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AOSS 321, Winter 2009 Earth System Dynamics Lecture 6 & 7 1/27/2009 1/29/2009 Christiane Jablonowski cjablono@umich.edu 734-763-6238 Eric Hetland ehetland@umich.edu 734-615-3177 What are the fundamental forces in the Earth’s system? • Surface forces: – Pressure gradient force – Viscous force • Body force: Gravitational force • Apparent forces: Centrifugal and Coriolis • Electromagnetic force (important for charged particles, e.g. in the ionosphere, not important for the neutral atmosphere) • Total Force is the sum of all of these forces. Viscous force(1): Friction • In a fluid there is friction which resists the flow. • It is dissipative, and if the fluid is not otherwise forced, it will slow the fluid and bring it to rest. • Away from boundaries in the atmosphere this frictional force is often small, and it is often ignored. • Close to the boundaries, however, we have to consider friction. Viscous force (2) velocity ≡ u in m/s Ocean Land Biosphere velocity must be 0 at surface Viscous force (3) velocity ≡ u in m/s Ocean Land Biosphere Velocity is zero at the surface; hence, there is some velocity profile. Viscous force (4) (How do we think about this?) z Moving plate with velocity u0 u(h) = u0 h u(z) u(0) = 0 u0 u(0) u(z) z h Linear velocity profile! Viscous force (5) (How do we think about this?) The drag on the moving plate is the same of as the force required to keep the plate moving. It is proportional to the area (A), proportional to the velocity u0 of the plate, and inversely proportional to the distance h between the plates; hence, u(h) = u0 h u(z) u0 F A h u(0) = 0 Proportional usually means we assume linear relationship. This is a model based on observation, and it is an approximation. The constant of proportionality assumes some physical units. What are they? Viscous force (6) (How do we think about this?) Recognize the u0/h can be represented by ∂u/∂z u(h) = u0 h u(z) u(0) = 0 u F A z Force per unit area is F/A is defined as shearing stress (t). Like pressure the shearing stress is proportional to area. Viscous force (7) & Shearing stress tzx = μ(∂u/∂z) is the component of the shearing stress in the x direction due to the vertical (z) shear of the x velocity component (velocity u) u(h) = u0 h u(z) u(0) = 0 Fzx u t zx A z Fzx/A: Force per unit area with units kg m-1 s-2: hence, like pressure Viscous force (8) & Shearing stress Remember Newton’s third law: "For every action, there is an equal and opposite reaction." tzx = -(tzx0 + (∂tzx/∂z)Dz/2) Stress above a surface tzx = tzx0 + (∂tzx/∂z)Dz/2 z • x tzx0 Stress below a surface Front view of 3D box tzx = -(tzx0 - (∂tzx/∂z)Dz/2) tzx = tzx0 - (∂tzx/∂z)Dz/2 The opposing force introduces a ‘-’ sign! We are interested in the net force inside the volume. Viscous force (9) & Shearing stress (Do the same thing we did for pressure) stress below the upper boundary (inside the volume): t zx Dz t zx t zx0 z 2 mid point: (x0, y0, z0) . Dz Dy stress above the lower boundary (inside the volume): Dx tzx0 = stress at (x0, y0, z0) t zx Dz t zx t zx0 z 2 Viscous force (10) & Shearing stress Viscous force = shearing stress * area Area of side C: DyDx Viscous force below the upper boundary: C Dz . Dy D Dx Area of side D: DyDx FCzx t zx Dz t zx0 DyDx z 2 Viscous force above the lower boundary: t zx Dz FDzx t zx0 DyDx z 2 Viscous force (11) Net viscous force (inside a volume) in the x direction due to the vertical (z) shear of the x velocity component (u): Sum of the forces that act below the upper boundary C and above the lower boundary D on the fluid F F F zx Czx Dzx t zx DzDyDx z We want the viscous force per unit mass: Fzx 1 t zx m z Viscous force (12) (using definition of t) Recall the definition of the shearing stress: u t zx z It follows: Fzx 1 t zx m z 1 u ( ) z z Viscous force (13) μ is also called dynamic viscosity coefficient. Assume that μ is constant, then: Fzx u u 2 2 m z z 2 2 ≡ /kinematic viscosity coefficient Viscous force (14) Do the same for the shearing stress in x direction due to the horizontal (x and y) shear of the x velocity component (u): txx and tyx The total frictional (viscous) force Fr in the x direction is called Frx . It is the sum of all viscous forces acting in x: Frx u u u ( 2 2 2 ) m x y z 2 2 2 Viscous force (15) Do same for other directions of force N-S (y) direction: v v v ( 2 2 2 ) m x y z Fry 2 velocity vector: 2 Frz w w w ( 2 2 2 ) m x y z 2 vertical direction: 2 v (u,v,w) 2 2 Viscous force (16) Viscous force per unit mass written in vector form: F 2 (v ) m Kinematic viscosity coefficient Where Laplacian is operating on the velocity vector: v u,v,w Our surface forces dv 1 2 p (v ) dt other forces Here, we use the text’s convention that the velocity is v u,v,w Highs and Lows Motion initiated by pressure gradient Opposed by viscosity Summary: Surface forces • Pressure gradient force and viscous force are examples of surface forces. • They were proportional