Lect-03-P4 - M.Sc. in Computational Science Lecture 3 The...

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Unformatted text preview: M.Sc. in Computational Science Lecture 3 The Equations of Motion Fundamentals of Atmospheric Modelling ´ Peter Lynch, Met Eireann Mathematical Computation Laboratory (Opp. Room 30) Dept. of Maths. Physics, UCD, Belfield. January–April, 2004. 2 The Thin Atmosphere The atmosphere is infinitely thick but very thin! 90% of its mass lies within 10 km of the earth’s surface. The lowest layer of the atmosphere, where temperature decreases with height, is called the troposphere. It about 10 km in depth. The vertical extent of the large scale motion systems is very much smaller than their horizontal scales. Characteristic horizontal and vertical scales for synoptic systems are Digression Did you know? The ratio of the length to the breadth of a credit card is equal to the ratio of the Golden Section: √ 1+ 5 Aspect Length ≈ 1.618 . = = 2 Breadth Ratio It is allegedly the most aesthetically pleasing rectangular shape, and is found in numerous classical works of art. This ratio is ubiquitous throughout nature. It is closely associated with the Fibonacci sequence of numbers {1, 1, 2, 3, 5, 8, 13, 21, 34, . . . } , where each term is the sum of the preceding two terms. 4 L = 106 m = 1000 km H = 104 m = 10 km . This geometrical situation is intimately linked to the existence of hydrostatic balance. A typical grid-box of a numerical model might have dimensions 10 km × 10 km × 100 m, which has an aspect ratio of one hundred to one, comparable to that of a credit card! 3 Hydrostatic Balance For a fluid at rest, the pressure at a point depends on the weight of fluid vertically above that point. The pressure difference between two points on the same vertical line depends only on the weight of fluid between them. p (z + ∆z ) @ @ @ @ @ @ @ @ @ @ @ @ ? u @ @ @ @ @ @ For equilibrium, the net force must be zero: p (z + ∆z ) − p ( z ) · ∆x∆y ∆z + mg = 0 . ∆z This may be written ∂p · V + mg = 0 , ∂z or, dividing through by the volume, ∂p + ρg = 0 . ∂z ∆z This is the Hydrostatic balance equation. It implies an exact balance between the vertical pressure gradient and gravity. For an atmosphere at rest, hydrostatic balance holds exactly. 6 g ? p (z ) Force Upward on Box : Force Downward on Box : + [p(z ) · ∆x∆y ] − [p(z + ∆z ) · ∆x∆y + mg ] 5 6 Exercise: Vertical Pressure Gradient Hydrostatic Approximation Suppose the atmosphere is in a state of hydrostatic balance. Calculate approximately the pressure drop over a vertical distance of 100 m, assuming the density is constant at ρ = 1.2 kg m−3 and g = 9.8 m s−2. The hydrostatic equation gives ∆p + ρg = 0 . ∆z Substituting the numerical values gives ∆p = −∆zρg = −100 m × 1.2 kg m−3 × 9.8 m s−2 (negative, since pressure decreases upwards). Evaluating this gives |∆p| = 1176 kg m−1s−2 = 1176 Pa = 11.76 hPa . The hectoPascal, numerically equal to the millibar, is the pressure unit most commonly used in practice. Note that the assumption of constant density is unrealistic over large vertical distances. We will relax this assumption presently. 7 The hydrostatic approximation consists of assuming that balance between the vertical pressure gradient and gravity holds even when the fluid is in motion. For the large scale motions of the atmosphere and ocean, hydrostatic balance holds to a high degree of accuracy. Until recently, most numerical models used to predict atmospheric flow were hydrostatic. For these models, the vertical velocity is a diagnostic variable, deduced from the other dependent variables at each point in time. Non-hydrostatic models are now growing in popularity, particularly where spatial grids of a few kilometres are used. For these models, the vertical velocity is a prognostic variable, predicted in the same way as the other dependent variables. 