Space Propulsi Prof. manuel martinez- sanchez Lecture 18: Hall Thruster Efficiency For a given mass flow m and thrust f, we would like to minimize the running power P. Define a thruster efficiency (1) ere is the minimum required power. The actual power is V s。,°。°°· B p。,°°·引 Where Va is the accelerating voltage and Ia the current through the power supply (or anode current or also cathode current). Of the Ia current of electrons injected by the cathode a fraction IB goes to neutralize the beam and the rest iBs back-streams into the thruster Since no net current is lost to the walls, 16.522, Space P pessan Lecture 18 Prof. Manuel martinez Page 1 of 20
16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 1 of 20 16.522, Space Propulsion Prof. Manuel Martinez-Sanchez Lecture 18: Hall Thruster Efficiency For a given mass flow m i and thrust F, we would like to minimize the running power P. Define a thruster efficiency 2 F = 2m P ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ η i (1) where 2 F 2m ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ i is the minimum required power. The actual power is P=IVa a (2) Where Va is the accelerating voltage and Ia the current through the power supply (or anode current, or also cathode current). Of the Ia current of electrons injected by the cathode, a fraction IB goes to neutralize the beam, and the rest, IBS back-streams into the thruster. Since no net current is lost to the walls, a B BS I =I +I (3) } Va (IB) e (IB) i IBS IBS Ia (IB) e
V The thrust is due to the accelerated ions only these are created at locations along the thruster which have different potentials v(x), and hence accelerate to different speeds. Then F=cdm (4) wnere Suppose the part dmi of mi is created in the region where v decreases by dv and define an"ionization distribution function"f(v) by dm=-fv.Va or, with dm=-f( o) do From the definition, f() satisfies f(o)do=1 (7) Then, from(4)and(5), 16.522, Space P pessan Lecture 18 Prof. Manuel martinez Page 2 of 20
16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 2 of 20 The thrust is due to the accelerated ions only. These are created at locations along the thruster which have different potentials V(x), and hence accelerate to different speeds. Then F = cdmi ∫ i (4) where i 2eV c = m (5) Suppose the part dmi i of mi i is created in the region where V decreases by dV, and define an “ionization distribution function” f(V) by i a a i dm V dV = -f V V m ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ i i (6) or, with a V = V ϕ , ( ) i i dm = -f d m ϕ ϕ. i i From the definition, f (ϕ) satisfies ( ) 1 0 f d =1 ϕ ϕ ∫ (7) Then, from (4) and (5), V Va 0 x
2ev and hence the efficiency is m √Gf(o)do 2m VI Notice that the beam current IB is related to mi by IB=mi. We can therefore re- m( o f() d there each of the factors is less than unity and can be assigned a separate meaning (11) is the utilization factor",i.e. it penalizes neutral gas flow (12) the backstreaming efficiency" penalizes electron backstreaming (13)o f(o)do=ne, the"nonuniformity factor"is less than unity because of the nonuniform ion velocity It is clear that, since f(o)do=1, we want to put most of f(o )where Vo is greatest, namely, we want to produce most of the ionization near the inlet. In that case f()=8(o-1), and ne =1. A somewhat pesimistic scenario would be f(9)=1 d mi proportional to dr. i.e., ionization rate proportional to field strength In that 「√@×1×do 16.522, Space P pessan Lecture 18 Prof. Manuel martinez Page 3 of 20
