Lecture 17
Reactor Theory-II
Objectives
In this lecture you will learn the following
In this lecture we shall look at cylindrical reactor.
As the solution turns out to be a Bessel function, we shall briefly look at the nature of these
functions.
Finally we will show that the flux is more peaked for a cylindrical reactor than for a slab
reactor.
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Lecture 17
Reactor Theory-II
Reactor Theory
Now we shall focus our attention towards cylindrical and spherical reactors.
The governing equation shall continue to be
where,
2 would be different for cylinder and sphere.
The term for
Cylindrical Reactor
Let us now consider an infinite cylindrical reactor as shown
From last slide, we can write the governing equation as
2
+B2
= 0.
Note that we have replaced B m with B.
For the Cylindrical case, the simplified equation shall be
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Lecture 17
Reactor Theory-II
The solution for the above equation is Bessel’s Equation of order zero
For those who are not familiar with the Bessel Functions, let us digress a bit and understand
them.
Notice that when a cylinder is large, the curvature effects are small and hence the solution of
cylinder must look similar to the Cartesian case. Thus the functions J0 and Y 0 should be
similar to Sine and Cosine.
Bessel Functions
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Lecture 17
Reactor Theory-II
Properties of Bessel Function
Note that both J0 and Y 0 are oscillating functions with decaying amplitude.
The reason for the decay in amplitude can be attributed to the divergence of the coordinate in
r direction.
Similar to Sine and Cosine, they have multiple roots.
The first root of J0 is 2.405.
There are some properties of J0 and Y 0 which we shall describe below
Y 0(0) = -∞
J0(0) = 1
First root of J0 is 2.405, That is J0 (2.405) = 0.
Another property of J0 is that
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Lecture 17
Reactor Theory-II
Cylindrical Reactor
The boundary conditions are:
As Y 0 blows up at r = 0, hence A2 = 0
Thus the solution reduces to
Now applying the vanishing flux boundary condition, we get
Thus A1 = 0, which leads to the trivial solution.
For non-trivial fundamental solution,
or
Thus the fundamental solution is
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Lecture 17
Reactor Theory-II
A1 shall be similarly derived as done previously by using power.
Consider a reactor that is producing power per unit length as P'.
Consider a strip of thickness dr at a distance r from origin as shown.
The volume of the strip is 2πrdrL, where L is the length of the reactor.
Now we shall evaluate A1 as follows:
No. of fissions per second =
Energy released per second =
Total Power, P =
From tables J1(2.405) = 0.518.
Thus, the flux distribution is given by
As seen previously the power density will be maximum at the centre.
Hence orificing has to be resorted to.
We shall define the average flux such that power is conserved.
Since peak flux = A1
More peaked than for slab reactor.
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