IPhO 2015 · Theory problem 3

The Design of a Nuclear Reactor

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Uranium occurs in nature as \(\mathrm{UO}_2\) with only 0.720% of the uranium atoms being \({}^{235}\mathrm{U}\). Neutron induced fission occurs readily in \({}^{235}\mathrm{U}\) with the emission of 2-3 fission neutrons having high kinetic energy. This fission probability will increase if the neutrons inducing fission have low kinetic energy. So by reducing the kinetic energy of the fission neutrons, one can induce a chain of fissions in other \({}^{235}\mathrm{U}\) nuclei. This forms the basis of the power generating nuclear reactor (NR).

A typical NR consists of a cylindrical tank of height \(H\) and radius \(R\) filled with a material called moderator. Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel pins of natural \(\mathrm{UO}_2\) in solid form of height \(H\), are kept axially in a square array. Fission neutrons, coming outward from a fuel channel, collide with the moderator, losing energy and reach the surrounding fuel channels with low enough energy to cause fission (Figs I-III). Heat generated from fission in the pin is transmitted to a coolant fluid flowing along its length. In the current problem we shall study some of the physics behind the (A) Fuel Pin, (B) Moderator and (C) NR of cylindrical geometry.

Figure 1

A Fuel Pin

Data for \(\mathrm{UO}_2\) 1. Molecular weight \(M_w=0.270\ \mathrm{kg\,mol^{-1}}\) 2. Density \(\rho=1.060\times10^4\ \mathrm{kg\,m^{-3}}\)
3. Melting point \(T_m=3.138\times10^3\ \mathrm{K}\) 4. Thermal conductivity \(\lambda=3.280\ \mathrm{W\,m^{-1}K^{-1}}\)
A1 Consider the following fission reaction of a stationary \({}^{235}\mathrm{U}\) after it absorbs a neutron of negligible kinetic energy.

\({}^{235}\mathrm{U}+{}^{1}\mathrm{n}\ \longrightarrow\ {}^{94}\mathrm{Zr}+{}^{140}\mathrm{Ce}+2\,{}^{1}\mathrm{n}+\Delta E\)

Estimate \(\Delta E\) (in MeV) the total fission energy released. The nuclear masses are: \(m({}^{235}\mathrm{U})=235.044\ \mathrm{u}\); \(m({}^{94}\mathrm{Zr})=93.9063\ \mathrm{u}\); \(m({}^{140}\mathrm{Ce})=139.905\ \mathrm{u}\); \(m({}^{1}\mathrm{n})=1.00867\ \mathrm{u}\) and \(1\ \mathrm{u}=931.502\ \mathrm{MeV}\,c^{-2}\). Ignore charge imbalance.		 0.8
A2 Estimate \(N\) the number of \({}^{235}\mathrm{U}\) atoms per unit volume in natural \(\mathrm{UO}_2\). 0.5
A3 Assume that the neutron flux density, \(\varphi=2.000\times10^{18}\ \mathrm{m^{-2}s^{-1}}\) on the fuel is uniform. The fission cross-section (effective area of the target nucleus) of a \({}^{235}\mathrm{U}\) nucleus is \(\sigma_f=5.400\times10^{-26}\ \mathrm{m^2}\). If 80.00% of the fission energy is available as heat, estimate \(Q\) (in \(\mathrm{W\,m^{-3}}\)), the rate of heat production in the pin per unit volume. \(1\mathrm{MeV}=1.602\times10^{-13}\ \mathrm{J}\) 1.2
A4 The steady-state temperature difference between the center (\(T_c\)) and the surface (\(T_s\)) of the pin can be expressed as \(T_c-T_s=k\,F(Q,a,\lambda)\), where \(k=1/4\) is a dimensionless constant and \(a\) is the radius of the pin. Obtain \(F(Q,a,\lambda)\) by dimensional analysis. Note that \(\lambda\) is the thermal conductivity of \(\mathrm{UO}_2\). 0.5
A5 The desired temperature of the coolant is \(5.770\times10^2\ \mathrm{K}\). Estimate the upper limit \(a_u\) on the radius \(a\) of the pin. 1.0

B The Moderator

Consider the two dimensional elastic collision between a neutron of mass \(1\ \mathrm{u}\) and a moderator atom of mass \(A\ \mathrm{u}\). Before collision all the moderator atoms are considered at rest in the laboratory frame (LF). Let \(\vec v_b\) and \(\vec v_a\) be the velocities of the neutron before and after collision respectively in the LF. Let \(\vec v_m\) be the velocity of the center of mass (CM) frame relative to LF and \(\theta\) be the neutron scattering angle in the CM frame. All the particles involved in collisions are moving at nonrelativistic speeds.

