NOTES 6 (2024) Derivation of Navier-Stokes Equations PDF

DERIVATION OF NAVIER-STOKE EQUATION 1. Derivation – Reduction steps from Cauchy’s Equation of motion. 𝐷𝒗 𝜌𝒈 − ∇𝑃 + 𝜇∇2 𝒗 = 𝜌 𝐷𝑡 ↓ internal force 2. Applications for simple...

DERIVATION OF NAVIER-STOKE EQUATION 1. Derivation – Reduction steps from Cauchy’s Equation of motion. 𝐷𝒗 𝜌𝒈 − ∇𝑃 + 𝜇∇2 𝒗 = 𝜌 𝐷𝑡 ↓ internal force 2. Applications for simple flow problems – Analytical solution.  DETERMINATION OF FLOW PROPERTIES axe velocity ENGINEERING APPLICATIONS ↳ looks like flow rection Velocity varies at different locations in the pipe. x Flow in a pipe bend y Isothermal flow in a pipe How to determine the flow properties (velocity and pressure) of the flow at various locations (coordinates) in the system? Modelling of a flow system: The velocity field and pressure field can be estimated from equation of motions that are derived from the law of physics:. 1. Conservation of mass: Continuity equation. Energy dissipations 2. Linear momentum balance: Navier-Stokes equation Flow patterns CAUCHY’S EQUATION OF MOTION Operational form of the original form of Cauchy’s equation of motion: The original form of Cauchy’s equation of motions contains vectors of 2nd order tensors which are x-direction: complex to be solved analytically. 𝜕𝑃 𝜕 𝜕 𝜕 𝜕 𝜕 𝜕 𝜕 Original form: 𝜌𝑔𝑥 − + (𝜏𝑥𝑥 ) + (𝜏𝑦𝑥 ) + (𝜏𝑧𝑥 ) = 𝜌𝑣𝑥 + 𝜌𝑣𝑥 𝑣𝑥 + 𝜌𝑣𝑦 𝑣𝑥 + 𝜌𝑣𝑧 𝑣𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕 𝜌𝒈 − ∇𝑃 + ∇ 𝝉 = 𝜌𝒗 + ∇ 𝜌𝒗𝒗 𝜕𝑡 y-direction: 2nd order tensors 𝜕𝑃 𝜕 𝜕 𝜕 𝜕 𝜕 𝜕 𝜕 𝜌𝑔𝑦 − + (𝜏𝑥𝑦 ) + (𝜏𝑦𝑦 ) + (𝜏𝑧𝑦 ) = 𝜌𝑣𝑦 + 𝜌𝑣𝑥 𝑣𝑦 + 𝜌𝑣𝑦 𝑣𝑦 + 𝜌𝑣𝑧 𝑣𝑦 𝑣𝑥 𝑣𝑥 𝑣𝑦 𝑣𝑥 𝑣𝑧 𝑣𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝑣𝑥 𝑣𝑦 𝑣𝑦 𝑣𝑦 𝑣𝑧 𝑣𝑦 𝑣𝑥 𝑣𝑧 𝑣𝑦 𝑣𝑧 𝑣𝑧 𝑣𝑧 z-direction: To overcome the limitation, an alternative form of 𝜕𝑃 𝜕 𝜕 𝜕 𝜕 𝜕 𝜕 𝜕 𝜌𝑔𝑧 − + (𝜏𝑥𝑧 ) + (𝜏𝑦𝑧 ) + (𝜏𝑧𝑧 ) = 𝜌𝑣𝑧 + 𝜌𝑣𝑥 𝑣𝑧 + 𝜌𝑣𝑦 𝑣𝑧 + 𝜌𝑣𝑧 𝑣𝑧 Cauchy’s equation was introduced and is widely 𝜕𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 used. CAUCHY’S EQUATION OF MOTION