lecture18

# lecture18 - 16.522, Space Propulsion Prof. Manuel...

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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 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, aBB S I= I+ I (3) } V a (I B ) e (I B ) i I BS I BS I a (I B ) e

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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 i F= cdm i (4) where i 2eV c= m (5) Suppose the part i dm i of i m i is created in the region where V decreases by dV, and define an “ionization distribution function” f(V) by i aa i V dV =-f VV m ⎛⎞ ⎛ ⎞ ⎜⎟ ⎜ ⎟ ⎝⎠ ⎝ ⎠ i i (6) or, with a V = V ϕ , () i i d m ϕϕ . i i From the definition, ( ) f ϕ 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 ϕϕϕ 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 i m i by B i i e I= m m i i . We can therefore re- write (9) as 2 1 B a 0 i I m = f d I m ηϕ ϕ ϕ i i (10) where each of the factors is less than unity and can be assigned a separate meaning: (11) u i m 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 d = η , the “nonuniformity factor” is less than unity because of the nonuniform ion velocity It is clear that, since 1 0 d = 1 ϕϕ , 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 ϕδ ϕ , and Ε =1 η . A somewhat pesimistic scenario would be ( ) 1 ϕ , namely i dm 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 ϕ

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16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 4 of 20 Measurements [1] tend to indicate Ε ~ 0.6 - 0.9 η , 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.
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## This note was uploaded on 11/07/2011 for the course AERO 16.512 taught by Professor Manuelmartinez-sanchez during the Fall '05 term at MIT.

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lecture18 - 16.522, Space Propulsion Prof. Manuel...

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