Fundamentals of compressible and viscous flow analysis - Part II

Transcription

1 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

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3 Contents Contents 2 3 Energy conservation Deriving the energy conservation equation Integral or global form Local form Non-conservative global form Kinetic energy and internal energy conservation equations Case of a continuous-flow system Energy conservation equation for an ideal fluid and Bernoulli equation Flow modeling : provisional overview Thermodynamic analysis Reversible process and entropy Equations of state Equations of state for a calorifically perfect gas Conclusions on the flow governing equations Final overview of the governing equations Velocity measurement in a compressible flow Exercises and problems Exercise # 1 : Analysis of a steam turbine Exercise # 2 : Analysis of the draining of a tank Exercise # 3 : Analysis of the compressible flow in a Venturi Problem # 1 : pressure rise in a nuclear reactor containment D compressible flow analysis Some key quantities Sound speed and Mach number Exercise : computing the speed of sound Stagnation quantities or total quantities Sonic quantities D shockwaves Problem definition

4 4 CONTENTS Prandtl s relationship Physical considerations Shock relationships Total quantities and 1D shockwaves Rankine-Hugoniot s relationship Exercises and problems Exercise 1 : Mach number and compressibility effects Exercise # 2 : pressure and temperature at the nose of an airplane in transonic flight Problem # 1 : velocity measurement in a supersonic flow Quasi-1D compressible flows Physical model Governing equations Velocity-area relationship Influence coefficients Isentropic flow in a nozzle Flow analysis Mass flowrate Analysis of the flow through a converging nozzle General analysis Example of application Analysis of the flow through a diverging nozzle Analysis of flow regimes in a converging-diverging nozzle General analysis Example of application Exercises and problems Exercise # 1 : Flow in a choked nozzle Exercise #2 : Flow in a diverging nozzle Problem # 1 : analysis of a scramjet performance

5 Lecture 3 Energy conservation Contents 3.1 Deriving the energy conservation equation Integral or global form Local form Non-conservative global form Kinetic energy and internal energy conservation equations Case of a continuous-flow system Energy conservation equation for an ideal fluid and Bernoulli equation Flow modeling : provisional overview Thermodynamic analysis Reversible process and entropy Equations of state Equations of state for a calorifically perfect gas Conclusions on the flow governing equations Final overview of the governing equations Velocity measurement in a compressible flow Exercises and problems Exercise # 1 : Analysis of a steam turbine Exercise # 2 : Analysis of the draining of a tank Exercise # 3 : Analysis of the compressible flow in a Venturi Problem # 1 : pressure rise in a nuclear reactor containment Deriving the energy conservation equation Integral or global form Case of a material fluid domain The first principle of thermodynamics applied to the material domain D m (t) states the following physical principle :

6 6 Lecture #3 : Energy conservation The rate of variation of the total energy contained in D m (t) is equal to the sum of the work (per unit time) done by the external (surface and volume) forces applied to D m (t) and the heat quantity transferred through S m (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 to 1 2 U 2. The total specific energy of the fluid particle is given by : E = e U 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 : ρedv = ρedv ρ U 2 dv which makes clear ρe is a total energy per unit volume. The total energy contained or stored in the material domain D m (t) is computed as : ρedv D m(t) The rate of variation of this total energy is expressed as : D Dt ( ρedv ) and the energy conservation reads : D Dt ( D m(t) D m(t) ρedv ) = Ẇ + Q (3.1) where Ẇ and Q are respectively the work done, per unit time, by the external forces applied to D m (t) and the heat transferred through S m (t), per unit time. Using the Reynolds transport theorem (??) with χ = E, the LHS of (3.1) can be turned into : D ρe dv = ρ DE Dt Dt dv D m(t) D m(t)

7 3.1. Deriving the energy conservation equation 7 so that (3.1) can also be expressed as : D m(t) ρ DE Dt dv = Ẇ + Q (3.2) 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 : D m(t) (ρe) dv + ρe ( U t n) ds = Ẇ + Q (3.3) S m(t) Case of an arbitrary control volume In the case of an arbitrary control volume, the first law of thermodynamics applied to D a (t) states D Dt ( ρedv ) = Ẇ + Q ρe ( U W ) n ds D a(t) S a(t) where the last term in the RHS corresponds to the convective flux of total energy through the surface S a (t). Using again the Reynolds transport theorem to transform the LHS of the above equation yields : D a(t) ρ DE Dt dv + S a(t) ρe ( W U) n ds = Ẇ + Q ρe ( U W ) n ds S a(t) After simplification the non-conservative form (3.2) is recovered for an arbitrary control volume : D a(t) ρ DE Dt dv = Ẇ + Q (3.4) Developing the material derivative of the total energy and using mass conservation yields : ρ DE Dt = D(ρE) Dt E Dρ Dt = (ρe) t = (ρe) t + U (ρe) + ρe U + (ρe U) Using the flux-divergence theorem to transform the volume integral of (ρeu) eventually yields : (ρe) dv + ρe ( U V a(t) t n) ds = Ẇ + Q (3.5) S a(t) which generalizes (3.5) previously established for a material domain.

8 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 : Ẇ = T ( n) UdS + ρf UdV S D The stress vector T is such that (see (??)) : hence Ẇ = S T = σ n = p n + τ n pu n ds + ( τ n) U ds + ρf UdV S D The work done, per unit time, by the viscous forces can be rewritten using : ( τ n) U = τij n j U i = τ ji U i n j = ( τ U)j n j = ( τ U) n hence Ẇ = S pu n ds + ( τ U) n ds + ρf UdV (3.6) S D 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 : Q = q n ds = k T n ds = k T ds (3.8) S S S n Inserting (3.6) and (3.8) into (3.5) yields the following global or integral form of the enery conservation equation : (ρe) dv + ρe ( U t n) ds = pu n ds + ( τ U) n ds + ρf UdV + k T S S S D S n ds D

9 3.1. Deriving the energy conservation equation 9 Let us introduce the specific enthalpy h : h = e + p ρ The total enthalpy H can be defined from the specific enthalpy as : H = h U 2 Total enthalpy and total energy are related as follows : H = h U 2 = e + p ρ U 2 = E + p ρ Using total energy and total enthalpy, the above global form of the energy conservation equation can be rewritten as : D (ρe) t dv + ρh ( U n) ds = ( τ U) n ds + ρf UdV + k T ds (3.9) S S D S 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 = (ρψ U) + Ψ (ρ U) = (ρψ U) Ψ ρ t = (ρψu) (ρψ) t where the mass conservation has been used ( ρ t + (ρu) = 0) as well as the fact the gravity potential Ψ does not depend on the time t. The power of the body forces can thus be rewritten as : D ρf UdV = or else, using the flux-divergence theorem : ρf UdV = Inserting this identity into (3.9) yields : (ρe) (ρψ) dv + dv + ρh ( U t t n) ds+ D D D S D (ρψu) dv S D (ρψ) t dv ρψu (ρψ) n dv dv D t S ρψu n dv = ( τ U) n ds+ S S k T n ds

10 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) : E = e U 2 + Ψ = e U 2 + gz Consequently, the total enthalpy H also includes now this potential energy : H = E + p ρ = e + p ρ U 2 + Ψ = h U 2 + Ψ = h U 2 + gz With these extended definitions of E and H, the global form of the energy conservation equation reads : (ρe) dv + ρh ( U t n) ds = ( τ U) n ds + k T ds (3.10) S S S n D 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 : Local form D (ρe) t dv + ρh ( U n) ds = 0 (3.11) S 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 : D (ρe) t dv + D (ρh U) dv = D ( τ U) dv + Hence the local conservative vector form for the total energy conservation : (ρe) t D (k T ) dv + (ρh U) = ( τ U) + (k T ) (3.12) 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) t + (ρh U) = ( τ U) + k 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) t + (ρh U) = 0 (3.13)

11 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) t + (ρh U) = (ρe) t + (ρe U) + (p U) = ρ E t + E ρ t + ρ U E + E (ρu) + (p U) = ρ DE ( ) ρ Dt + E t + (ρu) + (p U) }{{} =0 = ρ DE Dt + (p U) In the case of an ideal fluid, the non-conservative form of the total energy conservation equation reads : ρ DE Dt = (p U) (3.14) 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 : or else, since (p U) = (p 1 U) : ρ DE Dt = (p U) + ( τ U) + (k T ) (3.15) ρ DE Dt = ( σ U) + (k T ) with σ the Cauchy stress tensor used to express the external surface forces (pressure forces and viscous forces) Non-conservative global form The previous non-conservative local form can be integrated over an arbitrary fluid domain D a (t) (with surface S a (t) moving at velocity W ) to yield : ρ DE Dt dv = ( σ U) dv + (k T ) dv D a(t) D a(t) Using again the Reynolds transport theorem (??) to transform the LHS of the above equation yields : ρ DE D a(t) Dt dv = D ρe dv ρe ( W Dt U) n ds D a(t) S a(t) It follows that : D ρe dv = Dt D a(t) D a(t) ( σ U) dv + D a(t) D a(t) (k T ) dv + ρe ( W U) n ds S a(t)

12 12 Lecture #3 : Energy conservation The first integral of the RHS can be rewritten using the flux-divergence theorem : ( σ U) dv = ( σ U) n ds D a(t) S a(t) Using the symmetry of the Cauchy stress tensor, it is possible to rewrite : ( σ U) n = ( σ U)i n i = σ ij U j n i = σ ji n i U j = ( σ n)j U j = ( σ n) U = T U with T the stress vector (pressure stress and viscous stress). Consequently : S a(t) ( σ U) n ds = S a(t) T U ds = P S ext with P S ext the power of the external surface forces applied to the fluid domain. Note, from its definition, the alternative form of P S ext : Remembering that : Pext S = D a(t) D a(t) (pu) dv + ( τ U) dv D a(t) (k T ) dv = S a(t) k T n ds = Q the non-conservative form of the total energy conservation equation reads : D ρe dv = Pext S + Dt Q + ρe ( W U) n ds D a(t) where it must be also remembered the total energy contains the potential Ψ associated with the body forces. For a material domain D m (t) ( W = U) it is possible to write : S a(t) ( D ρ e + 1 ) U Dt D m(t) Ψ dv = Pext S + Q (3.16) 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 ρ U Dt D m(t) Ψ dv + D ρe dv = Pext S + Dt Q D m(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.

13 3.1. Deriving the energy conservation equation 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 : The LHS is immediately rewritten as : ρu D U Dt = U p + U ( ) τ ρu Ψ ρ U D U Dt = ρ D Dt ( ) 1 U 2 2 while the power of the body forces can be recast, since Ψ does not depend on t, as : It follows that : ρ D Dt ρ U Ψ = ρ DΨ Dt ( ) 1 U Ψ = U p + U ( ) τ 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 Dt Ψ dv = U p ( ) dv + U τ dv Note that : D m(t) D m(t) U p = (p U) p U D m(t) Besides, one can also write : ( ) U τ τ ij = U i = (τ iju i ) U i τ ij = (τ jiu i ) U i τ ij = x j x j x j x j x ( τ U i U) τij j x j The term τ ij U i x j can be also expressed as : U i τ ij = 1 ( ) U i U j τ ij + τ ji x j 2 x j x i Using the symmetry of the viscous stress tensor τ, that is τji = τ ij, one can also write : U i τ ij = 1 ( ) ( U i U j 1 Ui τ ij + τ ij = τ ij + U ) j = τ ij D ij x j 2 x j x i 2 x j x i }{{} =D ij

14 14 Lecture #3 : Energy conservation The scalar product of two second-order tensors a and b is defined as : b a : = a ij b ij = a ij b ij using Einstein s notation on repeated indices. Therefore : τ ij U i x j = τ : D Going back to the initial development, it comes : i,j ( ) U τ = ( τ U) τ D : Replacing U p and U ( ) τ in the mechanical energy conservation equation with the above expressions yields : D Dt D m(t) ( ) 1 ρ U Ψ dv = (pu) dv + p U dv D m(t) D m(t) + ( τ U) dv D τ : dv D m(t) D m(t) The power of the external surface forces applied to the material domain Pext S is easily identified in the RHS so that the mechanical energy conservation equation can be also expressed as : ( ) D 1 ρ U Dt Ψ dv = Pext S + p U dv D τ : dv D m(t) The internal power P int is defined as : P int = D m(t) p U dv }{{} compression / expansion + D m(t) D m(t) τ : D dv }{{} viscous dissipation = D m(t) D m(t) σ : D dv (3.17) It is the power required to deform the fluid particles contained in the material domain. P int 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 = D τ : dv which is an always positive term, expressing D m(t) the heating of the flow through the degradation of mechanical energy into thermal energy.

