Pls solve the Navier-Stokes Equations. Thanks​

Final Results:Velocity profile:[tex]\boxed{ u(y) = \frac{1}{2\mu} \frac{dp}{dx} (y^2 - hy) }[/tex]Maximum velocity:[tex]\boxed{ u_{\text{max}} = -\frac{1}{8\mu} \frac{dp}{dx} h^2 }[/tex]Volume flow rate per unit width:[tex]\boxed{ Q = -\frac{1}{12\mu} \frac{dp}{dx} h^3 }[/tex]______________________________________Pressure varies along x-axis only: p = p(x)[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 = -\frac{dp}{dx} + \mu \frac{d^2u}{dy^2}[/tex]2. Rearranging:[tex]\mu \frac{d^2u}{dy^2} = \frac{dp}{dx}[/tex]Important: [tex]\(\frac{dp}{dx}\)[/tex]3. Integrate once:[tex]\mu \frac{du}{dy} = \frac{dp}{dx} y + C_1[/tex][tex]{where \: (C_1) \: = integration \: constant.}[/tex]4. Integrate again:[tex]\mu u(y) = \frac{dp}{dx} \frac{y^2}{2} + C_1 y + C_2[/tex]5. Apply boundary conditions:At the plates:At (y=0), (u=0) (no slip)At (y=h), (u=0) (no slip)First BC (at (y=0):[tex] \small{0 = \frac{dp}{dx} \frac{(0)^2}{2} + C_1 (0) + C_2\quad \Rightarrow \quad C_2 = 0} [/tex]Second BC (at (y=h):[tex]0 = \frac{dp}{dx} \frac{h^2}{2} + C_1 h[/tex][tex]Solve \: for \: \(C_1\):[/tex][tex]C_1 = -\frac{dp}{dx} \frac{h}{2}[/tex]6. Final velocity profile:[tex]\mu u(y) = \frac{dp}{dx} \left( \frac{y^2}{2} - \frac{h}{2}y \right)[/tex]Then divide both sides by μ:[tex]u(y) = \frac{1}{\mu} \frac{dp}{dx} \left( \frac{y^2}{2} - \frac{h}{2}y \right)[/tex]or cleaner:[tex]u(y) = \frac{1}{2\mu} \frac{dp}{dx} \left( y^2 - hy \right)[/tex]The final velocity profile for Plane Poiseuille flow is:[tex] \: \: \: \: \: \boxed{ u(y) = \frac{1}{2\mu} \frac{dp}{dx} \left( y^2 - hy \right) }[/tex]Parabolic profile. Maximum speed at the center [tex]y = h/2\).[/tex]Zero speed at walls y = 0 and y = h.7. Find Maximum Velocity u_maxThe velocity profile:[tex] \: \: \: \: \: \: \: \: \: u(y) = \frac{1}{2\mu} \frac{dp}{dx} \left( y^2 - hy \right)[/tex][tex]\frac{du}{dy} = \frac{1}{2\mu} \frac{dp}{dx} (2y - h)[/tex][tex]Set \: \(\frac{du}{dy} = 0\):[/tex][tex] \: \: \: \: 2y - h = 0 \quad \Rightarrow \quad y = \frac{h}{2}[/tex][tex]Thus, \: maximum \: at \: \(y = \frac{h}{2}\).[/tex]Substitute back into [tex]\(u(y)\):[/tex][tex]u_{\text{max}} = \frac{1}{2\mu} \frac{dp}{dx} \left( \left( \frac{h}{2} \right)^2 - h\left( \frac{h}{2} \right) \right)[/tex]Simplify:[tex]u_{\text{max}} = \frac{1}{2\mu} \frac{dp}{dx} \left( \left( \frac{h}{2} \right)^2 - h\left( \frac{h}{2} \right) \right)[/tex]Thus:[tex]u_{\text{max}} = -\frac{1}{8\mu} \frac{dp}{dx} h^2[/tex]8. Find Volume Flow Rate Q per unit width[tex] \: \: \: \: \: \: \: \: \: \: \: \: Q = \int_0^h u(y) \, dy[/tex]Substitute u(y):[tex]Q = \frac{1}{2\mu} \frac{dp}{dx} \int_0^h (y^2 - hy) \, dy[/tex]Compute the integral:[tex]\int_0^h y^2 \, dy = \left. \frac{y^3}{3} \right|_0^h = \frac{h^3}{3}[/tex][tex]\int_0^h hy \, dy = h\left. \frac{y^2}{2} \right|_0^h = h\left( \frac{h^2}{2} \right) = \frac{h^3}{2}[/tex]Thus:[tex]Q = \frac{1}{2\mu} \frac{dp}{dx} \left( \frac{h^3}{3} - \frac{h^3}{2} \right)[/tex]Simplify inside:[tex]\frac{h^3}{3} - \frac{h^3}{2} = -\frac{h^3}{6}[/tex]Thus:[tex]Q = -\frac{1}{12\mu} \frac{dp}{dx} h^3[/tex]

Answered by mjPcontiga | 2025-04-26