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 nonNewtonian 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 xdirection, 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 xdirection. Let uW be the xcomponent of velocity at the western face, and uE be the xcomponent 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 xdirection 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 lefthand face: p ( x ) · ∆y ∆z Net force on righthand face: t p (x + ∆x ) −p(x + ∆x) · ∆y ∆z Total pressure force in the xdirection: − p ( x + ∆ x ) − p ( x ) · ∆y ∆z @ @ ∆x  21 22 Total pressure force in xdirection: 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 xdirection 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 crosssectional 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 crosssection. 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 eachother 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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 Spring '07
 Osserman
 Math, Thermodynamics, Continuity, Force, Mass, dt ∆x dt

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