Substituting Equation (6) in Equation (5) yields
= (2.5 K/CVoo709
(7)
Hence, for a specified ratio of costs K/C and a given (desired) precision
(Vy ) of x, Equation (7) may be solved to yield the required aliquot
size w.
The number n of aliquots of size w is easily obtained from
Equation (6).
Note that this approach does not take inte account the
need to decide on the number of field samples to collect.
This is
discussed in a later section.
To illustrate the use of Equation 7, suppose K = $25 per aliquot, C = $150,
and V, = 0.01.
Then Equations (6) and (7) give n = 6 and w = 58.3
= 60 g.
Alternatively, if we can afford only one aliquot per sample,
i.e., suppose K = C = $25, then w = 410.9 =~ 400 g.
This illustrates
that as the number of aliquots (mn) decreases, the aliquot size (w) must
increase if the precision level V, is held constant.
To see this, we
note from Equation (7) that w= (4, 5/nv,)t- 09 Also, for a given (desired)
precision V,, the total analysis cost per sample (C) will be minimized
when n = 1, in which case w = (2.5/s%) 1-99,
Conversely, if costs C and
K are fixed, then V, will be minimized by using the largest possible
aliquot size.
V,
A
This can be seen by expressing Equation (5) as
= 2.5 K/Cw
0.92
(8)
Figure 3 gives values of V, for w between 1 and 100 g using Equation (8)
for the two cost situations of C = K and C/K = 4, i.e., for n = 1 and 4,
respectively.
For either situation, Va decreases very rapidly for w
between 1 and 25.
Results for Pu
Now we consider determining aliquot size and number of aliquots for Pu
analyses.
The average (median) Pu to Am ration at NS-201 was estimated
by Gilbert et al.
(1977) to be 11.2.
By considering 11.2 to be free of
error, we obtain Var(Pu) = (11.2)2 Var(Am).
Taking the square root of
both sides of this equation yields s(Pu) = 11.2 s(Am), where s( ) denotes
the standard deviation of the quantity in parentheses.
Multiplying
Equation (3) by 11.2 gives
s(Pu) ~ 17.7 wor,
(9)
Hence, the variance of the mean Pu concentration vi, ) based on n aliquots
of size w is approximately
{s(Pu) ] 2 in = 313/nw
0.92
(10)
To simplify results below, we use the theoretical value of ~-0.50 obtained
by Grant and Pelton (1973) rather than the -0.46 in Equation 9.
This
gives
V, = 313/nw .
(11)
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