Lect-02-P4 - M.Sc. in Computational Science Fundamentals of...

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Unformatted text preview: M.Sc. in Computational Science Fundamentals of Atmospheric Modelling ´ Peter Lynch, Met Eireann Mathematical Computation Laboratory (Opp. Room 30) Dept. of Maths. Physics, UCD, Belfield. January–April, 2004. Lecture 2 The Continuity Equation 2 Geophysical Fluid Dynamics Geophysical Fluid Dynamics (GFD) is the study of the dynamics of the fluid systems of the earth and planets. The principal fluid systems in which we are interested are the atmosphere and the oceans. Inland waters such as lakes and rivers, and glaciers and lava systems as well as ground water and the molten outer core of the earth could technically be included in GFD, but will not be considered further. The basis of GFD lies in the principles of conservation of momentum, mass and energy. These are expressed mathematically in Newton’s equations of motion for a continuous medium, the equation of continuity and the thermodynamic energy equation. We will first express the equations of motion in an inertial framework F = ma . Then they will be transformed to a non-Newtonian coordinate system which is fixed with respect to the earth, and rotating with it. But first we must consider some preliminary issues. 3 4 Two Ways to Describe Fluid Flow Eulerian: Stay put and watch the flow Lagrangian: Drift along, see where you go. The independent variables are the space and time coordinates, r = (x, y, z ) and t. The dependent variables are the velocity, pressure, density and temperature, V = (u, v, w), p, ρ and T . Further variables are needed for a fuller treatment, e.g. humidity q in the atmosphere and salinity s in the ocean. Each variable is a function of both position and time. For example, We must consider variations with respect to space and time. p = p(x, y, z, t) . Eulerian: Stay put and watch the flow We denote the change of pressure with time at a fixed point by the Eulerian (or partial) derivative: ∂p x, y and z fixed. ∂t Lagrangian: Drift along, see where you go. We denote the change of pressure with time following the flow by the Lagrangian (or material or total) derivative: dp parcel of fluid fixed. dt p = p(x, y, z, t) 5 6 Digression Partial Derivatives Euler and Lagrange Derivatives Eulerian or Local Change Stand on a bridge, hang a thermometer into the stream. The temperature you measure is at a fixed location. The change in temperature is local, given by the partial time derivative ∂T Change at a fixed location. ∂t Lagrangian or Material Change Float on a raft, hang a thermometer into the ocean. The temperature you measure is at a point moving with the current. The change in temperature is given by the total time derivative dT Change for a material parcel. dt 8 Partial derivatives were first introduced by the French mathematician Jean Le Rond d’Alembert (1717–1783) in connection with his meteorological studies. 7 Connection: ∂p/∂t ⇐⇒ dp/dt The pressure is a function of both space and time: Exercise: Local v. Material Change. Suppose the flow is purely in the x-direction, given by u = a sin(kx − ωt) where the amplitude a, wavenumber k and frequency ω are constants. We can also write u as u = a sin k (x − ct) where c = ω/k is the phase speed of the wave. Calculate the local and total time derivative of u. How do ∂u/∂t and du/dt change if u −→ 2u? p = p(x(t), y (t), z (t), t) . The total variation, following the flow, is given by the chain rule: ∂p ∂p dx ∂p dy ∂p dz dp = + · + · + · dt ∂t ∂x dt ∂y dt ∂z dt ∂p ∂p ∂p ∂p = +u· +v· +w· ∂t ∂x ∂y ∂z ∂p = + V · p. ∂t This is true for all variables, so we have d( ) ∂ ( ) = +V· dt ∂t ( ). 