16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 3 of 20 ( ) 1 a i 0 i 2eV F = m f d m ϕϕϕ ∫ i (8) and hence the efficiency is ( ) 2 2 1 a i 0 a a i 2eV m f d m = 2mV I ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ϕϕϕ ⎝ ⎠ ⎝ ⎠ η ∫ i i (9) Notice that the beam current IB is related to mi i by B i i e I= m m i i . We can therefore rewrite (9) as ( ) 2 1 B a 0 mi I = f d I m ⎛ ⎞⎛ ⎞⎛ ⎞ η ϕϕϕ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ∫ i i (10) where each of the factors is less than unity and can be assigned a separate meaning: (11) u mi m ≡ η i i is the “utilization factor”, i.e., it penalizes neutral gas flow. (12) B a a I = I η , the “backstreaming efficiency” penalizes electron backstreaming. (13) ( ) 2 1 Ε 0 f d = ⎛ ⎞ ⎜ ⎟ ϕϕϕ η ⎝ ⎠ ∫ , the “nonuniformity factor” is less than unity because of the nonuniform ion velocity It is clear that, since ( ) 1 0 f d =1 ϕ ϕ ∫ , we want to put most of f (ϕ) where ϕ is greatest, namely, we want to produce most of the ionization near the inlet. In that case f = -1 () ( ) ϕ δϕ , and ηΕ = 1. A somewhat pesimistic scenario would be f =1 (ϕ) , namely dmi dx i proportional to dV - dx , i.e., ionization rate proportional to field strength. In that case 1 2 2 E 0 2 4 = ×1×d = = 3 9 ⎛ ⎞ ⎛ ⎞ η ϕϕ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ∫
Measurements tend to indicate ne w 0.6-0.9, which means that ionization tends to occur early in the channel. This is to be expected, because that is where the backstreaming electrons have had the most chance to gain energy by falling"up the potential The factor n, = is related to the ionization fraction. Putting mi=neCA, m=m+mn=(nc +ncn)A, (11) 1+(%) Since ci /cn is large(cn w neutral speed of sound, i.e., a few hundred m/sec, while C:= gIsp-20,000m/sec), nu can be high even with ne /nn no more than a few percent. Datall show n, ranging from 40% to 90% The factor ne = IB/I requires some discussion. Most of the ionization is due to the backstreaming electrons, so that we are not really free to drive IB towards Ia (Ies =I-I). What we need to strive for is (a) Conditions which favor creation of as many ions as possible per backstreaming electron, and (b) Minimization of ion-electron losses to the walls, once they are created This can be quantified as follows: Let b be the number of secondary electrons(and of ions) produced per backstreaming electron and let a be the fraction of these new e-i pairs which is lost by recombination on walls. Then, per backstreaming electron, (1-a)B ions make it to the beam, and an equal number of cathode electrons are used to neutralize them therefore I_(1-a)阝 I31+(1-a)阝 Clearly, we want B>>1 and a>1)implies lengthening the electron path by means of the applied radial magnetic field, and also using accelerating potentials which are not too far from 5/2 times the range of energies where ionization is most efficient(typically 30-80 Volts). This last condition creates some difficulties with heavy ions, which require higher accelerating potentials for a given exit speed The condition a < 1 implies minimization of insulation surfaces on which the recombination can take place and arrangement of the electric fields such that ions 16.522, Space P pessan Lecture 18 Prof. Manuel martinez
16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 4 of 20 Measurements[1] tend to indicate ηΕ ~ 0.6 - 0.9 , which means that ionization tends to occur early in the channel. This is to be expected, because that is where the backstreaming electrons have had the most chance to gain energy by “falling” up the potential. The factor u mi = m η i i is related to the ionization fraction. Putting e i m = n c A , i i n ( ) ei nn m=m+m = n c +n c A i i ii , e i n n u e i n n n c n c = n c 1+ n c ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ η ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ (11) Since i n c c is large (cn ~ neutral speed of sound, i.e., a few hundred m/sec, while i sp c gI 20,000m/sec ∼ ), ηu can be high even with n n e n no more than a few percent. Data[1] show ηu ranging from 40% to 90%. The factor ηE Ba =I I