es equation for j, and then we will calculate the velocity field from the potential. To do this we first need to realize that in order to satisfy the boundary condition at infinity, j cosq, therefore, we set:j(r,q) = f(r) cos q1.12Substituting this into Laplaces equation, eq 1.9, becomes a differential equation for f:1/r d/dr(r.df/dr) f/r = 0.1.13To solve this equation, we will guess that the solution f(r) = r, and see if we can find an appropriate value of . Substituting, we find = +/- 1, so that our solution is:j(r,q) = ( Ar + B/r) cos q.1.14using the boundary condition at infinity, (eq. 1.10) we have:A = -U.1.15And applying the boundary condition at the surface of the cylinder (eq. 1.11) we have:B = -U.R1.16So our final solution of j is:j(r,q) = -U. (r + R/r)cosq1.17The components of the velocity are:nr(r,q) = -j/r = U (1- R/r) cosq1.18nq(r,q) = -1/r j/q = -U (1 + R/r) sinq1.19In this case, the streamlines are symetric on the upstream and downstream sides of the cylinder (as shown in the figure 1)So the point is now to determine how does the pressure vary around the cylinder ? First the square of the velocity on the surface of the cylinder is:n(R,q) = nr(R,q) + nq(R,q)1.20As for r = R: nr(R,q) = 0n(R,q) = nq(R,q)n(R,q) = (-U.2.sinq)n(R,q) = 4Usinq1.21We can now use Bernouillis equation; since the flow is irrotational, the total pressure as constant is the same everywhere in the fluid. In particular, at infinity, the total pressure is: rU + po = cst1.22where po is the ambient pressure (which can be choose to be zero without any loss of generality). The pressure on the surface of the cylinder is then:for rU + po = n + p(R,q)1.23 rU + po = (4Usinq) + p(R,q)1.24so:p(R,q) = rU(1-4sinq)1.25From this pressure, we can calculate the net force acting on the cylinder. First, we have to recall that the pressure at the surface of the cylinder is the force per unit area acting on the cylinder; therefore the force dF acting upon an element of...
