Concept explainers
(a)
The minimum thickness of the paraffin shield.
(a)
Answer to Problem 10.1P
The minimum thickness of the paraffin shield is 44 cm.
Explanation of Solution
Given:
Source emits,
The fast neutron flux,
Attenuation coefficient,
Build up factor for hydrogenous shields, B = 5
Formula used:
Where,
t = thickness of paraffin shield
Calculation:
The shield's surface
Using the method of iteration,
First, we assume a small value of t, and then calculating the resulting value
Conclusion:
Att = 44 cm results in
(b)
Thermal neutron leakage atthe surface of the field.
(b)
Answer to Problem 10.1P
The thermal neutron leakage is
Explanation of Solution
Given:
Point source emit, S = 107 n/sec
Radius, R = 44 cm
Diffusion coefficient, D = 0.381 cm
thermal diffusion length, L = 3cm
Formula used:
The thermal neutron leakage
Calculation:
The thermal neutron leakage:
Conclusion:
The thermal neutron leakage is
(c)
The gamma ray dose rate, due to the hydrogen capture gammas, atthe surface of the shield.
(c)
Answer to Problem 10.1P
Explanation of Solution
Given:
Point source emit, S = 107 n/sec
Radius, R = 44 cm
The flux of fast neutron,
The flux of thermal neutron,
Formula used:
The concentration of gamma emitter
The source strength
Where,
fi = fraction of the transformations that yield a photon of the ith energy, and
Ei = energy of the ith photon, MeV
Where,
The dose rate :
Calculation:
Concentration of gamma emitter
The source strength
Linear energy absorption coefficient for paraffin
Inserting these values into the dose rate equation, we have
Conclusion:
Gamma ray dose rate is
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