IV. Compressible flow of inviscid fluids

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1 IV. Compressible flow of inviscid fluids Governing equations for n = 0, r const: + (u )=0 t u + ( u ) u= p t De e = + ( u ) e= p u+ ( k T ) Dt t p= p(, T ), e=e (,T )

2 Alternate forms of energy equation Equation for enthalpy h = e + p/r h p + ( u ) h= +( u ) p+ ( k T ) t t (derivation similar to what we did in Ch. 0) Equation for negligible heat conduction and perfect gas Start with De = p u Dt

3 For perfect ideal gas, e=e(t )=c v T, p= R T In this case De De DT DT = =c v Dt DT Dt Dt From continuity equation, ( u )= u+ u= t D u= +( u ) = t Dt ( )

4 Insert De DT =c v Dt Dt and D u= Dt Into the energy equation - DT p D cv = Dt Dt Recall that (for ideal gas) T = p/(rr) ( ) D p p D cv = Dt R Dt cv Dp D p D p = R Dt Dt Dt ( )

5 cv Dp c v p D p D = + R Dt R Dt Dt Multiply both parts by R/(cv p) ( ) Dp D R D D R = + = + p Dt Dt c v Dt Dt cv Note that R = cp - cv, so R c v +c p c v c p + = = =γ cv cv cv With that, Dp γ D, or = p Dt Dt D γ (log p log )=0 Dt

6 Integrate for the same fluid volume: p γ =const Isentropic law (inviscid fluid, no heat transfer)

7 .. Propagation of infinitesimal disturbances (a.k.a. linear acoustics) Assumptions D setting (x-axis only) Perfect gas initially at rest No heat conduction Small disturbance propagates in x-direction Governing equations Continuity (D): + (u )=0 + ( u )=0 t t x

8 Momentum (D) u u u p + ( u ) u= p +u = t t x x Energy (perfect gas, no heat transfer) p γ =const For isentropic gas, p = p(r), so p dp = x d x Rewrite continuity and momentum with this - u +u + =0 t x x u u dp +u + =0 t x d x

9 Assumptions for linearization Hydrostatic value p= p0 + p ' =0+' u=u ' Assume perturbed (') values to be small, plug into governing equations... Product: nd order r0 = const r0 = const u' 0+' )+u ' 0+' )+( 0+' ) =0 ( ( t x x Product: nd order ' u ' +0 =0 t x

10 Product: nd order Not a function of x u ' u ' d ( p 0+ p ' ) ( 0 + ' ) +u ' + =0 t x d x Product: nd order Note that d ( p 0+ p ' ) d dp = d p= p0, d p + p' d = 0 ( + p = p0, =0 ) ' = + ' = 0 +' / = Thus after linearization u ' d p ' + =0 t 0 d 0 x

11 Differentiate first equation (continuity) in t ' u' +0 =0 x t t Take second equation (momentum), multiply by r0, differentiate in x u ' d p ' 0 + =0 t x d 0 x Subtract the two equations to eliminate the crossdifferential term ' d p ' =0 d t 0 x

12 Now differentiate continuity in x ' u' +0 =0 t x x...and momentum in t u ' d p ' + =0 0 d 0 x t t Multiply first equation by... dp 0 d 0 Then subtract first equation from the second

13 u' d p u' =0 d t 0 x Let dp a 0= d 0 Speed of sound Equations can be rewritten as 0 0 ' tt a ' xx=0 u ' tt a u ' xx=0 D'Alembert's equations (wave equations) General solution for r' (same form for u') ' = f ( x a 0 t )+g ( x+a 0 t ) Wave traveling right Wave traveling left

14 For this theory (linear acoustics) to work, must have: ' p' u',, 0 p0 a0 Speed of sound in perfect gas evaluate using isentropic energy equation p p0 γ p0 p= γ γ= γ, 0 0 dp γ γ p0 γ γ p 0 =γ γ= γ= p d 0 0 Also for ideal gas p = rrt, so dp γ = p=γ R T d

15 Thus for the speed of sound we can write p0 a 0= γ RT 0 = γ 0

16 .. Propagation of finite disturbances Start with governing equations from previous section before linearization... u +u + =0 t x x u u dp +u + =0 t x d x p γ =const From previous section, u = u(r), p = p(r) only (both also depend on initial conditions)

17 If u = u(r), we can also rewrite that as r = r(u) Assuming the same dependence for finiteamplitude case (r = r(u), p = p(r)), use chain rule for derivatives in governing equations d u = x du x d u = t d u t p d p d u = x d du x Continuity equation becomes... d u d u u +u + =0 du t du x x ( ) d u u u +u + =0 du t x x

