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

Question

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.

Accepted Answer

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

[latex]\frac{∂}{∂x}(kx)+\frac{∂}{∂y}(-ky)+\frac{∂}{∂z}(0)=k-k=0[/latex]

Next we substitute the velocity components and pressure into the Euler equations for inviscid ﬂow in Cartesian coordinates, Eqs. 11.16a–c:

[latex]σ_{xx} =−p−\frac{2}{3} µ(∇• u)+2µ\frac{\partial u}{\partial x}[/latex] (11.6a)

[latex]σ_{yy} =−p−\frac{2}{3} µ(∇• u)+2µ\frac{\partial v}{\partial y}[/latex] (11.6b)

[latex]σ_{zz} =−p−\frac{2}{3} µ(∇• u)+2µ\frac{\partial w}{\partial z}[/latex] (11.6c)

[latex]\begin{aligned} ho\left(\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}+w \frac{\partial u}{\partial z} ight) &= ho f_x-\frac{\partial p}{\partial x} \ ho\left(\frac{\partial v}{\partial t}+u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}+w \frac{\partial v}{\partial z} ight) &= ho f_y-\frac{\partial p}{\partial y} \ ho\left(\frac{\partial w}{\partial t}+u \frac{\partial w}{\partial x}+v \frac{\partial w}{\partial y}+w \frac{\partial w}{\partial z} ight) &= ho f_z-\frac{\partial p}{\partial z}\end{aligned}[/latex]

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

[latex]ρ[0+(kx)(k)+(−ky)(0)+0]=-\frac{∂}{∂x}\left[p_0-\frac{ ho k^2}{2}(x^2+y^2) ight]=\frac{ ho k^2}{2}(2x)[/latex]

[latex]ρ[0+(kx)(0)+(−ky)(-k)+0]=-\frac{∂}{∂y}\left[p_0-\frac{ ho k^2}{2}(x^2+y^2) ight]=\frac{ ho k^2}{2}(2y)[/latex]

[latex]0=-\frac{∂}{∂z}\left[p_0-\frac{ ho k^2}{2}(x^2+y^2) ight]=0[/latex]

After simplifying we ﬁnd

[latex]ρ(kx)(k) =\frac{ ho k^2}{2}(2x),\ ρ(−ky)(−k) =\frac{ ho k^2}{2}(2y),\ and\ 0=0[/latex]

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.

Question 11.8: The steady, 2D inviscid ﬂow of an incompressible ﬂuid shown ...

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