Actual form of the alternative form of Cauchy’s equation of motion: To eliminate the 2nd order stress tensor, the equation is modified into an alternative form: x-direction: Alternative form: 𝜕𝑃 𝜕 𝜕 𝜕 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜌𝑔𝑥 − + (𝜏𝑥𝑥 ) + (𝜏𝑦𝑥 ) + (𝜏𝑧𝑥 ) = 𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝒗 𝜌𝒈 − ∇𝑃 + ∇ 𝝉 = 𝜌 + 𝒗 ∇𝒗 𝜕𝑡 y-direction: To solve the Cauchy’s equations, the degree of 𝜕𝑃 𝜕 𝜕 𝜕 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 freedom should be zero which depends on the 𝜌𝑔𝑦 − + (𝜏𝑥𝑦 ) + (𝜏𝑦𝑦 ) + (𝜏𝑧𝑦 ) = 𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 number of equations and unknowns: Degree of Freedom Pressure, 𝑃 z-direction: 10 unknowns and 3 Independent Velocities: 𝑣𝑥 , 𝑣𝑦 , 𝑣𝑧 independent 𝜕𝑃 𝜕 𝜕 𝜕 𝜕𝑣𝑧 𝜕𝑣𝑧 𝜕𝑣𝑧 𝜕𝑣𝑧 equations. 𝜌𝑔𝑧 − + (𝜏𝑥𝑧 ) + (𝜏𝑦𝑧 ) + (𝜏𝑧𝑧 ) = 𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 Stress tensors: 𝜏𝑥𝑥 , 𝜏𝑦𝑥 , 𝜏𝑧𝑥 , 𝜏𝑦𝑦 , 𝜏𝑧𝑦 , 𝜏𝑧𝑧 REDUCTION TO NAVIER-STOKES EQUATIONS Stress tensors To reduce the number of unknowns, the components of stress tensor (𝜏𝑥𝑥 , 𝜏𝑦𝑥 , 𝜏𝑧𝑥 , 𝜏𝑦𝑦 , 𝜏𝑧𝑦 and 𝜏𝑧𝑧 ) can be written in terms of the velocity Writing the stress tensors as function of strain rates: components (𝑣𝑥 , 𝑣𝑦 and 𝑣𝑧 ). Replacing stress with velocity 𝜏𝑥𝑥 𝜏𝑥𝑦 𝜏𝑥𝑧 2𝜇𝜀𝑥𝑥 2𝜇𝜀𝑥𝑦 2𝜇𝜀𝑥𝑧 (constitutive equation) 𝜏𝑖𝑗 = 𝜏𝑦𝑥 𝜏𝑦𝑦 𝜏𝑦𝑧 = 2𝜇𝜀𝑦𝑥 2𝜇𝜀𝑦𝑦 2𝜇𝜀𝑦𝑧 𝜏𝑧𝑥 𝜏𝑧𝑦 𝜏𝑧𝑧 2𝜇𝜀𝑧𝑥 2𝜇𝜀𝑧𝑦 2𝜇𝜀𝑧𝑧 Relating the viscous stress tensors with strain rates (constitutive equation): To represent the stress tensors in terms of Notes 𝜏𝑖𝑗 = 2𝜇𝜀𝑖𝑗 velocity components, the strain rate should Note: This relation is only be related to velocity. valid for incompressible where: and Newtonian fluid. 𝑖 − 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑛𝑑 𝑗 − 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 STRAIN RATE TENSOR RELATION TO VELOCITY Strain rate to velocity Two types of strain rate for incompressible deformations: Linear strain rates: 1. Linear strain rate: 𝜕𝑣𝑥 𝜕𝑣𝑦 𝜕𝑣𝑧 ‘rate of increase in length Va 𝜀𝑥𝑥 = 𝜀𝑦𝑦 = 𝜀𝑧𝑧 = 𝜕𝑥 𝜕𝑦 𝜕𝑧 per unit length’ Shear strain rates: 2. Shear strain rate 𝑣𝑦 1 𝜕𝑣𝑥 𝜕𝑣𝑦 1 𝜕𝑣𝑥 𝜕𝑣𝑧 𝜀𝑥𝑦 = + 𝜀𝑥𝑧 = + 2 𝜕𝑦 𝜕𝑥 2 𝜕𝑧 𝜕𝑥 ‘half of the rate of decrease 𝑣𝑦 of the angle between two ∝ 1 𝜕𝑣𝑦 𝜕𝑣𝑥 1 𝜕𝑣𝑦 𝜕𝑣𝑧 𝜀𝑦𝑥 = + 𝜀𝑦𝑧 = + 𝑣𝑥 2 𝜕𝑥 𝜕𝑦 2 𝜕𝑧 𝜕𝑦 initially perpendicular lines ↓ 𝑣𝑥 that intersect at the point’ blueinitial 1 𝜕𝑣𝑧 𝜕𝑣𝑥 1 𝜕𝑣𝑧 𝜕𝑣𝑦 𝜀𝑧𝑥 = + 𝜀𝑧𝑦 = + 2 𝜕𝑥 𝜕𝑧 2 𝜕𝑦 𝜕𝑧 STRAIN RATE TENSOR RELATION TO VELOCITY Stress tensors to velocity The strain rate tensors as a function of velocity components: Hence, the stress tensors as a function of velocity 𝜕𝑣𝜕𝑣 𝑥 𝑥 𝜕𝑣𝑥𝑥 𝜕𝑣 11 𝜕𝑣 𝜕𝑣𝑦𝑦 11 𝜕𝑣 𝜕𝑣𝑥𝑥 𝜕𝑣𝜕𝑣𝑧 𝜀𝑥𝑥 𝜀𝑥𝑥= = 𝜀.1111 𝑥𝑦 = + + 𝜀𝑥𝑧 = 1111 ++ 𝑧 components can be written as 𝜕𝑥𝜕𝑥 22 𝜕𝑦 𝜕𝑦 𝜕𝑥 𝜕𝑥 22 𝜕𝑧𝜕𝑧 𝜕𝑥 𝜕𝑥 1 1 𝜕𝑣𝑦 𝜕𝑣𝑥 𝜕𝑣𝑦𝜕𝑣𝑦 1 1𝜕𝑣𝜕𝑣 𝑦 𝜕𝑣𝜕𝑣 𝑧 𝜀𝑖𝑗 = 𝜀𝑦𝑥 = + 𝜀𝑦𝑦 = 𝜀 = 𝑦 + + 𝑧 1111122 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑦 𝑦𝑧 2 𝜕𝑧 2 𝜕𝑧 𝜕𝑦 𝜕𝑦 𝜏𝑖𝑗 = 2𝜇𝜀𝑖𝑗 1 𝜕𝑣𝑧 𝜕𝑣𝑥 1 1𝜕𝑣𝑧𝜕𝑣 𝜕𝑣𝑦𝜕𝑣𝑦 𝜕𝑣𝜕𝑣 𝑧 𝑧 1 𝜕𝑣 +𝑧 𝜕𝑣𝑥 𝜀𝑧𝑦 2= 𝜕𝑦 + + 𝑧 𝜀 = 𝜀𝑧𝑥2= 𝜕𝑥 + 𝜕𝑧 2 𝜕𝑦 𝜕𝑧 𝜕𝑧 𝑧𝑧 𝜕𝑧 𝜕𝑧 2 𝜕𝑥 𝜕𝑧 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑦 𝜕𝑣𝑥 𝜕𝑣𝑧 2𝜇 𝜇 + 𝜇 + 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑧 𝜕𝑥 𝜀𝑥𝑥 𝜀𝑥𝑦 𝜀𝑥𝑧 𝜕𝑣𝑦 𝜕𝑣𝑥 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑧 𝜏𝑖𝑗 = 𝜇 + 2𝜇 𝜇 + from 𝜀𝑖𝑗 = 𝜀𝑦𝑥 𝜀𝑦𝑦 𝜀𝑦𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑧 𝜕𝑦 𝜀𝑧𝑥 𝜀𝑧𝑦 𝜀𝑧𝑧 𝜕𝑣𝑧 𝜕𝑣𝑥 𝜕𝑣𝑧 𝜕𝑣𝑦 𝜕𝑣𝑧 𝜇 + 𝜇 + 2𝜇 𝜕𝑥 𝜕𝑧 𝜕𝑦 𝜕𝑧 𝜕𝑧 REDUCTION TO NAVIER-STOKES EQUATION FOR X - DIRECTION Stress tensors in x - directions Substituting the stress tensors (in terms of velocity) in Cauchy’s equations: 𝜕𝑃 𝜕 𝜕 𝜕 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑦 𝜕𝑣𝑥 𝜕𝑣𝑧 𝜌𝑔𝑥 − + (𝜏𝑥𝑥 ) + (𝜏𝑦𝑥 ) + (𝜏𝑧𝑥 ) = 𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 2𝜇 𝜇 + 𝜇 + 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑧 𝜕𝑥 𝜕𝑣𝑦 𝜕𝑣𝑥 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑧 𝜏𝑖𝑗 = 𝜇 + 2𝜇 𝜇 + 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑧 𝜕𝑦 𝜕𝑃 𝜕 𝜕𝑣𝑥 𝜕 𝜕𝑣𝑦 𝜕𝑣𝑥 𝜕 𝜕𝑣𝑧 𝜕𝑣𝑥 𝜕𝑣𝑧 𝜕𝑣𝑥 𝜕𝑣𝑧 𝜕𝑣𝑦 𝜕𝑣𝑧 𝜌𝑔𝑥 − + 