9 The Eulerian time derivative is a linear function of u: ∂u ∂ a sin(kx − ωt) = = −ω · a cos(kx − ωt) . ∂t ∂t x constant The Lagrangian time derivative is du ∂u ∂u = +u = −ω · a cos(kx − ωt) + u[k · a cos(kx − ωt)] . dt ∂t ∂x Note that du/dt is a nonlinear function of the amplitude a: du = k sin(kx − ωt) cos(kx − ωt) a2 − ω cos(kx − ωt) a . dt 10 Conservation of Mass Air is neither created nor destroyed. Therefore, the total mass must remain constant. Moreover, the mass of an identifiable parcel of air must remain unchanged with time. The mathematical expression of mass conservation is the Continuity Equation. To illustrate the two methods of describing fluid flow, we will derive the continuity equation in both Eulerian and Lagrangian forms. We must then show that the two forms are equivalent. Influx and Outflow @ @ @@ @@ @ @@ @@ @@ @@@ @@@ @ @@@ @ @@@@ @@@@ @ @@@@ Slab @@@@ @@@@ @@@ @@@@ Moves @@@@ @@@@ @@@@ @@@@ in at @@@@ @ @@@@ @@@@ @@ West @@@@ @@@@ @@@@ @ - @@@@ @@@@ @@@@ @ @@@@ uW @@@@ @@@@ @@@@ @ @@ @@@@ @@@@ @@@@ @@@ @@@@ @@@ @@@ @@@ @@ @@ @ @@ @@ @ @ @  @@ @@ @@ @@ Eulerian Formulation Consider a cubic region of dimensions ∆x = ∆y = ∆z , fixed in space. Air flows freely through the region. uE - Slab Moves out at East ∆x - The change of mass of the air in the cube must equal the net flux of mass into or out of the region. For simplicity, consider flow in the x-direction. Let uW be the x-component of velocity at the western face, and uE be the x-component of velocity at the eastern face. 12 11 Total mass of air in the box (density × volume): M = ρ · ∆ x ∆ y ∆ z = ρV Change of mass in time ∆t (volume is fixed): ∂M ∂ρ ∆M = ∆t = ∆t · V . ∂t ∂t Influx at western face (density × slab volume): ρW (uW ∆t)∆y ∆z = (ρu)W ∆t · ∆y ∆z Outflow at eastern face (density × slab volume): ρE(uE∆t)∆y ∆z = (ρu)E∆t · ∆y ∆z Net flow F into the box (influx − outflow) in time ∆t: (ρu)E − (ρu)W ∆ t · ∆ x ∆y ∆ z ∆x But ∆M = F , so the quantities in red must be equal: ∂ρ (ρu)E − (ρu)W ∂ (ρu) =− ≈− ∂t ∆x ∂x F = [(ρu)W − (ρu)E]∆t · ∆y ∆z = − 13 Thus, for flow only in the x-direction we have ∂ (ρu) ∂ρ =− ∂t ∂x However, there is also flow through the front and back faces, and through the top and bottom of the box. Symmetry arguments lead us immediately to the result ∂ (ρu) ∂ (ρv ) ∂ (ρw) ∂ρ =− + + ∂t ∂x ∂y ∂z This may be written using the divergence operator as ∂ρ + ∂t · ρV = 0 . This is the Eulerian form of the continuity equation. It is one of the fundamental equations of atmospheric dynamics. 14 Lagrangian Formulation We consider a parcel of air, of mass M , contained in a cube.1 But we allow the cube to move with the flow. The mass of the parcel does not change with time. Total mass of air in the box (density × volume): M = ρ · ∆ x ∆ y ∆ z = ρV Change of mass in time ∆t must be zero: d log M dM dM = 0. ∆t = 0 so =0 so ∆M = dt dt dt Since log M = log ρ + log ∆x + log ∆y + log ∆z , this means that 1 dρ 1 d∆x 1 d∆y 1 d∆z + + + =0 (∗) ρ dt ∆x dt ∆y dt ∆z dt But now notice that ∆x = xE − xW so that 1 d∆x 1 d(xE − xW ) 1 dxE dxW u − uW ∂u = = − =E ≈ ∆x dt ∆x dt ∆x dt dt ∆x ∂x The assumption that the parcel is initially cubic is purely for mathematical simplicity. We can relax it and consider an arbitrary parcel of mass M . 15 1 Substituting in (*) we get 1 dM 1 dρ ∂ u ∂v ∂w =0= + + + M dt ρ dt ∂x ∂y ∂z or, rearranging terms, dρ ∂ u ∂v ∂w + + = 0. +ρ dt ∂x ∂y ∂z Using vector operators, this is dρ +ρ dt · V = 0. This is the Lagrangian form of the continuity equation. We recall the Eulerian form, derived above: ∂ρ + ∂t · ρV = 0 . 16 The two forms look different, but must be equivalent. dρ +ρ ·V =0 dt Lagrangian Form ∂ρ + · ρV = 0 ∂t Eulerian Form ( ). ·V. Incompressibility For an incompressible fluid, the volume of a parcel remains unchanged. Thus, the material density is constant following the flow: dρ/dt = 0. Thus, the continuity equation reduces to · V = 0. The assumption of incompressibility is a natural one for the ocean. For the atmosphere, it is less obviously reasonable. Indeed, many atmospheric phenomena depend on compressibility. However, the essential large scale dynamics can be successfully modelled by an incompressible fluid. The benefit of assuming incompressibility is that we get a closed system without having to consider the thermodynamics explicitly. For compressible flow, we would have to have another equation for ρ, the thermodynamic equation. But this introduces the temperature T , and yet another equation, the equation of state, is required. 