Physics

Physics · modern-physics

Problem

A.4 The steady-state temperature difference between the center (Tc)\left(T_{c}\right) and the surface (Ts)\left(T_{s}\right) of the pin can be expressed as Tc−Ts=kF(Q,a,λ)T_{c}-T_{s}=k F(Q, a, \lambda) where k=1/4k=1 / 4 is a dimensionless constant and aa is the radius of the pin. Obtain F(Q,a,λ)F(Q, a, \lambda) by dimensional analysis. Given constants: Uranium occurs in nature as UO2\mathrm{UO}_{2} with only 0.720%0.720 \% of the uranium atoms being 235U{ }^{235} \mathrm{U}. Neutron induced fission occurs readily in 235U{ }^{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 235U{ }^{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 HH and radius RR filled with a material called moderator. Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel pins of natural UO2\mathrm{UO}_{2} in solid form of height HH, 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. <image_1> # 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 UO2\mathrm{UO}_{2}1. Molecular weight Mw=0.270 kg mol−1M_{w}=0.270 \mathrm{~kg} \mathrm{~mol}^{-1}2.Density ρ=1.060×104 kg m−3\rho=1.060 \times 10^{4} \mathrm{~kg} \mathrm{~m}^{-3}
3. Melting point Tm=3.138×103 KT_{m}=3.138 \times 10^{3} \mathrm{~K}4.Thermal conductivity λ=3.280 W m−1 K−1\lambda=3.280 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}
Context question: A.1 Consider the following fission reaction of a stationary 235U{ }^{235} \mathrm{U} after it absorbs a neutron of negligible kinetic <br> energy. <br> 235U+1n→94Zr+140Ce+21n+ΔE\qquad{ }^{235} \mathrm{U}+{ }^{1} \mathrm{n} \rightarrow{ }^{94} \mathrm{Zr}+{ }^{140} \mathrm{Ce}+2{ }^{1} \mathrm{n}+\Delta E <br> m(94Zr)=93.9063u;m(140Ce)=139.905u;m(1n)=1.00867m\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 1u=931.502MeVc−21 \mathrm{u}=931.502 \mathrm{MeV} \mathrm{c}{ }^{-2}. Ignore charge <br> imbalance. Context answer: ΔE=208.684\Delta E=208.684 Context question: A.2 Estimate NN the number of 235U{ }^{235} \mathrm{U} atoms per unit volume in natural UO2\mathrm{UO}_{2}. Context answer: =1.702×1026=1.702 \times 10^{26} Context question: A.3 Assume that the neutron flux ϕ=2.000×1018 m−2 s−1\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 235U{ }^{235} \mathrm{U} nucleus is σf=5.400×10−26\sigma_{f}=5.400 \times 10^{-26} m2\mathrm{m}^{2}. If 80.00%80.00 \% of the fission energy is available as heat, estimate QQ (in Wm−3\mathrm{W} \mathrm{m}^{-3} ) the rate of heat production in the pin per unit volume. 1MeV=1.602×10−13 J1 \mathrm{MeV}=1.602 \times 10^{-13} \mathrm{~J}. Context answer: 4.917×1084.917\times 10^8

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