Gasdynamics 1-D compressible, inviscid, stationary, adiabatic flows

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1 Gasdynamics 1-D compressible, inviscid, stationary, adiabatic flows 1st law of thermodynamics ρ const Kontrollfläche 1 2 m u 2 u 1 z Q 12 +P 12 = ṁ } h 2 h {{} 1 Enthalpy Q (u2 2 u2 1 }{{} ) + g(z } 2 {{ z 1 } ) kin. Energy pot. Energy P 12

2 Gasdynamics for Q 12 +P 12 = 0 (i.e.: adiabatic nozzle flow) potential energy is neglected h 2 h (u2 2 u2 1 ) = 0 with h 0 = h+u 2 /2 (Index 0 for stagnation condition) h 02 h 01 = 0 ; h 0 = const for perfect gases γ = const. γ(= κ) = c p c v (= 1.4, for air) c p : specific heat capacity at constant pressure c v : spezific heat capacity at constant volume dh = c p dt

3 Gasdynamics Energy equation for 1-d adiabatic flow without mechanical energy: in dimensionless form c p T 0 = c p T u2 T 0 T = 1+ γ 1 2 u 2 γrt ;c p = γ γ 1 R Speed of sound: a = γrt Machnaumber: Ma = u/a T 0 T = 1+ 1 (γ 1) Ma2 2 for ideal gases with γ = konst, Q = P = 0

4 Example Determine for an isentrropic flowfield (γ = 1.4 ) a) the critical temperature ratio b) the critical pressure ratio c) the limit if the critical Mach number Ma = u/a for Ma! 1. stat. flowfield with ρ const perfect gas, Q = P = 0; γ = const. T 0 T = 1+ 1 u2 (γ 1) 2 γrt = 1+ 1 (γ 1)Ma2 2 isentropic flow: s =const. ( ) γ ( ) p T γ 1 ρ γ = = p 0 ρ 0 T 0

5 Example p and ρ only depend on the initial stagnation state and the Mach number Definition critical value T = T(T 0, Ma) p = p(p 0, Ma) ρ = ρ(ρ 0, Ma) ρ 0 = ρ 0 (p 0,T 0 ) T = T(Ma = 1) p = p(ma = 1) ρ = ρ(ma = 1)

6 Example = T 0 T = (γ 1)c2 = γ +1 2 isentropic: P 0 p = ( ) γ ρ0 ρ = ( γ +1 2 ) γ γ 1 T,p,ρ only depend on T 0,p 0,γ Definition critical Mach number Ma = u c = u γrt = u c 0 T T = u c 0 γ +1 2 Advantage: Only one value in the quotien is variable Ma u

7 Example There is a limit Ma for Ma From the energy equation: Ma 2 = 2 γ 1 (T 0 T 1) ; c pt 0 = c p T u2 Expansion into Vacuum: u umax ; T 0 Ma, while u is finite.

8 Example Relation between Ma and Ma Ma 2 = u2 c 2 = u2 c 2 c 2 c 2 c c 2 = = Ma 2T T = Ma 2T T T 0 = Ma 2 2 γ +1 T = (1+ 12 ) (γ 1)Ma2 (1+ 12 (γ 1)Ma2 )

9 = Ma 2 = (γ +1)Ma2 2+(γ 1)Ma 2 = f(ma,γ)

10 Example T T 0 (Ma ), p p 0 (Ma ), ρ ρ 0 (Ma ) Expansion into Vacuum: T 0 ; Ma ; Ma γ+1 γ 1 3 Ma* Ma* max Ma* Ma

11 19.5 A turbine engine sucks air from the atmosphere. Immediately before the compressor the pressure is p 1. p 0 = 10 5 N/m 2 T 0 = 287 K p 1 = N/m 2 A = mm 2 R = 287 Nm/(kgK) γ = 1.4 Compute the mass flux flowing through the engine!