8 The Equation of State We are familiar from elementary physics with Boyle’s Law and Charles’ Law of gases. They are special cases of the Equation of State for a perfect gas: pV = nR∗T where R∗ = 8314 J K−1 kmol−1 is the universal gas constant and n is the number of kilomoles of gas (a kmole is the molecular weight in kg). The mean molecular weight of air is µ ≈ 29. Thus, m = µn. Dividing by the volume, we get the equation of state Table 1: Main Constitutents of the Atmosphere Gas Percentage Mol. Wt. Nitrogen N2 80% 28 Oxygen O2 20% 32 Air 100% 29 The atmosphere is composed primarily of nitrogen (80%) and oxygen (20%), so the mean molecular weight of air is about 29. Other constituents, such as carbon dioxide and methane, are vitally important for radiative balance, but their concentrations are quite small. Water occurs in all three phases, and is enormously important. However, we will be concentrating on the large-scale dynamics of the atmosphere and will largely ignore water, as it introduces great complexity. 10 p = RρT where R = R∗/µ = 287 J K−1 kmol−1 is the gas constant for dry air. 9 Vertical Variation of Pressure Let’s consider an isothermal atmosphere at rest. Let the constant temperature be T0. The hydrostatic equation and the equation of state are ∂p + gρ = 0 , p = RρT0 . ∂z Combining these we have ∂p p dp g dz dz = −g , so =− =− , ∂z RT0 p RT0 H where we define the scale-height by H = RT0/g . We integrate over the range p0 = p(0) to p = p(z ) to get p z log =− p0 H or Since ρ = p/RT0, density also decreases exponentially with height. The scale height is easily computed. Suppose T0 = 265 K. Then RT0 287 × 265 = = 7755 m = 7.755 km , H= g 9.8066 so the scale height is about 8 km. Pressure decreases by a factor of 1/e over this height. Exercise Relax the assumption of an isothermal atmosphere: Assume that temperature decreases linearly with height T = T0 + γz where the lapse rate, γ = ∂T /∂z < 0, is constant. Calculate the dependence of p on height. p(z ) = p0 exp(−z/H ) . Thus, pressure decreases exponentially with height. 11 12 Exercise: Constant Lapse-rate Answer: Assume that temperature decreases linearly with height T = T0 + γz where the lapse rate γ = ∂T /∂z < 0 is constant. Combining the hydrostatic equation and the equation of state as before, we get dp g dz g dz =− =− . p RT RT0 1 + γz/T0 Integrating this yields log so that p p0 =− T0 γz log 1 + γH T0 = log 1 + γz T0 −T0 /γH Exercise: Mass of Air Column Consider a vertical column of air, of cross-section A and infinite height. Suppose the pressure at the bottom of the column is p0. What is the mass M of the column? If we knew the density, we could calculate the mass using ∞ M= V ρ dV = A 0 ρ dz . However, there is a simpler way: the force of the column on the surface below it is given by two quantities, which must be equal: p0 × A = M × g , so that 2 M = p0A/g . p = p0 γz 1+ T0 −T0/γH For a unit cross-section, A = 1 m , the total column mass is M = p0/g ≈ 104 kg or ten tonnes! More Work: Recall the TV screen. The force was 12 kN for an area A = 0.12 m2. Show that this result is consistent with the above. 14 [More Work: Show that this reduces to the previous result when γ → 0. Use limn→∞(1 + x/n)−n = exp(−x).] 13 Equivalent Incompressible Fluid What is the depth of a layer of incompressible fluid having the same bottom pressure as the atmosphere? The mass of a vertical column of fluid of constant density ρ0 and depth h is h M =A 0 ρ0 dz = Aρ0h . The force exerted by the column on the surface below is M g . Since pressure is force-per-unit-area, the pressure is p0 = gρ0h so, for given pressure p0, the depth is p0 h= . gρ0 But the scale height of the atmosphere is RT0 p0 H= = , g gρ0 so we see that the depth h equals the scale height H . 