Following 22. a reactor shutdown /Jo-+ 0), show from Eq. (10.33) that the time tm for 135Xeo reach a maximum is given by t
Solution:
First we solve for the 135Xeand 1351ransients following a reactor shutdown t at t = 0 to zero flux. The 135Xe nd 1351 ecayequations, Eqs. (10.30) and a d (10.31) become for t > 0
~ dt
= ->'II(t),
dX(t) = AII(t) -AxX(t) dt
(PIO.4)
The solution of the simple radioactive decayequation, Eq.(PIO.3), gives I(t) = 1(0)exp[->'lt]. Substitution of this result in Eq. (P10.4) and solving (seethe general solution of Eqs. (5.51) and (5.52)) X(t) = e-Axt
t1 A/I(t')eAxt'dt' X(O) X(O)e-Axt +
[e-A/t-e->.xt]
(PIO.5)
= ~~ +
AX -Al
These two results for the post shutdown transients X(t) and I(t), can also be obtained directly from Eqs. (10.32) and (10.33) by setting l/Jo= 0 in these equations. To find the time tmaxfollowing the shutdown at which X (t) reachesa maximum, we differentiate Eq. (PIO.5) and set the result to zero at t = tmax. The result
0 = t->'IX(O)
> + >..x '1 1(0) ">,X->'1
eXp[-Axtmax]-~~
A~I(O)
eXp[-A[tmax].
Multiply by eA[tmax nd collect terms to obtain a exp[-(AI -AX)tmax] =
= (1 -r)r"""j(O) X(O) + r.
July 24, 2002
(PIO.3)