Fundamentals of Compressible and Viscous Flow Analysis

Fundamentals of compressible and viscous ﬂow analysis - Part II Lectures 3, 4, 5

Instantaneous and averaged temperature contours in a shock-boundary layer interaction.

Taken from (Pasquariello et al., Journal of Fluid Mechanics 2017).

Christophe Corre (LMFA - ECL)

January 19, 2018 2 Contents

Contents 2

3 Energy conservation5 3.1 Deriving the energy conservation equation...... 5 3.1.1 Integral or global form...... 5 3.1.2 Local form...... 10 3.1.3 Non-conservative global form...... 11 3.1.4 Kinetic energy and internal energy conservation equations...... 13 3.1.5 Case of a continuous-flow system...... 18 3.1.6 Energy conservation equation for an ideal fluid and Bernoulli equation. 21 3.1.7 Flow modeling : provisional overview...... 22 3.2 Thermodynamic analysis...... 23 3.2.1 Reversible process and entropy...... 23 3.2.2 Equations of state...... 25 3.2.3 Equations of state for a calorifically perfect gas...... 29 3.3 Conclusions on the flow governing equations...... 34 3.3.1 Final overview of the governing equations...... 34 3.3.2 Velocity measurement in a compressible flow...... 36 3.4 Exercises and problems...... 39 3.4.1 Exercise # 1 : Analysis of a steam turbine...... 39 3.4.2 Exercise # 2 : Analysis of the draining of a tank...... 40 3.4.3 Exercise # 3 : Analysis of the compressible flow in a Venturi...... 43 3.4.4 Problem # 1 : pressure rise in a nuclear reactor containment...... 46

4 1D compressible ﬂow analysis 65 4.1 Some key quantities...... 65 4.1.1 Sound speed and Mach number...... 65 4.1.1.1 Exercise : computing the speed of sound...... 69 4.1.2 Stagnation quantities or total quantities...... 70 4.1.3 Sonic quantities...... 73 4.2 1D shockwaves...... 75 4.2.1 Problem deﬁnition...... 75 4 CONTENTS

4.2.2 Prandtl’s relationship...... 75 4.2.3 Physical considerations...... 77 4.2.4 Shock relationships...... 78 4.2.5 Total quantities and 1D shockwaves...... 80 4.2.6 Rankine-Hugoniot’s relationship...... 84 4.3 Exercises and problems...... 85 4.3.1 Exercise 1 : Mach number and compressibility effects...... 85 4.3.2 Exercise # 2 : pressure and temperature at the nose of an airplane in transonic flight...... 87 4.3.3 Problem # 1 : velocity measurement in a supersonic flow...... 89

5 Quasi-1D compressible flows 95 5.1 Physical model...... 95 5.1.1 Governing equations...... 95 5.1.2 Velocity-area relationship...... 97 5.1.3 Influence coefficients...... 99 5.2 Isentropic flow in a nozzle...... 102 5.2.1 Flow analysis...... 102 5.2.2 Mass flowrate...... 105 5.3 Analysis of the flow through a converging nozzle...... 106 5.3.1 General analysis...... 106 5.3.2 Example of application...... 108 5.4 Analysis of the flow through a diverging nozzle...... 111 5.5 Analysis of flow regimes in a converging-diverging nozzle...... 122 5.5.1 General analysis...... 122 5.5.2 Example of application...... 128 5.6 Exercises and problems...... 138 5.6.1 Exercise # 1 : Flow in a choked nozzle...... 138 5.6.2 Exercise #2 : Flow in a diverging nozzle...... 140 5.6.3 Problem # 1 : analysis of a scramjet performance...... 142 Lecture 3 Energy conservation

Contents 3.1 Deriving the energy conservation equation...... 5 3.1.1 Integral or global form...... 5 3.1.2 Local form...... 10 3.1.3 Non-conservative global form...... 11 3.1.4 Kinetic energy and internal energy conservation equations...... 13 3.1.5 Case of a continuous-flow system...... 18 3.1.6 Energy conservation equation for an ideal fluid and Bernoulli equation 21 3.1.7 Flow modeling : provisional overview...... 22 3.2 Thermodynamic analysis...... 23 3.2.1 Reversible process and entropy...... 23 3.2.2 Equations of state...... 25 3.2.3 Equations of state for a calorifically perfect gas...... 29 3.3 Conclusions on the flow governing equations...... 34 3.3.1 Final overview of the governing equations...... 34 3.3.2 Velocity measurement in a compressible flow...... 36 3.4 Exercises and problems...... 39 3.4.1 Exercise # 1 : Analysis of a steam turbine...... 39 3.4.2 Exercise # 2 : Analysis of the draining of a tank...... 40 3.4.3 Exercise # 3 : Analysis of the compressible flow in a Venturi...... 43 3.4.4 Problem # 1 : pressure rise in a nuclear reactor containment...... 46

3.1 Deriving the energy conservation equation

3.1.1 Integral or global form Case of a material ﬂuid domain

The ﬁrst principle of thermodynamics applied to the material domain Dm(t) states the following physical principle : 6 Lecture #3 : Energy conservation

The rate of variation of the total energy contained in Dm(t) is equal to the sum of the work (per unit time) done by the external (surface and volume) forces applied to Dm(t) and the heat quantity transferred through Sm(t) (per unit time).

In order to turn this statement into an equation, the first question to answer is : what is the energy of a fluid particle ? From the (local) energy of the fluid particle, it will be easy to express the energy contained in the material domain by integrating over this domain. The total (specific, that is per unit mass) energy of a fluid particle can be decomposed into : -• an internal (specific) energy e coming from the molecular motion. In a gas, even at rest, each molecule displays a random motion and also rotates and vibrates. The motion of electrons also adds an electronic energy to the molecule. The (specific) internal energy e of a fluid particle is the sum of these energies for all the atoms or molecules present in the fluid particle (elementary fluid volume). -• a specific kinetic energy linked to the macroscopic fluid motion with velocity U~ and equal 1 to U~ 2. 2 The total specific energy of the fluid particle is given by :

1 E = e + U~ 2 2 The total energy of the fluid particle is obtained by multiplying the previous specific energy (or energy per unit mass) by the mass of the fuid particle, given by ρdV where dV is the elementary volume of the fluid particle. Hence the total energy of a fluid particle of elementary volume dV is given by : 1 ρEdV = ρedV + ρU~ 2dV 2 which makes clear ρE is a total energy per unit volume. The total energy contained or stored in the material domain Dm(t) is computed as :

ρEdV ˆDm(t) The rate of variation of this total energy is expressed as : D ( ρEdV ) Dt ˆDm(t) and the energy conservation reads : D ( ρEdV ) = W˙ + Q˙ (3.1) Dt ˆDm(t) where W˙ and Q˙ are respectively the work done, per unit time, by the external forces applied to Dm(t) and the heat transferred through Sm(t), per unit time. Using the Reynolds transport theorem (??) with χ = E, the LHS of (3.1) can be turned into : D DE ρE dV = ρ dV DtˆDm(t) ˆDm(t) Dt 3.1. Deriving the energy conservation equation 7 so that (3.1) can also be expressed as :

DE ρ dV = W˙ + Q˙ (3.2) ˆDm(t) Dt

If the form (??) of the transport theorem is used, the LHS of (3.1) is expressed using a conservative form so that the conservative form of the energy conservation equation reads :

∂(ρE) dV + ρE (U~ · ~n) dS = W˙ + Q˙ (3.3) ˆDm(t) ∂t ˆSm(t)

Case of an arbitrary control volume

In the case of an arbitrary control volume, the ﬁrst law of thermodynamics applied to Da(t) states D ( ρEdV ) = W˙ + Q˙ − ρE (U~ − W~ ) · ~ndS Dt ˆDa(t) ˆSa(t) where the last term in the RHS corresponds to the convective ﬂux of total energy through the surface Sa(t). Using again the Reynolds transport theorem to transform the LHS of the above equation yields :

DE ρ dV + ρE (W~ − U~ ) · ~ndS = W˙ + Q˙ − ρE (U~ − W~ ) · ~ndS ˆDa(t) Dt ˆSa(t) ˆSa(t)

After simpliﬁcation the non-conservative form (3.2) is recovered for an arbitrary control volume :

DE ρ dV = W˙ + Q˙ (3.4) ˆDa(t) Dt

Developing the material derivative of the total energy and using mass conservation yields :

DE D(ρE) Dρ ρ = − E Dt Dt Dt ∂(ρE) = + U~ · ∇~ (ρE) + ρE∇~ · U~ ∂t ∂(ρE) = + ∇~ · (ρEU~ ) ∂t

Using the ﬂux-divergence theorem to transform the volume integral of ∇~ · (ρEU~ ) eventually yields : ∂(ρE) dV + ρE (U~ · ~n) dS = W˙ + Q˙ (3.5) ˆVa(t) ∂t ˆSa(t) which generalizes (3.5) previously established for a material domain. 8 Lecture #3 : Energy conservation

Work of the external forces The work per unit time of the surface and volume forces applied to the ﬂuid domain D (of surface S) is given by : W˙ = T~(~n) · UdS~ + ρf~ · UdV~ ˆS ˆD The stress vector T~ is such that (see (??)) : −→ −→ T~ = −→σ · ~n = −p~n + −→τ · ~n hence −→ W˙ = −pU~ · ~ndS + (−→τ · ~n) · U~ dS + ρf~ · UdV~ ˆS ˆS ˆD The work done, per unit time, by the viscous forces can be rewritten using :

−→ −→ −→ ( τ · ~n) · U~ = τijnjUi = τjiUinj = ( τ · U~ )jnj = ( τ · U~ ) · ~n hence −→ W˙ = −pU~ · ~ndS + (−→τ · U~ ) · ~ndS + ρf~ · UdV~ (3.6) ˆS ˆS ˆD

Heat transferred per unit time The heat transferred per unit time to the volume D can be the result of two diﬀerent mechanisms :

-• heating of the volume by a radiation process

-• heat transferred through the surface S of the ﬂuid domain because of temperature gradients in the ﬂow. This process corresponds to heat conduction.

In this course, only heat conduction will be taken into account. The quantity ~q denotes the local heat conduction flux q per unit surface and unit time. ~q is linked to the temperature T of the flow using Fourier constitutive law : ~q = −k∇~ T (3.7) where k denotes the thermal conductivity of the fluid, which depends on the temperature T in the general case. The elementary heat flux through an infinitesimal fraction dS of the control surface S is given by −~q · ~ndS, where the − sign comes from the outward orientation of the unit vector ~n with respect to the fluid domain and the fact the heat received by / added to the system (fluid volume) is counted positively. The total heat flux through S per unit time is given by : ∂T Q˙ = − ~q · ~ndS = k∇~ T · ~ndS = k dS (3.8) ˆS ˆS ˆS ∂n Inserting (3.6) and (3.8) into (3.5) yields the following global or integral form of the enery conservation equation :

∂(ρE) −→ ∂T dV + ρE (U~ · ~n) dS = −pU~ · ~ndS + (−→τ · U~ ) · ~ndS + ρf~ · UdV~ + k dS ˆD ∂t ˆS ˆS ˆS ˆD ˆS ∂n 3.1. Deriving the energy conservation equation 9

Let us introduce the speciﬁc enthalpy h :

p h = e + ρ

The total enthalpy H can be deﬁned from the speciﬁc enthalpy as :

1 H = h + U~ 2 2

Total enthalpy and total energy are related as follows :

1 p 1 p H = h + U~ 2 = e + + U~ 2 = E + 2 ρ 2 ρ

Using total energy and total enthalpy, the above global form of the energy conservation equation can be rewritten as :

∂(ρE) −→ ∂T dV + ρH (U~ · ~n) dS = (−→τ · U~ ) · ~ndS + ρf~ · UdV~ + k dS (3.9) ˆD ∂t ˆS ˆS ˆD ˆS ∂n

The power of the body forces can be integrated into the conservation equation in the case where the body forces can be expressed using a potential. If f~ = ~g, the gravitation vector, this potentiel is Ψ = gz so that ~g = −∇~ Ψ. Since ρf~ = −ρ∇~ Ψ, it is possible to write the product ρf~ · U~ as follows : ρf~ · U~ = −ρU~ · ∇~ Ψ

= −∇~ · (ρΨU~ ) + Ψ∇~ · (ρU~ )

∂ρ = −∇~ · (ρΨU~ ) − Ψ ∂t ∂(ρΨ) = −∇~ · (ρΨU~ ) − ∂t ∂ρ where the mass conservation has been used ( + ∇~ · (ρU~ ) = 0) as well as the fact the gravity ∂t potential Ψ does not depend on the time t. The power of the body forces can thus be rewritten as :

∂(ρΨ) ρf~ · UdV~ = − ∇~ · (ρΨU~ ) dV − dV ˆD ˆD ˆD ∂t or else, using the ﬂux-divergence theorem :

∂(ρΨ) ρf~ · UdV~ = − ρΨU~ · ~ndV − dV ˆD ˆS ˆD ∂t Inserting this identity into (3.9) yields :

∂(ρE) ∂(ρΨ) −→ ∂T dV + dV + ρH (U~ · ~n) dS+ ρΨU~ · ~ndV = (−→τ · U~ ) · ~ndS+ k dS ˆD ∂t ˆD ∂t ˆS ˆS ˆS ˆS ∂n 10 Lecture #3 : Energy conservation

It is then natural to change the deﬁnition of the total energy in order to include into the total energy not only the internal energy and the kinetic energy but also the potentiel energy (of gravitation in the case where Ψ = gz):

1 1 E = e + U~ 2 + Ψ = e + U~ 2 + gz 2 2

Consequently, the total enthalpy H also includes now this potential energy :

p p 1 1 1 H = E + = e + + U~ 2 + Ψ = h + U~ 2 + Ψ = h + U~ 2 + gz ρ ρ 2 2 2

With these extended deﬁnitions of E and H, the global form of the energy conservation equation reads : ∂(ρE) −→ ∂T dV + ρH (U~ · ~n) dS = (−→τ · U~ ) · ~ndS + k dS (3.10) ˆD ∂t ˆS ˆS ˆS ∂n

Case of an ideal fluid In the case of an ideal fluid, with zero viscosity (µ = 0) and zero heat conductivity (k = 0), the energy conservation equation (3.10) simplifies into :

∂(ρE) dV + ρH (U~ · ~n) dS = 0 (3.11) ˆD ∂t ˆS

3.1.2 Local form The local form of the total energy conservation equation is readily derived from (3.10) using the ﬂux-divergence theorem. Indeed, the surface integrals can be systematically turned into volume integrals so as to yield :

∂(ρE) −→ dV + ∇~ · (ρH U~ ) dV = ∇~ · (−→τ · U~ ) dV + ∇~ · (k∇~ T ) dV ˆD ∂t ˆD ˆD ˆD Hence the local conservative vector form for the total energy conservation :

∂(ρE) −→ + ∇~ · (ρH U~ ) = ∇~ · (−→τ · U~ ) + ∇~ · (k∇~ T ) (3.12) ∂t

In the case where the dependence of the heat conductivity k on the temperature ﬁeld T can be neglected so that k can be assumed constant, the equation can also be written as :

∂(ρE) −→ + ∇~ · (ρH U~ ) = ∇~ · (−→τ · U~ ) + k ∆T ∂t −→ For an ideal ﬂuid with zero viscosity (µ = 0 hence −→τ = 0) and zero heat conductivity (k = 0), the local form of the total energy conservation equation reads :

∂(ρE) + ∇~ · (ρH U~ ) = 0 (3.13) ∂t 3.1. Deriving the energy conservation equation 11

Both equations (3.12) and (3.13) can be expressed in an alternative non-conservative form using the material derivative of E and the continuity equation. It is indeed possible to write : ∂(ρE) ∂(ρE) + ∇~ · (ρH U~ ) = + ∇~ · (ρE U~ ) + ∇~ · (p U~ ) ∂t ∂t ∂E ∂ρ = ρ + E + ρU~ · ∇~ E + E ∇~ · (ρU~ ) + ∇~ · (p U~ ) ∂t ∂t

DE ∂ρ = ρ + E + ∇~ · (ρU~ ) +∇~ · (p U~ ) Dt ∂t | {z } =0

DE = ρ + ∇~ · (p U~ ) Dt In the case of an ideal ﬂuid, the non-conservative form of the total energy conservation equation reads : DE ρ = −∇~ · (p U~ ) (3.14) Dt In the case of a viscous ﬂuid, with heat conductivity k(T ), the non-conservative form of the total energy conservation equation takes the form :

DE −→ ρ = −∇~ · (p U~ ) + ∇~ · (−→τ · U~ ) + ∇~ · (k∇~ T ) (3.15) Dt

−→ or else, since ∇~ · (pU~ ) = ∇~ · (p 1 · U~ ):

DE −→ ρ = ∇~ · (−→σ · U~ ) + ∇~ · (k∇~ T ) Dt −→ with −→σ the Cauchy stress tensor used to express the external surface forces (pressure forces and viscous forces).

3.1.3 Non-conservative global form

The previous non-conservative local form can be integrated over an arbitrary ﬂuid domain Da(t) (with surface Sa(t) moving at velocity W~ ) to yield : DE −→ ρ dV = ∇~ · (−→σ · U~ ) dV + ∇~ · (k∇~ T ) dV ˆDa(t) Dt ˆDa(t) ˆDa(t) Using again the Reynolds transport theorem (??) to transform the LHS of the above equation yields : DE D ρ dV = ρE dV − ρE (W~ − U~ ) · ~ndS ˆDa(t) Dt DtˆDa(t) ˆSa(t) It follows that : D −→ ρE dV = ∇~ · (−→σ · U~ ) dV + ∇~ · (k∇~ T ) dV + ρE (W~ − U~ ) · ~ndS DtˆDa(t) ˆDa(t) ˆDa(t) ˆSa(t) 12 Lecture #3 : Energy conservation

The ﬁrst integral of the RHS can be rewritten using the ﬂux-divergence theorem : −→ −→ ∇~ · (−→σ · U~ ) dV = (−→σ · U~ ) · ~ndS ˆDa(t) ˆSa(t)

Using the symmetry of the Cauchy stress tensor, it is possible to rewrite :

−→ −→ −→ −→ ( σ · U~ ) · ~n = ( σ · U~ )ini = σijUjni = σjiniUj = ( σ · ~n)jUj = ( σ · ~n) · U~ = T~ · U~ with T~ the stress vector (pressure stress and viscous stress). Consequently :

−→ −→ ~ ~ ~ S ( σ · U) · ~ndS = T · U dS = Pext ˆSa(t) ˆSa(t)

S with Pext the power of the external surface forces applied to the ﬂuid domain. Note, from its S deﬁnition, the alternative form of Pext :

−→ S ~ ~ ~ −→ ~ Pext = − ∇ · (pU) dV + ∇ · ( τ · U) dV ˆDa(t) ˆDa(t)

Remembering that : ∇~ · (k∇~ T ) dV = k∇~ T · ~ndS = Q˙ ˆDa(t) ˆSa(t) the non-conservative form of the total energy conservation equation reads :

D S ˙ ~ ~ ρE dV = Pext + Q + ρE (W − U) · ~ndS DtˆDa(t) ˆSa(t) where it must be also remembered the total energy contains the potential Ψ associated with the body forces. For a material domain Dm(t)(W~ = U~ ) it is possible to write :

D 1 ~ 2 S ˙ ρ e + U + Ψ dV = Pext + Q (3.16) DtˆDm(t) 2

The variation of the total energy contained in a material domain (sum of the internal, kinetic and potential energy) is equal to the power of the external surface forces applied to the material domain plus the heat added to the domain (per unit time).

The total energy can be decomposed into the mechanical energy (sum of the kinetic energy and the potential energy) on one hand and the internal energy on the other hand. Equation (3.16) also states : D 1 ~ 2 D S ˙ ρ U + Ψ dV + ρe dV = Pext + Q DtˆDm(t) 2 DtˆDm(t)

Therefore, if a conservation equation can be derived for the kinetic energy, it will be possible to also ﬁnd a conservation equation for the internal energy. 3.1. Deriving the energy conservation equation 13

3.1.4 Kinetic energy and internal energy conservation equations Kinetic energy conservation equation A conservation equation for kinetic energy (and mechanical energy) can be readily derived from the momentum conservation equation, by taking the scalar product between U~ and this equation (see also Section ?? where such a scalar product was performed to derive the Bernoulli equation. We apply the same strategy here, assuming only the body force f~ can be expressed as f~ = −∇~ Ψ but remaining in a general framework otherwise (no assumptions are made on incompressibility or ﬂow steadiness). Starting from the non-conservative local form of the momentum equation (??) and taking the scalar product of this equation with the velocity vector U~ yields :

DU~ −→ ρU~ · = −U~ · ∇~ p + U~ · ∇~ · −→τ − ρU~ · ∇~ Ψ Dt The LHS is immediately rewritten as :

DU~ D 1 ρU~ · = ρ U~ 2 Dt Dt 2 while the power of the body forces can be recast, since Ψ does not depend on t, as : DΨ −ρU~ · ∇~ Ψ = −ρ Dt It follows that : D 1 −→ ρ U~ 2 + Ψ = −U~ · ∇~ p + U~ · ∇~ · −→τ Dt 2 The LHS is the variation of the mechanical energy along the path of a ﬂuid particle. Integrating over a material domain and using the Reynolds transport theorem with W~ = U~ yields : D 1 −→ ρ U~ 2 + Ψ dV = − U~ · ∇~ p dV + U~ · ∇~ · −→τ dV DtˆDm(t) 2 ˆDm(t) ˆDm(t) Note that : U~ · ∇~ p = ∇~ · (pU~ ) − p∇~ · U~ Besides, one can also write :

−→ ∂τij ∂(τijUi) ∂Ui ∂(τjiUi) ∂Ui −→ ∂Ui U~ · ∇~ · τ = Ui = − τij = − τij = ∇~ · ( τ · U~ ) − τij ∂xj ∂xj ∂xj ∂xj ∂xj ∂xj

∂Ui The term τij can be also expressed as : ∂xj ∂Ui 1 ∂Ui ∂Uj τij = τij + τji ∂xj 2 ∂xj ∂xi −→ Using the symmetry of the viscous stress tensor τ , that is τji = τij, one can also write : ∂Ui 1 ∂Ui ∂Uj 1 ∂Ui ∂Uj τij = τij + τij = τij + = τijDij ∂xj 2 ∂xj ∂xi 2 ∂xj ∂xi | {z } =Dij 14 Lecture #3 : Energy conservation

−→ −→ The scalar product of two second-order tensors −→a and b is deﬁned as : −→ −→ −→ X a : b = aijbij = aijbij i,j using Einstein’s notation on repeated indices. Therefore : −→ ∂Ui −→ −→ τij = τ : D ∂xj Going back to the initial development, it comes :

−→ −→ −→ −→ U~ · ∇~ · −→τ = ∇~ · (−→τ · U~ ) − −→τ : D

−→ Replacing U~ ·∇~ p and U~ · ∇~ · −→τ in the mechanical energy conservation equation with the above expressions yields :

D 1 ρ U~ 2 + Ψ dV = − ∇~ · (pU~ ) dV + p∇~ · U~ dV DtˆDm(t) 2 ˆDm(t) ˆDm(t)

−→ −→ −→ + ∇~ · (−→τ · U~ ) dV − −→τ : D dV ˆDm(t) ˆDm(t)

S The power of the external surface forces applied to the material domain Pext is easily identiﬁed in the RHS so that the mechanical energy conservation equation can be also expressed as : −→ −→ D 1 ~ 2 S ~ ~ −→ −→ ρ U + Ψ dV = Pext + p∇ · U dV − τ : D dV DtˆDm(t) 2 ˆDm(t) ˆDm(t)

The internal power Pint is deﬁned as : −→ −→ −→ −→ −→ −→ −Pint = − p∇~ · U~ dV + τ : D dV = σ : D dV (3.17) ˆDm(t) ˆDm(t) ˆDm(t) | {z } | {z } compression / expansion viscous dissipation

It is the power required to deform the fluid particles contained in the material domain. −Pint is the sum of : -• the power required to compress (∇~ · U~ < 0) or to expand (∇~ · U~ > 0) the fluid particle. Remember formula (??) expressing the relative volume variation of a fluid particle as the divergence of the velocity vector. In a compression the volume of the fluid particle decreases hence ∇~ · U~ < 0 while in an expansion this volume increases so that ∇~ · U~ > 0. This power is positive for a compression (taken by the flow) and negative for an expansion (given by the flow). −→ −→ -• the viscous dissipation Φ = −→τ : D dV which is an always positive term, expressing ˆDm(t) the heating of the flow through the degradation of mechanical energy into thermal energy. 3.1. Deriving the energy conservation equation 15

The mechanical energy conservation equation can therefore be summarized as : D 1 ~ 2 S ρ U + Ψ dV = Pext + Pint (3.18) DtˆDm(t) 2

Physically, (3.18) can also be interpreted in the following form : S D 1 ~ 2 Pext = ρ U + Ψ dV − Pint DtˆDm(t) 2 showing the power supplied by the external surface forces applied to the material domain is used on one hand to increase the mechanical energy stored in the domain and on the other hand to deform the ﬂuid particles contained in the material domain. If a compression is taking place, −Pint is strictly positive since :

−Pint = − p∇~ · U~ dV + Φ (3.19) ˆDm(t) |{z} | {z } viscous dissipation always >0 >0 for compression

S so that the external power Pext is consumed in part by the internal power (compression of the flow and viscous dissipation) and in part to increase the mechanical energy of the material domain. If an expansion is taking place, the internal power contributes to the increase of mechanical energy through the term involving p∇~ · U~ while the viscous dissipation remains a net loss. In the case of an inviscid flow, the viscous dissipation Φ reduces to zero so that the internal power corresponds to the flow compression or expansion. S In the case where no external power Pext is provided to the material domain, the evolution of the mechanical energy stored in the material domain depends only on the internal power. Assuming for instance an incompressible flow, the internal power Pint reduces to −Φ, a negative quantity corresponding to the friction losses of the system. Equation (3.18) expresses in that case the fact the mechanical energy of the material domain is progressively lost through friction losses. As will be seen next, this mechanical energy is converted into heat and contributes to the increase of the internal energy of the material domain.

Internal energy conservation equation Substracting the mechanical energy conservation equation (3.18) from the total energy conservation equation (3.16) immediately yields the internal energy conservation equation :

D ρe dV = −Pint + Q˙ (3.20) DtˆDm(t) showing that the internal energy of the material domain Dm(t) evolves because of the internal power −Pint, sum of the always positive dissipation Φ and the compression or expansion power, and because of the heat provided to or extracted from the material domain. The conservation of total energy (3.16) can therefore be usefully rewritten in such a way the conservation of mechanical energy and internal energy are made visible : D 1 ~ 2 S ˙ U + Ψ + ρe dV = Pext + Pint−Pint + Q (3.21) DtˆDm(t) 2 16 Lecture #3 : Energy conservation

Figure 3.1: Energy conversion mechanisms for a material ﬂuid domain. where the choice of the blue and red text indicates the respective balance of :

-• the mechanical energy fraction (kinetic and potential energy) of the total energy, varying because of the external power of the surface forces and the internal power

-• the internal energy fraction of the total energy, varying because of the oppostive of the internal power and the quantity of heat brought to the material domain.

The internal power Pint can be further detailed using deﬁnition (3.19) so that :

D 1 ~ 2 S ~ ~ ~ ~ ˙ U + Ψ + ρe dV = Pext + p∇ · U dV − Φ− p∇ · U dV + Φ + Q DtˆDm(t) 2 ˆDm(t) ˆDm(t) (3.22) In conclusion, the total energy stored in the material domain varies only because of :

-• the external power of the surface forces applied to the material domain, this power being negative when it is extracted from the fluid flow as is the case for a turbine or being positive when it is supplied to the fluid flow as is the case for a pump

-• the heat quantity supplied to or extracted from the domain.

Within this total energy balance, it is useful to distinguish the internal energy transfer mechanisms taking place inside the material domain. The total energy varies because both the mechanical energy and the internal energy vary. The mechanical energy variation is driven by :

-• the external power of the surface forces applied to the material domain, which can be positive (case of a pump for instance) or negative (case of a turbine for instance)

-• the internal power of compression or expansion of the ﬂuid ﬂow, which can be positive (case of an expansion) or negative (case of a compression) 3.1. Deriving the energy conservation equation 17

-• the viscous dissipation which always contribute to decrease the mechanical energy (losses)

In the same time, the internal energy variation is driven by :

-• the internal power of compression or expansion of the fluid flow, which tends to decrease the internal energy in the case of an expansion (flow cooling) and to decrease the internal energy in the case of a compression (flow heating)

-• the viscous dissipation which always contribute to increase the mechanical energy (heating of the ﬂow by friction)

Case of an adiabatic incompressible viscous ﬂow

For an incompressible viscous flow, the expansion / compression term becomes zero since ∇·~ U~ = 0 while Φ > 0. Moreover, since the flow is adiabatic, Q˙ = 0. The energy balance (3.22) simplifies into : D 1 ~ 2 S U + Ψ + ρe dV = Pext − Φ+Φ DtˆDm(t) 2 or equivalently : D S ρE dV = Pext DtˆDm(t) The variation of the total energy of a material domain for such a flow is driven by the power of the external surface forces. Consequently, if a passive system is considered such as a pipe in which the viscous fluid is flowing, where no power is supplied to or extracted from the flow, S Pext = 0 and the total energy of the material domain is conserved. However, the nature of the energy changes over time since the dissipation is such that Φ > 0. The total energy remains constant but with a decrease of the mechanical energy and an increase of the internal energy. Typically, if the potential energy is negligible, the decrease of the mechanical energy corresponds to a decrease of the kinetic energy (the velocity is decreased) because of the viscous dissipation (friction losses on the pipe wall). In the same time, the internal energy increases because of this same viscous dissipation : the flow is actually heated by the friction losses.

Case of an adiabatic inviscid compressible flow For an inviscid flow, the viscous dissipation Φ becomes equal to zero (no viscous stresses). Moreover, the flow being adiabatic, Q˙ = 0 so that the energy balance (3.22) simplifies into : D 1 ~ 2 S ~ ~ ~ ~ U + Ψ + ρe dV = Pext + p∇ · U dV − p∇ · U dV DtˆDm(t) 2 ˆDm(t) ˆDm(t) or equivalently : D S ρE dV = Pext DtˆDm(t) The variation of the total energy of the material domain is driven again by the power of the external surface forces since Q˙ = 0. Consequently, if a passive system is considered where no S power is supplied to or extracted from the flow, Pext = 0 and the total energy of the material domain is conserved. However, here again, the nature of the energy changes over time since the material domain can be expanded or compressed depending on the sign of ∇~ · U~ . The 18 Lecture #3 : Energy conservation total energy remains constant but with a decrease or a decrease of the mechanical energy and a corresponding increase or decrease of the internal energy. If a compression is taking place, ∇~ · U~ < 0, so that the mechanical energy is decreased. If the potential energy is negligible, the mechanical energy decrease corresponds to a kinetic energy and a velocity decrease. In the same time, the internal energy increases : compression induces a heating of the flow. If an expansion is taking place, ∇~ · U~ > 0, so that the mechanical energy is increased. If the potential energy is negligible, the mechanical energy increase corresponds to a kinetic energy and a velocity increase. In the same time, the internal energy decreases : expansion induces a cooling of the flow.

Local form of the internal energy conservation equation The local form of the internal energy conservation equation can be either deduced from its global form or from the local form of the total energy and mechanical energy conservation equations. Using the global form :

D ρe dV = − p∇~ · U~ dV + Φ + Q˙ DtˆDm(t) ˆDm(t) and taking into account the Reynolds theorem for a material domain as well as the deﬁnitions of the viscous dissipation Φ and the added heat Q˙ yields :

De −→ −→ ρ dV = − p∇~ · U~ dV + −→τ : D dV + ∇~ (k∇~ T ) dV ˆDm(t) Dt ˆDm(t) ˆDm(t) ˆDm(t)

The local form of the internal energy conservation equation is immediately obtained :

De −→ −→ ρ = −p∇~ · U~ + −→τ : D + ∇~ (k∇~ T ) (3.23) Dt

Note that in the case of the incompressible flow of an ideal fluid (µ = 0 and k = 0), this equation reduces to : De = 0 Dt expressing the fact the internal energy of a fluid particle is conserved along the path of a fluid particle. For a non-ideal fluid, the internal energy of the fluid particle varies because of the viscous dissipation (which always contributes to the increase of e along the trajectory of the fluid particle) and of the heat conduction. Finally for a compressible flow of a non-ideal fluid, the internal energy also varies because of the compression / expansion process taking place in the flow.

