HW18-7c - NL -1 + Lh2o(l -f)B~ -1 + 8.1(1- O.6074)B~ To...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
10-5 From Eq. (10.2) 687 0-235 1 = a~35 + a.{f20(NH20 /N235) = 681 + 0.666(1/1.5 x 10-3) = 0.6014 Hence we have koo = fp1/1 ~ 111 = (2.068)(0.6014) = 1.2561. (c) The critical core buckling for a sphere is B~ = (7r / R)2 where R is the radius needed to make the core critical, i.e., keff = 1. In terms of the buckling, the p~ is found from Eq. (10.6) n/ - e -B~'" -27B2 rNL -= e c and the ~i{t is obtained from combining Eqs. (10.4) and (10.5) as 1 -= 1h 1 R --.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: NL -1 + Lh2o(l -f)B~ -1 + 8.1(1- O.6074)B~ To produce a critical configuration we must have 1 + 3.180B~' Problem 10-7 """' ~ '-'-... u ~ 1.0 Rcrlt=36.12 cm 0.90 30 32 34 36 Radius R (cm) 38 40 July 24, 2002 Solve this equation by trial and error for B; or, equivalently, for R. The critical radius is found to be 36.122 cm which yields a value of keff = 1.00001. The variation of keff with core radius is shown in the plot below....
View Full Document

This note was uploaded on 02/01/2010 for the course ECE 4130 taught by Professor Cady during the Spring '06 term at Cornell.

Ask a homework question - tutors are online