Question 11.9: The streamlines for the 2D, inviscid, constant density flow o...
The streamlines for the 2D, inviscid, constant density flow over a cylinder are shown in Figure 11.8. The streamfunction for this flow is given in cylindrical coordinates by ψ(r, θ) = U∞r(1− R²/r²) sin θ, where U∞ is the freestream velocity and R is the cylinder radius. If the body force is neglected, the pressure distribution is given by p(r, θ) = p∞+ 1 2ρU^2 _∞ [1−(1− R2/r2)2 −4(R2/r2) sin2 θ]. Show that the velocity field in this case is described by
v_r=U _∞\left(1-\frac{R^2}{r^2}\right)\cos \theta , v_θ=U _∞\left(1-\frac{R^2}{r^2}\right)\sin \theta , and vz = 0
and that the continuity and Euler equations are satisfied. Comment on the boundary conditions.
The streamfunction for a 2D constant density or incompressible flow that is described in cylindrical coordinates by velocity components (vr, vθ, 0) is defined by Eqs. 10.74a and 10.74b as
v_r=\frac{1}{r}\frac{\partial \psi }{d\theta } and v_θ=-\frac{\partial \psi }{dr}
Thus we can calculate the velocities from
v_r=\frac{1}{r}\frac{\partial \psi }{d\theta }=\frac{1}{r}\frac{\partial }{d\theta }\left[U _∞r\left(1-\frac{R^2}{r^2}\right)\sin \theta\right]=U _∞\left(1-\frac{R^2}{r^2}\right)\cos \theta
v_θ=-\frac{\partial \psi }{dr}=-\frac{\partial}{dr} \left[U _∞r\left(1-\frac{R^2}{r^2}\right)\sin \theta\right]
=-U _∞ \sin \theta \left[\left(1-\frac{R^2}{r^2}+r\left(\frac{2R^2}{r^3} \right) \right)\sin \theta\right]
=-U _∞ \left(1+\frac{R^2}{r^2} \right) \sin \theta
We see that this agrees with the velocity components given in the problem statement.
Although we know that a streamfunction guarantees that the continuity equation is satisfied, we can check these velocity components by substituting them into that equation in cylindrical coordinates, Eq. 11.4b, (1/r)[∂(rvr)/∂r] + (1/r)(∂vθ/∂θ) + ∂vz/∂z = 0. The left hand side of this equation is
\begin{aligned}\frac{1}{r} & \frac{\partial}{\partial r}\left[r U_{\infty}\left(1-\frac{R^2}{r^2}\right) \cos \theta\right]+\frac{1}{r} \frac{\partial}{\partial \theta}\left[-U_{\infty}\left(1+\frac{R^2}{r^2}\right) \sin \theta\right]+(0) \\&=\frac{U_{\infty}}{r}\left(1-\frac{R^2}{r^2}\right) \cos \theta+U_{\infty}\left(\frac{2 R^2}{r^3}\right) \cos \theta-\frac{U_{\infty}}{r}\left(1+\frac{R^2}{r^2}\right) \cos \theta+(0)=0\end{aligned}
To see if the Euler equations are satisfied, we will substitute these velocity components and the pressure into Eqs. 11.17a–11.17c. Writing only the nonzero terms in these equations yields
\rho\left(\frac{\partial v_r}{\partial t}+v_r \frac{\partial v_r}{\partial r}+\frac{v_\theta}{r} \frac{\partial v_r}{\partial \theta}+v_z \frac{\partial v_r}{\partial z}-\frac{v_\theta^2}{r}\right) =\rho f_r-\frac{\partial p}{\partial r} (11.17a)
\rho\left(\frac{\partial v_\theta}{\partial t}+v_r \frac{\partial v_\theta}{\partial r}+\frac{v_\theta}{r} \frac{\partial v_\theta}{\partial \theta}+v_z \frac{\partial v_\theta}{\partial z}+\frac{v_r v_\theta}{r}\right) =\rho f_\theta-\frac{1}{r} \frac{\partial p}{\partial \theta} (11.17b)
\rho\left(\frac{\partial v_z}{\partial t}+v_r \frac{\partial v_z}{\partial r}+\frac{v_\theta}{r} \frac{\partial v_z}{\partial \theta}+v_z \frac{\partial v_z}{\partial z}\right) =\rho f_z-\frac{\partial p}{\partial z} (11.17c)
\rho\left(v_r \frac{\partial v_r}{\partial r}+\frac{v_\theta}{r} \frac{\partial v_r}{\partial \theta}-\frac{v_\theta^2}{r}\right) =-\frac{\partial p}{\partial r}, \rho\left(v_r \frac{\partial v_\theta}{\partial r}+\frac{v_\theta}{r} \frac{\partial v_\theta}{\partial \theta}+\frac{v_r v_\theta}{r}\right) =-\frac{1}{r} \frac{\partial p}{\partial \theta}
and
0 =-\frac{\partial p}{\partial z}
Since the pressure given in the problem statement is only a function of r and θ, the last equation is satisfied. Although we leave the details of the final step as an exercise for the interested reader (and we suggest the use of a symbolic mathematics code) substituting the velocities and pressure into the remaining two equations shows that these are also satisfied. On the cylinder, r = R, and we find vr = 0, vθ = − 2U∞ sin θ, and vz = 0. So the no-penetration condition is satisfied, but the fluid slips along the surface with a θ velocity of vθ = − 2U∞ sin θ.