3.1.5 Case of a continuous-flow system Let us consider the continuous-flow system schematically displayed in Fig.3.2. The envelope surface Σ0 is a solid wall supposed both fixed and adiabatic hence such that U~ = 0 and ~q ·~n = 0 on Σ0. The surface Σq is a fixed heat transfer surface such that U~ = 0 and ~q · ~n 6= 0 on Σq. The surface Σm is a moving adiabatic solid surface hence such that U~ 6= 0 and ~q · ~n = 0. The 3.1. Deriving the energy conservation equation 19

no-slip condition for a viscous ﬂuid states the velocity on Σm is non-zero, equal to the velocity of the moving surface; however the wall being assumed solid and not porous there is no velocity going into it so that we also have U~ · ~n = 0 on Σm. The general global form (3.10) of the total energy conservation equation applies on the ﬂuid

Figure 3.2: Application of the total energy conservation equation to a continuous-flow system. The fluid enters the control volume through section S1 and leaves through section S2.Σ0 is a fixed adiabatic surface guiding the flow from S1 to S2.Σq is a fixed non-adiabatic surface through which the flow can be heated or cooled. Σm is a moving adiabatic surface which can correspond typically to the surface of turbine or pump blades and make the system an active one. domain D limited by the inlet section S1, the outlet section S2 and the solid surfaces (adiabatic or not, fixed or not) Σ0,Σq,Σm reads : ∂(ρE) −→ ∂T dV + ρH (U~ · ~n) dS = (−→τ · U~ ) · ~ndS + k dS ˆD ∂t ˆS ˆS ˆS ∂n where the surface S of D is such that S = S1 ∪S2 ∪Σ0 ∪Σq ∪Σm. Note the total energy includes 1 ~ 2 p the potentiel energy so that E = e + 2 U + Ψ and H = E + ρ . −→ −→ Assuming the flow steady and using (−→τ · U~ ) · ~n = (−→τ · ~n) · U~ , we can rewrite : −→ ∂T ρH (U~ · ~n) dS = (−→τ · ~n) · U~ dS + k dS ˆS ˆS ˆS ∂n

Since S = S1 ∪ S2 ∪ Σ0 ∪ Σq ∪ Σm, the available boundary conditions on the various surfaces must be used so as to compute the LHS and RHS integrals. Since Σ0,Σq are ﬁxed solid walls, U~ = 0 on these surfaces so that :

ρH (U~ · ~n) dS = ρH (U~ · ~n) dS ˆS ˆS1∪S2∪Σm The power of the viscous forces (first integral on the RHS) and the heat flux (second integral on the RHS) can be considered as negligible with respect to the enthalpy flux on the inlet and 20 Lecture #3 : Energy conservation

outlet sections S1 and S2. Therefore it remains to compute the RHS integrals for Σ0 ∪ Σm ∪ Σq. Since only Σq is a non-adiabatic solid surface, it follows that : ∂T ∂T k dS = k dS ˆS ∂n ˆΣq ∂n

Finally, the surfaces Σ0 and Σq being fixed, the power of the viscous forces reduces to : −→ −→ (−→τ · ~n) · U~ dS = (−→τ · ~n) · U~ dS ˆS ˆΣm The total energy conservation equation eventually reads : −→ ∂T ρH (U~ · ~n) dS = ρH (U~ · ~n) dS + (−→τ · ~n) · U~ dS + k dS ˆS1∪S2 ˆΣm ˆΣm ˆΣq ∂n | {z } | {z } =W˙ 0 =Q˙ where W˙ 0 denotes the sum of the power extracted from the flow or supplied to the flow by the moving surface and the power of the viscous forces on Σm and where Q˙ denotes the quantity of heat supplied or extracted through Σq. Let us denote N~1 the unit normal vector to S1 aligned with the inflow direction and N~2 the unit normal vector to S2 aligned with the outflow direction. Obviously, ~n = −N~1 on S1 and ~n = N~2 on S2. Let us also introduce the mass flow rate through section Si (i = 1 or i = 2) :

m˙ i = ρ(U~ · N~i) dS ˆSi If the ﬂow is assumed quasi-uniform in the inlet and outlet sections, it is possible to write :

ρH(U~ · N~i) dS ≈ Him˙ i ˆSi where Hi denotes the average value of the total enthalpy in section Si. Therefore, the total energy balance can be expressed in the form :

0 H2 m˙ 2 − H1 m˙ 1 = W˙ + Q˙

Denoting ∆(·) = (·)2 − (·)1, we can also write :

∆(mH ˙ ) = W˙ 0 + Q˙

The variation of the product between the mass flowrate and the total enthalpy between the inlet and the outlet of the continuous flow system is therefore produced by the power extracted or supplied by the moving parts in the flow and the associated (viscous) losses plus the heat supplied or extracted through thermal exchange surfaces.

Detailing the expression of the total enthalpy (in the case where Ψ = gz the gravity potential) yields : p 1 ∆[m ˙ (e + + U~ 2 + gz)] = W˙ 0 + Q˙ ρ 2 3.1. Deriving the energy conservation equation 21

If no moving parts are present in the ﬂow and if there is not heat addition to the ﬂow :

∆[mH ˙ t] = 0

Moreover, mass conservation ensures in that case the conservation of the mass ﬂowrate hence m˙ 1 =m ˙ 2 =m ˙ so that ∆(mH ˙ ) =m ˙ ∆H. The total enthalpy for the passive (Σm = ∅) and adiabatic (Σq = ∅) is therefore conserved between the inlet and outlet of the system so that :

p 1 p 1 e + + U~ 2 + gz = e + + U~ 2 + gz ρ 2 1 ρ 2 2

3.1.6 Energy conservation equation for an ideal ﬂuid and Bernoulli equation

If the ﬂuid is assumed ideal, it means the viscous stresses can be neglected because the viscosity is assumed zero. It also implies the heat conduction is neglected because of a zero heat conductivity. Therefore, as already stated by Equation (3.11), the total energy conservation equation reduces to : ∂(ρE) dV + ρH (U~ · ~n) dS = 0 ˆD ∂t ˆS or, in the local form (3.13): ∂(ρE) + ∇~ · (ρH U~ ) = 0 ∂t 1 ~ 2 Note that in these expressions E = e + 2 U + Ψ with Ψ = gz if the body forces reduce to the gravitational forces. Assuming a steady ﬂow, the local form of the total energy conservation reduces to :

∇~ · (ρH U~ ) = 0 or ρU~ · ∇~ H + H∇~ · (ρ U~ ) = 0 Using mass conservation for a steady (compressible) ﬂow, ∇~ · (ρU~ ) = 0 and identifying the convective derivative with the material derivative for a steady ﬂow yields :

DH = 0 Dt

The total enthalpy is constant along a streamline in a steady flow of an ideal fluid. In other words : p 1 e + + U~ 2 + gz = cst along a streamline (3.24) ρ 2 for the steady flow of an ideal fluid. In the general case of a compressible flow, the internal energy e varies along a streamline. Indeed, the general form (3.23) of the internal energy conservation equation (written for a real fluid) states : De −→ −→ ρ = −p∇~ · U~ + −→τ : D + ∇~ (k∇~ T ) Dt 22 Lecture #3 : Energy conservation

For an ideal ﬂuid, this equation reduces to :

De ρ = −p∇~ · U~ Dt For the compressible steady flow of an ideal fluid, the internal energy e increases or decreases along a streamline depending on the sign of ∇~ · U~ . For a compression, ∇~ · U~ < 0 hence e is p 1 increasing (heating of the flow) and the quantity + U~ 2 + gz decreases accordingly. For an ρ 2 p 1 expansion, ∇~ · U~ > 0 hence e is decreasing (cooling of the flow) and the quantity + U~ 2 + gz ρ 2 increases accordingly. For the incompressible steady flow of an ideal fluid, the internal energy e is constant along a streamline since ∇~ · U~ = 0 so that the total energy equation states :

p 1 + U~ 2 + gz = cst along a streamline (3.25) ρ 2

The Bernoulli equation valid for a steady inviscid incompressible ﬂow with body forces expressed using a potential is thus recovered. A key diﬀerence between the ”incompressible” Bernoulli equation (3.25) and the ”compressible” Bernoulli equation (3.24) is the number of unknown quantities varying along the streamline. For the incompressible case, ρ is also constant along a streamline so that, assuming the elevation z is known, the varying quantities are the static pressure p and the velocity magnitude U. This property has been previously used to achieve velocity measurement from pressure measurements (static and total pressure) using a Pitot tube. In the compressible case, the unknown quantities varying along a streamline are the internal energy e, the static pressure p, the density ρ and the velocity magnitude U which makes the analysis more complex.

3.1.7 Flow modeling : provisional overview Still focusing on the analysis of an ideal fluid flow, let us review the available governing equations in their local form, completing the previous overview performed in Section ?? with the energy conservation equation derived in the present lecture. Mass conservation, momentum conservation and total energy conservation read in their conservative local form written for an ideal fluid (neglecting the body forces) :

 ∂ρ ~ ~  + ∇ · (ρU) = 0  ∂t    ∂(ρU ) ∂(ρU U ) ∂p i + i j = − ∂t ∂x ∂x  j i    ∂(ρE)  + ∇~ · (ρH U~ ) = 0 ∂t p Since (with neglected body forces), E = e + 1 U~ 2 and since H = E + , the unknown variables 2 ρ in this set of governing equations are : 3.2. Thermodynamic analysis 23

-• the d components of velocity for a d-dimensional ﬂow

-• the density ρ since we are interested in compressible ﬂow analysis,

-• the pressure p,

-• the internal energy e, that is d+3 unknown quantities for d+2 governing equations. The quantities ρ, p and e are state variables which define the thermodynamic state of the flow. In what follows, the fluid particle or the material domain Dm(t) will be considered as a thermodynamic system which achieves a local thermodynamic equilibrium. This equilibrium assumption implies the thermodynamic state of the flow can be described using two state variables only. Keeping for instance p and ρ, it means the value of the local internal energy e(~x,t) can be expressed from the values of the local pressure p(~x,t) and the local density ρ(~x,t). Such a relationship between thermodynamic state variables is known as an Equation of State (EoS in short). In order to close the above system of governing equations describing the compressible flow of an ideal fluid, we therefore need one EoS, which will be provided by the thermodynamic description of the flow. In the general case of the compressible flow of a viscous or real fluid, the set of governing equations read :

 ∂ρ ~ ~  + ∇ · (ρU) = 0  ∂t    ∂(ρU ) ∂(ρU U ) ∂p µ ∂(∇ · U~ ) i + i j = − + µ∆U + ∂t ∂x ∂x i 3 ∂x  j i i    ∂(ρE) −→  + ∇~ · (ρH U~ ) = ∇~ · (−→τ · U~ ) + ∇~ · (k∇~ T ) ∂t With respect to the ideal fluid case, there is one extra state variable : the temper ature T . Hence the d + 2 governing equations involve d + 4 unknown quantites (d velocity components and the 4 thermodynamic state variables ρ, p, e and T ). In that case, two EoS are needed to close the set of governing equations and obtain a fully defined description of the fluid flow.

3.2 Thermodynamic analysis

This section will not develop in detail the thermodynamic analysis of a system. Only the key concepts which are useful for the analysis of compressible ﬂows will be reviewed.

3.2.1 Reversible process and entropy Concept of reversibility Let us consider a fluid particle or a set of fluid particles (material domain) exchanging energy with its surroundings. This exchange process is said to be reversible if it can be ”inverted” so that the fluid particle and its surroundings return to their initial state. No real process in a fluid flow is perfectly reversible. All real fluid flows display for instance irreversible effects due to viscous friction. However, if we think a little more deeply about these viscosity effects and remember the analysis performed for high-Reynolds number flows, we can 24 Lecture #3 : Energy conservation consider a fluid particle evolving outside the boundary layer region is not significantly affected by the irreversible effects of the viscous friction so that the evolution of this fluid particle could be considered as in fact reversible. As will be made clear in the next lecture, another source of irreversibility exists even for an inviscid flow where no viscous effects are taking place : this new source of irreversibility is the occurrence of discontinuities or shockwaves in the flow. Thus, a fluid particle outside the boundary layer in a real fluid flow will evolve in a quasi-reversible way provided it does not cross a shock inside the inviscid flowfield. The great importance of a reversible flow assumption will become obvious once introduce in the next paragraph the concept of flow entropy.

Concept of entropy The second principle or second law of thermodynamics states that : -• there exists a state variable s, called (specific) entropy, which is given for a reversible process, by the relationship : δq ds = ( ) (3.26) T rev where δq is the heat transferred to the thermodynamical system in a reversible way. This elementary heat input δq (or output depending on the fact heat is provided to the flow or extracted from the flow) can be related to the internal energy e using the first principle or law of thermodynamics : de = δq − pdv where pdv is the elementary work of the pressure forces. The thermodynamic state variable v denotes the specific volume, i.e. v = 1/ρ. The product pdv corresponds to the quantity (pressure times elementary surface), thus elementary pressure force, multiplied by an elementary displacement to obtain the elementary work of the pressure force. Combining the definition of entropy and the first principle yields the Gibbs relationship between the thermodynamic state variables temperature, entropy, internal energy, pressure and density : 1 p T ds = de + pdv = de + pd( ) = de − dρ (3.27) ρ ρ2

-• every state change in a thermodynamic system generates an entropy variation which can be decomposed as follows : δq ds = ( ) + ds (3.28) T rev irrev where the contribution dsirrev corresponds to irreversible processes, such as viscous friction, irreversible heat transfer, shockwaves and is such that :

dsirrev ≥ 0 (3.29)

Consequently, it can be concluded : -• for an adiabatic process or ﬂow evolution, that is a process or a ﬂow evolution taking place without heat transfer (δq = 0) : ds = dsirrev ≥ 0 3.2. Thermodynamic analysis 25

-• for a reversible adiabatic process or ﬂow evolution :

ds = 0

A reversible adiabatic process or flow evolution is therefore an isentropic process or flow evolution, in which the specific entropy s remains constant throughout the flow. In the flow of an ideal fluid, the viscosity effects and the heat transfer can be considered as non-existent so that such a flow is both adiabatic and with irreversibilities which are limited to the occurrence of shockwaves inside the flow. In the following lecture, the shockfree flow of an ideal fluid will be therefore adiabatic and irreversible that is isentropic. Let us emphasize once again there are no real processes / flows which are prefectly isentropic. However, it is possible to assume a flow isentropic and predict, under this assumption, the key features of the flow. If this prediction proves reliable, this will validate the isentropic flow assumption while if the prediction is far from observation it will imply the flow irreversibilities must be taken into account.

3.2.2 Equations of state Since Gibbs relationship (3.27) states : p de = T ds − pdv = T ds + dρ ρ2 the canonical form of the equation of state (EoS) deﬁning e is e(s, ρ) or e(s, v) with s and v (or ρ) the so-called natural variables associated with the internal energy e. Similarly the speciﬁc enthalpy h is such that :

dh = d(e + pv) = de + pdv + vdp = T ds + vdp hence the canonical form of the EoS for h is h(s, p) with s and p the natural variables associated with enthalpy. If the expressions e(s, v) or h(s, p) are available, it is possible to deduce : -• the thermal EoS which links the pressure p, the density ρ and the temperature T -• lthe caloric EoS which links the internal energy e with the temperature T and the density ρ (or the enthalpy h with T and p). Starting from de = T ds − pdv, we wish to write the variation of internal energy de as a function of the variation of state variables which can be easily measured, or at least which are known to be accessible through measurement, namely the temperature T and the pressure p. Therefore, we are looking for a relationship between de and dT , dp. Since the natural variables for e are s and v, this means we must first express entropy and specific volume as a function of T and p, that is find : s = s(T, p) and v = v(T, p). We start by assuming the consequence of such a relationship on the total differential ds and dv : ∂s ∂s ds = dT + dp ∂T p ∂p T and ∂v ∂v dv = dT + dp ∂T p ∂p T 26 Lecture #3 : Energy conservation

Reinjecting in the Gibbs relationship which provides de leads to : " # ∂s ∂v ∂s ∂v de = T − p dT + T − p dp ∂T p ∂T p ∂p T ∂p T

The partial derivatives appearing in this development can be expressed using important coefficients in the analysis of compressible flows. Let C denote the specific heat coefficient linking the elementary quantity of heat received (in a reversible way) per unit mass of the fluid for a temperature variation dT : δq = C dT . Since the second principle of thermodynamics expresses ∂s that δq = T ds, it follows C = T . Let us consider more specifically the entropy variation in- ∂T duced by a temperature variation when the second state variable, upon which entropy depends, is fixed. We define : -• the specific heat coefficient at constant pressure :

∂s Cp = T ∂T p

-• the speciﬁc heat coeﬃcient at constant volume :

∂s Cv = T ∂T v

The compressibility coefficient expresses the local variation of volume induced by a local variation ∂v of pressure. It is therefore proportional to and, again, this coefficient is more precisely defined ∂p for a specific type of transformation or flow : isothermal flow (T = constant) or isentropic flow (s = constant), ... We define : -• the isothermal compressibility coefficient :

1 ∂v βT = − v ∂p T

-• the isentropic compressibility coeﬃcient :

1 ∂v βs = − v ∂p s

1 Note the quantity 1/(ρβ) has the dimension of the square of a velocity : [√ ] = [||U~ ||] = m s−1. ρβ We will see later the speed of sound in a flow is built upon this quantity. Let us finally introduce the coefficient of thermal expansion, which expresses the relative volume variation of the flow induced by a temperature variation taking place at constant pressure :

1 ∂v α = v ∂T p 3.2. Thermodynamic analysis 27

Most of the partial derivatives appearing in the above expression of de can be expressed using the coefficients Cp, βT and α which have just been introduced. We can directly write for instance dv using the thermal expansion coefficient and the isothermal compressibility coefficient :

dv = αvdT − βT vdp

Note that for a liquid, the typical values of α and β are much smaller than the typical values reached in a gas so that the relative volume variation (dv/v) as a function of dT and dp can be generally considered as negligible for a liquid flow. ∂s The partial derivative requires a specific treatment in order to be expressed using the ∂p T coefficient of thermal expansion α. Let us first recall other thermodynamic potentials can be introduced, on top of the internal (specific) energy e and the (specific) enthalpy h. Namely, we can define the (specific) free energy f and (specific) free enthalpy as g : f = e − T s and g = h − T s . The total differential of the free enthalpy is computed as follows :

dg = dh − T ds − sdT = d(e + pv) − T ds − sdT = de + pdv + vdp − T ds − sdT

From Gibbs relationship, de − T ds + pdv = 0 hence :

dg = −sdT + vdp therefore g = g(T, p) and dg can be expressed as :

∂g ∂g dg = dT + dp ∂T p ∂p T

By immediate identiﬁcation, it follows that :

∂g ∂g −s = , v = ∂T p ∂p T The theorem of Schwarz on mixed second partial derivatives allows to write :

∂2g ∂2g = ∂T ∂p ∂p∂T which yields one of the so-called Maxwell identities :

∂s ∂v = − (3.30) ∂p T ∂T p

Using (3.30), the total diﬀerential de can ﬁnally be expressed as :

de = (Cp − αpv)dT + (βT p − αT )vdp (3.31)

Before we detail the important particular case of a perfect or ideal gas and obtain explicit expressions for all the coeﬃcients (Cp, Cv, βT , βs, α), we can establish a few general identities which connect these coeﬃcients. 28 Lecture #3 : Energy conservation

Relationship between Cp and Cv Considering the canonical form s = s(T, v) we can deduce :

∂s ∂s ds = dT + dv ∂T v ∂v T

Since dv = αvdT − βT vdp, ds can also be developed into : ∂s ∂s ∂s ds = + αv dT − βT v dp ∂T v ∂v T ∂v T Multiplying this relationship by T , identifying with the total differential associated with s(T, p) and taking into account the definition of the coefficients Cp and Cv yields :

∂s Cp = Cv + αvT ∂v T Another Maxwell’s identity, obtained from df = −sdT − pdv, takes the form :

∂s ∂p = (3.32) ∂v T ∂T v so that : ∂p Cp − Cv = αvT ∂T v Considering now p = p(T, v), it is possible to write :

∂p ∂p dp = dT + dv ∂T v ∂v T Therefore, for a transformation (ﬂow evolution) taking place at constant pressure, that is such that dp = 0, we have : ∂v ∂p ∂p = − / ∂T p ∂T v ∂v T or else ∂p α = (αv)/(βT v) = ∂T v βT

Inserting in the above relationship between Cp and Cv yields the general identity for the difference of the specific heat coefficients : α2 Cp − Cv = vT (3.33) βT The ratio of the specific heat coefficients is defined as :

∂s

Cp ∂T p = Cv ∂s

∂T v 3.2. Thermodynamic analysis 29

Starting from p = p(s, T ) we can write the total diﬀerential of p as :

∂p ∂p dp = ds + dT ∂s T ∂T s and deduce from this identity written for an isobaric (p = constant) process :

∂s ∂p ∂p = − / ∂T p ∂T s ∂s T

In a similar way, we can establish :

∂s ∂v ∂v = − / ∂T v ∂T s ∂s T

Using these identities to express the ratio Cp/Cv yields :

∂p ∂v ∂p × C ∂T ∂s ∂v p = s T = s C ∂p ∂v ∂p v × ∂s T ∂T s ∂v T or else C β p = T (3.34) Cv βs We will see in the next section how these general relationships simplify in the case of an ideal gas.

3.2.3 Equations of state for a calorifically perfect gas From now on, in this course, air will be systematically considered as a perfect gas or an ideal gas. A perfect gas is such the intermolecular forces can be neglected yielding a simple relationship between the macroscopic state variables. The perfect gas approximation for air provides a very good description of the real properties displayed by air for a wide range of temperature and density - very large or very low temperature and density conditions require a more complex thermodynamic description but such configurations will not be met in the present course. An ideal or perfect gas satisfies the following thermal Equation of State :

p = p(ρ, T ) = ρrT (3.35)

R where r is the gas constant. The gas constant is computed as r = with R the universal M constant of perfect gases and M the molar mass of the gas. The universal constant R is such that R = 8.314472 JK−1 mol−1 or else R = 8314.472 JK−1 kmol−1. Since air is made of 21% of oxygen and 79% of nitrogen, with respective molar mass 32 g/mol and 28.02 g/mol or else 32 and 28.02 kg/kmol, it follows the molar mass of air is equal to 28.97 kg kmol−1 so that the gas constant r for air is equal to r = 287 J kg−1 K−1. 30 Lecture #3 : Energy conservation

Using (3.35) or pv = rT or else v = rT/p, it is immediate to compute α and βT for a perfect gas and to deduce Cp − Cv. We have : ∂v rT v = − 2 = − ∂p T p p 1 ∂v hence βT = − = 1/p. v ∂p T Similarly, we also have : ∂v r = ∂T p p 1 ∂v r 1 hence α = = = . v ∂T p pv T After calculation and simpliﬁcation, we obtain :

de = CvdT

Theoretically, Cv depends both on T and v. Let us prove that, for a perfect gas, Cv depends only on T . This means the partial derivative of Cv with respect to v is zero. From the deﬁnition of Cv, we can write : ∂C ∂ ∂s ∂2s ∂ ∂s v = T = T = T ∂v T ∂v ∂T v ∂v∂T ∂T ∂v T Using then Maxwell’s identity (3.32) this partial derivative can also be expressed as :

2 ∂Cv ∂ p = T 2 (3.36) ∂v T ∂T v Since p = rT/v for a perfect gas, p depends linearly on T when v is fixed and the second derivative of p with respect to T for v fixed is therefore zero. It can be therefore concluded that for a perfect gas : ∂Cv = 0 ⇒ Cv = Cv(T ) ∂v T This establishes the so-called first law of Joule which specifies the internal energy of a perfect gas depends only on temperature : e = e(T ). It is then possible to derive from de = Cv(T )dT the following expression for the internal energy :

e(T ) = C (T )dT ˆ v For a calorifically perfect gas, the specific heat coefficient at constant volume is constant. Such a behaviour is observed in practice for air when the flow temperature remains typically below 600 K. For a calorifically perfect gas, we have :

e(T ) = Cv(T − T0) + e0 = CvT + cste (3.37)

In the case of a perfect gas, the relationship (3.33) between Cp and Cv becomes : α2 1 pv Cp − Cv = vT = vT 2 p = βT T T 3.2. Thermodynamic analysis 31 hence Cp − Cv = r (3.38) This relationship, known as Mayer’s formula, expresses the difference between the specific heat coefficient at constant pressure and the specific heat coefficient at constant volume is equal to the gas constant r. For a calorifically perfect gas, since Cv = constant, it follows that Cp is also constant.

The enthalpy equation of state for a perfect gas is obtained from : p h = e + = e(T ) + rT = C T + cste + rT ρ v hence, using (3.38):

h(T ) = CpT + cste (3.39) with Cp = constant for a caloriﬁcally perfect gas. This is the second Joule’s law, stating the enthalpy of a perfect gas depends only on temperature.

The thermodyamic description of a (calorifically) perfect gas is completed with the calculation of the ratio of the specific heat coefficients Cp/Cv using (3.34). In order to compute the coefficient βs appearing in this identity, the variation of the specific volume with the pressure during an isentropic evolution must be expressed. To do so, we go back to Gibbs’ relationship and take into account the thermal EoS and the caloric EoS : dv T ds = de + pdv = C dT + rT v v or else dT dv ds = C + r v T v

Let us integrate this relationship between two states 1 and 2, taking into account Cv = constant for a caloriﬁcally perfect gas :

2 2 2 dT dv T2 v2 ds = s2 − s1 = Cv + r = Cv log( ) + r log( ) ˆ1 ˆ1 T ˆ1 v T1 v1 The thermal EoS yields : p ρrT ρ T 2 = 2 = 2 × 2 p1 ρrT1 ρ1 T1 and besides v2/v1 = ρ1/ρ2. The entropy diﬀerence s2 − s1 can therefore also be expressed as :

p2 ρ1 ρ2 s2 − s1 = Cv log( × ) − r log( ) p1 ρ2 ρ1

p2 ρ2 = Cv log( ) − (Cv + r) log( ) p1 ρ1

p2 ρ2 = Cv log( ) − Cp log( ) p1 ρ1 32 Lecture #3 : Energy conservation

Let us introduce γ, the ratio of the specific heat coefficient, which is a constant for a calorifically perfect gas : C γ = p (3.40) Cv

For air, a diatomic gas, γ = 1.4 . Going back to the calculation of the entropy diﬀerence and inserting γ, it follows that :

p2 ρ2 s2 − s1 = Cv log( ) − γCv log( ) p1 ρ1 γ p2 ρ2 = Cv log( ) − Cv log( ) p1 ρ1 p2 p1 = Cv log γ − Cv log γ ρ2 ρ1 The expression of the speciﬁc entropy for a perfect gas takes therefore the form :

p s = C ln (3.41) v ργ

An isentropic flow (s = constant) of a calorifically perfect gas is therefore such the following identity applies anywhere in the flow :

p = cste (3.42) ργ

The identity (3.42) is a key tool in the analysis of internal and external compressible flows, as will be seen in the next lectures. From a practical viewpoint, it is important to keep in mind the conditions of applicability for this identity : the flow must be isentropic. Any flow of an ideal fluid can be considered as isentropic as long as no discontinuities (shockwaves) occur in the flow. If a fluid particle crosses a discontinuity, a (positive) production of irreversible entropy takes place for this particle and the entropy level increases. For the flow of a real (viscous and heat conducting) fluid, other sources of irreversibility hence entropy creation are the viscous effects taking place in the boundary layer region and in the wake region. Outside these regions where a creation of irreversible entropy is taking place, the flow can be considered inviscid and thus isentropic if shockfree.

Pratical use of the isentropic ﬂow identity Between two points 1 and 2 located in the same region of isentropic ﬂow, it is possible to write :

p1 p2 γ = γ ρ1 ρ2 that is : p ρ γ 2 = 2 (3.43) p1 ρ1 3.2. Thermodynamic analysis 33

Knowing the static pressure at both points and the density at one of the point, the density at the other point can be deduced from (3.43).

Using the thermal EoS p = ρrT , it is also possible to write :

γ p T γ−1 2 = 2 (3.44) p1 T1

Knowing the static pressure at both points and the temperature at one of the point, the temperature at the other point can be deduced from (3.44).

Finally, combining the identities (3.43) and (3.44) yields :

1 ρ T γ−1 2 = 2 (3.45) ρ1 T1

Other useful identities for a perfect gas

Since pvγ = constant = C in the isentropic evolution of a perfect gas, it is possible to write :

∂v ∂(C/p)1/γ 1 = = − ∂p s ∂p γpv

Consequently, the isentropic compressibility coeﬃcient βs can be computed as follows :

1 β = (3.46) s γp

It can be checked the formula (3.34) stating that Cp/Cv = βT /βs yields Cp/Cv = (γp)/p for a perfect gas hence Cp/Cv = γ that is the definition or identity (3.40). Combining the definition of the ratio of the specific heat coefficients with Mayer’s identity, we obtain the following expressions for Cp and Cv in the case of a calorifically perfect gas :

r γr Cv = ,Cp = (3.47) γ − 1 γ − 1

Since γ = 1.4 for a diatomic gas such as air and since r = 287 J kg−1 K−1 for air, an immediate −1 −1 −1 −1 calculation yields Cv = 717.5 J kg K and Cp = 1004.5 J kg K . These constant values can be used for airﬂow analysis with local temperatures below 600 K typically.

Note also the thermal and caloriﬁc EoS can be combined to get rid of the temperature and obtain a relationship between p, ρ and e :

p = (γ − 1)ρe (3.48) 34 Lecture #3 : Energy conservation

Limitations of the perfect gas assumption

It has been experimentally checked for ﬂows of air that the quantity pv/(rT ) deviates only very weakly from unity (less than a 1% deviation) as long as the pressure and temperature levels are not too high (nor too low). For very high pressure levels, the collisions between molecules become more frequent and neglecting these collisions, as is done in the perfect gas model, is no longer a relevant assumption. Similarly, for low temperature levels, the gas molecules move more slowly which makes it possible for the inter-molecular forces to act more eﬃciently. The validity of the perfect gas model is also reduced in that case. In order to accurately describe the gas behaviour for such thermodynamic conditions, a more complex EoS is needed, such as for instance van der Waals EoS, taking the form :

rT a ρrT p = − = − aρ2 (3.49) v − b v2 1 − bρ where the constant quantities a and b depend on the gas under study. Note that for a = b = 0 the perfect gas EoS is recovered. Another way to describe the fact the gas behaviour deviates from the perfect gas description is to introduce a so-called compressibility factor Z deﬁned as :

pv Z = (3.50) rT

For a perfect gas, Z = 1 while for a so-called real gas, a typical observation is Z < 1 for high pressure levels or low temperature levels. For low pressure levels and high temperatures levels, it is observed that Z > 1. The behaviour of a real gas can be described by writing :

pv = ZrT or p = ZρrT (3.51) with a proper definition for Z. During for instance an hypersonic flight, very high temperatures are reached in the flow, which lead to chemical reactions modifying the chemical air composition hence its molar mass. In such flight conditions, M decreases and therefore r increases. This is often described as ”real gas effects” even though the flow remains that of a mixture of perfect gases with occurrence of chemical reactions.

3.3 Conclusions on the ﬂow governing equations

3.3.1 Final overview of the governing equations

To conclude on the derivation of the ﬂow models, we can go back to Section 3.1.7 and take into account the thermodynamic description of a perfect gas.

Isentropic ﬂow of an ideal ﬂuid with a perfect gas EoS

If air is considered as an ideal fluid, that is non-viscous (inviscid) and non-heat conducting, and if it is also thermodynamically described using the perfect gas EoS, the isentropic model of a 3.3. Conclusions on the flow governing equations 35 compressible flow of air reduces to :  ∂ρ  + ∇~ · (ρU~ ) = 0  ∂t    ∂(ρU ) ∂(ρU U ) ∂p i + i j = − (3.52)  ∂t ∂xj ∂xi    p  = constant = Sref  ργ System (3.52) is indeed a closed model with for instance the static pressure p expressed as a function of the density ρ using the isentropic identity, yielding eventually the d + 1 mass and momentum conservation equations with d + 1 unknown, density ρ and d components of velocity U~ .

General flow of an ideal fluid with a perfect gas EoS In the general case of an inviscid adiabatic flow not necessarily isentropic, the Euler system of governing equations take the following conservative form :  ∂ρ  + ∇~ · (ρU~ ) = 0   ∂t    ∂(ρU~ ) −→ + ∇~ · (ρU~ ⊗ U~ + p δ ) = 0  ∂t    ∂(ρE)  + ∇~ · (ρH U~ ) = 0  ∂t −→ −→ where δ denotes the unit tensor such that ( δ )ij = δij with δij the Kronecker symbol. p Since (neglecting body forces), E = e + 1 U~ 2 and H = E + , the system is closed with an 2 ρ EoS linking the internal energy e, the static pressure p and the density ρ. For a perfect gas, we have p = (γ − 1)ρe hence the system is closed with d + 2 partial differential equations for d + 2 independent unknowns, which can be either the primitive variables ρ, U~ , p or the conservative variables ρ, ρU~ and ρE. We will see later, when numerically solving the set of governing equations, why the conservative variables are especially useful.

General flow of a real fluid with a perfect gas EoS In the general case of the compressible flow of a viscous or real fluid, the set of governing equations read :  ∂ρ ~ ~  + ∇ · (ρU) = 0  ∂t    ∂(ρU ) ∂(ρU U ) ∂p µ ∂(∇ · U~ ) i + i j = − + µ∆U + ∂t ∂x ∂x i 3 ∂x  j i i    ∂(ρE) −→  + ∇~ · (ρH U~ ) = ∇~ · (−→τ · U~ ) + ∇~ · (k∇~ T ) ∂t 36 Lecture #3 : Energy conservation and is closed for a perfect gas with the thermal and calorific equations of state :

p = ρrT and e = Cv T leaving again a set of d + 2 partial diﬀerential equations with d + 2 primitive unknowns (ρ, U~ , p) or d + 2 conservative variables (ρ, ρU~ , ρE).