15 3.1. Deriving the energy conservation equation 15 The mechanical energy conservation equation can therefore be summarized as : ( ) D 1 ρ U Dt Ψ dv = Pext S + P int (3.18) D m(t) Physically, (3.18) can also be interpreted in the following form : Pext S = D ( ) 1 ρ U Dt Ψ dv P int D m(t) 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, P int is strictly positive since : P int = p U dv + }{{} Φ (3.19) D } m(t) {{} viscous dissipation always >0 >0 for compression so that the external power P S ext 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. In the case where no external power P S ext 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 P int 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 = P int + Dt Q (3.20) D m(t) showing that the internal energy of the material domain D m (t) evolves because of the internal power P int, 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 Dt D m(t) [( ) 1 U Ψ ] + ρe dv = Pext S + P int P int + Q (3.21)

16 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 P int can be further detailed using definition (3.19) so that : [( ) ] D 1 U Dt D m(t) Ψ + ρe dv = Pext S + p U dv Φ p U dv + Φ + Q D m(t) D m(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)

17 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 U Dt D m(t) Ψ + ρe dv = Pext S Φ+Φ or equivalently : D ρe dv = Pext S Dt D m(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, P S ext = 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 U Dt D m(t) Ψ + ρe dv = Pext S + p U dv p U dv D m(t) D m(t) or equivalently : D ρe dv = Pext S Dt D m(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 power is supplied to or extracted from the flow, Pext S = 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 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 Dt U dv + Φ + Q D m(t) D m(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 : ρ De Dt dv = p U dv + D τ : dv + (k T ) dv D m(t) D m(t) D m(t) D m(t) The local form of the internal energy conservation equation is immediately obtained : ρ De Dt = p U + τ : D + (k T ) (3.23) Note that in the case of the incompressible flow of an ideal fluid (µ = 0 and k = 0), this equation reduces to : De Dt = 0 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 Case of a continuous-flow system Let us consider the continuous-flow system schematically displayed in Fig.3.2. The envelope surface Σ 0 is a solid wall supposed both fixed and adiabatic hence such that U = 0 and q n = 0 on Σ 0. The surface Σ q is a fixed heat transfer surface such that U = 0 and q n 0 on Σ q. The surface Σ m is a moving adiabatic solid surface hence such that U 0 and q n = 0. The

19 3.1. Deriving the energy conservation equation 19 no-slip condition for a viscous fluid states the velocity on Σ m 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 Σ m. 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 S 1 and leaves through section S 2. Σ 0 is a fixed adiabatic surface guiding the flow from S 1 to S 2. Σ q is a fixed non-adiabatic surface through which the flow can be heated or cooled. Σ m 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 S 1, the outlet section S 2 and the solid surfaces (adiabatic or not, fixed or not) Σ 0, Σ q, Σ m reads : (ρe) dv + ρh ( U t n) ds = ( τ U) n ds + k T n ds D S where the surface S of D is such that S = S 1 S 2 Σ 0 Σ q Σ m. Note the total energy includes the potentiel energy so that E = e U 2 + Ψ and H = E + p ρ. Assuming the flow steady and using ( τ U) n = ( τ n) U, we can rewrite : ρh ( U n) ds = ( τ n) U ds + k T n ds S S Since S = S 1 S 2 Σ 0 Σ q Σ m, the available boundary conditions on the various surfaces must be used so as to compute the LHS and RHS integrals. Since Σ 0, Σ q are fixed solid walls, U = 0 on these surfaces so that : ρh ( U n) ds = ρh ( U n) ds S S 1 S 2 Σ m 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 S S S

20 20 Lecture #3 : Energy conservation outlet sections S 1 and S 2. Therefore it remains to compute the RHS integrals for Σ 0 Σ m Σ q. Since only Σ q is a non-adiabatic solid surface, it follows that : k T S n ds = k T Σ q n ds Finally, the surfaces Σ 0 and Σ q being fixed, the power of the viscous forces reduces to : ( τ n) U ds = ( Σ τ n) U ds m S The total energy conservation equation eventually reads : ρh ( U n) ds = ρh ( U n) ds + ( τ n) U ds S 1 S 2 Σ m Σ }{{ m } =Ẇ + Σ q k T n ds } {{ } = Q where Ẇ 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 Σ m and where Q denotes the quantity of heat supplied or extracted through Σ q. Let us denote N 1 the unit normal vector to S 1 aligned with the inflow direction and N 2 the unit normal vector to S 2 aligned with the outflow direction. Obviously, n = N 1 on S 1 and n = N 2 on S 2. Let us also introduce the mass flow rate through section S i (i = 1 or i = 2) : ṁ i = ρ( U N i ) ds S i If the flow is assumed quasi-uniform in the inlet and outlet sections, it is possible to write : S i ρh( U N i ) ds H i ṁ i where H i denotes the average value of the total enthalpy in section S i. Therefore, the total energy balance can be expressed in the form : Denoting ( ) = ( ) 2 ( ) 1, we can also write : H 2 ṁ 2 H 1 ṁ 1 = Ẇ + Q (ṁh) = Ẇ + 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 : [ṁ(e + p ρ U 2 + gz)] = Ẇ + Q

21 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 : [ṁh t ] = 0 Moreover, mass conservation ensures in that case the conservation of the mass flowrate hence ṁ 1 = ṁ 2 = ṁ so that (ṁh) = ṁ H. The total enthalpy for the passive (Σ m = ) and adiabatic (Σ q = ) is therefore conserved between the inlet and outlet of the system so that : ( e + p ρ + 1 ) ( U gz = e + p 1 ρ + 1 ) U gz 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 : or, in the local form (3.13) : D (ρe) t (ρe) t dv + ρh ( U n) ds = 0 S + (ρh U) = 0 Note that in these expressions E = e U 2 + 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 ρ U H + H (ρ U) = 0 Using mass conservation for a steady (compressible) flow, (ρ U) = 0 and identifying the convective derivative with the material derivative for a steady flow yields : DH Dt = 0 The total enthalpy is constant along a streamline in a steady flow of an ideal fluid. In other words : e + p ρ U 2 + gz = cst along a streamline (3.24) 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 : ρ De Dt = p U + τ : D + (k T )

22 22 Lecture #3 : Energy conservation For an ideal fluid, this equation reduces to : ρ De Dt = p U 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 increasing (heating of the flow) and the quantity p ρ U 2 + gz decreases accordingly. For an expansion, U > 0 hence e is decreasing (cooling of the flow) and the quantity p ρ U 2 + gz 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 ρ U 2 + gz = cst along a streamline (3.25) 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 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 t + (ρ U) = 0 (ρu i ) t + (ρu iu j ) x j = p x i (ρe) t + (ρh U) = 0 Since (with neglected body forces), E = e U 2 and since H = E + p ρ, the unknown variables in this set of governing equations are :

23 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 D m (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 t + (ρ U) = 0 (ρu i ) t + (ρu iu j ) x j = p + µ U i + µ ( U) x i 3 x i (ρe) t + (ρh U) = ( τ U) + (k 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 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 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 : ds = ( δq T ) rev (3.26) 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 : T ds = de + pdv = de + pd( 1 ρ ) = de p dρ (3.27) ρ2 - every state change in a thermodynamic system generates an entropy variation which can be decomposed as follows : ds = ( δq T ) rev + ds irrev (3.28) where the contribution ds irrev corresponds to irreversible processes, such as viscous friction, irreversible heat transfer, shockwaves and is such that : Consequently, it can be concluded : ds irrev 0 (3.29) - for an adiabatic process or flow evolution, that is a process or a flow evolution taking place without heat transfer (δq = 0) : ds = ds irrev 0

25 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 Equations of state Since Gibbs relationship (3.27) states : de = T ds pdv = T ds + p ρ 2 dρ 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 : and ds = dv = ( s T ) dt + p ( ) v dt + T p ( ) s dp p T ( ) v dp p T

26 26 Lecture #3 : Energy conservation Reinjecting in the Gibbs relationship which provides de leads to : de = [ T ( ) ( ) ] [ s v p dt + T T p T p ( ) s p p T ( ) ] v dp 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 that δq = T ds, it follows C = T s. Let us consider more specifically the entropy variation induced by a temperature variation when the second state variable, upon which entropy depends, T is fixed. We define : - the specific heat coefficient at constant pressure : C p = T - the specific heat coefficient at constant volume : C v = T ( ) s T p ( ) s T v The compressibility coefficient expresses the local variation of volume induced by a local variation of pressure. It is therefore proportional to v 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 : β T = 1 v - the isentropic compressibility coefficient : β s = 1 v ( ) v p T ( ) 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 : α = 1 v ( v T ) p

27 3.2. Thermodynamic analysis 27 Most of the partial derivatives appearing in the above expression of de can be expressed using the coefficients C p, β T 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 β T 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 : dg = ( ) g dt + T p ( ) g dp p T By immediate identification, it follows that : ( ) g s = T p, v = ( ) g p T The theorem of Schwarz on mixed second partial derivatives allows to write : 2 g T p = 2 g p T which yields one of the so-called Maxwell identities : ( ) s p T ( ) v = T p (3.30) Using (3.30), the total differential de can finally be expressed as : de = (C p αpv)dt + (β T 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 (C p, C v, β T, β s, α), we can establish a few general identities which connect these coefficients.

28 28 Lecture #3 : Energy conservation Relationship between C p and C v Considering the canonical form s = s(t, v) we can deduce : ds = ( ) s dt + T v ( ) s dv v T Since dv = αvdt β T vdp, ds can also be developed into : [( ) ( ) s s ds = + αv T v v T ] dt β T v ( ) s dp 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 C p and C v yields : ( ) s C p = C v + αvt v T Another Maxwell s identity, obtained from df = sdt pdv, takes the form : so that : ( ) s v T = C p C v = αvt Considering now p = p(t, v), it is possible to write : dp = ( ) p dt + T v ( ) p T v ( ) p T v ( ) p dv v T (3.32) 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 = (αv)/(β T v) = α T v β T Inserting in the above relationship between C p and C v yields the general identity for the difference of the specific heat coefficients : The ratio of the specific heat coefficients is defined as : ( ) s C p T p = ( ) C v s T C p C v = vt α2 β T (3.33) v

29 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 : In a similar way, we can establish : ( ) s = T p ( ) p / T s ( ) p s T ( ) ( ) ( ) s v v = / T v T s s T Using these identities to express the ratio C p /C v yields : ( ) ( ) ( ) p v p C p T s s T v = ( ) ( ) = ( ) s C v p v p s T v or else T s T C p C v = β T β s (3.34) We will see in the next section how these general relationships simplify in the case of an ideal gas 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) where r is the gas constant. The gas constant is computed as r = 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 = J K 1 mol 1 or else R = J K 1 kmol 1. Since air is made of 21% of oxygen and 79% of nitrogen, with respective molar mass 32 g/mol and g/mol or else 32 and kg/kmol, it follows the molar mass of air is equal to kg kmol 1 so that the gas constant r for air is equal to r = 287 J kg 1 K 1.

30 30 Lecture #3 : Energy conservation for a per- Using (3.35) or pv = rt or else v = rt/p, it is immediate to compute α and β T fect gas and to deduce C p C v. We have : ( ) v = rt p T p 2 = v p hence β T = 1 ( ) v = 1/p. v p T Similarly, we also have : ( ) v = r T p p hence α = 1 ( ) v = r v T p pv = 1 T. After calculation and simplification, we obtain : de = C v dt Theoretically, C v depends both on T and v. Let us prove that, for a perfect gas, C v depends only on T. This means the partial derivative of C v with respect to v is zero. From the definition of C v, we can write : ( Cv v ) T = v [ ( ) ] s T = T 2 s T v v T = T T ( ) s v T Using then Maxwell s identity (3.32) this partial derivative can also be expressed as : ( ) ( Cv 2 ) p = T v T 2 T v (3.36) 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 C v = C v (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 = C v (T )dt the following expression for the internal energy : e(t ) = C v (T )dt 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 ) = C v (T T 0 ) + e 0 = C v T + cste (3.37) In the case of a perfect gas, the relationship (3.33) between C p and C v becomes : C p C v = vt α2 = vt 1 β T T 2 p = pv T

31 3.2. Thermodynamic analysis 31 hence C p C v = 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 C v = constant, it follows that C p is also constant. The enthalpy equation of state for a perfect gas is obtained from : h = e + p ρ = e(t ) + rt = C vt + cste + rt hence, using (3.38) : h(t ) = C p T + cste (3.39) with C p = 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 C p /C v using (3.34). In order to compute the coefficient β s 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 : or else T ds = de + pdv = C v dt + rt dv v dt ds = C v T + r dv v Let us integrate this relationship between two states 1 and 2, taking into account C v = constant for a calorifically perfect gas : 2 The thermal EoS yields : 1 ds = s 2 s 1 = C v 2 1 dt T + r 2 p 2 p 1 = ρrt 2 ρrt 1 = ρ 2 ρ 1 T 2 T 1 1 dv v = C v log( T 2 T 1 ) + r log( v 2 v 1 ) and besides v 2 /v 1 = ρ 1 /ρ 2. The entropy difference s 2 s 1 can therefore also be expressed as : s 2 s 1 = C v log( p 2 p 1 ρ 1 ρ 2 ) r log( ρ 2 ρ 1 ) = C v log( p 2 p 1 ) (C v + r) log( ρ 2 ρ 1 ) = C v log( p 2 p 1 ) C p log( ρ 2 ρ 1 )

32 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 : γ = C p C v (3.40) For air, a diatomic gas, γ = 1.4. Going back to the calculation of the entropy difference and inserting γ, it follows that : s 2 s 1 = C v log( p 2 p 1 ) γc v log( ρ 2 ρ 1 ) = C v log( p 2 p 1 ) C v log( = C v log ( p2 ρ γ 2 ) C v log ( ρ2 ρ 1 ( p1 ) γ ) ρ γ 1 ) The expression of the specific entropy for a perfect gas takes therefore the form : ( ) p s = C v ln ρ γ (3.41) 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 : p 1 ρ γ 1 = p 2 ρ γ 2 that is : ( p2 p 1 ) = ( ρ2 ρ 1 ) γ (3.43)

