The momentum integral equation (von Kármán) simplifies the PDE into an ODE.
uθ=−1rsinθ𝜕ψ𝜕ru sub theta equals negative the fraction with numerator 1 and denominator r sine theta end-fraction partial psi over partial r end-fraction
Time-averaged Navier-Stokes (RANS) introduces the Reynolds stress tensor (\rho \overlineu_i' u_j').
Using the chain rule, compute the partial derivatives:
Given the far-field boundary condition, we guess a separable solution of the form: advanced fluid mechanics problems and solutions
−η2tdudη=ν(14νtd2udη2)negative the fraction with numerator eta and denominator 2 t end-fraction the fraction with numerator d u and denominator d eta end-fraction equals nu open paren the fraction with numerator 1 and denominator 4 nu t end-fraction the fraction with numerator d squared u and denominator d eta squared end-fraction close paren Multiply both sides by to isolate the terms:
𝜕u𝜕x+𝜕v𝜕y=0partial u over partial x end-fraction plus partial v over partial y end-fraction equals 0
ρ(𝜕u𝜕t+u⋅∇u)=−∇p+μ∇2u+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 bold f — The source of non-linearity and chaos (turbulence). Viscous term: — The "internal friction" that smooths out flow. 2. Advanced Problem Scenario: Creeping Flow (Stokes Flow) The Problem: Consider a tiny spherical particle (radius
Convert the normal Mach number back to total downstream Mach number M2cap M sub 2 The momentum integral equation (von Kármán) simplifies the
), the inertial terms in the Navier-Stokes equations can be neglected (Stokes flow / creeping flow).
If you are working on a specific scenario, let me know if you would like to explore , adjust the boundary conditions , or examine non-Newtonian fluid behaviors . Share public link
2f′′′+ff′′=02 f triple prime plus f f double prime equals 0 Step 1: Relate Velocity Components to the Stream Function
ρ(u𝜕u𝜕x+v𝜕u𝜕y)=−𝜕p𝜕x+μ(𝜕2u𝜕x2+𝜕2u𝜕y2)rho open paren u partial u over partial x end-fraction plus v partial u over partial y end-fraction close paren equals negative partial p over partial x end-fraction plus mu open paren partial squared u over partial x squared end-fraction plus partial squared u over partial y squared end-fraction close paren Substitute Viscous term: — The "internal friction" that smooths
Potential flow allows the linear addition of independent velocity potentials. Combine three distinct configurations: a uniform flow, a doublet (to model the solid cylinder cylinder), and a free vortex (to model the circulation).
I can provide step-by-step calculations if you have a specific in mind.
0=−𝜕p𝜕x+μd2udy20 equals negative partial p over partial x end-fraction plus mu d squared u over d y squared end-fraction