Energy conservation for an ideal fluid with a perfect gas EoS We have previously established that for a steady adiabatic flow of an ideal fluid, total energy conservation is equivalent to the conservation of total enthalpy H along the flow streamlines. In particular, for a flow with uniform upstream conditions, H is constant everywhere in the flow which is then said to be homenthalpic (same (total) enthalpy everywhere). The general conservation of total enthalpy (along a streamline or everywhere in the flow) yields (when neglecting gravitational forces) : 1 h + U~ 2 = constant 2 In the case of a perfect gas flow, the equation of state (3.39) for the specific enthalpy allows to write : 1 C T + U~ 2 = constant (3.53) p 2 Using the thermal equation of state, p = ρrT , and expressing the specific heat coefficient at constant pressure as a function of r and γ yields the equivalent relationship :

γ p 1 + U~ 2 = constant (3.54) γ − 1 ρ 2

3.3.2 Velocity measurement in a compressible flow Let us go back the Pitot tube previously studied in Section ?? in the context of incompressible flows and displayed in Fig.3.3. Let us remind the reader this probe is equipped with both a total pressure measurement orifice (point A) and a static pressure measurement orifice (point B). The Pitot tube is inserted in a compressible flow of air, considered as an ideal gas. The generalized ”compressible Bernoulli equation” can be applied between points M and A on one hand and between points N and B on the other hand so that the following relationships are obtained : pM 1 2 pA 1 2 eM + + UM = eA + + UA ρM 2 ρA 2

pN 1 2 pB 1 2 eN + + UN = eB + + UB ρN 2 ρB 2 Points N and M are suﬃciently close to one another to ensure equality between the velocity and the thermodynamic state variables at point N and point M. Therefore, one can write :

pA 1 2 pB 1 2 eA + + UA = eB + + UB ρA 2 ρB 2 3.3. Conclusions on the ﬂow governing equations 37

Figure 3.3: Pitot probe for velocity measurement. or else using the EoS e = e(p, ρ) for a perfect gas :

γ pA 1 2 γ pB 1 2 + UA = + UB γ − 1 ρA 2 γ − 1 ρB 2

Since A is a stagnation point, UA = 0 and pA = (p0)A the total pressure measured at point A, hence :

γ (p0)A γ pB 1 2 = + UB (3.55) γ − 1 ρA γ − 1 ρB 2 Contrarily to the incompressible flow analysis, this relationship does not give immediate access to the velocity at point B because, even though the static pressure pB is also known from measurement, two other unknown quantities appear in the relationship : the density ρA and the density ρB. The flow can be assumed isentropic between point M and A (inviscid flow so no viscous effects contributing to an entropy increase) as well as between point N and B assuming the flow is taking place at a sufficiently high Reynolds number for the boundary layer to be thin and the streamline connecting N to B to be located outisde this boundary layer. Therefore, it is possible to write : pA pM pN pB γ = γ = γ = γ ρA ρM ρN ρB hence the densities at point A, B and the measured pressures are related by :

(p0)A pB γ = γ ρA ρB or else 1/γ (p0)A ρA = ρB (3.56) pB

Hence if ρB is known then ρA can be computed using the above expression where the pressures are known from measurement. Therefore, it remains to determine ρB in order to obtain an expression for the velocity UB. Since pB is measured, the measurement of the temperature TB at point B using a temperature probe allows to compute ρB from the EoS :

pB ρB = (3.57) rTB 38 Lecture #3 : Energy conservation

In a compressible flow, the velocity at a point is obtained from the measurement of a total pressure, a static pressure and a (static) temperature. In an incompressible flow, the measurement of a total pressure and a static pressure is sufficient. Inserting the relationships (3.56) into the compressible Bernoulli equation (3.55) yields :

" 1/γ # 2 2γ 1 pB pB UB = (p0)A × × − γ − 1 ρB (p0)A ρB

Factoring pB/ρB yields : " 1 # 1− γ 2 2γ pB (p0)A UB = − 1 γ − 1 ρB pB

Finally, using the EoS at point B (3.57), the formula giving the velocity UB as a function of the measured pressures and temperature for the compressible ﬂow under study reads :

s s γ−1 comp 2γrTB (p0)A γ UB = − 1 (3.58) γ − 1 pB

This relationship can be compared with the formula previously established in the context of an incompressible ﬂow analysis : s 2(p − p ) U = 0A B B ρ To make the comparison easier, the constant density ρ in the ”incompressible” formula can be computed using the EoS of a perfect gas at point B : ρ = ρB = pB/(rTB). The incompressible formula can then be expressed as :

s inc p (p0)A UB = 2rTB − 1 (3.59) pB

The ratio of the compressible and incompressible formulae depend only on the ratio between the measured total pressure and the measure static pressure, say η = (p0)A/pB :

s γ−1 comp γ UB η − 1 inc = UB η − 1 The higher the velocity to measure, the larger the diﬀerence between the total pressure at point A and the static pressure at point B hence the higher the ratio η and the larger the diﬀerence of prediction between the compressible and the incompressible formulae.

Let us consider a flow of air and let us assume the density has been computed in the air at rest from the measured pressure (corresponding to a stagnation pressure since the fluid is at rest) and the measured temperature. Let us assume the pressure p0 in the fluid at rest is p0 = 101325 P a 3 and the temperature is T0 = 300 K. The density ρ is thus given by ρ = p0/(rT0) ≈ 1.177 kg/m . Let us now assume the flow incompressible and exploit the measured pressures at A and B for various (unknown) flow velocities. A first test is performed for which (p0)A = 101325 P a and pB = 100619 P a. A second test is performed for which (p0)A = 101325 P a and pB = 66471 P a. 3.4. Exercises and problems 39

Applying formula (3.59) with ρ = 1.177 kg/m3 and the measured pressures yields an estimate for the velocity UB such that UB = 34.7 m/s in the ﬁrst test and UB = 283 m/s in the second test.

During the same tests, the temperature has been measured at point B : for the first test TB = 299.4 K while for the second test TB = 266 K. Using the more accurate formula (3.59) (more accurate because it takes into account the fact the density is not constant in the flow), the velocity UB is estimated at UB = 34.7 m/s in the first test and UB = 261.5 m/s in the second test. For the low velocity first test-case using the incompressible or the compressible more formula yields almost no difference in the velocity estimate. For the high-speed second test-case the difference is far from being negligible since the incompressible formula overestimates the velocity UB by 10% with respect to the compressible prediction.

3.4 Exercises and problems

3.4.1 Exercise # 1 : Analysis of a steam turbine

Question

Steam enters a turbine with a velocity of 30 m/s and a specific enthalpy of 3348 kJ/kg (see Fig. 3.4). The steam leaves the turbine as a mixture of vapor and liquid having a velocity of 60 m/s and a specific enthalpy of 2550 kJ/kg. If the flow through the turbine is adiabatic and changes in elevation are negligible, determine the work output involved per unit mass of steam through-flow.

Figure 3.4: Schematic view of the steam turbine with inﬂow and outﬂow conditions. 40 Lecture #3 : Energy conservation

Answer Using the analysis of Section 3.1.5 with a constant mass flowratem ˙ through the turbine we can write : p 1 m˙ ∆[(e + + U~ 2 + gz)] = W˙ 0 + Q˙ ρ 2 where W˙ 0 is the power extracted from the steam through-flow by the turbine and Q˙ is the head added to the system per unit time, equal to zero in the present cas since the flow is assumed adiabatic. Let us introduce wshaft the work output per unit mass of stream through-flow; it is such that : 0 W˙ =m ˙ wshaft Therefore, the total energy conservation yields : p 1 ∆[(e + + U~ 2 + gz)] = w ρ 2 shaft or 1 ∆h + ∆(U~ 2) + ∆(gz) = w 2 shaft The changes in elevation being negligible, we can write : 1 w = ∆h + ∆(U~ 2) shaft 2 An immediate numerical application yields : 1 w = (2250 − 3348) × 103 + ∆(602 − 302) = −1098 × 103 + 1350 = −1096.65 kJ/kg shaft 2 This quantity is negative since it is extracted from the fluid. The net work output per unit mass of steam through-flow is 1096.65 kJ/kg. Note that in this particular example the change in kinetic energy remains small with respect to the change in specific enthalpy, which is often true for applications involving steam turbines. Note also the mass flowratem ˙ must be known in order to determine the power output W˙ shaft.

3.4.2 Exercise # 2 : Analysis of the draining of a tank This exercise is another application of the ”incompressible Bernoulli equation” and could therefore be treated using the concepts available at the end of Lecture 2. It is proposed at the end of the present Lecture 3 in order to emphasize the diﬀerence between the ”incompressible Bernoulli equation” and the general ”compressible Bernoulli equation” used in the next exercise.

Question Let us consider the tank configuration displayed in Fig. 3.5. The tank sectional area is denoted AT while A0 denotes the orifice sectional area, assumed much smaller than AT : A0 << AT . The tank is initially filled with water up to a level h0 above the orifice located at the bottom of the tank. The static pressure surrounding the tank is the atmospheric pressure patm. At t = 0 the orifice is open and the tank starts to drain or to empty. The flow is slightly unsteady due to the lowering of the water level in the tank but this effect is small if AT >> A0 which is precisely the assumption we made. Viscous effects are negligible 3.4. Exercises and problems 41 everywhere away from the walls of the tank.

• Estimate how long it takes for the tank to empty. Give an analytical estimate for the draining time and compute it for h0 = 20 cm, a cylindrical tank of radius RT = 2.5 cm and an oriﬁce of radius R0 = 2 mm.

Figure 3.5: Flow through a sharp-edged oriﬁce at the bottom of a tank ﬁlled with water.

Answer Let h(t) be the height of the liquid in the tank above the orifice at time t. Assuming a quasi- steady and incompressible flow, neglecting the viscous effects and considering the only body force is the gravitational force, the Bernoulli equation can be applied along a streamline connecting a point 1 at the liquid free surface in the tank and a point 2 in the outflow orifice section : p 1 p 1 1 + U 2 + gz = 2 + U 2 + gz ρ 2 1 1 ρ 2 2 2

The static pressure p1 at the free surface is equal to patm. The static pressure in the jet forming at the oriﬁce is assumed uniform so that, neglecting the surface tension between the water and the surrounding air the pressure p2 is also equal to patm. The diﬀerence between the elevation z1 and the elevation z2 is approximately equal to h(t) the height of the liquid level in the tank. The Bernoulli equation takes therefore the form : 1 1 U 2 + gh(t) = U 2 2 1 2 2 The unknown velocities can be related to h(t) as follows : 42 Lecture #3 : Energy conservation

dh -• the velocity of the ﬂuid particle at point 1 is equal to − dt -• the mass conservation applied to the control volume limited by the tank wall, the free surface and the oriﬁce section yields :

ρU1AT = ρU2A0

AT hence U2 = U1. A0 Replacing in the Bernoulli equation yields : ! 1 dh2 A 2 1 − T + gh(t) = 0 2 dt A0 or v 1 dh u 2g = −u p dt u 2 h(t) t AT 2 − 1 A0 where the minus sign expresses the fact the water level in the liquid tank decreases over time. The mathematical problem to solve in order to find h(t) is an ordinary differential equation of the form : 1 dh = −C ph(t) dt v u 2g with the constant C given by C = u . The function h(t) satisfies the initial condition u 2 t AT 2 − 1 A0 h(0) = h0 and the final condition h(tf ) = 0 where tf is the draining time we are looking for. Integrating this equation between t = 0 and the current time t yields :

t dh t √ = − C dt ˆ0 h ˆ0 hence h it 2ph(t) = −Ct 0 so that p p 2 h(t) − 2 h0 = −Ct

When t = tf the liquid level above the oriﬁce is such that h(tf ) = 0, hence : p 2 × 0 − 2 h0 = −Ctf

The draining time is therefore given by :

s s 2 2h0 AT tf = 2 − 1 g A0 3.4. Exercises and problems 43 or else, using the radius of the tank and oriﬁce cross-sectional area and taking also into account the assumption AT /A0 >> 1 : s 2 2h0 RT tf ≈ g R0

For the proposed numerical application, RT /R0 = 2.5/0.2 = 12.5 so that : s 2h t = 156.25 0 f g

2 With an initial liquid height h0 = 0.2 m (and taking g = 9.81 m/s ), the draining time of the tank is equal to 31.55 s.

3.4.3 Exercise # 3 : Analysis of the compressible flow in a Venturi The flow in a Venturi has been studied in ?? under the assumption of an incompressible flow. In the present exercise, we assume a more general compressible flow and proceed to analyze the Venturi device with the available tools for compressible flow analysis.

Questions Let us consider again the Venturi device displayed in Fig. 3.6. It is made of a tube with hor- izontal axis Ox and such that the tube section varies smoothly (continuously) and displays a throat (section with minimum area). Air flows through the tube and it is desired to compute the mass flowrate of this flow. Let us denote A1 a section upstream of the throat and let us index 1 the physical quantities relative to this section. Similarly, the throat section is denoted A2 with associated quantities measured in this section indexed 2. The flow is also assumed quasi-1D, that is the flow properties are assumed uniform in each section (but taking possibly different values from one section to the other). Let us denote β = A2/A1 the area ratio of sections 2 to 1. Assume the flow is steady and inviscid.

• Show the mass ﬂowrate in the nozzle can be expressed as a function of p1, p2, ρ1, A2 and β.

The expansion factor of the Venturi device is defined as the ratio between the mass flowrate measured for an incompressible flow assumption and the mass flowrate measured while accounting for density variation. • Give the expression of this expansion factor using the result established in the previous question and the result obtained in ?? for an incompressible flow analysis.

Answers

For a steady compressible flow of an ideal fluid, the energy conservation simplifies into H = p 1 γ p 1 constant along a streamline, with H the specific total enthalpy, H = e+ + U 2 = + U 2 ρ 2 γ − 1 ρ 2 (neglecting the body forces). States 1 and 2 are therefore linked by the relationship :

γ p1 1 2 γ p2 1 2 + U1 = + U2 (3.60) γ − 1 ρ1 2 γ − 1 ρ2 2 44 Lecture #3 : Energy conservation

Figure 3.6: Flow in a Venturi.

A uniform flow is assumed in each section hence pi, ρi, Ui are respectively the uniform static pressure, density and velocity in section Ai. Moreover, if the inviscid flow of the perfect gas is assumed isentropic (no shockwaves in the Venturi nozzle), the relationship p/ργ = constant holds everywhere in the flow so that it is possible to write : p1 p2 γ = γ ρ1 ρ2 or else 1 ρ p γ 2 = 2 (3.61) ρ1 p1

Finally, mass conservation between section A1 and A2 can be also expressed as the conservation of the mass ﬂowrate through the nozzle, that is :

ρ1U1A1 = ρ2U2A2 Taking into account (3.61) to express the density ratio as a function of the pressure ratio and introducing the geometric factor β yields : 1 ρ2 p2 γ U1 = βU2 = β U2 (3.62) ρ1 p1 Injecting (3.62) into the generalized Bernoulli equation (3.60) for compressible ﬂows, we can eliminate U1 and obtain : 2 ! γ 1 2 2 p2 γ p1 p2 ρ1 U2 1 − β = 1 − 2 p1 γ − 1 ρ1 p1 ρ2 p Using again the isentropic ﬂow identity (3.61) and denoting η = 2 the static pressure ratio p1 between sections A2 and A1, the velocity U2 is given by :

1 2γ p1 γ−1 2 γ U2 = 2 1 − η 1 − β2η γ γ − 1 ρ1 3.4. Exercises and problems 45

The (constant) mass ﬂowratem ˙ is therefore such that :

2 A 2 2γ p1 γ−1 2 2 2 2 2 2 γ γ m˙ = ρ2U2 A2 = 2 ρ1η 1 − η 1 − β2η γ γ − 1 ρ1 or else a formula involving as requested p1, p2 (through η), ρ1, β and A2 :

r r A2 2γ √ 2 γ−1 m˙ = p1 ρ1 η γ 1 − η γ (3.63) q 2 γ − 1 1 − β2 η γ

In order to compute the ratio between the compressible formula (3.63) yieldingm ˙ comp and the previously established incompressible formula we slightly modify this latter, taken from ?? :

A2 p m˙ inc = ρ2U2A2 = 2ρ1(p1 − p2) p1 − β2

Introducing the pressure ratio η, we can write :

A2 √ √ p m˙ inc = 2 ρ1 p1 1 − η p1 − β2

Dividing the formula (3.63) valid for a general compressible ﬂow with the above formula valid under the assumption of incompressible ﬂow, we obtain : v γ−1 ! s r u γ 2 m˙ comp γ u 2 1 − η 1 − β γ = × tη × 2 m˙ inc γ − 1 1 − η 1 − β2η γ

For a fixed gemeotry of the Venturi, β is fixed. If the flowing gas is air, γ = 1.4. The expansion facotr is a function of the pressure ratio η only :

m˙ comp = Y (η) m˙ inc

For Venturi geometries such that the diameter ratio between sections A2 and A1 become less than 0.25, the factor β2 become small with respect to unity since it varies as the fourth power of this diameter ratio. It is then possible to approximate the expansion factor with the simpliﬁed formula : v u γ−1 ! r γ u 2 1 − η γ Y (η) ≈ × tη γ γ − 1 1 − η

This expression of the expansion factor is plotted as a function of η in Fig.3.7. It can be observed the difference between the incompressible and compressible formula for computing the mass flowrate becomes rapidly significant when the ratio of the measured pressures increases. In other words, a correct estimate of the mass flowrate using the Venturi device for compressible flows requires the application of the ”compressible formula”, which properly accounts for the variation of ρ, p and U in the nozzle. 46 Lecture #3 : Energy conservation

Expansion factor vs static pressure ratio 1.8 sqrt(1.4/0.4)*sqrt((x**(2/1.4))*(1-(x**(0.4/1.4)))/(1-x))

1.7

1.6

1.5

1.4 Y(eta)

1.3

1.2

1.1

1 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 eta

Figure 3.7: Evolution of Y (η) with η in a Venturi where β2 << 1.

3.4.4 Problem # 1 : pressure rise in a nuclear reactor containment

Questions

All forthcoming Lectures will be devoted to the analysis of flows in the aerospace context. How- ever the governing equations which have been established for a general compressible flow are also useful for many other engineering applications. In the present problem, you are asked to analyze the pressure rise in a nuclear reactor containment during an accidental scenario where a large breach appears in the circulation pipe of the primary refrigerant. This problem is taken from the book Thermohydraulique des réacteurs, written by Jean-Marc Delhaye and published in 2008 by EDP Sciences in the Collection GénieAtomique.

Let us consider a nuclear reactor containment with volume V = 49400 m3 containing, in nominal ◦ operating conditions, air at a pressure p0 = 1 bar and at a temperature T0 = 20 C. During an accidental scenario, a large breach appears on the main pipe in which water vapor circulates and this induces a pressure rise in the containment because of the injection of water vapor in the containment through the breach. We wish to determine the maximum pressure applied to the containment, under various assumptions. The mixture of air and water vapor in the containment can be modeled as a perfect gas (overheated vapor) for temperature in the range 200 ◦ to 250 ◦ and pressure in the range 1 bar to 5 bar. This mixture is such the ratio of the specific heat capacities is γ = Cp/Cv = 1.3, with a −1 −1 constant pressure specific heat capacity Cp = 2.08 kJ kg K . The specific enthalpy h is such that h(T ) = CpT + h0 with h0 = 1900 kJ/kg. 3.4. Exercises and problems 47

Case of a constant vapor mass ﬂowrate at the breach, without condensation on the containment walls The following law of variation is assumed for the mass ﬂowrate at the breach : m˙ in0 = 3500 kg/s for 0 < t ≤ tin = 20 s m˙ in = (3.64) 0 for t > tin

For 0 < t ≤ tin, the speciﬁc enthalpy of the vapor injected into the containment through the breach is assumed such that hin = hin0 = 2900 kJ/kg.

• Determine (and plot) the time evolution of the (volume averaged) pressure and the (volume averaged) temperature in the containment. Show that the maximum (averaged) pressure is equal to 5.25 bar for this ﬁrst scenario and that the maximum (averaged) temperature reaches 241◦C.

Case of a linearly decreasing vapor mass flowrate at the breach, without condensation on the containment walls The law of variation of the mass flowrate at the breach is now assumed of the form : m˙ in0(1 − t/tin) for 0 < t ≤ tin = 20 s m˙ in = (3.65) 0 for t > tin withm ˙ in0 = 7000 kg/s. For 0 < t ≤ tin, the specific enthalpy of the water vapor injected into the containment is still assumed such that hin = hin0 = 2900 kJ/kg.

• Determine (and plot) the time evolution of the (volume averaged) pressure and the (volume averaged) temperature in the containment under these new assumptions. Show that the maximum (averaged) pressure is also equal to 5.25 bar for this second scenario and that the maximum (averaged) temperature also reaches 241◦C.

Case of a constant mass flowrate at the breach, with condensation on the containment walls The assumptions of the first scenario (3.64) are retained for the flow conditions at the breach. However, in this third scenario, we account for the fact a liquid film develops on the walls of the containment because of the condensation of the overheated vapor. From a modelling viewpoint, the development of this liquid film is assumed to produce a mass flowratem ˙ cond leaving the control volume associated with the volume of overheated vapor and such that :   m˙ cond0 = 1200 kg/s for 0 < t ≤ tin = 20 s m˙ cond = m˙ cond0(tcond − t)/(tcond − tin) for tin < t ≤ tcond = 50 s (3.66)  0 for t > tcond A specific enthalpy is associated with the phase change taking place when the overheated vapor turns into the liquid film. This specific enthalpy, to be used to compute the enthalpy flux out of the control volume associated with the overheated vapor, is such that hfg = 2180 kJ/kg.

• Determine (and plot) the time evolution of the (volume averaged) pressure and the (volume averaged) temperature in the containment under these new assumptions. Show that the 48 Lecture #3 : Energy conservation maximum (averaged) pressure is equal to 4.54 bar for this third scenario and that the maximum (averaged) temperature reaches 467◦C.

Recommendation : the control volume in which the conservation equations will be applied must be judiciously selected. Using the suggested control volume displayed in Fig.3.8, you should be able to describe in a uniﬁed way the three scenarios detailed above.

Figure 3.8: Schematic view of the nuclear reactor containment and the associated control volume for the superheated vapor.

Answers

In order to describe the evolution of the pressure and temperature in the containment as a function of time, let us write the mass balance and the energy balance for the suggested ﬁxed control volume displayed in Fig. 3.8. Let us denote V0 this volume and S0 its surface. Note V0 is not a closed volume but an open system since a mass / momentum / energy ﬂux takes place through the breach. Going back to the general form (??) of the global mass conservation written for an arbitrary control volume :

d ρ dV = − ρ(U~ − W~ ) · ~ndS dtˆDa(t) ˆSa(t)

we can write in the case of the ﬁxed volume V0 (W~ = ~0) :

d ρ dV = − ρU~ · ~ndS (3.67) dtˆV0 ˆS0 3.4. Exercises and problems 49

Let us denote m(t) the mass of fluid (vapor) in the containment at time t. By definition of the flow density, we can write m(t) = ρdV so that (3.67) also reads : ˆV0

dm(t) = − ρU~ · ~ndS dt ˆS0 and expresses the fact the mass of fluid contained in V0 varies as a function of the mass flowrate entering or leaving the fluid domain. The surface S0 is made of the containment surface and of the breach surface through which a mass flow enters the domain. Let us denote Sin the surface of the breach since it is an inlet boundary for vapor into the containment. The mass flowrate of fluid entering the containment through the breach is given by :

m˙ in(t) = − ρU~ · ~ndS ˆSin where the minus sign comes from the fact the normal ~n is the outward-pointing unit normal vector. In the case where no condensation is taking place at the wall,m ˙ in(t) is the only massflow entering the volume V0. If condensation is taking place on the containment walls, we must also account for a vapor mass flowrate leaving the volume V0 to condense on the containment wall and form a liquid film. Let us denotem ˙ cond(t) this mass flowrate of vapor leaving V0 to condense on the containment wall. The mass balance (3.67) can eventually be expressed as :

dm = ˙m (t) − ˙m (t) (3.68) dt in cond withm ˙ cond(t) = 0 for case #1 and #2 andm ˙ cond(t) 6= 0 for case # 3. The identity (3.68) can then be used to derive the vapor mass variation in V0 depending on the inlet massﬂow and condensation massﬂow associated with the three scenarii or cases under study.

The energy balance is expressed using the general global form of the energy conservation law, which reads as follows for an arbitrary ﬂuid domain :

d ( ρEdV ) = W˙ + Q˙ − ρE (U~ − W~ ) · ~ndS dt ˆDa(t) ˆSa(t)

In the case of the ﬁxed control volume V0 (W~ = ~0) it comes :

d ( ρEdV ) = W˙ + Q˙ − ρE U~ · ~ndS dt ˆV0 ˆS0

Neglecting the power of body forces and viscous constraints yields the following expression of W˙ (power of the pressure constraints only) :

W˙ = − pU~ · ~ndS ˆS0 50 Lecture #3 : Energy conservation

Assuming Q˙ = 0 (no thermal losses through the boundary S0) yields the following form of the energy balance : d ( ρEdV ) = − pU~ · ~ndS − ρEU~ · ~ndS dt ˆV0 ˆS0 ˆS0 It is then natural to introduce the speciﬁc total enthalpy H such that H = E + p/ρ so that the energy balance can be recast as :

d ( ρEdV ) = − ρHU~ · ~ndS dt ˆV0 ˆS0

This relationship expresses the fact the total energy of the fluid (vapor) contained in the control volume V0 varies as a function of the total enthalpy flux entering or leaving the volume. This enthalpy flux is an inlet flux at the breach surface Sin. An outlet flux also exists when condensation takes place on the wall of the containment reactor. If condensation is supposed not to occur, the inlet total enthalpy flux reads :

− ρHU~ · ~ndS = − ρHU~ · ~ndS ˆS0 ˆSin Assuming the kinetic energy contribution to the specific total enthalpy can be neglected with respect to the specific enthalpy contribution h so that Hin, the specific total enthalpy at the breach, is such that Hin ≈ hin so that the inlet flux of total enthalpy can be expressed as :

− ρHU~ · ~ndS =m ˙ in × hin ˆSin If condensation is taking place on the containment wall, a total enthalpy outflow is taking place through S0 which can be expressed using the specific enthalpy hfg associated with the phase change (from vapor to liquid) and the condensation mass flowrate so that the total enthalpy flux reads :

− ρH~v · ~ndS =m ˙ in(t) × hin − m˙ cond(t) × hfg ˆSin

Let us recallm ˙ cond(t) = 0 for scenarii #1 and #2 whilem ˙ cond(t) 6= 0 for scenario # 3. Inserting this expression for the total enthalpy ﬂux in the energy balance and expressing ρE = ρH − p yields : d ( (ρH − p)dV ) =m ˙ in × hin − m˙ cond(t) × hfg dt ˆV0

As previously done at the breach, it is assumed that at any point in the control volume V0 the kinetic energy contribution to the total enthalpy remains negligible with respect to the speciﬁc enthalpy (thermal) contribution so that :

1 H = h(T ) + ~v2 ≈ h(T ) = C T + h 2 p 0 Therefore, the quantity ρH − p appearing the LHS of the energy balance can also be expressed as : Cp 1 ρH − p = ρh − p = ρC T + ρh − p = p − p + ρh = p + ρh p 0 r 0 γ − 1 0 3.4. Exercises and problems 51 where the perfect gas thermal equation of state p = ρrT has been used as well as the deﬁnition of the speciﬁc heat at constant pressure Cp = γr/(γ −1). The LHS volume integral of the energy balance, which expresses the variation of total energy in the volume V0, can be computed as : d 1 d d (ρH − p)dV = pdV + h0 ρdV dtˆV0 γ − 1 dtˆV0 dt ˆV0 | {z } =m(t)

The averaged pressure p(t) in the containment reactor at time t is deﬁned as :

V0p(t) = pdV ˆV0

Therefore the variation of total energy in the volume V0 can be computed as :

d V0 dp(t) dm(t) (ρH − p)dV = + h0 dtˆV0 γ − 1 dt dt

Since this variation is also equal to the total enthalpy ﬂux it comes :

V dp(t) dm(t) 0 + h =m ˙ × h − m˙ (t) × h γ − 1 dt 0 dt in in cond fg

The mass balance (3.68) can then be used to express the mass variation term in the energy balance : V dp(t) 0 + h [m ˙ (t) − m˙ (t)] =m ˙ × h − m˙ (t) × h γ − 1 dt 0 in cond in in cond fg

Since, whatever the scenario, hin = hin0 for t ∈ [0, tin] andm ˙ in(t) = 0 for t > tin, we can also write : V dp(t) 0 = ˙m (t) × (h − h ) − ˙m (t) × (h − h ) (3.69) γ − 1 dt in in0 0 cond fg 0 Identitty (3.69) can then be successively applied to the three scenarii under study to compute the respective evolution of the averaged containment reactor pressure in each case.

To determine the time evolution of the (averaged) temperature in the containment reactor, let us use the thermal equation of state when writing the deﬁnition of the average pressure from the local pressure : 1 1 p = pdV = ρrT dV V0 ˆV0 V0 ˆV0 Let us assume a quasi-uniform temperature in the containment reactor, equal to the average temperature T . Under this assumption, it is possible to write : 1 1 p ≈ ρrT dV = rT m(t) V0 ˆV0 V0 hence p(t)V T(t) = 0 (3.70) rm(t) 52 Lecture #3 : Energy conservation

Identity (3.70) can be applied for the three scenarii under study in order to compute the average temperature evolution in the containment reactor.

Before proceeding to the detailed analysis of each scenario and computing the corresponding maximum averaged pressure and temperature, we wish to write normalized or non-dimensional versions of the balance equations (3.68) and (3.69). This will make clearer what are the driving terms for mass and pressure evolution in the containment reactor. The initial uniform state of the fluid in the containment reactor is a natural reference state to retain for performing the non-dimensionalization of the balance equations. The approximation retained in the analysis is to suppose the initial fluid is already made of vapor (and not air only); the flow dynamics is fast enough for this approximation to be acceptable. Let us denote m0 the initial mass of vapor in the reactor. The mass flowrate at the breach can be made non-dimensional using the quantity m0/tin. Whatever the scenario considered, it is possible to write : m˙ in tin m˙ in0 tin t t = Ψin = βinΨin m0 m0 tin tin with the non-dimensional quantity βin, non-dimensional reference inlet mass flowrate at the breach, defined as :

m˙ in0 tin βin = m0

t We can also introduce the non-dimensional time tˆ normalized by tin : tˆ= . The non- tin dimensional mass ﬂowrate at the breach at any time reads :

m˙ in(t) tin = βinΨin(tˆ) m0

Similarly, we introduce the non-dimensional reference condensation mass ﬂowrate :

m˙ cond0 tcond βcond = m0

t and the non-dimensional time t˜= so that the non-dimensional condensation mass ﬂowrate tcond at any time reads :

m˙ cond(t) tcond = βcondΨcond(t˜) m0

With these choices, the three scenarii are now described by the non-dimensional functions Ψin and Ψcond :

-• case #1 : constant mass ﬂowrate at the breach over tin and no condensation.

1 for 0 < tˆ≤ 1 Ψ (tˆ) = , Ψ (t˜) = 0 (3.71) in 0 for tˆ > 1 cond 3.4. Exercises and problems 53

-• case #2 : linearly decreasing mass ﬂowrate at the breach over tin and no condensation.

1 − tˆ for 0 < tˆ≤ 1 Ψ (tˆ) = , Ψ (t˜) = 0 (3.72) in 0 for tˆ > 1 cond

-• case #3 : constant mass ﬂowrate at the breach over tin and condensation over tcond. In this scenario the condensation mass ﬂowrate made non-dimensional using (m0/tcond) involves a non-dimensional time η = tin/tcond .

 1 for 0 < t˜≤ η 1 for 0 < tˆ≤ 1  Ψ (tˆ) = , Ψ (t˜) = (1 − t˜)/(1 − η) for η < t˜≤ 1 in 0 for tˆ > 1 cond  0 for t˜ > 1 (3.73)

The dimensional mass balance (3.68) is readily turned into a non-dimensional form. Since :

dm =m ˙ in(t)dt − m˙ cond(t)dt or else m(t) d = βinΨin(tˆ)dtˆ− βcondΨcond(t˜)dt˜ m0 an immediate integration from m = m0 and t = 0 to m = m(t) and t yields :

m(t) ˆt ˜t = 1 + βin Ψin(ˆt)dˆt − βcond Ψcond(˜t)d˜t (3.74) m0 ˆ0 ˆ0

The above deﬁnitions (3.71), (3.72) and (3.73) can then be successively applied to (3.74) in order to obtain m(t) for each scenario.