18 For the momentum equation... u u dp d u +u + =0 t x d d u x From continuity, ( ) u u d u d u u +u = / = t x du x d x Plug that into momentum, move pressure term to the right, lose the - signs d u u dp d u = d x d du x d u dp d = d d du

19 ( ) du dp = d d Let Then du dp =± d d dp a= d du a =±, d For small amplitude, a a0 (speed of sound) du d =± a Use these expressions to alter the momentum equation... a ± a u u dp d u +u + =0 t x d d u x

20 For a forward-propagating wave ( +), u u u +u +a =0 t x x Rewrite this as u u +( u+a ) =0 t x A general solution of this equation (forward wave) Any differentiable function (particular solution from IC) u= f ( x ( u+a ) t ) Functions of x,t

21 Use the polytropic equation to rewrite a in terms of speed of sound a0 p p0 γ= γ 0 ( ) ( ) dp d γ p0 γ p0 = γ =γ γ 0 0 d d dp γ p 0 a= = γ γ 0 d From previous section, / p0 = γ 0 0 p0 a 0= γ 0 ( ) γ

22 Thus a=a0 0 ( ) Recall that we had... γ du a =± d For a forward wave (+), d d u=a Rewrite this using the expression with a0 d u=a 0 0 ( ) γ a0 d = γ 0 γ d

23 d u= a0 (γ )/ 0 (γ 3 )/ d Integrate this in r to get a0 (γ )/ u= +const (γ )/ γ 0 When r r0, u 0: 0= a0 +const γ Thus const= a0 γ

24 Rewrite the expression for u a0 (γ )/ u= a0 (γ )/ γ 0 γ ( u= a0 0 γ Recall that ( ) a=a0 0 γ ( ) Thus a 0 γ u= a a 0 ) ( γ )

25 The same expression with respect to a: γ a= u+a0 The general solution was of the form... u= f ( x ( u+a ) t ) Wave speed For wave speed of the forward wave, ( ) γ γ+ U =u+a=u + +a 0=a 0+ u The faster the local speed, the faster the wave wants to go!

26 Evolution of a finite-amplitude velocity disturbance t = t u x

27 Evolution of a finite-amplitude velocity disturbance t = t u x

28 Evolution of a finite-amplitude velocity disturbance t = t3 u x

29 Evolution of a finite-amplitude velocity disturbance t = t4 u Oops... multivalued velocity and pressure... cannot have this! x

30 Note: this problem arises only for a positivepressure perturbation (a negative-pressure finite perturbation, a.k.a. rarefaction wave, will disperse nicely)

31 What really happens to finite-strength positivepressure perturbation in gas... p = ps, r = rs, u = us Piston velocity (less than U) p = p0, r = r0, u = 0 Shock front Shock front speed U Ernst Mach Mach number M = U/a0 > Here continuum approximation breaks down!

32 Explosion of Tsar-Bomba (AH60, ~60 megaton thermonuclear device), Kurchatov, Khariton, Sakharov

33 Explosion of Tsar-Bomba (AH60, ~60 megaton thermonuclear device), Kurchatov, Khariton, Sakharov

34 M65 atomic cannon test (Upshot-Knothole test series, 953)

35 M65 atomic cannon test (Upshot-Knothole test series, 953)

36 .3. Rankine-Hugoniot equations Normal shock p = p, r = r, u = u p = p, r = r, u = u Shock front Reference frame moves with shock front (discontinuity in the framework of continuum theory, width ~ mean free path of a molecule) (as opposed to crazy mad shock) (velocity normal to shock front) Pierre Henri Hugoniot (85-887) Conservation equations u = u Mass Momentum jump on the interface due to pressure difference u + p= u + p Momentum fluxes Momentum

37 Energy equation u u +c p T = +c p T Rewrite enthalpy per unit mass cpt using ideal gas equation p = rrt: c p / cv p p p γ p c p T =c p =c p = = R (c p c v ) c p / c v γ With that, energy equation becomes u γ p u γ p + = + γ γ

38 Divide momentum equation by mass equation... u + p= u + p u = u p p u + =u + u u Rearrange p p p p u u = = u = u u (mass equation) Multiply this by (u+u) ( ) p p p p u u u = u +u )= + ( u u