B1 The collision in LF is shown schematically, where \(\theta_L\) is the scattering angle (Fig-IV). Sketch the collision schematically in CM frame. Label the particle velocities for 1, 2 and 3 in terms of \(\vec v_b\), \(\vec v_a\) and \(\vec v_m\). Indicate the scattering angle \(\theta\). 1.0
Figure 2
B2 Obtain \(v\) and \(V\), the speeds of the neutron and moderator atom in the CM frame after collision, in terms of \(A\) and \(v_b\). 1.0
B3 Derive an expression for \(G(\alpha,\theta)=E_a/E_b\), where \(E_b\) and \(E_a\) are the kinetic energies of the neutron, in the LF, before and after the collision respectively and \(\alpha\equiv[(A-1)/(A+1)]^2\). 1.0
B4 Assume that the above expression holds for \(\mathrm{D}_2\mathrm{O}\) molecule. Calculate the maximum possible fractional energy loss \(f_l\equiv(E_b-E_a)/E_b\) of the neutron for the \(\mathrm{D}_2\mathrm{O}\) (\(20\ \mathrm{u}\)) moderator. 0.5

C The Nuclear Reactor

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 \(k_1=1.021\times10^{-2}\ \mathrm{m}\) 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
C2 The fuel channels are in a square arrangement (Fig-III) with the nearest neighbour distance \(0.286\ \mathrm{m}\). The effective radius of a fuel channel (if it were solid) is \(3.617\times10^{-2}\ \mathrm{m}\). Estimate the number of fuel channels \(F_n\) in the reactor and the mass \(M\) of \(\mathrm{UO}_2\) required to operate the NR in steady state. 1.0

\(^{1}\) 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.

Official English text of the problem.

Answers

  • A1: \(\Delta E\simeq208.7\ \mathrm{MeV}\).
  • A2: \(N\simeq1.70\times10^{26}\ \mathrm{m^{-3}}\).
  • A3: \(Q\simeq4.92\times10^8\ \mathrm{W\,m^{-3}}\).
  • A4: \(\displaystyle F(Q,a,\lambda)=\frac{Qa^2}{\lambda}\), so \(\displaystyle T_c-T_s=\frac{Qa^2}{4\lambda}\).
  • A5: \(a_u\simeq8.27\times10^{-3}\ \mathrm{m}=8.27\ \mathrm{mm}\).
  • B1: CM velocity labels: \(1:\vec v_b-\vec v_m\), \(2:\vec v_a-\vec v_m\), \(3:-\vec v_m\). The angle between the initial and final neutron velocities is \(\theta\); the two outgoing particles move in opposite directions.
  • B2: \(\displaystyle v=\frac{A}{A+1}v_b,\qquad V=\frac{1}{A+1}v_b\).
  • B3: \(\displaystyle G(\alpha,\theta)=\frac{1+\alpha+(1-\alpha)\cos\theta}{2}\).
  • B4: \(\displaystyle f_{l,\max}=\frac{80}{441}\simeq0.1814=18.14\%\).
  • C1: \(\displaystyle R=2.405\sqrt{\frac{3k_1}{2k_2}}\simeq3.175\ \mathrm{m}\), \(\displaystyle H=\pi\sqrt{\frac{3k_1}{k_2}}\simeq5.865\ \mathrm{m}\).
  • C2: \(F_n\simeq387\) fuel channels; \(M\simeq9.89\times10^4\ \mathrm{kg}\), or \(98.9\) tonnes of \(\mathrm{UO}_2\).

