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 aa P=IV (2) Where V a is the accelerating voltage and I a the current through the power supply (or anode current, or also cathode current). Of the I a current of electrons injected by the cathode, a fraction I B goes to neutralize the beam, and the rest, I BS back-streams into the thruster. Since no net current is lost to the walls, aBBS I=I+I (3) } V a (I B ) e (I B ) i I BS I BS I a (I B ) e 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 iF= cdm ∫ i (4) where i 2eV c= m (5) Suppose the part idm i of im i is created in the region where V decreases by dV, and define an “ionization distribution function” f(V) by i aa i dm V dV =-f VV m ?? ???? ?? ???? ?? ???? ?? i i (6) or, with a V = V ?, () i i dm =-f d m ??. i i From the definition, ( )f ? satisfies () 1 0 fd=1?? ∫ (7) Then, from (4) and (5), V V a 0 x 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 ??? ∫ i (8) and hence the efficiency is () 2 2 1 a i 0 aa i 2eV m f d m = 2mVI ?? ?? ??? ?? ?? ?? ?? η ∫ i i (9) Notice that the beam current I B is related to im i by B i i e I= m m i i . We can therefore re- write (9) as () 2 1 B a 0 i Im = f d I m ?? ???? ?? η?? ???? ?? ???? ?? ∫ i i (10) where each of the factors is less than unity and can be assigned a separate meaning: (11) u im 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 = ?? ??? η ?? ?? ∫ , the “nonuniformity factor” is less than unity because of the nonuniform ion velocity It is clear that, since () 1 0 f d=1?? ∫ , we want to put most of ( )f ? where ? is greatest, namely, we want to produce most of the ionization near the inlet. In that case () ( )f= -1?δ? , and Ε =1η . A somewhat pesimistic scenario would be ( )f=1? , namely idm dx i proportional to dV - dx , i.e., ionization rate proportional to field strength. In that case 2 2 1 E 0 24 =×1×d== 39 ?? ?? η?? ?? ?? ?? ?? ∫ 16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 4 of 20 Measurements [1] tend to indicate Ε ~ 0.6 - 0.9η , 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 im = m η i i is related to the ionization fraction. Putting ei im=ncA , i ()n ei nn im=m+m = nc +nc A iii , e i nn u e i nn n c nc = n c 1+ nc ???? ???? ???? η ???? ???? ???? (11) Since in cc is large (c n ~ neutral speed of sound, i.e., a few hundred m/sec, while isp c gI 20,000m/sec～null ), u η can be high even with en nnno more than a few percent. Data [1] show u η ranging from 40% to 90%. The factor EBa =I Iη requires some discussion. Most of the ionization is due to the backstreaming electrons, so that we are not really free to drive B I towards () aBS a B II=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 β be the number of secondary electrons (and of ions) produced per backstreaming electron, and let αbe the fraction of these new e-ipairs which is lost by recombination on walls. Then, per backstreaming electron, ()1-αβions make it to the beam, and an equal number of cathode electrons are used to neutralize them. Therefore () () B a a 1- I == I1+1- αβ η αβ (12) Clearly, we want >>1β and <<1α . The first ( )>>1β 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 <<1α implies minimization of insulation surfaces on which the recombination can take place, and arrangement of the electric fields such that ions 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. 22 nd Electric Propulsion Conference, Viareggio, Italy, Oct. 1991. 