12 19.5 m = ρv 1 A Ablösung

13 19.5 Separation Losses p 0 not constant Q+P t = 0 h 0 = konst. separation no isentropic flow = irreversible change from kinetic to inner energy Remark: For isentropic flow momentum and energy equation are equivalent 1-D momentum equation ρv dv dx = dp dx vdv = 1 ρ dp Definition of entropy: T ds = dh dp ρ (2nd law of thermodynamics)

14 19.5 Isentropic flow: dh = dp ρ ( ) d v 2 2 = dh d ( h+ v2 2 ) = 0 = h + v2 2 = h 0 = konst For isentropic flows the momentum equation gives no new information against the energy equation

15 19.5 Momentum in x-direction di x dt = ρ 1 v 2 1 A 0 = (p 0 p 1 )A ρ 1 v 2 1 = p 0 p 1 momentum c p T 0 = c p T v2 1 energy unknown: ρ 1,v 1 express the unknows using the known values (p 1,p 0 ) ρ 1 = p 1 RT 1 v 1 = Ma 1 γrt1 new unknown: T 1, Ma 1

16 19.5 momentum: p 0 p 1 = ρ 1 v 2 1 = ρ 1v 2 1 p 1 p 1 = v2 1 γ p 1 ρ 1 γp 1 = = γp 1 Ma 2 1 = Ma 2 1 = 1 γ ( ) p0 1 p 1 energy: c p T 0 = c p T v2 1 ; c p = γ γ 1 R γrt 0 = γrt 1 + γ 1 v = T 0 = 1+ 1 T 1 2 (γ 1)Ma2 1 = T 1 = T (γ 1)Ma2 1

17 ṁ = ρ 1 v 1 A = p 1 RT 0 γrt0 1 γ ( ) p γ 1 p 1 2 ( ) p0 1 p 1 A

18 19.5 ṁ = f(t 0,p 0,ρ 0 ) = 1.41 kg s ( ) Machnumber in the pipe Ma 2 1 = γ 1 p0 p 1 = 0.5. (In a lossfree flow 1 the pressure ratio would be the value of a pipe Machnumber of Ma = 0.67.) compared to problem 7.8): incompressible flow mit ρ = ρ 0 = p 0 RT 0 ṁ = ρva = ρ pa = 1.6 kg s

19 19.3 Air is flowing isentropically (γ = 1.4) from a large and frictionless supported container through a well rounded nozzle into the open air. a) Determine the dimensionless thrust F s /p 0 A D for the pressure ratios p a /p 0 = 1; 0.6; 0.2; 0! b) What are these values for an incompressible fluid?

20 19.3 a) Outflow from a large container (perfect gas) Forces from momentum equation di dt = t (ρ v)dτ + ρ v( v n)da = F p + }{{} F R =0 τ A + F s + F vol }{{} =0 di x dt = ρ ev 2 ea D = F s +(p a p e )A D

21 19.3 F s p 0 A D = ρ ev 2 e p 0 + p e p a p 0 what is: ρ e,v e,p e expand with γp e F s p 0 A D = γρ ev 2 e γp e p e p 0 + p e p 0 p a p 0 = = p e p 0 γ Ma 2 e + p e p 0 p a p 0 Energie equation: c p T 0 = c p T v2 T 0 T = 1+ 1 (γ 1)Ma2 2

22 = Ma 2 = 2 γ 1 ( ) T0 T 1 = 2 γ 1 ( p0 p )γ 1 γ 1

23 19.3 = F s p 0 A D = p e p 0 2γ γ 1 ( p0 p e )γ 1 γ 1 + ( pe p ) a p 0 p 0 Pressure distribution p e =?? p 0 2 possibilities: v e < a e subsonic flow v e = a e sonic flow v e > a e supersonic is impossible, since the exit cross section is the smallest cross section.

