compressible flow: some preliminary aspects · pdf file02.06.2013 · v15$...

V15 Aerodynamics II 1

Compressible Flow: Some Preliminary Aspects

Aerospace Engineering, International School of Engineering (ISE) Academic year : 2014-2015 (January – May, 2015)

Jeerasak Pitakarnnop , Ph.D.

[email protected] [email protected]

Content

– Introduction – Brief Reviews of Thermodynamics – Governing Equations for Inviscid,

Comprssible Flow (Integral Form) – Definition of Total (Stagnation) Conditions – Some Aspects of Supersonic Flow: Shock

Wave


High-Speed Compressible Flow

•  2 important aspects – Compressible Flow à Variable Density Flow

“All real substance are compressible to some greater or lesser extent: when we squeeze or press on them,

their density will change.” – High-Speed Flow à High Energy Flow “When velocity of high-speed flow is decreased, some of Kinetic

Energy is lost and appear as an increase in internal energy, Hence, increase the temperature of gas.”


THERMODYNAMICS


Reviews of Thermodynamics

•  A high speed flow is high energy flow – As Kinetic Energy = (1/2)mv2

– K.E. per unit mass = (1/2)v2 “When velocity of high-speed flow is decreased, some of Kinetic

Energy is lost and appear as an increase in internal energy, Hence, increase the temperature of gas.”

•  In high-speed flow, energy transformation and temperature changes are “important”.

“Thermodynamics” (see lecture note)


Equa8on of State and Kine8c Theory of Gases

Assump8ons for enable using the kine8c theory of gases are: 1.  Intermolecular forces between the molecules is negligibly small. 2.  The volume of molecules that occupy a space is negligibly small and ignored.

Equa8on of State may be derived from the kine8c theory of gas:

P = ρRTRRM

≡Where the gas constant (R) M is the molecular weight

8314.JR

kmol K=

Gas is Air


Ideal Gas and Perfect Gas


Following dis8nc8on can be made: •  Ideal Gas: holds p = ρRT, and Cp – Cv = R without assump8on of p and T dependence. •  Semi-‐Perfect Gas (Thermally Perfect Gas): holds p = ρRT, and Cp – Cv = R with Cp and Cv are a func8on of T. •  Perfect Gas (Calorically Perfect Gas): holds p = ρRT, and Cp – Cv = R. with Cp and Cv are constant •  Imperfect Gas (Real Gas): doesn’t hold p = ρRT, Cv and Cp are func8ons of p and T and Cp – Cv ≠ R.

Specific Heat for a Perfect Gas and Calorically Perfect Gas

Specific heats at constant pressure and volume:

p

v

dh C dTde C dT

≡

≡In general

( )

( )p p

v v

C C TC C T

=

=

There is oYen a simplifying assump8on of constant specific heats, which is a valid approxima8on to gas behavior in a narrow temperature range

0

0

1500

p p

v v

C ConstC Const

=

=


1st Law of Thermodynamics Statement of conserva8on of energy for a system of fixed mass m.

δq = de +δw!" #$ δ = “path dependent” 2

1 212

2 11

rev

q q

de e e

w pdv

δ

δ

=

= −

=

∫

∫

Flow work on fluid created by the pressure forces at reversible


de = δq − −δw( )

Work done by the surrounding on C.V. (-‐)

Work done by the C.V. on surrounding (+)

2nd Law of Thermodynamics and Entropy

d s ≥ δqT+d sgen


« A condiAon that tells us which direc&on a process will take place. »

The dissipa8ve phenomena always increase the entropy:

d sgen ≥ 0> 0 : Irreversible Process = 0 : Reversible Process < 0 : Impossible

δq = 0If,

Then,

d s ≥ 0

AdiabaAc Process

1st and 2nd Law combina8on: Gibbs Equa8on

The pressure forces within the fluid perform reversible work, and the viscous stresses account for dissipated energy of the system (into heat).

