Geoscience Reference
In-Depth Information
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EXAMPLE 23.91
Problem:
A total of 18,100,000 gal of water was filtered during a filter run. If backwashing used
74,000 gal of this product water, what percent of the product water was used for backwashing?
Solution:
Backwash water(gal)
Water fi
74,000 gal
18,100,000 gal
Backwash water
=
×=
100
×
100
= .%
04
ltered (gal)
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EXAMPLE 23.92
Problem:
A total of 11,400,000 gal of water was filtered during a filter run. If backwashing used
48,500 gal of this product water, what percent of the product water was used for backwashing?
Solution:
Backwash water(gal)
Water fi
48,500 gal
11,400,000 gal
Backwash water
=
×=
100
×
100
= .%
043
ltered (gal)
23.6.10 p
erCent
m
udball
v
olume
Mudballs are heavier deposits of solids near the top surface of the medium that break into pieces
during backwash, resulting in spherical accretions of floc and sand (usually less than 12 in. in diam-
eter). The presence of mudballs in the filter media is checked periodically. The principal objection
to mudballs is that they diminish the effective filter area. To calculate the percent mudball volume
we use Equation 23.77:
Mudball volume(mL)
Totals
(23.77)
%Mudball volume
=
×100
ample volume(mL)
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EXAMPLE 23.93
Problem:
A 3350-mL sample of filter media was taken for mudball evaluation. The volume of water
in the graduated cylinder rose from 500 mL to 525 mL when mudballs were placed in the cylinder.
What is the percent mudball volume of the sample?
Solution:
First determine the volume of mudballs in the sample:
525 mL - 500 mL = 25 mL
Then calculate the percent mudball volume:
Mudball volume(mL)
Totals
25 mL
3350 mL
%Mudball volume
=
×= ×=
100
100
0.5%
ample volume(mL)
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EXAMPLE 23.94
Problem:
A filter is tested for the presence of mudballs. The mudball sample has a total sample
volume of 680 mL. Five samples were taken from the filter. When the mudballs were placed in 500
mL of water, the water level rose to 565 mL. What is the percent mudball volume of the sample?
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