2𝜇 + 𝜇 + + 𝜇 + 𝜇 + 𝜇 + 2𝜇 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 =𝜌 𝜕𝑡 + 𝑣𝑥 𝜕𝑥 + 𝑣𝑦 𝜕𝑦 + 𝑣𝑧 𝜕𝑧 where: 𝜕𝑣𝑥 𝜏𝑥𝑥 = 2𝜇 For Newtonian fluid: 𝜕𝑥 𝜕𝑣𝑦 𝜕𝑣𝑥 𝜕𝑃 𝜕 2 𝑣𝑥 𝜕 𝜕𝑣𝑦 𝜕𝑣𝑥 𝜕 𝜕𝑣𝑧 𝜕𝑣𝑥 𝜏𝑦𝑥 = 𝜇 + 𝜌𝑔𝑥 − + 2𝜇 + 𝜇 + + 𝜇 + 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑥 2 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑧 𝜕𝑣𝑥 =𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 𝜏𝑧𝑥 = 𝜇 + 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑧 REDUCTION TO NAVIER-STOKES EQUATION FOR X - DIRECTION 𝜕𝑃 𝜕 2 𝑣𝑥 𝜕 𝜕𝑣𝑦 𝜕𝑣𝑥 𝜕 𝜕𝑣𝑧 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜌𝑔𝑥 − + 2𝜇 + 𝜇 + + 𝜇 + = 𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 Modifications 𝜕𝑥 𝜕𝑥 2 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 Expanding the second order differentiation term and rearranging the terms on the left: 2𝜇 =𝜇 +𝜇 𝜕𝑥 2 𝜕𝑥 2 𝜕𝑥 2 𝜕 𝑣𝑦 𝜕 𝑣𝑦 𝜕𝑃 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜕 𝜕𝑣𝑦 𝜕 𝜕𝑣𝑥 𝜕 𝜕𝑣𝑧 𝜕 𝜕𝑣𝑥 𝜇 =𝜇 𝜌𝑔𝑥 − +𝜇 + 𝜇 + 𝜇 + 𝜇 + 𝜇 + 𝜇 𝜕𝑦 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑥 2 𝜕𝑥 2 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑧 𝜕𝑃 𝜕 2 𝑣𝑥 𝜕 𝜕𝑣𝑥 𝜕 𝜕𝑣𝑦 𝜕 2 𝑣𝑥 𝜕 𝜕𝑣𝑧 𝜕 2 𝑣𝑥 𝜌𝑔𝑥 − +𝜇 +𝜇 +𝜇 +𝜇 +𝜇 +𝜇 𝜕𝑥 𝜕𝑥 2 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 2 𝜕𝑥 𝜕𝑧 𝜕𝑧 2 𝜕𝑃 𝜕 𝜕𝑣𝑥 𝜕𝑣𝑦 𝜕𝑣𝑧 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜌𝑔𝑥 − +𝜇 + + + + + = 𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 REDUCTION TO NAVIER-STOKES EQUATION FOR X - DIRECTION 𝒙 − 𝒅𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏: 𝜕𝑃 𝜕 𝜕𝑣𝑥 𝜕𝑣𝑦 𝜕𝑣𝑧 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜌𝑔𝑥 − +𝜇 + + + + + = 𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 Writing for other cartesian coordinate directions: 𝒚 − 𝒅𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏: 𝜕𝑃 𝜕 𝜕𝑣𝑥 𝜕𝑣𝑦 𝜕𝑣𝑧 𝜕 2 𝑣𝑦 𝜕 2 𝑣𝑦 𝜕 2 𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜌𝑔𝑦 − +𝜇 + + + + + = 𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 𝜕𝑦 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝒛 − 𝒅𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏: 𝜕𝑃 𝜕 𝜕𝑣𝑥 𝜕𝑣𝑦 𝜕𝑣𝑧 𝜕 2 𝑣𝑧 𝜕 2 𝑣𝑧 𝜕 2 𝑣𝑧 𝜕𝑣𝑧 𝜕𝑣𝑧 𝜕𝑣𝑧 𝜕𝑣𝑧 𝜌𝑔𝑧 − +𝜇 + + + + + = 𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 REDUCTION TO NAVIER-STOKES EQUATION - SIMPLIFICATIONS Navier-Stokes equation Simplifications are performed to simplify the equation 0 of motions and reduce the steps to solve the equations. 𝜕𝑃 𝜕 𝜕𝑣𝑥 𝜕𝑣𝑦 𝜕𝑣𝑧 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜌𝑔𝑥 − +𝜇 + + + + + 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 From continuity equation, for incompressible fluid: 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 =𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 0 𝜕𝜌 𝜕 𝜕 𝜕 + (𝜌𝑣𝑥 ) + (𝜌𝑣𝑦 ) + (𝜌𝑣𝑧 ) = 0 𝜕𝑃 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜌𝑔𝑥 − +𝜇 + + = 𝜌 + 𝑣 + 𝑣 + 𝑣 𝑥 𝑦 𝑧 𝜕𝑥 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑣𝑥 𝜕𝑣𝑦 𝜕𝑣𝑧 𝜕𝑥 + 𝜕𝑦 + 𝜕𝑧 =0 Similarly, for 𝑦 and 𝑧 – directions: 𝜕𝑃 𝜕 2 𝑣𝑦 𝜕 2 𝑣𝑦 𝜕 2 𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 In compact notation: ∇ 𝒗=0 𝜌𝑔𝑦 − 𝜕𝑦 +𝜇 𝜕𝑥 2 + 𝜕𝑦 2 + 𝜕𝑧 2 = 𝜌 𝜕𝑡 + 𝑣𝑥 𝜕𝑥 + 𝑣𝑦 𝜕𝑦 + 𝑣𝑧 𝜕𝑧 𝜕𝑃 𝜕 2 𝑣𝑧 𝜕 2 𝑣𝑧 𝜕 2 𝑣𝑧 𝜕𝑣𝑧 𝜕𝑣𝑧 𝜕𝑣𝑧 𝜕𝑣𝑧 𝜌𝑔𝑧 − +𝜇 + + = 𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 𝜕𝑧 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 NAVIER-STOKES EQUATION 𝑪𝒐𝒎𝒑𝒂𝒄𝒕 𝒏𝒐𝒕𝒂𝒕𝒊𝒐𝒏𝒔 COMPACT FORM 2 𝜕2 𝜕2 𝜕2 ∇ = 2+ 2+ 2 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝐷 𝜕 𝜕𝑃 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 is defined as +𝒗∙𝛁 𝜌𝑔𝑥 − +𝜇 + + = 𝜌 + 𝑣 + 𝑣 + 𝑣 𝐷𝑡 𝜕𝑡 𝜕𝑥 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝜕𝑡 𝑥 𝜕𝑥 𝑦 𝜕𝑦 𝑧 𝜕𝑧 Navier-Stokes solution 𝜕𝑃 𝜕 2 𝑣𝑦 𝜕 2 𝑣𝑦 𝜕 2 𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜌𝑔𝑦 − +𝜇 + + = 𝜌 + 𝑣 + 𝑣 + 𝑣 𝜕𝑦 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝜕𝑡 𝑥 𝜕𝑥 𝑦 𝜕𝑦 𝑧 𝜕𝑧 Variables/unknowns: 𝜕𝑃 𝜕 2 𝑣𝑧 𝜕 2 𝑣𝑧 𝜕 2 𝑣𝑧 𝜕𝑣𝑧 𝜕𝑣𝑧 𝜕𝑣𝑧 𝜕𝑣𝑧 𝑃, 𝑣𝑥 , 𝑣𝑦 and 𝑣𝑧 𝜌𝑔𝑧 − +𝜇 + + = 𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 𝜕𝑧 