CSU ATS601 Fall 2015 2 Equations of Motion In this section, we will derive the six full equations of motion in a non-rotating, Cartesian coordinate system. 2.1 Six equations of motion (non-rotating, Cartesian coordinates) For reference, the six non-rotating equations of fluid motion in Cartesian coordinates. There are 5 prognostic equations, i.e. provide information on how the fluid will evolve, and 1 diagnostic equation, i.e. describing the instantaneous state of the fluid. Continuity Equation; Mass Conservation (1): D⇢ + ⇢ ( v) = 0 (2.1) Dt r· or (2.2) @⇢ + (⇢v) = 0 (2.3) @t r· Momentum (Navier-Stokes) Equations; Momentum Conservation (3): Dv p =-r - Φ + ⌫ 2v (2.4) Dt ⇢ r r Thermodynamic Equation; Energy Conservation (1): D✓ ✓ Q˙ p R/cp = , where ✓ = T 0 (2.5) Dt T c p p ✓ ◆ Equation of State (1): p = f(✓, ⇢), e.g. p = ⇢RT (2.6) 2.2 Continuity (conservation of mass) Mass is absolutely conserved in classical mechanics, and so, the conservation of mass is arguably our first building block in describing the evolution of a fluid. 2.2.1 an Eulerian perspective Consider an infinitesimal cube control volume ∆V = ∆x∆y∆z. Fluid moves into and out of the cube at its faces. E. A. Barnes 10 updated 18:35 on Friday 25th September, 2015 CSU ATS6011.2 The Mass Continuity Equation 9 Fall 2015 Fig. 1.1 Mass conservation in a cubic Eulerian control volume. The accumulationface, ofinto fluid the control within volume, the volume is due to flow in the x-direction can be written as .⇢u/ .⇢u/ ıyız@(⇢u) ∆y∆z[(⇢u)(x, y, z)-(⇢ux)(xArea+ ∆x, y, z)]x = - | ∆x∆y∆z.(1.23) (2.7) ⇥ D @x x,y,z where u is the component of velocity in the x-direction, and the subscript x here denotes Similarly, in thethe y-direction coordinate of the argument. A small distance to the right the flow out of the control @(⇢v) volume is - ∆x∆y∆z (2.8) .⇢@uy/x ıxıyız: (1.24) C and z-directionThus, the accumulation of fluid within the control volume, due to motion in the x- direction only, is @(⇢w) - ∆x∆y∆z. (2.9) @z @.⇢u/ ıyızŒ.⇢u/x .⇢u/x ıxç ıxıyız: (1.25) Note the minus signs - this is because having moreC leaveD the@ rightx face compared to the left face (a pos- itive derivative)To thiswould must imply be added a mass the effectsdivergence of motion, or, in mass the y leaving- and z-directions, the cube. namely Thus, the minus sign is the accumulation within the control volume. @.⇢v/ @.⇢w/ ıxıyız: (1.26) @y C @z Thus, the total net accumulation of mass within the control volume can be written as This net accumulation of fluid must@(⇢u be) accompanied@(⇢v) @( by⇢w a) corresponding increase of fluid mass within the control-∆V volume. This+ is + (2.10) @x @y @z @ ✓ @⇢◆ In order to have mass conservation. (somethingDensity Volume we are/ assumingıxıyız here,; that is, it does(1.27) not fall out of the @t ⇥ D @t equations themselves)because the the volume net accumulation is constant. Thus, must because be balanced mass isby conserved, a corresponding (1.25), (1.26) increase and of fluid mass within the control(1.27) volume. give That is, @⇢ @.⇢u/ @.⇢v/ @.⇢w/ ıxıyız 0: (1.28) @(∆V⇢) @@⇢t C @x C @(@⇢yu)C @@(z⇢v) D@(⇢w) = ∆V =-∆V + + (2.11) @t @t @x @y @z Because the control volume is arbitrary the✓ quantity in square brackets◆ must be zero using the factzero that and the wevolume have theof themass cube continuity does not equation: change in time. Moving all of the terms to the left hand @⇢ side, .⇢v/ 0: (1.29) @⇢ @@(⇢t uCr) @(⇢v)D @(⇢w) ∆V + + + = 0 (2.12) @t @x @y @z ✓ ◆ Because the control volume is arbitrary, the quantity in the parentheses must be zero, and thus, the continuity equation is @⇢ @(⇢u) @(⇢v) @(⇢w) @⇢ + + + = + (⇢v)=0. (2.13) @t @x @y @z @t r· E. A. Barnes 11 updated 18:35 on Friday 25th September, 2015 CSU ATS601 Fall 2015 2.2.2 a Lagrangian perspective One come perform a comparable derivation in terms of a fixed fluid parcel to derive the continuity equation from a Lagrangian perspective, but we will not do that here. However, we take a moment to derive the advective/Lagrangian form of continuity starting from 2.13. Using our vector formulas, we know that ( a)= a + a . (2.14) r· r· ·r Thus, @⇢ @⇢ + (⇢v)= + ⇢ v + v ⇢ = 0 (2.15) @t r· @t r· ·r Using the definition of the material derivative (1.11), this leads to the Lagrangian form of mass continuity: D⇢ + ⇢ v = 0. (2.16) Dt r· Note that if the flow is incompressible, then ⇢ = constant and starting from 2.15, the continuity equation becomes @⇢ + ⇢ v + v ⇢ = 0 (2.17) @t r· ·r 0 + ⇢ v + 0 = 0 (2.18) r· and since ⇢ is constant (2.19) v = 0 (2.20) r· Note that the first numerical weather forecast was of the (seemingly simple) equation: D(f+⇣) Dt = 0. 