33 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 : ( p2 p 1 ) = ( T2 T 1 ) γ γ 1 (3.44) 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 : ( ρ2 ρ 1 ) = ( T2 T 1 ) 1 γ 1 (3.45) Other useful identities for a perfect gas Since pv γ = constant = C in the isentropic evolution of a perfect gas, it is possible to write : ( ) v = (C/p)1/γ p s p = 1 γpv Consequently, the isentropic compressibility coefficient β s can be computed as follows : β s = 1 γp (3.46) It can be checked the formula (3.34) stating that C p /C v = β T /β s yields C p /C v = (γp)/p for a perfect gas hence C p /C v = 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 C p and C v in the case of a calorifically perfect gas : C v = r γ 1, C p = γr γ 1 (3.47) Since γ = 1.4 for a diatomic gas such as air and since r = 287 J kg 1 K 1 for air, an immediate calculation yields C v = J kg 1 K 1 and C p = J kg 1 K 1. 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 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 : p = rt v b a v 2 = ρrt 1 bρ aρ2 (3.49) 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 : Z = pv rt (3.50) 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 Final overview of the governing equations To conclude on the derivation of the flow models, we can go back to Section 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

35 3.3. Conclusions on the flow governing equations 35 compressible flow of air reduces to : ρ t + (ρu) = 0 (ρu i ) t + (ρu iu j ) x j = p x i (3.52) p ρ γ = constant = S ref 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 t + (ρu) = 0 (ρ U) t + (ρ U U + p δ ) = 0 (ρe) t + (ρh U) = 0 where δ denotes the unit tensor such that ( δ )ij = δ ij with δ ij the Kronecker symbol. Since (neglecting body forces), E = e + 1 2U 2 and H = E + p, the system is closed with an 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 t + (ρu) = 0 (ρu i ) t + (ρu iu j ) x j = p + µ U i + µ ( U) x i 3 x i (ρe) t + (ρh U) = ( τ U) + (k T )

36 36 Lecture #3 : Energy conservation and is closed for a perfect gas with the thermal and calorific equations of state : p = ρrt and e = C v 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, ρ U, 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) : h U 2 = constant In the case of a perfect gas flow, the equation of state (3.39) for the specific enthalpy allows to write : C p T U 2 = constant (3.53) 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 : Velocity measurement in a compressible flow γ p γ 1 ρ + 1 U 2 2 = constant (3.54) 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 : e M + p M + 1 ρ M 2 U M 2 = e A + p A + 1 ρ A 2 U A 2 e N + p N ρ N U 2 N = e B + p B ρ B U 2 B 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 : e A + p A ρ A U 2 A = e B + p B ρ B U 2 B

37 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 p A + 1 γ 1 ρ A 2 U A 2 = p B γ + 1 γ 1 ρ B 2 U B 2 Since A is a stagnation point, U A = 0 and p A = (p 0 ) A the total pressure measured at point A, hence : γ (p 0 ) A = γ p B + 1 γ 1 ρ A γ 1 ρ B 2 U B 2 (3.55) 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 p B is also known from measurement, two other unknown quantities appear in the relationship : the density ρ A and the density ρ B. 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 : p A ρ γ = p M A ρ γ = p N M ρ γ = p B N ρ γ B hence the densities at point A, B and the measured pressures are related by : (p 0 ) A ρ γ A = p B ρ γ B or else ( ) (p0 ) 1/γ A ρ A = ρ B (3.56) p B Hence if ρ B is known then ρ A can be computed using the above expression where the pressures are known from measurement. Therefore, it remains to determine ρ B in order to obtain an expression for the velocity U B. Since p B is measured, the measurement of the temperature T B at point B using a temperature probe allows to compute ρ B from the EoS : ρ B = p B rt B (3.57)

38 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 : U 2 B = 2γ γ 1 [ (p 0 ) A 1 ρ B ( pb (p 0 ) A ) 1/γ p B ρ B ] Factoring p B /ρ B yields : U 2 B = 2γ γ 1 p B ρ B [ ((p0 ) A p B ) 1 1 γ 1 ] Finally, using the EoS at point B (3.57), the formula giving the velocity U B as a function of the measured pressures and temperature for the compressible flow under study reads : U comp B = 2γrT B γ 1 ((p0 ) A p B ) γ 1 γ 1 (3.58) This relationship can be compared with the formula previously established in the context of an incompressible flow analysis : 2(p 0A p B ) U 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 = ρ B = p B /(rt B ). The incompressible formula can then be expressed as : U inc B = 2rT B (p 0 ) A p B 1 (3.59) 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 = (p 0 ) A /p B : U comp B η γ 1 γ 1 UB inc = 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 p 0 in the fluid at rest is p 0 = P a and the temperature is T 0 = 300 K. The density ρ is thus given by ρ = p 0 /(rt 0 ) kg/m 3. 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 (p 0 ) A = P a and p B = P a. A second test is performed for which (p 0 ) A = P a and p B = P a.

39 3.4. Exercises and problems 39 Applying formula (3.59) with ρ = kg/m 3 and the measured pressures yields an estimate for the velocity U B such that U B = 34.7 m/s in the first test and U B = 283 m/s in the second test. During the same tests, the temperature has been measured at point B : for the first test T B = K while for the second test T B = 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 U B is estimated at U B = 34.7 m/s in the first test and U B = 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 U B by 10% with respect to the compressible prediction. 3.4 Exercises and problems 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 40 Lecture #3 : Energy conservation Answer Using the analysis of Section with a constant mass flowrate ṁ through the turbine we can write : ṁ [(e + p ρ U 2 + gz)] = Ẇ + Q where Ẇ 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 w shaft the work output per unit mass of stream through-flow; it is such that : Ẇ = ṁ w shaft Therefore, the total energy conservation yields : or [(e + p ρ U 2 + gz)] = w shaft h ( U 2 ) + (gz) = w shaft The changes in elevation being negligible, we can write : An immediate numerical application yields : w shaft = h ( U 2 ) w shaft = ( ) ( ) = = kj/kg This quantity is negative since it is extracted from the fluid. The net work output per unit mass of steam through-flow is 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 flowrate ṁ must be known in order to determine the power output Ẇshaft 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 The tank sectional area is denoted A T while A 0 denotes the orifice sectional area, assumed much smaller than A T : A 0 << A T. The tank is initially filled with water up to a level h 0 above the orifice located at the bottom of the tank. The static pressure surrounding the tank is the atmospheric pressure p atm. 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 A T >> A 0 which is precisely the assumption we made. Viscous effects are negligible

41 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 h 0 = 20 cm, a cylindrical tank of radius R T = 2.5 cm and an orifice of radius R 0 = 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 quasisteady 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 ρ U gz 1 = p 2 ρ U gz 2 The static pressure p 1 at the free surface is equal to p atm. 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 p 2 is also equal to p atm. The difference between the elevation z 1 and the elevation z 2 is approximately equal to h(t) the height of the liquid level in the tank. The Bernoulli equation takes therefore the form : 1 2 U gh(t) = 1 2 U 2 2 The unknown velocities can be related to h(t) as follows :

42 42 Lecture #3 : Energy conservation - the velocity of the fluid particle at point 1 is equal to dh dt - the mass conservation applied to the control volume limited by the tank wall, the free surface and the orifice section yields : hence U 2 = A T A 0 U 1. Replacing in the Bernoulli equation yields : or 1 2 ρu 1 A T = ρu 2 A 0 ( ) ( dh 2 ( ) ) 2 AT 1 + gh(t) = 0 dt A 0 1 dh h(t) dt = 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 h(t) dt = C with the constant C given by C = 2g A 2. The function h(t) satisfies the initial condition T 1 A 2 0 h(0) = h 0 and the final condition h(t f ) = 0 where t f is the draining time we are looking for. Integrating this equation between t = 0 and the current time t yields : t 0 dh = h t 0 A 2 T A 2 0 2g C dt hence [ 2 ] t h(t) = Ct 0 so that 1 2 h(t) 2 h 0 = Ct When t = t f the liquid level above the orifice is such that h(t f ) = 0, hence : The draining time is therefore given by : h 0 = Ct f t f = 2h 0 g A 2 T A 2 0 1

43 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 A T /A 0 >> 1 : ( ) 2h 2 0 RT t f g R 0 For the proposed numerical application, R T /R 0 = 2.5/0.2 = 12.5 so that : 2h 0 t f = g With an initial liquid height h 0 = 0.2 m (and taking g = 9.81 m/s 2 ), the draining time of the tank is equal to s 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 It is made of a tube with horizontal 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 A 1 a section upstream of the throat and let us index 1 the physical quantities relative to this section. Similarly, the throat section is denoted A 2 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 = A 2 /A 1 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 p 1, p 2, ρ 1, A 2 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 = constant along a streamline, with H the specific total enthalpy, H = e+ p ρ U 2 = γ p γ 1 ρ U 2 (neglecting the body forces). States 1 and 2 are therefore linked by the relationship : γ p γ 1 ρ 1 2 U 1 2 = γ p γ 1 ρ 2 2 U 2 2 (3.60)

44 44 Lecture #3 : Energy conservation Figure 3.6: Flow in a Venturi. A uniform flow is assumed in each section hence p i, ρ i, U i are respectively the uniform static pressure, density and velocity in section A i. Moreover, if the inviscid flow of the perfect gas is assumed isentropic (no shockwaves in the Venturi nozzle), the relationship p/ρ γ = constant holds everywhere in the flow so that it is possible to write : p 1 ρ γ = p 2 1 ρ γ 2 or else ρ 2 ρ 1 = ( p2 p 1 ) 1 γ (3.61) Finally, mass conservation between section A 1 and A 2 can be also expressed as the conservation of the mass flowrate through the nozzle, that is : ρ 1 U 1 A 1 = ρ 2 U 2 A 2 Taking into account (3.61) to express the density ratio as a function of the pressure ratio and introducing the geometric factor β yields : U 1 = ρ ( ) 1 2 p2 γ βu 2 = β U2 (3.62) ρ 1 Injecting (3.62) into the generalized Bernoulli equation (3.60) for compressible flows, we can eliminate U 1 and obtain : ( ( ) 2 ) 1 2 U β 2 p2 γ = γ ( p 1 1 p ) 2 ρ 1 p 1 γ 1 ρ 1 p 1 ρ 2 p 1 Using again the isentropic flow identity (3.61) and denoting η = p 2 between sections A 2 and A 1, the velocity U 2 is given by : U γ p ( ) 1 = 1 η γ 1 1 β 2 η 2 γ γ γ 1 ρ 1 p 1 the static pressure ratio

45 3.4. Exercises and problems 45 The (constant) mass flowrate ṁ is therefore such that : ṁ 2 = ρ 2 2U 2 2 A 2 2 = A 2 2 ρ 2 1η 2 2γ p ( ) 1 β 2 η 2 γ 1 1 η γ 1 γ γ γ 1 ρ 1 or else a formula involving as requested p 1, p 2 (through η), ρ 1, β and A 2 : ṁ = A 2 1 β 2 η 2 γ 2γ p1 ρ 1 η 2 γ γ 1 ) (1 η γ 1 γ (3.63) In order to compute the ratio between the compressible formula (3.63) yielding ṁ comp and the previously established incompressible formula we slightly modify this latter, taken from?? : ṁ inc = ρ 2 U 2 A 2 = A 2 1 β 2 2ρ1 (p 1 p 2 ) Introducing the pressure ratio η, we can write : ṁ inc = A 2 1 β 2 2 ρ1 p 1 1 η Dividing the formula (3.63) valid for a general compressible flow with the above formula valid under the assumption of incompressible flow, we obtain : ṁ comp γ = ṁ inc γ 1 2 η γ (1 η γ 1 γ 1 η ) 1 β 2 1 β 2 η 2 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 : ṁ comp ṁ inc = Y (η) For Venturi geometries such that the diameter ratio between sections A 2 and A 1 become less than 0.25, the factor β 2 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 : ) γ Y (η) γ 1 2 (1 η γ 1 γ η γ 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 46 Lecture #3 : Energy conservation 1.8 Expansion factor vs static pressure ratio sqrt(1.4/0.4)*sqrt((x**(2/1.4))*(1-(x**(0.4/1.4)))/(1-x)) Y(eta) eta Figure 3.7: Evolution of Y (η) with η in a Venturi where β 2 << 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. However 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 Génie Atomique. Let us consider a nuclear reactor containment with volume V = m 3 containing, in nominal operating conditions, air at a pressure p 0 = 1 bar and at a temperature T 0 = 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 = C p /C v = 1.3, with a constant pressure specific heat capacity C p = 2.08 kj kg 1 K 1. The specific enthalpy h is such that h(t ) = C p T + h 0 with h 0 = 1900 kj/kg.

