6.19
Component Concentration
Equiv. Weight
(mg/meq)
Concentration
(meq/L)
Concentration
(mg/L CaCO3)
Ca2+ 90 mg/L 20 4.500 225.0
Mg2+ 30 mg/L 12.2 2.4596 123.0
HCO3
–165 mg/L 61.0 2.705 135.3
pH 7.5 – – –
348.0
135.3
0
225.0
0
Ca2+
HCO3–
CH
Mg2+
Ca–CH
NCH
Mg–NCH
Ca–NCH
348.0
135.3
0
225.0
0
Ca2+
HCO3–
CH
Mg2+
Ca–CH
NCH
Mg–NCH
Ca–NCH
()
[]
[][]
3
4–
7–
5.7
1
–
3
aq2 CaCO mg/L 19.14 M 10 1.914
M 10 4.47
L
mol
10
g 61.0
mol
L
g
0.165
HHCO
CO =×=
×
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
==
−
+
K
Table P6.19 Components, lime and soda ash dosage, and solids generated
Component (eq’n.) Concentration
(mg/L
CaCO3)
Lime
(mg/L
CaCO3)
Soda ash
(mg/L
CaCO3)
CaCO3(s)
(mg/L
CaCO3)
Mg(OH)2(s)
(mg/L
CaCO3)
CO2(aq) (6.37) 19.14 19.14 0 19.14 0
Ca-CH (6.38) 135.3 135.3 0 270.6 0
Mg-CH (6.39) 0 0 0 0 0
Ca-NCH (6.40) 90.00 0 90.00 90.00
Mg-NCH (6.41) 123.0 123.0 123.0 123.0 123.0
Excess 20
Totals 297.4 213.0
6.20 The solution for this problem is the solution for problem 6.6 in the 2nd edition
Solutions Manual.
6.21
Component Concentration
Equiv. Weight
(mg/meq)
Concentration
(meq/L)
Concentration
(mg/L CaCO3)
Ca2+ 95 mg/L 20 4.75 238
Mg2+ 26 mg/L 12.2 2.13 107
HCO3
–160 mg/L 61.0 2.62 131
pH 7.0 – – –
6.88
2.62
0
4.75
0
Ca2+
HCO3–
CH
Mg2+
Ca–CH
NCH
Mg–NCH
Ca–NCH
6.88
2.62
0
4.75
0
Ca2+
HCO3–
CH
Mg2+
Ca–CH
NCH
Mg–NCH
Ca–NCH
()
[]
[][]
meq/L 1.17 M 10 5.868
M 10 4.47
L
mol
10
g 61.0
mol
L
g
0.160
HHCO
CO 4–
7–
7
1
–
3
aq2 =×=
×
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
==
−
+
K
Table P6.21 Components, lime and soda ash dosage, and solids generated
Component (eq’n.) Concentration
(meq/L)
Lime
(meq/L)
Soda ash
(meq/L)
CaCO3(s)
(meq/L)
Mg(OH)2(s)
(meq/L)
CO2(aq) (6.37) 1.17 1.17 0 1.17 0
Ca-CH (6.38) 2.62 2.62 0 5.24 0
Mg-CH (6.39) 0 0 0 0 0
Ca-NCH (6.40) 2.13 0 2.13 2.13 0
Mg-NCH (6.41) 2.13 2.13 2.13 2.13 2.13
Excess 0.400
Totals 6.32 4.26 10.67 2.13
6.22 The solution for this problem is the solution for problem 6.7 in the 2nd edition
Solutions Manual.
6.23
Component Concentration
Equiv. Weight
(mg/meq)
Concentration
(meq/L)
Concentration
(mg/L CaCO3)
Ca2+ 40.0 mg/L 20 2.00 100
Mg2+ 10.0 mg/L 12.2 0.820 41.0
HCO3
–110 mg/L 61.0 1.80 90.2
pH 7.0 (assumed) – – –
2.82
1.80
0
2.00
0
Ca2+
HCO3–
CH
Mg2+
Ca-CH
NCH
Mg–NCH
Ca-NCH
2.82
1.80
0
2.00
0
Ca2+
HCO3–
CH
Mg2+
Ca-CH
NCH
Mg–NCH
Ca-NCH
()
[]
[][]
meq/L 0.807 M 10 4.034
M 10 4.47
L
mol
10
g 61.0
mol
L
g
0.110
HHCO
CO 4–
7–
7
1
–
3
aq2 =×=
×
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
==
−
+
K
Table P6.23 Components, lime and soda ash dosage, and solids generated
Component (eq’n.) Concentration
(meq/L)
Lime
(meq/L)
Soda ash
(meq/L)
CaCO3(s)
(meq/L)
Mg(OH)2(s)
(meq/L)
CO2(aq) (6.37) 0.807 0.807 0 0.807 0
Ca-CH (6.38) 1.80 1.80 0 3.60 0
Mg-CH (6.39) 0 0 0 0 0
Ca-NCH (6.40) 0.200 0 0.200 0.200 0
Mg-NCH (6.41) 0.820 0.820 0.820 0.820 0.820
Excess 0.400
Totals 3.83 1.02 5.43 0.82
c. Sludge generated:
6.24
6.25
Qf = 5 ×106 L/d Qp = 3 ×106 L/d
Cf = 1,500 mg/L Cp = 75 mg/L
Cc = 3,640 mg/L
Qc = 2 ×106 L/d
b. Unit operation function based on an overall aim of removing BOD:
6.27 Both primary and secondary wastewater treatment are designed to remove
biodegradable organic matter (BOM) and the superset of total solids. Primary treatment
6.28 – 6.29 The solutions for these problems are the solutions for problems 6.9 – 6.10 in
the 2nd edition Solutions Manual.
6.30
Q = 30 m3/d
a. Assuming X0 = 0, the microbe mass balance yields equation 6.57:
b. From equation 6.60 and 6.51,
6.31
well-mixed
aeration lagoon
Q, S0, X0Q, S, X
V, S, X, rg’, rsu
l = 60 m
w = 5 m
h = 2 m
Q = 30 m3/d
S0 = 350 mg/L BOD5
S = 20 mg/L BOD5
Ks = 100 mg/L BOD5
kd = 0.10 d-1
μ
m
= 1.6
d
-1
a. The lagoon’s hydraulic detention time is:
c. The lagoon’s daily oxygen demand is:
6.32
Q, S0QsQe, Xe, S0
WAS
Qw, Sw, Xw
RAS
Qr, Sr, Xr
V, X, S P.C.
Act. Sludge Basin
S.C.
Q = 0.300 m3/s X = 2,100 mg VSS/L Xr = 10,000 mg VSS/L
θ
c = 9.0 d S0 = 220 mg BOD5/L
dVSS mg
BOD mg
0.52 5
⋅
=
M
F
b. Noting that Xr = Xw, the cell retention time can be used to calculate the WAS
flow rate.
d. First, write a microbial mass balance around the secondary clarifier (S.C.).
6.33 Q, C0Qe, Ce
Qs, Cs
Q = 20 MGD
Qs = 0.070 MGD
C0 = 800 mg TS/L
Ce = (1 – 0.18)C0 = 656 mg TS/L
a. A mass balance around the primary clarifier yields:
b. The mass of solids removed annually is: