CEE 6560
Answer Key Problem 1 (Mass Balances)
Sept. 4, 2009
Mass loading of settleable and suspended solids = SL
SL = (Q)(So) = (1.51 x 107 liters/day)(800 mg/L) = 1.21 x 104 kg/day.
PRIMARY CLARIFIER
Solids removed in primary clarifier = 0.3(SL) = 3.62 x103 kg/day
These solids would all be in the primary clarifier underflow.
Concentration of solids in primary underflow (Pc) = 55,000 mg/L. Volume of underflow
(UP) required to make 3.62 x 103 kg/day of solids to a concentration of 55,000 mg/L is
given by:
kg
liters
day
 6.59 x 104
mg
day
55000
liter
3.62 x 103
UP =
ALUM ADDITION
100 mg/L of Al2 (SO4 ) 14H2O (Alum) = 100mg/594, 000 mg/mole = 3.37 x 10-4 moles
Each mole of alum generates 2 moles of Al(OH)3 or 2 x 3.37 x10-4 x 78 gm = 26.26
mg/liter. Influent flow to the secondary clarifier contains 26.26 mg/L of alum plus
suspended solids not removed in the primary clarifier.
Thus influent concentration = 26.26 mg/L + 0.7 (800 mg/L) = 586.26 mg/liter.
SECONDARY CLARIFIER
Underflow from primary clarifier is small relative to influent flow,
6.59 x 104 liters / day
(
 7.82 x 103 ). So assume flow into secondary clarifier (after
7
1.51 x 10 liters / day
coagulation/flocculation) still equals 1.51 x 107 liters/day.
60% of the influent solids (586.26 mg/liter) are removed in the secondary clarifier. So total
solids removed are:
0.6 (586.26 mg/liter)(1.51 x107 liter/day) = 5.31 x 103 kg/day.
1
Desired secondary clarifier underflow solids concentration is 45,000 mg/L. Underflow
required is:
US = (5.31 x 103 kg/day)/45 kg/liter = 1.18 x 105 liters/day.
Again, this flow is small compared to the influent flow rate. (1.18 x 105 liter/day)/(1.51 x
107 liter/day = 7.82 x 10-3). So assume the flow rate remains at 1.51 x 107 liter/day to the
filter.
FILTER
Solids flow to the filter equals 40% of solids flow to secondary clarifier (since 60% of
solids removed in secondary clarifier). Solids loading to filter (SLF) is given by:
0.4 (586.26 mg/liter)(1.51 x107 liter/day) = 3.54 x 103 kg/day.
85% of these solids are removed on the filter and subsequently captured in the filter
backwash.
Total solids removed in filter per day = 0.85 x 3.54 x 103 kg/day = 3.01 x 103 kg/day.
Total backwash flow per day = (1.58 x 104 liters/day)(10 min) = 1.58 x 105 liters.
Concentration of solids in backwash water is:
FC = 3.01 x 103
kg
mg
/1.58 x 104 liters/day = 1.91 x 104
day
liter
TOTALS
Process
Primary
Secondary
Filter
Total
Solids Removed ( kg/day)
3.62 x103
5.31 x 103
3.01 x 103
1.20 x 104
UnderFlow or backwash (L/day)
6.59 x 104
1.18 x 105
1.57 x105
3.41 x 105
Solids concentration in combined clarifier plus backwash flows:
C=
1.20 x 104 kg
mg
 3.50 x 104

5
3.41 x 10 liters
liter
2