A similar non-dimensional approach can be applied to the energy balance. Since the initial mass m0 of vapor in the reactor is thus such that :

m0 = ρ0V0 where ρ0 satisﬁes p0 = ρ0rT0, this yields m0 r T0 = p0V0 which can also be rewritten as :

V 0 p = m C T (γ − 1) 0 0 v 0 with Cv = r/(γ − 1). The energy balance (3.69) can be rewritten as : V0 p(t) p0 d =m ˙ in(t) × (hin0 − h0) dt − m˙ cond(t) × (hfg − h0) dt γ − 1 p0 or else p(t) m0CvT0 d =m ˙ in(t) × (hin0 − h0) dt − m˙ cond(t) × (hfg − h0) dt p0 54 Lecture #3 : Energy conservation

Dividing by m0CvT0 allows to reintroduce the non-dimensional quantities βin, βcond as well as the non-dimensional functions Ψin,Ψcond and the non-dimensional times tˆ, t˜ : p(t) (hin0 − h0) (hfg − h0) d = βinΨin(tˆ) × dtˆ− βcondΨcond(t˜) × dt˜ p0 CvT0 CvT0 It is then natural to deﬁne the following non-dimensional quantities :

hin0 − h0 hfg − h0 αin = , αcond = (3.75) CvT0 CvT0 so that : p(t) d = βinαinΨin(tˆ) dtˆ− βcondαcondΨcond(t˜) dt˜ p0

An immediate integration from p = p0 and t = 0 to p(t) and t yields :

p(t) ˆt ˜t = 1 + αinβin Ψin(ˆt)dˆt − αcondβcond Ψcond(˜t)d˜t (3.76) p0 ˆ0 ˆ0

The deﬁnitions (3.71), (3.72) and (3.73) for Ψin and Ψcond can then be successively applied to (3.76) in order to obtain p(t) for each scenario.

The identity (3.70) is used to express the non-dimensional averaged temperature T /T0 : T (t) p(t) m V p = × 0 × 0 0 T0 p0 m(t) rT0m0

Since p0 = ρ0rT0 and m0 = ρ0V0, it follows that : p(t) T (t) p = 0 T0 m(t) m0 Using (3.74) and (3.76) yields :

ˆt ˜t 1 + αinβin Ψin(ˆt)dˆt − αcondβcond Ψcond(˜t)d˜t T(t) ˆ0 ˆ0 = ˆ ˜ (3.77) T0 t t 1 + βin Ψin(ˆt)dˆt − βcond Ψcond(˜t)d˜t ˆ0 ˆ0

The deﬁnitions (3.71), (3.72) and (3.73) for Ψin and Ψcond can then be successively applied to (3.77) in order to obtain T (t) for each scenario.

tˆ t˜ ˆ ˆ ˜ ˜ Let us now compute the non-dimensional factors Ψin(t)dt and Ψcond(t)dt for each sce- ˆ0 ˆ0 nario and deduce the corresponding evolutions of the mass, averaged pressure and averaged temperature. 3.4. Exercises and problems 55

-• case #1 : constant mass ﬂowrate at the breach over tin and no condensation. Using (3.71) yields for tˆ≤ 1 : tˆ tˆ Ψin(tˆ)dtˆ= 1 dtˆ= tˆ ˆ0 ˆ0 tˆ When tˆ = 1, Ψin(tˆ)dtˆ is precisely equal to 1 and keeps this value for tˆ > 1. Hence we ˆ0 can write in a general way : tˆ Ψin(tˆ)dtˆ= min (t,ˆ 1) (3.78) ˆ0 Since no condensation occurs in this scenario, we also have :

t˜ t˜ Ψcond(t˜)dt˜= 0 dt˜= 0 (3.79) ˆ0 ˆ0 Inserting these values in (3.74) yields :

m(t) = 1 + βin min (ˆt, 1) (3.80) m0 Note that going back to a dimensional formula yields : m(t) m˙ t t = 1 + in0 in min ( , 1) m0 m0 tin

that is, multiplying by m0 :

m(t) = m0 +m ˙ in0 min (t, tin) (3.81) The physical significance of (3.80) or of its above dimensional form is therefore straightforward : the mass of vapor in the containment reactor increases linearly in time when t < tin, the coefficient of increase being the mass flowrate through the breachm ˙ in0. The mass reaches a maximum value m0 +m ˙ in0tin at t = tin and keeps this value for t > tin since there is no longer a vapor massflow entering through the breach after time tin. Let us emphasize the interest of the non-dimensional formula (3.80) is that it was established using a non-dimensional approach which will be applied in a straightforward way to the more complex scenarii #2 and # 3. 3 A first numerical application can be performed. We have V0 = 49400 m and, assuming va- 5 por (and not air) is initially in the containment reactor, ρ0 = p0/(rT0) with p0 = 10 P a, T0 = 293.15 K and r = (γ − 1)Cp/γ with γ = 1.3 and Cp = 2080 J/kg/K so that 3 r = 480 J/kg/K for vapor. We compute ρ0 = 0.71 kg/m and m0 = ρ0V0 = 35107.17 kg. Since the vapor mass flowrate at the breach ism ˙ in0 = 3500 kg/s in this scenario with tin = 20 s, the mass of vapor in the reactor containment at any time t is given by : m(t) = 35107.17 + 3500 min (t, 20)

so that the maximum (constant) mass for t ≤ 20 s is mmax = 105107.17 kg .

Using directly (3.78) and (3.79) in the non-dimensional energy balance (3.76) yields :

p(t) = 1 + αinβin min (ˆt, 1) (3.82) p0 56 Lecture #3 : Energy conservation

The dimensional form is readily obtained : hin0 − h0 m˙ in0 tin t p(t) = p0 + p0 min , 1 CvT0 m0 tin

and can be simpliﬁed using p0 = ρ0rT0 and m0 = ρ0V0 to yield :

γ − 1 p(t) = p0 + (hin0 − h0)m ˙ in0 min (t, tin) (3.83) V0

The maximum averaged pressure is reached at time t = tin and is equal to :

(γ − 1)(hin0 − h0)m ˙ in0tin pmax = p0 + V0

The pressure increase is directly related to the enthalpy injected into the containment 6 reactor through the breach. A numerical application with hin0 = 2.9 × 10 J/kg, h0 = 6 1.9 × 10 J/kg and the values of p0,m ˙ in0, tin, V0, γ already used to compute the mass evolution yields : 5 5 5 pmax = 10 + 4.25 × 10 = 5.25 × 10 P a

or pmax = 5.25 bar as suggested in the question.

Combining (3.80) and (3.82) to compute the non-dimensional average temperature evolution according to (3.77) yields :

T (t) 1 + α β min (t,ˆ 1) = in in T0 1 + βin min (t,ˆ 1)

hin0 − h0 6 Since αin = or, numerically, αin = 10 /((2080/1.3) × 293.15) that is αin = CvT0 2.132 > 1, the averaged temperature is increasing with time for t < tin until it reaches at time t = tin a maximum value T max such that :

1 + αinβin T max = T0 × 1 + βin

With αin = 2.132 and βin = 3500 × 20/35107.17 that is βin = 2, we obtain :

◦ T max = T0 × 1.7547 = 514.4 K = 241.2 C

which is the suggested maximum temperature in the question. Plots for m(t), p(t) and T (t) will be provided once the two other scenarii analyzed.

-• case #2 : linearly decreasing mass flowrate at the breach over tin and no condensation. The only difference between scenario #2 and scenario #1 is the fact the injection mass flowrate at the breach decreases in time between t = 0 and t = tin. Using (3.72) yields for tˆ≤ 1 : tˆ tˆ tˆ tˆ2 tˆ2 Ψin(tˆ)dtˆ= (1 − tˆ) dtˆ= tˆ− = tˆ− ˆ0 ˆ0 2 0 2 3.4. Exercises and problems 57

tˆ When tˆ= 1, Ψin(tˆ)dtˆ is precisely equal to 1/2 and keeps this value for tˆ > 1. Hence we ˆ0 can write in a general way :

 tˆ2  tˆ− if 0 ≤ tˆ≤ 1 tˆ  2 Ψ (tˆ)dtˆ= f (tˆ) (3.84) ˆ in in 0  1  if tˆ > 1 2

Since no condensation occurs in this scenario, we still have (3.79). Inserting these values in (3.74) and (3.76) yields :

m(t) = 1 + βinfin(ˆt) (3.85) m0

p(t) = 1 + αinβinfin(ˆt) (3.86) p0

In this scenario #2, sincem ˙ in0 = 7000 kg/s, the initial vapor injection mass ﬂowrate through the breach has been multiplied by a factor 2 with respect to the previous scenario #1 so that βin = 4. The non-dimensional factor αin remains unchanged : αin = 2.132. Therefore, we can also write :

m(t) = 1 + 4 fin(ˆt) m0

and p(t) = 1 + 8.528 fin(ˆt) p0

For t > tin = 20 s, the mass remains constant, equal to its maximum value that is, since fin = 1/2 for tˆ > 1, mmax = 3m0 which is the same value as in scenario #1. The averaged pressure is also, in this scenario, an increasing function of t and reaches its maximum value for t = tinj, this maximum value being equal to pmax = (1 + 8.528/2)p0 = 5.25 bar, which is the same value as in scenario # 1. The temperature evolution is such that :

T(t) 1 + 8.528 f (ˆt) = in T0 1 + 4 fin(ˆt)

that is an increasing function of tˆuntil tˆ= 1. The maximum averaged temperature reached at t = tin = 20 s remains constant for t > tin, equal to T max = (5.25/3)T0 = 1.75 T0 = 513 K = 241◦C, that is the same maximum value as in scenario # 1.

-• case #3 : constant mass ﬂowrate at the breach over tin and condensation over tcond. The non-dimensional strategy will be particularly fruitful for the analysis of this last scenario. Since the vapor injection through the breach is the same as the one retained in case #1, 58 Lecture #3 : Energy conservation

we still have (3.78) and for numerical application βin = 2 again. In this scenario, the condensation mass ﬂowrate must be taken into account with :  ˜  1 for 0 < t ≤ η Ψcond(t˜) = (1 − t˜)/(1 − η) for η < t˜≤ 1  0 for t˜ > 1 and η = tin/tcond. For the case under study, η = 20/50 = 0.4. If 0 ≤ t˜≤ η then : t˜ t˜ Ψcond(t˜)dt˜= 1 dt˜= t˜ ˆ0 ˆ0 If η ≤ t˜≤ 1 then :

˜ t˜ η t˜ 1 − t˜ 1 t˜2 t Ψcond(t˜)dt˜= 1 dt˜+ dt˜= η + t˜− ˆ0 ˆ0 ˆη 1 − η 1 − η 2 η so that t˜ t˜− η 1 t˜2 − η2 Ψcond(t˜)dt˜= η + − ˆ0 1 − η 2 1 − η If t˜ > 1, there is now contribution to the integral so that :

t˜ 1 − η 1 12 − η2 1 Ψcond(t˜)dt˜= η + − = η + ˆ0 1 − η 2 1 − η 2 We can write in a general way :

 t˜ if 0 ≤ t˜≤ η    2 2 t˜  t˜− η 1 t˜ − η Ψ (t˜)dt˜= f (t˜) η + − if η ≤ t˜≤ 1 (3.87) ˆ cond cond 1 − η 2 1 − η 0    1  (1 + η) if t˜ > 1 2

The non-dimensional coeﬃcients βcond and αcond are respectively such that :

m˙ cond0 tcond – βcond = that is βcond = (1200 × 50)/35107.17 or βcond = 1.709. m0 hfg − h0 6 6 – αcond = that is αcond = (2.18 × 10 − 1.9 × 10 )/(1600 × 293.15) or αcond = CvT0 0.597.

In this scenario # 3, the mass evolution (3.74) is such that : m(t) = 1 + 2 min (ˆt, 1) − 1.709fcond(˜t) m0 or else, using t˜= ηtˆ with η = 0.4 :

m(t) = 1 + 5 min (˜t, η) − 1.709fcond(˜t) m0 3.4. Exercises and problems 59

Due to the deﬁnition (3.87), we can conclude the mass of vapor in the containment reactor increases as : m(t) = 1 + 3.291t˜ m0

when t˜≤ η that is t ≤ tin. For tin ≤ t ≤ tcond, the variation of vapor mass is such that :

m(t) t˜− η 1 t˜2 − η2 = 1 + 5η − 1.709 η + − m0 1 − η 2 1 − η that is, with η = 0.4 : m(t) = 3.22785 − 2.8484t˜+ 1.4242t˜2 m0

for t˜ ∈ [η, 1] that is t ∈ [tin, tcond]. The vapor mass is therefore decreasing between tin and tcond since there is no longer a vapor injection through the breach while there is still a vapor condensation taking place on the wall of the containment reactor. The maximum value of the vapor mass in the reactor is therefore reached when t˜= η or t = tin and given by :

mmax = (1 + 5η − 1.709η) × m0 = 2.3164m0 = 81322.25 kg

The ﬁnal constant value of the vapor mass in the reactor is reached at time t = tcond and is such that : m 1 final = 1 + 5η − 1.709 × (1 + η) = 1.8037 m0 2

or mfinal = 63322.8 kg.

Inserting (3.78) and (3.87) into (3.76) yields :

p(t) = 1 + αinβin min (t,ˆ 1) − αcondβcondfcond(t˜) p0 or else p(t) αinβin = 1 + min (˜t, η) − αcondβcondfcond(˜t) p0 η Replacing each non-dimensional factor with its numerical value yields :

p(t) 2.132 × 2 = 1 + min (t,˜ 0.4) − 0.597 × 1.709 × fcond(t˜) p0 0.4 or else p(t) = 1 + 10.66 min (˜t, 0.4) − 1.02027 × fcond(˜t) p0

Because of the deﬁnition (3.87) for fcond(t˜), the averaged pressure reaches therefore its maximum value at time t = tin, corresponding to t˜= η before decreasing between tin and tcond. This maximum value is given by :

p max = 4.84 p0 60 Lecture #3 : Energy conservation

which corresponds to the suggested value in the question since p0 = 1 bar. Note also the ﬁnal constant value of the averaged pressure is reached at time t = tcond corresponding to t˜= 1 and is such that : p 1 final = 1 + 10.66 × 0.4 − 1.02027 × (1 + 0.4) = 4.55 p0 2 The evolution of the averaged temperature is given by (3.77), that is for scenario # 3 :

α β 1 + in in min (˜t, η) − α β f (˜t) T(t) η cond cond cond = T β 0 1 + in min (˜t, η) − β f (˜t) η cond cond

The averaged temperature at time t = tin, that is t˜= η, is such that :

T (t ) 1 + α β − α β η in = in in cond cond T0 1 + βin − βcondη

The averaged temperature at time t = tcond, that is t˜= 1, is such that : 1 T (t ) 1 + αinβin − αcondβcond(1 + η) in = 2 T 1 0 1 + β − β (1 + η) in 2 cond

With αinβin = 4.264, βin = 2 and αcondβcond = 1.02, βcond = 1.709, η = 0.4 the numerical application yields :

T (t ) 1 + 4.264 − 1.02 × 0.4 4.856 in = = = 2.096 T0 1 + 2 − 1.709 × 0.4 2.3164

◦ that is T (tin) = 614.55 K = 341.4 C.

T (t ) 1 + 4.264 − 1.02 × 0.7 4.55 in = = = 2.52 T0 1 + 2 − 1.709 × 0.7 1.8037 ◦ that is T (tcond) = 739.5 K = 466.4 C, which corresponds (accounting for the rounding errors) to the suggested value in the question. Physically, the decrease of the averaged pressure is lower than the decrease of the vapor mass between tin and tcond so that the averaged temperature keeps increasing between tin and tcond to reach its maximum (constant) value at t = tcond : T max = 739.5 K. The vapor mass, averaged pressure and averaged temperature corresponding to the three studied scenarii are respectively plotted in Fig.3.9, Fig.3.10 and Fig.3.11. 3.4. Exercises and problems 61

110000

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30000 0 5 10 15 20 25 30 35 40 45 50 evol m scenario1 evol m scenarioscenario2 evol m scenario3

Figure 3.9: Time evolution of the vapor mass in the reactor containment. Mass is expressed in kg, time is expressed in s. Black line : scenario #1 (constant injection until t = 20 s, no condensation). Blue line : scenario #2 (linearly decreasing injection until t = 20 s, no condensation). Green line : scenario # 3 (constant injection until t = 20 s, wall condensation until t = 50 s). 62 Lecture #3 : Energy conservation

550000

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100000 0 5 10 15 20 25 30 35 40 45 50 evol p scenario1 evol p scenario2 evol p scenario3

Figure 3.10: Time evolution of the averaged pressure in the reactor containment. Averaged pressure is expressed in P a, time is expressed in s. Black line : scenario #1. Blue line : scenario #2. Green line : scenario # 3. 3.4. Exercises and problems 63

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250 0 5 10 15 20 25 30 35 40 45 50 evol T scenario1 evol T scenario2 evol T scenario3

Figure 3.11: Time evolution of the averaged temperature in the reactor containment. Averaged temperature is expressed in K, time is expressed in s. Black line : scenario #1. Blue line : scenario #2. Green line : scenario # 3. 64 Lecture #3 : Energy conservation Lecture 4 1D compressible ﬂow analysis

Contents 4.1 Some key quantities...... 65 4.1.1 Sound speed and Mach number...... 65 4.1.2 Stagnation quantities or total quantities...... 70 4.1.3 Sonic quantities...... 73 4.2 1D shockwaves...... 75 4.2.1 Problem definition...... 75 4.2.2 Prandtl’s relationship...... 75 4.2.3 Physical considerations...... 77 4.2.4 Shock relationships...... 78 4.2.5 Total quantities and 1D shockwaves...... 80 4.2.6 Rankine-Hugoniot’s relationship...... 84 4.3 Exercises and problems...... 85 4.3.1 Exercise 1 : Mach number and compressibility effects...... 85 4.3.2 Exercise # 2 : pressure and temperature at the nose of an airplane in transonic flight...... 87 4.3.3 Problem # 1 : velocity measurement in a supersonic flow...... 89

4.1 Some key quantities

4.1.1 Sound speed and Mach number Physical considerations Let us consider the ambient air. It is made of molecules in motion. Let us assume a small perturbation is introduced into this medium in the form of a pointwise energy input. This energy is absorbed by the molecules next to the energy source and these molecules move now with an increased velocity. This faster molecules interact with other molecules, located further from the initial source of perturbation and modify the velocity of these molecules through this interaction. The interaction between molecules goes on with these latter molecules interacting 66 1D compressible ﬂow analysis again with molecules more distant from the source so that the energy initially introduced into the medium is transmitted through the medium. This energy wave moves with a velocity which can be related of course with a molecular velocity, when working in the fraemwork of so-called gas kinetics analysis. In this lecture, we work with a more macroscopic approach and are therefore interested in the eﬀects produced by this wave. The variation of energy created by this wave yields a pressure variation but also a density variation, a temperature variation. These variations remain weak since the initial energy input is a perturbation and not an energy peak corresponding for instance to an explosion. If an observer takes place in the medium where the wave is propagating, his/her ear will perceive in particular the weak pressure variation as a sound, hence the name of sound wave or acoustic wave given to this low-intensity wave. The propagation velocity of this wave is the speed of sound.

Expressing the speed of sound

Let us denote a the speed of sound and let us link this quantity with the local flow properties : pressure p, density ρ, temperature T . In an absolute frame of reference, the flow configuration is illustrated in Fig.4.1 (a). Let us consider now a frame of reference associated with the moving wave : the flow in front of the wave is then going into the direction of the wave (steady / fixed in its own referential) and we suppose the uniform state of air in this region is characterized by a pressure p, a density ρ and a temperature T (see Fig. 4.1 (b)). Through the wave, small variations of the flow properties take place so that the state behind the wave is characterized by a perturbed density ρ + dρ, a perturbed temperature T + dT , a perturbed pressure p + dp and a perturbed velocity a + da where the quantities d· are infinitesimal variations. In this local frame of reference attached to the sound wave, the whole flow can be considered as steady.

In the following analysis, air is considered as an ideal ﬂuid (inviscid ﬂuid, devoid of viscosity

a a + da p a p + dp ρ ρ + dρ T T + dT onde acoustique

(a) (b) Figure 4.1: Propagation of an acoustic wave. (a) In an absolute frame of reference. (b) In a frame of reference attached to the acoustic wave. and heat conductivity). The ﬂow is thus described by the (compressible) Euler equations, the integral or global form of which has been introduced in the previous lectures. For a steady ﬂow, 4.1. Some key quantities 67 the mass, momentum and energy conservation equations thus read :

 ~  ρU · ~ndS = 0  ˆS    ρU~ (U~ · ~n)dS = −p~ndS ˆ ˆ  S S    ~ ~  ρE(U · ~n)dS = − p(U · ~n)dS ˆS ˆS where~n denotes the outward-pointing unit normal vector to the control surface S and where the body forces (gravitational forces) have been neglected. Let us apply these conservation equations to the control volume displayed in Figure 4.2. The following equations governing the 1D ﬂow of an ideal ﬂuid are readily obtained :

 ρ1u1 = ρ2u2    2 2 ρ1u1 + p1 = ρ2u2 + p2 (4.1)    ρ1u1H1 = ρ2u2H2 where ui denotes the velocity (single component for the 1D flow) in state i. The (specific) total 1 2 enthalpy H is defined by H = E + p/ρ with the (specific) total energy E such that E = e + 2 u and the (specific) internal energy e related to the pressure and the density by p = (γ − 1)ρe in the case of an ideal or perfect gas.

Let us apply the set of equations (4.1) to the two states on each part of the acoustic wave :

u p u p 1 1 ρ1 2 2 ρ2

Figure 4.2: Control volume for the analysis of a steady flow made of two constant states separated by a steady wave. u1 = a, ρ1 = ρ and p1 = p on one side, u2 = a + da, ρ2 = ρ + dρ and p2 = p + dp on the other side. Injecting these states into the two first relationships of (4.1) (mass and momentum conservation) and neglecting the second-order terms (of the form d · ×d·) it can be established that : dp a2 = dρ (see also the following exercise for the derivation of this identity). In fact, the variations of the flow properties being assumed weak, the flow transformation from state 1 to state 2 can be considered as a reversible process. This flow transformation being also adiabatic (there is not 68 1D compressible flow analysis heat transfer phenomena, the fluid being assumed ideal) , the acoustic wave can be considered as an isentropic phenomenon, so that the sound velocity also reads :

∂p a2 = ( ) ∂ρ s where the notation (·)S means the variation is computed for an isentropic transformation. The isentropic compressibility coeﬃcient has been introduced in the previous lecture and deﬁned as :

1 ∂v 1 ∂ρ βs = − = v ∂p s ρ ∂p s

Consequently, it is possible to write : 1 a2 = ρβs 1 and since it was established in the previous lecture that, for a perfect gas, β = it followd s γp eventually the expression of the local velocity of sound as a function of the state variables p and ρ reads as : rγp a = (4.2) ρ or else, making use of the thermal Equation of State (EoS) valid for a perfect gas (that is p = ρrT ): a = pγrT (4.3)

Note : at sea level, with T ≈ 293K, r = 287J/kg − K and γ = 1.4 the sound speed is equal to a ≈ 340m/s.

Mach number deﬁnition and ﬂow regimes

The speed of sound being well defined, it is now possible to introduce another key physical quantity for the analysis of compressible flows : the Mach number, ratio between the local flow velocity and the local sound velocity : u M = a For a 1D flow where the velocity field u is such that u(x, t) and the sound velocity field a is such that a(x, T ), the Mach number M is also a flowfield : M(x, t). A flow will be locally subsonic when the local Mach number is below 1, M < 1, while it will be said supersonic when M > 1. Various flow regimes for the flow over a fixed obstacle for instance can be defined according to the value of the far-field Mach number M∞ and the value of the local Mach number :

-• if the far-field flow is subsonic, that is M∞ = U∞/a∞ < 1, and if the local Mach number remains below 1 everywhere in the flow then the overall flow regime will be called subsonic. 4.1. Some key quantities 69

-• if the far-field flow is subsonic but if the local Mach number goes above 1 at some location in the flow, then the overall flow will be called transonic. This is the flow configuration typically encountered for flows over large commercial airplanes where the cruise flight Mach number based on the plane velocity and the speed of sound at high altitude (around 10000 m) lies between 0.8 and 0.9.

-• if the far-field flow is supersonic, that is M∞ > 1, then the overall flow will be called supersonic, even though, as will be seen next, some local subsonic regions may appear in the flow depending on the occurrence of shockwaves in the flow.

The Mach number is the key similarity parameter for the study of compressible flows; it is crucial to perform wind tunnel experiments on reduced scale models which reproduce the Mach number encountered in the real full-scale flow. Note the geometric scale does not appear in the Mach number (compressibility effects) while it is of course crucial in the Reynolds number (viscous effects). In order to reproduce in a wind tunnel, at ambient temperature T∞ ≈ 300K, a flow taking place at Mach number M∞ = 2, the flow must be put in motion with a velocity u∞ ≈ 2 × 340 = 680 m/s that is roughly 2500 km/h. The same Mach number can be reached for a lower velocity by lowering the speed of sound through the decrease of the flow temperature ◦ T . With√ a low temperature of −150 C, that is T = 123 K, the speed of sound goes down to a = γrT = 222 m/s and consequently Mach number 2 is reached for a flow velocity reduced to U∞ ≈ 444 m/s that is about 1600 km/h. This idea is applied in cryogenic wind tunnels, where the lowering of the temperature also impacts the fluid viscosity hence the Reynolds number and the viscous similarity. An example of cryogenic wind tunnel is the European Transonic Wind Tunnel (see the website https://www.etw.de). This same idea was also used in 1947 to perform the first supersonic test-flight : the Bell-XS1 prototype aircraf was mounted on a carrier aircraft and brought to high altitude before its thruster was turned on so as to benefit from the lower speed of sound at high altitude (low temperature) and be able to achieve a supersonic flight with a limited aircraft velocity.

4.1.1.1 Exercise : computing the speed of sound dp • Derive the identity a2 = . dρ

Let us start from the 1D Euler equations written to relate states 1 and 2 on both sides of the sound wave :  ρ1u1 = ρ2u2  2 2 ρ1u1 + p1 = ρ2u2 + p2  ρ1u1H1 = ρ2u2H2

State 1 is selected as the reference state u1 = a, ρ1 = ρ and p1 = p while state 2 corresponds to the perturbed state induced by the acoustic wave propagation : u2 = a + da, ρ2 = ρ + dρ and p2 = p + dp. Injecting these quantities into the mass conservation identity yields :

ρa = ρa + ρda + adρ + H.OT. where H.O.T. stands for ”Higher Order Terms”, namely in the present analysis the second order term da · dρ, which can be considered as negligible with respect to the ﬁrst-order perturbations 70 1D compressible ﬂow analysis involving da and dρ. It is thus possible to rewrite : ρda + adρ = 0. Working in a similar way on the momentum conservation equations yields : 2ρada + a2dρ + dp = 0 Replacing in this identity ρda with −adρ leads to the identity : dp a2 = dρ

4.1.2 Stagnation quantities or total quantities Let us consider a flow with so-called static properties u, p, T (p : static pressure, T : static temperature). The word ”static” makes reference to the fact these quantities are precisely measured where the fluid is ”standing” ( ”static” derives from the latin verb ”stare” meaning ”to stand”). This is to make the difference with other states of the flow which can be interesting to manipulate as reference states but which can remain virtual and not be actually achieved in the flow. Among these reference states, we will introduce now the stagnation state or state where the fluid is at rest as well as, a bit later, the sonic state where the local Mach number is equal to unity. In order to define the stagnation quantities or total quantities associated with a fluid particle, let us imagine a fluid particle with local / static properties p, T and velocity u which is brought to rest (or stagnation state) in an isentropic way. In other words, the fluid particle goes through an isentropic process from the state defined by u, p, T to a stagnation state defined by a zero velocity (fluid at rest) and a stagnation or total pressure p0, a stagnation or total temperature T0. Sometimes, to underline the isentropic process leading to the stagnation state from the local state, these total quantities are denoted pi, Ti. These stagnation quantities can be computed as a function of the local / static quantities using again the 1D Euler equations (4.1) between both states. The energy conservation, takin into account the mass conservation, reduces to the conservation of total enthalpy in the 1D flow. Therefore, the total enthalpy H computed using the local state is equal to the total enthalpy H0 computed using the stagnation state :

H = H0 It was established in the previous lecture, the specific total enthalpy is such that (neglecting body forces) : 1 p 1 p p 1 γ p 1 H = h + U~ 2 = e + + U~ 2 = + + U~ 2 = + U~ 2 2 ρ 2 (γ − 1)ρ ρ 2 γ − 1 ρ 2 Since the speed of sound for a perfect gas is such that a2 = γp/ρ, the specific total enthalpy can be expressed as a function of a2 and U~ 2 = u2 (for a 1D flow) as follows :

a2 1 H = + u2 (4.4) γ − 1 2 Alternatively, the specific total enthalpy can also be expressed as a function of the temperature T and the velocity u. Since a2 = γrT and since the spefici heat capacity at constant pressure Cp is defined as Cp = γr/(γ − 1) for a perfect gas, it is also possible to write : 1 H = C T + u2 (4.5) p 2 4.1. Some key quantities 71

where we can identify CpT as the speciﬁc enthalpy for a perfect gas. Since the ﬂow velocity u is zero in the stagnation state, the total enthalpy H0 computed for the stagnation state depends only on the temperature T0 in this stagnation state, also known as the stagnation or total temperature :

H0 = CpT0

Since H0 is a characteristic constant of an adiabatic flow, so is the total temperature T0. In an adiabatic flow, the total temperature T0 is constant throughout the flow. Physically, this means that any fluid particle in the flow which is brought to rest through an adiabatic process (no need for the process to be reversible) will reach the same total or stagnation temperature T0. In the context of propulsion system, the stagnation temperature is the typical temperature of the fluid at rest in the reservoir connected with the nozzle ensuring the thrust. This stagnation or total temperature remains constant throughout the flow as along as it remains adiabatic that is as long as no heat is added to or extracted from the flow. Using the definition (4.5) for the specific total enthalpy yields : u2 C T + = C T (4.6) p 2 p 0 This identity can also be expressed as a non-dimensional relationship between the ratio of the total temperature to the static temperature and the Mach number. Dividing (4.6) by CpT and identifying a2 = γrT then M 2 = u2/a2 yields the very important relationship :

T γ − 1 0 = 1 + M 2 (4.7) T 2

Let us insist again on the fact this identity is valid for any adiabatic flow of an inviscid fluid. In particular, it is valid for an inviscid flow in which shockwaves are taking place, as along as the flow remains free of heat transfer. The total temperature of a fluid particle crossing a shockwave remains unchanged downstream of the shock with respect to its value upstream of the shock. The total temperature remains constant in an irreversible flow as long as the flow remains adiabatic. A change in the total temperature is produced by heat being added to the flow (increase of the total temperature) or being taken from the flow (decrease of the total temperature).

The stagnation pressure or total pressure p0 must be defined however under the assumption of an isentropic flow (hence a flow which is both adiabatic and reversible, that is free of shockwaves for the inviscid flows we will study). This is also true for the total or stagnation density ρ0. We have seen previously the isentropic flow of a perfect gas is such that :

p = cste ργ

Besides, the thermal Equation of State for a perfect gas is also satisfied both for the local/static state (p = ρrT ) and for the stagnation state (p0 = ρ0rT0). Let us consider two distinct local states in an isentropic flow, denoted respectively with indices 1 and 2. The fluid particle in state 1 reaches isentropically a stagnation state (p0)1,(ρ0)1 and (T0)1 while the fluid particle in state 2 reaches isentropically a stagnation state (p0)2,(ρ0)2 and (T0)2. Note that the flow being isentropic it is reversible and adiabatic; the adiabaticity property is sufficient to ensure (T0)1 = (T0)2 = T0. 72 1D compressible flow analysis

Since (p0)1 = (ρ0)1rT0 and (p0)2 = (ρ0)2rT0, taking the ratio of the stagnation pressures (p0)1 and (p0)2 yields : (p ) (ρ ) (T ) 0 1 = 0 1 × 0 1 (p0)2 (ρ0)2 (T0)2 | {z } =1 so that (p ) (ρ ) 0 1 = 0 1 (4.8) (p0)2 (ρ0)2 Besides, the isentropic ﬂow assumptions allows to write :

(p0)1 (p0)2 γ = γ (ρ0)1 (ρ0)2 hence, reorganizing : (p ) (ρ ) γ 0 1 = 0 1 (4.9) (p0)2 (ρ0)2

It can be deduced from (4.8) and (4.9) that (p0)1/(p0)2 = 1 and (ρ0)1/(ρ0)2 = 1. Since (p0)1 = (p0)2 and (ρ0)1 = (ρ0)2 whatever the local states 1 and 2 in the isentropic flow, it can be concluded the stagnation or total pressure p0 is constant in an isentropic flow and so is the stagnation or total density ρ0. In the context of propulsion system, the stagnation or total pressure is the typical pressure of the fluid at rest in the reservoir connected with the nozzle ensuring the thrust. This stagnation or total pressure remains constant throughout the flow as along as the flow is both adiabatic and reversible (that is isentropic). For inviscid flows, reversibility means no shockwaves occurr in the flow. When a fluid particle crosses a shockwave, its total pressure varies (as well as its total density) and takes a new constant value in the isentropic flow downsteam of the shockwave, which is distinct from the constant value in the isentropic flow upstream of the shockwave. Writing the thermal equation of state both for the local/static state and the stagnation state as well as the entropy conservation between the local/static state and the stagnation state yields the following relationship between the ratio total to static for the pressure p0/p and the temperature T0/T : γ p T 0 = 0 γ − 1 p T Taking into account the identity (4.7), we obtain the very important relationship between the total pressure, static pressure and local Mach number for an isentropic flow : γ p γ − 1 0 = 1 + M 2 γ − 1 (4.10) p 2

We derive in a similar way : 1 ρ γ − 1 0 = 1 + M 2 γ − 1 (4.11) ρ 2 Let us emphasize again that in a ﬂow displaying a shockwave which separates two isentropic ﬂow regions, the total pressure and total density are constant upstream of the shock, respectively 4.1. Some key quantities 73

equal to (p0)up,(ρ0)up say, and also constant downstream of the shock, respectively equal to (p0)down,(ρ0)down say. But the downstream values are distinct from the upstream values because of the entropy increase when a ﬂuid particle crosses the shockwave. Therefore, identities (4.10) and (4.11) can be applied respectively upstream and downstream of the shockwave, using respectively (p0)up,(ρ0)up and (p0)down,(ρ0)down as constant stagnation properties.

