WMOD00297

• Title: Hygroscopic Seeding in Oklahoma
• Date: 10/31/1971
• State: OK
• Weather Modification - Doc Type: Report

1. Assume that on the average one can extract 1. 0 cm of water
from the atmosphere - - say a third of the average water vapor
over an area at any time. Per section of land, this means 21. 1
acre-feet of water, requiring 21. 1 kilograms of seeding material
if we use the one kg per acre-foot factor just discussed.

2. Find out how many particles per cubic meter can enter the cloud
before they begin interfering with each others growth. First one can
look to a simple steady state case, without considering any micro-
physics details, and calculate from mass cO!2'ervation that 31 drops/m3
can exit at a size of 5 mm.diameter (6.5' 10 ~s each) from a cloud
with an average liquid water content of 2 gm/m and an upcurrent of
5 ml sec. Actually each cell is transient, and the early drops will
extract more of the water per drop than the later drops which will be
encountering a depleted cloud volume. Thus we estimate that even 10
times higher concentrations of hygroscopic material can be useful,
and still more particles will not be harmful. As a rule of thumb, when
the cloud is long lived and/ or the Langmuir chain reaction is operating,
concentrations of some tens or hundreds per m 3 should be used, while
for short lived clouds the concentration should be some hundreds per m3
or even l/liter. MacCready (1959) has examined the particle interference
problem and the desired seeding concentrations for AgI seeding of super-
cooled clouds. The study showed that as long as the particles fell back
down through the upcurrent, the required embryo concentrations for
maximizing rainfall increased as upcurrents decreased, and for the rather
dry clouds investigated a concentration of 100/m3 was a reasonable number
to aim for.

3. Now relate this to a particular cloud for a cloud of dimensions 4x4x4 km 3,
having an average 2 gm/m3 of liquid water. There are about 100 acre
feet of water condensed. With a 106 multiplier factor, 100 kg of seeding
material would be needed to extract this water -- a 100 second release at
1 kg/sec. At one 20 pm particle per liter, however, 256 kg of material
would be required. From these ball-park estimates one sees that one or
two passes under a medium size cloud can release enough particles,
although the particles must diffuse properly if the operation is to be
efficient.

4. For a large volume approach to seeding, consider a 100 km x 100 km
area, 2 km deep. To put one 20 p.m particle in each liter of this volume,
assuming perfect diffusion, one needs 80 metric tons of material, a
large but not impossible amount of material to provide. If one puts
out 10 pm particles, the need is only for 10 tons, which is within the
payload of the C-97. At 1 kg/second, 10 tons can be released in less
than 3 hours.

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