A submarine at a depth of 1000 meters.?

Question

Say I took a reading on contact with the outside of the core while it was operating and was 200R/hr. Now say the distance from the core to the hull is 30 feet and the hull is 2 feet of steel. How much exposure would a fish get in 1 minute it was 5 feet from the outside of the hull.

For the sake of this problem, say that 4 inches of steel will decrease the intensity by a factor of 10.

Oh, and 24 inches of water will do the same as 4 inches of steel.

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Answer 1

Say I took a reading on contact with the outside of the core while it was operating and was 200R/hr. Now say the distance from the core to the hull is 30 feet and the hull is 2 feet of steel. How much exposure would a fish get in 1 minute it was 5 feet from the outside of the hull.

For the sake of this problem, say that 4 inches of steel will decrease the intensity by a factor of 10. Oh, and 24 inches of water will do the same as 4 inches of steel.

What you have is diminuation of the radiation R by th = hull thickness, dw = distance in water, dh = distance to hull from core. You claim a "factor of 10" diminuation by 4 inches of hull and 24 inches of water. I'll use that, but it's not clear that is really the case. All radiation, atomic, light, EM, etc. dimishes from its source by 1/r^2 where r is the distance from the source in addition to the shielding factor (k) based on the material. Anyway, using your numbers.

The diminuation by the salt water is 1/((10/2)*5) = 1/25;where 10/2 is the factor of ten per two feet and 5 is the five feet of water the fish is from the hull. In the hull, we have 1/(10/(1/3))*2 = 1/60; where 10/(1/3) is the factor of ten per 1/3 feet (4 in) and 2 is two feet of steel in the hull. That leaves the 30 ft of air gap between the hull and core. Here, we invoke the 1/r^2 rule because there is virtually no shielding factor in air. Thus R' = R (r0/r)^2; where we set r0 = 1 ft. at the core source and r = 30 ft at the inner hull surface. Thus R' = R/900 and R' is the radiation in a time period at the inner hull.

Therefore, R(fish) = (R/900)*(1/25 + 1/60) = (R/900)*(60 + 25)/1500 = 85R/(900*1500) = 200*85/(900*1500) = (4/30)*(85/900) ~ .012 Roentgen/hr which is about what the fish endures 5 ft from the hull. So, over 1/60 of an hour (1 min), our fish and chips plate would have been exposed to about .012(1/60) ~ .0002 Roentgen.

Not to worry, it's been a while since I worked in nuke weapons design, but I seem to recall 450 R as the whole body lethal threshold for humans. So I doubt, even for a fish. .0002 Roentgen would do much harm...to the fish or the eater of the fish.