Advanced Fluid Mechanics: Problems and Solutions

Subject: Fluid Dynamics & Hydraulics Level: Senior Undergraduate / Graduate Focus: Navier-Stokes Applications, Dimensional Analysis, and Boundary Layers

Part 5: Computational Approaches to Challenging Problems

When analytical methods fail, advanced problems require CFD. But "solutions" are not just numbers—they require verification and validation.

The Solution

Step 1: Simplify the Navier-Stokes Equations We start with the incompressible Navier-Stokes equation for the x-momentum: $$ \rho \left( \frac\partial u\partial t + u \frac\partial u\partial x + v \frac\partial u\partial y \right) = -\frac\partial P\partial x + \mu \left( \frac\partial^2 u\partial x^2 + \frac\partial^2 u\partial y^2 \right) $$

Given the assumptions:

  • Steady: $\frac\partial u\partial t = 0$
  • Fully developed: $\frac\partial u\partial x = 0$
  • Parallel flow: $v = 0$ (no vertical velocity component).

The equation reduces to a simple balance between pressure and viscous forces: $$ 0 = -\fracdPdx + \mu \fracd^2 udy^2 $$ (Note: Partial derivatives become total derivatives as $u$ depends only on $y$.)

Step 2: Integrate the Differential Equation Rearranging gives: $$ \fracd^2 udy^2 = \frac1\mu \fracdPdx $$

Integrate once with respect to $y$: $$ \fracdudy = \frac1\mu \fracdPdx y + C_1 $$

Integrate a second time: $$ u(y) = \frac12\mu \fracdPdx y^2 + C_1 y + C_2 $$

Step 3: Apply Boundary Conditions

  1. No-slip at bottom plate ($y=0$): $u(0) = 0$. $$ 0 = 0 + 0 + C_2 \implies C_2 = 0 $$
  2. No-slip at top plate ($y=B$): $u(B) = U$. $$ U = \frac12\mu \fracdPdx B^2 + C_1 B $$ $$ C_1 = \fracUB - \frac12\mu \fracdPdx B $$

Step 4: Final Velocity Profile Substitute $C_1$ and $C_2$ back into the equation: $$ u(y) = \fracU yB - \frac12\mu \left(-\fracdPdx\right) (By - y^2) $$ (Here, we typically define a favorable pressure gradient as negative, so we swap signs for clarity).

Step 5: Condition for Zero Net Flow The flow rate per unit width is $Q = \int_0^B u(y) dy$. $$ Q = \int_0^B \left[ \fracU yB + \frac12\mu \fracdPdx (By - y^2) \right] dy $$ $$ Q = \fracU B2 + \frac12\mu \fracdPdx \left[ \fracB y^22 - \fracy^33 \right]_0^B $$ $$ Q = \fracUB2 + \frac12\mu \fracdPdx \left( \fracB^32 - \fracB^33 \right) $$ $$ Q = \fracUB2 + \fracB^312\mu \fracdPdx $$

For $Q = 0$: $$ \fracUB2 = - \fracB^312\mu \fracdPdx $$ $$ \fracdPdx = \frac6\mu UB^2 $$ This implies an adverse pressure gradient is required to exactly counteract the shear-driven flow from the moving plate.

Part I: Theoretical and Conceptual Problems

Part III: Dimensional Analysis and Similitude

Part 2: Exact Solutions to the Navier-Stokes Equations

The Navier-Stokes equations represent the holy grail of fluid mechanics. Most advanced problems cannot be solved exactly, but a few canonical problems yield to analytical methods. These solutions serve as validation benchmarks for CFD and provide deep physical insight.

Problem 3: Stability and Transition – Orr–Sommerfeld Equation

Problem:
For a parallel shear flow ( U(y) ), small disturbances of streamfunction ( \psi = \phi(y) e^i(\alpha x - \omega t) ) satisfy the Orr–Sommerfeld equation:
[ (U - c)(\phi'' - \alpha^2 \phi) - U'' \phi = \frac-i\alpha Re (\phi'''' - 2\alpha^2 \phi'' + \alpha^4 \phi) ] Explain the physical meaning of each term for inviscid (( Re \to \infty )) case, and derive the Rayleigh inflection point criterion.

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