Nuclear Engineering

Fission Reaction Mechanism, Energy in Fission Reaction and Basic Formulation on Fission Reactor Copy

Let us first to understand the mechanism of nuclear fission process based on simple liquid drop picture of a nucleus.

Bohr and Wheeler explained the fission process considering the liquid-drop model, where nucleus resembles a drop of liquid in the respect that each constituent particle interacts equally with its nearest neighbors. In a liquid droplet, the total energy of the molecules inside the droplet is not sufficient enough to overcome the forces that hold the molecules in the form of a droplet. However, if an external force is applied so that it is set into oscillation, the system passes through a series of stages, of which the most significant ones are shown in Fig. m3.1. This drop is at first spherical (S-1) and then elongated into an ellipsoid (S-2). Although the volume remains constant, the surface area has increased. But provided the volume energy exceeds the surface energy, the drop will return to its original form. However, if the deforming force is sufficiently large, the drop will acquire a shape similar to a dumb-bell as in S-3. In this state, the surface energy will generally exceed the volume energy, which provides the cohesive force for the liquid drop. Consequently, the drop will not return to its initial shape but will rather split into two droplets. These will, at first, be somewhat deformed, as in S-4, but finally they will become spherical (S-5).

FIGURE m3.1 Schematic representation of nuclear fission process based on liquid drop model.

For further understanding, readers are suggested to go through literature.^2,1

Now let us calculate energy production in nuclear fission reaction and also establish basic equations related to power generation through this reaction.

Let us consider the following fission reaction

The fission fragments decay to their radioactive isotopes through β-decay in the following manner:

and

The total mass before fission is	The isotopic mass after fission is

Mass difference = 0.223 amu. Therefore, Q = 0.223 X 931 = 208 MeV.

In MKS unit, Q = 208 X 1.6 X 10^-13 = 3.32 X 10^-11 joules.

Considering U of mass equal to 1 gm, the energy released is (6.02 X 1023/235) X 3.32 X 10^-11 = 8.50 X 10¹⁰ joules, which is indeed very large. In the unit of power, 1 gm U generates 8.50 X 10¹⁰watt -second i.e. 2.36 X 10⁴ kw-hr. Therefore, one gm-atom U produces a power of 5.54 X 10⁶ kw-hr.

Let us see how many fission reactions are required to generate 1 Watt power.

Example m3.3: When a U-235 nucleus undergoes fission, 200 MeV energy is released. How many fissions per sec are needed for generating power of 1 Watt?

Solution:

1 Watt = 1 joule per sec = 6.242 X 10¹² MeV/s

No. of fission needed per sec =

Next important issue related to the realization of generation of power from a reactor is to satisfy the condition of criticality. In a finite size reactor, some neutrons are lost by leaking out. The criticality condition is then defined as the effective multiplication factor, which is basically the ratio of number of neutrons resulting from fission in each generation to the total number lost by both absorption and leakage in the preceding generation. Let, for fissile material, which is a combination of U-238 and U-235, at an instant there be N0 fast neutrons available to produce fission. A few of them may produce fission in U-238 and hence the number of neutrons increases by a factor ν and become N₀ν. These neutrons must be slowed down before they produce fission in U-235. The slowing down process is done in moderator (discussed later), which reduces the energy of the fast neutrons to thermal energy. Considering p to be the fraction slowed down, the available neutrons are N₀νp. Of these again, a fraction f may succeed in producing fission in U-235 before being lost by diffusion or otherwise. Therefore the number of U-235 nuclei undergoing fission is N₀νpf. If at each fission of U-235, fast neutrons are generated to start the cycle again, the total number of neutrons after one cycle is

The ration N/N₀ (=νpfe) is also known as reproduction factor, k. The requirement of criticality in a finite system is thus k = 1; in these circumstances a steady-state fission chain would be possible. If the conditions are such that k < 1, the chain would be convergent and gradually die out. In each generation, more neutrons would be lost in one way or another than are produced by fission, and so the neutron density, and hence the fission rate would decrease steadily. Such system is known to be subcritical. Finally, if k > 1, the chain is divergent and the system is known to be supercritical. More neutrons are produced than are lost in each generation, so that both neutron population and fission rate increase continuously.

How the power of a reactor is calculated?³

Fission rate depends on two quantities, namely macroscopic cross section Σ_f for the process and flux Φ of monoenergetic neutrons.

Where

N is the number of fissile nuclei per cm³, and σ_f cm² is the fission cross section; n is the neutron density, i.e., neutron per cm³, and v is the neutron velocity in cm² per sec. In a reactor of volume V cm³, the fission rate will be VΣ_f Φ. Since it requires a fission rate of 3.1 X 10¹⁰ fissions per sec to produce 1 watt of power (Ex.5.1), then the power of the nuclear reactor can be obtained using the following equation:

If the mass of fissile material (U-235) is g, then we can write

Therefore,

If the mass of U-235 is expressed as w lb, then the above equation becomes,

Therefore, it is clear that the power of the reactor depends on the neutron flux. Therefore, the design of a nuclear reactor depends on efficient way of optimizing neutron number in the reactor. In the next two lectures I am going to discuss the various components of a fission reactor and controlling neutron flux at different stages.

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Fission Reaction Mechanism, Energy in Fission Reaction and Basic Formulation on Fission Reactor Copy

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