13.021 { Marine Hydrodynamics, Fall 2004 Lecture 4 Copyright °c 2004 MIT - Department of Ocean Engineering, All rights reserved. 13.021 - Marine Hydrodynamics Lecture 4 Introduction Governing Equations so far: Knowns Number of Equations Number of Unknowns ½ Continuity(conservation of 1 vi 3 mass) Fi Euler (conservation of 3 ¿ij 6 3 of 9 eliminated by momentum) symmetry 4 9 Therefore, some constitutive relationships are needed to relate vi to ¿ ij. 1.7 Newtonian Fluid 1. Consider a fluid at rest (vi ´ 0). Then according to Pascal's Law: ¿ij = ¡ps±ij (Pascal's Law) 2 3 ¡ps 0 0 ¿ = 4 0 ¡ps 0 5 » 0 0 ¡ps where ps is the hydrostatic pressure and ±ij is the Kroenecker delta function, equal to 1 if i = j and 0 if i 6= j. 2. Consider a fluid in motion. The fluid stress is de¯ned as: ¿ij = ¡p±ij +¿ ^ij 1 where p is the thermodynamic pressure and¿ ^ij is the dynamic stress, which is related to the velocities empirically. Experiments with a wide class of fluids, "Newtonian" fluids, obtain that: ¿^ij ¼ linear function of the rate of strain ´ velocity gradient z µ ¶ }| µ ¶{ z}|{ @ @X @ @X @u = k @t @x @x @t @x | {z } m u @uk i.e.¿ ^ij ¼ ®ijkm i; j; k; m = 1; 2; 3 | {z } @xm 34=81 empirical coe±cients (constants for Newtonian fluids) Note that the shear stress is proportional to the rate of strain. τ ij NewtonianFluid Fluid ∂ uk ∂ xm For isotropic fluids, this reduces to: µ ¶ µ ¶ @u @u @u ¿^ = ¹ i + j + ¸ l ; ij @x @x @x j i | {zl } r¢~v where the fluid properties are: ² ¹ - (coe±cient of) dynamic viscosity. ² ¸ - bulk elasticity, `second' coe±cient of viscosity 2 For incompressible flow, @ul = 0. Therefore, for an incompressible, isotropic, Newtonian fluid the viscous @xl stress¿ ^ij is given as µ ¶ @ui @uj ¿^ij = ¹ + @xj @xi 1.8 Navier-Stokes equations Equations Number of Unknowns Number of Equations Unknowns Euler 3 ui 3 continuity 1 p 1 constitutive 6 (symmetry) ¿ij 6 (Newtonian) 10 10 | {z } closure Substitute the equation for the stress tensor µ ¶ @ui @uj ¿ij = ¡p±ij + ¹ + @xj @xi for a Newtonian fluid into Euler's equation: Dui @¿ij ½ = Fi + Dt @xj where µ ¶ @¿ @p @ @u @u ij = ¡ + ¹ i + j @x @x @x @x @x j i | j {zj i } 2 @uj @ ui + @ @xj @xj @xi @x |{z}j 0 and @uj = 0 due to continuity. Finally, @xj 3 2 Dui @ui @ui 1 @p @ ui 1 = + uj = ¡ + º + Fi Tensor form Dt @t @xj ½ @xi @xj@xj ½ D~v @~v 1 1 = + ~v ¢ r~v = ¡ rp + ºr2~v + F~ Vector form Dt @t ½ ½ ¹ 2 where º ´ ½ denoted as the Kinematic viscosity [ L =T ]. ² Navier-Stokes equations for incompressible, Newtonian fluids Number of Number of Equations Unknowns continuity 1 p 1 Navier-Stokes 3 (symmetry) ui 3 4 4 1.9 Boundary Conditions 1. Kinematic Boundary Conditions: Speci¯es kinematics (position, velocity, . ) On a solid boundary, velocity of the fluid = velocity of the body. i.e. velocity continuity. ~v = ~u "no-slip" boundary condition where ~v is the fluid velocity at the body and ~u is the body surface velocity ² ~v ¢ n^ = ~u ¢ n^ no flux { continuous flow ² ~v ¢ t^= ~u ¢ t^ no slip { ¯nite shear stress G v G u 4 2. Dynamic Boundary Conditions: Speci¯es dynamics ( pressure, sheer stress, . ) Stress continuity: 0 p = p + pinterface 0 ¿ij = ¿ij + ¿ij interface The most common example of interfacial stress is surface tension. τ p interface , ij interface τ ' 'p ij τ p ij Surface Tension ² Notation: § [Tension force / Length] ´ [Surface energy / Area]. ² Surface tension is due to the inter molecular forces attraction forces in the fluid. ² At the interface of two fluids, surface tension implies in a pressure jump across the interface. § gives rise to ¢p across an interface. ² For a water/air interface: § = 0.07 N/m. This is a function of temperature, impurities etc. ² 2D Example: dθ dθ dθ cos ¢ ¢p ¢ Rdθ = 2§sin ¼ 2§ 2 | {z2} | {z2} ¼1 dθ ¼ 2 § ) ¢p = R Higher curvature implies in higher pressure jump at the interface. 5 p θ p’=p+ ∆p d /2 Σ R Σ dθ ² 3D Example: Compound curvature µ ¶ 1 1 ¢p = + § R1 R2 where R1 and R2 are the principle radii of curvature. 1.10 Body Forces { Gravity ² Conservative force: F~ = ¡r' for some '; where ' is the force potential. I Z 2 Z 2 ~ ~ F ¢ d~x = 0 or F ¢ d~x = ¡ r' ¢ d~x = '(~x1) ¡ '(~x2) 1 1 ² A special case of a conservative force is gravity. F~ = ¡½gk;^ with gravitational potential: ' = ½gz ! F~ = ¡r' = r(¡½gz) = ¡½gk;^ where ¡½gz is the hydrostatic pressure ps = ¡½gz = ¡'. 6 Navier-Stokes equation: D~v ½ = ¡rp + F~ + ½ºr2~v Dt = ¡rp ¡ r½gz + ½ºr2~v = ¡r (p + ½gz) + ½ºr2~v; but p ¡ ps = pd and ps = ¡½gz, where p is the total pressure and pd is the dynamic pressure. Therefore, D~v ½ = ¡rp + ½ºr2~v Dt d ² Presence of gravity body force is equivalent to replacing total pressure p by dynamic pressure pd in the Navier-Stokes(N-S) equation. ² Solve the N-S equation with pd, then calculate p = pd + ps = pd ¡ ½gz. 7.