requires some discussion. Most of the ionization is due to the backstreaming electrons, so that we are not really free to drive BI towards I I =I -I a BS a B ( ) . What we need to strive for is (a) Conditions which favor creation of as many ions as possible per backstreaming electron, and (b) Minimization of ion-electron losses to the walls, once they are created. This can be quantified as follows: Let β be the number of secondary electrons (and of ions) produced per backstreaming electron, and let α be the fraction of these new e-ipairs which is lost by recombination on walls. Then, per backstreaming electron, ( ) 1 - α β ions make it to the beam, and an equal number of cathode electrons are used to neutralize them. Therefore ( ) ( ) B a a I 1 - = = I 1+ 1- α β η α β (12) Clearly, we want β >> 1and α > 1) implies lengthening the electron path by means of the applied radial magnetic field, and also using accelerating potentials which are not too far from 5/2 times the range of energies where ionization is most efficient (typically 30-80 Volts). This last condition creates some difficulties with heavy ions, which require higher accelerating potentials for a given exit speed. The condition α << 1implies minimization of insulation surfaces on which the recombination can take place, and arrangement of the electric fields such that ions
are not directly accelerated into walls. This is difficult to achieve without detailed surveys of equipotential surfaces Reference Komurasaki, K, Hirakowa, M. and Arakawa, Y, IEPC paper 91-078. 22 Electric Propulsion Conference, Viareggio, Italy Oct. 1991 1-D Model of hall thruster M=-1 Drift (not included) Anode pre-sheath ionization Define「e=n(v。<0,sor。<0) T=neVi (2) Tn=nvn v,=constant=SKT w (3) 1. Conservation of particles n=n vion dx (4) 16.522, Space P pessan Lecture 18 Prof. Manuel martinez Page 5 of 20
16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 5 of 20 are not directly accelerated into walls. This is difficult to achieve without detailed surveys of equipotential surfaces. Reference: Komurasaki, K, Hirakowa, M. and Arakawa, Y., IEPC paper 91-078. 22nd Electric Propulsion Conference, Viareggio, Italy, Oct. 1991. 1-D Model of Hall Thruster Define Γe ee =n v ( ) v < 0, so < 0 e e Γ (1) Γi ei =n v ( ) + or - (2) Γn nn =n v w n i 5 kT v constant 3 m (3) 1. Conservation of particles e i n e ion d d d = =- =n dx dx dx Γ Γ Γ ν (4) Drift Zone SHEATH (not included) Sonic point Ionization region M = -1 Tw x Anode Pre-sheath Diffusion
R,(T R=Ooe/i* kte).e (6) E KT and,forx,o=3.6×1039m2,E1=eV,V=12.1V From(4), two first integrals ;-I。=IA= constant (8) T+rn=rm=constant (9) Conservation a 2. Ion Momentum Equation The force per unit volume on the ion gas is eEn(from the axial electric field E) There is also a"pick-up"drag due to ionization For each ionization event, a new ion is"incorporated"to the ion population(of velocity vi), jumping from the neutral velocity vn; this gives a drag-ne onm(v, -vn). We then have mv, dv=eE-mvion (v, -vn) (11) 3. Electron Momentum Equation Consider first only classical electron collisions(say, with neutrals). The vector equation of motion of electrons, including electric force, magnetic force and collisional"drag"(and neglecting inertia)is 16.522, Space P pessan Lecture 18 Prof. Manuel martinez Page 6 of 20