18 We recall the relationship between the time derivatives: d( ) ∂ ( ) = +V· dt ∂t We also note the vector identity · ρV = V · ρ+ρ Substituting in the Lagrangian form, we get: ∂ρ dρ +ρ ·V = +V· ρ +ρ ·V dt ∂t ∂ρ = + V· ρ+ρ ·V ∂t ∂ρ + · ρV = 0 . = ∂t Thus the equivalence of the two forms is established. QED 17 Exercise Open Windows Answer: Area of the frame (window): A = 1 m × 1 m = 1 m2 . Distance along wind in 10 seconds d = V × t = 10 m s−1 × 10 s = 100 m . Volume of air flowing through the frame: d × A = 100 m × 1 m2 = 100 m3 . Thus, 100 cubic metres of air flow through the frame in ten seconds. The volume of the room is V = 5 m × 5 m × 4 m = 100 m3 . Thus, the air flowing in through the window in ten seconds equals the volume initially within the room. Since the volume of the room is fixed, the mass of air within it must double. Thus, the density must also double. If the temperature remains constant, the pressure will double too! This result is surprising. We would not expect wind flowing through an open window to cause such huge change in pressure. What has been overlooked? 19 20 Consider a square frame of dimensions 1 m×1 m. Suppose the wind blows through the frame with speed 10 m s−1. Compute the volume of air which flows through the frame in ten seconds. Now imagine the frame is that of an open window in an otherwise closed room of dimensions 5 m×5 m×4 m. If initially the pressure in the room equals the external pressure, by what proportion will the pressure increase in ten seconds, assuming that air continues to flow in at a constant rate? Is the result physically reasonable? If not, discuss what important physical factors may have been neglected. Can you deduce a more reasonable value of the pressure increase. Pressure Force Consider a cubic box of air, of dimension ∆x × ∆y × ∆z = V . The pressure acts normally on each face of the cube. Pressure on Box @ @ @ @ @ @ @ @  d The Forces on a Parcel of Air p (x ) @ @ Net force on left-hand face: p ( x ) · ∆y ∆z Net force on right-hand face: -t p (x + ∆x ) −p(x + ∆x) · ∆y ∆z Total pressure force in the x-direction: − p ( x + ∆ x ) − p ( x ) · ∆y ∆z @ @  ∆x - 21 22 Total pressure force in x-direction: p (x + ∆x ) − p (x ) − p ( x + ∆ x ) − p ( x ) · ∆y ∆z = − · ∆x ∆ y ∆ z ∆x But ∆x∆y ∆z = V , so the force per unit volume is: ∂p p (x + ∆x ) − p (x ) ≈− − ∆x ∂x A parcel of mass m has volume V = m/ρ, so a unit mass has volume 1/ρ. The pressure force per unit mass in the x-direction is thus 1 ∂p − ρ ∂x Similar arguments apply in the y and z directions. So, the vector force per unit mass due to pressure is 1 ∂p 1 ∂p 1 ∂p 1 Fp = − ,− ,− = − p. ρ ∂x ρ ∂y ρ ∂z ρ This force acts in the direction of lower pressure. 23 Exercise pressure. Amazing Pressure This problem is to give you an impression of the surprising strength of atmospheric Consider a 20” television CRT screen, of width 16” and height 12”. Assume the atmospheric pressure is 105 Pa, and that there is a perfect vacuum within the tube. • What is the force on the screen due to air pressure? • How does this force compare to that of a fat man (100kg) standing on the screen: (a) much less; (b) similar; (c) much greater? 