Subfield: Modern Physics

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.

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# Schematic sketch of the Nuclear Reactor (NR) 

Fig-I: Enlarged view of a fuel channel (1-Fuel Pins)

Fig-II: A view of the NR (2-Fuel Channels)

Fig-III: Top view of NR (3-Square Arrangement of Fuel Channels and 4-Typical Neutron Paths).

Only components relevant to the problem are shown (e.g. control rods and coolant are not shown).

## A Fuel Pin

| Data <br> for $\mathrm{UO}_{2}$ | 1. Molecular weight $M_{w}=0.270 \mathrm{~kg} \mathrm{~mol}^{-1}$ | 2. | Density $\rho=1.060 \times 10^{4} \mathrm{~kg} \mathrm{~m}^{-3}$ |
| :---: | :--- | :--- | :--- |
|  | 3. Melting point $T_{m}=3.138 \times 10^{3} \mathrm{~K}$ | 4. | Thermal conductivity $\lambda=3.280 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ |
Context question:
A.1 Consider the following fission reaction of a stationary ${ }^{235} \mathrm{U}$ after it absorbs a neutron of negligible kinetic <br> energy. <br> $\qquad{ }^{235} \mathrm{U}+{ }^{1} \mathrm{n} \rightarrow{ }^{94} \mathrm{Zr}+{ }^{140} \mathrm{Ce}+2{ }^{1} \mathrm{n}+\Delta E$ <br> $m\left({ }^{94} \mathrm{Zr}\right)=93.9063 \mathrm{u} ; m\left({ }^{140} \mathrm{Ce}\right)=139.905 \mathrm{u} ; m\left({ }^{1} \mathrm{n}\right)=1.00867$ u and $1 \mathrm{u}=931.502 \mathrm{MeV} \mathrm{c}{ }^{-2}$. Ignore charge <br> imbalance.
Context answer:
\boxed{$\Delta E=208.684 $}


Context question:
A.2 Estimate $N$ the number of ${ }^{235} \mathrm{U}$ atoms per unit volume in natural $\mathrm{UO}_{2}$.
Context answer:
\boxed{$=1.702 \times 10^{26} $}


Context question:
A.3 Assume that the neutron flux $\phi=2.000 \times 10^{18} \mathrm{~m}^{-2} \mathrm{~s}^{-1}$ on the fuel is uniform. The fission cross-section (effective area of the target nucleus) of a ${ }^{235} \mathrm{U}$ nucleus is $\sigma_{f}=5.400 \times 10^{-26}$ $\mathrm{m}^{2}$. If $80.00 \%$ of the fission energy is available as heat, estimate $Q$ (in $\mathrm{W} \mathrm{m}^{-3}$ ) the rate of heat production in the pin per unit volume. $1 \mathrm{MeV}=1.602 \times 10^{-13} \mathrm{~J}$.
Context answer:
\boxed{$4.917\times 10^8$}


Question: A.4 The steady-state temperature difference between the center $\left(T_{c}\right)$ and the surface $\left(T_{s}\right)$ 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.

Solution:
The dimensions of $T_{c}-T_{s}$ is temperature. We write this as $T_{c}-T_{s}=[K]$. Once can similarly write down the dimensions of $Q, a$ and $\lambda$. Equating the temperature to powers of $Q, a$ and $\lambda$, one could state the following dimensional equation:

$K=Q^{\alpha} a^{\beta} \lambda^{\gamma}$

$=\left[M L^{-1} T^{-3}\right]^{\alpha}[L]^{\beta}\left[M L^{1} T^{-3} K^{-1}\right]^{\gamma}$

This yields the following algebraic equations

$\gamma=-1$ equating powers of temperature

$\alpha+\gamma=0$ equating powers of mass or time. From the previous equation we get $\alpha=1$

Next $-\alpha+\beta+\gamma=0$ equating powers of length. This yields $\beta=2$.

Thus we obtain $T_{c}-T_{s}=\frac{Q a^{2}}{4 \lambda}$ where we insert the dimensionless factor $1 / 4$ as suggested in the problem. No penalty if the factor $1 / 4$ is not written.

Final answer: $T_{c}-T_{s}=\frac{Q a^{2}}{4 \lambda}$