Question 11.8: The steady, 2D inviscid flow of an incompressible fluid shown ...
The steady, 2D inviscid flow of an incompressible fluid shown in Figure 11.7 is described by u = kx, v = −ky, w = 0, and p(x, y) = p0 − ρ(k²/2)(x² + y²), where k is a constant, p0 is the pressure at the origin, and body forces have been neglected. This inviscid flow model for a constant density flow approaching a plane wall is referred to as plane stagnation point flow. Show that this flow satisfies the continuity and Euler equations, and comment on whether the no-slip, no-penetration conditions are or are not satisfied.
We will check that the continuity equation for an incompressible fluid is satisfied by substituting the velocity components into Eq. 11.4a, ∂u/∂x + ∂v/∂y + ∂w/∂z = 0, to obtain
\frac{∂}{∂x}(kx)+\frac{∂}{∂y}(-ky)+\frac{∂}{∂z}(0)=k-k=0
Next we substitute the velocity components and pressure into the Euler equations for inviscid flow in Cartesian coordinates, Eqs. 11.16a–c:
σ_{xx} =−p−\frac{2}{3} µ(∇• u)+2µ\frac{\partial u}{\partial x} (11.6a)
σ_{yy} =−p−\frac{2}{3} µ(∇• u)+2µ\frac{\partial v}{\partial y} (11.6b)
σ_{zz} =−p−\frac{2}{3} µ(∇• u)+2µ\frac{\partial w}{\partial z} (11.6c)
\begin{aligned}\rho\left(\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}+w \frac{\partial u}{\partial z}\right) &=\rho f_x-\frac{\partial p}{\partial x} \\\rho\left(\frac{\partial v}{\partial t}+u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}+w \frac{\partial v}{\partial z}\right) &=\rho f_y-\frac{\partial p}{\partial y} \\\rho\left(\frac{\partial w}{\partial t}+u \frac{\partial w}{\partial x}+v \frac{\partial w}{\partial y}+w \frac{\partial w}{\partial z}\right) &=\rho f_z-\frac{\partial p}{\partial z}\end{aligned}
To simplify, note that the body forces are zero by assumption, the flow is steady, the velocity components u and v are functions of only one variable and w = 0, and the pressure does not depend on z. Inserting the velocity components and pressure, we have
ρ[0+(kx)(k)+(−ky)(0)+0]=-\frac{∂}{∂x}\left[p_0-\frac{\rho k^2}{2}(x^2+y^2)\right]=\frac{\rho k^2}{2}(2x)
ρ[0+(kx)(0)+(−ky)(-k)+0]=-\frac{∂}{∂y}\left[p_0-\frac{\rho k^2}{2}(x^2+y^2)\right]=\frac{\rho k^2}{2}(2y)
0=-\frac{∂}{∂z}\left[p_0-\frac{\rho k^2}{2}(x^2+y^2)\right]=0
After simplifying we find
ρ(kx)(k) =\frac{\rho k^2}{2}(2x),\ ρ(−ky)(−k) =\frac{\rho k^2}{2}(2y),\ and\ 0=0
Thus, the Euler equations are also satisfied for this flow.
At the wall, y = 0; thus we find u = kx, v = 0, w = 0 on the wall. The fluid satisfies the no-penetration condition but slips with u = kx in the x direction.