47 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 : { ṁin0 = 3500 kg/s for 0 < t t ṁ in = in = 20 s (3.64) 0 for t > t in For 0 < t t in, the specific enthalpy of the vapor injected into the containment through the breach is assumed such that h in = h in0 = 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 : { ṁin0 (1 t/t ṁ in = in ) for 0 < t t in = 20 s (3.65) 0 for t > t in with ṁ in0 = 7000 kg/s. For 0 < t t in, the specific enthalpy of the water vapor injected into the containment is still assumed such that h in = h in0 = 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 flowrate ṁ cond leaving the control volume associated with the volume of overheated vapor and such that : ṁ cond = ṁ cond0 = 1200 kg/s for 0 < t t in = 20 s ṁ cond0 (t cond t)/(t cond t in ) for t in < t t cond = 50 s (3.66) 0 for t > t cond 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 h fg = 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 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 Let us denote V 0 this volume and S 0 its surface. Note V 0 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 dt W ) n ds D a(t) S a(t) we can write in the case of the fixed volume V 0 ( W = 0) : d dt V 0 ρ dv = S 0 ρ U n ds (3.67)

49 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 : V 0 dm(t) dt = ρu n ds S 0 and expresses the fact the mass of fluid contained in V 0 varies as a function of the mass flowrate entering or leaving the fluid domain. The surface S 0 is made of the containment surface and of the breach surface through which a mass flow enters the domain. Let us denote S in 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 : ṁ in (t) = ρu n ds S in 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, ṁ in (t) is the only massflow entering the volume V 0. If condensation is taking place on the containment walls, we must also account for a vapor mass flowrate leaving the volume V 0 to condense on the containment wall and form a liquid film. Let us denote ṁ cond (t) this mass flowrate of vapor leaving V 0 to condense on the containment wall. The mass balance (3.67) can eventually be expressed as : dm dt = ṁ in(t) ṁ cond (t) (3.68) with ṁ cond (t) = 0 for case #1 and #2 and ṁ cond (t) 0 for case # 3. The identity (3.68) can then be used to derive the vapor mass variation in V 0 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 dt ( ρedv ) = Ẇ + Q ρe ( U W ) n ds D a(t) S a(t) In the case of the fixed control volume V 0 ( W = 0) it comes : d dt ( ρedv ) = Ẇ + Q ρe U n ds V 0 S 0 Neglecting the power of body forces and viscous constraints yields the following expression of Ẇ (power of the pressure constraints only) : Ẇ = S 0 p U nds

50 50 Lecture #3 : Energy conservation Assuming Q = 0 (no thermal losses through the boundary S 0 ) yields the following form of the energy balance : d dt ( ρedv ) = pu nds ρeu n ds V 0 S 0 S 0 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 dt ( ρedv ) = ρhu n ds V 0 S 0 This relationship expresses the fact the total energy of the fluid (vapor) contained in the control volume V 0 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 S in. 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 n ds = ρhu n ds S 0 S in Assuming the kinetic energy contribution to the specific total enthalpy can be neglected with respect to the specific enthalpy contribution h so that H in, the specific total enthalpy at the breach, is such that H in h in so that the inlet flux of total enthalpy can be expressed as : ρhu n ds = ṁ in h in S in If condensation is taking place on the containment wall, a total enthalpy outflow is taking place through S 0 which can be expressed using the specific enthalpy h fg associated with the phase change (from vapor to liquid) and the condensation mass flowrate so that the total enthalpy flux reads : ρh v n ds = ṁ in (t) h in ṁ cond (t) h fg S in Let us recall ṁ cond (t) = 0 for scenarii #1 and #2 while ṁ cond (t) 0 for scenario # 3. Inserting this expression for the total enthalpy flux in the energy balance and expressing ρe = ρh p yields : d dt ( (ρh p)dv ) = ṁ in h in ṁ cond (t) h fg V 0 As previously done at the breach, it is assumed that at any point in the control volume V 0 the kinetic energy contribution to the total enthalpy remains negligible with respect to the specific enthalpy (thermal) contribution so that : H = h(t ) v2 h(t ) = C p T + h 0 Therefore, the quantity ρh p appearing the LHS of the energy balance can also be expressed as : ρh p = ρh p = ρc p T + ρh 0 p = Cp r p p + ρh 0 = 1 γ 1 p + ρh 0

51 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 C p = γr/(γ 1). The LHS volume integral of the energy balance, which expresses the variation of total energy in the volume V 0, can be computed as : d (ρh p)dv = 1 dt V 0 γ 1 d dt V 0 pdv + h 0 d dt V 0 ρdv }{{} =m(t) The averaged pressure p(t) in the containment reactor at time t is defined as : V 0 p(t) = pdv V 0 Therefore the variation of total energy in the volume V 0 can be computed as : d (ρh p)dv = V 0 dp(t) dm(t) + h 0 dt V 0 γ 1 dt dt Since this variation is also equal to the total enthalpy flux it comes : V 0 dp(t) dm(t) + h 0 = ṁ in h in ṁ cond (t) h fg γ 1 dt dt The mass balance (3.68) can then be used to express the mass variation term in the energy balance : V 0 dp(t) + h 0 [ṁ in (t) ṁ cond (t)] = ṁ in h in ṁ cond (t) h fg γ 1 dt Since, whatever the scenario, h in = h in0 for t [0, t in ] and ṁ in (t) = 0 for t > t in, we can also write : V 0 dp(t) = ṁ in (t) (h in0 h 0 ) ṁ cond (t) (h fg h 0 ) (3.69) γ 1 dt 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 : p = 1 pdv = 1 ρrt dv V 0 V 0 V 0 V 0 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 : p 1 V 0 V 0 ρrt dv = rt 1 V 0 m(t) hence T(t) = p(t)v 0 rm(t) (3.70)

52 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 m 0 the initial mass of vapor in the reactor. The mass flowrate at the breach can be made non-dimensional using the quantity m 0 /t in. Whatever the scenario considered, it is possible to write : ṁ in t in m 0 ( ) ( ) = ṁin0 t in t t Ψ in = β in Ψ in m 0 t in t in with the non-dimensional quantity β in, non-dimensional reference inlet mass flowrate at the breach, defined as : β in = ṁin0 t in m 0 The non- We can also introduce the non-dimensional time t normalized by t in : t = t. t in dimensional mass flowrate at the breach at any time reads : ṁ in (t) t in m 0 = β in Ψ in (t) Similarly, we introduce the non-dimensional reference condensation mass flowrate : β cond = ṁcond0 t cond m 0 and the non-dimensional time t = at any time reads : t t cond so that the non-dimensional condensation mass flowrate ṁ cond (t) t cond m 0 = β cond Ψ cond ( t) With these choices, the three scenarii are now described by the non-dimensional functions Ψ in and Ψ cond : - case #1 : constant mass flowrate at the breach over t in and no condensation. Ψ in (t) = { 1 for 0 < t 1 0 for t > 1, Ψ cond ( t) = 0 (3.71)

53 3.4. Exercises and problems 53 - case #2 : linearly decreasing mass flowrate at the breach over t in and no condensation. Ψ in (t) = { 1 t for 0 < t 1 0 for t > 1, Ψ cond ( t) = 0 (3.72) - case #3 : constant mass flowrate at the breach over t in and condensation over t cond. In this scenario the condensation mass flowrate made non-dimensional using (m 0 /t cond ) involves a non-dimensional time η = t in /t cond. Ψ in (t) = { 1 for 0 < t 1 0 for t > 1 1 for 0 < t η, Ψ cond ( t) = (1 t)/(1 η) for η < t 1 0 for t > 1 (3.73) The dimensional mass balance (3.68) is readily turned into a non-dimensional form. Since : dm = ṁ in (t)dt ṁ cond (t)dt or else ( ) m(t) d = β in Ψ in (t)dt β cond Ψ cond ( t)d t m 0 an immediate integration from m = m 0 and t = 0 to m = m(t) and t yields : m(t) m 0 = 1 + β in t 0 t Ψ in (t)dt β cond Ψ cond ( t)d t (3.74) 0 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. Since the ini- A similar non-dimensional approach can be applied to the energy balance. tial mass m 0 of vapor in the reactor is thus such that : m 0 = ρ 0 V 0 where ρ 0 satisfies p 0 = ρ 0 rt 0, this yields m 0 r T 0 = p 0 V 0 which can also be rewritten as : V 0 (γ 1) p 0 = m 0 C v T 0 with C v = r/(γ 1). The energy balance (3.69) can be rewritten as : ( ) V 0 p(t) γ 1 p 0 d = ṁ in (t) (h in0 h 0 ) dt ṁ cond (t) (h fg h 0 ) dt p 0 or else ( ) p(t) m 0 C v T 0 d = ṁ in (t) (h in0 h 0 ) dt ṁ cond (t) (h fg h 0 ) dt p 0

54 54 Lecture #3 : Energy conservation Dividing by m 0 C v T 0 allows to reintroduce the non-dimensional quantities β in, β cond as well as the non-dimensional functions Ψ in, Ψ cond and the non-dimensional times t, t : ( ) p(t) d = β in Ψ in (t) (h in0 h 0 ) dt β cond Ψ cond ( t) (h fg h 0 ) d t p 0 C v T 0 C v T 0 It is then natural to define the following non-dimensional quantities : so that : α in = h in0 h 0 C v T 0, α cond = h fg h 0 C v T 0 (3.75) ( ) p(t) d = β in α in Ψ in (t) dt β cond α cond Ψ cond ( t) d t p 0 An immediate integration from p = p 0 and t = 0 to p(t) and t yields : p(t) p 0 = 1 + α in β in t 0 t Ψ in (t)dt α cond β cond Ψ cond ( t)d t (3.76) 0 The definitions (3.71), (3.72) and (3.73) for Ψ in and Ψ cond 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 /T 0 : T (t) T 0 = p(t) p 0 m 0 m(t) V 0 p 0 rt 0 m 0 Since p 0 = ρ 0 rt 0 and m 0 = ρ 0 V 0, it follows that : Using (3.74) and (3.76) yields : T (t) T 0 = p(t) p 0 m(t) m 0 T(t) T 0 = t t 1 + α in β in Ψ in (t)dt α cond β cond Ψ cond ( t)d t 0 t t 1 + β in Ψ in (t)dt β cond Ψ cond ( t)d t (3.77) The definitions (3.71), (3.72) and (3.73) for Ψ in and Ψ cond can then be successively applied to (3.77) in order to obtain T (t) for each scenario. Let us now compute the non-dimensional factors t 0 Ψ in (t)dt and t 0 Ψ cond ( t)d t for each scenario and deduce the corresponding evolutions of the mass, averaged pressure and averaged temperature.

55 3.4. Exercises and problems 55 - case #1 : constant mass flowrate at the breach over t in and no condensation. Using (3.71) yields for t 1 : When t = 1, t can write in a general way : 0 t 0 Ψ in (t)dt = t 0 1 dt = t Ψ in (t)dt is precisely equal to 1 and keeps this value for t > 1. Hence we Since no condensation occurs in this scenario, we also have : t t Inserting these values in (3.74) yields : 0 0 Ψ in (t)dt = min (t, 1) (3.78) Ψ cond ( t)d t = t Note that going back to a dimensional formula yields : that is, multiplying by m 0 : m(t) m d t = 0 (3.79) m(t) m 0 = 1 + β in min (t, 1) (3.80) = 1 + ṁin0t in m 0 min ( t t in, 1) m(t) = m 0 + ṁ in0 min (t, t in ) (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 < t in, the coefficient of increase being the mass flowrate through the breach ṁ in0. The mass reaches a maximum value m 0 + ṁ in0 t in at t = t in and keeps this value for t > t in since there is no longer a vapor massflow entering through the breach after time t in. 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. A first numerical application can be performed. We have V 0 = m 3 and, assuming vapor (and not air) is initially in the containment reactor, ρ 0 = p 0 /(rt 0 ) with p 0 = 10 5 P a, T 0 = K and r = (γ 1)C p /γ with γ = 1.3 and C p = 2080 J/kg/K so that r = 480 J/kg/K for vapor. We compute ρ 0 = 0.71 kg/m 3 and m 0 = ρ 0 V 0 = kg. Since the vapor mass flowrate at the breach is ṁ in0 = 3500 kg/s in this scenario with t in = 20 s, the mass of vapor in the reactor containment at any time t is given by : m(t) = min (t, 20) so that the maximum (constant) mass for t 20 s is m max = kg. Using directly (3.78) and (3.79) in the non-dimensional energy balance (3.76) yields : p(t) p 0 = 1 + α in β in min (t, 1) (3.82)

56 56 Lecture #3 : Energy conservation The dimensional form is readily obtained : p(t) = p 0 + p 0 h in0 h 0 C v T 0 ṁ in0 t in m 0 ( ) t min, 1 t in and can be simplified using p 0 = ρ 0 rt 0 and m 0 = ρ 0 V 0 to yield : p(t) = p 0 + γ 1 V 0 (h in0 h 0 )ṁ in0 min (t, t in ) (3.83) The maximum averaged pressure is reached at time t = t in and is equal to : p max = p 0 + (γ 1)(h in0 h 0 )ṁ in0 t in V 0 The pressure increase is directly related to the enthalpy injected into the containment reactor through the breach. A numerical application with h in0 = J/kg, h 0 = J/kg and the values of p 0, ṁ in0, t in, V 0, γ already used to compute the mass evolution yields : p max = = P a or p max = 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) T 0 = 1 + α inβ in min (t, 1) 1 + β in min (t, 1) Since α in = h in0 h 0 C v T 0 or, numerically, α in = 10 6 /((2080/1.3) ) that is α in = > 1, the averaged temperature is increasing with time for t < t in until it reaches at time t = t in a maximum value T max such that : T max = T α inβ in 1 + β in With α in = and β in = / that is β in = 2, we obtain : T max = T = K = 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 t in 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 = t in. Using (3.72) yields for t 1 : t t [ Ψ in (t)dt = (1 t) dt = t t 2 ]t = t t

