Re: Selling Sewer Disposal

["Dale E. Boyce" <dale@radpro.uchicago.edu>]

Sue and others who have expressed interest,

The stuff I had on file had some site specific information
so I have distilled some of the information for this post.
Most of the information is taken from NCRP Report #45.  The
rainfall data comes from The CRC Handbook of Radiation Measuement
and Protection.  Note: the area of the earth is 1.5E8 Km^2.

         Earth Inventory in Curies      Inventory in Curies per Km^2
        ____________________________    ____________________________
Nuclide	Total	in Air	in Biosphere	Total	in Air	in Biosphere

C14	3.0E8	4.8E6	1.2E7		2.0	3.2E-2	7.9E-2
Cl36	4.5E5	2.7E-2	1.3E5		3.0E-3	------	8.6E-4
H3	3.4E7	1.4E5	9.2E6		2.2E-1	9.3E-4	6.1E-2
Na22	1.2E4	2.0E2	2.5E3		7.9E-5	1.3E-6	1.7E-5
P32	1.1E5	2.7E4	5.2E3		7.3E-4	1.8E-4	3.4E-5
P33	9.7E4	1.6E4	5.4E3		6.4E-4	1.1E-4	3.6E-5
S35	1.9E5	1.5E4	1.9E4		1.3E-3	1.0E-4	1.3E-4

	Equilibrium Activity Based on	The Area Required to Disperse
	Uniform Daily Acquisition and	Activity to the Concentration
	Discharge of 1 Curie per Year	of the Nuclide in the Biosphere

	Doo = 1/(1-exp(-lamda*1 day)) 	Doo / Concentration in Biosphere
		Curies				Km^2
	_____________________________	________________________________
C14		8.3E3				1.1E5
Cl36		4.3E5				5.0E8
H3		1.8E1				3.0E2
Na22		3.8E0				2.2E5
P32		5.5E-2				1.6E3
P33		1.0E-1				2.8E3
S35		3.5E-1				2.7E3

Now to apply the above data, take your annual receipt total, multiply
by the fraction of nuclide entering the aqueous stream, and apply the
decay factor for the time between acquisition and disposal.  Take
your result and multiply by the area in column 2 of table 2 and 
compare it to the area of you campus, city, and/or state.

Example:  We receive 2.5 Curies of S35 per year.  Let's say we hold
S35 aqueous waste for 1 year after the date a container is filled.
It takes 90 days to fill a typical container.  We'll estimate that
one half of the activity is disposed in the aqueous stream.

A = 2.5 Ci * 0.06 (decay) * 0.5 (fill decay) * 0.5 (fraction) = 0.0375

Area = 0.0375 * 2,700 = 101 Km^2

The area of our campus is about 1.5 Km^2, Chicago is about 1,500 Km^2,
Illinois is about 150,000 Km^2, and the earth is about 1.5E8 Km^2.
Chicago could then have about 15 generators releasing that quantity
of S35 without exceeding the S35 found in nature.  Illinois could
have about 1500 such generators.  In fact Illinois has about 1,000
licensees, but only a small fraction of the licensees are large enough
to approach our usage of S35.

I think that most large universities will find that their receipts
of major research isotopes will compare with the biosphere inventories
of their communities.  Now let's look at some rainfall concentrations.
Forty inches (~ 1 meter) of rainfall per year corresponds to 1E9 l/km^2.

				    Annual Rainfall Total Activity
                                ______________________________________
Nuclide	Rain Concentration	Campus Ci	Chicago Ci	Il. Ci
_______ __________________      _________       __________      ______
		Ci/liter
C14		1E-13*		1.5E-4		1.5E-1		15
H3 &		2E-14		3.0E-5		3.0E-2		3
		8E-12*		1.2E-2		12		1,200
		5E-11		7.5E-2		75		7,500
Na22		9E-15		1.4E-5		1.4E-2		1.4
P32		6E-13		9.0E-4		9.0E-1		90
S35		1E-12		1.0E-3		1		100

*  Concentration estimated from average surface sea water concentration
&  Tritium concentration apparently varies wildly in rainfall so I
included the median of surface seawater as an indicator of the mean.

My apologies for any typos or errors in this mail.  Let me know if
this helps or if you note in errors.

Dale E. Boyce, Director
dale@radpro.uchicago.edu