HW18-7c

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

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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 --.
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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....
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