𝜕2u𝜕y2=𝜕𝜕y(U∞f′′U∞νx)=U∞f′′′U∞νx=U∞2νxf′′′partial squared u over partial y squared end-fraction equals the fraction with numerator partial and denominator partial y end-fraction open paren cap U sub infinity end-sub f double prime the square root of the fraction with numerator cap U sub infinity end-sub and denominator nu x end-fraction end-root close paren equals cap U sub infinity end-sub f triple prime the fraction with numerator cap U sub infinity end-sub and denominator nu x end-fraction equals the fraction with numerator cap U sub infinity end-sub squared and denominator nu x end-fraction f triple prime Step 4: Substitute into Momentum Equation
(Assuming an ideal scenario where compressibility is ignored or the tunnel uses compressed air to increase density) : If we proceed with the calculation for
Substituting constants back into the equation: $$ u(y) = \fracUyh - \frach^22\mu\fracdpdx \left[ \fracyh - \left(\fracyh\right)^2 \right] $$ Note: The first term is simple Couette flow (linear), and the second term is Poiseuille flow (parabolic) induced by the pressure gradient.
ρ(𝜕u𝜕t+u⋅∇u)=−∇p+μ∇2u+(ζ+μ3)∇(∇⋅u)+frho open paren the fraction with numerator partial bold u and denominator partial t end-fraction plus bold u center dot nabla bold u close paren equals negative nabla p plus mu nabla squared bold u plus open paren zeta plus the fraction with numerator mu and denominator 3 end-fraction close paren nabla open paren nabla center dot bold u close paren plus bold f is fluid density is the velocity vector is thermodynamic pressure is dynamic viscosity is bulk viscosity represents body forces (like gravity) advanced fluid mechanics problems and solutions
In advanced applications, we rarely solve these in their complete form. Instead, we apply scaling analysis (non-dimensionalization) to eliminate negligible terms based on dimensionless parameters like the Reynolds number ( ), Mach number ( ), and Froude number ( 2. Problem Set 1: Exact Solutions for Viscous Flows
A viscous, incompressible fluid flows between two infinite parallel plates separated by distance
Given the far-field boundary condition, we guess a separable solution of the form: Problem Set 1: Exact Solutions for Viscous Flows
0=AJ0(kR)+P0iωρ⟹A=−P0iωρJ0(kR)0 equals cap A cap J sub 0 open paren k cap R close paren plus the fraction with numerator cap P sub 0 and denominator i omega rho end-fraction ⟹ cap A equals negative the fraction with numerator cap P sub 0 and denominator i omega rho cap J sub 0 open paren k cap R close paren end-fraction
𝜕2u𝜕y2=U∞2νxf′′′(η)partial squared u over partial y squared end-fraction equals the fraction with numerator cap U sub infinity end-sub squared and denominator nu x end-fraction f triple prime of open paren eta close paren Step 4: Substitute into the Momentum Equation Substitute these derivatives into
Using Bernoulli's equation between the pipe (1) and the nozzle exit (2), assuming horizontal flow and negligible losses: Mach number ( )
Micro- and nano-scale flows (rarefied and slip flows)
uz(r,t)=U(r)eiωtu sub z open paren r comma t close paren equals cap U open paren r close paren e raised to the i omega t power Substituting this into the differential equation yields:
The total drag consists of form drag (pressure) and skin friction drag (shear stress). Integrating the stress tensor components over the surface of the sphere (
Problem 3: Flow Past a Rotating Cylinder (The Magnus Effect)
Flow U_infty ---> ____________________________________ (Flat Plate) v = -v_0 (Suction ↓)