16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 6 of 20 Where νion n i e =nR T( ) (5) i e E - e kT e i 0 i kT R = c 1+2 e E ⎛ ⎞ σ ⎜ ⎟ ⎝ ⎠ (6) e e i 8 kT c = π m (7) and, for Xe, -20 2 σ0 i ii 3 .6 × 10 m , E = eV , V = 12 .1V -25 m = 2.2 × 10 kg i From (4), two first integrals : ie d ΓΓ Γ - = = constant (8) ΓΓ Γ in m + = = constant (9) Conservation Equations and a d m i I m = , = Ae Am Γ Γ i (10) 2. Ion Momentum Equation The force per unit volume on the ion gas is eEne (from the axial electric field E). There is also a “pick-up” drag due to ionization. For each ionization event, a new ion is “incorporated” to the ion population (of velocity vi ), jumping from the neutral velocity vn ; this gives a drag-n m v - v e ion i i n ν ( ) . We then have ( ) i i i i ion i n dv m v = eE - m v - v dx ν (11) 3. Electron Momentum Equation Consider first only classical electron collisions (say, with neutrals). The vector equation of motion of electrons, including electric force, magnetic force and collisional “drag” (and neglecting inertia) is
ⅴPQ=enE+vxB)- meVenneve (12) nn ce Project on x, y: dx =en(E+v B)-m v v (13) E=0, 0=-en(0-Vex B)-mevennev (14) From(14), Substitute in(13) ene-enb-cv-mv. nv en, - Ve In the Hall thruster plasma, ven <<o(low collisionality), so second term in parenthesis is neglected. The quantity -acts then as an effective collision frequency accounting for the magnetic effect 16.522, Space P pessan Lecture 18 Prof. Manuel martinez Page 7 of 20
16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 7 of 20 ( ) e e ∇ ν P = -en E + v ×B - m n v e e e en e G JG JG JG (12) ( ) e ν σ en n en =n c Project on x, y: ( ) e e x ey e en e ex dP = -en E + v B - m n v dx ν (13) e y P E = 0, 0 x ⎛ ⎞ ∂ ⎜ ⎟ = ⎝ ⎠ ∂ 0 = -en 0 - v B - m n v e ex e en e e ( ) y ν (14) From (14), c ey ex ex e en en eB v= v= v m ω ν ν (15) Substitute in (13): e c e x e ex e en e ex en dP = -en E - en B v - m n v dx ω ν ν 2 c e x e e ex en en = -en E - m n v + ⎛ ⎞ ω ⎜ ⎟ ν ν⎝ ⎠ (16) In the Hall thruster plasma, ν ω en c << (low collisionality), so second term in parenthesis is neglected. The quantity 2 c en ω ν acts then as an “effective collision frequency”, accounting for the magnetic effect 2 c e en " "= ω ν ν (17) y z B x vey vex
If there were several types of real collisions, such as e-n, and e-n,, then ven =ven. +ven in(17). We write the momentum equation(with P =ne kTe)as (n kTe )=-en Ex-mene veve In the SPT type of Hall thruster, the dominant scattering affect is actually not collisions, but the random deflections due to plasma turbulence ( anomalous diffusion, Bohm diffusion"). To see what modifications this introduces, consider a simple case with T. =constant and E=0. Equation(18) then gives mv dx (19) So that the diffusivity isD= ke, usIng (17), with more than one type of collision meve (20 If anomalous(Bohm) diffusion dominates it is known empirically that Bohm -C 16 If we liken this to collisionality effect, say, ven, then :B mea2venz,or ven,= @ Oc (21) Therefore if we want to account for both classical e-n collisions and bohm diffusion effects, we will define"ve"as(from(17) 4. Electron Energy Equation The convected enthalpy flux of the electron gas is r. skT.. Its divergence is the net rate of work done on this gas per unit volume, minus the work rate expended in ionization and excitation of neutrals -er Ex -ne vionE 16.522, Space P pessan Lecture 18 Prof. Manuel martinez Page 8 of 20
16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 8 of 20 If there were several types of real collisions, such as e-n1 ande-n2 , then en en en 1 2 νν ν = + in (17). We write the momentum equation (withP = n kT e ee ) as ( ) e e ex ee ee d n kT = -en E - m n v dx ν ( ) v v e ex ≡ (18) Γe In the SPT type of Hall thruster, the dominant scattering affect is actually not collisions, but the random deflections due to plasma turbulence (“anomalous diffusion, Bohm diffusion”). To see what modifications this introduces, consider a simple case with T =e constant and E =0 x . Equation (18) then gives e e e e e kT dn = - m dx Γ ν (19) So that the diffusivity is e e e kT D = m ν . Using (17), with more than one type of collision, ( ) 1 2 e 2 en en e c kT D= + m ν ν ω (20) If anomalous (Bohm) diffusion dominates, it is known empirically that e Bohm B B kT 