1-D Model of Hall Thruster Define eee =nvΓ () ee v<0, so <0Γ (1) iei =nvΓ ()+ or - (2) nnn =nvΓ w n i kT5 vconstant 3m nullnull (3) 1. Conservation of particles e in eion d dd ==-=n dx dx dx Γ ΓΓ ν (4) Drift Zone SHEATH (not included) Sonic point Ionization region M = -1 T w x Anode Pre-sheath D if fu s io n 16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 6 of 20 Where ( ) ion n i e =nR Tν (5) i e E - kTe e i0 i kT R= c 1+2 e E ?? σ ?? ?? (6) e e i kT8 c= mπ (7) and, for X e , -20 2 0ii 3.6×10 m , E = eV, V = 12.1Vσ null -25 i m = 2.2×10 kg From (4), two first integrals : ie d -= =constantΓΓ Γ (8) in m += =constantΓΓ Γ (9) Conservation Equations and a dm i I m =, = Ae Am ΓΓ i (10) 2. Ion Momentum Equation The force per unit volume on the ion gas is e 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 i v ), jumping from the neutral velocity n v ; this gives a drag ( ) eion i i n -n m v - vν . We then have () i ii iion i n dv mv =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 16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 7 of 20 () ee ee ene P=-en E+v×B-m v?ν nullnullnullnullnullnullnull (12) () e en n en =n cν σ Project on x, y: () e e x ey e en e ex dP =-en E +v B -m nv dx ν (13) e y P E=0, 0 x ??? = ?? ? ?? ( ) eexeney 0=-en 0-v B -m nvν (14) From (14), c ey ex ex een en eB v= v= v m ω νν (15) Substitute in (13): ec ex e ex eeneex en dP = -en E - en B v - m n v dx ω ν ν 2 c ex eeex en en =-enE -mnv + ??ω ν ?? ν ?? (16) In the Hall thruster plasma, en c <<νω (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 v ey v ex 16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 8 of 20 If there were several types of real collisions, such as 1 e-n and 2 e-n , then 12 en en en =+νννin (17). We write the momentum equation (with eee P=nkT) as () ee ex eeee d nkT =-enE -mnv dx ν ( ) eex vv≡ (18) e Γ 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 e T=constant and x E=0. Equation (18) then gives ee e ee kT dn =- mdx Γ ν (19) So that the diffusivity is e ee kT D= m ν . Using (17), with more than one type of collision, () 12 e en en2 ec kT D= + m νν ω (20) If anomalous (Bohm) diffusion dominates, it is known empirically that e Bohm B B kT 1 D=D eB 16 ?? αα ?? ?? nullnull. If we liken this to collisionality effect, say, 2 en ν , then 2 ee Ben2 c kT kT = eB me αν ω , or 2 en B c =ναω (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 ναω (22) 4. Electron Energy Equation The convected enthalpy flux of the electron gas is ee 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: ' ee exeioni d5 kT = -e E - n E dx 2 ?? ΓΓν ?? ?? (23) 16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 9 of 20 Where ' i E is roughly 2-3 times the actual ionization energy i E to include the radiative losses due to excitation by electron impact, followed by prompt photon emission. Notice that, since e eion d n= dx Γ ν this can also be written as ' eei ex d5 kT +E = -e E dx 2 ???? ΓΓ ???? ???? (24) 5. Solving for Derivatives We combine here for clarity, the main equations: e ix ion e d dd ==-= dx dx dx Γ ΓΓ ν (25) () i ii x iion i n dv mv =eE -mv v -v dx (26) () ee ex eee dnkT =-enE -m dx Γν (27) ' ee ex eioni d5 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 iei i eii eeeiion e iiin e 2E +5kTdv v555 kT - mv = m v v + kT +mv v - v - 3dx3 3 3 ? ? ?? νν ? ? ?? ?? ? ? (29) () ' 2 eii eeeeeioniin e dn 2E +5kT55 kT - mv = m n v - n m 2v - v - 3dx3 3v ? ? ?? νν ? ? ?? ?? ? ? (30) () 2' 22e ii e i e eii eeeii iion ein ie dT mv -kT 2E +5kT522 kT -mv k =- m v mv -m kT 2v -v - 3dx3 3 3 ? ? ?? νν ? ? ?? ?? ? ? (31) () '2 iei eii x eeeii iion ein e 2E +5kTv55 5 kT - mv eE = m v mv +m kT 2v - v - 33 3 3 ? ? ?? νν ? ? ?? ?? ? ? (32) Here e eex e ν = ν = n Γ (generally negative). 