15 The Equations of Motion 16 Forces Acting on a Parcel of Air (1) Recall that the pressure force per unit mass is 1 Fp = − p . ρ It points towards low pressure. (2) The force due to gravity acts vertically downward, towards the centre of the earth. Per unit mass, it is: Equations in an Inertial Frame Independent Variables: Dependent Variables: Space and time , r and t V = (u, v, w), p, ρ and T . The basic equations of motion (a = F/m) are: dV 1 = − p + g + Ff (1) dt ρ where the total, material or Lagrangian derivative is d ∂ = +V· dt ∂t which measures the change with time of a variable moving along with the fluid flow. The continuity equation, representing the conservation of mass, may be written in Lagrangian form: dρ + ρ · V = 0. dt If the fluid is incompressible, it is especially simple: ·V =0 (2) 18 g∗ = −g k . The star on g∗ will be explained below. (3) The force of friction acts in a direction opposite to the velocity of the flow. We could model it as Ff = −ν 2 V, or Ff = −κV . 17 The friction coefficient κ will depend on position and, perhaps, on velocity. A Complete System If we assume incompressible, inviscid flow, equations (1) and (2) comprise a system of four equations for the four variables (u, v, w; p): 1 dV =− p+g dt ρ ·V =0 Written out in (local) cartesian coordinates, they are ∂ ∂ ∂ 1 ∂p ∂ +u +v +w u=− ∂t ∂x ∂y ∂z ρ ∂x ∂ ∂ ∂ ∂ 1 ∂p +u +v +w v=− ∂t ∂x ∂y ∂z ρ ∂y ∂ ∂ ∂ ∂ 1 ∂p +u +v +w w=− −g ∂t ∂x ∂y ∂z ρ ∂z ∂u ∂v ∂w + + = 0. ∂x ∂y ∂z 19 Rotating Coordinate Frames Theorem: Consider a vector A fixed in a frame which is rotating with constant angular velocity Ω. Then the rate of change of A is dA = Ω × A. dt (see, e.g., Synge and Griffith, pg. 278.) The vector rotates through an angle Ω∆t in time ∆t. The projection of A on the Ω-axis does not change The projection of A in the X-Y-plane is A sin θ. It does not change in magnitude, but its direction changes (see Figure). We have ∆A = (ΩA sin θn) · ∆t ˆ where n is a unit vector perpendicular to both Ω and A. ˆ Thus dA = Ω × A. dt 20 Consider a vector A = A1 i + A2 j + A3 k . The rate of change of A in the rotating frame is dA dA1 dA2 dA3 i+ j+ k. = dt R dt dt dt A ¡ ! ¡ ¡ ¡ ¡ ¡ ¡ $ A sin θ $$$ $ $$$ $$$ $ b ˆ Ω∆t ∆A ˆ ˆˆˆ ˆˆˆ ˆˆˆ ˆ A sin θ Z Ω T θ ¡ ¡ ¡ ¨ The rate of change of A in the inertial frame is dA dA1 dA2 dA3 di dj dk = i+ j+ k + A1 + A2 + A 3 dt I dt dt dt dt dt dt . Y ¨¨ X Vector in Rotating Coordinates Horizontal Projection The changes in the unit vectors are (by the above theorem) di dj dk = Ω × i; = Ω × j; = Ω × k. dt dt dt Therefore, dA dA1 dA2 dA3 = i+ j+ k + Ω× (A1i + A2j + A3k) . dt I dt dt dt 21 22 Thus, the relationship between the relative and absolute rates of change of A is: Exercise If A = r0 is a point fixed in the rotating frame, the velocity in the absolute frame is V = Ω × r0 . Find the absolute velocity of a point on the earth’s surface (i) at the Equator, and (ii) at 60◦ North. dA dt = I dA dt + Ω × A. R ( ∗) Now let A be the position vector r. Since (dr/dt)I = VI and (dr/dt)R = VR, we get: VI = VR + Ω × r , which relates the relative velocity to that in the inertial frame: Inertial Relative Velocity = + Velocity Velocity of Frame 23 The rotation of the earth (assuming solar day ≈ siderial day) is: 2π Ω = 1 rev. per day = rad/sec = 7.29 × 10−5 s−1 24 × 60 × 60 The radius of the earth is 2 × 107 m ≈ 6.37 × 106 m . a= π (i) At the equator, φ = 0◦ and θ = 90◦, so that sin θ = 1 and Ω × r0 = Ωa = (7.29 × 10−5 s−1) × (6.37 × 106 m) = 4.64 × 102 m s−1 ∼ 1000 m.p.h. (ii) At φ = 60◦, we have θ = 30◦, so that sin θ = 0.5, and the value of the velocity due to the earth’s rotation is half that at the equator. 