# HW20-22page1 - + X(O) = X(O)e-Axt + ~~ [e-A/t -e->.xt]...

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22. Following a reactor shutdown «/Jo -+ 0), show from Eq. (10.33) that the time tm for 135Xe to reach a maximum is given by Solution: First we solve for the 135Xe and 1351 transients following a reactor shutdown at t = 0 to zero flux. The 135Xe and 1351 decay equations, Eqs. (10.30) and (10.31) become for t > 0 ~ dt = ->'II(t), dX(t) dt (PIO.4) The solution of the simple radioactive decay equation, Eq.(PIO.3), gives I(t) = 1(0) exp[->'lt]. = AII(t) -AxX(t) Substitution of this result in Eq. (P10.4) and solving (see the general solution of Eqs. (5.51) and (5.52)) X(t) = e-Axt t1 A/I(t')eAxt'dt'
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Unformatted text preview: + X(O) = X(O)e-Axt + ~~ [e-A/t -e->.xt] AX -Al (PIO.5) 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 tmax following the shutdown at which X (t) reaches a maximum, we differentiate Eq. (PIO.5) and set the result to zero at t = tmax. The result >..x >'1 1(0) A~I(O) eXp[-Axtmax] -~~ eXp[-A[tmax]. 0 = t->'IX(O) + ">,X ->'1 Multiply by eA[tmax and collect terms to obtain X(O) + r. = (1 -r)r"""j(O) exp[-(AI -AX)tmax] = July 24, 2002...
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## This note was uploaded on 02/01/2010 for the course ECE 4130 taught by Professor Cady during the Spring '06 term at Cornell University (Engineering School).

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