gas dynamics -...

Gas Dynamics

PHY 688: Astrophysical Fluids and Plasmas

Gas Dynamics

● Gas dynamics is the study of compressible fow

– Typically arising when Mach number is ~O(1)

● Very diferent fow phenomena can result when compressibility is important

– Shocks

– Rarefatons

● We’ll mostly ignore viscous efects here

The Equaton of State

● Choudhuri assumes an ideal gas throughout—we’ll point out some diferences with a general gas along the way

– R is the specifc gas constant

– Usually in stellar astrophysics, we cannot assume an ideal gas

● Internal energy (monatomic gas):

● 1st law of thermodynamics relates the heat added to a system to the change in internal energy and the work done on the system

– A useful quantty to work with is the specifc heat

– with

– cx is the amount of heat, dq, that must be added to a system, at constant x, to raise T by a unit amount.

● Specifc heat at constant volume:

– Then

● Note, for constant cv, we can write

● Next cp. We treat T and p as independent variables.

– From specifc enthalpy: h = e + p/ρ

– or

– giving

● Assuming ideal gas:

● Rato of specifc heats,

– This difers for diferent gases● The internal energy is split between translatonal, rotatonal,

vibratonal, etc. degrees of freedom.● The nature of the gas gives the number of degrees of freedom.● Air (diatomic molecules) is γ = 1.4

● We already saw entropy

● Meaning of γ:

– In general, a large γ means that adiabatc compression yields a large pressure increase. A small γ means adiabatc compression gives a small pressure increase.

– A special case is isothermal fow. Here, compression is not allowed to heat the fuid, so γ = 1 (this corresponds to infnite d.o.f)

● General gas:

– Entropy relaton:

● It’s common in gas dynamics to use a gamma-law equaton of state

Linear Acoustcs(Shore Ch. 9)

● Consider isentropic fow, so the fow is described by

– Assume that the fow is at rest, and consider perturbatons about some constant background

● insertng these into the fow equatons, and keeping to frst order, we have

● We can eliminate δu to arrive at:

– now, when the fow is isentropic,: p(ρ, s) = p(ρ)

● Defning

– we can write this as a second order wave equaton

● The soluton to this is simply lef and right going waves, propagatng at speed c.

– Here, ±c are the characteristc speeds of the linear acoustcs system. They are the speed at which informaton propagates.

● As we will see, when we add back in the energy (or entropy) equaton, for the full Euler equatons, there will be a third speed

Sound Speed

● We defne the sound speed of the fuid as

– This is the speed at which disturbances propagate with in the linear acoustcs equatons.

● For the ideal gas law, the sound speed is

– For the stellar equaton of state it is:

Sound Speed(Choudhuri Ch. 6)

● Going back to the wave equaton,

– Consider evoluton of a Fourier mode

– we get

● The phase velocity is vp = ω/k = c

● The group velocity is vg = ∂kω = c

– Sound waves are non-dispersive.

– Note we assumed a uniform medium

Conservatve Form

● The Euler equatons can be writen as

– with

– and

● This can be writen in quasi-linear form

– where A is the Jacobian.

Conservatve System

● Express fux vector in terms of

Conservatve Form

● The Jacobian is

● The eigenvalues are

– These are the characteristc speeds of the system—the speeds at which informaton propagates.

– They tell us a lot about the structure of the soluton, and we will exploit them when developing our numerical methods.

Primitve Variable Form

● We derived the Euler equatons in terms of density, momentum, and pressure, but these are not quanttes that we have a real physical intuiton for.

● It is sometmes useful to consider instead density, velocity, and pressure.

● The density equaton is unchanged from the conservatve form,

● Working out the derivatves in the momentum equaton

– Subtractng of the contnuity equaton, we get the momentum equaton in non-conservatve form,

– and fnally,

● You derived a pressure equaton on your homework:

● For a general EOS, this is:

● As with the conservatve set of equatons, we can write this system in quasi-linear form:

– Note the Jacobian here is much simpler than that in conserved form.

● These equatons describe the same physical system as the conservatve form, so they should have the same eigenvalues.

Acoustc waves to Shocks

● We showed that for small perturbatons in the fow,

● the Euler equatons reduce to a single acoustc wave equaton,

● In deriving this expression, we neglected any second order terms, in partcular, we neglected

– This nonlinear term can have dramatc consequences when the perturbed velocity is no longer small.

– In fact, when the amplitude of the waves is not small, it does not make sense to do the perturbatonal analysis.

Shocks

● Consider the one-dimensional momentum equaton,

– There are three variables here

– We need to include the mass and energy equatons and the equaton of state.

● We can gain some insight into the role of the non-linearity by considering a simpler equaton:

– This is sometmes known as the inviscid Burgers’ equaton.

Shocks

● To solve this, consider the curves

– the total tme derivatve of this is:

– This is just the equaton we are trying to solve, so u is constant along the curves dx/dt = u

– We call these the characteristc curves.

– Along these curves the PDE becomes an ODE.

Shocks

● A quantty that is constant along a characteristc curve is called a Riemann invariant.

● In this simple example, u is a Riemann invariant.

– Since dx/dt = u, the characteristc curves are straight lines.

● These tell us that u associated with a fuid element does not change as that element moves.

● Consider the velocity profle:

Shocks

● If we just move along the characteristc curves, the functon becomes multple-valued

Shocks

● This is clearly unphysical: u(x) cannot be multvalued

● The physical soluton is to place a discontnuity there—a shock wave.

– Since the soluton is no longer smooth, the PDE is invalid, and we need to consider the equatons in integral form.

Shock position

Shocks(Laney Ch. 2)

● We can stll solve the PDEs

– Treatng them as shorthand for the integral equatons

– Using jump conditons to relate the change in fuid variables across discontnuites.

● Such solutons to the PDE are called weak solutons

● The jump conditons are given by the Rainkine-Hugoniot relatons

– These can be obtained from the integral form of the equatons (see, e.g., LeVeque 11.8):

– Here, S is the shock speed.

Initally Discontnuous Data(Toro Ch. 2)

● We can get a beter feel for nonlinear waves by looking at inital discontnuites.

● Consider two states initally separated by a jump at an interface.

– Here, the characteristc speeds on the lef are greater than those on the right.

– Immediately, the characteristcs will intersect, creatng a shock.

● The shock speed is such that λ(uL) > S > λ(uR), where λ is the characteristc speed.

– This is called the entropy conditon (shock can only arise in compressive region)

● Now let's consider an alternate case,

● Here, the characteristcs diverge

● It is incorrect to put a shock between the states

– Inital discontnuity did not arise as the result of compression.

– Shock here would violate the entropy conditon.

● Proper soluton here is a rarefacton wave.

tailhead

● A rarefacton wave is a nonlinear (expansion) wave that smoothly connects the lef and right states.

– Head of the rarefacton moves at the speed λ(uR)

– The tail of the rarefacton moves at the speed λ(uL)

● General conditon for a rarefacton wave is λ(uL) < λ(uR)

● Both rarefactons and shocks are present in the solutons to the Euler equatons.

– Both of these waves are nonlinear.

● In general a third type of wave can exist, a contact discontnuity: λ(uL) = λ(uR)

● We can look at wave steepening in Burgers’ equaton numerically:

● And rarefactons...

● Graphically, we can look at these diferent waves in terms of the characteristc speeds.

– The elementary wave solutons are:

Shock wave Contact Discontinuity

Rarefaction

Rankine-Hugoniot Conditons(LeVeque Ch. 11)

● The shock speed is determined by the states immediately to the lef and right of the shock

● We can get jump conditons for a general conservaton law

– Consider a short tme interval, t to t + Δt.

– In this tme, the shock speed is essentally constant.

● Apply the integral form of the conservaton law in this interval:

● Integratng in tme:

● Now, if the intervals are small, then q is constant, and we have

– or taking the limit

● Now shock speed is defned in terms of conservatve fuxes and variables:

● For Burger's equaton the shock speed is

● Shock speed can vary in tme (as states do)

● This methodology holds for systems of conservaton laws, but then q and f(q) are vectors

● Recall when the data is not smooth, the diferental equaton does not hold.

– This means the soluton may not be unique.

● Consider Burgers’ equaton:

● and the same equaton multplied by 2u

● The solutons to these diferental equatons are identcal. However, these conservaton laws yield diferent shocks speeds.

● For the original equaton:

– and

● The trouble here is that our manipulaton in deriving the second equaton (multplying by 2u) is not valid for discontnuous data.

– Only the integral form of the equatons is truly valid with discontnuous data.

Euler Equaton Jump Conditons(Toro Ch. 2)

● Back to the Euler equatons—what are the jump conditons?

● Consider a right moving shock, and transform to the frame of the shock

Shock Jump Conditons

● We need to work in conservatve form:

– These can be combined to yield the shock speed.

● First we can fnd the density jump (blackboard):

● We want shock speed in terms of the pressure jump.

– We will make use of this relatonship when we solve the Euler equatons later in class.

● Blackboard...

● Now we switch back to the statonary frame of reference

● Dividing through by the sound speed ahead of the shock, we have

● We can solve the pressure relaton for the shock speed:

● then we have

– The sign was chosen such that the shock is supersonic.

● In the future, we will want an expression giving the post-shock velocity, u* in terms of the pressure jump.

– You'll work this out on your homework.

Sedov Blast Wave

● Put a large amount of energy in a small volume—blast wave forms

– Shock moves outwards, material evacuated behind

– Evoluton is self-similar

● Soluton discovered in 1940s

– Sought as a model for air explosion of nuclear weapons

– Usually referred to as Sedov-Taylor blast wave afer two independent papers

Sedov Blast Wave

● Soluton: dump energy E into ambient medium w/ density ρ1

– Ambient pressure negligible

● Only combinaton of t, E, and ρ1 that is length is:

● Assume (shown later) self-similar evoluton, r(t) must evolve as λ

– Each shell labeled with ξ

Sedov Blast Wave

● Shock front labeled ξ0

– Shock radius:

– Shock velocity:

● Velocity drops with tme

Sedov Blast Wave

● Statonary frame:

– Shock propagates w/ velocity vs

– ρ2, v2, p2 are conditons just inside shock front

● Shock reference frame:

– Material comes in w/ velocity −vs

– Material leaves w/ velocity −vs + v2

● Jump conditons (M 1)≫ :

Sedov Blast Wave

● Picking appropriate scalings, we can construct dimensionless variables

● Introducing these into the Euler equatons gives a set of coupled nonlinear ODEs

– Only a functon of ξ—self-similarity is found

● Soluton method:

– Guess value for ξ0

– Integrate from ξ0 to 0

– Check if we conserved energy:

Sedov Blast Wave

● Soluton is used commonly as a test problem for hydrodynamics codes

Applicaton: Jets in Astrophysics(Choudhuri Ch. 6)

● Let's look at fow through a channel of varying cross-sectonal area.

– This is a one-dimensional problem.

– If the fow is steady, then conservaton of mass gives

● If we consider the fow to be adiabatc, then p ∝ ργ, and we get

Applicaton: Jets in Astrophysics

● The momentum equaton becomes

– from the contnuity equaton, we get:

– and we can eliminate the density derivatve:

– where we have defned the Mach number, M = v/c

● Subsonic fows, M < 1: dv/dx and dA/dx have opposite signs.

– Making the pipe narrow results in acceleratng fow.

