Съдържание

IPhO 2015, Теория — Задача 3.3. The Nuclear Reactor

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Всички материали от IPhO 2015

The Design of a Nuclear Reactor

Footnote 1

Joseph Amal Nathan (BARC) and Vijay A. Singh (ex-National Coordinator, Science Olympiads) were the principal authors of this problem. The contributions of the Academic Committee, Academic Development Group and the International Board are gratefully acknowledged.

Условие

To operate the NR at any constant neutron flux $\psi$ (steady state), the leakage of neutrons has to be compensated by an excess production of neutrons in the reactor. For a reactor in cylindrical geometry the leakage rate is $k_1\,[(2.405/R)^2 + (\pi/H)^2]\,\psi$ and the excess production rate is $k_2\,\psi$ . The constants $k_1$ and $k_2$ depend on the material properties of the NR.

C1 Consider a NR with $k_1 = 1.021\times10^{-2}$ m and $k_2 = 8.787\times10^{-3}\ \mathrm{m^{-1}}$ . Noting that for a fixed volume the leakage rate is to be minimized for efficient fuel utilization, obtain the dimensions of the NR in the steady state. [1,5 т.]

C2 The fuel channels are in a square arrangement (Fig-III) with the nearest neighbour distance 0.286 m. The effective radius of a fuel channel (if it were solid) is $3.617\times10^{-2}$ m. Estimate the number of fuel channels $F_n$ in the reactor and the mass $M$ of UO $_2$ required to operate the NR in steady state. [1 т.]

Отговори

Решение

Solution: $R = 3.175$ m, $H = 5.866$ m.

Detailed solution: For constant volume $V = \pi R^2H$ ,

\frac{d}{dH}\left[\left(\frac{2.405}{R}\right)^2 + \left(\frac{\pi}{H}\right)^2\right] = 0,

\frac{d}{dH}\left[\frac{2.405^2\pi H}{V} + \frac{\pi^2}{H^2}\right] = \frac{2.405^2\pi}{V} - 2\frac{\pi^2}{H^3} = 0,

gives $\left(\frac{2.405}{R}\right)^2 = 2\left(\frac{\pi}{H}\right)^2$ .

For steady state,

1.021\times10^{-2}\left[\left(\frac{2.405}{R}\right)^2 + \left(\frac{\pi}{H}\right)^2\right]\Psi = 8.787\times10^{-3}\ \Psi.

Hence $H = 5.866$ m [Acceptable Range (5.870 to 5.890)]

$R = 3.175$ m [Acceptable Range (3.170 to 3.180)].

Alternative Non-Calculus Method to Optimize

Minimisation of the expression $\left(\dfrac{2.405}{R}\right)^2 + \left(\dfrac{\pi}{H}\right)^2$ , for a fixed volume $V = \pi R^2H$ :

Substituting for $R^2$ in terms of $V,\ H$ we get $\dfrac{2.405^2\pi H}{V} + \dfrac{\pi^2}{H^2}$ ,

which can be written as, $\dfrac{2.405^2\pi H}{2V} + \dfrac{2.405^2\pi H}{2V} + \dfrac{\pi^2}{H^2}$ .

Since all the terms are positive applying AMGM inequality for three positive terms we get

\frac{\frac{2.405^2\pi H}{2V} + \frac{2.405^2\pi H}{2V} + \frac{\pi^2}{H^2}}{3} \ge \sqrt[3]{\frac{2.405^2\pi H}{2V}\times\frac{2.405^2\pi H}{2V}\times\frac{\pi^2}{H^2}} = \sqrt[3]{\frac{2.405^4\pi^4}{4V^2}}.

The RHS is a constant. The LHS is always greater or equal to this constant implies that this is the minimum value the LHS can achieve. The minimum is achieved when all the three positive terms are equal, which gives the condition $\dfrac{2.405^2\pi H}{2V} = \dfrac{\pi^2}{H^2} \Rightarrow \left(\dfrac{2.405}{R}\right)^2 = 2\left(\dfrac{\pi}{H}\right)^2$ .

For steady state,

1.021\times10^{-2}\left[\left(\frac{2.405}{R}\right)^2 + \left(\frac{\pi}{H}\right)^2\right]\Psi = 8.787\times10^{-3}\ \Psi.

Hence $H = 5.866$ m [Acceptable Range (5.870 to 5.890)]

$R = 3.175$ m [Acceptable Range (3.170 to 3.180)].

Note: Putting the condition in the RHS gives the minimum as $\dfrac{\pi^2}{H^2}$ . From the condition we get $\dfrac{\pi^3}{H^3} = \dfrac{2.405^2\pi^2}{2V} \Rightarrow \dfrac{\pi^2}{H^2} = \sqrt[3]{\dfrac{2.405^4\pi^4}{4V^2}}$ .

Note: The radius and height of the Tarapur 3 & 4 NR in Western India is 3.192 m and 5.940 m respectively.

Solution: $F_n = 387$ and $M = 9.892\times10^{4}$ kg.

Detailed solution: Since the fuel channels are in square pitch of 0.286 m, the effective area per channel is $0.286^2\ \mathrm{m^2} = 8.180\times10^{-2}\ \mathrm{m^2}$ .

The cross-sectional area of the core is $\pi R^2 = 3.142\times(3.175)^2 = 31.67\ \mathrm{m^2}$ , so the maximum number of fuel channels that can be accommodated in the cylinder is the integer part of $\frac{31.67}{0.0818} = 387$ .

Mass of the fuel=387×Volume of the rod×density

= 387\times(\pi\times0.03617^2\times5.866)\times10600 = 9.892\times10^{4}\mathrm{kg}.

$F_n = 387$ [Acceptable Range (380 to 394)]

$M = 9.892\times10^{4}$ kg [Acceptable Range (9.000 to 10.00)]

Note 1: (Not part of grading) The total volume of the fuel is $387\times(\pi\times0.03617^2\times5.866) = 9.332\ \mathrm{m^3}$ . If the reactor works at 12.5 % efficiency then using the result of a-(iii) we have that the power output of the reactor is $9.332\times4.917\times10^{8}\times0.125 =$ 573 MW.

Note 2: The Tarapur 3 & 4 NR in Western India has 392 channels and the mass of the fuel in it is $10.15\times10^{4}$ kg. It produces 540 MW of power.

Оригинал в Архива: 3.pdf · официални решения: 3_sol.pdf