In the channel ﬂow of a constant density, constant viscosity ﬂuid shown in Figure 11.5, suppose the complete description of the ﬂow is given in Cartesian coordinates by the velocity ﬁeld u = {[h²(p1 − p2)]/2µL}[1−(y/h)²], v = 0, and w = 0, and the pressure ﬁeld p(x) = p1 + [(p2 − p1)/L](x − x1).

Question

In the channel ﬂow of a constant density, constant viscosity ﬂuid shown in Figure 11.5, suppose the complete description of the ﬂow is given in Cartesian coordinates by the velocity ﬁeld u = {[h²(p₁ − p₂)]/2µL}[1−(y/h)²], v = 0, and w = 0, and the pressure ﬁeld p(x) = p₁ + [(p₂ − p₁)/L](x − x₁). Here p₁ and p₂are the pressures at the indicated locations, and the slight hydrostatic variation in pressure across the channel has been ignored. Show that this ﬂow satisﬁes the constant density, constant viscosity Navier–Stokes equations in Cartesian coordinates, with the body force neglected.

Accepted Answer

We will ﬁrst substitute the three velocity components into the continuity equation for an incompressible ﬂuid, Eq. 11.4a,

[latex]\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0[/latex] (11.4a)

then substitute the velocity components and pressure into the constant density, constant viscosity forms of the Navier–Stokes equations, Eqs. 11.10a–11.10c. Since we have u = u(y) only, with v and w zero, the continuity equation, ∂u/∂x + ∂v/∂y + ∂w/∂z = 0, is satisﬁed by inspection. Terms in Eqs. 11.10a–11.10c that contain the velocity components v and w are zero, as are time derivatives and spatial derivatives of u with respect to x and z. Writing only the remaining nonzero terms, and setting the body force terms to zero, we ﬁnd

[latex]\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 \rho }{\partial x}+\mu \left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}\right)[/latex] (11.10a)

[latex]\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 \rho }{\partial y}+\mu \left(\frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 v}{\partial y^2}+\frac{\partial^2 v}{\partial z^2}\right)[/latex] (11.10b)

[latex]\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 \rho }{\partial z}+\mu \left(\frac{\partial^2 w}{\partial x^2}+\frac{\partial^2 w}{\partial y^2}+\frac{\partial^2 w}{\partial z^2}\right)[/latex] (11.10c)

[latex]0=-\frac{\partial \rho }{\partial x}+\mu \frac{\partial^2 u}{\partial y^2},[/latex] [latex]0=-\frac{\partial \rho }{\partial y},[/latex] and [latex]0=-\frac{\partial \rho }{\partial z}[/latex]

The last two equations are consistent with the given pressure distribution. Substituting the x velocity component and pressure into the ﬁrst pressure equation yields

[latex]0=-\frac{\partial }{\partial x}\left(p_1+\left(\frac{p_2-p_1}{L}\right) (x_2-x)\right)+\mu \frac{\partial}{\partial y}\left[\frac{\partial}{\partial y}\left(\left(\frac{h^2\left(p_1-p_2\right) }{2\mu L}\right)\left[1-\left(\frac{y}{h} \right)^2 \right]\right)\right] [/latex]

[latex]0=-\left(\frac{p_2-p_1}{L}\right)+\mu \left(\frac{h^2\left(p_1-p_2\right) }{2\mu L}\right)\frac{\partial}{\partial y}\left(\frac{\partial}{\partial y}\left[1-\left(\frac{y}{h} \right)^2 \right]\right) [/latex]

[latex]0=-\left(\frac{p_2-p_1}{L}\right)+\mu \left(\frac{h^2\left(p_1-p_2\right) }{2\mu L}\right)\left(\frac{-2}{h^2} \right)=0[/latex]

We see that the velocity and pressure do satisfy the appropriate forms of the continuity and Navier–Stokes equations. It is also straightforward to show that the velocity ﬁeld satisﬁes the no-slip, no-penetration conditions at the channel walls.

Question 11.6: In the channel ﬂow of a constant density, constant viscosity...

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