– This makes sense intuitvely

● Supersonic fows, M > 1: dv/dx and dA/dx have the same sign

– In a region of increasing cross-secton, the fow will accelerate!

– Supersonic fow behaves very diferently than subsonic.

● Now consider the possibility of the fow being subsonic upon entering and supersonic at exit.

● How do we achieve this?

● We need M = 1 at some point in the pipe.

– The only way to get this is by having dA/dx = 0 somewhere in the pipe.

– Therefore the pipe needs to be shaped as above.● This is called a de Laval nozzle.● This geometry is at the core of every rocket engine.

converging fow diverging fow

Longitudinal secton of RD-107 rocket engine (Tsiolkovsky State Museum of the History of Cosmonautcs)

Albina-belenkaya/Wikipedia

Blandford & Rees (1974) suggested that this could explain extra-galactc jets. The light gas in the jet is produced near the center of the galaxy as pushes its way through the denser interstellar medium. The pressure of the medium can for a nozzle that accelerates the material to supersonic speeds.

“If the relatvistc plasma escapes along two cylindrical 'channels', whose cross-sectonal radii are r(R), the 'nozzle' forms at a radius comparable with the pressure scale-height for the gas, ... The only general requirement is that the external pressure should be high enough to make the nozzle radius r* much lass than R*.”

Blandford and Rees

● Astrophysical jets are more complicated than this

● Observatons suggest that this mechanism cannot explain all jets

● Magnetc felds likely play a role in the confnement

Applicaton: The Solar Wind(Choudhuri Ch. 6)

● Recall: we worked out a statc model for the solar corona

– Found T → 0 as r → ∞, but p → fnite constant value

– Soluton: corona is not in equilibrium

Applicaton: The Solar Wind(Choudhuri Ch. 6)

● Parker (1958) also worked out a spherical model for the solar wind

– Magnetc felds make the problem somewhat non-spherical

● In steady spherical fow, the contnuity equaton is

– Assume the gravitatonal acceleraton to come from a central mass

– Assume that we are isothermal, then p = ρc2 and

Applicaton: The Solar Wind

● Diferentatng the contnuity equaton gives:

– which we can combine with

– to give

● Collectng terms, we have

● If v = c then the lef hand side is zero:

– so the radius at which we can go sonic is

● Defning ξ = r/rc and M = v/c,

● What do the solutons look like?

● Notce that at ξ = 1, the righthand side wants to be 0

– For dM2/dξ to be non-zero, we must have M → 1● ξ = 1 is a sonic point

– Otherwise, dM2/dξ = 0, and we’ll be a maximum or minimum w/ some M2 ≠ 1

● If M2 → 1 with ξ ≠ 1, then dM2/dξ → ∞

● The soluton is

– C is the integraton constant.

– Only two solutons are transonic.● Parker picked V, since it is

subsonic at the solar surface.● V is a good match to

observatons.

● Numerical integraton of

requires using l’Hôpital’s rule

– We fnd

– We can integrate outward (inward) from ξ = 1 ± ε to fnd the transonic soluton

● We can work out the form for an adiabatc gas (instead of isothermal)

– Reduces to the isothermal case for γ = 1

Applicaton: Bondi Accreton

● Interestngly, the same model applies to spherical accreton.

– “The problem to be discussed may be defied as follows: A star of mass M is at rest ii ai iifiite cloud of gas, which at iifiity is also at rest aid of uiiform deisity ρ∞ aid pressure p∞. The motoi of the gas is spherically symmetrical aid steady, the iicrease ii mass of the star beiig igiored so that the feld of force is uichaigiig.” —Boidi (1952)

● Bondi considered the case of an adiabatc gas,

Applicaton: Bondi Accreton

● If we assume isothermal, we arrive at the same soluton as in the solar wind problem.

● Bondi (1952) suggested that the correct soluton is the transonic soluton that is zero at infnity and supersonic at the surface of the star.

(from Bondi 1952)

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