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Fluid mechanics formula sheet, Cheat Sheet of Fluid Mechanics

Formula sheet in Strain rates tensor, vorticity tensor, the material derivative, the balance equation of linear momentum and internal energy, dynamic viscosity and complex function.

Typology: Cheat Sheet

2021/2022

Uploaded on 02/07/2022

mjforever
mjforever 🇺🇸

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Download Fluid mechanics formula sheet and more Cheat Sheet Fluid Mechanics in PDF only on Docsity! 31446 Mechanics of fluids: Formula sheet to be used at written examinations ◮The ǫ− δ identity reads ǫinmǫmjk = ǫminǫmjk = ǫnmiǫmjk = δijδnk − δikδnj ◮Strain rate tensor, vorticity tensor ∂vi ∂xj = 1 2     ∂vi ∂xj + ∂vi ∂xj 2∂vi/∂xj + ∂vj ∂xi − ∂vj ∂xi =0     = 1 2 ( ∂vi ∂xj + ∂vj ∂xi ) + 1 2 ( ∂vi ∂xj − ∂vj ∂xi ) = Sij +Ωij ◮The vorticity vector is computed as ω = ∇× v ωi = ǫijk ∂vk ∂xj ◮The material derivative ρ dΨ dt = ρ ∂Ψ ∂t + ρvj ∂Ψ ∂xj where Ψ = vi, u, T, k, v′iv ′ j . . . ◮The balance equation for mass dρ dt + ρ ∂vi ∂xi = 0 ◮The balance equation for linear momentum ρ dvi dt = ∂σji ∂xj + ρfi ◮The balance equation for internal energy ρ du dt = σji ∂vi ∂xj − ∂qi ∂xi ◮The equation for kinetic energy reads ρ dk dt = ∂viσji ∂xj − σji ∂vi ∂xj + ρvifi ◮The constitutive law for Newtonian viscous fluids σij = −pδij + 2µSij − 2 3 µSkkδij , σij = −pδij + τij qi = −k ∂T ∂xi ◮Viscosity 1 µ: dynamic viscosity ν: kinematic viscosity (ν = µ/ρ) ◮The continuity equation and the Navier-Stokes equation for incompressible flow with constant viscosity read (conservative form, p denotes the hydrostatic pressure, i.e. p = 0 if vi = 0) ∂vi ∂xi = 0 ρ ∂vi ∂t + ρ ∂vivj ∂xj = − ∂p ∂xi + µ ∂2vi ∂xj∂xj ◮The Navier-Stokes equation for incompressible flow with constant viscosity read (non-conservative form) ρ ∂vi ∂t + ρvj ∂vi ∂xj = − ∂p ∂xi + µ ∂2vi ∂xj∂xj The viscous stress tensor then reads τij = 2µSij = µ ( ∂vi ∂xj + ∂vj ∂xi ) ◮The equation for internal energy reads ρ du dt = −p ∂vi ∂xi + 2µSijSij − 2 3 µSkkSii Φ + ∂ ∂xi ( k ∂T ∂xi ) ◮Streamfunction, Ψ; potential, Φ v1 = ∂Ψ ∂x2 , v2 = − ∂Ψ ∂x1 vk = ∂Φ ∂xk ◮The Rayleigh problem η = x2 2 √ νt , f = v1 V0 d2f dη2 + 2η df dη = 0 ◮Blasius solution ξ = ( V1,∞ νx1 )1/2 x2, Ψ = (νV1,∞x1) 1/2 g 1 2 gg′′ + g′′′ = 0 ◮The Navier-Stokes (different form of the convective term) ∂vi ∂t + ∂k ∂xi − εijkvjωk = −1 ρ ∂P ∂xi + ν ∂2vi ∂xj∂xj + fi 2
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