Fluid Mechanics #12 - Momentum Equations: From Integral to Local Form | Real Fluids

Video Description In this lecture (Section I.6), we derive the Momentum Equations in their local form from the Momentum Theorem. Starting from the integral form, we introduce the viscous stress tensor and apply the Divergence and Gradient Theorems to transform surface integrals into volume integrals. By equating the integrands, we obtain the first local form of the Momentum Equations. We then derive the second local form by expanding the derivatives of products, leading to the fundamental equation linking density, acceleration, and forces per unit volume. We also introduce the concept of driving pressure for incompressible fluids and discuss the special case of a perfect fluid (Euler's Equations). Detailed Structure (Chapters) 1. Introduction and Objective Determine the local form of the Momentum Theorem (valid at every point and every instant). Starting Point: The Momentum Theorem in integral form, projected along Oxi (Equation I.16.1). 2. Introducing Viscous Stresses: Reminder of the viscous friction stress expression: (fv)i = τij nj Substitution into the integral equation (Equations I.92 and I.93). 3. Transition to Volume Integrals Theorems Used: a) Divergence Theorem applied to the last terms of the first and second members. b) Gradient Theorem applied to the pressure term. Transformation: Converting surface integrals into volume integrals (Equations I.95 to I.97). Remark: The integrals are now taken over the same volume v. 4. First Local Form of the Momentum Equations Grouping Terms: Since the integrals are over the same domain, the integrands can be equated. Resulting Equations: Equation (I.100): ∂/∂t (ρVi) + ∂/∂xj (ρVi Vj) = ρ gi − ∂p/∂xi + ∂τij/∂xj Equation (I.101): ∂/∂t (ρVi) + ∂/∂xj (ρVi Vj) = ρ gi + ∂σij/∂xj Remark: For a perfect fluid, equation (I.100) corresponds to the first expression of Euler's Equations. 5. Second Local Form of the Momentum Equations Expansion: Developing the derivatives of products in the first member (Equations I.102 and I.103). Simplification: Using the continuity equation (E.C. = 0) to eliminate terms. Fundamental Result: Equation (I.104): ρ (dVi/dt) = ρ gi − ∂p/∂xi + ∂τij/∂xj Equation (I.105): ρ (dVi/dt) = ρ gi − ∂p/∂xi + ∂σij/∂xj 6. Physical Interpretation Newton's Second Law: The first member [ρ (dVi/dt)] represents mass × acceleration per unit volume. Forces per Unit Volume: The second member groups the resultant of external forces (gravity, pressure, viscous friction). Nature of Forces: Surface forces appear as spatial derivatives (gradient of pressure, divergence of stress tensors). 7. Special Case: Incompressible Fluid (ρ = const) Grouping Pressure and Gravity: Introducing the Driving Pressure p* = p + ρ g z Final Equation (I.109): ρ (dVi/dt) = - ∂p*/∂xi + ∂τij/∂xj Interpretation: Only the gradient of driving pressure produces fluid acceleration. Viscous friction forces are resistive and cannot generate motion on their own. 8. Remark on Perfect Fluids Equation (I.104) reduces to the second expression of Euler's Equations: ρ (dVi/dt) = - ∂p*/∂xi ρ (dVi/dt) = ρ gi - ∂p/∂xi Playlist Context This video is part of the playlist "5. Incompressible Fluid Mechanics". It follows the lecture on stresses in real fluids and provides the foundation for the Navier-Stokes equations. Useful Links: Previous Session (Session 9):    • Fluid Mechanics #9 - Drag and Lift of an A...   Watch the Full Playlist on "Incompressible Fluid Mechanics" here:    • 5. Incompressible Fluid Mechanics   Watch the Full Playlist on "Perfect Fluid Mechanics" here:    • 1. Mécanique des Fluides Parfaits   Tags fluid mechanics, real fluid, stresses, viscosity, pressure, stress tensor, viscous friction, Couette flow, shear stress, Kronecker delta, continuum mechanics, physics, civil engineering, process engineering, undergraduate, graduate, exam prep, tutorial Hashtags:#FluidMechanics #RealFluid #StressTensor #viscosity #pressure #ViscousFriction #CouetteFlow #ShearStress #ContinuumMechanics #physics #engineering