24 Relative Acceleration Recall from (*) above that dA dA = + Ω × A. dt I dt R Now let A be the absolute velocity VI = VR + Ω × r: dVR dΩ × r dVI = + + Ω × V R + Ω × (Ω × r ) dt I dt R dt R dVR = + 2Ω × VR + Ω × (Ω × r) . dt R The term Ω × (Ω × r) is called the centrifugal acceleration. Since it depends only on position, it can be combined with the gravitational acceleration to give an apparent gravitational attraction g = g − Ω × (Ω × r ) . This is a small adjustment to the true gravitational acceleration. 25 Exercise: Centrifugal Acceleration Calculate the magnitude of the centrifugal acceleration at the Equator and compare it to the magnitude of the gravitational acceleration. First note that Ω × (Ω × r) = −Ω2R , where R is the projection of r on the equatorial plane. Thus, near the earth’s surface, the magnitude of the centrifugal acceleration is Ω2a = (7.29 × 10−5 s−1)2 × (6.37 × 106 m) = 3.18 × 10−2 m s−2 Now, comparing with true gravity, the percentage correction is Ω2a × 100 ≈ 0.3% . g The centrifugal acceleration is responsible for the flattened form of the earth, which assumes an oblate spheroidal shape. Do you lose weight when you travel to the Tropics? If so, how much? 26 The Coriolis Acceleration The term 2Ω × V is called the Coriolis acceleration. It varies linearly with the speed V and is perpendicular to the velocity V Exercises (1) Calculate the deflection of a golf ball travelling for 10 seconds at 10 m/s. Assume a latitude of 60◦N, and make reasonable assumptions to simplify the problem. (2) Suppose the pressure at Cork is 1014 hPa and at Sligo is 1008 hPa. Take the distance between Cork and Sligo to be 330 km. Assume the isobars are east-west, and assume that Cork and Sligo are on the same meridian. Calculate the acceleration due to the pressure gradient (assume ρ = 1.2 kg m−3). What wind speed would give a Coriolis acceleration of the same magnitude (take 2Ω sin φ = 10−4 s−1)? [For further problems, see Holton, Chapter 1.] The Coriolis acceleration is of primary importance for the dynamics of the atmosphere. It is a dominant factor in all large-scale weather systems. Since 2Ω × V has no component in the direction of motion, it does no work. However, once the air is moving, it is subject to the deflecting effect of this term. This is why the atmospheric flow is predominantly rotational in character. 27 28 Alternative Treatment of Rotation Let (X, Y, Z ) be the coordinates in a fixed frame, and (x, y, z ) be those in a rotating frame. Suppose the rotating frame spins about the Z -axis with angular velocity Ω. The coordinates are related by x cos Ωt − sin Ωt 0 X Y = sin Ωt cos Ωt 0 y . 0 0 1 z Z If we differentiate with respect to time, we get ˙ X ˙ x − Ωy cos Ωt − sin Ωt 0 ˙ ˙ Y = sin Ωt cos Ωt 0 y + Ωx . ˙ ˙ 0 0 1 z Z If we differentiate once again, we get ¨ X ¨ ˙ x − 2Ωy − Ω2x cos Ωt − sin Ωt 0 ¨ ¨ ˙ Y = sin Ωt cos Ωt 0 y + 2Ωx − Ω2y . ¨ 0 0 1 ¨ z Z 29 The equations of motion in non-rotating coordinates are ¨ ¨ ¨ X = FX , Y = FY , Z = FZ . Substituting from the matrix equation, we get ˙ ˙ (¨ − 2Ωy − Ω2x) cos Ωt − (¨ + 2Ωx − Ω2y ) sin Ωt = FX x y 2x) sin Ωt + (¨ + 2Ωx − Ω2y ) cos Ωt = F ˙ ˙ (¨ − 2Ωy − Ω x y Y ¨ z = FZ ¨ ¨ Solving for the terms with x and y , we get ¨ ˙ x − 2Ωy − Ω2x = cos ΩtFX + sin ΩtFY ≡ Fx ¨ ˙ y + 2Ωx − Ω2y = cos ΩtFY − sin ΩtFX ≡ Fy ¨ z = FZ ≡ Fz Thus, the rotation introduces additional terms: ˙ ˙ The terms −2Ωy and 2Ωx are the Coriolis acceleration. 2x and Ω2y are the centrifugal acceleration. The terms Ω 30 ...
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