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)sinθ, where U∞ is the freestream velocity and R is the cylinder radius. If the body force is neglected,

Question

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 2ρ $U^2 _\infty$ [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 v_z = 0

and that the continuity and Euler equations are satisﬁed. Comment on the boundary conditions.

[latex]v_r=U _∞\left(1-\frac{R^2}{r^2} ight)\cos heta , [/latex] [latex]v_θ=U _∞\left(1-\frac{R^2}{r^2} ight)\sin heta , [/latex] and v_z = 0

and that the continuity and Euler equations are satisﬁed. Comment on the boundary conditions.

Accepted Answer

The streamfunction for a 2D constant density or incompressible ﬂow that is described in cylindrical coordinates by velocity components (v_r, v_θ, 0) is deﬁned by Eqs. 10.74a and 10.74b as

[latex]v_r=\frac{1}{r}\frac{\partial \psi }{d heta }[/latex] and [latex]v_θ=-\frac{\partial \psi }{dr}[/latex]

Thus we can calculate the velocities from

[latex]v_r=\frac{1}{r}\frac{\partial \psi }{d heta }=\frac{1}{r}\frac{\partial }{d heta }\left[U _∞r\left(1-\frac{R^2}{r^2} ight)\sin heta ight]=U _∞\left(1-\frac{R^2}{r^2} ight)\cos heta[/latex]

[latex]v_θ=-\frac{\partial \psi }{dr}=-\frac{\partial}{dr} \left[U _∞r\left(1-\frac{R^2}{r^2} ight)\sin heta ight][/latex]

[latex]=-U _∞ \sin heta \left[\left(1-\frac{R^2}{r^2}+r\left(\frac{2R^2}{r^3} ight) ight)\sin heta ight][/latex]

[latex]=-U _∞ \left(1+\frac{R^2}{r^2} ight) \sin heta [/latex]

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)[∂(rv_r)/∂r] + (1/r)(∂v_θ/∂θ) + ∂v_z/∂z = 0. The left hand side of this equation is

[latex]\begin{aligned}\frac{1}{r} & \frac{\partial}{\partial r}\left[r U_{\infty}\left(1-\frac{R^2}{r^2} ight) \cos heta ight]+\frac{1}{r} \frac{\partial}{\partial heta}\left[-U_{\infty}\left(1+\frac{R^2}{r^2} ight) \sin heta ight]+(0) \&=\frac{U_{\infty}}{r}\left(1-\frac{R^2}{r^2} ight) \cos heta+U_{\infty}\left(\frac{2 R^2}{r^3} ight) \cos heta-\frac{U_{\infty}}{r}\left(1+\frac{R^2}{r^2} ight) \cos heta+(0)=0\end{aligned}[/latex]

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

[latex] ho\left(\frac{\partial v_r}{\partial t}+v_r \frac{\partial v_r}{\partial r}+\frac{v_ heta}{r} \frac{\partial v_r}{\partial heta}+v_z \frac{\partial v_r}{\partial z}-\frac{v_ heta^2}{r} ight) = ho f_r-\frac{\partial p}{\partial r}[/latex] (11.17a)

[latex] ho\left(\frac{\partial v_ heta}{\partial t}+v_r \frac{\partial v_ heta}{\partial r}+\frac{v_ heta}{r} \frac{\partial v_ heta}{\partial heta}+v_z \frac{\partial v_ heta}{\partial z}+\frac{v_r v_ heta}{r} ight) = ho f_ heta-\frac{1}{r} \frac{\partial p}{\partial heta}[/latex] (11.17b)

[latex] ho\left(\frac{\partial v_z}{\partial t}+v_r \frac{\partial v_z}{\partial r}+\frac{v_ heta}{r} \frac{\partial v_z}{\partial heta}+v_z \frac{\partial v_z}{\partial z} ight) = ho f_z-\frac{\partial p}{\partial z}[/latex] (11.17c)

[latex] ho\left(v_r \frac{\partial v_r}{\partial r}+\frac{v_ heta}{r} \frac{\partial v_r}{\partial heta}-\frac{v_ heta^2}{r} ight) =-\frac{\partial p}{\partial r},[/latex] [latex] ho\left(v_r \frac{\partial v_ heta}{\partial r}+\frac{v_ heta}{r} \frac{\partial v_ heta}{\partial heta}+\frac{v_r v_ heta}{r} ight) =-\frac{1}{r} \frac{\partial p}{\partial heta}[/latex]

and
[latex]0 =-\frac{\partial p}{\partial z}[/latex]

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 v_r= 0, v_θ = − 2U_∞sin θ, and v_z = 0. So the no-penetration condition is satisﬁed, but the ﬂuid slips along the surface with a θ velocity of v_θ = − 2U_∞sin θ.

Question 11.9: The streamlines for the 2D, inviscid, constant density ﬂow o...

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