[ RadSafe ] [RADSAFE] gamma modeling from large sources

Fri Jan 2 18:28:00 CST 2009

Some time ago I was working on trying to model exposure from a complex
source (the pipes and tanks of a waste water treatment plant).  After a
fair amount of work I concluded that sometimes there is nothing useful
that can be done with modeling, and if you really want to know the
answer you have to measure it, and if you want to know the dose that
people who are moving around the source are getting, you have to do a
"tag and release" program.  If the situation is complex enough, all that
modeling will do is give you greater confidence in a wrong answer.  

There are a couple of sayings I've run across that I think pertain:

* All models are wrong, and some are useful. 

* I'd rather be right in a broad range than wrong with a precise number.

* If it was easy to figure out, the Business Administration majors would
do it.

-----Original Message-----
From: radsafe-bounces at radlab.nl [mailto:radsafe-bounces at radlab.nl] On
Behalf Of Dale Boyce
Sent: Friday, January 02, 2009 3:24 PM
To: Nick Tsurikov; radsafe at radlab.nl
Subject: Re: [ RadSafe ] [RADSAFE] gamma modeling from large sources

Nick,

I can almost recreate this formula. The flux from a a circle on a disk
source at a point x on the source axis is given by multiplying the area
of the ring by the flux per unit area. and dividing by r^2. In angular
coordinates this is

2*pi*flux*sin(theta)/cos(theta)^2

this integrates to 2*pi*flux/cos(theta)

cos(theta) is 1 at a radius of 0

cos(theta) is x/sqrt(x^2+r^2)

so evaluating from 0 to the radius r of the source you get

2*pi*flux( x/x - x/sqrt(x^2 + r^2))

or

2*pi*flux(1 - x/sqrt(x^2 +r^2))

note that r^2 is area/pi

note also that the 2*pi*flux divides out when evaluating for x= 0
relative to some positive value of x

So the main difference is the substituion of HxW/A for r^2

Dale

----- Original Message -----
From: "Nick Tsurikov" <nick.tsurikov at gmail.com>
To: <radsafe at radlab.nl>
Sent: Saturday, December 13, 2008 4:32 PM
Subject: [ RadSafe ] [RADSAFE] gamma modeling from large sources

Greetings from Western Australia!

The help will be very much appreciated in the following:

I'm using the following formula to estimate the gamma dose rate from
large
sources for some years now.  Works more or less OK but I have no
recollection as to where I've got it from, so I'm a bit curious:

a)      Does anyone knows the reference for this one? and -

b)      Are there other similar equations that may be used?

The simple equation that I use looks like that:

Dose rate (@ X) = Dose rate (@ surface) * (1-X/SQRT[X^2 + H*W/pi])

X - distance

H - height

W - width

With "normal" numbers - say, the surface level is 10 microSv/hour and
the
"source" is twelve bags with a mineral concentrate, three vertical rows
of
four 1x1 metre bags each.  Then, the predicted level would be:

Dose rate (@ 1 m) = 10 * (1 - 1/SQRT[1*1 + 3*4/3.142]) = 10 * (1 -
1/SQRT[4.82]) = 5.4 microSv/hr

Dose rate (@ 2 m) = 10 * (1 - 2/SQRT[2*2 + 3*4/3.142]) = 10 * (1 -
2/SQRT[7.52]) = 2.7 microSv/hr

Dose rate (@ 4 m) = 10 * (1 - 4/SQRT[4*4 + 3*4/3.142]) = 10 * (1 -
4/SQRT[19.82]) = 1.0 microSv/hr

Dose rate (@ 8 m) = 10 * (1 - 8/SQRT[8*8 + 3*4/3.142]) = 10 * (1 -
8/SQRT[67.82]) = 0.3 microSv/hr

But if we'll double the amount of bags and have six rows of eight bags:

Dose rate (@ 1 m) = 10 * (1 - 1/SQRT[1*1 + 6*8/3.142]) = 10 * (1 -
1/SQRT[16.28]) = 7.5 microSv/hr

Dose rate (@ 2 m) = 10 * (1 - 2/SQRT[2*2 + 6*8/3.142]) = 10 * (1 -
2/SQRT[19.28]) = 5.4 microSv/hr

Dose rate (@ 4 m) = 10 * (1 - 4/SQRT[4*4 + 6*8/3.142]) = 10 * (1 -
4/SQRT[31.28]) = 2.8 microSv/hr

Dose rate (@ 8 m) = 10 * (1 - 8/SQRT[8*8 + 6*8/3.142]) = 10 * (1 -
8/SQRT[79.28]) = 1.0 microSv/hr

Etc,etc...

Generally, the inverse square law works with not a very significant
difference from a 'point source', but things usually getting quite
different
when the "source" is A LOT of bags like that, or, about 12 sea
containers...

I have also found that the equation above is quite useful to estimate
the
gamma level from a large stockpile (at the height of 1 meter) - it
pretty
much stabilises to a level of about 80-90% of a surface one when the
size of
the area reaches about 30x30 meters in size.

There are some equations in the literature but they would be a bit too
complicated to pass along to someone like a mining/processing site
radiation
safety officer - the one above is quite simple and the estimate can be
easily done by simply measuring the gamma surface level and estimating
the
surface in square meters (without getting through the whole thing of
MeV,
different isotopes, etc, etc)...

Any information relevant to this issue will be very much appreciated.

Kind regards

Nick Tsurikov

Western Australia
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