4.1.3 Sonic quantities Another interesting state of reference for the study of compressible flows is the sonic state, defined as follows. Let us consider a fluid particle with local properties ρ, p, T , u, a, M... Note that 2 thermodynamic state variables, for instance p and T , and the velocity u are sufficient to define all the other quantities (density, speed of sound, Mach number). Let us now imagine this fluid particle is brought adiabatically to the sonic state, defined as a state where the velocity of the fluid particle is equal to the speed of sound associated with the fluid particle (computed from the temperature of the fluid particle) : M = 1. In this sonic state, the fluid particle displays a pressure denoted p∗, a density denoted ρ∗, a temperature denoted√T∗. The velocity u∗ of this fluid particle at sonic state is equal to a∗ (since M = 1) with a∗ = γrT∗. Let us also define for future use the characteristic Mach number as : u M∗ = (4.12) a∗

Note carefully M∗ is the ratio of the local (actual) velocity u of the ﬂuid particle and the sonic (virtual) speed of sound a∗ associated with the ﬂuid particle. The ratio of u∗ to a∗ does not require to introduce a notation for the corresponding Mach number which is equal to unity. Hence the notation M∗ stands for the ratio u/a∗, the usefulness of which will be soon established. The sonic variables can be computed from the local variables u, p, T by applying the Euler equations (4.1) between the current local state and the sonic state. We focus on the calculation of the sonic speed of sound a∗ and write therefore the energy conservation in the form :

H = H∗

Going back to the deﬁnition (4.4) of the speciﬁc total enthalpy we obtain :

a2 1 a2 1 H = + u2 = H = ∗ + u2 γ − 1 2 ∗ γ − 1 2 ∗

Since u∗ = a∗ by deﬁnition of the sonic state, it follows that :

γ + 1 γ − 1 2 γ − 1 a2 = a2 − u2 or a2 = a2 + u2 (4.13) 2 ∗ 2 ∗ γ + 1 γ + 1

Dividing next (4.13) by u2 yields :

2 (γ + 1)M 2 M 2 = or M 2 = (4.14) γ + 1 ∗ 2 + (γ − 1)M 2 [ 2 ] − (γ − 1) M∗

It is also important to notice the following properties : 74 1D compressible ﬂow analysis

-• if M∗ = 1 then M = 1 and reversely if M = 1 then M∗ = 1.

-• similarly, if M∗ < 1 then M < 1 and reversely

-• if M∗ > 1 then M > 1 and reversely. These properties will be useful in the next Section when analyzing a 1D shockwave.

Exercise : total quantities, sonic quantities Let us consider the flow of air over a wing model in a wind tunnel. The measurements performed at a point M in the flow yields a local velocity U = 221.85 m/s, a local static pressure p = 0.9 atm and a local static temperature T = 250 K. • Assuming the flow isentropic, compute p0, T0, p∗, T∗ and a∗. Use r = 287 J kg−1 K−1 and γ = 1.4 for air.

The constant total temperature is related to the local temperature T and local Mach number M by the following identity : T γ − 1 0 = 1 + M 2 = f(M) T 2 The local Mach number is not directly available but can be computed from√ the local velocity and the local temperature. The local speed of sound is indeed equal to a = γrT = 316.9 m/s so that M = U/a = 0.7. Knowing the static temperature and the Mach number at one location in the isentropic flow makes it possible to find the value of the constant total temperature (which would remain constant in an adiabatic but not reversible flow) using the above identity. An immediate calculation yields T0 = 1.098 T = 274.5 K. The constant total pressure is computed from the static pressure p and the ratio static to total for the temperature using the isentropic flow identity :

γ p T γ−1 0 = 0 p T hence p0 = 1.387 p = 1.25 atm. The sonic temperature is related to the total temperature applying the general above identity for the total to static temperature ratio as a function of the Mach number in the particular case where the Mach number is equal to unity : M = 1. It comes : T γ + 1 0 = f(1) = T∗ 2 This identity being valid in any adiabatic flow, we observe the sonic temperature remains also constant in an adiabatic flow - it is also observed when the flow displays shockwaves. The numerical application for air (γ = 1.4) yields T∗ = 0.833T0 and for the case under study, using the computed value of T0, we find T∗ = 229 K. The flow being assumed isentropic, the sonic pressure can be computed applying again the isentropic flow identity : γ p T γ−1 0 = 0 p∗ T∗ hence p∗ = 0.528 p0 = 0.66 atm. √ −1 Finally the sonic speed of sound a∗ is given by a∗ = γrT∗ hence a∗ = 317m s . 4.2. 1D shockwaves 75

4.2 1D shockwaves

4.2.1 Problem definition We consider now the case of a flow in which a sudden finite variation of the fluid properties takes place across a propagating wave instead of the infinitesimal variation considered and observed for the variation of the fluid properties across an isentropic acoustic wave. Such a wave will be referred to as a discontinuity wave or shockwave. Working again in the frame of reference attached to this propagating wave, the configuration under study is of the form displayed in Fig. 4.3. We assume state 1 (upstream state with respect to the wave) is known and we wish to compute the state 2 downstream of the shockwave. The steady Euler equations apply again to this configuration : with 3 independent unknown quantities, such as ρ2, u2 and p2 for 3 equations, it is possible to solve system (4.1) and find quantities (·)2 as a function of quantities (·)1. However such a brute force approach can be advantageously replaced with a more clever strategy taking advantage of the previously defined sonic quantities. Etat 1 Etat 2 ρ u p T M a ρ u p T M a 1 1 1 1 1 1 2 2 2 2 2 2

Figure 4.3: Discontinuous ﬂow properties of a ﬂuid when crossing a 1D shockwave. ”Etat” stands for ”(Physical) State”. The 3 independant variables ρ, u, p allow to compute T (from the thermal EoS), a (from T ), M (from u and a).

4.2.2 Prandtl’s relationship Let us consider the momentum conservation equation derived from the application of the global or integral form of the momentum conservation principle on a control volume enclosing the discontinuity and such that its left face lies in the region of state 1 while its right face lies in the region of state 2 (see Fig. 4.2 with a shockwave instead of an acoustic wave) :

2 2 ρ1u1 + p1 = ρ2u2 + p2 Introducing the speed of sound through a2 = γp/ρ (written for states 1 and 2) and using (4.13) to express a1 as a function of u1 and a∗ and similarly a2 as a function of u2 and a∗, the following Prandtl’s identity is obtained (see the exercise proposed below for the detailed proof of this identity) : 2 a∗ = u1u2 (4.15) Introducing the characteristic Mach number we can also write : 1 (M∗)2 = (4.16) (M∗)1

This identity is particularly interesting since it indicates that if (M∗)1 > 1 then (M∗)2 < 1 and similarly if (M∗)1 < 1 then (M∗)2 > 1. Since the local Mach number and the characteristic 76 1D compressible ﬂow analysis

Mach number share the same variation with respect to unity, we can also state that if M1 < 1 then M2 > 1 while if M1 > 1 then M2 < 1. Therefore two physical configurations only are admissible for a shockwave to take place within the flow : either the flow is subsonic upstream of the shockwave and becomes supersonic downstream of the shockwave or, reversely, the flow is supersonic upstream of the shockwave and becomes subsonic downstream of the shockwave. Both configurations are illustrated in Fig. 4.4. As will be explained next, only the first configuration, namely the supersonic flow becoming subsonic through the shockwave, is physically admissible.

M> 1 M < 1 M < 1 M > 1 1 2 1 2 ou

Figure 4.4: Flow conﬁgurations that can a priori take place on both sides of a shockwave. ”Ou” stands for ”or” and underline both options can be envisioned based on the mass, momentum and energy conservation laws only. To discriminate between both options, the second principle of thermodynamic must be called upon.

Exercise : deriving the Prandtl’s relationship • Use the Euler equations linking states 1 and 2 on both sides of the shockwave to derive the Prandtl’s identity : 2 a∗ = u1u2 Let us start the proof from the conservation of momentum between 1 and 2 :

2 2 p1 + ρ1u1 = p2 + ρ2u2

2 2 The pressure pi (i = 1, 2) can be related to ai using the deﬁnition of the sound speed : ai = γpi/ρi so that : 2 2 a1 a2 ρ1u1(u1 + ) = ρ2u2(u2 + ) γu1 γu2 which is readily simpliﬁed into :

2 2 a1 a2 u1 + = u2 + γu1 γu2 when taking into account the mass conservation ρ1u1 = ρ2u2. We can also deduce from (4.13) 2 2 2 an expression for ai as a function of a∗ and ui : γ + 1 γ − 1 a2 = a2 − u2 i 2 ∗ 2 i Replacing into the above identity yields :

2 2 γ + 1 a∗ γ + 1 a∗ (u1 + ) = (u2 + ) 2γ u1 2γ u2 4.2. 1D shockwaves 77 or else, after simpliﬁcation and reorganization :

2 a∗(u2 − u1) = (u2 − u1)u1u2

Since it is assumed u1 6= u2 (finite variation of the flow properties through the shockwave), the above equality can be simplified by the difference (u2 − u1) yielding the Prandtl’s identity.

4.2.3 Physical considerations Let us consider an obstacle positioned in a subsonic flow (see Fig. 4.5(a)) and let us go back to our initial explanation regarding the formation and the propagation of an acoustic wave. In the present configuration, the first fluid particles meeting the obstacle see their velocity modified when they collide with the obstacle’s wall. This modification or perturbation is transmitted to the surrounding fluid, in particular in the upstream direction. Since the flow is subsonic, the transmission velocity of this perturbation (that is the speed of sound a) is larger than the velocity of the flow u. Therefore the information regarding the wall presence is transmitted to the fluid particles in the upstream fluid which, being ”informed” about the downstream wall, are able to modify their trajectories so as to get round the obstacle as illustrated in Fig. 4.5(b).

Let us now assume the incident ﬂow is supersonic that is M∞ > 1. Since a∞ < u∞, the

M < 1

(a) (b)

Figure 4.5: Obstacle in a subsonic flow. (a) Flow configuration. (b) Typical trajectories of the fluid particles. acoustic waves informing the upstream fluid about the obstacle are no longer able to propagate upstream at a velocity faster than the velocity of the incoming fluid particles. These acoustic waves tend to form a shockwave slightly upstream of the obstacle (see Fig. 4.6). Upstream of this shockwave the flow is uniform with far-field upstream conditions while downstream of the shockwave (and slightly upstream of the obstacle) the flow becomes subsonic and the bypassing process observed in the subsonic case is recovered. Note the flow downstream of the shockwave is not subsonic everywhere but only in flow regions where the shock is 1D or normal that is normal to the incoming flow direction : these regions are located close to the leading edge of the blunt body.

Based on these physical considerations, we can assume the only physically possible flow configuration for a 1D steady shockwave is the one described on the left side of Fig. 4.4 that is an upstream supersonic flow, M1 > 1, becoming subsonic downstream of the shockwave, M2 < 1. 78 1D compressible flow analysis

choc

M>1 M<1

Figure 4.6: Obstacle in a supersonic ﬂow.

As will be seen a bit later, the second principe or law of thermodynamics is the key reason for this ﬂow conﬁguration to be the only physically relevant one. Only the case M1 > 1 is indeed consistent with an increase of entropy when crossing the shockwave. This result will be proved later in Section 4.2.5, once the shock relationships or jump relationships available to compute the ratio between two physical quantities upstream and downstream of the shock.

4.2.4 Shock relationships

From a quantitative viewpoint, a relationship between the downstream Mach number M2 and the upstream Mach number M1 is easily obtained from (4.14) and (4.16):(4.14) provides the relationship between the characteristic Mach number M∗ and the local Mach number M while (4.16) links (M∗)1 to (M∗)2). An immediate calculation yields :

γ − 1 2 1 + ( )M1 M 2 = 2 (4.17) 2 γ − 1 γM 2 − ( ) 1 2

This first relationship is a jump relationship which defines how the state downstream of the shock can be computed from the state upstream of the 1D shock also called straight shock or normal shock because the flow direction is orthogonal to the shock structure. For instance, considering a flow of air (γ = 1.4) with a straight shock such that M1 = 2, it is found that M2 ≈ 0.577. The relationship (4.17) is tabulated and appendices with such shock tables will be provided along the lecture notes.

The upstream Mach number M1 is an extremely interesting parameter to express the jump of physical quantities through the shock, that is the ratio say ϕ2/ϕ1 where ϕ can be the static pressure, static temperature, density, velocity, ... If a relationship of the form ϕ2/ϕ1 = f(M1) is derived, it makes possible to compute the value ϕ2 downstream of the shock when the values of ϕ1 and M1 are available. The density jump and the velocity jump through the shock can be 4.2. 1D shockwaves 79 computed as follows. The mass conservation yields the identity :

ρ1u1 = ρ2u2 which can be recast into the following form :

2 2 ρ2 u1 u1 u1 2 = = = 2 = (M∗)1 ρ1 u2 u1u2 a∗ where the Prandtl’s relationship (4.15) has been used. Using (4.14) which links the characteristic Mach number to the local Mach number yields :

2 u1 ρ2 (γ + 1)M1 = = 2 (4.18) u2 ρ1 2 + (γ − 1)M1

Since M1 > 1 is a necessary condition for the shock to take place, the RHS quantity in (4.18) is greater than unity so that it can be concluded the density increases through the shock (the shock is an highly compressive mechanism) while the velocity of a ﬂuid particle decreases through the shock.

The pressure jump p2/p1 through the shock is also called the shock strength. It can be obtained from the momentum conservation equation in (4.1). After some calculation (see the exercise below for the detailed calculation), the following identity is obtained :

p2 2γ 2 = 1 + (M1 − 1) (4.19) p1 γ + 1

As expected this quantity is larger than unity : the pressure increases when the ﬂow crosses the shockwave. The shock compresses the ﬂow. Finally, the temperature jump is obtained using the thermal equation of state p = ρrT and the previously computed density jump and pressure jump. It is found that temperature increases when crossing the shock.

It is interesting to study the limit values of the jump relationships which have just been derived. When M1 tends to 1, all the ratios of the form ϕ2/ϕ1 tend to 1 : there is no longer a discontinuity in the flow, the shock is said to be evanescent. When M1 → ∞, it can be observed both the pressure jump and the temperature jump also tend γ−1 to infinity while the density jump tends to the finite value γ+1 , that is 6 for γ = 1.4. The down- q γ−1 stream Mach number M2 tends to the limit value 2γ , hence 0.378 for γ = 1.4. Considering a point of the reentry trajectory into the atmosphere of an hypersonic vehicle, the upstream Mach number can be as high as M1 = M∞ = 20 with T∞ = 280 K. The direct application of the above jump relationships yields a temperature behind the straight part of the bow shock developing upstream of the vehicle blunt nose such that T2 ≈ 22000 K ! This type of value is never reached in practice (fortunately because it would not be possible to design an efficient thermal shield for such a high temperature level) because in an hypersonic flow the initial assumption of a calorifically perfect gas is no longer valid. Real gas effects including chemical reactions occur which tend to consume a part of the energy and thus to lower the temperature reached at the nose of the vehicle. 80 1D compressible flow analysis

Exercise : pressure jump through a 1D shockwave

• Show the pressure jump p2/p1 through a 1D shockwave is given by :

p2 2γ 2 = 1 + (M1 − 1) p1 γ + 1

The momentum conservation written between states 1 and 2 on both sides (respectively upstream and downstream) of the shockwave takes the form :

2 2 p1 + ρ1u1 = p2 + ρ2u2 so that p2 − p1 1 2 2 = (ρ1u1 − ρ2u2) p1 p1 and p ρ u2 u 2 = 1 + 1 1 (1 − 2 ) p1 p1 u1 2 Using the relationship (4.18) which provides u2/u1 as a function of M1 :

2 u2 2 + (γ − 1)M1 = 2 u1 (γ + 1)M1 and also identifying : 2 2 ρ1u1 u1 2 = γ 2 = γM1 p1 a1 it is possible to rewrite the pressure jump as :

2 2 p2 2 (γ + 1)M1 − 2 − (γ − 1)M1 = 1 + γM1 ( 2 ) p1 (γ + 1)M1 hence, after a ﬁnal simpliﬁcation, the expected identity.

4.2.5 Total quantities and 1D shockwaves As far as total or stagnation variables are concerned, the flow configuration to deal with is summarized in Fig. 4.7. Since the total temperature T0 remains constant in an adiabatic flow, it remains in particular constant when the fluid particle crosses the shockwave. Hence the identity (T0)1 = (T0)2 = T0 where(T0)1 (respectively (T0)2) denotes the constant total temperature associated to the upstream (respectively downstream) isentropic flow.

Exercise : using the total temperature Let us consider the missile flying at supersonic speed displayed in Fig. 4.8. • Determine an approximation of the temperature at the nose of the missile. The nose of the missile is a stagnation point for the flow so that the temperature at the noseTnose is the stagnation or total temperature (T0)2 associated to state 2 downstream of the detached shock developing in front of the missile in supersonic flight. This total temperature 4.2. 1D shockwaves 81

Etat 1 Etat 2

Grandeurs statiques : Grandeurs statiques : S p S p T M 1 1T 1M 1 2 2 2 2

Grandeurs d’arrêt associées Grandeurs d’arrêt associées S (évolution isentropique) S2 (évolution isentropique) 1 p T (u=0) p0T 0 (u=0) 010 1 2 2 Figure 4.7: Static and total or stagnation) quantities on both sides of the 1D shockwave (state 1 is the upstream state while state 2 is the downstream state).

choc ‘‘1D’’

M =2 inf Missile T inf

Niveau de la mer

Figure 4.8: Missile ﬂying at supersonic speed above sea level. 82 1D compressible ﬂow analysis

(T0)2 is equal to the total temperature upstream of the shock (T0)1 since the ﬂow is adiabatic - the variation of entropy through the shockwave has no impact on the total temperature, while it modiﬁes of course the total pressure. The total temperature (T0)1 can be computed using formula (4.7) so that :

γ − 1 2 Tnose = (T0)2 = (T0)1 = (1 + M )T∞ 2 ∞ Assuming the (warm) sea level temperature around 300K, it comes Tnose = 540K or Tnose ≈ ◦ 270 C.

The total pressure (p0)1 associated with the isentropic flow upstream of the shockwave differs from the total pressure (p0)2 associated with the isentropic flow downstream of the shockwave (the same is true for the total density values : (ρ0)1 6= (ρ0)2). Let us use the second law of thermodynamics to relate the variation of the total pressure through the shock to the variation (increase) of specific entropy through the same shock. The well-known Gibbs relationship states : p de = T ds + dρ ρ2 or else, for a calorifically perfect gas such that p = ρrT and de = CvdT : dT dρ ds = C − r v T ρ This relationship can be integrated between states 1 and 2 on each side of the shock wave so that : T2 ρ2 s2 − s1 = Cv log − r log T1 ρ1 or else, using again the perfect gas thermal equation of state and the Mayer’s relationship Cp − Cv = r : T2 p2 s2 − s1 = Cp log − r log (4.20) T1 p1 where log denotes the natural logarithm (also denoted sometimes ln). Since (4.20) is valid for any states with entropy s1 and s2 it is valid in particular for the stagnation states on both sides of the shockwave so that : (T0)2 (p0)2 s2 − s1 = Cp log − r log (T0)1 (p0)1

Now, since total temperature is conserved through the shock, (T0)2/(T0)1 = 1 so that the relationship simpliﬁes into : (p0)2 s2 − s1 = −r log (p0)1 According to the second law of thermodynamics, the irreversible entropy variation throught the shockwave, s2 −s1, is necessarily a positive quantity since the irreversible entropy increases along (p ) the trajectory of a ﬂuid particle. This implies the ratio of total pressure 0 2 is less than unity (p0)1 that is (p0)2 < (p0)1 . The shockwave creates a total pressure loss. This loss is itself directly 4.2. 1D shockwaves 83 related to the entropy increase through the shock. The total pressure loss through the shock can be computed from the previously established relationships. Indeed, the total pressure ratio (p ) 0 2 can be be expressed as the following product : (p0)1

(p ) (p ) p p 0 2 = 0 2 × 2 × 1 (p0)1 p2 p1 (p0)1

2 The first term of the product is a function of M2 according to the isentropic flow identity (4.10) which, when applied to the isentropic flow downstream of the shockwave, yields : γ (p0)2 γ − 1 2 γ − 1 = 1 + M2 p2 2 2 This ratio can also be expressed as a function of M1 since the Mach number values on both sides of the shockwave are linked by the identity (4.17):

γ − 1 2 1 + ( )M1 M 2 = 2 2 γ − 1 γM 2 − ( ) 1 2 2 The third term of the product is also a function of M1 according again to the isentropic ﬂow identity (4.10) written for the isentropic ﬂow upstream of the shockwave : γ (p0)1 γ − 1 2 γ − 1 = 1 + M1 p1 2 Finally, the second term, corresponding to the static pressure jump through the shockwave, is given by the relationship (4.19):

p2 2γ 2 = 1 + (M1 − 1) p1 γ + 1 Gathering all these relationships, the variation of total pressure across the 1D shock can be expressed as the following function of the upstream Mach number M1 :

1 γ − γ−1 − γ−1 (p0)2 2γ 2 2 1 = 1 + (M1 − 1) 1 − (1 − 2 ) (4.21) (p0)1 γ + 1 γ + 1 M1 A few comments can be made regarding the identity (4.21): -• this formula giving the ratio of total pressures on both side of a 1D shock becomes a bit complex to compute so that it will be quite useful to be able to use tabulated values for (4.21) (see the appendices of the full lecture notes). -• formula (4.21) is of high practical use when analyzing compressible flows with shocks. Indeed, once the downstream total pressure has been computed from the upstream value of the total pressure and the upstream Mach number, this constant value can then be used everywhere in the downstream isentropic flow to compute using (4.10) for instance the local pressure from the total pressure and the local Mach number or the local Mach number from the local / static pressure and the total pressure. 84 1D compressible flow analysis

-• also, it must be noted that (4.21) could be applied for M1 > 1 as well as for M1 < 1. However, if M1 < 1 the total pressure ratio (p0)2/(p0)1 computed using (4.21) is larger than unity. Since the second law of thermodynamics forbids an increase of total pressure through the shockwave and imposes a decrease of the total pressure, this means that the necessary condition for the shockwave to occur is M1 > 1. This complements the physical considerations made in (4.2.3) to explain why a shockwave can appear in a supersonic flow, becoming subsonic after the 1D shock, while it cannot be formed in a subsonic flow which would become supersonic once the shock crossed. The governing equations admit both solutions but only the configuration ”upstream supersonic flow / downstream subsonic flow” is physically admissible for the case of a 1D steady shock.

4.2.6 Rankine-Hugoniot’s relationship

The Rankine-Hugoniot’s equation, or Hugoniot’s equation, links the static pressure jump p2/p1 through a 1D or normal shock to the corresponding density jump ρ2/ρ1. Since both the pressure 2 ratio and the density ratio are known as functions of the upstream Mach number M1 (or M1 ), this Mach number can be eliminated between both identities in order to obtain the desired relationship. It is easy to establish the following identity :

ρ (γ + 1)p + (γ − 1)p 2 = 2 1 ρ1 (γ + 1)p1 + (γ − 1)p2 or else p (γ + 1)ρ − (γ − 1)ρ 2 = 2 1 p1 (γ + 1)ρ1 − (γ − 1)ρ2 The usual form retained for the relationship between the density ratio and the static pressure ratio through a 1D shockwave is :

γ + 1 ρ 2 − 1 p γ − 1 ρ 2 = 1 (4.22) p γ + 1 ρ2 1 − γ − 1 ρ1

This relationship characterizes the non-isentropic flow compression mechanism induced by the shockwave. It is interesting to compare it with the isentropic flow compression which satisfies the isentropic flow identity so that :

p ρ γ 2 = 2 (4.23) p1 ρ1

Both formulae (4.22) and (4.23) are plotted in Fig. 4.9 - more precisely the static pressure ratio p2/p1, greater than unity through a compression, is plotted versus the density ratio ρ1/ρ2 less than unity through the compression since ρ2/ρ1 is also greater than unity through the compression. The limit value of the ratio ρ2/ρ1 when the pressure jump p2/p1 through the shockwave tends to inﬁnity is such that : ρ γ + 1 lim 2 = p 2 →∞ ρ1 γ − 1 p1 4.3. Exercises and problems 85

that is ρ2/ρ1 → 6 or ρ1/ρ2 → 0.16667 for γ = 1.4. Figure (4.9) illustrates how the (non-isentropic) shockwave is a more intense compression process than an isentropic compression. For the same density ratio ρ1/ρ2 say of 0.25 (that is ρ2/ρ1 = 4) the corresponding pressure increase in the isentropic compression is p2/p1 ≈ 7 while it reaches 11.5 through the 1D shockwave, a compression strenght (value of the pressure ratio) multiplied by a bit more than 1.6 with respect to the isentropic process. Naturally the price to pay for this increased compression is a loss of total pressure - while the isentropic process induces no variation in the total pressure.

20 Compression par choc droit (2.17) Compression isentropique (2.18)

10 p2 / p1

0 0.2 0.4 0.6 0.8 1 rho1 / rho2

Figure 4.9: Plot of the Rankine-Hugoniot’s relationship (4.22) and of the isentropic relationship (4.23). The continuous red curve corresponds to (4.22) while the dashed green line corresponds to (4.23). Note the limiting value ρ1/ρ2 = 1/6 for the Rankine-Hugoniot plot.

4.3 Exercises and problems

4.3.1 Exercise 1 : Mach number and compressibility eﬀects

Question

In an isentropic flow of a perfect gas, the ratio between the stagnation or total density ρ0 (constant in the isentropic flow) and the local density ρ is expressed as the following function of 86 1D compressible flow analysis the local Mach number M : 1 ρ γ − 1 γ−1 0 = 1 + M 2 ρ 2 • Explain, using this identity, why it is a customary engineering practice to consider a flow must be analyzed as a compressible flow once the Mach number exceeds 0.3 while it can be considered as incompressible while the Mach number remains below 0.3.

Answer

The relationship between the total density to static density ratio ρ0/ρ and the local Mach number, valid throughout an isentropic flow, is plotted in Fig. 4.10 (for γ = 1.4). It can be checked the variation of local density with respect to the reference stagnation value (value of the fluid particle at rest) does not exceed 5% as long as the local Mach number remains below 0.3. This explains the rationale behind this rule of thumb. A close-up on the curve shows the variation of ρ with respect to ρ0 does not exceed 1% as long as M remains below 0.14. Depending on the level of approximation which is accepted in the analysis, a more stringent requirement such as M > 0.1 could also alternatively be retained for the flow to be consider as compressible.

The same conclusions can naturally be drawn from the curve displayed in Fig. 4.11) where

Figure 4.10: Total to local/static density ratio ρ0/ρ (y coordinate) as a function of the local Mach number (x coordinate) for an isentropic ﬂow of perfect gas (γ = 1.4). the local Mach number M (y coordinate) is plotted as a function of the density ratio ρ0/ρ (x coordinate), according to the identity : v u " γ−1 # u 2 ρ0 M = t − 1 γ − 1 ρ 4.3. Exercises and problems 87

Figure 4.11: Local Mach number M as a function of the total to static density ratio ρ0/ρ for an isentropic ﬂow of perfect gas (γ = 1.4).

4.3.2 Exercise # 2 : pressure and temperature at the nose of an airplane in transonic ﬂight

Questions

An airplane is flying with velocity U = 270 m/s at an altitude of 10000 m where the air temperature is T = 224.3 K and the pressure is p = 250 hP a. Air is described as a perfect gas with γ = 1.4. A stagnation point A is located at the nose of the airplane (UA = 0, see Fig.4.12) and we wish to compute the pressure and the temperature at point A, that is the total pressure (p0)A and the total temperature (T0)A, assuming the flow is isentropic between the upstream farfield flow region and the stagnation point A. 1) • Compute the speed of sound for this high-altitude flow and compare this value with the speed of sound for standard conditions at ground level (T = 15◦C, p = 1013hP a). Deduce from the speed of sound the flight Mach number of the airplane.

2) • Working along the streamline which goes to the stagnation point A, determine TA as a function of the upstream ﬂow conditions and compute its numerical value. Does the quantity TA have a physical signiﬁcance on other streamlines ?

3) • Determine pA as a function of the upstream conditions and compute its numerical value. Does this stagnation pressure have a physical signiﬁcance on other streamlines than the one going to the stagnation point ? 88 1D compressible ﬂow analysis

Figure 4.12: Flow conﬁguration for the airplane ﬂying at high altitude.

Answers √ 1) Using a = γrT with r = 287 J/kg/K for air, γ = 1.4 for air and T = 224.3 K yields a ≈ 300 m/s. At ground level in standard conditions, T = 288.15 K so that aground = 340.5 /s. Since U = 270 m/s and a = 300 m/s, the Mach number of the upstream flow with respect to the airplane, which is also the Mach number of the plane flying through still air, is equal to M = 0.9. The plane is typically flying in transonic regime. The flow becomes locally supersonic in some regions along the airplane surface and shockwaves are forming. Upstream farfield conditions (in the frame of reference attached to the airplane) can also be denoted with ∞ index : T∞ = 224.3 K, U∞ = 270 m/s, a∞ = 300 m/s, M∞ = 0.9, p∞ = 25000 P a.

2) We have previously established that for a compressible flow of an inviscid fluid the conservation of energy is expressed as the conservation of specific total enthalpy H along a streamline (see (3.24)) : U 2 h + = constant along a streamline 2 or, since h = CpT for a perfect gas :

U 2 C T + = constant along a streamline p 2 with Cp = γr/(γ − 1). Writing this relationship (in the frame of reference attached to the airplane) between the upstream ﬂow, where T = T∞ = 224.3 K, U∞ = 270 m/s, and the stagnation point, where T = TA = T0 the constant total temperature for the ﬂow and UA = 0, yields : U 2 C T + ∞ = C T p ∞ 2 p 0 We can also write this expression in the form (4.7):

T0 γ − 1 2 = 1 + M∞ T∞ 2 4.3. Exercises and problems 89

so that TA = T0 = 1.162 × T∞ = 260.6 K. The total temperature T0 is directly related to the total ”total enthalpy” H0 since H0 = CpT0. In the present flow configuration, the upstream flow conditions being uniform the total enthalpy H is the same for any streamline originating from the upstream farfield flow so that the flow is homenthalpic (H = constant everywhere) with the total enthalpy H at any point equal to the stagnation (or total) total enthalpy H0.

3) The flow being isentropic, the identity (4.10) can be applied at the farfield : γ p0 γ − 1 2 γ − 1 = 1 + M∞ p∞ 2 with p0 the constant total pressure for the isentropic flow under study, also equal to the pressure 3.5 at the stagnation point A. Since p∞ = 250 hP a, p0 = (1.162 ) × p∞ = 1.69 × p∞ = 422.8 hP a. p We have previously established that s = C log (plus a constant reference value, which v ργ cancels out when computing the entropy difference between two states in the flow). Using the thermal equation of state, it is straighforward to establish the alternative expression for (specific) entropy : ! T s = Cp log 1− 1 p γ This identity can also be written at the stagnation state where s remains unchanged (since the fluid particle is assumed to go isentropically from its local state to the stagnation state) while T becomes T0 and p becomes p0 :   T0 s = Cp log  1  1− γ p0

For a given total enthalpy or a given total temperature T0 (which will be constant throughout an adiabatic flow, which can include irreversible processes such as shockwaves), the total pressure p0 is therefore directly related to the entropy. For the isentropic flow under study, the entropy is the same along every streamline so that the total pressure is also the same along every streamline hence constant throughout the flow. If a shockwave takes place in the flow, it is an irreversible process and the entropy of a fluid particle increases when it crosses the shockwave. Consequently the total pressure of the fluid particle also varies when crossing the shockwave : the total pressure goes down, expressing a total pressure loss. For a compressible flow the total pressure is equivalent to the head for an incompressible flow.

4.3.3 Problem # 1 : velocity measurement in a supersonic flow Questions Let us consider a space shuttle during its reentry phase in the atmosphere. The flow around the shuttle is supersonic. On the nose of the shuttle, a Pitot tube measures the stagnation pressure at point A (see Fig. 4.13). A temperature probe provides the temperature at the same point. In the supersonic regime, a detached bow shock develops upstream of the Pitot tube. A pressure 90 1D compressible flow analysis

Figure 4.13: Mesure de vitesse en rgime supersonique par un tube de Pitot. probe positioned at point B provides the static pressure upstream of this shockwave.

1/ • Introduce states 1 and 2 respectively upstream and downstream of the detached shock and clearly identify the known and unknown quantities.

1 2

Figure 4.14: State 1 is the (uniform) state upstream of the shock. State 2 is the state just downstream of the shock, where the bow shock can be considered as close to a locally 1D shock.

2/ • Show that the shuttle velocity can be expressed as a function of the sole upstream Mach number.

3/ • Explain how the value of the upstream Mach number can be determined from the total pressure downstream of the shock and the static pressure upstream of the shock. Deduce the process for computing the desired velocity from the performed measurements. Note : tabulated values of the ratio (downstream total pressure) / (upstream static pressure) are provided in Table 4.1. 4.3. Exercises and problems 91

4/ • Compute the shuttle velocity for the following measured values : static pressure at point B equal 2.83 P a; total pressure at point A equal to 92.42 P a; total temperature at point A equal to 1200 K. At this point of the reentry trajectory, the gas constant r for air is equal to 280 J kg−1 K−1.

M1 M2 p02 /p1 4.50 0.42355 26.539 4.60 0.42168 27.710 4.70 0.41992 28.907 4.80 0.41826 30.130 4.90 0.41670 31.379 5.00 0.41523 32.653 5.10 0.41384 33.954 5.20 0.41252 35.280 5.30 0.41127 36.631 5.40 0.41009 38.009 5.50 0.40897 39.412

Table 4.1: 1D shockwave relationships (γ = 1.4)

Answers

1/ Assuming a quasi-uniform state upstream of the detached shockwave, the pressure p1 upstream of the shock is obtained from the measurement made at point B. Downstream of the shock, the total temperature (T0)2 and total pressure (p0)2 associated to state 2 are known from the measurements made at point A. The shuttle velocity is given by u1. Therefore, we are looking for a way to compute u1 from the known values p1,(p0)2 and (T0)2.

2/ The velocity u1 can be computed from the Mach number M1 and the speed of sound a1, which is known provided the temperature T1 is known : p u1 = a1M1 = γrT1M1

The upstream temperature T1 is not directly known but it can be computed from the total temperature associated with the upstream state 1 and the Mach number M1. Since the total temperature is conserved in an adiabatic ﬂow, T01 = T02 . It follows that :

(T0)2 γ − 1 2 = 1 + M1 T1 2 so that : p M1 u1 = γrT0 (4.24) 2 r γ − 1 1 + M 2 2 1 92 1D compressible ﬂow analysis

With (T0)2 available, u1 can be computed provided the upstream Mach number M1 is known.

3/ The remaining problem to solve is to ﬁnd the upstream Mach number M1 from the measured pressures, namely the downstream total pressure p02 and the upstream static pressure p1. Let us form the ratio p02 /p1 and try to express this non-dimensional ratio as a function of the Mach number M : 1 p p p 02 = 02 · 2 p1 p2 p1 For the downstream isentropic ﬂow we can write : γ γ p02 T02 γ − 1 γ − 1 2 γ − 1 = = 1 + M2 p2 T2 2

The downstream Mach number M2 can be expressed as a function of the upstream Mach number M1 using the 1D shock relationship :

γ − 1 2 1 + ( )M1 M 2 = 2 2 γ − 1 γM 2 − ( ) 1 2

p02 so that can be expressed as a function of M1 only. p2 The static pressure jump through the 1D shockwave reads :

p2 2γ 2 = 1 + (M1 − 1) p1 (γ + 1)

A few simple calculations yield the following identity for the ratio between p02 and p1 : 1   γ + 1 2 γ − 1 M1 p02 γ + 1 2  2  = M   (4.25) p 2 1 2γ γ − 1 1  M 2 −  γ + 1 1 γ + 1

Inverting the identity (4.25) to obtain M1 for the known value of the pressure ratio (p0)2/p1 is not straightforward. However, tabulated values are provided for (4.25) so that it becomes easy to estimate M1 from this know ratio. Once M1 is know, its value can be inserted in (4.24), along with the measured value of (T0)2, to obtain u1.

4/ Since p02 /p1 ≈ 32.65, Table 4.1 provides M1 = 5. It comes : √ 5 −1 u1 = 1.4 × 280 × 1200 × √ = 1400 m s ( ≈ 5000 km / h) 1 + 0.2 × 25 Remark : In the limit of (very) large Mach number values, formula (4.24) giving the velocity simplifies into : r 2γr u = T = p2C T 1 γ − 1 02 p 02 4.3. Exercises and problems 93 and depends only on the total temperature. Note however that for very large Mach numbers real gas effects occur in the flow which make the Euler equations model no longer sufficient to correctly describe the flow physics. If the above approximate formula is used in the present case, it yields u1 = 1533 m/s, a value which differs by almost 10% from the correct value. 94 1D compressible flow analysis Lecture 5 Quasi-1D compressible flows

Contents 5.1 Physical model...... 95 5.1.1 Governing equations...... 95 5.1.2 Velocity-area relationship...... 97 5.1.3 Influence coefficients...... 99 5.2 Isentropic flow in a nozzle...... 102 5.2.1 Flow analysis...... 102 5.2.2 Mass flowrate...... 105 5.3 Analysis of the flow through a converging nozzle...... 106 5.3.1 General analysis...... 106 5.3.2 Example of application...... 108 5.4 Analysis of the flow through a diverging nozzle...... 111 5.5 Analysis of flow regimes in a converging-diverging nozzle...... 122 5.5.1 General analysis...... 122 5.5.2 Example of application...... 128 5.6 Exercises and problems...... 138 5.6.1 Exercise # 1 : Flow in a choked nozzle...... 138 5.6.2 Exercise #2 : Flow in a diverging nozzle...... 140 5.6.3 Problem # 1 : analysis of a scramjet performance...... 142

5.1 Physical model

5.1.1 Governing equations Let us consider the compressible flow (from left to right) in the nozzle (varying section tube) displayed in Fig. 5.1. In the framework of a quasi-1D description, the flow properties are assumed uniform in a given section, at axial position x with area A(x) : velocity u(x), pressure is p(x), density ρ(x), temperature T (x), speed of sound a(x), Mach number M(x). In other words, the physical state in each section of the nozzle is the average physical state over the section A(x). 96 Quasi-1D compressible flows in nozzles

The governing equations of the inviscid ﬂow are the Euler equations which can be applied

M A(x)

x Figure 5.1: General conﬁguration for a quasi 1D ﬂow on the control volume displayed in Fig.5.2 to obtain the following mass, momentum and total energy conservation equations :

 ρ u A = ρ u A  1 1 1 2 2 2  A2 p A + ρ u2A + pdA = p A + ρ u2A (5.1) 1 1 1 1 1 ˆ 2 2 2 2 2  A1  H1 = H2 where index 1 denotes the physical state in the upstream section of area A1 and index 2 denotes the physical state in the upstream section of area A2. If we assume the control volume is N

Etat 1 v . N = 0 Etat 2

v . N = 0

Figure 5.2: Control volume on which the integral form of the Euler equations is applied. infinitesimal, that is state 1 is defined as u, p, T and section area A while state 2 is defined as u + du, p + dp, T + dT and section area A + dA, the following differential form of the governing equations is obtained :   d(ρuA) = 0 dp = −ρudu (5.2)  dh + udu = 0 5.1. Physical model 97

where the (specific) enthalpy h is such that h = e + p/ρ and h = CpT for a perfect gas. Let us show next how the differential identities (5.2) can be combined in order to obtain a very useful relationship between velocity variation and section variation to analyze 1D flows with varying section.

5.1.2 Velocity-area relationship d(ρuA) The identity d(ρuA) = 0 can also be rewritten as = 0 so that an immediate expansion ρuA yields : dρ du dA + + = 0 (5.3) ρ u A Since the ﬂow is assumed isentropic, the speed of sound a is such that :

dp a2 = dρ

dρ u2 du Using the momentum equation dp = −ρudu, we obtained = − or ρ a2 u

dρ du = −M 2 (5.4) ρ u

Using (5.4) to express dρ/ρ in (5.3) yields the following important relationship between the section variation and the velocity variation in the nozzle, also called velocity-area relationship :

dA du = (M 2 − 1) (5.5) A u

This identity is important for a preliminary qualitative analysis of compressible ﬂows in a nozzle.

Case of a subsonic ﬂow : M < 1

u augmente u diminue

M < 1 M < 1

Figure 5.3: Evolution of the ﬂow velocity for a subsonic ﬂow in a varying-section nozzle. ”u augmente” stands for ”u increases” while ”u dimininue” stands for ”u decreases”.

dA du According to (5.5), quantities and are of opposite sign in that case : a decrease in the A u section (dA/A < 0) induces an acceleration of the flow (velocity increase, du/u > 0). Reversely, an increase in the section (dA/A > 0) induces a decrease in the velocity that is a flow deceleration. Such a subsonic / low-speed behaviour (see Fig. 5.3) is consistent with our daily experience 98 Quasi-1D compressible flows in nozzles which deals mainly with incompressible / low-speed flows. When the flow Mach number is low, the (isentropic) inviscid flow is such that the density ρ is almost constant so that mass conservation which is strictly equivalent to mass flowrate conservation, that is ρuA = constant, is almost equivalent to volume flowrate conservation, that is uA = constant. This relationship, valid for low-speed flows only, yields immediately the velocity evolution displaying in Fig. 5.3. When the flow becomes supersonic, ρ displays significant variation through the flow so that the mass flowrate conservation ρuA = constant can no longer be simplified into a volume flowrate conservation.

Case of a supersonic ﬂow : M > 1

u diminue u augmente

M > 1 M > 1

Figure 5.4: Evolution of the ﬂow velocity for a supersonic ﬂow in a varying-section nozzle. ”u diminue” stands for ”u decreases” while ”u augmente” stands for ”u increases”.

dA du According to (5.5), quantities and are of the same sign in that case This leads to a A u supersonic flow behaviour which is the inverse of the subsonic behaviour (see Fig. 5.4): -• a supersonic flow decelerates, that is du/u < 0, in a converging nozzle (such that dA/A < 0) -• a supersonic flow accelerates, that is du/u > 0, in a diverging nozzle (such that dA/A > 0)

Case where M = 1

M = 1

u augmente M < 1 M > 1

col Figure 5.5: Isentropic expansion of a gas in a converging-diverging nozzle. ”u augmente” stands for ”u increases” while ”col” stands for ”throat”.

dA If M = 1 then (5.5) indicates that = 0, which means section A reaches an extremum. In A other words, the section A in which the Mach number is sonic, M = 1, is either a section of 5.1. Physical model 99 minimum area or a section of maximum area. Taking into account the previous subsonic and supersonic analysis, the only possible case is that of a sonic section A of minimum area. This, if the sonic state M = 1 is reached in a section of a varying-section nozzle, it will be necessarily reached in a throatof the nozzle, a throat being a section of minimum area. Note that this does not mean the Mach number is necessarily equal to unity at a throat. If for instance, the flow remains subsonic in the nozzle, a velocity extremum (maximum) will take place at the throat with a Mach number wich remains strictly below 1 at the throat. What must be remembered is that if the sonic state is reached in a nozzle (we will see later when this flow configuration can occur) then the sonic state is reached at a nozzle’s throat. It can also be deduced that in order to isentropically expand a gas, accelerating this gas from a subsonic velocity to a supersonic velocity, it is required to use a converging-diverging nozzle configuration such as the one described in Fig. 5.5.

Example of a rocket engine

poussée

chambre de combustion M < 1 M > 1

Figure 5.6: Flow acceleration for a rocket nozzle, from the combustion chamber (”chambre de combustion”) to the ambient, in order to create thrust (”pousse”).

Such an acceleration through a converging-diverging nozzle is used in a rocket engine in order to produce thrust (see Fig.5.6).

Similarly, a gas can be isentropically compressed by decelerating it from the supersonic regime to the subsonic regime through a converging-diverging nozzle conﬁguration. Such a converging- diverging geometry is also called a De Laval nozzle in reference to the Swedish engineer who ﬁrst introduced it at the end of the 19th century (see Fig. 5.7).

5.1.3 Inﬂuence coeﬃcients Formula (5.5) can also be expressed as :

du 1 dA = (5.6) u M 2 − 1 A

The quantity 1/(M 2 − 1)) which links the velocity variation with the section variation is an influence coefficient : it expresses the influence of the section variation on the velocity variation. Similar influence coefficients can be computed for other flow quantities : pressure p, density ρ, temperature T , Mach number M. 100 Quasi-1D compressible flows in nozzles

M = 1

u diminue M > 1 M < 1

col

Figure 5.7: Isentropic compression of a gas in a converging-diverging nozzle. ”u diminue” stands for ”u decreases” and ”col” stands for ”throat”.

In an isentropic ﬂow, pressure and density variations are such that dp = a2dρ. Since the local 1D momentum conservation reads dp = −ρudu, it follows that :

dρ du = −M 2 ρ u or else, using (5.6): dρ M 2 dA = (5.7) ρ 1 − M 2 A

We note that density and velocity have opposite variations, with a relative variation for density stronger than the relative variation of velocity in a supersonic flow. The influence coefficient for pressure is obtained using the isentropic identity :

dp a2dρ γp dρ = = dρ = γ p p ρp ρ

We note this expression can also be directly obtained by diﬀerentiating the relationship p/ργ = constant valid for the isentropic ﬂow of an ideal gas. Pressure and density have similar variations : they both increase or both decrease. Using (5.7), the pressure variation is computed as : dp γM 2 dA = (5.8) p 1 − M 2 A

The energy conservation equation written in diﬀerential form yields, using dh = CpdT = γr dT : γ − 1 CpdT + udu = 0

Dividing this identity by a2 = γrT and introducing the Mach number M = u/a yields :

1 dT du + M 2 = 0 γ − 1 T u 5.1. Physical model 101

Flow field Subsonic flow M < 1 Supersonic flow M > 1 u increases decreases M increases decreases p decreases increases ρ decreases increases T decreases increases

Table 5.1: Evolution of the ﬂow properties in the main ﬂow direction (increasing x) for a dA converging nozzle ( < 0). A

Flow field Subsonic flow M < 1 Supersonic flow M > 1 u decreases increases M decreases increases p increases decreases ρ increases decreases T increases decreases

Table 5.2: Evolution of the ﬂow properties in the main ﬂow direction (increasing x) for a dA diverging nozzle ( > 0). A

Using (5.6) provides the temperature variation as a function of the section variation :

dT (γ − 1)M 2 dA = (5.9) T 1 − M 2 A

Note the temperature variation is similar to the density and pressure variation : p, ρ and T increase and decrease simultaneously. The Mach number variation is obtained by coming back to the very deﬁnition of this quantity : u M = = u(γrT )−1/2 a Diﬀerentiating yields : dM du 1 dT = − M u 2 T and using (5.6), (5.9) provides :

γ − 1 1 + M 2 dM 2 dA = (5.10) M M 2 − 1 A

Comparing the relationships (5.6) and (5.10) shows that the velocity u and the Mach number M vary in the same way : they both increase or decrease simultaneously. The similar variations of u, M on one hand and ρ, p, T on the other hand are summarized in Tables 5.1 and 5.2. 102 Quasi-1D compressible ﬂows in nozzles

5.2 Isentropic ﬂow in a nozzle

5.2.1 Flow analysis Let us consider the isentropic flow in a nozzle and let us look for a relationship between the area A(x) of the nozzle section located at position x and the local Mach number M(x) in this section. More precisely, since the Mach number is a non-dimensional quantity, let us look for A a relationship of the form = f(M), where Aref denotes the area of a reference section. Aref Let us choose as reference section a sonic section, denoted A∗, that is a section in which the local Mach number is equal to unity. The sonic state can be actually reached in the nozzle, in which case the section A∗ is an actual section but it can also remain a virtual state if the flow remains either subsonic or supersonic everywhere in the nozzle. The flow properties in the sonic section are the pressure p∗, the temperature T∗, the density ρ∗. The speed of sound and the local velocity in the sonic section are such that u∗ = a∗. The sonic section is also often called the critical section. Expressing mass conservation between the local state and the sonic state yields :

ρuA = ρ∗u∗A∗ so that : A ρ u ρ u a ρ 1 = ∗ · ∗ = ∗ · ∗ · ∗ = ∗ · A∗ ρ u ρ a∗ u ρ M∗

The characteristic Mach number M∗ = u/a∗ can be expressed as a function of the local Mach number using the identity (4.14). In order to link ρ∗/ρ to the Mach number, another interesting state of reference is introduced, namely the stagnation state : ρ ρ ρ ∗ = ∗ · 0 ρ ρ0 ρ where ρ0 denotes the stagnation or total density. The ﬂow being assumed isentropic, relationships (4.7) and (4.11)) yield : 1 1 ρ T γ − 1 0 = ( 0 )γ − 1 = (1 + M 2)γ − 1 = f(M) ρ T 2 When this identity is written for the sonic state (M = 1) it comes :

1 ρ γ + 1 0 = f(M = 1) = ( )γ − 1 ρ∗ 2

The following relationship between the section (more precisely the ratio A/A∗) and the Mach number is obtained :

(γ + 1) A 1 2 γ − 1 = (1 + M 2) 2(γ − 1) (5.11) A∗ M (γ + 1) 2

This relationship, know as the area-Mach number identity, is essential for the analysis of ﬂows in nozzles. 5.2. Isentropic ﬂow in a nozzle 103

A Note it can also be seen as a relationship of the form M = f( ). Thus, for an isentropic ﬂow, A∗ the Mach number in a section of the nozzle is entirely determined by the ratio between the local section area and the critical or sonic section area. Since the ratio total pressure to static pressure and the Mach number are linked by the relationship (4.10) in an isentropic ﬂow :

γ p γ − 1 0 = (1 + M 2)γ − 1 p 2

p A it follows that a relationship of the form = g( ) also exists for the isentropic ﬂow in a p0 A∗ nozzle. Combining the above identity with (5.11) yields :

v u γ+1 A uγ − 1 2 γ−1 1 = u 2 γ+1 (5.12) A∗ u 2 γ + 1 p γ p γ t − p0 p0

A The graphical display of the relationships M = f corresponding to (5.11) and p/p0 = A∗ A A f corresponding to (5.12) is proposed in Figure 5.8. Equivalently, the plots = f(M) A∗ A∗ A p and = f are displayed in Figure 5.9. Several important observations can be made A∗ p0 regarding these graphical displays :

-• the critical or sonic section A∗ is the section of minimal area in an isentropic ﬂow. If the sonic state is actually reached in the nozzle, it is necessarily reached at the throat so that A∗ is known and equal to Athr with thr standing for throat (or Acol with ”col” standing for ”throat” in French).

-• for a given value of the area ratio A/A∗, there is two possible values for the Mach number or the static to total pressure ratio hence also two possible values for the local static pressure in the section. One of the values corresponds to a subsonic flow while the other one corresponds to a supersonic flow. One solution or the other is selected depending on the boundary conditions for the nozzle flow as will be analyzed in the following sections.

-• Figures 5.8 and 5.9 will be used in the following analysis of nozzle flows, especially for a qualitative understanding of the flow behavior. For a more quantitative analysis, tabulated values of the relationships (5.11) and (5.12) will be used. Such tables are provided in Appendix A for the flow of an ideal gas with γ = 1.4. It must be emphasized again that a given value of the area ratio A/A∗ in Table II Isentropic flow with varying section corresponds to two distinct values of the Mach number M (and correspondingly the pressure ratio p/p0), one subsonic and the other supersonic. For instance A/A∗ = 1.558673 corresponds to M = 0.41 and p/p0 = 0.890711 if the flow regime in the nozzle section is subsonic while A/A∗ = 1.555256 corresponds to M = 1.90 and p/p0 = 0.149240 if the flow regime in the nozzle section is supersonic. 104 Quasi-1D compressible flows in nozzles

4 1 Branche subsonique 0.9 3.5 0.8

3 0 Branche supersonique 0.7 2.5 0.6

2 0.5

0.4 1.5

Nombre de Mach M Mach de Nombre 0.3

1 p/p pression de Rapport 0.2 0.5 0.1 Branche subsonique Branche supersonique

0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 Rapport de sections A/A* Rapport de sections A/A*

Figure 5.8: Relationship between the local Mach number M and the section or area ratio A/A∗ (left) or between the pressure ratio p/p0 and the area ratio A/A∗ (right). ”Branche subsonique” stands for ”Subsonic branch” while ”Branche supersonique” stands for ”Supersonic branch”.

12 10

10 8

8 6

6 4 section sonique A / A* A / A*

4 M < 1 section sonique 2 M > 1 p/p0=0.5283 1 2 0

0 0 1 2 3 4 0 0.2 0.4 0.6 0.8 1 Nombre de Mach M Rapport p/p0

Figure 5.9: Relationship between the area ratio A/A∗ and the local Mach number (left) or the local static to total pressure ratio p/p0 (right). 5.2. Isentropic ﬂow in a nozzle 105

5.2.2 Mass flowrate Mass conservation through the nozzle yields the conservation of the mass flowrate : m˙ = ρuA = constant The constant mass flowrate can also be expressed using the constant stagnation state associated with the isentropic flow in the nozzle : ρ a m˙ = ρuA = ρaMA = MAρ0a0 ρ0 a0 where quantities with index 0 denote stagnation quantities. Using the thermal equation of state for an ideal gas, p = ρrT , in the stagnation√ state yields p0 =√ρ0rT0. The definition of the speed of sound provides, for an ideal gas, a = γrT hence a0 = γrT0. The constant mass flowrate can also be expressed as : r p T0 p0 p m˙ = γrT0MA p0 T rT0

Using eventually the relationships (4.7) and (4.10) valid for isentropic ﬂows, the ratios p0/p and T0/T can be expressed as functions of the local Mach number M so that the mass ﬂowrate is given by the following expression depending on the local Mach number M(x) in section A(x) and the stagnation pressure p0 and temperature T0 :

γ+1 r − 2(γ−1) γ γ − 1 2 m˙ = p0 M 1 + M A (5.13) rT0 2

The ratiom/A ˙ is therefore given by :

γ+1 r − 2(γ−1) m˙ γ γ − 1 2 = p0 M 1 + M A rT0 2 and depends only on the Mach number M since the stagnation or total properties p0, T0 are constant for the isentropic flow under study. Since the mass flowratem ˙ is a flow constant for the nozzle flow, the ratiom/A ˙ is maximum when the section A reaches a minimum area value. From a previous analysis, we know this minimum value corresponds to the sonic section A∗ where the Mach number M reaches 1. The maximum mass flowrate which can pass through the nozzle is therefore such that : v u γ + 1 r u m˙ max γ u 2 γ − 1 = p0 t A∗ rT0 γ + 1 Denoting Γ the constant value : v u u γ + 1 u 2 Γ = tγ γ − 1 γ + 1 such that Γ(γ = 1.4) = 0.06847, the maximum mass flowrate of air passing through a nozzle is given by : p0 m˙ max = √ ΓA∗ (5.14) rT0 106 Quasi-1D compressible flows in nozzles

As long as the sonic state is not reached in the flow, the sonic section A∗ remains a virtual or fictitious section and the real section of minimal area Amin is greater than the critical section A∗. For a nozzle such that M = 1 is reached at the throat, the relationship (5.14) can also be rewritten as : p0 m˙ max = 0.6847 √ Athroat (5.15) rT0 The maximum mass flowrate going through a nozzle in which the sonic state is reached (at the throat, necessarily), is given by the identity (5.15) and depends only on the area of the throat section and the stagnation conditions in the reservoir feeding the nozzle. Once the sonic state is reached, that is once the nozzle is choked, the mass flowrate reaches its maximum value and further increasing the pressure difference between the reservoir and the ambient in which the nozzle discharges will have no impact on the mass flowrate which will keep its maximum value.

5.3 Analysis of the ﬂow through a converging nozzle

5.3.1 General analysis Let us consider (see Fig. 5.10) the ﬂow through a converging nozzle connecting a reservoir where the gas is at rest, hence the pressure and temperature take the respective stagnation or total values p0, T0, to a medium (space for instance) where the ambient pressure takes the value ps. The pressure ps is also called the discharge pressure. For the ﬂow to take place, the discharge pressure ps must be necessarily lower than the reservoir pressure p0. Let us also denote Ae the area of the exit section of the nozzle. Note the pressure pe in the exit section Ae is not necessarily equal to the ambient or discharge pressure ps as will be explained next.

Let us first assume the ambient pressure ps is slightly below the reservoir pressure p0. This small pressure difference induces a low-speed flow in the nozzle : the flow velocity or Mach number increases from 0 in the reservoir and a low value in the inlet section (since of large area) up to a still modest value in the exit section. In the same time, the pressure decreases from p0 in the reservoir down to pe in the exit section, the value pe being equal to the value of the ambient or discharge pressure ps : pe = ps. The flow is then subsonic throughout the nozzle so that the sonic state is not reached at the throat (section of minimum area) of the nozzle which is the exit section in the case of a converging nozzle. The virtual sonic area as well as the exit Mach number Me can be computed from the ratio between the total pressure p0 and the exit pressure pe known from the value of the discharge pressure ps. For the isentropic flow under study, we can write : γ p0 p0 γ − 1 2 γ − 1 = = 1 + Me ps pe 2 so that : v u  γ − 1  u u 2 p0 γ Me = u  − 1 (5.16) tγ − 1  ps 

When the discharge pressure is lowered (we assume the reservoir pressure or stagnation pressure p0 remains unchanged throughout the analysis), the increase in the pressure diﬀerence between 5.3. Analysis of the ﬂow through a converging nozzle 107

Figure 5.10: General configuration for a converging nozzle. ”Conditions de réservoir” stands for ”Stagnation conditions”. ”Géométriedu convergent” stands for ”Converging nozzle geometry”. ”Pression de décharge” stands for ”Discharge pressure”. ”Section de sortie” stands for ”Exit section”. the reservoir and the ambient leads to a faster flow in the nozzle hence a larger Mach number increase. This flow regime is illustrated in Fig.5.11. The limiting configuration corresponds to the value of ps for which the Mach number in the exit section reaches unity : the exit section becomes a sonic or critical section in that case. Note that Mach number and pressure evolution in the converging nozzle can be followed on Fig.5.8 or 5.9. Looking for instance at Fig.5.9, the Mach number increases along the subsonic branch, up to the point where M = 1 while the pressure decreases along the sonic branch down to the point where p/p0 = 0.5283. Let us recall this value is such that : γ γ p γ − 1 γ + 1 0 = 1 + 12 γ − 1 = γ − 1 = (1.2)3.5 = 1.892929 p 2 2 hence p0/p = 0.5283. Two distinct flow regimes in the converging nozzle can therefore be considered depending on the value of the discharge pressure with respect to the pivotal value 0.5283 p0 :

-• if ps > 0.5283p0 then the sonic state is not reached in the exit section. The ﬂow accelerates from the inlet to the outlet but remains subsonic. The pressure decreases (a subsonic expansion takes place in the nozzle) down to the ambient or discharge pressure ps. The mass ﬂowrate passing through the nozzle increases when ps decreases and is given by :

γ+1 r − 2(γ−1) γ γ − 1 2 m˙ = p0 Me 1 + Me Ae (5.17) rT0 2 108 Quasi-1D compressible ﬂows in nozzles

Figure 5.11: Evolution of the pressure (left) and the Mach number (right) in the diverging nozzle as a function of the discharge pressure ps.”ps d´ecroissante” stands for ”decreasing ps”. 3.5

-• if ps ≤ 0.5283 p0 then the isentropic ﬂow in the converging nozzle reaches the sonic state in the exit section so that the mass ﬂowrate also reaches its maximum value given by the previous formula applied with Me = 1 that is :

p0 m˙ max = √ ΓAe (5.18) rT0 with Γ = 0.6847 for γ = 1.4. The nozzle flow is said to be choked. The exit pressure pe is equal to the critical value 0.5283 p0 and a multi-dimensional expansion wave appears immediately downstream of the exit section in order to further decrease the pressure from pe down to ps < pe. When ps is lowered, below the critical value 0.5283 p0, the flow in the converging nozzle no longer varies : it is frozen from the reservoir to the sonic exit section and the mass flowrate keeps itsabove maximum value depending only on p0, T0 and the exit section area Ae. Only the expansion wave outside the nozzle is further modified by the decrease of the ambient or discharge pressure ps in order to lower the pressure from pe = 0.5283 p0 down to ps

5.3.2 Example of application

Let us consider the converging nozzle displayed in Fig. 5.12. The stagnation pressure p0 is assumed equal to p0 = 2 bar while the stagnation temperature is assumed equal to 300 K. The 2 2 inlet section area A1 is equal to 0.5 m while the outlet section area A2 = Ae is equal to 0.1 m . Figure 5.13 displays the evolution of the Mach number distribution in this converging nozzle for 3 successive values of the discharge pressure ps : ps = 1.9 bar, ps = 1.5 bar and ps = 1.1 bar. As expected, the acceleration of the subsonic flow in the converging nozzle is observed. The discharge pressure below which the flow in the converging nozzle becomes choked, with the sonic state reached in the exit section, is such that ps = 0.5283 × 2 = 1.0566 bar. It can be checked on Fig. 5.13 the Mach number reaches a value slightly below unity in the exit section when ps = 1.1 bar, that is when the discharge pressure is slightly above its pivotal value. 5.3. Analysis of the flow through a converging nozzle 109

Figure 5.12: Example of geometry for a converging nozzle (linear decrease of the section area A(x) with x). In the following application, p0 = 2 bar and T0 = 300 K.

0.8

0.6

Mach 0.4

0.2

0 0 0.2 0.4 0.6 0.8 1 x/L

Figure 5.13: Mach number distribution in the converging nozzle with stagnation pressure p0 = 2 bar. Blue curve : discharge or ambient pressure ps = 1.9 bar; green curve ; ps = 1.5 bar; red curve : ps = 1.1 bar. 110 Quasi-1D compressible ﬂows in nozzles

For ps = 1.1 bar, applying the formula (5.19) which gives Me as a function of ps/p0 yields Me = 0.965 (γ = 1.4). Using Me and knowing Ae as well as the stagnation conditions, the mass flowratem ˙ can be computed for ps = 1.1 bar thanks to formula (5.17). It is found that m˙ = 46.62 kg/s, a value very close to the maximum mass flowrate reached when the sonic state is achieved in the exit section and given by (5.18) :m ˙ max = 46.67 kg/s. Similar calculations of the mass flowrate can be performed for ps = 1.9 bar and ps = 1.5 bar.

Similarly, Fig.5.14 displays the distribution of the pressure ratio p/p0 in the converging nozzle for 3 successive values of the discharge pressure : ps = 1.9 bar, ps = 1.5 bar and ps = 1.1 bar. The expected pressure decrease is observed, which corresponds to a (subsonic) expansion from the stagnation pressure in the reservoir through the inlet pressure pi in the inlet section Ai and down to the exit pressure pe, equal to the ambient or discharge pressure as long as the ﬂow remains subsonic in the exit section. Note the pressure in the inlet section varies only slightly when the discharge pressure goes down from ps = 1.9 bar to ps = 1.1 bar. The inlet pressure pi or the ratio pi/p0 is easily obtained by application of the identity (5.11), that is the relationship between the local Mach number and the area ratio A/A∗, even though the sonic section area A∗ is not directly known as long as the ﬂow is not sonic at the nozzle exit. Once Me (< 1) is computed in the exit section from the ratio pe/p0, equal to ps/p0 (as long again as the nozzle is not choked), the value of the sonic section area A∗ can be computed since Ae is also known. The ratio between the inlet section area and the sonic section area Ai/A∗ can then be computed and the inlet Mach number Mi follows as well as the pressure ratio pi/p0. For ps = 1.9 bar, we obtain ps/p0 = 0.95. Table II of appendix ?? provides, in its subsonic part, 2 Ms = 0.27 and Ae/A∗ = 2.2385 hence A∗ = 0.04467 m . It follows that Ai/A∗ = 11.192 hence, reading again the subsonic part of Table II, an inlet Mach number Mi between 0.05 and 0.06 and a pressure ratio pi/p0 between 0.998 and 0.997 (these values are well in line with the M(x) and ps(x)/p0 distributions plotted in Fig. 5.13 and 5.14. For ps = 1.1 bar, we compute ps/p0 = 0.55 and the subsonic part of Table II provides a corre- 2 sponding value of Ms close to 0.96 and Ae/A∗ ≈ 1.001 so that A∗ = 0.0999 m (the virtual sonic section area depends on the value of the discharge pressure while it will become equal to the actual exit section area once the sonic state is reached at the converging nozzle exit). lt follows that Ai/A∗ = 5.005 hence, from reading the subsonic part of Table II, an inlet Mach number Mi between 0.11 and 0.12 and a ratio pi/p0 between 0.991 and 0.9899 - these values are again consistent with the distributions plotted in Fig. 5.13 and 5.14.

Finally, let us analyze the evolution of the exit Mach number Me and the mass flowratem ˙ as a function of the pressure ratio ps/p0. These respective evolutions are plotted in Fig.5.15 and 5.16. The exit Mach number Me first increases when the pressure ratio ps/p0 decreases, in accordance with the physical intuition stating a larger pressure difference induces a stronger flow acceleration. The flow remains subsonic in the exit section until ps reaches the critical value 0.5283 p0; for this specific value, we have Me = 1 in the exit section and this is the last value of the discharge or ambient pressure which is equal to the exit pressure pe. For values of the discharge pressure ps below the critical value 0.5283 p0, the exit section remains sonic (Me remains equal to 1) with a flow which remains frozen in the converging nozzle, equal to the unique isentropic solution with A∗ = Ae. The flow regime in the nozzle is a choked flow. The mass flowrate increases with Me hence with the decrease of the ratio ps/p0 until it reaches its maximum value, depending only on stagnation or total conditions and the exit section area 5.4. Analysis of the flow through a diverging nozzle 111

0.95

0.9

0.85

0.8

p/p0 0.75

0.7

0.65

0.6

0.55 0 0.2 0.4 0.6 0.8 1 x/L

Figure 5.14: Pressure (p/p0) distribution in the converging nozzle with stagnation pressure p0 = 2 bar. Blue curve : discharge or ambient pressure ps = 1.9 bar; green curve ; ps = 1.5 bar; red curve : ps = 1.1 bar.

