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Question 9.10: Apply appropriate software to Eqs. (9.37) through (9.42) for......

Apply appropriate software to Eqs. (9.37) through (9.42) for a uranium fueled reactor. Use the following parameters, which are typical of a large pressurized water reactor: Λ = 50 × 10^{-6}  s , α = -4.2 × 10^{-5} /°C   M_ƒc_ƒ = 32  MW/°C ,  τ = 4.5  s , With an initial steady state power of 10 MW, make a plot of the power for

a. step reactivity increase of 10 cents,
b. a step reactivity increase of 20 cents
c. a step reactivity decrease of. 10 cents,
d. a step reactivity decrease of 20 cents.

Step-by-Step
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β1 := 0.00021          \operatorname{c1}:=\mathbb{\beta 1}\cdot {\frac{\operatorname{P0}}{(\lambda1\cdot\Lambda)}}     \mathrm{c}2:=\beta 2.{\frac{\mathrm{P}0}{(\lambda2\cdot\Lambda)}}

β2 := 0.00142

β3 := 0.00128            \operatorname{c3}:=\mathbb{\beta 3}\cdot {\frac{\operatorname{P0}}{(\lambda3\cdot\Lambda)}}    \mathrm{c}4:=\beta 4.{\frac{\mathrm{P}0}{(\lambda4\cdot\Lambda)}}

β4 := 0.00257

β5 := 0.00075            \operatorname{c5}:=\mathbb{\beta 5}\cdot {\frac{\operatorname{P0}}{(\lambda5\cdot\Lambda)}}    \mathrm{c}6:=\beta 6.{\frac{\mathrm{P}0}{(\lambda6\cdot\Lambda)}}

β6 := 0.00027

Sample calculations shown for $0.10 below using the stiff differential equations solver Radau (Mathcad) to obtain the vector Y.

\mathrm{Rf}:={\frac{\tau}{\mathrm{MfCf}}}=0.141                             dollars := 0.1

Rf.WCP = 8.438                                       α1 = -4.2 × 10^{-5}

Tf0 := Rf.P0 + Ti = 551.406

\mathrm y := \begin{pmatrix}P0 \\ c1 \\ c2 \\ c3 \\ c4 \\ c5 \\c6 \\ \mathrm {Tf0} \end{pmatrix}

 

\mathrm{DD(t,y)} := \begin{bmatrix}\frac{\bigg[\mathrm {dollars}\beta + \alpha 1 \cdot (\mathrm {y_7 – Tf0}) – \beta \bigg ]}{\Lambda} \cdot y_0 + \lambda1 \cdot y_1 + \lambda2 \cdot y_2 + \lambda3 \cdot y_3 + \lambda4 \cdot y_4 + \lambda5 \cdot y_5 + \lambda6 \cdot y_6 \\ \begin{matrix}\left({\frac{\beta1}{\Lambda}}\right)\mathbf{y}_0-\lambda1.\mathrm{y_1} \\ \left({\frac{\beta2}{\Lambda}}\right)\mathbf{y}_0-\lambda2.\mathrm{y_2} \\ \left({\frac{\beta3}{\Lambda}}\right)\mathbf{y}_0-\lambda3.\mathrm{y_3}\\ \left({\frac{\beta4}{\Lambda}}\right)\mathbf{y}_0-\lambda4.\mathrm{y_4} \\ \left({\frac{\beta5}{\Lambda}}\right)\mathbf{y}_0-\lambda5.\mathrm{y_5} \\ \left({\frac{\beta6}{\Lambda}}\right)\mathbf{y}_0-\lambda6.\mathrm{y_6} \\ \frac{y_0}{\mathrm{MfCf}} – \frac{(y_7 – \mathrm{Tf0})}{\tau}\end{matrix} \end{bmatrix}
Y:=\mathrm{R a d a u}(y,0,800,1000,\mathrm{DD})

n := 0 .. 1000

22 d
23 d

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