57 3.4. Exercises and problems 57 When t = 1, t can write in a general way : 0 Ψ in (t)dt is precisely equal to 1/2 and keeps this value for t > 1. Hence we t 0 Ψ in (t)dt = f in (t) ( t t 2 ) if 0 t 1 if t > 1 Since no condensation occurs in this scenario, we still have (3.79). Inserting these values in (3.74) and (3.76) yields : (3.84) m(t) m 0 = 1 + β in f in (t) (3.85) p(t) p 0 = 1 + α in β in f in (t) (3.86) In this scenario #2, since ṁ 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 β in = 4. The non-dimensional factor α in remains unchanged : α in = Therefore, we can also write : m(t) m 0 = f in (t) and p(t) p 0 = f in (t) For t > t in = 20 s, the mass remains constant, equal to its maximum value that is, since f in = 1/2 for t > 1, m max = 3m 0 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 = t inj, this maximum value being equal to p max = ( /2)p 0 = 5.25 bar, which is the same value as in scenario # 1. The temperature evolution is such that : T(t) T 0 = f in(t) f in (t) that is an increasing function of t until t = 1. The maximum averaged temperature reached at t = t in = 20 s remains constant for t > t in, equal to T max = (5.25/3)T 0 = 1.75 T 0 = 513 K = 241 C, that is the same maximum value as in scenario # 1. - case #3 : constant mass flowrate at the breach over t in and condensation over t cond. 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 58 Lecture #3 : Energy conservation we still have (3.78) and for numerical application β in = 2 again. condensation mass flowrate must be taken into account with : 1 for 0 < t η Ψ cond ( t) = (1 t)/(1 η) for η < t 1 0 for t > 1 In this scenario, the and η = t in /t cond. For the case under study, η = 20/50 = 0.4. If 0 t η then : If η t 1 then : so that t 0 Ψ cond ( t)d t = t 0 t 0 η 0 Ψ cond ( t)d t = t 1 d t + η t 0 1 d t = t 1 t 1 η d t = η η Ψ cond ( t)d t = η + t η 1 η 1 2 If t > 1, there is now contribution to the integral so that : t We can write in a general way : 0 Ψ cond ( t)d t = η + 1 η 1 η 1 2 t 2 η 2 1 η 1 2 η 2 1 η = η [ t t 2 ] t 2 η t if 0 t η t 0 Ψ cond ( t)d t = f cond ( t) η + t η 1 η 1 t 2 η η if η t (1 + η) if t > 1 (3.87) The non-dimensional coefficients β cond and α cond are respectively such that : β cond = ṁcond0 t cond m 0 that is β cond = ( )/ or β cond = α cond = h fg h 0 C v T 0 that is α cond = ( )/( ) or α cond = In this scenario # 3, the mass evolution (3.74) is such that : m(t) m 0 = min (t, 1) 1.709f cond ( t) or else, using t = ηt with η = 0.4 : m(t) m 0 = min ( t, η) 1.709f cond ( t)

59 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) = t m 0 when t η that is t t in. For t in t t cond, the variation of vapor mass is such that : ( m(t) = 1 + 5η η + t η m 0 1 η 1 t 2 η 2 ) 2 1 η that is, with η = 0.4 : m(t) m 0 = t t 2 for t [η, 1] that is t [t in, t cond ]. The vapor mass is therefore decreasing between t in and t cond 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 = t in and given by : m max = (1 + 5η 1.709η) m 0 = m 0 = kg The final constant value of the vapor mass in the reactor is reached at time t = t cond and is such that : m final = 1 + 5η (1 + η) = m 0 2 or m final = kg. Inserting (3.78) and (3.87) into (3.76) yields : p(t) p 0 = 1 + α in β in min (t, 1) α cond β cond f cond ( t) or else p(t) p 0 = 1 + α inβ in η min ( t, η) α cond β cond f cond ( t) Replacing each non-dimensional factor with its numerical value yields : p(t) = min ( t, 0.4) f cond ( t) p or else p(t) p 0 = min ( t, 0.4) f cond ( t) Because of the definition (3.87) for f cond ( t), the averaged pressure reaches therefore its maximum value at time t = t in, corresponding to t = η before decreasing between t in and t cond. This maximum value is given by : p max p 0 = 4.84

60 60 Lecture #3 : Energy conservation which corresponds to the suggested value in the question since p 0 = 1 bar. Note also the final constant value of the averaged pressure is reached at time t = t cond corresponding to t = 1 and is such that : p final = ( ) = 4.55 p 0 2 The evolution of the averaged temperature is given by (3.77), that is for scenario # 3 : 1 + α inβ in min ( t, η) α T(t) cond β cond f cond ( t) η = T β in η min ( t, η) β cond f cond ( t) The averaged temperature at time t = t in, that is t = η, is such that : T (t in ) T 0 = 1 + α inβ in α cond β cond η 1 + β in β cond η The averaged temperature at time t = t cond, that is t = 1, is such that : T (t in ) 1 + α in β in 1 = 2 α condβ cond (1 + η) T β in 1 2 β cond(1 + η) With α in β in = 4.264, β in = 2 and α cond β cond = 1.02, β cond = 1.709, η = 0.4 the numerical application yields : that is T (t in ) = K = C. T (t in ) = = T = T (t in ) = = 4.55 T = 2.52 that is T (t cond ) = K = 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 t in and t cond so that the averaged temperature keeps increasing between t in and t cond to reach its maximum (constant) value at t = t cond : T max = 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.

61 3.4. Exercises and problems evol m scenario1 evol m scenario2 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 62 Lecture #3 : Energy conservation 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.

63 3.4. Exercises and problems 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 64 Lecture #3 : Energy conservation

65 Lecture 4 1D compressible flow analysis Contents 4.1 Some key quantities Sound speed and Mach number Stagnation quantities or total quantities Sonic quantities D shockwaves Problem definition Prandtl s relationship Physical considerations Shock relationships Total quantities and 1D shockwaves Rankine-Hugoniot s relationship Exercises and problems Exercise 1 : Mach number and compressibility effects Exercise # 2 : pressure and temperature at the nose of an airplane in transonic flight Problem # 1 : velocity measurement in a supersonic flow Some key quantities 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 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 onde acoustique a p ρ T a + da p + dp ρ + dρ T + dt (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,

67 4.1. Some key quantities 67 the mass, momentum and energy conservation equations thus read : S S S ρ U nds = 0 ρu( U n)ds = S ρe( U n)ds = p nds S p( U n)ds 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 : ρ 1 u 1 = ρ 2 u 2 ρ 1 u p 1 = ρ 2 u p 2 ρ 1 u 1 H 1 = ρ 2 u 2 H 2 where u i denotes the velocity (single component for the 1D flow) in state i. The (specific) total enthalpy H is defined by H = E + p/ρ with the (specific) total energy E such that E = e u2 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. (4.1) Let us apply the set of equations (4.1) to the two states on each part of the acoustic wave : u p ρ u 2 p ρ 1 2 Figure 4.2: Control volume for the analysis of a steady flow made of two constant states separated by a steady wave. u 1 = a, ρ 1 = ρ and p 1 = p on one side, u 2 = a + da, ρ 2 = ρ + dρ and p 2 = 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 : a 2 = dp 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 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 : a 2 = ( p 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 : Consequently, it is possible to write : β s = 1 v ( ) v = 1 p s ρ a 2 = 1 ρβ s ( ) ρ p s and since it was established in the previous lecture that, for a perfect gas, β s = 1 it followd γp eventually the expression of the local velocity of sound as a function of the state variables p and ρ reads as : γp 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 = γrt (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 : M = u 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.

69 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 m) lies between 0.8 and 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 = 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 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 Exercise : computing the speed of sound Derive the identity a 2 = dp dρ. Let us start from the 1D Euler equations written to relate states 1 and 2 on both sides of the sound wave : ρ 1 u 1 = ρ 2 u 2 ρ 1 u p 1 = ρ 2 u p 2 ρ 1 u 1 H 1 = ρ 2 u 2 H 2 State 1 is selected as the reference state u 1 = a, ρ 1 = ρ and p 1 = p while state 2 corresponds to the perturbed state induced by the acoustic wave propagation : u 2 = a + da, ρ 2 = ρ + dρ and p 2 = 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 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 + a 2 dρ + dp = 0 Replacing in this identity ρda with adρ leads to the identity : a 2 = dp dρ 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 p 0, a stagnation or total temperature T 0. Sometimes, to underline the isentropic process leading to the stagnation state from the local state, these total quantities are denoted p i, T i. 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 H 0 computed using the stagnation state : H = H 0 It was established in the previous lecture, the specific total enthalpy is such that (neglecting body forces) : H = h U 2 = e + p ρ U 2 = p (γ 1)ρ + p ρ + 1 U 2 2 = γ γ 1 p ρ + 1 U 2 2 Since the speed of sound for a perfect gas is such that a 2 = γp/ρ, the specific total enthalpy can be expressed as a function of a 2 and U 2 = u 2 (for a 1D flow) as follows : H = a2 γ u2 (4.4) Alternatively, the specific total enthalpy can also be expressed as a function of the temperature T and the velocity u. Since a 2 = γrt and since the spefici heat capacity at constant pressure C p is defined as C p = γr/(γ 1) for a perfect gas, it is also possible to write : H = C p T u2 (4.5)

71 4.1. Some key quantities 71 where we can identify C p T as the specific enthalpy for a perfect gas. Since the flow velocity u is zero in the stagnation state, the total enthalpy H 0 computed for the stagnation state depends only on the temperature T 0 in this stagnation state, also known as the stagnation or total temperature : H 0 = C p T 0 Since H 0 is a characteristic constant of an adiabatic flow, so is the total temperature T 0. In an adiabatic flow, the total temperature T 0 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 T 0. 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 : C p T + u2 2 = C pt 0 (4.6) 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 C p T and identifying a 2 = γrt then M 2 = u 2 /a 2 yields the very important relationship : T 0 T = 1 + γ 1 M 2 (4.7) 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 p 0 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 ρ 0. We have seen previously the isentropic flow of a perfect gas is such that : p ρ γ = cste 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 (p 0 = ρ 0 rt 0 ). 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 (p 0 ) 1, (ρ 0 ) 1 and (T 0 ) 1 while the fluid particle in state 2 reaches isentropically a stagnation state (p 0 ) 2, (ρ 0 ) 2 and (T 0 ) 2. Note that the flow being isentropic it is reversible and adiabatic; the adiabaticity property is sufficient to ensure (T 0 ) 1 = (T 0 ) 2 = T 0.

72 72 1D compressible flow analysis Since (p 0 ) 1 = (ρ 0 ) 1 rt 0 and (p 0 ) 2 = (ρ 0 ) 2 rt 0, taking the ratio of the stagnation pressures (p 0 ) 1 and (p 0 ) 2 yields : (p 0 ) 1 = (ρ 0) 1 (T 0) 1 (p 0 ) 2 (ρ 0 ) 2 (T 0 ) }{{ 2 } =1 so that Besides, the isentropic flow assumptions allows to write : hence, reorganizing : (p 0 ) 1 (p 0 ) 2 = (ρ 0) 1 (ρ 0 ) 2 (4.8) (p 0 ) 1 (ρ 0 ) γ 1 (p 0 ) 1 (p 0 ) 2 = = (p 0) 2 (ρ 0 ) γ 2 ( ) (ρ0 ) γ 1 (4.9) (ρ 0 ) 2 It can be deduced from (4.8) and (4.9) that (p 0 ) 1 /(p 0 ) 2 = 1 and (ρ 0 ) 1 /(ρ 0 ) 2 = 1. Since (p 0 ) 1 = (p 0 ) 2 and (ρ 0 ) 1 = (ρ 0 ) 2 whatever the local states 1 and 2 in the isentropic flow, it can be concluded the stagnation or total pressure p 0 is constant in an isentropic flow and so is the stagnation or total density ρ 0. 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 p 0 /p and the temperature T 0 /T : p 0 p = ( ) γ T0 γ 1 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 : ( p 0 p = 1 + γ 1 ) γ M 2 γ 1 2 (4.10) We derive in a similar way : ( ρ 0 ρ = 1 + γ 1 ) 1 M 2 γ 1 2 (4.11) 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

73 4.1. Some key quantities 73 equal to (p 0 ) up, (ρ 0 ) up say, and also constant downstream of the shock, respectively equal to (p 0 ) down, (ρ 0 ) 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 (p 0 ) up, (ρ 0 ) up and (p 0 ) down, (ρ 0 ) down as constant stagnation properties 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 : M = u a (4.12) 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 : H = a2 γ u2 = H = a2 γ u2 Since u = a by definition of the sonic state, it follows that : a 2 = γ Dividing next (4.13) by u 2 yields : a 2 γ 1 u 2 or a 2 = 2 2 γ + 1 a2 + γ 1 γ + 1 u2 (4.13) M 2 = 2 [ γ + 1 M 2 ] (γ 1) or M 2 = (γ + 1)M (γ 1)M 2 (4.14) It is also important to notice the following properties :