to the area of the surface of our particle of atmosphere. • They are independent of the mass of the particle of atmosphere. • They depend on characteristics of the particle of atmosphere; characteristics of the flow. Body forces • Body forces act on the center of mass of the parcel of fluid. • Magnitude of the force is proportional to the mass of the parcel. • The body force of interest to dynamic meteorology is gravity. Newton’s Law of Gravitation Gm1m2 r F 2 r r Newton’s Law of Gravitation: Any two elements of mass in the universe attract each other with a force proportional to their masses m1 and m2 and inversely proportional to the square of the distance r between the centers of the mass. The force acts along the line joining the particles and has the magnitude proportional to G, the universal gravitational constant. Gravitational Force Gravitational force for dynamic meteorology GMm r F 2 r r Newton’s Law of Gravitation: M = mass of Earth m = mass of air parcel r = distance from center (of mass) of Earth to parcel force directed down, towards Earth, hence ‘-’ sign G = gravitational constant (6.673 × 10-11 m3 kg-1 s-2 ) Gravitational force per unit mass Force per unit mass exerted on the atmosphere by the gravitational attraction of the earth: F GM r * g 2 m r r r m M g* = gravitational acceleration Adaptation to dynamical meteorology GM r * r * r g 2 g0 g0 r r a r r a z * 0 This is the gravitational acceleration at mean sea level. a: radius of the Earth z: height above the mean sea level For z a g g * * 0 Gravitational force per unit mass F GM a r 2 2 m a r r 2 r a * g0 2 (a z) r r 1 * g0 2 (1 z /a) r 2 Our momentum equation so far 2 dv 1 a r 2 * p (v ) g0 2 dt (a z) r + other forces Material derivative of the velocity vector v Pressure gradient force Viscous force Gravitational force Apparent forces • Due to the fact that we live on a rotating planet • Centrifugal force • Coriolis force Back to Basics: Newton’s Laws of Motion • Law 1: Bodies in motion remain in motion with the same velocity, and bodies at rest remain at rest, unless acted upon by unbalanced forces. • Law 2: The rate of change of momentum of a body with time is equal to the vector sum of all forces acting upon the body and is the same direction. • Law 3: For every action (force) there is and equal and opposite reaction. Back to basics: A couple of definitions • Newton’s laws assume we have an “inertial” coordinate system; that is, and absolute frame of reference – fixed, absolutely, in space. • Any motion relative to the coordinate system fixed in space is known as inertial motion. • Velocity is the change in position of a particle (or parcel). It is a vector and can vary either by a change in magnitude (speed) or direction. Non-inertial coordinate systems Unit 6,frames 16-21: http://www.atmos.washington.edu/2005Q1/101/CD/MAIN3.swf • The Earth rotates, it is an accelerating frame of reference, hence: • Our coordinate system ‘north, south, east, west’ is a non-inertial, non-absolute coordinate system. • The ‘north’ direction in Ann Arbor is different from the ‘north’ direction in San Francisco or Sydney, Australia. Two coordinate systems Can describe the velocity and forces (acceleration) in either coordinate system. dx dx ' z y x dt or dt One coordinate system related to another by: x' Ax (t ) x Bx (t ) y C x (t ) z y ' Ay (t ) x By (t ) y C y (t ) z z ' Az (t ) x Bz (t ) y C z (t ) z Example: Velocity (x’ direction) x' Ax (t)x Bx (t)y Cx (t)z dx' dx dy dz Ax (t) Bx (t) Cx (t) dt dt dt dt d(Ax (t)) d(Bx (t)) d(Cx (t)) x y z dt dt dt So we have the velocity relative to the coordinate system and the velocity of one coordinate system relative to the other. This velocity of one coordinate system relative to the other leads to apparent forces. They are real, observable forces to the observer in the moving coordinate system. Two coordinate systems z’ axis is the same as z, and there is rotation of the x’ and y’ axis z’ z y y’ x x’ Example: One coordinate system related to another by x' x cos(t) y sin( t) y' x sin( t) y cos(t) z' z 2 2 2 with 7.292 105 s1 T sidereal day 86164 s We also use the symbol for . is the angular speed of rotation of the earth. T is time needed to complete rotation. On Earth it takes a sidereal day = 86164 s. Acceleration (force) in rotating coordinate system d 2 x' dx 2 dy 2 cos(t) sin( t) 2 dt dt dt dx dy 2 sin( t) cos(t) dt dt 2 (x cos(t) y sin( t)) The apparent forces that are proportional to rotation and the velocities in the inertial system (x,y,z) are called the Coriolis forces. The apparent forces that are proportional to the square of the rotation and position are called centrifugal forces. Acceleration (force) in rotating coordinate system d 2 x' dx 2 dy 2 cos(t) sin( t) 2 dt dt dt dx dy 2 sin( t) cos(t) 2 (x cos(t) y sin( t)) dt dt d 2 y' dx 2 dy 2 sin( t) cos(t) 2 dt dt dt dx dy 2 cos(t) sin( t) 2 (x sin( t) y cos(t)) dt dt d 2 z' d 2 z 2 2 dt dt Coriolis forces Centrifugal forces