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### HW20-22page1

Course: ECE 4130, Spring 2008
School: Cornell
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Word Count: 208

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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 &gt; 0 ~ dt = -&gt;'II(t), dX(t) = AII(t) -AxX(t) dt (PIO.4) The solution of the simple...

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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)
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10-15where r =Ax/ A[ ~ 0.724. Take the logarithm of both sides and solve for tmax to get the desired result, namely tmax= -:&gt;;~ 1 In r 1 + (1 -r)I(O)' ] X(O)cfw_[(P10.6)Now consider under what conditions there is a rise in 135Xe ollowing the shutf do
Cornell - ECE - 4130
July 24, 2002
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