Tds de pdv= + It looks as if we have s u b s 8 t u t e d t h e reversible forms of heat and work into the first law to obtain the Gibbs equa8on

Enthalpy: combina8on 2 forms of fluid energy; internal energy (thermal energy) and flow work (pressure energy)

h e pvdh de pdv vdp≡ +

= + +

Tds dh vdp= −


δq = de +δw!" #$

2nd Law of Thermodynamics: Entropy Change in Process

Tds dh vdp= −

From Gibbs Equa8on:

p vdh C dT de RdT C dT RdT≡ = + = +where

22

2 111

ln

p p

p

dT v dT dPds C dP C RT T T P

PdTs s s C RT P

= − = −

− ≡ Δ = −∫

2 22 1

1 1

ln lnT Ps s Cp RT P

− = −

22

2 111

ln TCpT

φ φ≡ −∫ Tabulated thermodynamic func8on φ


Constant specific heats, which is a valid approxima8on to gas behavior in a narrow temperature range

p p

v v

C ConstC Const

=

=

Calorically Perfect Gas

Specific Heat and The Ra8o

Think about p vdh C dT de RdT C dT RdT≡ = + = +

1

1, ,1 1

p v

p v

p vp v

v v

C C RC CR RC C R C R C RC C

γγ

γ γ

= +

= +

+≡ = = =

− −

One can obtain

The ra8o of specific heats is related to the degrees of freedom of the gas molecules, n, via:

2nn

γ+

= Diatomic gas at ‘normal’ temperature

5 2 1.45

γ+

= =

Diatomic gas •  at normal Temp.,

5 degrees of freedom: •  3 transla8onal mo8on •  2 rota8onal mo8on

•  600 K < Temp. < 2000 K, vibra8on mode ac8vated: n = 6, γ = 1.33

•  > 2000K, n = 7, γ = 9/7 = 1.29

0

0

1500

?hotγ =?coldγ =


Specific Heat and The Ra8o

Specific Heat at Constant Pressure (Cp) in kJ/kg.K and Specific Heat RaAo γ


To analyze the performance

of Ideal Engine one can assume every components are working under the isentropic process EXCEPT ‘Combustor’

Isentropic Process Isentropic Flow for a Calorically Perfect

ds =C pdTT

− R dPP

, Isentropic Process: ds = 0

From Gibbs Equa8on:

Using the Perfect Gas Law:

P2P1=

ρ2ρ1

!

"##

$

%&&

γ

=T2T1

!

"##

$

%&&

γγ−1

0

0

1500


P2P1=T2T1

!

"##

$

%&&

CPR=T2T1

!

"##

$

%&&

γγ−1

Exercise1: Isentropic Flow for a Calorically Perfect

2 2

1 1

PP

γρρ

⎛ ⎞= ⎜ ⎟⎝ ⎠

From the Isentropic Flow for Perfect Gas:

Show that dp Pd

γρ ρ=


Stagna8on State We define the stagna8on state of a gas as the state reached in decelera8ng a flow to rest reversibly and adiabaAcally and without any external work

h0 = ht ≡ h +V 2

2(The total energy of the gas does not change in the decelera8on process)

Calorically Perfect Gas

ht ≡ h +V 2

2⇒ Tt =T +

V 2

2CP

2th

3th

1th


Steady, Adiaba8c, Inviscid flow: ht = const.

ht 1 = ht 2 , h1 +V1

2

2= h2 +

V22

2

Stagna8on State and Local Mach Number of a Calorically Perfect Gas

1. Based on the stagna8on enthalpy, proof that the total to stagna8on temperature ra8o can be derived as:

2112

tT MT

γ −⎛ ⎞= + ⎜ ⎟⎝ ⎠

2. What is the assump8on to have the rela8on of the total to stagna8on pressure ra8o as:

1211

2tP MP

γγγ −⎡ ⎤−⎛ ⎞= + ⎜ ⎟⎢ ⎥

⎝ ⎠⎣ ⎦

3. Show that the density ra8o of a calorically perfect gas in isentropic process is:

11

2112

t Mγρ γ

ρ

−⎡ ⎤−⎛ ⎞= + ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ 2tT

3tT1tT


Definition of Total (Stagnation) Conditions

•  Static Pressure (P): Pressure you feel when you ride along with the gas at the local flow velocity.

•  Stagnation Pressure (P0): Pressure existing at a point in the flow where V = 0.

•  The value temp. of the fluid element after it has been brought to rest is T0, for calorically perfect gas h0 = cpT0.


Definition of Total (Stagnation) Conditions

•  For inviscid, steady, adiabatic flow: h + V 2/2 = const.

•  Definition of total enthalpy (at a pt. if the fluid were brought to rest adiabatically) h0 = h + V 2/2

•  Adiabatic flow h0 = const.

•  Calorically perfect gas T0 = const.