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 eii 5 kT - mv 2 which would appear in the denominator. It becomes zero when e iis i kT5 v=v = 3m (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” e kT , while the ions, more massive, provide the inertia, i 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 iis v=v(actually, if one of them is zero, the others will also be, at iis 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 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 ι is v=v. In this case, the derivatives (including E x ) are locally infinite, although one can show that they behave as s 1 x-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 the end condition ( ) iise v=v T . 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 a I 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 kT 0 3 vx=0=- 5m (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 16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 11 of 20 equations. However, we can assume the integration constants m Γ and d Γ are prescribed, which leaves as in need of two additional boundary conditions. [Comment: Prescribing both, m Γ and d Γ becomes redundant once full ionization is reached; a more realistic prescription would be m Γ and the total applied voltage V. However, most of V develops in the supersonic near-plume, outside the channel, and is not captured in this analysis.] The two missing conditions must play the role of connectors to the outside plume. In an ordinary gas flow, conditions upstream of the sonic point would be fully decoupled from those downstream. But in this problem we are dealing with two counter flowing streams, and electrons, in particular, do carry information from the outside back to the plasma in the channel. The most obvious part is that the mean energy() e T of the backstreaming electrons entering the channel must increase (linearly?) as the total voltage V increases. We therefore prescribe ( ) e T L as a kind of proxy for the applied voltage. A more subtle effect is that the fraction of the electrons emitted by the downstream cathode which do backstream to the channel must depend on details of the outside region comprising the cathode itself and the near plume. Pending a model of this region, we also prescribe ( ) e d LΓ Γ . 7. Anatomy of the Discharge Since analytical full solution is out of the question, it is useful to analyze the different regions in the thruster according to the dominant mechanisms in them. Numerical integration is, of course possible, and detailed results are presented in Reference (1) (for the choked-exit case) and Reference (2) (for the smooth sonic passage case). Unsteady effects have also been considered in Reference (3). Figure 1, from Reference (1) shows a useful categorization, 16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 12 of 20 which has resulted from a combination of basic considerations, detailed numerical solutions, and examination of experimental data. We will in the following sections offer partial analyses for several of these regions. 8. The Presheath and the Diffusion Zone In the region B-C-D, ionization is very weak, because e T has fallen to a low value as the electrons lose energy in the ionization layer. Setting ion =0ν , we can form the ratio () () i v × Equation 29 Equation 31 : i e i n Sketch of the channel: L is the channel length; r 1 and r 2 are the inner and outer radius, respectively; , and are the axial flows of neutral gas, ions, and electrons, respectively; B(x) is the profile of the radial magnetic field; A is the anode, E is the channel exit, B is the transition from the space-charge sheath to the quasineutral channel, and S (that can coincide with E) is the sonic point for ions. Three regions are usually distinguished in the quasineutral channel: the anode presheath BC, the diffusion zone CD, and the ionization zone DE. n i e r 1 r 2 A B D L B(x) Plume S EC 16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 13 of 20 ii i e vdv 5 =- m kdT 2 or 2 ii e mv 5 + kT = constant 22 (35) At x=0 (point B), we use (34): () () () ee e 15 5 10 kT 0 + kT 0 = constant constant = kT 0 23 2 3 → or () 2 ii ee mv4 T= T 0- 35k (36) This can now be substituted into (29): () e ee ii ee ee2 iii kT 0 v 44 - m dv = m dx = m dx 35v ?? Γ νν ?? Γ ?? Since e Γ and i Γ are constant in this region (no ionization), we can integrate again (assuming e ν null constant as well): () e e ii ee ii 0 kT 0 44 --mv=C+mx 3v 5 ??