24 19.3 1)v e < a e = The influence of the ambient pressure porpagates upstream(w abs < 0) v e ambient pressure is impressed w abs a e Druckstörung für p a > p : p e = p a

25 Ma 2 e = 2 γ 1 ( p0 p a )γ 1 γ 1

26 19.3 2)v e = a e = The flow in the nozzle is not influenced by the downstream boundary condition ( w abs = 0) the pressure decrease in the environent is not felt in the nozzle. w abs = 0 a e v e ambient pressure is not felt for p a p : p e = p Mae = 1 Te = T

27 19.3 In the throat section (A D ) the Mach number is not larger than 1 The compensation of the pressure difference via Expansions fans Mach sche Linien bilden einen Fächer Expansion fans are without losses The flow is two- or three dimensional

28 19.3 Sketch pe /p 0 1 M < p*/p 0 M = p*/p 0 Aussendruck p /p a

29 19.3 Conclusion p a p : F s = (p e p a )A D + ρ e vea 2 D = γp e }{{} Ma 2 e ) f( p e p 0 p a < p : F s = (p p a )A D + ρ v 2 A D }{{} const. forp 0 =const. further increase is defined by the pressure difference p e p 0.

30 19.3 b) inkompressible fluid c 2 = since dρ = 0 w abs ( ) p ρ s The external pressure is impressed independently of v e. Bernoulli ρ = const. 1 2 ρ ve 2 v }{{} 0 2 = (p 0 p e ) = ρve 2 = 2(p 0 p e ) =0

31 19.3 since ṁ ink > ṁ kompr F s = ρ eve 2 + p e p a p 0 A D p 0 p 0 =0 }{{} = 2 p ( 0 p a = 2 1 p ) a ( p 0 ) p 0 Fs > p 0 A D compressibel

32 Shocks p 0, T 0 Kessel mit Ruhezustand Austritt ins Freie engster Querschnitt Cross section velocity relation: du u = 1 da 1 Ma 2 A with Ma = 1 da = 0 Ma = 1 can only occur in the throat cross section

33 Shocks 1 p/p a b c d Ma Überschall d x 1 Unterschall c b a x

34 Shocks a) subsonic flow b) sonic flow in the smallest section, then deceleration c) supersonic flow behind the throat, no information upstream p is decreased and then increases across the shock d) isentropic supersonic flow

35 Shocks Computation across the normal shock VS 1 2 Ma > 1 Ma < 1 Entropy increase : s 2 > s 1 isentropic equation cannot be used

36 Shocks Continuity, momentum, energy equations Ma 1 = 1 Ma 2 p 2 = 1 + 2γ p 1 γ +1 (Ma ) The pressure ratio increases with the Machnumber.

37 19.6 Air is flowing from a large reservoir through a well rounded nozzle into the open air. At the exit cross section A 1 a normal shock develops. a) Compute the mass flux b) Sketch the distribution of the static pressure along the nozzle axis.

38 19.6 A 1 = m 2 T 0 = 287 K A = 0.01 m 2 p a = 10 5 N m 2 J R = 287 kg K Hint: p 2 = 1+ 2γ p 1 γ +1 (M2 1 1) A A* 3,0 2,0 1, ,5 1,0 1,5 2,0 2,5 M

39 19.6 ṁ = ρ 1 u 1 A 1 = ρ 1 p 0 u 1 T 1 γrt0 A ρ 0 RT 0 γrt1 T 1 0 M 1 > 1, M 2 < 1; p 2 = p a 0 1 isentropic flow A A 1 = (from diagram) M 1 = 2 p 1 = p γ = N/m 2 γ +1 (M2 1 1)

40 T 1 = T 0 1+ γ 1 2 M2 1

41 19.6 ( ) γ T0 p 0 = p 1 T 1 γ 1 = p1 (1+ γ 1 M ) γ γ 1 = N/m 2 ṁ = 1 1+ γ 1 M γ+1 2(γ 1) M 1 p 0 RT0 γa1 = 4.43 kg/s

42 19.6 b)

Introduction to Chemical Engineering Thermodynamics. Chapter 7. KFUPM Housam Binous CHE 303

Introduction to Chemical Engineering Thermodynamics Chapter 7 1 Thermodynamics of flow is based on mass, energy and entropy balances Fluid mechanics encompasses the above balances and conservation of momentum