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 Equation of motions: Writing the equation using compact notations: Three Navier-Stokes equations 𝜕𝑃 𝐷𝑣𝑥 𝜌𝑔𝑥 − + 𝜇∇2 𝑣𝑥 = 𝜌 𝜕𝑥 𝐷𝑡 𝐷𝒗 𝐷𝒗 2 𝜌𝒈 − ∇𝑃 + 𝜇∇ 𝒗 = 𝜌 𝜌𝒈 − ∇𝑃 + 𝜇∇2 𝒗 = 𝜌 𝜕𝑃 𝐷𝑣𝑦 𝐷𝑡 𝐷𝑡 𝜌𝑔𝑦 − + 𝜇∇2 𝑣𝑦 = 𝜌 𝜕𝑦 𝐷𝑡 For Cartesian coordinate Continuity equation 𝜕𝑃 𝐷𝑣𝑧 𝜌𝑔𝑧 − + 𝜇∇2 𝑣𝑧 = 𝜌 𝜕𝑧 𝐷𝑡 ∇ 𝒗=0 NAVIER-STOKES EQUATION APPLICATIONS In the derivation of Navier-Stokes equations, assumptions Navier-Stokes solution were made which then governed the applicability of the equations: 𝜕𝑃 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜌𝑔𝑥 − +𝜇 + + = 𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 𝜕𝑥 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 1. Newtonian fluids: Constant kinematic viscosity, 𝜇 𝜕𝑃 𝜕 2 𝑣𝑦 𝜕 2 𝑣𝑦 𝜕 2 𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 2. Incompressible fluids: Constant density, 𝜌 𝜌𝑔𝑦 − +𝜇 + + = 𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 𝜕𝑦 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 For non-Newtonian fluid, the viscous friction terms need 𝜕𝑃 𝜕 2 𝑣𝑧 𝜕 2 𝑣𝑧 𝜕 2 𝑣𝑧 𝜕𝑣𝑧 𝜕𝑣𝑧 𝜕𝑣𝑧 𝜕𝑣𝑧 to be modified to include the rheology of the fluid. 𝜌𝑔𝑧 − +𝜇 + + = 𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 𝜕𝑧 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 APPLICATION OF NAVIER-STOKES EQUATION - EXAMPLE SOLUTION Consider a steady, two-dimensional, incompressible velocity field with 𝒗 = 𝑣𝑥 , 𝑣𝑦 = 𝑎𝑥 + 𝑏 𝑖 + −𝑎𝑦 + 𝑐𝑥 𝑗, where 𝑎, 𝑏 and 𝑐 Governed how the problem are constants: 𝑎 = 0.05 𝑠 −1 , 𝑏 = 1.5 𝑚Τ𝑠 and 𝑐 = 0.35 𝑠 −1. Assumptions: will be solved Determine the pressure in the flow as a function of 𝑥 and 𝑦. Flow is steady and incompressible. The velocity field is given as: Fluid has constant properties. 𝑣𝑥 = 𝑎𝑥 + 𝑏 Flow is two-dimensional. 