CONSERVATION PRINCIPLES



Gas Flow Regimes: Molecular to Continuum Flow

Kn = 10 Kn = 0.1 Kn = 0.01 Kn 0

Analytical Methods Variational Methods

Discrete Velocity Methods

Integro-moment MethodDSMC

Analytical MethodsTypical CFD schemesTest Particle Monte Carlo

Boltzman Equation without collisions Boltzman Equation Navier-Stokes

+ slip BC. Navier-Stokes Euler

Free molecular regime

Transition regime Slip-flow regimeViscous Inviscid

Continuum regime

[1] Dimitris Valougeorgis (2007), Solution of vacuum flows via kinetic theory, 51st IUVSTA Workshop on Modern Problems and Capability of Vacuum Gas Dynamics.


Governing Equations

MicroscopicTheory

MesoscopicTheory

Macroscopic (Continuum)Theory

BurnettNavier-StokesEuler Super-

BurnettGrad’s 13 moments

Grad’s 26 moments

Molecular Dynamics

Boltzmann Equation

Grad’s moments methodKinetic Theory: Hilbert et Chapman-Enskog analysis

Direct Simulation Boltzmann

Turbulence Modeling

QHD & QGD0th

Ord

er

1st O

rder

2nd O

rder

Hig

her O

rder

Direct Simulation

Molecular Models Continuum Models

Kinetic Models

Integral Forms of the Governing Equations for Inviscid, Compressible Flow

•  Incompressible flow obeys purely mechanical laws and doesn’t require thermodynamic considerations (P and V are unknown).

•  Compressible flow, ρ is variable à additional unknown. – Additional energy equation is needed. – Energy energy requires e which is related to T.

Primary Dependent Variable “P, V, ρ, e and T”

V15 24 Aerodynamics II

Mathematical Model

To determine previous values, following principles are applied: • Equation of state • Conservation of mass • Conservation of momentum • Conservation of energy


Governing Equation Integral & Differential Forms

•  Conservation of Mass: – Mass Can be Neither Create Nor Destroy.

Rate of Increase of mass of fluid in C.V.

Rate of mass enters C.V.

Rate of mass leaves C.V. = -‐


∂ρ∂t

= −∇⋅ ρu( )

∂∂t

ρ dVV∫∫∫ = − ρV

S∫∫ ⋅d S


•  Conservation of Momentum: – The Time Rate of Change of Momentum of a Body

Equal to Net Force Exerted on It.

Net force on gas in C.V. in direc8on considered

Rate of increase of momentum indirec8on

considered of fluid in C.V.

Rate momentum leaves C.V. in direc8on considered

= -‐ Rate

momentum enter C.V. in direc8on considered

+



•  Conservation of Momentum:


∂∂t

ρV dVV∫∫∫ = ρf dV

V∫∫∫ − P

S∫∫ d S− ρV ⋅d S( )

S∫∫ V

∂ ρux( )∂t

+∇⋅ ρuxV( ) = −∂P∂x

+ ρ f x

∂ ρu y( )∂t

+∇⋅ ρu yV( ) = −∂P∂y

+ ρ f y

∂ ρuz( )∂t

+∇⋅ ρuzV( ) = −∂P∂z

+ ρ f z


•  Conservation of Energy: – Energy Can be Neither Create Nor Destroy, It Can

Only Change in Form.

Rate of change of the energy of the fluid as it flows through C.V.

Rate of heat added to the fluid inside

C.V. from surrounding

Rate of work done on the fluid inside

C.V. = +



•  Conservation of Energy:


∂∂t

ρ e +V2

2+ gz

"

#$

%

&'dV +

V∫∫∫ ρ e +V

2

2+ gz

"

#$

%

&'

S∫∫ V ⋅d S =

Q + Wshaft + Wviscous − PS∫∫ V ⋅d S+ ρ f ⋅V( )dV

V∫∫∫

∂∂t

ρ e +V2

2+ gz

"

#$

%

&'

(

)*

+

,-+∇⋅ ρ e +

V 2

2+ gz

"

#$

%

&'V

(

)*

+

,-=

Q + Wetc + ρ q −∇⋅ P V( )+ ρ f ⋅V( )

Engine Thrust

•  pe < p∞ : Nozzle is overexpanded (only supersonic). •  pe = p∞ : Nozzle is perfect expanded (all subsonic). •  pe > p∞ : Nozzle is unexpanded (only sonic & supersonic).


Tuninstall = ( m∞+ m f )ve − m∞

v∞+ ( pe − p∞)Ae

Concorde’s Variable Exhaust Nozzle.

compressible flow: some preliminary aspects · pdf file02.06.2013 · v15$...

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