lecture22

lecture22 - 16.522, Space Propulsion Prof. Manuel...

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16.522, Space Propulsion Lecture 22 Prof. Manuel Martinez-Sanchez Page 1 of 8 16.522, Space Propulsion Prof. Manuel Martinez-Sanchez Lecture 22: A Simple Model For MPD Performance-onset It is well known that rapidly pulsed current tends to concentrate near the surface of copper conductors forming a “skin”. A similar effect occurs when current flows through a highly conductive and rapidly moving plasma: current tends to concentrate near the entrance and exit of the channel. The reason is the appearance of a strong back EMF which tends to block current over most of the channel’s length. This is most easily seen if we “unwrap” the annular chamber of an MPD thruster into a rectangular 1-D channel. Ampère’s law: 0 1 j= B µ × GJ G (1) In our case x =l , x G so y z 0 dB 1 jj = + dx µ and calling y B- B , 0 1dB j=- dx µ (2) Ohm’s law (ignoring Hall effect) is () E+u±×±B σ GG G J G (3)
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16.522, Space Propulsion Lecture 22 Prof. Manuel Martinez-Sanchez Page 2 of 8 or, using zy x V E E = , B -B , u = u H ≡≡ () j= E-uB σ (4) Combining (2) and (4), 0 dB =- dx σµ (5) The flow velocity u evolves along x according to the momentum equation (ignoring pressure forces) x du dP m+ A= j B A = j B w H dx dx × i GJ G (6) neglect for now Substitute (2) into (6): 2 00 du 1 dB wH d B m= - B w H = - dx dx dx 2 ⎛⎞ ⎜⎟ µµ ⎝⎠ i (7) Integrate: 2 2 0 0 B B mu+wH =mu +wH 22 ii neglect 0 0 B-B wH u= 2 m µ i (8) Putting this in Equation (5), 0 dB wH E- B B -B dx 2m ⎡⎤ ⎢⎥ σµ µ ⎣⎦ i (9) If we approximate the conductivity σ as a constant, this can be integrated as 0 B 0 B 0 0 dB x= wH B(B -B ) σµ µ i (10) This integral can actually be calculated analytically, but the resulting expression is not very transparent. It is more useful to examine its behavior qualitatively. The
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16.522, Space Propulsion Lecture 22 Prof. Manuel Martinez-Sanchez Page 3 of 8 denominator in the integrand is the driving field (applied field E, minus back emf, uB). The field B 0 at x=0 is a measure of the current I, because integrating (2) between x=0 and x=1 gives L 00 0 0 0 BI I jdx = = B = ww µ µ (11) On the other hand, carrying (10) all the way to x=L, gives 0 B 0 22 0 0 0 dB L= wH E- B(B -B ) 2m σµ µ i (12) where, once I and m i are specified, only E remains as an unknown. This is then the equation for voltage V=EH. Consider conditions where the maximum value of the back emf uB reaches almost the level E. This means that the integrand will be very
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lecture22 - 16.522, Space Propulsion Prof. Manuel...

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