1. No category

"Ingegneria delle acque reflue - Trattamento e riuso 4/ed" - Metcalf & Eddy
titolo originale: "Wastewater Engineering - Treatment and Reuse 4/e"
Copyright © - The McGraw-Hill Companies
2
CONSTITUENTS IN WASTEWATER
PROBLEM
2-1
Problem Statement - See text, page 139
Solution
1.
Set up a computation table to determine the sum of milliequivalents per liter
for both cations and anions for water C, for example.
Concentration
Cation
mg/meq
meq/L
Anion
mg/meq
mg/L
meq/L
Ca2+
20.04
190.2
9.49
HCO3-
61.02
260.0
4.26
Mg2+
12.15
84.1
6.92
SO42-
48.03
64.0
1.33
Na+
23.00
75.2
3.27
Cl-
35.45
440.4
12.41
K+
39.10
5.1
0.13
NO3-
62.01
35.1
0.58
0.2
0.01
CO32-
30.0
1.00
-
19.82
-
19.58
Fe2+
Sum
2.
mg/L
Concentration
-
Sum
-
Check the accuracy of the cation-anion balance using Eq. (2-5).
Ê Â cations - Â anions ˆ
Percent difference = 100 x Á
˜
Ë Â cations + Â anions ¯
Ê 19.82 - 19.58 ˆ
Percent difference = 100 x Á
˜ = - 0.6 %(ok)
Ë 19.82 + 19.58 ¯
PROBLEM
2-2
Problem Statement - See text, page 140
Solution
1.
Determine the ionic strength of the wastewater using Eq. (2-11)
a.
Prepare a computation table to determine the summation term in Eq.
(2-10) using the data for water C in Problem 2-1
2-1
"Ingegneria delle acque reflue - Trattamento e riuso 4/ed" - Metcalf & Eddy
titolo originale: "Wastewater Engineering - Treatment and Reuse 4/e"
Copyright © - The McGraw-Hill Companies
Conc., C,
mg/L
C x 103,
mole/L
Ca2+
190.2
4.75
4
19.00
Mg2+
84.1
3.46
4
13.84
Na+
75.2
3.27
1
3.27
K+
5.1
0.13
1
0.13
Fe2+
0.2
-
4
-
Ion
CZ2 x
103
260.0
4.26
1
4.26
SO42-
64.0
0.67
4
2.68
440.4
12.42
1
12.42
NO3-
35.1
0.57
1
0.57
CO32-
30.0
0.50
4
2.00
Sum
58.17
Determine the ionic strength for the concentration C using Eq. (2-11)
I=
2.
Z2
HCO3-
Cl-
b.
Chapter 2 Constituents In Wastewater
1
1
C i Z 2i = (58.17 x 10 - 3 ) = 29.09 x 10 - 3
Â
2
2
Determine the activity coefficients for monovalent and divalent ions
a.
For monovalent ions
log g = –
0.5 (Z i) 2 I
1+ I
= –
0.5 (1) 2 29.09 x 10 -3
1+
29.09 x 10
-3
= – 0.0728
g = 0.846
b.
For divalent ions
log g = –
0.5 (Z i) 2 I
1+ I
= –
0.5 (2 ) 2 29.09 x 10 -3
1+
g = 0.511
PROBLEM
2-3
Problem Statement - See text, page 140
2-2
29.09 x 10
-3
= – 0.2913
"Ingegneria delle acque reflue - Trattamento e riuso 4/ed" - Metcalf & Eddy
titolo originale: "Wastewater Engineering - Treatment and Reuse 4/e"
Copyright © - The McGraw-Hill Companies
Chapter 2 Constituents In Wastewater
Solution
1.
Determine the activity coefficients for monovalent and divalent ions for
water C from Problem 2-1
a. For monovalent ions
Ê
ˆ
I
- 0.3 I˜
log g = – 0.51 (Zi )2 Á
Ë1 + I
¯
= – 0.51(1)
È
29.09 x 10 -3
2Í
Í1 +
Î
29.09 x 10 -3
- 0.3 (29.09 x 10
-3
˘
)˙
˙
˚
= – 0.0597
g = 0.872
b. For divalent ions
log g = –
0.5 (Z i) 2 I
= –
1+ I
- 0.3I
0.5 (2 ) 2 29.09 x 10 -3
29.09 x 10 -3
1+
- 0.3(29.09 x 10 -3 ) = – 0.2913
g = 0.511
log g = – 0.51(2)
= – 0.2387
g = 0.577
PROBLEM
È
2Í
Í1 +
Î
29.09 x 10 -3
29.09 x 10 -3
- 0.3 (29.09 x 10
-3
2-4
Problem Statement - See text, page 140
Solution
1.
Estimate the TDS for water C from Problem 1 using Eq. (2-12)
I = 2.5 x 10-5 x TDS
where TDS = total dissolved solids, mg/L or g/m3
2-3
˘
)˙
˙
˚
"Ingegneria delle acque reflue - Trattamento e riuso 4/ed" - Metcalf & Eddy
titolo originale: "Wastewater Engineering - Treatment and Reuse 4/e"
Copyright © - The McGraw-Hill Companies
I x 105
TDS =
2.
2.5
=
(0.02909) x 105
2.5
Chapter 2 Constituents In Wastewater
= 1,164 mg / L
Estimate the TDS for water C from by summing the solids concentrations
Conc., C,
mg/L
Ion
Ca2+
190.2
Mg2+
84.1
Na+
75.2
K+
5.1
Fe2+
0.2
HCO3-
260.0
SO42-
64.0
Cl-
440.4
NO3-
35.1
CO32-
30.0
Sum
1184.3
The results from the two methods are remarkably close.
PROBLEM
2-5
Problem Statement - See text, page 140
Solution
1.
For organic carbon, for example
a.
For BOD
i.
Plan to analyze samples immediately; if samples must be held, store
at low temperature (4°C)
ii. In general, never use preservatives for BOD samples. If a
preservative must be used, it should be added to the bottle first
b.
For COD
i.
Use glass bottles for sample collection
2-4
"Ingegneria delle acque reflue - Trattamento e riuso 4/ed" - Metcalf & Eddy
titolo originale: "Wastewater Engineering - Treatment and Reuse 4/e"
Copyright © - The McGraw-Hill Companies
Chapter 2 Constituents In Wastewater
ii. If samples cannot be analyzed soon after collection, preserve
sample by acidification to pH £ 2.0 with sulfuric acid
c.
For TOC
i.
Use glass bottles for sample collection
ii. If samples cannot be analyzed soon after collection, preserve
sample by acidification to pH £ 2.0 with sulfuric acid
PROBLEM
2-6
Problem Statement - See text, page 140
Instructors Note: This problem should be assigned when BOD problems (2-20
through 2-36) are assigned.
Solution
1.
The range of expected BOD is 130 to 430 mg/L based on a mean value of
280 mg/L ± 150 mg/L.
2.
From Table 2-16, a 2 and 1 percent mixture would be suitable for BOD
values varying from 100-350 and 200-700 mg/L. Both percent mixtures
should be used.
3.
To be able to do some statistical analysis (e.g., student-t statistics) on the
laboratory results, a minimum of 3 samples would be needed at each
percentage mixture. To account for a possible outlier, the minimum number
of samples analyzed at each dilution should be 4.
4.
Because the size of the sample is so small, the amount of unseeded or
seeded dilution would be equal to the number of tests times 300 plus a 10
percent excess or a minimum of 100 mL.
PROBLEM
2-7
Problem Statement - See text, page 140
Solution
1.
Determine total solids for sample B.
2-5
CONTINUA...