[ RadSafe ] [RADSAFE] gamma modeling from large sources

Sun Jan 4 16:31:20 CST 2009

Thank you very much for everyone who replied
The situation is more clear now.  I agree that in some situations the
modeling is useless.  There is, however, no escape frm govt. requirements to
estimate doses prior to a project going ahead...
I did find quite useful (and FREE) program by the Environmental Instruments
Canada Internet site, for those interested:
http://members.shaw.ca/eic/Tools/
The JavaShield program is at the bottom of the page.
Kind regards and thanks again
Nick Tsurikov
Western Australia
http://calytrix.biz/

On Sat, Jan 3, 2009 at 9:28 AM, Brennan, Mike (DOH) <Mike.Brennan at doh.wa.gov
> wrote:

> 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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