2.3 Momentum equations The momentum equations describe how the momentum (or velocity mass) evolves in time in response ⇥ to both internal and imposed forces. Throughout this course, we will be mostly focused on imposed forces from outside of the fluid, rather than internal forces such as viscosity. Depending on the fluid of interest, there can be a wide range of forces acting on a fluid parcel. The basic starting point is Newton’s 2nd Law: the acceleration of an object depends on the force acting on the object divided by its mass. Or, put another way, F = ma. (2.21) E. A. Barnes 12 updated 18:35 on Friday 25th September, 2015 CSU ATS601 Fall 2015 From this, we can solve for the acceleration a of the fluid element as a function of f which is a force per unit volume: F F f = = = a (2.22) m ⇢V ⇢ However, the acceleration of the parcel is by definition the material derivative of the velocity of the parcel, and so: F f Dv = = a = @tv +(v )v (2.23) m ⇢ ⌘ Dt ·r The goal of deriving the momentum equations (Navier-Stokes equations) is to qauntify all of the forces f acting on the parcel. Here, we will describe the dominant forces acting on earth’s atmosphere and ocean: pressure and gravitational forces with a short comment on internal viscous forces. That is, Dv Fg Fp Fv = @tv +(v )v = + + (2.24) Dt ·r m m m 2.3.1 Gravity Newton’s law of gravity states that: GMm r Fg =- , (2.25) r2 r where the term on the far right is there to provide a direction pointing between the centers of mass of one object with mass M and a second object with mass m. G is the gravitational constant. For gravity on earth, M is the mass of the earth, and r = a + z where a is the radius of the earth. This results in an acceleration of mass m due to gravity ag of Fg fg 1 = =-g kˆ -gkˆ -g kˆ (2.26) m ⇢ 0 (1 + z/a)2 ⌘ ⇡ 0 where g = GM/(a2) 9.81 m/s2, and g = g /(1 + z/a)2. Note that we have assumed that z<<a. 0 ⇡ 0 It turns out it will be useful to define a quantity Φ =-GM/r =-GM/(a + z) called the gravitational potential, which allows us to write -gkˆ without the kˆ notation: -gkˆ =- Φ. (2.27) r Since Φ is the gradient of Φ, in Cartesian coordinates, r @ @ @ - Φ = ˆi + ˆj + kˆ (2.28) r @x @y @z E. A. Barnes 13 updated 18:35 on Friday 25th September, 2015 CSU ATS601 Fall 2015 From the definition of Φ we know that it is only a function of z, and thus, only changes in the local vertical direction. Thus, @ -GM - Φ = 0 + 0 + kˆ (2.29) r @z (a + z) ✓ ◆ @ 1 =-GM kˆ (2.30) @z (a + z) ✓ ◆ 1 =-GM kˆ (2.31) (a + z)2 GM 1 =- kˆ (2.32) a2 (1 + z/a)2 =-gkˆ (2.33) Thus, the acceleration due to the gravitational force is Fg =- Φ. (2.34) m r We have our first acceleration to add to the right-hand-side of our momentum equation: Dv =- Φ + ... (2.35) Dt r 2.3.2 Pressure gradient force Pressure p is a scalar field defines as the force per unit area (A). Pressure always acts normal to the face of the area (denotedn ˆ), that is, it always acts in the direction that makes a 90 degree angle with the surface of the area. Thus, the pressure force acting on an area is Fp = pAnˆ. p y z p∂x p x y z z z y y x x Consider a fluid element, say a cube, of volume ∆V = ∆x∆y∆z. If the pressure is the same on all sides of the cube, it is probably intuitive to you that the no net force will be felt. Now instead consider the case where the pressure force on the left-hand-side of the cube is FLHS = p∆y∆z (2.36) E. A. Barnes 14 updated 18:35 on Friday 25th September, 2015 CSU ATS601 Fall 2015 and the pressure force on the right-hand-side of the cube is FRHS =(p + ∆p)∆y∆z (2.37) due to a slightly different pressure of (p + ∆p). It is the pressure difference, ∆p that determines the net force felt on the cube.