Solution

A1

Since the initial kinetic energy is negligible, conservation of energy gives the released energy as the decrease in total rest energy:

\[ \Delta E= \left[ m({}^{235}\mathrm U)+m({}^{1}\mathrm n) -m({}^{94}\mathrm{Zr})-m({}^{140}\mathrm{Ce}) -2m({}^{1}\mathrm n) \right]c^2. \]

Using the supplied masses without any charge correction,

\[ \begin{aligned} \Delta m &=\left[235.044+1.00867 -\left(93.9063+139.905+2(1.00867)\right)\right]\mathrm u\\ &=0.22403\ \mathrm u. \end{aligned} \]

Therefore,

\[ \Delta E=0.22403(931.502)\ \mathrm{MeV} =208.684\ldots\ \mathrm{MeV}, \]

or

\[ \boxed{\Delta E\simeq208.7\ \mathrm{MeV}}. \]

A2

Each formula unit of \(\mathrm{UO}_2\) contains one uranium atom. The number of formula units per unit volume is

\[ n_{\mathrm{UO}_2}=\frac{\rho}{M_w}N_A, \]

where the standard Avogadro constant is \(N_A\simeq6.022\times10^{23}\ \mathrm{mol^{-1}}\).

The fraction of uranium atoms that are \({}^{235}\mathrm U\) is \(0.00720\). Hence

\[ \begin{aligned} N &=0.00720\,\frac{\rho N_A}{M_w}\\ &=0.00720\, \frac{1.060\times10^4}{0.270} (6.022\times10^{23})\ \mathrm{m^{-3}}\\ &=1.7022\ldots\times10^{26}\ \mathrm{m^{-3}}. \end{aligned} \]

Thus

\[ \boxed{N\simeq1.70\times10^{26}\ \mathrm{m^{-3}}}. \]

A3

The fission rate of one target nucleus in the prescribed uniform neutron flux is \(\varphi\sigma_f\). Therefore the fission rate per unit volume is

\[ r_f=N\varphi\sigma_f. \]

Numerically,

\[ \varphi\sigma_f =(2.000\times10^{18})(5.400\times10^{-26}) =1.080\times10^{-7}\ \mathrm{s^{-1}}, \]

and

\[ r_f=1.8384\ldots\times10^{19}\ \mathrm{m^{-3}s^{-1}}. \]

The heat deposited per fission is

\[ E_h=0.8000\,\Delta E =0.8000(208.684\ldots)(1.602\times10^{-13})\ \mathrm J. \]

Consequently,

\[ \begin{aligned} Q &=N\varphi\sigma_f E_h\\ &=4.9168\ldots\times10^8\ \mathrm{W\,m^{-3}}, \end{aligned} \]

giving

\[ \boxed{Q\simeq4.92\times10^8\ \mathrm{W\,m^{-3}}}. \]

A4

Let \(F\) have the dimensional form \(Q^x a^y\lambda^z\). Denoting dimensions of power, length and temperature by \(P,L,\Theta\),

\[ [Q]=PL^{-3},\qquad [a]=L,\qquad [\lambda]=PL^{-1}\Theta^{-1}. \]

Thus

\[ [F]=P^{x+z}L^{-3x+y-z}\Theta^{-z}. \]

For \(F\) to have the dimensions of temperature,

\[ x+z=0,\qquad -3x+y-z=0,\qquad -z=1. \]

These equations give

\[ x=1,\qquad y=2,\qquad z=-1. \]

The dimensionless numerical factor is already supplied as \(k\), so

\[ \boxed{F(Q,a,\lambda)=\frac{Qa^2}{\lambda}}, \qquad \boxed{T_c-T_s=\frac{Qa^2}{4\lambda}}. \]

A5

With uniform internal heat production, the center is the hottest part of the pin. Avoiding melting requires

\[ T_c\leq T_m. \]

Heat must flow from the pin surface to the coolant, so the surface temperature cannot be below the coolant temperature:

\[ T_s\geq T_{\mathrm{cool}}. \]

It follows that

\[ \frac{Qa^2}{4\lambda}=T_c-T_s \leq T_m-T_{\mathrm{cool}}. \]

The limiting radius is therefore

\[ a_u=\sqrt{\frac{4\lambda(T_m-T_{\mathrm{cool}})}{Q}}. \]

Using

\[ T_m-T_{\mathrm{cool}}=3138-577.0=2561\ \mathrm K, \]

we obtain

\[ a_u= \sqrt{\frac{4(3.280)(2561)}{4.9168\ldots\times10^8}} =8.2667\ldots\times10^{-3}\ \mathrm m. \]

Thus

\[ \boxed{a_u\simeq8.27\ \mathrm{mm}}. \]

This upper limit corresponds to the ideal limiting case \(T_s=T_{\mathrm{cool}}\). A finite temperature drop between the surface and coolant would reduce the permissible radius.