74 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 = m/s, a local static pressure p = 0.9 atm and a local static temperature T = 250 K. Assuming the flow isentropic, compute p 0, T 0, 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 0 T = 1 + γ 1 M 2 = f(m) 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 = 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 T 0 = T = 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 : ( p0 p ) = ( T0 T ) γ γ 1 hence p 0 = 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 0 = f(1) = γ + 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 = 0.833T 0 and for the case under study, using the computed value of T 0, we find T = 229 K. The flow being assumed isentropic, the sonic pressure can be computed applying again the isentropic flow identity : ( ) ( ) γ p0 T0 γ 1 = p hence p = p 0 = 0.66 atm. Finally the sonic speed of sound a is given by a = γrt hence a = 317m s 1. T

75 4.2. 1D shockwaves D shockwaves 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 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 ρ 2, u 2 and p 2 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. ρ 1 u 1 Etat 1 p 1 T 1 M 1 a 1 ρ 2 Etat 2 u 2 p 2 T 2 M 2 a 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) 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) : ρ 1 u p 1 = ρ 2 u p 2 Introducing the speed of sound through a 2 = γp/ρ (written for states 1 and 2) and using (4.13) to express a 1 as a function of u 1 and a and similarly a 2 as a function of u 2 and a, the following Prandtl s identity is obtained (see the exercise proposed below for the detailed proof of this identity) : Introducing the characteristic Mach number we can also write : a 2 = u 1 u 2 (4.15) (M ) 2 = 1 (M ) 1 (4.16) 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 76 1D compressible flow analysis Mach number share the same variation with respect to unity, we can also state that if M 1 < 1 then M 2 > 1 while if M 1 > 1 then M 2 < 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 As will be explained next, only the first configuration, namely the supersonic flow becoming subsonic through the shockwave, is physically admissible. M > 1 M < ou M < 1 M > 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 : a 2 = u 1 u 2 Let us start the proof from the conservation of momentum between 1 and 2 : p 1 + ρ 1 u 2 1 = p 2 + ρ 2 u 2 2 The pressure p i (i = 1, 2) can be related to a 2 i using the definition of the sound speed : a 2 i = γp i /ρ i so that : which is readily simplified into : ρ 1 u 1 (u 1 + a2 1 γu 1 ) = ρ 2 u 2 (u 2 + a2 2 γu 2 ) u 1 + a2 1 γu 1 = u 2 + a2 2 γu 2 when taking into account the mass conservation ρ 1 u 1 = ρ 2 u 2. We can also deduce from (4.13) an expression for a 2 i as a function of a2 and u 2 i : a 2 i = γ Replacing into the above identity yields : γ + 1 2γ (u 1 + a2 a 2 γ 1 u 2 i 2 ) = γ + 1 u 1 2γ (u 2 + a2 u 2 )

77 4.2. 1D shockwaves 77 or else, after simplification and reorganization : a 2 (u 2 u 1 ) = (u 2 u 1 )u 1 u 2 Since it is assumed u 1 u 2 (finite variation of the flow properties through the shockwave), the above equality can be simplified by the difference (u 2 u 1 ) yielding the Prandtl s identity 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, M 1 > 1, becoming subsonic downstream of the shockwave, M 2 < 1.

78 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 M 1 > 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 Shock relationships From a quantitative viewpoint, a relationship between the downstream Mach number M 2 and the upstream Mach number M 1 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 : M 2 2 = 1 + ( γ 1 )M1 2 2 γm1 2 (γ 1 ) 2 (4.17) 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 M 1 = 2, it is found that M The relationship (4.17) is tabulated and appendices with such shock tables will be provided along the lecture notes. The upstream Mach number M 1 is an extremely interesting parameter to express the jump of physical quantities through the shock, that is the ratio say ϕ 2 /ϕ 1 where ϕ can be the static pressure, static temperature, density, velocity,... If a relationship of the form ϕ 2 /ϕ 1 = f(m 1 ) is derived, it makes possible to compute the value ϕ 2 downstream of the shock when the values of ϕ 1 and M 1 are available. The density jump and the velocity jump through the shock can be

79 4.2. 1D shockwaves 79 computed as follows. The mass conservation yields the identity : which can be recast into the following form : ρ 1 u 1 = ρ 2 u 2 ρ 2 ρ 1 = u 1 u 2 = u2 1 u 1 u 2 = u2 1 a 2 = (M ) 2 1 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 : u 1 u 2 = ρ 2 ρ 1 = (γ + 1)M (γ 1)M 2 1 (4.18) Since M 1 > 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 p 2 /p 1 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 : p 2 = 1 + 2γ p 1 γ + 1 (M 1 2 1) (4.19) 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 M 1 tends to 1, all the ratios of the form ϕ 2 /ϕ 1 tend to 1 : there is no longer a discontinuity in the flow, the shock is said to be evanescent. When M 1, it can be observed both the pressure jump and the temperature jump also tend to infinity while the density jump tends to the finite value γ 1, that is 6 for γ = 1.4. The downstream Mach number M 2 tends to the limit value γ 1 2γ γ+1, hence 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 M 1 = 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 T 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 80 1D compressible flow analysis Exercise : pressure jump through a 1D shockwave Show the pressure jump p 2 /p 1 through a 1D shockwave is given by : p 2 = 1 + 2γ p 1 γ + 1 (M 1 2 1) The momentum conservation written between states 1 and 2 on both sides (respectively upstream and downstream) of the shockwave takes the form : p 1 + ρ 1 u 2 1 = p 2 + ρ 2 u 2 2 so that and p 2 p 1 p 1 = 1 p 1 (ρ 1 u 2 1 ρ 2 u 2 2) p 2 p 1 = 1 + ρ 1u 2 1 p 1 (1 u 2 u 1 ) Using the relationship (4.18) which provides u 2 /u 1 as a function of M 2 1 : and also identifying : u 2 u 1 = 2 + (γ 1)M 2 1 (γ + 1)M 2 1 ρ 1 u 2 1 p 1 = γ u2 1 a 2 1 it is possible to rewrite the pressure jump as : = γm 2 1 p 2 = 1 + γm1 2 ( (γ + 1)M (γ 1)M 1 2 p 1 (γ + 1)M1 2 ) hence, after a final simplification, the expected identity Total quantities and 1D shockwaves As far as total or stagnation variables are concerned, the flow configuration to deal with is summarized in Fig Since the total temperature T 0 remains constant in an adiabatic flow, it remains in particular constant when the fluid particle crosses the shockwave. Hence the identity (T 0 ) 1 = (T 0 ) 2 = T 0 where(t 0 ) 1 (respectively (T 0 ) 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 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 (T 0 ) 2 associated to state 2 downstream of the detached shock developing in front of the missile in supersonic flight. This total temperature

81 4.2. 1D shockwaves 81 Etat 1 Etat 2 Grandeurs statiques : S p T M S p Grandeurs d arrêt associées 1 0 (évolution isentropique) T (u=0) Grandeurs statiques : S p T M Grandeurs d arrêt associées S p T (évolution isentropique) (u=0) 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 T inf Missile Figure 4.8: Missile flying at supersonic speed above sea level. Niveau de la mer

82 82 1D compressible flow analysis (T 0 ) 2 is equal to the total temperature upstream of the shock (T 0 ) 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 (T 0 ) 1 can be computed using formula (4.7) so that : Tnose = (T 0 ) 2 = (T 0 ) 1 = (1 + γ 1 M 2 2 )T Assuming the (warm) sea level temperature around 300K, it comes Tnose = 540K or Tnose 270 C. The total pressure (p 0 ) 1 associated with the isentropic flow upstream of the shockwave differs from the total pressure (p 0 ) 2 associated with the isentropic flow downstream of the shockwave (the same is true for the total density values : (ρ 0 ) 1 (ρ 0 ) 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 : de = T ds + p ρ 2 dρ or else, for a calorifically perfect gas such that p = ρrt and de = C v dt : dt ds = C v T r dρ ρ This relationship can be integrated between states 1 and 2 on each side of the shock wave so that : ( ) ( ) T2 ρ2 s 2 s 1 = C v log r log or else, using again the perfect gas thermal equation of state and the Mayer s relationship C p C v = r : ( ) ( ) T2 p2 s 2 s 1 = C p log r log (4.20) where log denotes the natural logarithm (also denoted sometimes ln). Since (4.20) is valid for any states with entropy s 1 and s 2 it is valid in particular for the stagnation states on both sides of the shockwave so that : ( ) ( ) (T0 ) 2 (p0 ) 2 s 2 s 1 = C p log r log (T 0 ) 1 (p 0 ) 1 T 1 T 1 ρ 1 p 1 Now, since total temperature is conserved through the shock, (T 0 ) 2 /(T 0 ) 1 relationship simplifies into : ( ) (p0 ) 2 s 2 s 1 = r log (p 0 ) 1 = 1 so that the According to the second law of thermodynamics, the irreversible entropy variation throught the shockwave, s 2 s 1, is necessarily a positive quantity since the irreversible entropy increases along the trajectory of a fluid particle. This implies the ratio of total pressure (p 0) 2 (p 0 ) 1 is less than unity that is (p 0 ) 2 < (p 0 ) 1. The shockwave creates a total pressure loss. This loss is itself directly

83 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 (p 0 ) 1 can be be expressed as the following product : (p 0 ) 2 (p 0 ) 1 = (p 0) 2 p 2 p 2 p 1 p 1 (p 0 ) 1 The first term of the product is a function of M2 2 according to the isentropic flow identity (4.10) which, when applied to the isentropic flow downstream of the shockwave, yields : ( (p 0 ) 2 = p γ 1 M2 2 2 ) γ γ 1 This ratio can also be expressed as a function of M1 2 since the Mach number values on both sides of the shockwave are linked by the identity (4.17) : M 2 2 = 1 + ( γ 1 )M1 2 2 γm1 2 (γ 1 ) 2 The third term of the product is also a function of M1 2 according again to the isentropic flow identity (4.10) written for the isentropic flow upstream of the shockwave : ( (p 0 ) 1 = p γ 1 M1 2 2 ) γ γ 1 Finally, the second term, corresponding to the static pressure jump through the shockwave, is given by the relationship (4.19) : p 2 = 1 + 2γ p 1 γ + 1 (M 1 2 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 M 1 : [ (p 0 ) 2 = 1 + 2γ ] 1 [ (p 0 ) 1 γ + 1 (M 1 2 γ 1 1) 1 2 γ + 1 (1 1 ] γ γ 1 M1 2 ) A few comments can be made regarding the identity (4.21) : (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 84 1D compressible flow analysis - also, it must be noted that (4.21) could be applied for M 1 > 1 as well as for M 1 < 1. However, if M 1 < 1 the total pressure ratio (p 0 ) 2 /(p 0 ) 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 M 1 > 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 Rankine-Hugoniot s relationship The Rankine-Hugoniot s equation, or Hugoniot s equation, links the static pressure jump p 2 /p 1 through a 1D or normal shock to the corresponding density jump ρ 2 /ρ 1. Since both the pressure ratio and the density ratio are known as functions of the upstream Mach number M 1 (or M 2 1 ), this Mach number can be eliminated between both identities in order to obtain the desired relationship. It is easy to establish the following identity : or else ρ 2 ρ 1 = (γ + 1)p 2 + (γ 1)p 1 (γ + 1)p 1 + (γ 1)p 2 p 2 p 1 = (γ + 1)ρ 2 (γ 1)ρ 1 (γ + 1)ρ 1 (γ 1)ρ 2 The usual form retained for the relationship between the density ratio and the static pressure ratio through a 1D shockwave is : p 2 p 1 = γ + 1 ρ 2 1 γ 1 ρ 1 γ + 1 γ 1 ρ (4.22) 2 ρ 1 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 2 p 1 = ( ρ2 ρ 1 ) γ (4.23) Both formulae (4.22) and (4.23) are plotted in Fig more precisely the static pressure ratio p 2 /p 1, greater than unity through a compression, is plotted versus the density ratio ρ 1 /ρ 2 less than unity through the compression since ρ 2 /ρ 1 is also greater than unity through the compression. The limit value of the ratio ρ 2 /ρ 1 when the pressure jump p 2 /p 1 through the shockwave tends to infinity is such that : ρ 2 lim = γ + 1 p 2 ρ p 1 γ 1 1

85 4.3. Exercises and problems 85 that is ρ 2 /ρ 1 6 or ρ 1 /ρ 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 ρ 1 /ρ 2 say of 0.25 (that is ρ 2 /ρ 1 = 4) the corresponding pressure increase in the isentropic compression is p 2 /p 1 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) 15 p2 / p 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 ρ 1 /ρ 2 = 1/6 for the Rankine-Hugoniot plot. 4.3 Exercises and problems Exercise 1 : Mach number and compressibility effects Question In an isentropic flow of a perfect gas, the ratio between the stagnation or total density ρ 0 (constant in the isentropic flow) and the local density ρ is expressed as the following function of