Γ ν ?? Γ ?? (37) At x=0 () 0 e ii i kT 5 v=v 0=- 3m , giving () ii 8 C=- mv 0 5 . Then (37) can be written as () () () 2 eeeii ii i 5m xvv 21+ +1=0 v0 v08mv0 ?? ?? ?? Γν ?? ??? ?? Γ ?? ?? ?? ?? (38) Defining a “pre-sheath thickness” 16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 14 of 20 () ii i ps eee 0 mv 0 8 x= 5m ??Γ ?? Γν ?? (39) the solution to (38) is () 2 i ipsps v xx =1+ ± 1+ -1 v0 x x ?? ?? ?? (select lower sign) (40) ps 1 = x 1+ x ? Notice that for ps x>>x , the upper sign choice leads to () i ips v 2x v0 x → while the lower choice gives () ps i i x v v0 2x → . Since ( ) i v0<0, the first of these would imply a reverse ion flow which is decelerating towards sonic, and is therefore unphysical. For that reason, the lower sign has been selected. Since ( ) i 0 e i n= v Γ , we also have then () () () () 2 ii i e 2 ips ps ps ps 0v0 0 xx n= = 1+ + 1+ -1 v0 x x xx 1+ - 1+ -1 ? ? Γ??Γ ? ? ?? ? ? ???? ? ? ? ? ?? ?? (41) Returning to (36), we can now calculate also () 2 ps e 2 e ps ps x 1+ -1 x T 2 =1+ T0 3 xx 1+ + 1+ -1 ?? ?? ?? ?? ?? ?? (42) The electric field follows from (32), for example. It is actually easier to return to (26) and integrate it (with ion =0ν ) to () 22 iiii mv 0mv +e = 22 φ (43) 16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 15 of 20 and then, using (40), () 2 ps e 2 ps ps x 1+ -1 x kT 0 5 = 3e xx 1+ + 1+ -1 ?? ?? ?? φ ?? ?? ?? (44) and, by differentiation, () s ss e p x 2 22 ppp kT 0 5 3ex E= xxx 1+ 1 1+ + 1+ -1 ? ?? ?? ?? ?? ??? ?? ?? ?? ?? (45) The limits of these expressions when ps x >> x (namely, in the diffusion region), are simple: () ( ) () ps i i e ps i psi x 0 vx2 ; - ; n vxv0 Γ →∞ → → () () ( ) s 2 pee e x 3 e xkT 0 kT 0 T 45 5 ; ; E = - T0 3 6 e 12 e x →φ→ (46) 16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 16 of 20 0 4 1 x x ps1 -1 3 5 6 v i v i (0) ef kT e (0) T e (0) T e 16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 17 of 20 Notice how the electric field is very weak, except right near the anode, within the presheath, where the ions accelerate to sonic speed. Also, the temperature reaches a constant value outside the presheath, at 4/3 the sheath-edge value (this is +1 2 γ with 5 = 3 γ ). The electron density increases linearly in the diffusion region: ions need to advance towards the anode (in order to maintain neutrality), and since the field is too weak to move them against the collisional drag, a density gradient must appear so as to produce diffusion instead. Combining the asymptotic form ( ) () i e ips 0 2x n v0x Γ null with Equation (39) for ps x, we can write for the diffusion region () () e eee e 0 3 nmx 4kT 0 Γ νnull (47) 0 1 2 3 4 5 6 1 2 3 4 5 n e x x ps ( i / v i ) 0 16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 18 of 20 (NOTE: If B is non-uniform, so that ( ) ee =xνν , the factor e xν is simply replaced by x e 0 dxν ∫ ). 