Everything covered so far will also apply later to the arc in a flow. The difference is in how the heat is evacuated from the arc periphery. With no flow, this must be accomplished by heat conduction through the buffer gas. If kout denotes its thermal conductivity(probably much lower than ke), and if we ignore cylindrical effects, we must have equality of heat flux(per unit area) on both sides of the arc's edg 2 R R-R where Tw is the temperature of the constrictors wall, controlled externally. Substituting Te-Te from(8)into(6), I=R2ak xR(=) This is a quadratic equation for the arc radius Ra. To simplify algebra, introduce non- dimensional quantities y=nR√2k(-7) 2k (10) R d I=2 (11) Solving for ra, (12 which approaches I from below as I becomes large. We can now obtain other quantities of interest. From( 8), 2 T-T k r-r 16.522 Space Propulsion Lecture 11-12Everything covered so far will also apply later to the arc in a flow. The difference is in how the heat is evacuated from the arc periphery. With no flow, this must be accomplished by heat conduction through the buffer gas. If kout denotes its thermal conductivity (probably much lower than kc), and if we ignore cylindrical effects, we must have equality of heat flux (per unit area) on both sides of the arc’s edge: kc2 Tc − Te Ra = kout Te − Tw R − Ra (8) where Tw is the temperature of the constrictor’s wall, controlled externally. Substituting Tc-Te from (8) into (6), I = πRa 2ak ( )c 1 2 kout kc Ra R − Ra Te − Tw ( ) This is a quadratic equation for the arc radius Ra. To simplify algebra, introduce non￾dimensional quantities: I * = I Iref ; Iref = πR 2akc Te − Tw ( ) (9) λ = kout 2kc (10) and ra = Ra R (11) and so I * = λ ra 2 1 − ra (11) Solving for ra, ra = 2 1 + 1+ 4λ I * (12) which approaches 1 from below as I * becomes large. We can now obtain other quantities of interest. From (8), Tc − Te Te − Tw = 1 2 kout kc Ra R− Ra = λ r a 1− ra = 2λ 1 + 4λ I * −1 16.522, Space Propulsion Lecture 11-12 Prof. Manuel Martinez-Sanchez Page 4 of 18