86 86 1D compressible flow analysis the local Mach number M : ( ρ 0 ρ = 1 + γ 1 ) 1 M 2 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 ρ 0 /ρ and the local Mach number, valid throughout an isentropic flow, is plotted in Fig (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 ρ 0 does not exceed 1% as long as M remains below 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 γ 1 Figure 4.10: Total to local/static density ratio ρ 0 /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 ρ 0 /ρ (x coordinate), according to the identity : [ M = 2 (ρ0 ) γ 1 1] γ 1 ρ

87 4.3. Exercises and problems 87 Figure 4.11: Local Mach number M as a function of the total to static density ratio ρ 0 /ρ for an isentropic flow of perfect gas (γ = 1.4) 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 m where the air temperature is T = 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 (U A = 0, see Fig.4.12) and we wish to compute the pressure and the temperature at point A, that is the total pressure (p 0 ) A and the total temperature (T 0 ) 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 T A as a function of the upstream flow conditions and compute its numerical value. Does the quantity T A have a physical significance on other streamlines? 3) Determine p A 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 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 = K yields a 300 m/s. At ground level in standard conditions, T = K so that a ground = /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 = K, U = 270 m/s, a = 300 m/s, M = 0.9, p = 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)) : h + U 2 = constant along a streamline 2 or, since h = C p T for a perfect gas : C p T + U 2 2 = constant along a streamline with C p = γr/(γ 1). Writing this relationship (in the frame of reference attached to the airplane) between the upstream flow, where T = T = K, U = 270 m/s, and the stagnation point, where T = T A = T 0 the constant total temperature for the flow and U A = 0, yields : C p T + U 2 2 = C pt 0 We can also write this expression in the form (4.7) : T 0 = 1 + γ 1 T 2 M 2

89 4.3. Exercises and problems 89 so that T A = T 0 = T = K. The total temperature T 0 is directly related to the total total enthalpy H 0 since H 0 = C p T 0. 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 H 0. 3) The flow being isentropic, the identity (4.10) can be applied at the farfield : ( p 0 = 1 + γ 1 ) γ M 2 γ 1 p 2 with p 0 the constant total pressure for the isentropic flow under study, also equal to the pressure at the stagnation point A. Since p = 250 hp a, ( p 0 = )( ) p = 1.69 p = hp a. p We have previously established that s = C v log (plus a constant reference value, which 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 = C p log 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 T 0 and p becomes p 0 : ρ γ p 1 1 γ s = C p log T 0 For a given total enthalpy or a given total temperature T 0 (which will be constant throughout an adiabatic flow, which can include irreversible processes such as shockwaves), the total pressure p 0 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. p 1 1 γ 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 90 1D compressible flow analysis A B 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.

91 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 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. M 1 M 2 p 02 /p Table 4.1: 1D shockwave relationships (γ = 1.4) Answers 1/ Assuming a quasi-uniform state upstream of the detached shockwave, the pressure p 1 upstream of the shock is obtained from the measurement made at point B. Downstream of the shock, the total temperature (T 0 ) 2 and total pressure (p 0 ) 2 associated to state 2 are known from the measurements made at point A. The shuttle velocity is given by u 1. Therefore, we are looking for a way to compute u 1 from the known values p 1, (p 0 ) 2 and (T 0 ) 2. 2/ The velocity u 1 can be computed from the Mach number M 1 and the speed of sound a 1, which is known provided the temperature T 1 is known : u 1 = a 1 M 1 = γrt 1 M 1 The upstream temperature T 1 is not directly known but it can be computed from the total temperature associated with the upstream state 1 and the Mach number M 1. Since the total temperature is conserved in an adiabatic flow, T 01 = T 02. It follows that : so that : (T 0 ) 2 = 1 + γ 1 T 1 2 M 2 1 u 1 = M 1 γrt γ 1 M1 2 2 (4.24)

92 92 1D compressible flow analysis With (T 0 ) 2 available, u 1 can be computed provided the upstream Mach number M 1 is known. 3/ The remaining problem to solve is to find the upstream Mach number M 1 from the measured pressures, namely the downstream total pressure p 02 and the upstream static pressure p 1. Let us form the ratio p 02 /p 1 and try to express this non-dimensional ratio as a function of the Mach number M 1 : p 02 = p 0 2 p 1 p 2 For the downstream isentropic flow we can write : p 02 p 2 = ( T02 T 2 p2 p 1 ) γ γ 1 = (1 + γ 1 ) γ M2 2 γ 1 2 The downstream Mach number M 2 can be expressed as a function of the upstream Mach number M 1 using the 1D shock relationship : so that p 0 2 p 2 M 2 2 = 1 + ( γ 1 )M1 2 2 γm1 2 (γ 1 ) 2 can be expressed as a function of M 1 only. The static pressure jump through the 1D shockwave reads : p 2 = 1 + 2γ p 1 (γ + 1) (M 1 2 1) A few simple calculations yield the following identity for the ratio between p 02 and p 1 : p 02 = γ + 1 p 1 2 M 2 1 γ + 1 M γ γ + 1 M 2 1 γ 1 γ γ 1 (4.25) Inverting the identity (4.25) to obtain M 1 for the known value of the pressure ratio (p 0 ) 2 /p 1 is not straightforward. However, tabulated values are provided for (4.25) so that it becomes easy to estimate M 1 from this know ratio. Once M 1 is know, its value can be inserted in (4.24), along with the measured value of (T 0 ) 2, to obtain u 1. 4/ Since p 02 /p , Table 4.1 provides M 1 = 5. It comes : u 1 = = 1400 m s 1 ( 5000 km / h) Remark : In the limit of (very) large Mach number values, formula (4.24) giving the velocity simplifies into : 2γr u 1 = γ 1 T 0 2 = 2C p T 02

93 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 u 1 = 1533 m/s, a value which differs by almost 10% from the correct value.

94 94 1D compressible flow analysis

95 Lecture 5 Quasi-1D compressible flows Contents 5.1 Physical model Governing equations Velocity-area relationship Influence coefficients Isentropic flow in a nozzle Flow analysis Mass flowrate Analysis of the flow through a converging nozzle General analysis Example of application Analysis of the flow through a diverging nozzle Analysis of flow regimes in a converging-diverging nozzle General analysis Example of application Exercises and problems Exercise # 1 : Flow in a choked nozzle Exercise #2 : Flow in a diverging nozzle Problem # 1 : analysis of a scramjet performance Physical model Governing equations Let us consider the compressible flow (from left to right) in the nozzle (varying section tube) displayed in Fig 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 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 1 u 1 A 1 = ρ 2 u 2 A 2 A2 p 1 A 1 + ρ 1 u 2 1 A 1 + pda = p 2 A 2 + ρ 2 u 2 2A 2 (5.1) A 1 H 1 = H 2 where index 1 denotes the physical state in the upstream section of area A 1 and index 2 denotes the physical state in the upstream section of area A 2. 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 dh + udu = 0 (5.2)

97 5.1. Physical model 97 where the (specific) enthalpy h is such that h = e + p/ρ and h = C p T 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 Velocity-area relationship The identity d(ρua) = 0 can also be rewritten as d(ρua) ρua yields : = 0 so that an immediate expansion dρ ρ + du u + da A = 0 (5.3) Since the flow is assumed isentropic, the speed of sound a is such that : a 2 = dp dρ Using the momentum equation dp = ρudu, we obtained dρ ρ = du u2 a 2 u or dρ ρ = M 2 du u (5.4) 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 A = (M 2 1) du u (5.5) This identity is important for a preliminary qualitative analysis of compressible flows in a nozzle. Case of a subsonic flow : M < 1 u augmente M < 1 u diminue 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. According to (5.5), quantities da A and du are of opposite sign in that case : a decrease in the 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 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 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 M > 1 u augmente 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. According to (5.5), quantities da A and du are of the same sign in that case This leads to 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. If M = 1 then (5.5) indicates that da = 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

99 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 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 convergingdiverging geometry is also called a De Laval nozzle in reference to the Swedish engineer who first introduced it at the end of the 19 th century (see Fig. 5.7) Influence coefficients Formula (5.5) can also be expressed as : du u = 1 da M 2 1 A (5.6) 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.

100 100 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 = a 2 dρ. Since the local 1D momentum conservation reads dp = ρudu, it follows that : dρ ρ = M 2 du u or else, using (5.6) : dρ ρ = M 2 da 1 M 2 A (5.7) 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 p = a2 dρ p = γp dρ dρ = γ ρp ρ We note this expression can also be directly obtained by differentiating the relationship p/ρ γ = 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 p = γm 2 da 1 M 2 (5.8) A The energy conservation equation written in differential form yields, using dh = C p dt = γr γ 1 dt : CpdT + udu = 0 Dividing this identity by a 2 = γrt and introducing the Mach number M = u/a yields : 1 dt γ 1 T + M 2 du u = 0

101 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 converging nozzle ( da A < 0). 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 diverging nozzle ( da A > 0). Using (5.6) provides the temperature variation as a function of the section variation : dt T (γ 1)M 2 da = 1 M 2 A (5.9) 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 : M = u a = u(γrt ) 1/2 Differentiating yields : and using (5.6), (5.9) provides : dm M = dm M = du u 1 dt 2 T ( 1 + γ 1 ) M 2 2 M 2 1 da A (5.10) 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.

102 102 Quasi-1D compressible flows in nozzles 5.2 Isentropic flow in a nozzle 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 relationship of the form A = f(m), where A ref denotes the area of a reference section. A ref 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 : so that : ρua = ρ u A A A = ρ ρ u u = ρ ρ u a a u = ρ ρ 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 ρ0 ρ 0 ρ where ρ 0 denotes the stagnation or total density. The flow being assumed isentropic, relationships (4.7) and (4.11)) yield : 1 M 1 1 ρ 0 ρ = (T 0 T ) γ 1 γ 1 = (1 + M 2 ) γ 1 = f(m) 2 When this identity is written for the sonic state (M = 1) it comes : ρ 0 = f(m = 1) = ( γ ) γ 1 ρ 2 The following relationship between the section (more precisely the ratio A/A ) and the Mach number is obtained : A = 1 [ 2 A M (γ + 1) (1 + γ 1 ] (γ + 1) M 2 ) 2(γ 1) 2 (5.11) This relationship, know as the area-mach number identity, is essential for the analysis of flows in nozzles.

103 5.2. Isentropic flow in a nozzle 103 Note it can also be seen as a relationship of the form M = f( A A ). Thus, for an isentropic flow, 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 : p = (1 + γ 1 γ M 2 ) γ 1 2 p 0 it follows that a relationship of the form p = g( A ) also exists for the isentropic flow in a p 0 A nozzle. Combining the above identity with (5.11) yields : A A = γ 1 2 ( 2 ) γ+1 γ + 1 γ 1 1 ( p p 0 ) 2 ( ) γ+1 γ p γ p 0 (5.12) ( ) A The graphical display of the relationships M = f corresponding to (5.11) and p/p 0 = ( ) A A f corresponding to (5.12) is proposed in Figure 5.8. Equivalently, the plots A = f(m) A A and A ( ) p = f are displayed in Figure 5.9. Several important observations can be made A p 0 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 A thr with thr standing for throat (or A col 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/p 0 ), one subsonic and the other supersonic. For instance A/A = corresponds to M = 0.41 and p/p 0 = if the flow regime in the nozzle section is subsonic while A/A = corresponds to M = 1.90 and p/p 0 = if the flow regime in the nozzle section is supersonic.

104 104 Quasi-1D compressible flows in nozzles Branche subsonique Nombre de Mach M Branche supersonique Rapport de pression p/p Branche subsonique 0.1 Branche supersonique 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/p 0 and the area ratio A/A (right). Branche subsonique stands for Subsonic branch while Branche supersonique stands for Supersonic branch A / A* 6 A / A* 4 section sonique 4 2 section sonique M > 1 p/p0= M < 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/p 0 (right).

