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144 Chapter 4 baffle can be high. For these reasons, the ion current to the cathode, Ik , can no longer be neglected. The power into the plasma is given by Eq. (4.3-46), and the power out of the discharge is given by Pout= IpU++I*U*+Is(Vd+ i)+Ik(Vd+ i)+Ib i+Iia i+Ia e, (4.4-5) where i is the ion energy leaving the plasma, which is written here from Eq. (4.3-10) as TeV / 2 + , and e is the electron energy removed from the plasma, which is written from Eq. (4.3-9) as 2TeV + . Equating the power in to the power out again and solving for the discharge loss gives Ip + I* * T (Is+Ik) T V d U + U + + e V + 2 V d – V c + 2 + e V IbIb2Ib 2 d = The first current ratio, I p / Ib , is given by Eq. (4.3-55), and the second current Eq. (4.3-57), and the last current ratio is 1nke kTeAk Ik = 2 M = nkAsa , (4.4-7) Ib 1 n e kTe A T neAsTs 2iMss where nk is the plasma density at the cathode baffle. The discharge loss for Kaufman thrusters is then . (4.4-6) ratio, I* / Ib , is given by Eq. (4.3-56). The current ratio, Is / Ib , is given by Vd –Vc –2TeV Ip+I** T 1–TnA T Vd U+ U+ +eV+ s+ksa 2Vd–Vc+2 +eV IbIb 2 TsneAsTs 2 d= . Vd –Vc –2TeV (4.4-8) The plasma potential in Eq. (4.4-8) is found from solving Eq. (4.4-4): =kTeln –D neAas , e Id (4.4-9)PDF Image | Fundamentals of Electric Propulsion: Ion and Hall Thrusters
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