B1

Let the neutron mass be \(m\), so the moderator mass is \(Am\). Before collision the moderator is stationary in the LF. The CM velocity is therefore

\[ \vec v_m =\frac{m\vec v_b+Am\vec0}{(A+1)m} =\frac{\vec v_b}{A+1}. \]

A Galilean transformation to the CM frame subtracts \(\vec v_m\) from every LF velocity. Thus the required labels are

\[ \boxed{ \vec u_1=\vec v_b-\vec v_m,\qquad \vec u_2=\vec v_a-\vec v_m,\qquad \vec u_3=-\vec v_m }. \]

The sketch should have these features:

  • Before collision, neutron 1 approaches from the left, moving horizontally right with velocity \(\vec v_b-\vec v_m\).
  • Moderator 3 approaches from the right, moving horizontally left with velocity \(-\vec v_m\).
  • After collision, neutron 2 leaves the collision point with velocity \(\vec v_a-\vec v_m\). Mark \(\theta\) between this arrow and the original rightward neutron direction.
  • Moderator 4 leaves in the direction exactly opposite to neutron 2, because the total CM momentum is zero. Its velocity is \[ \vec u_4=-\frac{\vec u_2}{A}. \]

For example, if the outgoing neutron arrow points upward and right, the outgoing moderator arrow points downward and left.

B2

The initial CM speeds are

\[ u_{n,b}=v_b-v_m=\frac{A}{A+1}v_b, \qquad u_{M,b}=v_m=\frac{1}{A+1}v_b. \]

Conservation of momentum in the CM frame requires the final momenta to be equal and opposite. Their magnitudes therefore satisfy

\[ mv=AmV, \qquad\text{so}\qquad v=AV. \]

Because the collision is elastic, kinetic energy is conserved:

\[ \frac12 m u_{n,b}^2+\frac12 Am u_{M,b}^2 =\frac12 mv^2+\frac12 AmV^2. \]

After cancelling \(m/2\),

\[ \frac{A}{A+1}v_b^2=v^2+AV^2. \]

Substituting \(v=AV\) gives

\[ A(A+1)V^2=\frac{A}{A+1}v_b^2. \]

Hence

\[ \boxed{V=\frac{v_b}{A+1}}, \qquad \boxed{v=\frac{A}{A+1}v_b}. \]

The collision changes the velocity directions, but each particle’s speed in the CM frame remains unchanged.

B3

The neutron’s final LF velocity is

\[ \vec v_a=\vec v_m+\vec u_2. \]

The vector \(\vec v_m\) points along the initial neutron direction, while \(\vec u_2\) makes angle \(\theta\) with that direction. Consequently,

\[ v_a^2=v_m^2+v^2+2v_mv\cos\theta. \]

Using the results above,

\[ \frac{v_a^2}{v_b^2} =\frac{1+A^2+2A\cos\theta}{(A+1)^2}. \]

Since both energies refer to the same neutron in the LF,

\[ G(\alpha,\theta)=\frac{E_a}{E_b} =\frac{v_a^2}{v_b^2}. \]

Now

\[ \alpha=\frac{(A-1)^2}{(A+1)^2} \]

implies

\[ \frac{1+\alpha}{2} =\frac{A^2+1}{(A+1)^2}, \qquad \frac{1-\alpha}{2} =\frac{2A}{(A+1)^2}. \]

Therefore,

\[ \boxed{ G(\alpha,\theta) =\frac{1+\alpha+(1-\alpha)\cos\theta}{2} }. \]

An equivalent form is

\[ G(\alpha,\theta) =1-(1-\alpha)\sin^2\frac{\theta}{2}. \]

In particular, forward scattering gives \(G=1\), while backward scattering in the CM frame gives \(G=\alpha\).

