The Material Derivative the Equations Above Apply to a Fluid Element

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The Material Derivative the Equations Above Apply to a Fluid Element The Material Derivative The equations above apply to a fluid element which is a small “blob” of fluid that contains the same material at all times as the fluid moves. Figure 1. A fluid element, often called a material element. Fluid elements are small blobs of fluid that always contain the same material. They are deformed as they move but they are not broken up. Consider a property γ (e.g. temperature, density, velocity com- ponent) of the fluid element. In general, this will depend on the time, t, and on the position (x, y, z) of the fluid element at that time. So γ = γ (x, y, z, t) = γ (r, t) . Now suppose we move with the fluid element which has coordi- nates (x (t) , y (t) , z (t)) . In a small time δt, suppose that the element moves from (x, y, z) to (x + δx, y + δy, z + δz) (see Fig. 2). There will be a corre- sponding small change in γ, denoted by δγ. Figure 2. A fluid element moves from (x, y, z) to (x + δx, y + δy, z + δz) in time δt. It contains exactly the same material at the two times. ∂γ ∂γ ∂γ ∂γ δγ = δt + δx + δy + δz. ∂t ∂x ∂y ∂z The observed rate of change of γ for that fluid element will be dγ ∂γ ∂γ dx ∂γ dy ∂γ dz = + + + . dt ∂t ∂x dt ∂y dt ∂z dt The velocity of the fluid element is its rate of change of position dr dx dy dz = u = (u, v, w) = , , . dt dt dt dt Hence dγ ∂γ ∂γ ∂γ ∂γ ∂γ = + u + v + w = + u.∇γ. dt ∂t ∂x ∂y ∂z ∂t Note: • (i)u.∇ is defined by the Cartesian expansion ∂ ∂ ∂ u.∇ ≡ u + v + w . ∂x ∂y ∂z • (ii)dγ/dt is the rate of change of γ moving with the fluid element. • (iii)∂γ/∂t is the rate of change of γ at a fixed point in space. In fluid dynamics, the time rate of change for a fluid element is usually denoted by D/Dt. Thus Dγ ∂γ D ≡ + u.∇γ. (Definition of .) Dt ∂t Dt D/Dt is often called the material derivative or Lagrangian deriva- tive. In section 2.1 the volume of gas under consideration was in fact a fluid element in a flow and so we could divide through by dt in equation for adiabatic expansion contraction (equation (2.8)) and note that for a fluid element dP/dt = DP/Dt. Hence DP γP Dρ = . Dt ρ Dt The Continuity Equation (Conservation of Mass) Matter cannot be made or destroyed, and so the total mass of a fluid element must remain the same. Thus if the density of a fluid element decreases, its volume must expand accordingly. This expansion causes a divergence of the velocity field, giving the conservation equation Dρ + ρ∇.u = 0. Dt where ∂u ∂v ∂w divu = ∇ · u = + + ∂x ∂y ∂z Eq. (2.15) is called the continuity equation. A detailed deriva- tion of this equation can be found in most textbooks on fluid mechanics. If the density of the fluid is constant, approximately true of most liquids including water, then equation (2.15) re- duces to ∇.u = 0 and the fluid is said to be incompressible. Equations of Motion (Momentum Equations) The equation of motion (Newton’s second law) for a fluid ele- ment is: Rate of change of momentum of fluid element = Total force acting on it. The forces acting on a material element of volume δV are: • (a) The pressure gradient force, generated by differences in pressure. −∇P δV (2.17) • (b) The gravitational force, −ρδV ∇Φg (2.18) • (c) The viscous force, due to friction between fluid elements. For a Newtonian fluid (such as air or water) n 1 o ρν ∇2u + ∇ (∇.u) δV (2.19) 3 where 2 2 2 2 2 2 2 ∂ u ∂ u ∂ u ∂ u ∂ u ∂ u ∇ u = 2 + 2 + 2 = 2 + 2 + 2 ∂x1 ∂x2 ∂x3 ∂x ∂y ∂z and ν is a constant, called the kinematic viscosity. In other fluid mechanics courses you may have been introduced to the dynamic viscosity µ = ρν. The kinematic viscosity of air is approximately 1.4×10−5m2s−1 at atmospheric pressure and for water it is roughly 10−6m2s−1. The mass of a fluid element cannot change, therefore Du Du Rate of change of momentum = δm = ρ δV. Dt Dt Putting all these components together and dividing by ρδV we obtain Du 1 1 = − ∇P − ∇Φ + ν∇2u + ∇ (∇.u). Dt ρ g 3 In this form, the equation of motion is usually called the Navier-Stokes equation. Transformation to a Rotating Frame of Reference Newton’s laws of motion and therefore the Navier-Stokes equa- tion apply only in an inertial frame of reference. When con- sidering the flow of rotating bodies such as the Earth (or other rotating bodies such as the sun or galaxies), it is convenient to choose coordinate axes which rotate with the body e.g. fixed relative to the Earth. There are 2 reasons: • (i) avoids considering the large tangential velocity associated with the rotation of the Earth (or other body). • (ii) the atmosphere and oceans are observed from the rotat- ing frame of reference. Note: The Earth rotates about an axis through the poles with angular velocity Ω. |Ω| = 7.27 × 10−5s−1 = 2πday−1 and so we must modify the Navier-Stokes equation. Let uI = velocity in inertial frame of reference. uR = velocity in rotating frame of reference. By definition uI = uR + Ω × r. This can be written as dr dr = + Ω × r dt I dt R and the above is a special case of dA dA = + Ω × A dt I dt R where A is an arbitrary vector (see e.g. Gill’s book). Applying d to uI, we get dt I du d = (uR + Ω × r) + Ω × (uR + Ω × r) . dt I dt R Hence du du = + 2Ω × uR + Ω × (Ω × r) . dt I dt R Applying this to a fluid element (and dropping the subscript R for convenience), we get the Navier-Stokes equation in a rotating frame of reference: Du + 2Ω × u = −1∇P − Ω × (Ω × r) − ∇Φ Dt ρ h 2 1 i +ν ∇ u + 3∇ (∇.u) . 1 2 • (i) Ω×(Ω × r) = 2∇ (Ω × r) is the centripetal acceleration and can be combined with the gravitational potential to form the geo-potential 1 Φ = Φ − (Ω × r)2 (2.24) g 2 The gradient g = −∇Φ is the effective gravity corrected for the centrifugal acceleration. However, the corrections to g due to the centrifugal acceleration are very small (How small?) so that g is approximately constant (variations in g caused by the variation in the Earth’s radius are as impor- tant). • (ii) 2Ω × u is the Coriolis Force. Thus the appropriate equation of motion to use in a rotating frame of reference is Du 1 h 1 i + 2Ω × u = − ∇P + g + ν ∇2u + ∇ (∇.u) Dt ρ 3.
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