9. The Transition to the Ionization Layer As x increases away from the anode, e n cannot continue to increase indefinitely according to (47), because its maximum value must be of the order of mi,exit vΓ , assuming near-full ionization. The end of the diffusion layer will be marked by an increase in e T that will initiate ionization. Eventually, a forward field x E will develop as well, but the analysis allowing both e T and φto vary as well as ionization to occur, is too complicated. We analyze first the portion of the ionization layer in which e T rises, but x E remains negligible. In this region, the form (24) of the energy equation is useful: ' eei 5 kT +E = constant 2 ?? Γ ?? ?? Evaluate the constant in the diffusion region, where () ee 4 T =T0 3 and () ee =0ΓΓ : () () ' ei e ' e ei 10 kT 0 +E 3 = 5 0 kT +E 2 Γ Γ (48) In the diffusion layer, ' eii 33 kT << E E 10 4 ～ , and so, even though (48) indicates a change in e Γ as soon as e T increases, the change will be slight at first. We can then approximately integrate (27), with ex = constant and E = 0ν , to ee eee nkT -m xΓνnull (notice this is implied in Equation 47) (49) From (48), e eeion ' ee ei dT5 k dnν1 2dx - = = 5 dx kT +E 2 Γ ΓΓ and, using (49), 2 ee ' eeion e i kTdT xdx 2 mkT+E 5 ?? νν ?? ?? null (50) In the range where ei kT << E , the strong variability of ion ν with e T appears (see Equations 5-7) in the exponential ie -E kT e , if we write for now ie -E kT ion ceν null , then 16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 19 of 20 () () i e 33E 2 2 eekTee ee ion i ion i ion kT kTkTdT k11 Te dT - d - d cEE ? ? ?? ? ? ?? ννν ? ? ?? ? ? nullnull null where the last step ignores a weaker term due to () 3 e ion 1 dkT ν . With these approximations, (50) integrates to () () ee 3 e2 ' eeioni e i 4 TT0 3 2kT L 2 mEkT+E 5 ?? ?? ?? νν ?? ???? null null (51) where L is the channel length (mainly its diffusion part), and T e , K are evaluated in the diffusion region, which contributes the most to the integral. Given the length L, the B field, etc, Equation (51) determines ( ) e T 0 , and therefore ( ) i v0and other quantities required in the previous section. Figure (2) shows numerical results of integrating the full set of differential equations, as well as (dotted) those obtained using the approximations of the last two sections. Notice in particular that the rise of e T , once started, is quite rapid, and leads to peak values MAX ei TE～ , at which ionization is rapid. The main body of the ionization layer is therefore very thin. A quick argument for sonic exit to the ionization layer In (29), assume (to be verified) that () eeei e ion MAX 55 mvv<<kT 33 νν. Notice the bracket in (29) is >0 (since e v<0). Then, given the inequality above, the only way the RHS can cross zero if () ion ion MAX <<νν. The zero-crossing occurs at a smooth sonic point (where the LHS = 0). On the subsonic side of this, LHS>0, so this must be where the ion ... > 0ν ?? ?? term dominates; conversely, on the supersonic side, ion ...ν ?? ?? must become negligible. This means by the time the sonic point is reached, ion ν has fallen to nearly zero, which can only be because n n has been depleted - full ionization. 16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 20 of 20 References: Ahedo, E., and M. Martinez-Sanchez. “AIAA 98-8788, One–dimensional Plasma Structure in Hall Thrusters.” 34th AIAA/ASME/SAE/ASEE Joint Propulsion Conference and Exhibit. Cleveland, OH, July 13-15, 1998, E.T.S.I. Aeronáuticos Universidad Politécnica, Madrid, Spain and Massachusetts Institute of Technology, Cambridge, MA. M * 1 0.5 0 -0.5 -1 0 1 2 x/l * * 8 6 4 2 0 0 1 2 x/l * n / n * i / * * 0.8 0.6 0.4 0.2 0 -0.2 0 1 2 x/l * * / T * T 1.5 1 0.5 0 0 1 2 x/l * * Plasma profiles for a channel with choked exit, no wall losses (l*/h d / m = 1 , T eB / E i = 0.10, iB / m = -0.01, * = m . Solid and dashed lines correspond to the exact and the approximate solution, respectively. Asterisks in figures correspond toand point B. 0), uniform B-field, c / * = 1, a b c d ( n ) / ( n * T * ) T 1.5 1 0.5 0 0 1 2 x/l * e / T * q * 0.5 0 -0.5 -1 -1.5 0 1.5 x/l * 0.5 1 f