HW19-16-18

# HW19-16-18 - a core power cannot decrease any faster than...

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10-12 Principles of Nuclear Reactors Chap. 10 What is the asymptotic period resulting from a reactivity insertion of (a) 0.08\$ and (b) -0.08\$ ? Solution: By definition p(\$) == (keff -l)/(flkeff). Solving this for keff and substituting p(\$) = :1:0.08\$ we find Thus for p(\$) = :f:0.08(\$), we have 8k = :f:0.000520. Finally, from Eq. (10.17) and from {3r = 0.083 s for 235U, we find the two asymptotic periods are {:Jr ok 0.083 :f:0.000520 T,= = :f:160 s. = 17. Following a reactor scram, in which all the control rods are inserted into a power reactor, how long is it before the reactor power decreases to 0.0001 of the steady-state power prior to shutdown? Solution: On page 281 it is discussed that, for large negative reactivity insertions into
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Unformatted text preview: a core, power cannot decrease any faster than the decay rate of the longest lived delayed neutron precursors. For 235U the mean lifetime of the longest lived precursors is about 80 s. Thus, the smallest asymptotic period for a very subcritical reactor is T = -80 s. Asymptotically, the power decays exponen-tially, i.e., P(t) = P(O) exp[tjT]. Solving this equation for t, and ignoring any prompt drop in power, the time to decrease the power to O.OOOlP(O) is Then from Eq. (10.16), which can be written as P(t) = P(O) exp[tjT], ( P(t) ) = (85 t=Tln P(O) = 785 s = 13.1 min. July 24, 2002 ( 166 W ) ;) In wow:...
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