39 Divide mass equation by ru u = u Use this to replace u/u in energy equation p p u u = + =( p p ) + ( ) ( ) Rewrite an earlier form of energy equation to get ( γ p γ p u u = γ γ )

40 With a little tidying up... ( ) p p γ =( p p ) + γ ( ) Multiply by r γ p p =( p p ) + γ ( ) ( ) Collect terms with density ratio, solve for it γ+ p+ p u γ = u = γ+ p + p γ From energy eq. RankineHugoniot equations

41 .4. Conditions for normal shock waves Across the shock, the flow is not isentropic log p γ =const log This part cannot be realized - violates nd law of thermodynamics p p

42 With more algebra (using the expression we had for velocity difference...) u u=a * Prandtl-Meyer relation We will be able to use it, but first... How exactly does the second law of thermodynamics apply at the shock front?

43 Calorically perfect gas (Appendix E.5) Perfect gas equation of state p= R T Specific heat at constant volume [ ] [ dq e h h C v= = = + v dt v T T p ][ ] p T v (Enthalpy h = e + pv) Specific heat at constant pressure [ ] dq C p= dt [ h e e = = + +p T T v p ][ ] v T p

44 Perfect gas equation of state p= R T pv=rt pdv+vdp=r dt dh=d (e+ pv )=de+ pdv+vdp dh=de+rdt h e =R T T From previous slide, e h C v=, C p= T T Thus R=C p C v

45 Can further show that for perfect gas e=e(t )= C v dt +const h=h (T )= C p dt +const Gas is called calorically perfect if C p =const, C v =const Then for calorically perfect gas e=e(t )=C v T +const h=h (T )=C p T +const

46 Second law of thermodynamics Uniqiely determined by (Appendix E.8) the state of the system Entropy s thermodynamic variable of state Consider a system in equilibrium state A By adding heat Q to the system, we move it to another equilibrium state B B Q= A dq Introduce entropy s so that the change in s between equilibrium states A and B is B s B s A= A dq T Evaluated for reversible process, i.e. change is so slow that system remains in thermodynamic equilibrium

47 Statement of the second law of thermodynamics For any spontaneous process, the entropy change is non-negative B s B s A A dq T Evaluated for reversible process Consider calorically perfect gas s B s A=[ C p log T Rlog p ]B [ C p log T R log p ] A

48 Now let states and correspond to ideal gas before and after the shock, then s s =Δ s= =[ C p log T R log p ] [C p log T R log p ] = T p =C p log R log T p For ideal gas, temperature T = p/(rr), so p p Δ s=c p log R log p p Δ s=c p ( ) p p log +log Rlog p p

49 p Δ s=( C p R) log C p log p p Δ s=c v log C p log p p Δs =log γ log Cv p State can be reached from state by some kind of quasi-equilibrium, slow isentropic (I) process, then s = s and [ ] p Δs =0=log γ log Cv p I

50 We arrive at state by shock acceleration (nearly instant, thus definitely not isentropic) use subscript RH [] p Δs =log γ log Cv p RH From previous slide, use [] [ ] [ ] p log =γ log p I Δs =γ log γ log Cv I RH

51 From second law of thermodynamics, ( [] []) Δ s=c v γ log log I log >0 RH p γ =const log Here Ds would be negative p p

52 Some interesting corollaries A shock wave is thermodynamically realizable if p log 0, log 0 p Recall that, from Rankine-Hugoniot equations, u = u Thus u = u

53 Definition for speed of sound from acoustics section p0 a 0= γ RT 0 = γ 0 Let a = gp/r, use that to rewrite energy equation yet again u a u a + = + γ γ Use * subscript to denote the case when u = a (M =, sonic flow) and * * u a γ+ + =a * + = a* γ γ ( γ ) ( )

54 With more algebra (using the expression we had for velocity difference...) u u=a * Prandtl-Meyer relation From the second law of thermodynamics, we had u = u Rewrite this as * u u or u u a

55 Earlier we had energy equation in the form... u a u a + = + γ γ We formally introduced case from the righthand side of the same energy equation if M =, u = a = u = a * * u a γ+ + = a* γ ( γ ) Take energy equation in the form u a γ+ + = a* γ ( γ )

56 Divide both parts by u a γ+ a + = u γ ( γ ) u * M * a ( ) (γ ) = + M (γ ) γ+ u * a ( ) M (γ )+ (γ ) = γ+ u M (γ )

57 Flip it over... * u a = M (γ+) M (γ )+ M ( γ+) M (γ )+ M M + M Moreover, from Prandtl-Meyer relation it follows that M

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