1 D=D eB 16 ⎛ ⎞ α α⎜ ⎟ ⎝ ⎠ . If we liken this to collisionality effect, say, en2 ν , then 2 e e B en 2 c kT kT = eB me α ν ω , or en B c 2 ν αω = (21) Therefore, if we want to account for both, classical e-n collisions and Bohm diffusion effects, we will define e " " ν as (from (17) 2 c e en B c = + ω ν ν αω (22) 4. Electron Energy Equation The convected enthalpy flux of the electron gas is e e 5 kT 2 Γ . Its divergence is the net rate of work done on this gas per unit volume, minus the work rate expended in ionization and excitation of neutrals: ' e e e x e ion i d 5 kT = -e E - n E dx 2 ⎛ ⎞ Γ Γν ⎜ ⎟ ⎝ ⎠ (23)
Where Ei is roughly 2-3 times the actual ionization energy Ei to include the radiative losses due to excitation by electron impact, followed by prompt photon emission Notice that, since ne ion dx - e this can also be written as 5 2 5. Solving for Derivatives We combine here for clarity, the main equations: dx m.v. ee -mv (26) d(noTe) = -ene-m t dx (27) d E (28) It is just a matter of algebra to solve for each of the gradients separately( including the potential gradient -E==). The results are (30m) vn)-V 2E +5kT =5mevevevi+vion kTe +mv(v (29) 5 dx 3meneveve-ne vion ) m,(2 2V.-v n) 2E 5kT (30) 3v。 3kTe-mve k dx=.3m v,mv?-myion3kTe(2v, -V) mv2-kT。2E+5kT miVe mv eEx=5mevevemv? +mvionI5kTe(2v, -vn) V2 2E +5kT (32) Here vo=-.(generally negative) 16.522, Space P pessan Lecture 18 Prof. Manuel martinez Page 9 of 20
16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 9 of 20 Where ' Ei is roughly 2-3 times the actual ionization energy Eito include the radiative losses due to excitation by electron impact, followed by prompt photon emission. Notice that, since e e ion d n = dx Γ ν this can also be written as ' e e i ex d 5 kT + E = -e E dx 2 ⎡ ⎤ ⎛ ⎞ ⎢ ⎥ Γ Γ ⎜ ⎟ ⎣ ⎦ ⎝ ⎠ (24) 5. Solving for Derivatives We combine here for clarity, the main equations: e i x ion e d d d = =- = n dx dx dx Γ Γ Γ ν (25) ( ) i i i x i ion i n dv m v = eE - m v v - v dx (26) ( ) e e ex e ee d n kT = -en E - m dx Γ ν (27) ' e e e x e ion i d 5 kT = -e E - n E dx 2 ⎛ ⎞ ΓΓν ⎜ ⎟ ⎝ ⎠ (28) It is just a matter of algebra to solve for each of the gradients, separately (including the potential gradient x -E = x ∂φ ∂ ). The results are ( ) ' 2 i i i e e i i e e e i ion e i i i n e 5 55 dv v 2E + 5kT kT - m v = m v v + kT +m v v - v - 3 dx 3 3 v 3 ⎛ ⎞ ⎡ ⎤ ⎜ ⎟ ν ν ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ (29) ( ) ' 2 e ie e i i e e e e e ion i i n e 5 5 dn 2E + 5kT kT - m v = m n v - n m 2v - v - 3 dx 3 3v ⎛ ⎞ ⎡ ⎤ ⎜ ⎟ ν ν ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ (30) ( ) 2 ' 2 2 e ii e i e e i i e e e i i i ion e i n i e 5 22 dT m v - kT 2E + 5kT kT - m v k = - m v m v - m kT 2v - v - 3 dx 3 3 m v 3 ⎛ ⎞ ⎡ ⎤ ⎜ ⎟ ν ν ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ (31) ( ) 2 ' 2 2 i i e e i i x e e e i i i ion e i n e 55 5 v 2E + 5kT kT - m v eE = m v m v +m kT 2v - v - 3 3 3 v3 ⎛ ⎞ ⎡ ⎤ ⎜ ⎟ ν ν ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ (32) Here e e ex e ν = ν = n Γ (generally negative)
The most important feature of these equations is the factor = kT -mv which would appear in the denominator. It becomes zero when which is the ion- sonic wave speed, an acoustic wave in which both ions and electrons, undergo compressions and expansions; they are coupled to each other electrostatically and since electrons are hotter, they provide the restoring force kTe, while the ions, more massive, provide the inertia, m Because of this, the gas can accelerate across this speed (of the order of 3000-4000 m/s in Xenon) in one of two modes: (a) Smoothly, if the right-hand sides of all of Equations(29-32)are zero when v,=vs(actually if one of them is zero, the others will also be, at v=v). This imposes an internal condition on the differential equations, to