24 Answer: First, the force due to atmospheric pressure: p = 105 Pa = 105 N m−2 A = 16” × 12” ≈ 40 cm × 30 cm = 0.40 m × 0.3 m = 0.12 m2 F = pA = 105 × 0.12 = 12 × 103 N = 12 kN . Next, the force due to the fat man: m = 102 kg , g = 10 m s−2 F = mg = 102 × 10 = 103 N = 1 kN . Conclusion: Air Pressure 12 Fat Men Exercise “The Spring of Air” This expression was used by Robert Boyle for the tendency of air to resist compression. Consider a cylinder with a piston which is free to move. Suppose the fat man stands on the piston. Assume that the temperature remains constant. Derive expressions for the pressure increase ∆p within the cylinder; the change in volume ∆V ; the work W done by the man in compressing the air; show that W = |∆p∆V |. Calculate numerical values of ∆p, ∆V and W assuming the initial volume is V = 1 m3, the initial pressure is p = 105 Pa and the fat man weighs m = 102 kg. Consider two values of the cross-sectional area of the cylinder: (a) A = 1 m2; (b) A = 10−2 m2. 26 ∼ This should convince you that differences in pressure can result in significant forces. The pressure gradient force is dominant in atmospheric dynamics. 25 Answer: Without the fat man, the forces on the top and bottom of the piston are F↑ = pA F↓ = pA . Naturally, they are equal. After he hops on board, the forces are: F↑ = (p + ∆p)A These must also be equal, hence mg ∆p = . A By Boyle’s Law, the product of pressure and volume is constant. Thus (p + ∆p)(V + ∆V ) = pV which, rearranging terms, leads to ∆V = − V ∆p p + ∆p F↓ = pA + mg . The fixed values are V = 1 m3, p = 105 Pa and m = 102 kg. We consider two values of A. (a) A = 1 m2. Therefore the height of the air column is h = V /A = 1 m. mg 102 × 10 ∆p = = = 103 Pa . A 1 The percentage change in pressure is 100∆p/p = 1%. The volume change is V ∆p 1 × 103 1 × 103 ∆V = − =− 5 ≈− = −10−2 m3 . p + ∆p (10 + 103) 105 The percentage volume decrease is 100|∆V |/V = 1%. The work done is W = ∆p · ∆V = 103 × 10−2 = 10 J . (b) A = 10−2 m2. Therefore the height of the air column is h = V /A = 102 m. mg 102 × 10 ∆p = = = 105 Pa . A 10−2 The percentage change in pressure is 100∆p/p = 100%. The volume change is ∆V = − 1 × 105 V ∆p =− 5 = −0.5 m3 . p + ∆p (10 + 105) 28 The work done by the man is force multiplied by distance. But the distance is ∆h = ∆V /A, so mg W = mg × ∆h = ∆V = ∆p · ∆V A which completes the first part of the answer. 27 The percentage volume decrease is 100|∆V |/V = 50%. The work done is W = ∆p · ∆V = 105 × 0.5 = 5 × 104 J = 50 kJ which is 5000 times greater than in the previous case. Remarks: • The above results imply that we can compress air to an arbitrary state by using a cylinder of sufficiently small cross-section. Is this physically reasonable? • The work done in compressing the gas must go somewhere. Where does it go? • Compare two cases: (a) Isothermal, where the cylinder is immersed in a bath of water held at a constant temperature; (b) Adiabatic, where the cylinder is insulated, so that no heat enters or leaves. Force of Gravity Newton’s law of gravity states that two bodies of mass m1 and m2 attract each-other with a force given by mm F = G 12 2 d where d the distance between them. The constant G is the universal gravitational constant. Near the earth’s surface, a parcel of air of mass m is attracted towards the earth with a force GM F =m 2 a where M is the mass of the earth and a its radius. We define the acceleration due to gravity by GM g= 2 a 29 30 The acceleration due to gravity can be evaluated as follows: G = 6.672×10−11 , M = 5.974×1024 , a = 6.375×106 =⇒ g = 9.807 (all values are in SI units). So, roughly, g ≈ 10 m s−2. The force due to gravity acts vertically downward, towards the centre of the earth. If k is a unit vector pointing upward, we may write it: Digression Definition of Metre A simple way to remember the earth’s radius. • The metre is defined in terms of the size of the earth. • The distance from the equator to the pole is 10 million metres. • Thus, the circumference of the earth is 4 × 107 m. • Thus, the radius of the earth is a= 4 × 107 ≈ 6.366 × 106 m . 2π Fg = −mg k . 31 32 ...
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This note was uploaded on 01/31/2011 for the course MATH 21A taught by Professor Osserman during the Spring '07 term at UC Davis.

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