B4

The fractional energy loss is

\[ f_l=1-G =(1-\alpha)\sin^2\frac{\theta}{2}. \]

It is largest for CM backscattering, \(\theta=\pi\). Thus

\[ f_{l,\max}=1-\alpha =1-\left(\frac{A-1}{A+1}\right)^2 =\frac{4A}{(A+1)^2}. \]

Following the problem’s approximation that the whole \(\mathrm D_2\mathrm O\) molecule is a target of mass \(20\ \mathrm u\), set \(A=20\):

\[ \boxed{ f_{l,\max}=\frac{4(20)}{21^2} =\frac{80}{441} \simeq0.1814=18.14\% }. \]

C1

For a nonzero steady neutron flux, production and leakage must balance:

\[ k_1\left[\frac{2.405^2}{R^2}+\frac{\pi^2}{H^2}\right]\psi =k_2\psi. \]

Cancelling \(\psi\),

\[ \frac{2.405^2}{R^2}+\frac{\pi^2}{H^2} =\frac{k_2}{k_1}. \tag{1} \]

We first determine the shape that minimizes leakage at fixed volume. Write

\[ \mathcal V=\pi R^2H,\qquad b=2.405, \]

so that

\[ H=\frac{\mathcal V}{\pi R^2}. \]

The geometric factor in the leakage rate becomes

\[ S(R)=\frac{b^2}{R^2}+\frac{\pi^4R^4}{\mathcal V^2}. \]

Its derivative vanishes when

\[ -\frac{2b^2}{R^3} +\frac{4\pi^4R^3}{\mathcal V^2}=0. \]

Rearranging and using \(\mathcal V=\pi R^2H\),

\[ \boxed{\frac{b^2}{R^2}=2\frac{\pi^2}{H^2}}. \tag{2} \]

This stationary point is a minimum, since

\[ S''(R)=\frac{6b^2}{R^4} +\frac{12\pi^4R^2}{\mathcal V^2}>0. \]

The optimum aspect ratio is therefore

\[ \frac{H}{R}=\frac{\sqrt2\,\pi}{b} =1.84735\ldots. \]

Combining (1) and (2),

\[ 3\frac{\pi^2}{H^2}=\frac{k_2}{k_1}, \]

which gives

\[ \boxed{H=\pi\sqrt{\frac{3k_1}{k_2}}}, \qquad \boxed{R=b\sqrt{\frac{3k_1}{2k_2}}}. \]

Numerically,

\[ H=\pi\sqrt{\frac{3(1.021\times10^{-2})} {8.787\times10^{-3}}} \simeq\boxed{5.865\ \mathrm m}, \]

\[ R=2.405\sqrt{\frac{3(1.021\times10^{-2})} {2(8.787\times10^{-3})}} \simeq\boxed{3.175\ \mathrm m}. \]

C2

Let the square-lattice spacing be \(d=0.286\ \mathrm m\). Neglecting boundary corrections, each fuel channel occupies one square cell of cross-sectional area \(d^2\). Therefore,

\[ F_n\simeq\frac{\pi R^2}{d^2}. \]

Substitution gives

\[ F_n\simeq \frac{\pi(3.17507\ldots)^2}{(0.286)^2} =387.19\ldots. \]

Hence the estimated number is

\[ \boxed{F_n\simeq387\text{ fuel channels}}. \]

With effective solid-fuel radius \(r_f=3.617\times10^{-2}\ \mathrm m\), the volume of \(\mathrm{UO}_2\) in one channel is \(\pi r_f^2H\). Its mass is therefore \(\rho\pi r_f^2H\), and the total fuel mass is

\[ M=F_n\rho\pi r_f^2H. \]

Using the unrounded area estimate,

\[ \begin{aligned} M &\simeq(387.19\ldots)(1.060\times10^4) \pi(3.617\times10^{-2})^2(5.86547\ldots)\ \mathrm{kg}\\ &=9.8942\ldots\times10^4\ \mathrm{kg}. \end{aligned} \]

Thus

\[ \boxed{M\simeq9.89\times10^4\ \mathrm{kg} =98.9\text{ tonnes of }\mathrm{UO}_2}. \]

The solution was checked against the official solution.