supplement the boundary conditions. The difficulty is that one does not know a-priory where(in x) this condition will occur. It is also difficult o integrate numerical through this point because each derivative is of the -form One needs to use l Hospitals rule to extract the finite ratio (two values normally) (b) Abruptly, if the right-hand sides are nonzero when v, =vs. In this case the derivatives(including Ex) are locally infinite, although one can show that they behave as功/√×-x, and so this infinity is integrable.This can only happen at the open end of the channel, just as with a normal open gas pipe discharging into a vacuum In this case, we impose th end condition V,=Vis (Te) Notice that condition(b) can also occur at the inlet(x=0). Infact, it does occur. This restrict the electron capture to the required I, level this same sheath will then o is because the anode will develop a negative sheath(electron repelling) in order to attract ions, which will therefore enter it at their sonic velocity(a form of Bohms sheath-edge criterion): 6. Boundary Conditions So, in this device we have two sonic points, one(reversed)at inlet, and one (forward )either at the exit plane or somewhere in the channel. This provides either two boundary conditions, or one(Equation (34)) plus one internal condition of smooth sonic passage. Looking at Equations(25-28) we count 6 differential 16.522, Space P pessan Lecture 18 Prof. Manuel martinez Page 10 of 20
16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 10 of 20 The most important feature of these equations is the factor 2 e ii 5 kT - m v 2 which would appear in the denominator. It becomes zero when e i is i 5 kT v =v = 3 m (33) which is the ion-sonic wave speed, an acoustic wave in which both, ions and electrons, undergo compressions and expansions; they are coupled to each other electrostatically, and since electrons are hotter, they provide the “restoring force” kTe , while the ions, more massive, provide the inertia, mi . Because of this, the gas can accelerate across this speed (of the order of 3000-4000 m/s in Xenon) in one of two modes: (a) Smoothly, if the right-hand sides of all of Equations (29-32) are zero when v =v i is (actually, if one of them is zero, the others will also be, at v =v i is ). This imposes an internal condition on the differential equations, to supplement the boundary conditions. The difficulty is that one does not know a-priory where (in x) this condition will occur. It is also difficult to integrate numerical through this point, because each derivative is of the 0 0 form. One needs to use L’ Hospital’s rule to extract the finite ratio (two values normally). (b) Abruptly, if the right-hand sides are nonzero when v =v ι is . In this case, the derivatives (including Ex) are locally infinite, although one can show that they behave as 1 x-xs , and so this infinity is integrable. This can only happen at the open end of the channel, just as with a normal open gas pipe discharging into a vacuum. In this case, we impose the end condition v =v T i is e ( ) . Notice that condition (b) can also occur at the inlet (x=0). Infact , it does occur. This is because the anode will develop a negative sheath (electron repelling) in order to restrict the electron capture to the required aI level. This same sheath will then attract ions, which will therefore enter it at their sonic velocity (a form of Bohm’s sheath-edge criterion): ( ) ( ) e i i 3 kT 0 v x=0 =- 5 m (34) 6. Boundary Conditions So, in this device we have two sonic points, one (reversed) at inlet, and one (forward) either at the exit plane or somewhere in the channel. This provides either two boundary conditions, or one (Equation (34)) plus one internal condition of smooth sonic passage. Looking at Equations (25-28) we count 6 differential