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Summary of the Equations of Fluid Dynamics

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Summary of the Equations of Fluid Dynamics Summary of the Equations of Fluid Dynamics Reference: Fluid Mechanics, L.D. Landau & E.M. Lifshitz 1 Introduction Emission processes give us diagnostics with which to estimate important parameters, such as the density, and magnetic field, of an astrophysical plasma. Fluid dynamics provides us with the capability of understanding the transport of mass, momentum and energy. Normally one spends more than a lecture on Astrophysical Fluid Dynamics since this relates to many areas of astrophysics. In following lectures we are going to consider one principal application of astrophysical fluid dynamics – accretion discs. Note also that magnetic fields are not included in the following. Again a full treatment of magnetic fields warrants a full course. 2 The fundamental fluid dynamics equations The equations of fluid dynamics are best expressed via conservation laws for the conservation of mass, mo- mentum and energy. Fluid Dynamics 1/22 2.1 Conservation of mass n Consider the rate of change of mass within a fixed volume. This i changes as a result of the mass flow through the bounding surface. v ∂ V i ∫ ρdV = –∫ρv n dS ∂t i i S V S Using the divergence theorem, ∂ ∂ Control volume for as- ∫ ρdV +0∫ ()ρv dV = ∂t ∂x i sessing conservation of V V i mass. ∂ρ ∂ ⇒ ∫ ------ + ()ρv dV = 0 ∂t ∂x i V i The continuity equation Since the volume is arbitrary, ∂ρ ∂ ()ρ -----∂- +0∂ vi = t xi Fluid Dynamics 2/22 2.2 Conservation of momentum Consider now the rate of change of momentum within a vol- ume. This decreases as a result of the flux of momentum ni through the bounding surface and increases as the result of body forces (in our case gravity) acting on the volume. Let V Π ijn j S Π ij = Flux of i component of momentum in the j direction and f i = Body force per unit mass then ∂ ∫ ρv dV = –∫Π n dS + ∫ ρ f dV ∂t i ij j i V S V Π Π th There is an equivalent way of thinking of ij, which is often useful, and that is, ijn jdS is the i component of the force exerted on the fluid exterior to SS by the fluid interior to . Fluid Dynamics 3/22 Again using the divergence theorem, ∂ ∂Π ∫ ()ρv + ij dV = ∫ ρ f dV ∂t i ∂x i V j V ∂ ∂Π ⇒ρ()ρ ij ∂ vi + ∂ = f i t x j Gravity For gravity we use the gravitational potential ∂φ G f i = –∂ xi For a single gravitating object of mass M GM φ = –--------- G r Fluid Dynamics 4/22 and for a self-gravitating distribution ∇2φ π ρ G = 4 G ρ()x ′ ⇒ φ = –G ∫ -------------------i d3x′ G x – x ′ V′ i i where G is Newton’s constant of gravitation. Π Expressions for ij The momentum flux is composed of a bulk part plus a part resulting from the motion of particles moving with () respect to the centre of mass velocity of the fluid vi . For a perfect fluid (an approximation often used in as- trophysics), we take p to be the isotropic pressure, then Π ρ δ ij = viv j + p ij The equations of motion are then: ∂ ∂ ∂φ ()ρ ()ρ δ ρ G ∂ vi + ∂ viv j + p ij = – ∂ t x j xi ∂ ∂ ∂ ∂φ ⇒ ()ρ ()ρ p ρ G ∂ vi + ∂ viv j = – ∂ – ∂ t x j xi xi Fluid Dynamics 5/22 There is also another useful form for the momentum equation derived using the continuity equation. ∂ ∂ ∂ρ ∂v ∂ ∂v ()ρ ()ρ ρ i ()ρρ i ∂ vi + ∂ viv j = vi∂ ++∂ vi∂ v j +v j∂ t x j t t x j x j ∂ρ ∂ ∂v ∂v ()ρ ρ i ρ i = vi ∂ + ∂ v j + ∂ + v j∂ t x j t x j ∂v ∂v ρ i ρ i = ∂ + v j∂ t x j Hence, another form of the momentum equation is: ∂v ∂v ∂ ∂φ ρ i ρ i p ρ G ∂ + v j∂ = – ∂------- – ---------∂ t x j xi xi On dividing by the density ∂ ∂ ∂φ vi vi 1 ∂p G ∂ + v j∂ = – ρ---∂------- – ---------∂ t x j xi xi Fluid Dynamics 6/22 Differentiation following the motion (), This is a good place to introduce differentiation following the motion. For a function fxi t , the variation of f following the motion of a fluid element which has coordinates () xi = xi t is given by: df ∂ f ∂ f dxi ∂ f ∂ f ----- ==∂ + ∂ ------- ∂ + vi∂ dt t xi dt t xi Hence, the momentum equation can be written compactly as dv ∂ ∂φ ρ i p ρ G ------- = – ∂ – ∂ dt xi xi Fluid Dynamics 7/22 2.3 Thermodynamics m Before going on to consider the consequences of the conservation of S energy, we consider the thermodynamics of a comoving volume ele- U ment. (See the figure at left.) V p Element of fluid and the variables used to describe its state. Define: m = Mass of element ε = Internal energy density per unit volume P = pressure (as above) s = entropy per unit mass T = temperature (in degrees Kelvin) Fluid Dynamics 8/22 We have the following quantities for the volume element: mε U ==Total internal energy ------ρ- S ==Entropy ms m V ==volume ---ρ- The second law of thermodynamics tell us that the change in entropy of a mass of gas is related to changes in other thermodynamic variables as follows: kTdS= dU+ pdV ε ⇒ () m m kTd ms = d------ρ- + pd---ρ- ε 1 1 ()ε + p ⇒ kTds ==d --- + pd --- ---dε – -----------------dρ ρ ρ ρ ρ2 ()ε ⇒ ρ ε + p ρ kTds = d – ----------------ρ -d Fluid Dynamics 9/22 Specific enthalpy A commonly used thermodynamic variable is the specific enthalpy: ε + p h = -----------ρ - In terms of the specific enthalpy, the equation ε 1 kTds= dρ--- + pdρ--- becomes ε + p p 1 dp kTds== d-----------ρ - – d--ρ- + pdρ--- dh – -----ρ- For a parcel of fluid following the motion, we obtain, after dividing by the time increment of a volume element, ds dε ()ε + p dρ ρkT----- = ----- – ----------------------- dt dt ρ dt ds dh 1dp kT----- = ------ – --------- dt dt ρ dt Fluid Dynamics 10/22 The fluid is adiabatic when there is no transfer of heat in or out of the volume element: dε ()ε + p dρ ----- –0----------------------- = ds dt ρ dt kT----- = 0 ⇒ dt dh 1dp ------ –0--------- = dt ρ dt The quantities ds, dε,dp etc. are perfect differentials, and these relationships are valid relations from point to point within the fluid. Two particular relationships we shall use in the following are: ∂s ∂ε ∂ρ ρkT = – h ∂t ∂t ∂t ∂ ∂ ∂ ρ s ρ h p kT∂ = ∂ – ∂ xi xi xi 2.3.1 Equation of state The above equations can be used to derive the equation of state of a gas in which the ratio of specific heats γ ⁄ (= cp cv ) is a constant. Consider the following form of the entropy, internal energy, pressure relation: ()ε ρ ε + p ρ kTds= d – ----------------ρ -d Fluid Dynamics 11/22 In a perfect gas, ρkT p = ----------µ - mp where µ is the mean molecular weight and p = ()εγ – 1 ⇒γεε + p = Hence, γε µ ()εγ ε ρ mp – 1 ds= d – ----ρ-d ε γ ⇒ µ ()γ d ρ mp – 1 ds = -----ε – ρ---d ⇒εγρµ ()γ () mp – 1 ss– 0 = ln – ln ε ⇒µ----- ==exp[]m ()γ – 1 ()ss– exp[]µm ()γ – 1 s ργ p 0 p (We can discard the s0 since the origin of entropy is arbitrary.) Fluid Dynamics 12/22 We therefore have, εµ[]()γ × ργ = exp mp – 1 s ()µγ []ρ()γ × γ p = – 1 exp mp – 1 s = Ks()ργ The function Ks() is often referred to as the pseudo-entropy. For a completely ionised monatomic gas γ = 53⁄ . 2.4 Conservation of energy Take the momentum equation in the form: ∂v ∂v ∂ ∂φ ρ i ρ i P ρ G ∂ + v j∂ = – ∂------- – ---------∂ t x j xi xi and take the scalar product with the velocity: ∂v ∂v ∂ ∂φ ρ i ρ i P ρ G vi∂ + v jvi∂ = – vi∂------- – vi---------∂ t x j xi xi ∂ ∂ ∂ ∂φ ⇒ ρ 1 ρ 1 P ρ G ∂ ---vivi + v j∂ ---vivi = – vi∂ – vi---------∂ t 2 x j 2 xi xi Fluid Dynamics 13/22 That is, ∂ ∂ ∂ ∂φ ρ 1 2 ρ 1 2 p ρ G ∂ ---v + v j∂ ---v = – vi∂ – vi---------∂ t 2 x j 2 xi xi ρρ Before, we used the continuity equation to move the and v j outside the differentiations. Now we can use the same technique to move them inside and we recover the equation: ∂ ∂ ∂ ∂φ 1ρ 2 1ρ 2 p ρ G ∂ --- v + ∂ --- v v j = – vi∂ – vi---------∂ t 2 x j 2 xi xi The aim of the following is to put the right hand side into some sort of divergence form. Consider first the term ∂ ∂ ∂ P ρ s ρ h –vi∂ = kTvi∂ – vi∂ xi xi xi ∂ ∂ ρ ds ρ s ρ h = kT----- – kT∂ – vi∂ dt t xi ∂ε ∂ρ ∂ ρ ds ρ h = kT----- – ∂ + h∂ – vi∂ dt t t xi Fluid Dynamics 14/22 ∂ρ We now eliminate the term using continuity, viz ∂t ∂ρ ∂ ()ρ ∂ = –∂ vi t xi and we obtain ∂ ∂ε ∂ ∂ p ρ ds ()ρ ρ h –vi∂ = kT----- – ∂ – h∂ vi – vi∂ xi dt t xi xi ∂ε ∂ ρ ds ()ρ = kT----- – ∂ – ∂ hvi dt t xi Fluid Dynamics 15/22 The term ∂φ ∂ ∂ ρ G ()ρφ φ ()ρ – vi∂ = – ∂ Gvi + G∂ vi xi xi xi ∂ ∂ρ ()φρφ = –∂ Gvi – G∂ xi t ∂ ∂ ∂φ ()ρφ ()ρρφ G = –∂ Gvi – ∂ G + ∂ xi t t When the gravitational potential is constant in time, ∂φ ∂φ ∂ ∂ G ⇒ ρ G ()ρφ ()ρφ ∂ = 0 – vi∂ = –∂ Gvi – ∂ G t xi xi t Hence, the energy equation ∂ ∂ ∂ ∂φ 1ρ 2 1ρ 2 P ρ G ∂ --- v + ∂ --- v v j = – vi∂ – vi---------∂ t 2 x j 2 xi xi Fluid Dynamics 16/22 becomes ∂ ∂ ∂ε ∂ 1ρ 2 1ρ 2 ρ ds ()ρ ∂ --- v + ∂ --- v v j = kT----- – ∂ – ∂ hvi t 2 x j 2 dt t xi ∂ ∂ ()ρφ ()ρφ –∂ Gvi – ∂ G xi t Bringing terms over to the left hand side: ∂ ∂ 1ρ 2 ερφ 1ρ 2 ρ ρφ ρ ds ∂ --- v ++ G + ∂ --- v v j ++hvj Gv j = kT----- t 2 x j 2 dt When the fluid is adiabatic ds ρkT----- = 0 dt and we have the energy equation for a perfect fluid: ∂ ∂ 1ρ 2 ερφ 1ρ 2 ρ ρφ ∂ --- v ++ G +0∂ --- v v j ++hvj Gv j = t 2 x j 2 Fluid Dynamics 17/22 The total energy per unit volume is 1 E ==---ρv2 +ερφ + Kinetic + internal + gravitational energy 2 G and the energy flux is 1 1 F ==---ρv2v ++ρhv ρφ v ρ ---v2 ++h φ v Ei, 2 i i G i 2 i = Kinetic + enthalpy + gravitational fluxes 3 Fluids with viscosity In most astrophysical contexts we do not have to consider molecular viscosity since it is generally small.
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