Ae. For an ambient pressure below the critical value 0.5283 p0, the mass flowratem ˙ keeps its maximum value. The flow is no longer modified in the converging nozzle; only the supersonic expansion adapting the flow from the exit sonic state to the ambient state keeps evolving, outside the nozzle, when ps keeps decreasing until vacuum (ps = 0) is reached.

5.4 Analysis of the ﬂow through a diverging nozzle

Let us consider the diverging nozzle geometry displayed in Fig. 5.17. Let us analyze the various flow regimes that can occur for the flow through this nozzle geometry depending on the value of the discharge or ambient pressure ps, when this pressure ps takes values between p0 and the vacuum (ps = 0). If the inlet flow is subsonic, we know a subsonic flow in a diverging nozzle decelerates so that the flow becomes even more subsonic and reaches a subsonic exit Mach number Me such that : v u  γ − 1  u u 2 p0 γ Me = u  − 1 (5.19) tγ − 1  ps 

with the exit pressure pe equal to the discharge pressure ps. The subsonic flow inside the diverging nozzle adapts itself to the ambient pressure (in the same way as the subsonic flow inside the converging nozzle adapts itself to the ambient pressure as long as the sonic state is not reached in the exit section). From now on, we will assume the flow entering the diverging nozzle is supersonic and, for the sake of numerical application, we will assume the inlet Mach number in the inlet section Ai is such that Mi = 1.5.

The supersonic isentropic ﬂow in the nozzle is entirely known. Indeed, from Mi = 1.5 the ratio Ai/A∗ = 1.176167 can be deduced (reading the supersonic part of Table II in Appendix 112 Quasi-1D compressible ﬂows in nozzles

0.8

0.6

0.4

0.2 Mach Me dans la section d’éjection d’éjection section la Medans Mach 0 0 0.2 0.4 0.6 0.8 1 ps/p0

Figure 5.15: Evolution of the exit Mach number of the converging nozzle as a function of the pressure ratio ps/p0.

0.8

0.6

0.4

0.2

débit massique / débit massiquemaximal débit massique/ débit 0 0 0.2 0.4 0.6 0.8 1 ps/p0

Figure 5.16: Evolution of the mass ﬂowrate through the converging nozzle as a function of the pressure ratio ps/p0. 5.4. Analysis of the ﬂow through a diverging nozzle 113

Figure 5.17: Example of geometry for a diverging nozzle (linear increase of the section area A(x) with x). In the following application, p0 = 2 bar and T0 = 300 K.

A) hence the area of the sonic section associated with this isentropic flow since Ai is known : 2 A∗ = 0.085 m . The pressure ratio pi/p0 can also be read in Table II or obtained by applying the general isentropic formula between the total to static pressure ratio and the local Mach number. We obtain either way pi/p0 = 0.2724. so that pi = 0.5448 bar assuming a stagnation pressure equal to p0 = 2 bar. Since the flow is supersonic in the diverging nozzle, the flow accelerates from the inlet section Ai to the outlet section Ae, that is M increases and p/p0 decreases. The flow is a supersonic (isentropic) expansion. Since A∗ has been computed, the exit Mach number Me can be computed from the ratio Ae/A∗ : Ae/A∗ = 5.88 hence, reading the supersonic part of Table II in Appendix A, Me ≈ 3.35 and pe/p0 ≈ 0.016. The flow accelerates isentropically from Mi = 1.5 to Me = 3.35 and in the same time the static pressure decreases from pi = 0.5448 bar down to pe = 0.032 bar.

The key question is now to determine for which ambient conditions (which ambient or discharge pressure ps) such an isentropic supersonic flow can actually occur in the diverging nozzle. We must check whether this isentropic solution is indeed consistent with a given value of the discharge pressure ps. A pivotal value for ps is the value of the exit pressure pe corresponding to the isentropic solution in the nozzle, that is 0.032 bar. If ps lies in the interval [0 bar, 0.032 bar[, that is if the ambient pressure ps is less than the exit pressure when the flow in the nozzle corresponds to the (unique) supersonic isentropic expansion, then an expansion occurs at the nozzle exit (outside the nozzle) to further lower the pressure from pe = 0.032 bar down to ps. This multidimensional supersonic expansion takes place outside the nozzle; it deviates the flow leaving the nozzle away from the nozzle axis and the deviation angle is such the corresponding pressure decrease adjusts the pressure from pe to ps. The distribution of the Mach number M and the pressure ratio p/p0 for the converging nozzle are plotted respectively in Fig. 5.18 and 5.19 in the case where ps is low enough to ensure the supersonic isentropic expansion develops in the diverging nozzle, followed by a multidimensional supersonic expansion outside the nozzle. Note the flow in the nozzle remains unchanged when ps takes any value between0.032 bar and 0 bar; only the multidimensional expansion outside the 114 Quasi-1D compressible flows in nozzles

nozzle evolves with ps. If the ambient pressure is precisely equal to 0.032 bar then the diverging nozzle is said to be adapted (to the ambient).

3.5

2.5 Mach

1.5 0 0.2 0.4 0.6 0.8 1 x/L

Figure 5.18: Case of an ambient pressure ps below the isentropic solution exit pressure pe. Mach number distribution vs non-dimensional position x/L.

If the ambient pressure ps is greater than the value 0.032 bar, the isentropic supersonic flow in the nozzle remains feasible as long as a compression mechanism exists outside the nozzle to increase the pressure from the exit value pe = 0.032 bar up to the ambient value ps. For a range of values of ps we will detail next, this compression can be ensured by oblique shockwaves developing from the exit section, outside the nozzle. These oblique shockwaves compress the flow from the exit section where the pressure is equal to pe = 0.032 bar to the larger value of the ambient pressure ps; this compression deviates the supersonic flow towards the axis of the nozzle. The limiting state of this flow regime corresponds to the case where the oblique shockwaves become a 1D (or normal) shock located precisely in the exit section of the diverging nozzle (see Fig.5.20).

The value of the ambient pressure ps corresponding to the occurrence of a 1D shock in the exit section of the diverging nozzle can be easily computed. Indeed, the isentropic ﬂow upstream of the 1D shock is known until the state immediately upstream of the shock which corresponds to the isentropic supersonic solution in the exit section, Me = 3.35 and pe = 0.032 bar. The ambient pressure being precisely the static pressure downstream of the 1D shock, it is computed either using Table I (Shock relationships) in Appendix A or applying the pressure jump formula :

ps 2γ 2 = 1 + (Me − 1) pe γ + 1

We find the static pressure jump is such that ps/pe = 12.92625 so that ps = 0.4136 bar. For the diverging nozzle under study, the flow regime ”isentropic expansion in the nozzle and oblique shocks outside the nozzle to adapt the flow from the exit pressure to the ambient pressure” exists for ps between 0.032 bar and 0.4136 bar. Figures 5.21 and 5.22 display respectively the Mach number distribution and the pressure ratio p/p0 distribution in the diverging nozzle for ps = 0.4136 bar. The flow corresponds to the isentropic supersonic solution until the state 5.4. Analysis of the flow through a diverging nozzle 115

0.3

0.2 p/p0

0.1

0 0 0.2 0.4 0.6 0.8 1 x/L

Figure 5.19: Case of an ambient pressure ps below the isentropic solution exit pressure pe. Pressure ratio p/p0 distribution vs non-dimensional position x/L.

Figure 5.20: Diverging nozzle. Flow regime corresponding to the supersonic isentropic expansion inside the nozzle with outside adaptation to the ambient pressure ps through oblique shockwaves. From left to right : increasing strenght of the compression mechanism (increasing pressure jump from pe = 0.032 bar to an increasing value of ps). Limiting case : 1D shock in the nozzle exit section. 116 Quasi-1D compressible ﬂows in nozzles

immediately upstream of the exit section where Me = 3.35 and pe = 0.032 bar. Through the 1D shock in the exit section, the pressure jumps to ps = 0.4136 bar immediately downstream of the shock. Meanwhile, the Mach number decreases from Me = 3.35 to a downstream value Ms which can be obtained either from the shock table I or using the shock relationship between the upstream state 1 (or M1 = Me) and the downstream state 2 (or M2 = Ms):

γ − 1 2 1 + ( )M1 M 2 = 2 2 γ − 1 γM 2 − ( ) 1 2

For γ = 1.4 and Me = 3.5, we obtain Ms = 0.457.

3.5

2.5

2 Mach 1.5

0.5

0 0.2 0.4 0.6 0.8 1 x/L

Figure 5.21: Diverging nozzle. Flow regime corresponding to the supersonic isentropic expansion inside the nozzle except right in the exit section where a 1D shock occurs, adapting the pressure from pe = 0.032 bar to ps = 0.4136 bar. Mach number distribution vs x/L.

If the ambient pressure is larger than 0.4136 bar and lower than p0 = 2 bar, yet another flow regime is reached in the diverging nozzle for which it is no longer possible to conserve the isentropic supersonic expansion inside the nozzle. Physically, for ps in the range [0.4136, 2] bar, a compression mechanism must take place inside the nozzle in order to adapt the flow to the ambient pressure. The typical pressure distribution inside the nozzle for this flow regime is displayed in Figure 5.23 for ps = 1 bar. The isentropic supersonic flow develops in the diverging nozzle until a given section (located between the inlet and the exit section) in which a 1D or normal shock occurs which strongly compresses the supersonic flow, turning it into a subsonic flow as seen in the previous Lecture devoted to 1D compressible flows. Downstream of this 1D shock, the subsonic flow decelerates in the remainder of the diverging nozzle and the pressure increases, adding a supplementary (isentropic) compression to the compression through the shockwave. The location of the 1D shock in the nozzle is such that the compression through 5.4. Analysis of the flow through a diverging nozzle 117

0.3

0.2 p/p0

0.1

0 0 0.2 0.4 0.6 0.8 1 x/L

Figure 5.22: Diverging nozzle. Flow regime corresponding to the supersonic isentropic expansion inside the nozzle except right in the exit section where a 1D shock occurs, adapting the pressure from pe = 0.032 bar to ps = 0.4136 bar. Pressure ratio p/p0 distribution vs x/L. the shock and the isentropic compression downstream of the shock adapts the nozzle ﬂow to the ambient pressure ps : the exit pressure pe is precisely equal to ps. In the case where ps = 1 bar, it is observed in Figure 5.23 the normal shock stands around the section x = 0.29 m. Let us now explain how this shock location can be predicted from the known value of the ambient pressure ps, stagnation conditions and nozzle geometry.

The problem to solve is summarized in Fig.5.24:

-• the inlet conditions of the flow are known (Mi is known in section Ai) as well as the stagnation or total flow conditions in the reservoir. The total pressure p0 in the reservoir is the constant total pressure associated with the isentropic flow upstream of the 1D shock, also denoted (p0)up (in French (p0)aval). The total temperature T0 is constant everywhere in the adiabatic flow, upstream and downstream of the 1D shockwave.

-• the ambient pressure ps is also known. In the present case, we assume ps = 1 bar. The pressure pe in the exit section is equal to ps. The Mach number Me in the exit section is no longer immediately available since it can still be computed from the total to static pressure ratio in this exit section but this ratio is now deﬁned as (p0)down/pe = (p0)down/ps with the downstream total pressure (p0)down unknown at this stage :

v u  γ − 1  u u 2 (p0)down γ Me = u  − 1 tγ − 1  ps  118 Quasi-1D compressible ﬂows in nozzles

Figure 5.23: Flow in the diverging nozzle for the regime ”shock in the nozzle”. Ambient pressure equal to ps = 1 bar. Upstream of the 1D shock the ﬂow is isentropic with an associated total pressure equal to p0 the stagnation pressure in the reservoir. Downstream of the 1D shock, the ﬂow is also isentropic but with an associated total pressure strictly lowr than p0 because of the total pressure loss through the shock.

In order to compute Me, the total pressure (p0)down downstream of the 1D shock must be computed.

-• the 1D or normal shock is assumed to take place in a section x = xc, corresponding to a section area Ac (the section area A(x) is supposed to vary linearly between Ai at the inlet x = 0 to Ae at the exit x = L). Finding Ac will therefore immediately allow to determine the location xc.

In order to compute the location of the 1D shock in the diverging nozzle, let us use mass conservation to link the flow conditions upstream and downstream of the shockwave. The mass flowrate can be computed in any section of area A where the Mach number is equal to M using the formula : γ+1 r − 2(γ−1) γ γ − 1 2 m˙ = p0 M 1 + M A rT0 2 where p0 = (p0)up is the total pressure associated to the isentropic flow region where the section A is located. Thus if A is taken to be the inlet section Ai, the total pressure p0 is the upstream total pressure (p0)up equal to the reservoir stagnation pressure. An immediate calculation yields m˙ = 39.68 kg/s. This same mass flowrate can also be computed using its expression in the exit section :m ˙ = ρeueAe or else : pe aeMeAe =m ˙ rTe wherem ˙ has been put in the right-hand-side to emphasize the fact it is a known quantity. The exit pressure pe is equal to the ambient pressure ps and the speed of sound in the exit section 5.4. Analysis of the flow through a diverging nozzle 119

Figure 5.24: Flow conﬁguration in the case where ps = 1 bar. Flow regime such that a 1D shock takes place inside the nozzle to adapt the ﬂow to the ambient pressure. (p0)aval and (p0)amont denote respectively the total pressure upstream (amont) or downstream (aval) of the 1D shockwave.

√ can be computed as : ae = γrTe. The exit temperature Te can be linked with the exit Mach number Mach Me using the identity :

T0 γ − 1 2 = 1 + Me Te 2 with the reservoir temperature T0 constant throughout the adiabatic ﬂow in the nozzle. It is thus possible to write : r √ T0 ps γ Me √ Ae =m ˙ Te rT0 or else 1/2 √ γ − 1 2 m˙ rT0 Me 1 + Me = √ 2 ps γAe with the known quantity in the right-hand-side (RHS) denoted K, such that : √ m˙ rT0 K = √ ps γAe

Taking the square of the above identity, a second-order equation is obtained for the unknown 2 variable X = Me . The positive root is selected to provide an explicit expression for the exit Mach number Me as a function of the constant K and the speciﬁc heat ratio γ :

v us u 1 2K2 1 M = t + − e (γ − 1)2 (γ − 1) (γ − 1) 120 Quasi-1D compressible ﬂows in nozzles

For γ = 1.4 and r = 287 J/kg/K an immediate numerical application yields :

v us u m˙ 2T t 0 Me = 1.5811 1 + 164 2 2 − 1 psAe

For the case under study, with ps = 1 bar and the previously detailed values of T0,m ˙ and Ae, the exit Mach number is found equal to Me = 0.196. The Mach distribution in the nozzle for ps = 1 bar is plotted in Fig.5.25. This distribution corresponds to the static pressure distribution displayed in Fig.5.24. It can be checked the exit Mach number Me corresponds indeed to the predicted value Me ≈ 0.2. It remains to explain how the shock location can be eventually from this information on the exit Mach number.

2.5

1.5 Mach 1

0.5

0 0 0.2 0.4 0.6 0.8 1 x/L

Figure 5.25: Mach distribution in the case where ps = 1 bar. Flow regime such that a 1D shock takes place inside the nozzle to adapt the ﬂow to the ambient pressure.

Since Me is know, the section area ratio Ae/(A∗)down can be computed. The sonic section (A∗)down is relative to the isentropic ﬂow downstream of the shockwave. Similarly, (A∗)up denotes the sonic section relative to the isentropic ﬂow upstream of the shockwave. For Me = 0.196 a linear interpolation in Table II of Appendix A yields Ae/(A∗)down = 3.0226 hence (A∗)down = 2 0.1654 m . Similarly, in the inlet section Mi = 1.5 so that Ai/(A∗)up = 1.176 hence (A∗)up = 2 0.085 m . The ratio of the sonic section through the shock is equal to (A∗)down/(A∗)up = 1.945. As will be proved next, the sonic section increases through the shockwave.

It remains to explain how the knowledge of the sonic section ratio gives access to the total pressure ratio through the normal shock. It the ratio (p0)down/(p0)up is known then Table I of Appendix A allows to compute the Mach number Mc,up immediately upstream of the 1D shock. From this Mach number the section area ratio Ac/(A∗)up can be deduced so that Ac can be computed as well as the location xc of the shock. 5.4. Analysis of the ﬂow through a diverging nozzle 121

Let us write the mass conservation between the upstream sonic state and the downstream sonic state, respectively denoted with indices 1 (instead of up) and 2 (instead of down):

(ρ∗)1(u∗)1(A∗)1 = (ρ∗)2(u∗)2(A∗)2 √ By deﬁnition of a sonic state, u∗ = a∗ = γrT∗. Moroever the thermal equation of state written at the sonic state yieds p∗ = ρ∗rT∗. It comes :

(p∗)1 p (p∗)2 p γr(T∗)1(A∗)1 = γr(T∗)2(A∗)2 r(T∗)1 r(T∗)2

The sonic temperature and the total temperature are linked by :

T γ − 1 γ + 1 0 = 1 + × 12 = T∗ 2 2

Since the total temperature T0 is conserved through the shock, this means the sonic temperature is also conserved through the shock so that (T∗)1 = (T∗)2 hence the simpliﬁed form of the mass conservation between the upstream and downstream sonic states :

(p∗)1(A∗)1 = (p∗)2(A∗)2

The total pressure and the sonic pressure in an isentropic ﬂow are such that : γ p γ + 1 0 = γ − 1 p∗ 2 or, for γ = 1.4, p∗ = 0.5283 p0. The above simpliﬁed form of the mass conservation can thus be recast in the form : 0.5283 (p0)1(A∗)1 = 0.5283 (p0)2(A∗)2 or equivalently : (p0)1(A∗)1 = (p0)2(A∗)2 We can therefore conclude the total pressure loss through a 1D shock and the sonic section area jump through the same shock are such that :

(p ) (A ) 0 aval = ∗ amont (5.20) (p0)amont (A∗)aval

For the case under study, a numerical application using the previously computed sonic section area ratio yields : (p ) 1 0 down,2 = = 0.514 (p0)up,1 1.945 (p ) In the shock table I of Appendix A, this total pressure loss 0 2 through the shock corresponds (p0)1 to a Mach number immediately upstream of the 1D shock such that Mc,up ≈ 2.46. Reporting to Fig.5.25, we can check this value is indeed the one observed on the Mach distribution just upstream of the shockwave. From this value of Mc,up, we can deduce using Table II ”Isentropic 122 Quasi-1D compressible ﬂows in nozzles

ﬂow with varying section” of Appendix A the ratio Ac/(A∗)up : Ac/(A∗)up = 2.54. It follows 2 that Ac = 0.2159 m and since : x A + (A − A ) c = A i e i L c the shock location can be computed : xc/L = 0.28975. The accuracy of this prediction can be checked on Fig. 5.24 and Fig.5.25 as well as on Fig.5.26 which displays a close-up of the Mach distribution in the shock region.

2.5

1.5 Mach

0.5

0.2 0.25 0.3 0.35 x/L

Figure 5.26: Diverging nozzle. Mach number distribution vs x/L in the case ps = 1 bar. Close-up on the shock region.

Figures 5.27 and 5.28 display some distributions of the static pressure (normalized by the upstream total pressure p0) and of the Mach number for successive values (1, 0.8 and 0.6 bar) of the ambient pressure ps yielding a shockwave in the nozzle. The higher the ambient pressure ps, the closer the shock location to the inlet of the nozzle.

5.5 Analysis of ﬂow regimes in a converging-diverging nozzle

5.5.1 General analysis Let us consider the converging-diverging or De Laval nozzle displayed in Fig. 5.29 and let us analyze the various flow regimes through this nozzle when the reservoir or stagnation pressure p0 is kept fixed while the ambient pressure ps (downstream of the exit section Ae) takes different values between p0 and 0.

Let us first assume the ambient pressure is equal to the stagnation pressure p0. The exit pressure is also equal to ps in that case; in fact p = p0 everywhere and no flow is taking place through the nozzle : the fluid remains at rest (since dp = −ρudu, dp = 0 yields du = 0). Let us now slightly decrease the ambient pressure ps. The small pressure difference thus established creates a low-speed flow through the nozzle. In the converging part of the nozzle, the 5.5. Analysis of flow regimes in a converging-diverging nozzle 123

0.5

0.45

0.4

0.35

0.3

0.25

0.2 p/(p0)_amont

0.15

0.1

0.05

0 0 0.2 0.4 0.6 0.8 1 x/L

Figure 5.27: Distribution of the pressure ratio p/(p0)up in the diverging nozzle for ps = 1 bar (blue curve), ps = 0.8 bar (green curve) and ps = 0.6 bar (red curve).

2.5

1.5 Mach

0.5

0 0 0.2 0.4 0.6 0.8 1 x/L

Figure 5.28: Distribution of the Mach number in the diverging nozzle for ps = 1 bar (blue curve), ps = 0.8 bar (green curve) and ps = 0.6 bar (red curve). 124 Quasi-1D compressible ﬂows in nozzles

A/A très grand * pressionde sortie conditions ou d’éjection d’arrêt A c p p e 0 col

Figure 5.29: Typical conﬁguration of a De Laval nozzle. Ustream of the inlet section, the nozzle is connected with a reservoir where the stagnation pressure is p0. The throat section area is denoted Athr. The exit pressure pe is not necessarily equal to the ambient or discharge pressure ps downstream of the nozzle exit.

fluid is accelerated from 0 to the reservoir to a (small) maximum value at the throat where the flow remains subsonic. This subsonic flow decelerates in the diverging part of the nozzle and reaches a subsonic Mach number in the exit section which is determined by the pressure ratio ps/p0 also equal to pe/p0 using formula (5.19). This flow behaviour can also be envisioned using for instance the plot of Fig. 5.8 which displays the relationship between the local Mach number in a nozzle isentropic flow and the section area ratio A/A∗. The state describing the inlet flow lies initially (near the inlet) on the right part (small Mach number corresponding to the large section ratio Ai/A∗) of the bottom subsonic branch. While the flow develops in the diverging part of the nozzle, the Mach number slightly increases along the subsonic branch (moving from right to left) until it reaches a maximum value at the throat, this maximum value being still subsonic and the throat section area Ac being larger than the sonic or critical section area A∗. Downstream of the throat, the Mach number goes down along the subsonic branch (moving from left to right since the section is increasing) until it reaches the exit value (say (Me)1) determined by the static pressure (pe)1 = (ps)1 and the stagnation pressure p0. A similar description can be made for the static pressure : p decreases from the total pressure p0 to a minimum value at the throat of the nozzle then, downstream of the throat, the pressure increases to reach the value (pe)1 = (ps)1 in the exit section. This isentropic subsonic flow behaviour is displayed in Fig.5.30

Suppose now we decrease a bit more the ambient or discharge pressure ps. The pressure ratio p0/ps being a bit larger between the reservoir and the ambient the flow acceleration in the nozzle will be a bit stronger : the Mach number is going up a little higher along the subsonic branch until it reaches a maximum (subsonic) value at the throat while the static pressure goes down a little lower until it reaches a minimum value at the throat. Downstream of the throat, the Mach number decreases until it reaches the exit value (Me)2 determined by the pressure ratio p0/(pe)2 with the exit pressure (pe)2 equal to the ambient pressure (ps)2 (see again Fig. 5.30). Two flow configurations only have been reported in Fig.5.30, with (pe)2 < (pe)1 and (Me)2 > (Me)1, but there is of course an infinity of such isentropic subsonic solutions for the flow in the De Laval nozzle, which are controlled by the pressure ratio ps/p0 and the section area ratio 5.5. Analysis of flow regimes in a converging-diverging nozzle 125

Mach p/p 0 1 1 pe1 pe2

M e2 M e1 x x entrée col sortie entrée col sortie

Figure 5.30: Subsonic isentropic ﬂows in the De Laval nozzle.

Ac/A∗. If the ambient pressure ps is further reduced, a specific value (ps)3 will be reached such that the Mach number at the throat reaches the sonic state (see Fig.5.31). In the flow configuration displayed in Fig.5.31, the Mach number is just below 1 at the throat so that the Mach number goes back to a decreasing subsonic value downstream of the throat while the static pressure increases downstream of the throat until it reaches the value (pe)3 = (ps)3. The Mach number (Me)3 in the exit section is computed using the ratio p0/(ps)3. When the throat section becomes a sonic or critical section, Athr = A∗. The temperature at the throat is easily computed from the stagnation properties since : T γ − 1 γ + 1 0 = 1 + (M = 1)2 = T∗ 2 2 √ hence T∗ = 0.833T0 for γ = 1.4. Hence the velocity at the throat : u∗ = a∗ = γrT∗ = p 2γrT0/(γ + 1). The sonic pressure p∗ at the throat is such that : γ p T 0 = 0 γ − 1 p∗ T∗ hence pc = p∗ = 0.528p0 for γ = 1.4.

When going from the flow configuration 1 (ps = (ps)1) to the flow configuration 2 (ps = (ps)2) and next the flow configuration 3 (ps = (ps)3) , the mass flowratem ˙ increases :

m˙ 1 < m˙ 2 < m˙ 3 = ρ∗A∗u∗ sincem ˙ 3 = ρcAcuc = ρ∗A∗u∗. The mass ﬂowrate at the sonic throat is the maximum mass ﬂowrate which can go through the nozzle; it is given by (5.18):

p0 m˙ max = √ ΓAe rT0

Once the sonic state is reached at the throat, the nozzle is said to be choked. Decreasing the ambient pressure ps below (ps)3 will no longer modify the ﬂow in the converging part of the nozzle, which remains frozen, equal to the isentropic solution with A∗ = Athr and p0 equal to the 126 Quasi-1D compressible ﬂows in nozzles

Mach p/p 0 1 1

pe3 0.528 Me 3

x x entrée col sortie entrée col sortie

Figure 5.31: Flow in the De Laval nozzle just before the nozzle is started, that is just before a supersonic ﬂows starts to develop in the diverging part of the nozzle. ”Entre” stands for ”inlet”; ”col” stands for ”throat”; ”sortie” stands for ”exit”. reservoir pressure. The ﬂow in the diverging part of the nozzle keeps evolving for lower values of ps.

The value of the ambient pressure (ps)3 is also known as the starting pressure of the nozzle since it is the ambient pressure for which the flow starts to become supersonic in the diverging part of the nozzle. Indeed, if the pressure ps is slightly below (ps)3, the Mach number keeps increasing in the diverging part of the nozzle and the static pressure keeps decreasing. However, this isentropic supersonic solution can take place in the diverging part of the De Laval nozzle only for sufficiently low values of the ambient pressure ps. Indeed, if the isentropic supersonic solution applies in the diverging part of the nozzle, the exit pressure pe is low since resulting from a continuous flow expansion (subsonic then supersonic) from the reservoir to the exit section. By ”sufficiently low values” we mean therefore values of the ambient pressure ps which are compatible with an outside adaptation of the flow between the exit section of the nozzle where p = pe and the ambient where p = ps. This adaptation taking place outside the nozzle can be a compression or an expansion. If the ambient pressure is larger than the pressure resulting from a 1D shockwave in the exit section of the De Laval nozzle, then a shock wave occur in the diverging part of the nozzle in order to adapt the flow to the ambient pressure (see Fig.5.32 for a typical flow configuration in this regime).

Between the throat and the shock section, the (unique) isentropic supersonic solution defined by A/A∗ = A/Athr develops. Downstream of the shock, the flow becomes subsonic and decelerates in the diverging part of the nozzle. The shock is precisely located in such a way the pressure increase through the shock (static pressure jump) and the isentropic compression downstream of the shock leads to an exit pressure (pe)4 equal to the ambient pressure (ps)4 < (ps)3. Note the flow is also isentropic downstream of the shock but with a lower (constant) total pressure and a larger (constant and virtual) sonic section.

If the ambient pressure is further reduced, the shockwave moves toward the exit section. The limit of this ﬂow regime is reached when the 1D shock is located right in the exit section (see Fig. 5.34). 5.5. Analysis of ﬂow regimes in a converging-diverging nozzle 127

choc droit

col

Figure 5.32: Flow in a De Laval nozzle. Flow regime with a 1D shock or straight / normal shock (”choc droit” in French) taking place in the diverging part of the nozzle.

solution isentropique Mach p/p supersonique 0 1

pe4 1 0.528 Me 4

x x entrée colchoc sortie entrée col sortie

Figure 5.33: Flow in the De Laval nozzle when a 1D shock occurs in the diverging part of the nozzle.

col

Figure 5.34: Development of a normal shock in the exit section of the De Laval nozzle. 128 Quasi-1D compressible ﬂows in nozzles

For such a flow configuration, the isentropic supersonic solution given by (5.11) with A∗ = Athr applies until the exit section, providing the upstream state of the shock with Me,up = (Me)6 and pe,up = (pe)6. Note the upstream shock conditions are indexed 6 for reasons which will be clarified before the end of this flow regime analysis. Right in the exit section, of area Ae, a discontinuous variation of the flow properties takes place which can be computed using the 1D shock relationships previously established. The downstream Mach number Me,down and static pressure pe,down are respectively denoted (Me)5 and (pe)5. The flow configuration ”1D shock in the exit section” occurs precisely when the ambient pressure ps is equal to (pe)5.

choc droit en solution isentropique section d’ éjection Mach supersonique p/p 0 Me 6 1 saut de pression à travers le choc droit 1 0.528 pe5 Me 5 pe6 x x entrée col choc sortie entrée col sortie Figure 5.35: Flow in the De Laval nozzle with a normal shock right in the exit section.

When the ambient pressure ps goes below the value (pe)5, the adaptation between the exit section of the nozzle and the ambient is achieved through ﬂow mechanisms which develop outside the nozzle. The ﬂow solution inside the nozzle, including the exit section, is the unique isentropic supersonic expansion given by (5.11) with A∗ = Athroat. When the ambient pressure ps is exactly equal to (pe)6, the exit pressure corresponding to the isentropic supersonic expansion, the nozzle is said to be adapted.

When the ambient pressure ps lies in the interval ](pe)6, (pe)5[, a physical mechanism outside the nozzle ensures the recompression of the ﬂow from the value (pe)6 in the exit section to the ambient pressure ps. This recompression is less strong than the one corresponding to a 1D shock in the exit section (which would yield the value (pe)5 for the ambient pressure). This mechanism is an oblique shockwave attached to the nozzle exit (see Fig. 5.36).

Finally, when the ambient pressure ps is lower than (pe)6, a supersonic expansion ensures the ﬂow adaptation from the exit pressure (pe)6 down to the ambient pressure ps < (pe)6 (see Fig. 5.37).

5.5.2 Example of application Let us consider the De Laval nozzle described in Fig.5.38 and let us descrive the ﬂow through this converging-diverging nozzle when the discharge or ambient pressure ps varies from the reservoir or stagnation pressure p0 down to 0 (vacuum conditions). 5.5. Analysis of ﬂow regimes in a converging-diverging nozzle 129

p p < p < p e6 e6 a e5 col

Figure 5.36: Recompression of the ﬂow through an oblique shock attached to the nozzle exit section. The ambient pressure is pa = ps

p p < p e6 a e6 col

Figure 5.37: Adaptation to the ambient pressure for the supersonic ﬂow leaving the De Laval nozzle through a supersonic expansion outside the nozzle. The ambient pressure is pa = ps.

Figure 5.38: Geometrical features of the De Laval nozzle under study. The total length of the nozzle is L = 2 m. ”Milieu ambiant” stands for ”ambient” and ”pression de d´echarge” stands for ”Discharge or ambient pressure”. 130 Quasi-1D compressible ﬂows in nozzles

Some key or pivotal values can be computed for the ambient pressure ps. A first key-value is the discharge pressure for which the nozzle is started that is the sonic state is reached at the nozzle throat. Let us denote (ps)start this specific value (in the following figures, it also appears as (ps)amor where ”amor” stands for the French word ”amor¸cage”equivalent of the ”starting” pressure). If the discharge pressure ps remains above the value (ps)start (or (ps)amor) the flow remains subsonic everywhere in the nozzle, with a maximum (subsonic) velocity reached at the throat. An overview of the various flow regimes in a De Laval nozzle is displayed in Fig. 5.39. The value of the ambient or discharge pressure is reported at the top of the plot : from 0 on the far right up to p0 on the far left. This first ”subsonic flow” / ”écoulement subsonique” regime takes place when ps lies in the range [(ps)start, p0] or [(ps)amor, p0].

If the discharge pressure is below (ps)start then the flow in the converging part of the nozzle remains frozen : the flow is choked with a sonic throat through which the maximum mass flowrate passes. This maximum value is entirely determined by the reservoir conditions and the throat section. The flow in the diverging part of the nozzle still evolves when ps is further lowered. A normal shockwave appears in the diverging part of the nozzle when the discharge pressure ps is between (ps)start (or (ps)amor) and (ps)sie or (ps)cde. Index sie stands for ”shock in exit (section)” while, in French, cde stands equivalently for ”choc dans (la section d’) éjection” . The value (ps)sie or (ps)cde corresponds to the value of the discharge pressure such that the 1D shock lies precisely in the exit section of the nozzle (with the isentropic supersonic expansion developing until the exit section, immediately upstream of the shock). In Fig.5.39, this second flow regime appears on the second line of the plot.

