Lesson 4 Equations for a Nonrotating Fluid Objectives The state of the atmosphere or ocean can be defined by 5 variables – Fundamental Forces – u, v, w, P and T • Inertial Reference Frame • Recall, using the EOS ρ can be found from P and T – Gravitational Force – Pressure Gradient Force (PGF) We therefore need 5 independent equations – Friction (Viscosity) Force (with boundary conditions) to determine the • Non-inertial Reference Frame evolution of a fluid – Centrifugal Force – The Laws of motion (3 directions) – Coriolis Force – Conservation of Mass – Conservation of Mass – The 1st Law of Thermodynamics • Continuity Equation Fundamental Forces Law of Motion (N2L) In lesson 1 we made an assumption - that the Inertial frame of reference atmosphere and ocean were continuous in – A non-accelerating coordinate system structure and properties at our scale of – Follow the same parcel around - therefore interest the time derivative is a total derivative – This means they are Newtonian fluids and therefore = can use N2L, F m a , to describe acceleration N2L becomes: • The approach is slightly different than we saw in Physics DU – In Physics we dealt with discrete objects of constant mass F = ρδxδyδz – In Fluids we deal with a continuous field of matter with varying Dt density – where: DU ∂U = + (U •∇)U Dt ∂t Gravitational Force (g*) Pressure Gradient Force (PGF) Newton’s Law of Universal Gravitation P = F/A Every particle in the universe attracts every other particle wit h a force that is proportional to the product of their masses and inversely propor tional to the Force due to variation of pressure within square distance between them. The force acts along the line joi ning the two particles. a fluid P (x ,y ,z ) Mm 0 0 0 0 F ≡ g* = −G r g r 2 P P+ δP z δ z δy y m F M δ r g ≡ = − x g* G 2 r m a x F 1 ∂P ∂P ∂P 1 P = − ( iˆ + ˆj + kˆ ) = − ∇P ρ ∂ ∂ ∂ ρ M m x y z Pressure Gradient Force (PGF) Frictional (Viscous) Force Molecular friction Example – Accelerates (±) a body that is in contact with another – Consider a SLP of 1000.3 hPa from a body moving at a different speed station near the center of a low, with a – Molecules collide with those of another • If one object moves (has momentum: p = mv), the other temperature of 9 C. 500 km away a object will receive some of it and move second station reports a SLP of 1000.9 Eddies can exchange momentum and therefore hPa. What is the pressure gradient force? cause frictional forces • Hint: IGL is needed – Wind experiences a retarding force: F fr (ground) – The ground experiences an accelerating force • Free moving surface - waves, currents… Viscous Force Viscous Force Force on upper plate µAu ∂u A fluid is a state that continuously yields to F = 0 =µA any shear stress d ∂z µ Here a solid exerts a shear stress on a (“mu”): dynamic (or shear) viscosity fluid coefficient Property of the fluid This force must just equal the force z = d u0 exerted by the upper plate on the fluid u(l) = u 0 immediately below it (N3L) u(z) State of uniform motion u(0) = 0 Every horizontal layer of fluid must exert the z = 0 same force on the fluid below it Viscous Force Shear Stress The shear stress - τ (“tau”) - exerted on the Defined steady state flow - no fluid by the upper plate is proportional to the acceleration and therefore no net velocity of the plate and inversely proportional viscous force to the distance between the plates (d) – The shear stress (internal friction) does not The x-component of the shear stress can be cause a net force written: ∂u τ iˆ =µ ∂z So, there must be a gradient in the – It is a flux of momentum shear stress ∂τ 2 • Momentum per unit time per unit area 1 µ ∂ ∂µ ∂ u = ( ) = υ υ = µ / ρ ρ ∂z ρ ∂z ∂z ∂z2 Shear Stress Apparent Forces υ (“nu”) is the kinematic viscosity coefficient – Atmosphere: 1.46 x 10 -5 m2/s Non-inertial reference frame – Ocean: 1.0 x 10 -6 m2/s – An accelerating reference frame Shear stress is a vector – Uniform circular motion – Shear stress has nine components (3D) • Centrifugal force • Coriolis force ∂ 2 ∂ 2 ∂ 2 = υ∇ 2 ˆ = υ u + u + u F u Fshear i ( ) x ∂x 2 ∂y 2 ∂z2 = υ∇2 Fy v = υ∇ 2 Fz w Centrifugal Acceleration Effective Gravity Appears to deflect the moving body outward from center of curvature Resultant of gravitational and centrifugal – Equal and opposite to centripetal acceleration forces – Always perpendicular to Earth’s axis; directed Directed perpendicular to local tangent outward of Earth, not toward the center F cf = Ω2R m Coriolis Force Coriolis Acceleration The force that appears to act on a body moving relative to a rotating system, when The coriolis acceleration describes the viewed from that system tendency for fluid parcels to turn In the absence of other forces, a parcel Merry-Go-Round Example would accelerate as: DU = − Ω× 2 2 U 1 Ω Dt Coriolis Force By definition: CF does no work, it only 3 acts to change the direction of flow Coriolis Acceleration: Derivation Coriolis Acceleration: Derivation Recall the coriolis parameter: f = 2Ωsin φ dw = Ω φ 2 ucos φ dt du = Ω φ d = dv Horizontal 2 sin (a ) fv = −2Ωusin φ = − fu dt dt dt Vertical du dz = −2Ωcos φ( ) = −2Ωcos φw dt dt Coriolis Acceleration: Full 3-D Coriolis Acceleration du = fv − 2Ωcos φw Example dt – The Coriolis acceleration can be calculated using observations. dv = − fu – City X, has a latitude of 40N and reports a dt wind of 5 kts from the North. – Calculate the Coriolis acceleration dw = 2Ωu cos φ dt Basic Laws of Physics Conservation of Mass Conservation of: Cannot create or destroy mass - 1 kg of a – Mass fluid will always be 1 kg of that fluid • Leads to Continuity Equation – If fluid is compressible then the shape will • SO335 and SO414 change – Momentum • Leads to EOM Continuity Equation • SO414 – Provides relationship between horizontal and – Energy vertical motions • Leads to 1st Law of Thermodynamics • SO345 Parcel A Parcel B Continuity Equation Continuity Equation Consider: ρ – A cube fixed in space thru which a fluid flows Mass Flux at the center of the cube is u – Mass is continually advected through the Using Taylor's expansion about the sides by the fluid motion center, we get in/out flow per unit area • Net inflow of mass thru sides = Rate of through sides A and B accumulation inside the volume ∂ ∂ ∂x A ρ B ∂ ∂x – Mass flux = M ρu− (ρu) u ρu+ (ρu) ∂x 2 δz ∂x 2 ∂t z δy y δx x Continuity Equation Continuity Equation Can get the net rate of flow into the volume due to the x component of velocity Mass Divergence form of mass continuity Then, the net rate of mass inflow per unit ∂ρ ∂ ∂ ∂ volume for all directions is: = −[ (ρu)+ (ρv)+ (ρw)] = −∇⋅(ρv ) ∂ t ∂x ∂y ∂z ∂M ∂ ∂ ∂ = −[ ( ρu ) + ( ρv) + ( ρw )] δxδyδz ∂t ∂x ∂y ∂z Velocity Divergence form of mass continuity 1 dρ + ∇⋅ v = 0 ρ dt The Boussinesq Approximation Incompressibility dρ If density is constant , = 0 then the fluid dt Joseph Boussinesq (1842-1929) noted is incompressible that we can safely assume density is 1 dρ constant except when it is multiplied by + ∇⋅ v = 0 and ρ becomes: gravity in calculations of pressure dt ∇ ⋅ v = 0 The assumption will greatly simplify the – Density MUST be constant (incompressible) equations of motion! – Don’t consider vertical length scales so large that hydrostatic pressure causes density variations Incompressibility Example - Ocean We can easily use ∇ ⋅ v = 0 to learn We have an offshore wind, and must about vertical motions solve for the resultant upwelling (i.e. vertical velocity) Expanded: ∂u ∂v ∂w Assume incompressible flow (why is this + + = 0 a good assumption?) ∂x ∂y ∂z N v=10 cm/s or 100 km ∂u ∂v ∂w + = − 100 m ∂x ∂y ∂z Example - Atmosphere Notes Low-level convergence (1m/s each), solve Conservation laws applied to flow in the for the vertical velocity 100 m from the atmosphere and ocean lead to equations of surface if v = 0 and dx = 10 m. motion Flow in the ocean can be assumed to be incompressible except when describing sound Density can be assumed to be constant except * when multiplied by gravity – Boussinesq approximation z Conservation of mass leads to the continuity 100 m y equation, which has an especially simple form u u 1 2 for an incompressible fluid (why?) x dx.