Gilson and Voss - Voss Associates

Neutron and Gamma Shielding
Using the graph for neutron or gamma shielding estimation.
Neutron and Gamma Shielding
Using the graph for neutron or gamma shielding estimation.
Given:
Find:
2
A 5 MeV neutron source reading 12,000 n/cm -sec at
30 cm
the number of half-value layers to reduce the neutron
2
flux rate to 200 n/cm -sec at 30 cm
Given:
Find:
2
A 5 MeV neutron source reading 12,000 n/cm -sec at
30 cm
the number of half-value layers to reduce the neutron
2
flux rate to 200 n/cm -sec at 30 cm
2 2
1. Divide 200 n/cm -sec by 12,000 n/cm -sec = 0.0167
2. Locate 0.0167 on the vertical axis and move across to
where the slanted line crosses 0.0167, then move
vertically down to the “Number of Half-Value Layers”
horizontal axis, this value is approximately 5.9
2 2
1. Divide 200 n/cm -sec by 12,000 n/cm -sec = 0.0167
2. Locate 0.0167 on the vertical axis and move across to
where the slanted line crosses 0.0167, then move
vertically down to the “Number of Half-Value Layers”
horizontal axis, this value is approximately 5.9
Given:
Find:
60
A Co source reading 120 mrem/hr at 30 cm
the number of half-value layers to reduce the exposure
rate to 5 mrem/hr at 30 cm
Given:
Find:
60
A Co source reading 120 mrem/hr at 30 cm
the number of half-value layers to reduce the exposure
rate to 5 mrem/hr at 30 cm
1. Divide 5 mrem/hr by 120 mrem/hr = 0.042
2. Locate 0.042 on the vertical axis and see where it
crosses the slanted line, then move vertically down to
the “Number of Half-Value Layers” horizontal axis, this
is approximately 4.6
Pick a shielding material from the shielding tables and multiply
the number of half-value layers by the cm thickness in the
shielding table to obtain the thickness required.
1. Divide 5 mrem/hr by 120 mrem/hr = 0.042
2. Locate 0.042 on the vertical axis and see where it
crosses the slanted line, then move vertically down to
the “Number of Half-Value Layers” horizontal axis, this
is approximately 4.6
Pick a shielding material from the shielding tables and multiply
the number of half-value layers by the cm thickness in the
shielding table to obtain the thickness required.
Page 59
Page 59
Neutron and Gamma Shielding
Using the graph for neutron or gamma shielding estimation.
Neutron and Gamma Shielding
Using the graph for neutron or gamma shielding estimation.
Given:
Find:
2
A 5 MeV neutron source reading 12,000 n/cm -sec at
30 cm
the number of half-value layers to reduce the neutron
2
flux rate to 200 n/cm -sec at 30 cm
Given:
Find:
2
A 5 MeV neutron source reading 12,000 n/cm -sec at
30 cm
the number of half-value layers to reduce the neutron
2
flux rate to 200 n/cm -sec at 30 cm
2 2
1. Divide 200 n/cm -sec by 12,000 n/cm -sec = 0.0167
2. Locate 0.0167 on the vertical axis and move across to
where the slanted line crosses 0.0167, then move
vertically down to the “Number of Half-Value Layers”
horizontal axis, this value is approximately 5.9
2 2
1. Divide 200 n/cm -sec by 12,000 n/cm -sec = 0.0167
2. Locate 0.0167 on the vertical axis and move across to
where the slanted line crosses 0.0167, then move
vertically down to the “Number of Half-Value Layers”
horizontal axis, this value is approximately 5.9
Given:
Find:
60
A Co source reading 120 mrem/hr at 30 cm
the number of half-value layers to reduce the exposure
rate to 5 mrem/hr at 30 cm
Given:
Find:
60
A Co source reading 120 mrem/hr at 30 cm
the number of half-value layers to reduce the exposure
rate to 5 mrem/hr at 30 cm
1. Divide 5 mrem/hr by 120 mrem/hr = 0.042
2. Locate 0.042 on the vertical axis and see where it
crosses the slanted line, then move vertically down to
the “Number of Half-Value Layers” horizontal axis, this
is approximately 4.6
Pick a shielding material from the shielding tables and multiply
the number of half-value layers by the cm thickness in the
shielding table to obtain the thickness required.
1. Divide 5 mrem/hr by 120 mrem/hr = 0.042
2. Locate 0.042 on the vertical axis and see where it
crosses the slanted line, then move vertically down to
the “Number of Half-Value Layers” horizontal axis, this
is approximately 4.6
Pick a shielding material from the shielding tables and multiply
the number of half-value layers by the cm thickness in the
shielding table to obtain the thickness required.
Page 59
Page 59