Advanced Fluid Mechanics Problems And Solutions | PROVEN |

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A power-law fluid follows ( \tau = K \dot\gamma^n ) ( ( \dot\gamma = -\fracdudr ) ). Derive the velocity profile and volumetric flow rate for laminar flow in a circular pipe of radius ( R ).

𝜕u𝜕y=U∞f′′(η)𝜕η𝜕y=U∞f′′(η)U∞νxpartial u over partial y end-fraction equals cap U sub infinity end-sub f double prime of open paren eta close paren partial eta over partial y end-fraction equals cap U sub infinity end-sub f double prime of open paren eta close paren the square root of the fraction with numerator cap U sub infinity end-sub and denominator nu x end-fraction end-root advanced fluid mechanics problems and solutions

2f′′′+ff′′=02 f triple prime plus f f double prime equals 0 Step 5: Define Boundary Conditions Translate physical conditions into the similarity domain: Impermeable wall condition ( ): Freestream matching ( ): 3. Potential Flow Theory: Superposition of Elementary Flows

v=−𝜕ψ𝜕x=−[12νU∞xf(η)+νxU∞f′(η)𝜕η𝜕x]v equals negative partial psi over partial x end-fraction equals negative open bracket one-half the square root of the fraction with numerator nu cap U sub infinity end-sub and denominator x end-fraction end-root f of open paren eta close paren plus the square root of nu x cap U sub infinity end-sub end-root f prime of open paren eta close paren partial eta over partial x end-fraction close bracket If you need help resolving a specific fluid

This is typically implemented in CFD boundary conditions using Riemann solvers (e.g., Roe, HLLC) rather than manual shock polars, but the analytic solution provides essential validation.

At extremely low Reynolds numbers ((Re \ll 1)), inertia is negligible, and the Navier-Stokes equations reduce to the linear Stokes equations. For a sphere of radius (a) moving with velocity (U) in a viscous fluid, Stokes derived the famous drag force (F = 6\pi\mu a U). However, this solution fails to satisfy the boundary conditions at infinity uniformly. In two dimensions, the Stokes paradox states no steady solution exists. In three dimensions, the Stokes solution is valid only as a leading-order approximation. The question: How do we find the first inertial correction to the drag? For a sphere of radius (a) moving with

uθ=−𝜕ψ𝜕r=−U∞sinθ(1+R2r2)−Γ2πru sub theta equals negative partial psi over partial r end-fraction equals negative cap U sub infinity end-sub sine theta open paren 1 plus the fraction with numerator cap R squared and denominator r squared end-fraction close paren minus the fraction with numerator cap gamma and denominator 2 pi r end-fraction Evaluate components exactly at the boundary

back into the velocity equation and simplifying, we get the velocity profile:

Ma22=2+(γ−1)Ma122γMa12−(γ−1)cap M a sub 2 squared equals the fraction with numerator 2 plus open paren gamma minus 1 close paren cap M a sub 1 squared and denominator 2 gamma cap M a sub 1 squared minus open paren gamma minus 1 close paren end-fraction

( \fracG r2 = K \left( -\fracdudr \right)^n ) → ( -\fracdudr = \left( \fracG r2K \right)^1/n ).