If ps < (ps)cde or ps < (ps)sie the the flow in the De Laval nozzle corresponds to the isentropic expansion (subsonic in the converging part, supersonic in the diverging part) from the inlet section to the exit section included. The flow in the nozzle is then frozen both in the converging and the diverging part and only the flow outside the nozzle evolves when ps is lowered below (ps)cde or (ps)sie (see also the third line of the plot in Fig.5.39.

The last pivotal value for the discharge pressure is the so-called adaptation value, denoted (ps)adapt, which corresponds to the case where the discharge pressure is precisely equal to the exit pressure for the isentropic expansion solution in the nozzle. If ps = (ps)adapt the nozzle is said to be adapted : no specific mechanism is required to adapt the flow from pe to ps (see again the third line of the plot in Fig.5.39. If ps > (ps)adapt (but ps < (ps)cde or ps < (ps)sie), oblique shockwaves appear at the nozzle exit, oustide the nozzle, to compress the flow pe = (ps)adapt to ps. Reversely, if ps < (ps)adapt, expansion waves appear at the nozzle exit, outside the nozzle, to further expand the flow from pe = (ps)adapt down to ps. Let us now determine the pivotal values (ps)start,(ps)sie and (ps)adapt for the De Laval nozzle displayed in Fig.5.38. Just before the nozzle becomes started, the flow configuration in the nozzle is the following one : :

-• increase of the Mach number from the inlet section to the throat with a Mach number at the throat just below 1;

-• decrease of the Mach number from the throat to the exit section where th exit pressure pe is equal to the ambient pressure ps. 5.5. Analysis of ﬂow regimes in a converging-diverging nozzle 131

Figure 5.39: Overview of the various ﬂow regimes in the De Laval nozzle depending on the discharge pressure ps. 132 Quasi-1D compressible ﬂows in nozzles

Since the Mach is equal to 1 (say 1−) at the throat, it implies the sonic section is equal to the tho- 2 rat section, that is A∗ = Acol = 0.1 m . The state in the exit or ejection section can therefore be determined from the known section area ratio Ae/A∗ = 5. Reading the subsonic part of Table II in Appendix A and performing a linear interpolation between the values given in the table yields Me = 0.117 and pe/p0 = ps/p0 = 0.992. The nozzle is started for ps = (ps)start = 1.984 bar. For ps > 1.984 bar and ps < 2 bar, a subsonic ﬂow takes place in the nozzle. Note the nozzle start occurs for a value of the discharge pressure which is only slightly lower than the reservoir or stagnation pressure p0 because the section area ratio Ae/Athroat is large.

The second pivotal value for the discharge pressure is the pressure (ps)sie. To determine the state just upstream of the normal shock in the exit section, we use Ae/A∗ = Ae/Athr = 5 for the isentropic flow upstream of the shock (everywhere in the nozzle until the exit section where the 1D shock takes place) and look for Me,up in the supersonic part of Table II. We find Me,up = 3.174 and pe,up/p0 = 0.021 hence pe,up = 0.042 bar since the total pressure associated with the upstream isentropic flow is the reservoir pressure p0. The pressure (ps)sie = pe,down is obtained from the shock relationship :

pe,down 2γ 2 = 1 + (Me,up − 1) pe,up γ + 1 so that (ps)sie = pe,down = 11.587 × pe,up = 0.486 bar.

For a discharge pressure lower than 1.984 bar and larger than 0.486 bar the ﬂow regime in the De Laval nozzle is : frozen ﬂow in the converging part of the nozzle, sonic throat and normal shock in the diverging part of the nozzle. Figures 5.40 and 5.41 display the Mach number and pressure ratio p/(p0)up in the De Laval nozzle for various values of the discharge pressure in the range [0.5 bar, 1.8 bar]. The notation (p0)up is used to make clear the constant total pressure used for the normalization of the static (local) pressure p is the total pressure upstream of the shockwave, that is the reservoir total pressure. One can clearly observe the evolution of the shock location from a section near the throat when ps is only slightly below (ps)start to a section near the nozzle exit when ps is only slightly above (ps)sie.

For ps < 0.486 bar, the isentropic expansion solution (subsonic in the converging part of the nozzle, supersonic in the diverging part of the nozzle) is achieved everywhere in the nozzle, exit section included. It remains to compute the last pivotal value (ps)adapt in order to complete the quantitative description of the ﬂow regimes for the De Laval nozzle geometry under study. In fact, (ps)adapt has already been computed when looking for (ps)sie. It corresponds to the pressure pe,up in the exit section just upstream of the 1D shock located in the exit section. Therefore we have : (ps)adapt = 0.042 bar.

For ps between 0.486 bar and 0.042 bar, the isentropic solution in the nozzle is completed with oblique shockwaves outside the nozzle to recompress the ﬂow from pe = 0.042 bar to ps. For ps = 0.042 bar the nozzle is adapted. For ps < 0.042 bar, the isentropic solution in the nozzle is completed with supersonic expansion waves which lower the pressure from pe = 0.042 bar to ps. The isentropic (subsonic then supersonic) expansion in the nozzle is displayed in Fig.5.42 and 5.43 respectively for the Mach number distribution and the static pressure distribution (normalized by the upstream total pressure or reservoir pressure). 5.5. Analysis of ﬂow regimes in a converging-diverging nozzle 133

3 ps=1.8 bar ps=1.4 bar 2.5 ps=1 bar ps=0.8 bar ps=0.6 bar ps=0.5 bar 2

1.5 Mach

0.5

0 •0.4 •0.2 0 0.2 0.4 x/L

Figure 5.40: Distribution of the local Mach number versus x/L for diﬀerent values of the discharge or ambient pressure ps in the interval [0.486, 1.984] bar.

0.8 ps=1.8 bar ps=1.4 bar ps=1 bar 0.6 ps=0.8 bar ps=0.6 bar ps=0.5 bar p / p0 / p 0.4

0.2

•0.4 •0.2 0 0.2 0.4 x/L

Figure 5.41: Distribution of p/(p0)up versus x/L for diﬀerent values of the discharge or ambient pressure ps in the interval [0.486, 1.984] bar. 134 Quasi-1D compressible ﬂows in nozzles

2.5

1.5 Mach

0.5

0 •0.4 •0.2 0 0.2 0.4 x/L

Figure 5.42: Mach number distribution in the De Laval nozzle for ps strictly below (ps)sie = 0.486 bar.

Let us conclude this detailed analysis with the calculation of the ﬂow solution when ps = 0.8 bar that is a discharge pressure yielding a ﬂow regime such that a normal shock appears in the diverging part of the nozzle. The problem to solve is summarized in Fig.5.44: -• the nozzle geometry is supposed to be known as well as the reservoir conditions or total conditions upstream of the normal shock.

-• the ambient pressure is also known, equal here to ps = 0.8 bar. The exit pressure is equal to the ambient pressure : pe = ps = 0.8 bar.

-• the normal shock is assumed to take place in a section Ash or Ashock which must be determined. The section A(x) is assumed to decrease linearly between Ai and Athroat and to increase linearly between Athroat and Ae. Knowing Ash yields immediately the shock location xsh. As previously done when analyzing the flow in a diverging nozzle, the flow upstream of the shock and the flow downstream of the shock are linked by expressing the mass conservation principle which reduces for the quasi-1D flow under study to the conservation of the mass flowratem ˙ . For a started nozzle, this mass flowrate is given by :

p0 m˙ = √ ΓAthroat rT0

5 with Γ = 0.6847 for γ = 1.4. A numerical application with p0 = 2 bar = 2 × 10 P a and T0 = 300 K yieldsm ˙ = 46.67 kg/s. 5.5. Analysis of ﬂow regimes in a converging-diverging nozzle 135

0.8

0.6 p/p0 0.4

0.2

0 •0.4 •0.2 0 0.2 0.4 x/L

Figure 5.43: Pressure distribution p/(p0)up in the De Laval nozzle for ps strictly below (ps)sie = 0.486 bar.

Figure 5.44: Overview of the available information for the calculation of the shock position in the diverging part of the nozzle (with ps = 0.8 bar). 136 Quasi-1D compressible ﬂows in nozzles

The mass ﬂowrate can also be computed using the ﬂow state in the exit section :m ˙ = ρeueAe. As explained in the case of a diverging nozzle, this identity can be turned into a second-order 2 equation for the unknown variable X = Me . Selecting the positive root of this equation yields :

v us u m˙ 2T t 0 Me = 1.5811 1 + 164 2 2 − 1 psAe where γ = 1.4 and r = 287 J/kg/K have been used to compute the numerical factors. In the case 2 under study,m ˙ = 46.67 kg/s, ps = 0.8 bar, T0 = 300 K and Ae = 0.5 m so that the above formula yields Me = 0.287. This value is indeed the one reported at the exit in the plot of Fig. 5.45.

2.5

1.5 Mach

0.5

0 0 0.2 0.4 x/L

Figure 5.45: Mach number distribution in the De Laval nozzle for ps = 0.8 bar.

Introducing the total pressure (p0)down constant in the isentropic ﬂow downstream of the 1D shock, we can express the ratio between this downstream total pressure and the exit static presure pe = ps as a function of the exit Mach number :

γ (p0)down (p0)down γ − 1 2 γ − 1 = = 1 + Me pe ps 2

2 Replacing Me with its previously established expression and explicting the value of γ (γ = 1.4) and r (r = 287 J/kg/K) for air, the following formula is obtained for the total pressure loss 5.5. Analysis of ﬂow regimes in a converging-diverging nozzle 137 through the 1D shock :

 3.5 s 2 (p0)down ps m˙ T0 = × 0.0884 × 1 + 1 + 164 2 2  (p0)up (p0)up psAe which allows to compute the total pressure ratio as a function of the known quantities ps,(p0)up, m˙ , T0 and Ae. (p ) An immediate numerical application yields 0 down = 0.4236. From Table I (Shock rela- (p0)up tionships) in Appendix A, the corresponding upstream Mach number Msh,up = 2.70 is computed. Turning now to Table II (Isentropic flow in a varying-section nozzle) the section area ratio Ash/(A∗)up = 3.183 is computed so that Ashock = 3.183 × (A∗)up = 3.183 × (A∗)throat = 2 2 2 0.3183 m . Since A(x) is varying linearly with x, from Athroat = 0.1 m at x = 0 m to Ae = 0.5 m at x = 1 m we can write : 2x A(x) = A + (A − A ) × col e col L where L = 2 m is the total length of the nozzle. It follows that : 2x A = A + (A − A ) × choc choc col e col L hence x choc = 0.273 L It can be checked in Figure 5.45 the shock location is indeed consistent with this prediction. The close-up provided in Fig.5.46 further confirms this good agreement. Note on this close-up the computed shock is not stricly speaking a discontinuity : it is spread on a few grid cells of the discretization grid used for the simulation yielding this numerical simulation. This (very limited) spreading is due to the numerical or artificial dissipation used in the numerical solution of the quasi-1D Euler equations.

The shock location is computed again in the case where ps = 1.4 bar in order to get used to the methodlogy which has just been used in the case where ps = 0.8 bar. Since the nozzle is started for ps = 1.4 bar, the mass ﬂowrate is equal to the previously computed (maximum) value that is 2 m˙ = 46.67 kg/s. Since T0 = 300 K and Ae = 0.5 m for the problem under study, whatever the ambient pressure ps, the formula giving the ratio (p0)aval/(p0)amont can also be expressed as a function of ps only : s !3.5 (p0)down ps 0.042865 = × 0.0884 × 1 + 1 + 2 (p0)up (p0)up ps with ps and (p0)up expressed in bar (and not in P a). Thus, for ps = 1.4 bar, we obtain : (p ) 0 aval = 0.7135 (p0)amont

We can deduce from the shock table I that Mamont = 2.015 and from Table II we ﬁnd Ashock/Athroat = 1.709 (after a linear interpolation in the table). Consequently Ashock = 0.1709 and since 2x A(x ) = A = A + (A − A ) × shock shock shock throat e throat L 138 Quasi-1D compressible ﬂows in nozzles

2.5

1.5 Mach 1

0.5

0.23 0.24 0.25 0.26 0.27 0.28 0.29 0.3 0.31 0.32 0.33 0.34 x/L

Figure 5.46: De Laval nozzle. Mach number distribution for ps = 0.8 bar : close-up on the shock region.

we eventually obtain xshock/L ≈ 0.09. It can be checked in Fig.5.40 or 5.41 that this prediction is in excellent agreement with the numerical solution of the quasi-1D Euler equations in this De Laval nozzle.

5.6 Exercises and problems

5.6.1 Exercise # 1 : Flow in a choked nozzle

Questions Let us consider a De Laval nozzle with the geometric and operating features displayed in Fig. 5.47. The throat section area is known and so is the exit section area. The reservoir pressure is also known as well as the ambient or discharge pressure, indicated as pa in the ﬁgure but denoted ps = 1 bar from now on. We will also assume, if needed, the total temperature in the reservoir is T0 = 300 K.

• Describe as precisely as possible the ﬂow in the nozzle.

Answers

As a preliminary step, we can compute the key values of the ambient pressure : (ps)start, (ps)sie and (ps)adapt so as to determine the ﬂow regime occurring in the nozzle. Let us start with the calculation of (ps)adapt. This calculation supposes the nozzle is started and the ﬂow isentropic from the inlet to the exit section included. Therefore, the sonic section A∗ 2 is known since equal to the throat section Athroat = 1 cm . The ratio between the exit section 5.6. Exercises and problems 139

2 Ae = 11.71 cm

p = 5 bar 2 0 Ac = 1 cm p a = 1 bar col

Figure 5.47: Characteristics of the De Laval nozzle under study. Ac stands for the throat section Athroat.

and the sonic section is such that Ae/A∗ = 11.71. Assuming an isentropic ﬂow in the nozzle and a supersonic ﬂow in the exit section (isentropic expansion, subsonic in the converging part and supersonic in the diverging part), the supersonic Mach number corresponding to this value in Table II is Me = 4.10. The ratio between the exit pressure and the total pressure p0 is thus given by : pe γ − 1 γ 2 − γ−1 −2 = (1 + Me ) = 0.5769 × 10 p0 2

Since p0 = 5 bar it follows that pe = 0.028845 bar. The exit pressure computed with this assumption of isentropic expansion from the inlet to the outlet section of the De Laval nozzle is the pressure (ps)adapt, thus (ps)adapt = 0.028845 bar.

The other key value for ps is (ps)start. It is computed assuming again the sonic state is reached at the throat so that Ae/A∗ = 11.71. But in that case, the ﬂow becomes subsonic again in the diverging part of the nozzle so that the corresponding value of Me is obtained by reading the subsonic part of Table II. It comes Me ≈ 0.05 so that :

pe γ − 1 γ 2 − γ−1 = (1 + Me ) = 0.998 p0 2 hence pe = 4.99 bar in that case. The value of (ps)start is thus 4.99 bar, very close to the stagnation pressure because of the large ratio Ae/Athroat.

Finally the value (ps)sie (shock in the exit section) is computed using the data already gathered to compute (ps)adapt. Indeed, (ps)sie is the downstream pressure of a 1D shock with the upstream state deﬁned by pe = 0.028845 bar and Me = 4.10. Since the static pressure jump is such that :

(ps)sie 2γ 2 = 1 + (Me − 1) pe γ + 1 the numerical application yields (ps)sie/pe = 19.445 so that (ps)sie = 19.445 × 0.028845 = 0.56 bar.

Since the actual ambient pressure ps is equal to 1 bar, it is below (ps)start but above (ps)sie so that the ﬂow regime in the nozzle (see Fig.5.39) corresponds to a shockwave in the diverging 140 Quasi-1D compressible ﬂows in nozzles part of the nozzle.

In order to compute the area of the section where the shock is located, we compute first the (maximum) mass flowrate for the choked flow through the nozzle :

p0 m˙ = √ ΓAthroat rT0

5 −4 2 Since Γ = 0.6847 for γ = 1.4, p0 = 5 × 10 P a, Athroat = 10 m , r = 287 J/kg/K and T0 = 300 K, it comes :m ˙ = 0.1167 kg/s.

The total pressure ratio through the shock is such that :

 3.5 s 2 (p0)down ps m˙ T0 = × 0.0884 × 1 + 1 + 164 2 2  (p0)up (p0)up psAe

5 5 A numerical application with ps = 1 bar = 10 P a,(p0)up = 5 bar = 5 × 10 P a, T0 = 300 K, −4 2 m˙ = 0.1167 kg/s and Ae = 11.71 × 10 m yields (p0)down/(p0)up = 0.2086. The corresponding upstream Mach number Msh,up is obtained from Table I and found such that Msh,up ≈ 3.5. Note that this value is consistent with the value Me = 4.10 which would be reached at the exit if the supersonic isentropic solution applied everywhere in the diverging part of the nozzle. Looking up 2 Table II provides Ashock/(A∗)up = 6.7896 for Msh,up = 3.5. Since (A∗)up = Athroat = 1 cm , it 2 comes Ashock ≈ 6.79 cm (a plausible value, between Athroat and Ae). Since the law of variation for A(x) is not provided, we cannot go further and identify the shock location xshock such that A(xshock) = Ashock.

5.6.2 Exercise #2 : Flow in a diverging nozzle

Questions Let us consider the diverging nozzle with the geometric and operating features displayed in Fig. 5.48.

• Determine the Mach number M2 and M3 in sections 2 and 3, respectively immediately upstream and downstream of the 1D shock occuring in the diverging nozzle, the exit Mach number Me and the ratio pe/p1 where p1 is the static pressure in the inlet section.

Answers

The flow upstream of the shock is both supersonic and isentropic : it accelerates from the inlet section to the shock. Let us denote (A∗)up the (virtual) sonic or critical section associated with this isentropic flow. The flow downstream of the shock is both subsonic and isentropic : it decelerares from the shock to the exit section. Let us denote (A∗)down the (virtual) sonic section associated with this isentropic flow. The Mach number M2 can be computed from the section area ratio A2/(A∗)up, such that :

A2/(A∗)up = Achoc/(A∗)up = 2A1/(A∗)up 5.6. Exercises and problems 141

choc

A3 A1 A2

M 1=3 A = 3 A e 1

A = 2 A choc 1

Figure 5.48: Key features of the diverging nozzle under study. looking at Table II in its supersonic part. Since M1 is known (M1 = 3), the ratio A1/(A∗)up is know from Table II : A1/(A∗)up = 4.234569 so that A2/(A∗)up = 2 × 4.234569 = 8.46914. Interpolating in the supersonic part of Table II yields M2 = 3.738. The subsonic Mach number M3 is obtained either from the 1D shock relationships (see formula (4.17) with M2 the upstream Mach number and M3 the downstream Mach number) :

γ − 1 2 1 + ( )M2 M 2 = 2 3 γ − 1 γM 2 − ( ) 2 2 or from Table I. In both cases, we obtain M3 = 0.4427. Since M3 is known, we can compute A3/(A∗)down. Using Table II yields A3/(A∗)down = 1.4672. Ae The (subsonic) Mach number Me in the exit section is computed from : (A∗)down A A A 3A e = e 3 = 1 × 1.4672 = 1.5 × 1.4672 = 2.2 (A∗)down A3 (A∗)down 2A1

Reading the subsonic part of Table II then yields Me = 0.275. To compute the static pressure ratio pe/p1, let us pressure ratios which we know how to compute that is, typically, static to total pressure ratio (or total to static). Since sections 1 and e (exit section) are on both sides of the shockwave, the total pressure associated with each of these sections is not the same. Let us denote (p0)up the total pressure associated with the isentropic flow upstream of the shock and (p0)down the total pressure associated with the isentropic flow downstream of the shock. We can write : p p (p ) (p ) e = e · 0 down · 0 up p1 (p0)down (p0)up p1 The first ratio and the third ratio are directly known from the respective local Mach number γ γ−1 2 γ−1 Me and M1. Denoting f(M) = (1 + 2 M ) , we can write :

pe 1 (p0)up = and = f(M1) (p0)down f(Me) p1 142 Quasi-1D compressible ﬂows in nozzles

The second ratio can be directly read in the shock table I of Appendix A or computed from the Mach number M2 and M3. Indeed : (p ) (p ) p p 0 down = 0 down · 3 · 2 p0)up p3 p2 (p0)up

We can use again : p2/(p0)up = 1/f(M2) and (p0)down/p3 = f(M3). The static pressure ratio p3/p2 is given by the jump relationship (4.19) with M2 the upstream Mach number: p3 2γ 2 = 1 + (M2 − 1) = g(M2) p2 γ + 1 Gathering these various contributions we obtain :

pe 1 1 = · f(M3) · g(M2) · · f(M1) p1 f(Me) f(M2)

pe After calculation, the compression ratio of the diverging nozzle is found equal to = 6.045. p1

5.6.3 Problem # 1 : analysis of a scramjet performance A ramjet is a propulsion device which works in supersonic regime, compressing the air entering the air inlet thanks to shockwaves - thus avoiding the use of rotating parts as done in conven- tional turbojet. Fig. 5.49 displays the simplified configuration studied in the problem : the ramjet is made of an air inlet, a combustion chamber and an ejection nozzle. It is more precisely a scramjet because the combustion takes place in supersonic regime, the flow still being supersonic when entering the combustion chamber. The present problem is devoted to a rough study of such a scramjet system.

Chambre de Entree d’air Tuyere d’ejection combustion

0 1 2 3 4

Figure 5.49: Schematic view of the scramjet studied in the problem. ”Entréed’air” stands for ”Air inlet”. ”Chambre de combustion” stands for ”Combustion chamber”. ”Tuyèred’éjection” stands for ”Ejection nozzle”.

Assumptions : The flow is supersonic in the whole scramjet. It is described as a 1D, steady flow with viscous effets neglected everywhere. The fluid flowing through the scramjet is successively air and a mixture of air and fuel. To keep the analysis simple, the fluid is systematically described as an 5.6. Exercises and problems 143 ideal gas with constant specific heats, such that γ = 1.4 and r = 287 J kg−1 K−1.

Section 0 in Fig. 5.49 corresponds to the following upstream conditions : Mach number M0 = 11 (yes, you read correctly), static pressure p0 = 900 P a, static temperature T0 = 230 K. Note that in this problem the notations p0 and T0 are used to denote it static quantities (pressure and temperature) in the section indexed 0. The total pressure associated with the state in section 0 will be denoted (p0)0. The section area of the streamtube captured by the air inlet is such that 2 A0 = 5 m .

Between section 0 and section 1, located downstream of the inlet throat, the flow is slowed down by a series of oblique shocks (which are not detailed here and not displayed in Fig. 5.49). This series of reflecting oblique shocks induces a total pressure loss for the flow, such that (p0)1/(p0)0 = 0.5. The Mach number in section 1 is M1 = 4.5.

2 Between sections 1 and 2 the ﬂow is isentropic; the area of section 2 is such that A2 = 0.25 m .

The ﬂow evolution through the scramjet and the available data are summarized in Fig. 5.50.

Figure 5.50: Flow evolution in the scramjet and available data. ”Chocs obliques” stands for ”Oblique shocks” while ”Ecoulement isentropic” stands for ”Isentropic ﬂow”.

1) • Compute the ﬂow properties (static pressure, temperature, total pressure and total temperature, density, velocity, Mach number) in sections 0, 1 and 2 . • Give the mass ﬂowratem ˙ passing through the scramjet.

2) Let us consider now the combustion chamber. In this chamber, the fuel injection and its re- action with air produce a positive heat contribution to the flow such that ∆Q = 1.5×106J kg−1. The combustion chamber displays a diverging variable section, designed so as to ensure a quasi- uniform pressure in the chamber : the pressure increase coming from the combustion is balanced by the pressure decrease of the supersonic flow in the diverging nozzle. • Using these assumptions, write the mass, momentum and total energy balance between sections 2 and 3. • Deduce from these conservation laws the flow properties in section 3 (p3, T3,(p0)3,(T0)3, u3, M3) as well as the area A3 of section 3.

3) At the outlet of the combustion chamber, the flow is accelerated in an isentropic way through 2 a diverging nozzle with outlet section A4 = 5 m . • Compute the flow conditions in section 4. 144 Quasi-1D compressible flows in nozzles

4) The thrust T of the scramjet is given by :

T =m ˙ (u4 − u0) + (p4 − p0)A4.

The scramjet propulsive efficiency ηp is defined as : thrust power η = . p rate of heat added to the flow

• Give the expression of ηp as a function ofm ˙ , v0, T and ∆Q. • Compute ηp for the scramjet under study. 5.6. Exercises and problems 145

Answers 1) • Compute the ﬂow properties (static pressure, temperature, total pressure and total temperature, density, velocity, Mach number) in sections 0, 1 and 2 . • Give the mass ﬂowratem ˙ passing through the scramjet.

Section 0 : The identities T γ − 1 0 = 1 + M 2 T 2 and γ p T γ−1 0 = 0 p T can be applied in section 0 where the Mach number M0 is known, as well as the static pressure and temperature. It is thus possible to compute the upstream total pressure and total temperature : 6 (T0)0 = 5796 K,(p0)0 = 72.301 × 10 P a. Density is obtained from the ideal gas thermal equation of state :

p0 = ρ0rT0

−2 3 hence ρ0 = 1.3634 × 10 kg/m . The velocity u0 is obtained from : p u0 = a0M0 = γrT0M0 = 3343.97 m/s

Since ρ0, u0 and A0 are known, it is possible to compute the mass ﬂowratem ˙ 0 in section 0, which will be also the constant mass ﬂowratem ˙ throughout the scramjet :

m˙ 0 = ρ0u0A0 = 227.96 kg/s =m ˙

Section 1 : Since the loss of total pressure between section 0 and section 1 is known, we can 1 6 compute the total pressure associated with the flow in section 1 : (p0)1 = 2 (p0)0 = 36.15×10 P a. Since M1 = 4.5 and since γ γ−1 (p0)1 γ − 1 2 = 1 + M1 p1 2 5 it comes p1 = 1.249 × 10 P a. One can check that p1 > p0 : the flow is compressed between section 0 and section 1. The flow being adiabatic between sections 0 and 1, the total temperature is conserved : (T0)1 = (T0)0 = 5796 K. Since (T0)1 γ − 1 2 = 1 + M1 T1 2 it comes T1 = 1147.7 K. 3 The thermal EoS yields ρ1 = p1/(rT1) =√ 0.3792 kg/m . The velocity is such that u1 = a1M1 = γrT1M1 = 3055.9 m/s. The area of section 1 can be computed from the mass flowrate :

m˙ 1 = ρ1u1A1 =m ˙

2 hence A1 = 227.96/(0.3792 × 3055.9) = 0.1967 m . 146 Quasi-1D compressible ﬂows in nozzles

Section 2 : The flow being assumed isentropic between sections 1 and 2, the total pressure is 6 conserved : (p0)2 = (p0)1 = 36.15 × 10 P a and the total temperature remains also constant : (T0)2 = (T0)1 = (T0)0 = 5796 K. However, in order to compute the static pressure and the static temperature, respectively from the total pressure and the total temperature, the Mach number M2 is needed. Since section A2 is known, this Mach number can be computed provided the sonic section associated to the flow in section 2 is known. This sonic section, denoted (A∗)down, is in fact associated to the isentropic flow downstream of the oblique shock system taking place between sections 0 and 1. Therefore, we can write the Mach-area relationship in section 1 as A1/(A∗)down = f(M1) and use Table II of Appendix A to find the section area ratio corresponding to M1 = 4.5 : A1/(A∗)down = 16.56 −2 2 hence (A∗)down = 0.1967/16.56 = 1.188 × 10 m .

2 −2 2 Since A2 = 0.25 m and (A∗)down = 1.188 × 10 m , it comes A2/(A∗)down = 21.044. The supersonic ﬂow in section 1 accelerates in the diverging nozzle between section 1 and section 2 so that reading the supersonic part of Table II yields M2 ≈ 4.8. We can now compute the static pressure and temperature in section 2 :

−γ γ − 1 γ−1 p = (p ) × 1 + M 2 = 8.65 × 104 P a 2 0 2 2 2

We can check that p2 < p1 in agreement with the fact a supersonic expansion takes place between sections 1 and 2. Finally :

γ − 1 −1 T = (T ) × 1 + M 2 = 1033.5 K 2 0 2 2 2

3 The density is computed from the thermal EoS which yields ρ2 = 0.2916 kg/m . The velocity is obtained again from the Mach number and the temperature (giving the speed of sound) so that u2 = 3093 m/s.

2) • Write the mass, momentum and total energy balance between sections 2 and 3. • Deduce from these conservation laws the ﬂow properties in section 3 (p3, T3,(p0)3,(T0)3, u3, M3) as well as the area A3 of section 3.

The momentum balance written in integral form for a steady inviscid ﬂow on a ﬁxed control volume with a closed surface Σ reads :

ρU~ (U~ · ~n) dS = − p~ndS ˆΣ ˆΣ Since the pressure p is assumed uniform in the combustion chamber, this pressure uniformity implies : p~ndS = p ~ndS = 0 ˆΣ ˆΣ since Σ ~ndS = 0 when Σ is a closed surface. Since´ there is no ﬂow through the wall of the nozzle, the momentum balance reduces to :

ρU~ (U~ · ~n) dS = 0 ˆS2∪S3 5.6. Exercises and problems 147

with S2 (resp. S3) the section 2 (resp. 3) with area A2 (resp. A3). The momentum balance in the combustion chamber can be ﬁnally expressed as :

2 2 ρ2u2A2 = ρ3u3A3

Since ρ2u2A2 = ρ3u3A3 =m ˙ from mass conservation, it follows that u3 = u2 = 3093 m/s. 4 Since p is constant in the combustion chamber, we have also p3 = p2 = 8.65 × 10 P a. The energy balance is readily obtained using the general results established in 3.1.5 for a continuous flow system. Since there is no moving surface in the combustion chamber, W˙ 0 = 0 and therefore ∆H = Q˙ = ∆Q with ∆H = H3 − H2 the variation of total enthalpy between the inlet and the outlet of the combustion chamber and ∆Q the heat brought by the combustion (expressed for a unit mass since H is a specific total enthalpy). 1 2 Since the total enthalpy H is such that H = CpT + 2 u , we can write : 1 1 C T + u2 = C T + u2 + ∆Q p 3 2 3 p 2 2 2 or else, since u3 = u2 : ∆Q T3 = T2 + Cp −1 −1 Since Cp = 1004.5 J kg K ,√ it comes T3 = 1033.5 + 1493.28 = 2526.78 K. From u3 = 3093 m/s and a3 = γrT3 = 1007.6 m/s, we can compute M3 = 3.07. Using the Mach number value in section 3 and the values of the static pressure and temperature, the total pressure and total temperature can be computed : (T0)3 = 7290.3 K and 6 (p0)3 = 3.525 × 10 P a. 3 The density is obtained from p3 and T3 : ρ3 = p3/(rT3) = 0.119 kg/m . 2 The mass flowrate conservation yields A3 =m/ ˙ (ρ3u3) = 0.613 m .

3) • Compute the ﬂow conditions in section 4.

Between section 3 and section 4 the flow expands isentropically in a diverging nozzle. Since the area A4 of section 4 is known, the Mach number M4 in this section can be computed if the sonic section (A∗)out associated to this exit section is known. Since the flow has experienced a non-adiabatic transformation in the combustion chamber, between section 2 and section 3, the sonic section (A∗)out differs from the sonic section (A∗)down which was previously computed for the isentropic flow between sections 1 and 2, upstream of the combustion chamber. The sonic or critical section (A∗)out is also associated to section 3 and can be therefore computed from M3 and A3 which are both known. Alternatively, we can directly write : A A A 4 = 4 × 3 A∗ A3 A∗

The first ratio in the RHS can be computed since both A3 and A4 are known while the second ratio can be obtained from Table II since M3 is known. It comes : A 5 4 = × 4.5 = 36.7 A∗ 0.613 Since the flow in section 4 is supersonic (the flow expands between 3 and 4), reading the supersonic part of Table II yields : M4 = 5.5. 148 Quasi-1D compressible flows in nozzles

The ﬂow between sections 3 and 4 being isentropic, it follows that (p0)4 = (p0)3 and (T0)4 = (T0)3 so that, using these total values and the Mach number, we can compute the static quantities p4 = 3789 P a, T4 = 1034 K. Computing the speed of sound yields a4 = 644.5 m/s and using M4 yields u4 = 3545 m/s.

4) • Give the expression of the scramjet propulsive eﬃciency ηp as a function ofm ˙ , v0, T and ∆Q. • Compute ηp for the scramjet under study.

Since the thrust T of the scramjet is given by :

T =m ˙ (u4 − u0) + (p4 − p0)A4. and since the scramjet propulsive efficiency ηp is defined as : thrust power η = . p rate of heat added to the flow it is straightforward to define :

-• thrust power (in W ): T × u0 -• heat added to the ﬂow, per unit time, that is expressed in J/s = W :m ˙ × ∆Q since ∆Q is heat added to the ﬂow per unit mass.

Therefore, the propulsive eﬃciency reads : T × u η = 0 p m˙ ∆Q

Sincem ˙ , u0, u4, p0, p4 and A4 are known, the thrust can be computed : T = 59769.3 N. T u0 Hence the value of the propulsive eﬃciency : ηp = m˙ ∆Q = 0.59.