fact that the drag is zero is due to the symmetry of the pressure field about the cylinder--the fluid is pushing as hard on the upstream side of the cylinder as on the downstream side, so there are no unbalanced forces which would lead to drag. Whether or not this is paradoxical is a matter of opinion. If there were a drag on the cylinder, how would the energy be dissipated in a nonviscous fluid? To really understand drag, we need to go beyond the nonviscous approximation and treat real fluids.


gif 1 In electrostatics j would be the electrostatic potential, and the electric field would be given by E = - Ñj.

next up previous
Next: References Up: Steady flows of incompressible Previous: Applications of Bernoulli

Notes by A. Dorsey, revised by V. Celli, U. of Virginia
Thu Oct 12 01:02:02 EDT 1997