105 5.2. Isentropic flow in a nozzle Mass flowrate Mass conservation through the nozzle yields the conservation of the mass flowrate : ṁ = ρua = constant The constant mass flowrate can also be expressed using the constant stagnation state associated with the isentropic flow in the nozzle : ṁ = ρua = ρama = ρ ρ 0 a a 0 MAρ 0 a 0 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 p 0 = ρ 0 rt 0. The definition of the speed of sound provides, for an ideal gas, a = γrt hence a 0 = γrt 0. The constant mass flowrate can also be expressed as : ( p ṁ = p 0 ) T0 T p 0 rt 0 γrt0 MA Using eventually the relationships (4.7) and (4.10) valid for isentropic flows, the ratios p 0 /p and T 0 /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 p 0 and temperature T 0 : The ratio ṁ/a is therefore given by : ( γ ṁ = p 0 M 1 + γ 1 ) γ+1 M 2 2(γ 1) A (5.13) rt 0 2 ( ṁ γ A = p 0 M 1 + γ 1 rt 0 2 M 2 ) γ+1 2(γ 1) and depends only on the Mach number M since the stagnation or total properties p 0, T 0 are constant for the isentropic flow under study. Since the mass flowrate ṁ is a flow constant for the nozzle flow, the ratio ṁ/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 : Denoting Γ the constant value : ṁ max γ = p 0 A rt 0 Γ = γ ( 2 γ + 1 ( 2 γ + 1 ) γ + 1 γ 1 ) γ + 1 γ 1 such that Γ(γ = 1.4) = , the maximum mass flowrate of air passing through a nozzle is given by : ṁ max = p 0 rt0 ΓA (5.14)

106 106 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 A min 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 : ṁ max = p 0 rt0 A throat (5.15) 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 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 p 0, T 0, to a medium (space for instance) where the ambient pressure takes the value p s. The pressure p s is also called the discharge pressure. For the flow to take place, the discharge pressure p s must be necessarily lower than the reservoir pressure p 0. Let us also denote A e the area of the exit section of the nozzle. Note the pressure p e in the exit section A e is not necessarily equal to the ambient or discharge pressure p s as will be explained next. Let us first assume the ambient pressure p s is slightly below the reservoir pressure p 0. 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 p 0 in the reservoir down to p e in the exit section, the value p e being equal to the value of the ambient or discharge pressure p s : p e = p s. 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 M e can be computed from the ratio between the total pressure p 0 and the exit pressure p e known from the value of the discharge pressure p s. For the isentropic flow under study, we can write : so that : p 0 = p ( 0 = p s p e M e = 2 γ γ 1 Me 2 2 ( p0 p s ) γ 1 γ ) γ γ 1 1 (5.16) When the discharge pressure is lowered (we assume the reservoir pressure or stagnation pressure p 0 remains unchanged throughout the analysis), the increase in the pressure difference between

107 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. Géométrie du 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 The limiting configuration corresponds to the value of p s 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/p 0 = Let us recall this value is such that : ( p 0 p = 1 + γ 1 ) γ ( γ 1 γ + 1 = 2 ) γ γ 1 = (1.2) 3.5 = hence p 0 /p = 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 p 0 : - if p s > p 0 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 p s. The mass flowrate passing through the nozzle increases when p s decreases and is given by : ( γ ṁ = p 0 M e rt γ 1 Me 2 2 ) γ+1 2(γ 1) Ae (5.17)

108 108 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 p s. p s décroissante stands for decreasing p s if p s p 0 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 M e = 1 that is : ṁ max = p 0 rt0 ΓA e (5.18) with Γ = for γ = 1.4. The nozzle flow is said to be choked. The exit pressure p e is equal to the critical value p 0 and a multi-dimensional expansion wave appears immediately downstream of the exit section in order to further decrease the pressure from p e down to p s < p e. When p s is lowered, below the critical value p 0, 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 p 0, T 0 and the exit section area A e. Only the expansion wave outside the nozzle is further modified by the decrease of the ambient or discharge pressure p s in order to lower the pressure from p e = p 0 down to p s Example of application Let us consider the converging nozzle displayed in Fig The stagnation pressure p 0 is assumed equal to p 0 = 2 bar while the stagnation temperature is assumed equal to 300 K. The inlet section area A 1 is equal to 0.5 m 2 while the outlet section area A 2 = A e is equal to 0.1 m 2. Figure 5.13 displays the evolution of the Mach number distribution in this converging nozzle for 3 successive values of the discharge pressure p s : p s = 1.9 bar, p s = 1.5 bar and p s = 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 p s = = bar. It can be checked on Fig the Mach number reaches a value slightly below unity in the exit section when p s = 1.1 bar, that is when the discharge pressure is slightly above its pivotal value.

109 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, p 0 = 2 bar and T 0 = 300 K Mach x/l Figure 5.13: Mach number distribution in the converging nozzle with stagnation pressure p 0 = 2 bar. Blue curve : discharge or ambient pressure p s = 1.9 bar; green curve ; p s = 1.5 bar; red curve : p s = 1.1 bar.

110 110 Quasi-1D compressible flows in nozzles For p s = 1.1 bar, applying the formula (5.19) which gives M e as a function of p s /p 0 yields M e = (γ = 1.4). Using M e and knowing A e as well as the stagnation conditions, the mass flowrate ṁ can be computed for p s = 1.1 bar thanks to formula (5.17). It is found that ṁ = 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) : ṁ max = kg/s. Similar calculations of the mass flowrate can be performed for p s = 1.9 bar and p s = 1.5 bar. Similarly, Fig.5.14 displays the distribution of the pressure ratio p/p 0 in the converging nozzle for 3 successive values of the discharge pressure : p s = 1.9 bar, p s = 1.5 bar and p s = 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 p i in the inlet section A i and down to the exit pressure p e, 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 p s = 1.9 bar to p s = 1.1 bar. The inlet pressure p i or the ratio p i /p 0 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 M e (< 1) is computed in the exit section from the ratio p e /p 0, equal to p s /p 0 (as long again as the nozzle is not choked), the value of the sonic section area A can be computed since A e is also known. The ratio between the inlet section area and the sonic section area A i /A can then be computed and the inlet Mach number M i follows as well as the pressure ratio p i /p 0. For p s = 1.9 bar, we obtain p s /p 0 = Table II of appendix?? provides, in its subsonic part, M s = 0.27 and A e /A = hence A = m 2. It follows that A i /A = hence, reading again the subsonic part of Table II, an inlet Mach number M i between 0.05 and 0.06 and a pressure ratio p i /p 0 between and (these values are well in line with the M(x) and p s (x)/p 0 distributions plotted in Fig and For p s = 1.1 bar, we compute p s /p 0 = 0.55 and the subsonic part of Table II provides a corresponding value of M s close to 0.96 and A e /A so that A = m 2 (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 A i /A = hence, from reading the subsonic part of Table II, an inlet Mach number M i between 0.11 and 0.12 and a ratio p i /p 0 between and these values are again consistent with the distributions plotted in Fig and Finally, let us analyze the evolution of the exit Mach number M e and the mass flowrate ṁ as a function of the pressure ratio p s /p 0. These respective evolutions are plotted in Fig.5.15 and The exit Mach number M e first increases when the pressure ratio p s /p 0 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 p s reaches the critical value p0; for this specific value, we have M e = 1 in the exit section and this is the last value of the discharge or ambient pressure which is equal to the exit pressure p e. For values of the discharge pressure p s below the critical value p0, the exit section remains sonic (M e remains equal to 1) with a flow which remains frozen in the converging nozzle, equal to the unique isentropic solution with A = A e. The flow regime in the nozzle is a choked flow. The mass flowrate increases with M e hence with the decrease of the ratio p s /p 0 until it reaches its maximum value, depending only on stagnation or total conditions and the exit section area

111 5.4. Analysis of the flow through a diverging nozzle p/p x/l Figure 5.14: Pressure (p/p 0 ) distribution in the converging nozzle with stagnation pressure p 0 = 2 bar. Blue curve : discharge or ambient pressure p s = 1.9 bar; green curve ; p s = 1.5 bar; red curve : p s = 1.1 bar. A e. For an ambient pressure below the critical value p 0, the mass flowrate ṁ 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 p s keeps decreasing until vacuum (p s = 0) is reached. 5.4 Analysis of the flow through a diverging nozzle Let us consider the diverging nozzle geometry displayed in Fig 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 p s, when this pressure p s takes values between p 0 and the vacuum (p s = 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 M e such that : M e = 2 γ 1 ( p0 p s ) γ 1 γ 1 (5.19) with the exit pressure p e equal to the discharge pressure p s. 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 A i is such that M i = 1.5. The supersonic isentropic flow in the nozzle is entirely known. Indeed, from M i = 1.5 the ratio A i /A = can be deduced (reading the supersonic part of Table II in Appendix

112 112 Quasi-1D compressible flows in nozzles 1 Mach Me dans la section d éjection ps/p0 Figure 5.15: Evolution of the exit Mach number of the converging nozzle as a function of the pressure ratio p s /p 0. débit massique / débit massique maximal ps/p0 Figure 5.16: Evolution of the mass flowrate through the converging nozzle as a function of the pressure ratio p s /p 0.

113 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, p 0 = 2 bar and T 0 = 300 K. A) hence the area of the sonic section associated with this isentropic flow since A i is known : A = m 2. The pressure ratio p i /p 0 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 p i /p 0 = so that p i = bar assuming a stagnation pressure equal to p 0 = 2 bar. Since the flow is supersonic in the diverging nozzle, the flow accelerates from the inlet section A i to the outlet section A e, that is M increases and p/p 0 decreases. The flow is a supersonic (isentropic) expansion. Since A has been computed, the exit Mach number M e can be computed from the ratio A e /A : A e /A = 5.88 hence, reading the supersonic part of Table II in Appendix A, M e 3.35 and p e /p The flow accelerates isentropically from M i = 1.5 to M e = 3.35 and in the same time the static pressure decreases from p i = bar down to p e = bar. The key question is now to determine for which ambient conditions (which ambient or discharge pressure p s ) 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 p s. A pivotal value for p s is the value of the exit pressure p e corresponding to the isentropic solution in the nozzle, that is bar. If p s lies in the interval [0 bar, bar[, that is if the ambient pressure p s 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 p e = bar down to p s. 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 p e to p s. The distribution of the Mach number M and the pressure ratio p/p 0 for the converging nozzle are plotted respectively in Fig and 5.19 in the case where p s 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 p s takes any value between0.032 bar and 0 bar; only the multidimensional expansion outside the

114 114 Quasi-1D compressible flows in nozzles nozzle evolves with p s. If the ambient pressure is precisely equal to bar then the diverging nozzle is said to be adapted (to the ambient) Mach x/l Figure 5.18: Case of an ambient pressure p s below the isentropic solution exit pressure p e. Mach number distribution vs non-dimensional position x/l. If the ambient pressure p s is greater than the value 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 p e = bar up to the ambient value p s. For a range of values of p s 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 p e = bar to the larger value of the ambient pressure p s ; 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 p s 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, M e = 3.35 and p e = 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 : p s = 1 + 2γ p e γ + 1 (M e 2 1) We find the static pressure jump is such that p s /p e = so that p s = 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 p s between bar and bar. Figures 5.21 and 5.22 display respectively the Mach number distribution and the pressure ratio p/p 0 distribution in the diverging nozzle for p s = bar. The flow corresponds to the isentropic supersonic solution until the state

115 5.4. Analysis of the flow through a diverging nozzle p/p x/l Figure 5.19: Case of an ambient pressure p s below the isentropic solution exit pressure p e. Pressure ratio p/p 0 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 p s through oblique shockwaves. From left to right : increasing strenght of the compression mechanism (increasing pressure jump from p e = bar to an increasing value of p s ). Limiting case : 1D shock in the nozzle exit section.

116 116 Quasi-1D compressible flows in nozzles immediately upstream of the exit section where M e = 3.35 and p e = bar. Through the 1D shock in the exit section, the pressure jumps to p s = bar immediately downstream of the shock. Meanwhile, the Mach number decreases from M e = 3.35 to a downstream value M s which can be obtained either from the shock table I or using the shock relationship between the upstream state 1 (or M 1 = M e ) and the downstream state 2 (or M 2 = M s ) : M 2 2 = For γ = 1.4 and M e = 3.5, we obtain M s = ( γ 1 )M1 2 2 γm1 2 (γ 1 ) Mach 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 p e = bar to p s = bar. Mach number distribution vs x/l. If the ambient pressure is larger than bar and lower than p 0 = 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 p s 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 p s = 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

117 5.4. Analysis of the flow through a diverging nozzle p/p 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 p e = bar to p s = bar. Pressure ratio p/p 0 distribution vs x/l. the shock and the isentropic compression downstream of the shock adapts the nozzle flow to the ambient pressure p s : the exit pressure p e is precisely equal to p s. In the case where p s = 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 p s, stagnation conditions and nozzle geometry. The problem to solve is summarized in Fig.5.24 : - the inlet conditions of the flow are known (M i is known in section A i ) as well as the stagnation or total flow conditions in the reservoir. The total pressure p 0 in the reservoir is the constant total pressure associated with the isentropic flow upstream of the 1D shock, also denoted (p 0 ) up (in French (p 0 ) aval ). The total temperature T 0 is constant everywhere in the adiabatic flow, upstream and downstream of the 1D shockwave. - the ambient pressure p s is also known. In the present case, we assume p s = 1 bar. The pressure p e in the exit section is equal to p s. The Mach number M e 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 (p 0 ) down /p e = (p 0 ) down /p s with the downstream total pressure (p 0 ) down unknown at this stage : M e = 2 ( ) γ 1 (p0 ) down γ γ 1 p s 1

118 118 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 p s = 1 bar. Upstream of the 1D shock the flow is isentropic with an associated total pressure equal to p 0 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 p 0 because of the total pressure loss through the shock. In order to compute M e, the total pressure (p 0 ) down downstream of the 1D shock must be computed. - the 1D or normal shock is assumed to take place in a section x = x c, corresponding to a section area A c (the section area A(x) is supposed to vary linearly between A i at the inlet x = 0 to A e at the exit x = L). Finding A c will therefore immediately allow to determine the location x c. 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 ṁ = p 0 M rt γ 1 2 M 2 ) γ+1 2(γ 1) A where p 0 = (p 0 ) 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 A i, the total pressure p 0 is the upstream total pressure (p 0 ) up equal to the reservoir stagnation pressure. An immediate calculation yields ṁ = kg/s. This same mass flowrate can also be computed using its expression in the exit section : ṁ = ρ e u e A e or else : p e rt e a e M e A e = ṁ where ṁ has been put in the right-hand-side to emphasize the fact it is a known quantity. The exit pressure p e is equal to the ambient pressure p s and the speed of sound in the exit section