NE STUD 180 Midterm: ne180-fa2010-mt2-Morse-soln
Document Summary
Ti(0. 9 a) = 15 103(cid:16)1 (0. 9)2)(cid:17)0. 3. = 9114. 17ev ni(0. 9 a) = 1. 0 1020 m 3 = 1. 0 1014 cm 3 n i = = 2. 0 nti i mi( ci i)2 = 1. 8093 1017m 1 s 1: b. dt dr. Total ion heat flux = area q = 159. 86 kw: c. q = 1/(1 (2/3)(r/a)2) = 2. 17391. Neoclassical factor = qneo = 2q2 3/2 = 57. 5218. Since q = qneo dt (r) dr. At center, ti te = 15 10 = 5. 0 kev. E = 2. 0 1010t 3/2 e /ne with te in ev and mks density. = 2. 8 10 8zt 3/2 e(kev ) Then pe = 2 c 1 ne = 2. 59959 1031m 3 crit = 2. 87 1017 s 1 and, since 2 pe = nee2/(me 0), we have: b. = nmi = 108. 316 g cm 3 , / liq = 433. 264: c. Then fb = r/(6 + r) = 0. 142857 and y = 3. 39 1011j g 1.