𝑣𝑦 = −𝑎𝑦 + 𝑐𝑥 Gravity is negligible in both x and y directions. APPLICATION OF NAVIER-STOKES EQUATION - EXAMPLE 𝑹𝒆𝒍𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕𝒔 From the N-S equations for the directions involved, perform simplifications based on the assumptions made: Check the relations of velocity components x-direction: for further simplifications: 0 0 0 0 0 0 𝜕𝑃 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝜕𝑣𝑥 𝑣𝑥 = 𝑎𝑥 + 𝑏 𝑣𝑥 (𝑥) 𝜌𝑔𝑥 − +𝜇 + + = 𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 𝜕𝑥 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 no gravity steady 𝑣𝑥 is only a function of 𝑥 y-direction: 0 0 0 𝑣𝑦 = −𝑎𝑦 + 𝑐𝑥 𝑣𝑦 (𝑥, 𝑦) 0 𝜕𝑃 𝜕 2 𝑣𝑦 𝜕 2 𝑣𝑦 𝜕 2 𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜌𝑔𝑦 − +𝜇 + + = 𝜌 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 𝜕𝑦 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝑣𝑦 is a function of 𝑥 and 𝑦 APPLICATION OF NAVIER-STOKES EQUATION - EXAMPLE 𝑬𝒙𝒑𝒓𝒆𝒔𝒔𝒊𝒐𝒏𝒔 𝒇𝒐𝒓 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕𝒊𝒂𝒍 𝒕𝒆𝒓𝒎𝒔 Solve the simplified N-S equations: 𝑣𝑥 = 𝑎𝑥 + 𝑏 2 𝜕𝑃 𝜕 𝑣𝑥 𝜕𝑣𝑥 =𝜇 − 𝜌 𝑣𝑥 cause 𝜕 𝜕𝑥 𝜕𝑥 2 𝜕𝑥 se either 𝜕𝑥 𝜕 2 𝑣𝑥 𝜕𝑣𝑥 =0 =𝑎 𝜕𝑃 2 𝜕 𝑣𝑦 𝜕 𝑣𝑦 2 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑥 2 𝜕𝑥 =𝜇 + − 𝜌 𝑣𝑥 + 𝑣𝑦 𝜕𝑦 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑥 𝜕𝑦 𝜕 𝜕𝑥 Substituting the expressions for differential terms: 𝑣𝑦 = −𝑎𝑦 + 𝑐𝑥 𝜕𝑃 = 𝜌 −𝑎2 𝑥 − 𝑎𝑏 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑥 = −𝑎 =𝑐 𝜕𝑦 𝜕𝑥 𝜕𝑃 = 𝜌 −𝑏𝑐 + 𝑎2 𝑦 𝜕𝑦 𝜕 2 𝑣𝑦 𝜕 2 𝑣𝑦 =0 =0 𝜕𝑥 2 𝜕𝑦 2 APPLICATION OF NAVIER-STOKES M - > EQUATION - EXAMPLE > > - > - SOLUTION 𝜕𝑃 𝜕𝑃 = 𝜌 −𝑎2 𝑥 − 𝑎𝑏 and = 𝜌 −𝑏𝑐 + 𝑎2 𝑦 Substituting relation of 𝜕𝑃Τ𝜕𝑥: 𝜕𝑥 𝜕𝑦 𝑑ℎ Integration of either x or y components will yield 𝑃(𝑥, 𝑦) : = 𝜌 −𝑎2 𝑥 − 𝑎𝑏 𝑑𝑥 Integration To find ℎ 𝑥 , integrate 𝑑ℎΤ𝑑𝑥: 𝜕𝑃 = 𝜌 −𝑏𝑐 + 𝑎2 𝑦 න 𝜕𝑃 = න 𝜌 −𝑏𝑐 + 𝑎2 𝑦 𝜕𝑦 𝜕𝑦 New unknown 2 𝑎2 𝑥 2 ℎ 𝑥 = න 𝜌 −𝑎 𝑥 − 𝑎𝑏 𝑑𝑥 = 𝜌 − − 𝑎𝑏𝑥 + 𝐶 variable has 2 2 𝑃 𝑥, 𝑦 = 𝜌 න −𝑏𝑐 + 𝑎 𝑦 𝜕𝑦 = 𝜌 −𝑏𝑐𝑦 + 2 % 𝑎2 𝑦 2 + ℎ(𝑥) appeared Substituting ℎ 𝑥 to get 𝑃 𝑥, 𝑦 : To find ℎ(𝑥), differentiate 𝑃(𝑥, 𝑦) with respect to 𝑥 to get: 𝑎2 𝑦 2 𝑎2 𝑥 2 𝜕𝑃 𝑑ℎ 𝑃 𝑥, 𝑦 = 𝜌 −𝑏𝑐𝑦 + +𝜌 − − 𝑎𝑏𝑥 + 𝐶 = & 2 2 F 𝜕𝑥 𝑑𝑥 G C are different