For pion injection the total energy in the area will be 3100x6.3x10^7=2x10^11
Mev deposited for each pulse. This energy will be absorbed by the
inflector and magnet and yield secondary spray.
I will assume that the mass of the absorbing material is about 10kg. So,
the dose will be 2x10^10 Mev/kg
or 0.03 Gy/per injected pulse. This number can be compared with the
canonical (p 1268 of PDG) 3.5x10^9 minI particles
(versus 6.3x10^7 min I particles for pion injection) in carbon=1Gy.
1 Gy = 100 rad. So we expect about 3 rad/per
pulse during pion injection.
In order to work in rem (roentgen equivalent for man) we need to transform
Grays to sieverts...
1 Sv = 100 rem.
Sv = Gy x w_R.
Charged particle dosage
Where w_R = 3 on average for the mixture of protons, pions,
electrons and muon in the area of beam impact
The number of Sieverts in the beam will be 0.1 Sv/per pion pulse.
So, we expect that there will be about 10rem/pulse in the beam. Assuming that
things fall off geometrically and that the tops of the calos are 10" from the
beam point on average we expect about 0.01 to hit a sensor on top of the calo.
The rate at that point will be 0.10rem/pulse. In order to get up into the
potentially interesting region we will want at least 150 pulses or 15rem
absorbtion. This is in the ball park of the dose of 5 Rad total
exposure which is most useful for TLDs.
150 pulses will be about five minutes of continuous running.
Neutral dosage
There is a wide variation in the weighting factors for different energies of
neutrons. For the sake of averaging, we will assume that typical neutrons
have energies of about 1 Mev and thus w_R=10 for the canonical neutron in
this calculation.
Given that we expect a 6x10^8
neutrons/pulse and we will again use an absorbtion medium of 10Kg. we get
6x10^-5 Gy of neutron dosing in the canonical 1Kg. I will guess that
This number will be uniformly distributed
over a cloud which will surround the point of impact for some several feet.
I will use a typical neutron energy of 1 Mev for determing the number of Gys
and hence the total dose in rem.
So, from this we expect about 6x10^-4 Sv everywhere in the region around the
calorimeters, which yields about 60mrem/pulse
of neutrons everywhere in the area. For the canonical 5 minutes of running
time I expect there to be about 150 pulses or 9rem of exposure. In terms of
rads we have 6x10^-5Gy=6x10^-3Rad*150=0.9 Rad. Which is again right in the
correct ball park.
Summary
So the absorbed dose from charged particles should be about 0.10rem/pulse at
10" from the point of impact. This number will be falling
as r^2 as we move away from (all of) the points of impact.
The neutron absorbed dose will be about 0.06rem/pulse throughout the neutron
gas. Thus points which are about a foot away from the beam should see about
equal amounts of charged and neutral exposure with the charged component
falling away rapidly and the neutral component less rapidly depending on the
kinematics of the neutron gas.