Science of Water : Concepts and Applications

376 The Science of Water: Concepts and Applications
Example 10.92
Problem:
A total of 11,400,000 gal of water are fi ltered during a fi lter run. If 48,500 gal of product water are used
for backwashing, what percent of the product water is used for backwashing?
Solution:
backwash water, gal
Backwash water, % 100
water filtered, gal
48,500 gal
100
11,400,000 gal
0.43% backwash water
PERCENT MUD BALL VOLUME
Mud balls are heavier deposits of solids near the top surface of the medium that break into pieces
during backwash, resulting in spherical accretions (usually less than 12 in. in diameter) of fl oc and
sand. The presence of mud balls in the fi lter media is checked periodically. The principal objection
to mudballs is that they diminish the effective fi lter area. To calculate the percent mud ball volume
we use Equation 10.78:
Example 10.93
mud ball volume, mL
% Mud ball volume
100 (10.78)
total sample volume, mL
Problem:
A 3350-mL sample of fi lter media was taken for mud ball evaluation. The volume of water in the graduated
cylinder rose from 500 mL to 525 mL when mud balls were placed in the cylinder. What is the
percent mud ball volume of the sample?
Solution:
First determine the volume of mud balls in the sample:
Then calculate the percent mud ball volume:
525 mL500 mL 25 mL
mud ball volume, mL
% Mud ball volume
100
total sample volume, mL
25 mL
100
3350 mL
075 . %