Chapter 6 FE EXAM FORMATTED PROBLEMS
6-1.
Three parallel flow flocculation basins are to be used to treat a flow rate of 3 MGD. What
is the design volume, in cubic feet, of one of these tanks if the detention time is 30
minutes?
Parallel flow therefore,
Q
3  10 6 gal / d
 1 10 6 gals / d per tank
3 tanks
The volume of one tank is
 1day 
3
3
3
volume  tQ  (30 min)(110 6 gal / d )
(0.1337 ft / gal)  2,785 ft or 2,800ft
 1440 min 
6-2.
Determine the number of rapid sand filters to treat a flow rate of 75.7 x 103 m3/d if the
design loading rate is 300 m3/d∙m2. The maximum dimension is 7.5 m and the length to
width ratio is 1.2:1.
Assume a length of 7.5m, then
7.5
 6.25m
1.2
Area per filter  LW  (7.5m)(6.25m)  46.88m 2
W
75.7  10 3 m 3 / d
 252 m 2
300 m 3 / d  m 2
252m 2
No. of filters 
 5.38 or 6 filters
46.88m 2 / filter
Total Area required 
6-3.
Estimate the amount of lime, in tons/d, required to soften 5 MGD of water to the practical
solubility limits. The constituent concentrations are as follows:
CO2 = 0.44 meq/L
Alkalinity = 3.96 meq/L
Ca2+ = 4.76 meq/L
Cl- = 1.91 meq/L
Mg2+ = 1.11 meq/L
SO42- = 1.58 meq/L
Lime additions:
Lime = CO2 = 0.44 meq/L
Lime = Alkalinity = 3.96 meq/L
Lime = Mg2+ = 1.11 meq/L
Total Lime = 0.44 + 3.96 + 1.11 = 5.51 meq/L
Equivalent weight of lime:
MW of CaO = 56mg
56 mg
 28mg /meq
2
Converting meq/L to mg/L
(5.51meq/L)(28mg/meq )  154.28mg/L  154 .28 ppm
(154 .28 ppm)(5MGD)(8.34lbm / gal)
tons of lime per day 
 3.2tons / d
2000 lbm / ton
EW 
6-4.
Calculate the theoretical percent removal of a particle having a settling velocity of 0.25
cm/s settling in a horizontal flow clarifier with an overflow rate of 0.05 cm/s.
From Eq. 6-33.
%removal 
vs
0.25cm / s
(100 ) 
(100 %)  500 %
vo
0.05cm / s
Therefore, 100%