PRINCIPLES OF
WATER TREATMENT
by Howe, Hand, Crittenden, Trussell, and Tchobanoglous
HOMEWORK SOLUTION MANUAL
FOR
Chapter 3
Process Selection
PROBLEM 3-1
Problem Statement: Calculate rejection and log removal value for the following
filtration process (to be selected by instructor). Use the number of significant figures
necessary to correctly illustrate the removal being obtained.
Influent concentration (#/mL)
Effluent concentration (#/mL)
A
106
10
B
6.85 105
136
C
7.1 10
0.16
5
D
1.65 107
65
E
2.8
106
96
Instructor’s Notes: This problem is similar to Example 3-1. The solution is worked out
for Problem B and the answers for the remaining problems are included in a table at the
end of the solution.
Solution to Problem B:
1.
Calculate removal using Eq. 3-1:
R 1
2.
Ce
Ci
1
136 mL 1
6.85 105 mL 1
0.999801
Calculate the log removal value using Eq. 3-2:
LRV
C
log i
Ce
6.85 105 mL 1
log
136 mL 1
3.70
Solutions to all problems:
Rejection
Log rejection value
A
B
C
D
E
0.99999
5.00
0.999801
3.70
0.99999977
6.65
0.99999606
5.40
0.9999657
4.46
PROBLEM 3-2
Problem Statement: You work for a national environmental engineering consulting
firm and a potential client has called and said that a new contaminant has recently been
identified in their water supply. She wants your firm to identify what processes might be
able to remove the contaminant. For each contaminant (to be selected by your
instructor), suggest what processes might be used and explain how you arrived at your
answer.
Instructor’s Notes: Nearly all the compounds listed in the problem statement are on
the CCL3, EPA’s Contaminant Candidate List of compounds that are being considered
for regulation in the future. The instructor can assign other compounds, as well. The
information necessary to solve this problem is not in the textbook. The problem requires
the student to look up information on these contaminants in other literature, and to use
Homework Solution Manual
Principles of Water Treatment
Chapter 3 - Process Selection
Page 2 of 4
Version 1
the strategy described in the book to compare the properties of these contaminants with
the capabilities of the treatment processes. The processes presented below are only
suggestions and students may come up with other viable treatment technologies.
Solution:
Constituent
Type of compound
Possible treatment
processes
a.
acrolein
Synthetic organic
(aquatic herbicide and
industrial chemical)
Adsorption, advanced
oxidation, reverse osmosis
b.
calicivirus
Virus
Rapid granular filtration,
membrane filtration,
disinfection
c.
17-
Synthetic organic
(estrogenic hormone
used in veterinary and
human pharmaceuticals)
Adsorption, advanced
oxidation, reverse osmosis
d.
Mycobacterium avium
Bacterium
e.
Naegleria fowleri
Amoeba
f.
perchlorate
Inorganic compound
g.
plutonium-239
Radioactive inorganic
chemical
h.
Salmonella enterica
Bacteria
i.
strontium
Inorganic compound
j.
1,1,1,2tetrachloroethane
Synthetic organic
(industrial chemical)
k.
vanadium
Inorganic compound
l.
vinclozolin
Synthetic organic
(fungicide)
Homework Solution Manual
Principles of Water Treatment
Chapter 3 - Process Selection
Rapid granular filtration,
membrane filtration,
disinfection
Rapid granular filtration,
membrane filtration
Ion exchange, reverse
osmosis
Coagulation/filtration, reverse
osmosis
Rapid granular filtration,
membrane filtration,
disinfection
Ion exchange, reverse
osmosis
Adsorption, air stripping,
advanced oxidation, reverse
osmosis
Ion exchange, reverse
osmosis
Adsorption, advanced
oxidation, reverse osmosis
Page 3 of 4
Version 1
PROBLEM 3-3
Problem Statement: Pick a city in the United States and read the Consumer
Confidence Report (often called a Water Quality Report) provided by the water utility.
Answer a series of questions listed in the textbook about the CCR.
Instructor’s Notes: This problem requires the student to gather information about
water utilities from information typically found on the internet.
PROBLEM 3-4
Problem Statement: Calculate the specific energy consumption by the process or
system selected by the instructor (see page 44 in the textbook for the individual problem
statements).
Instructor’s Notes: This problem is similar to Example 3-3. The solution is worked out
for Problem A and the answers for the remaining problems are included in a table at the
end of the solution.
Solution for Problem A:
1.
Calculate the pressure produced by the pump using Eq. 3-4:
P
2.
gh
9.81 10 4 N/m2
Calculate the pump power using Eq. 3-3. Note that QF = QP.
QF P (8200 m3 /d)(98,100 N/m2 )
e
(0.85)(86,400 s/d)
10,950 N m/s 10,950 W = 10.95 kW
PW
3.
(1000 kg/m3 )(9.81 m/s2 )(10 m)
Calculate the specific energy consumption using Eq. 3-5:
PW
QP
E
(10.95 kW)(24 h/d)
8200 m3 /d
0.0320 kW/m3
Solutions to all problems:
2
Pressure, Pa (N/m )
Power, W (N m/s) (Eq. 3-3)
Rate of energy consumption, kWh/d
3
Specific energy consumption, kWh/m (Eq. 3-5)
Homework Solution Manual
Principles of Water Treatment
Chapter 3 - Process Selection
A
98,100
10,950
262.9
0.0320
B
7,500,000
908,400
21,800
4.40
C
26,500
0.3066
0.007358
0.00736
D
118,000
2.21E+08
5,297,000
0.00215
E
150,000
7940
190.6
0.00100
Page 4 of 4
Version 1
PRINCIPLES OF
WATER TREATMENT
by Howe, Hand, Crittenden, Trussell, and Tchobanoglous
HOMEWORK SOLUTION MANUAL
FOR
Chapter 4
Fundamental Principles of
Environmental Engineering
Note: If any errors are noted in this solution manual or in the textbook, please notify
Kerry Howe at howe@unm.edu
PROBLEM 4-1
Problem Statement: Using the principles of stoichiometry, (a) balance the reaction for
the coagulation of water with 50 mg/L of ferric sulfate, Fe2(SO4)3 9H2O, shown below,
(b) calculate the amount of Fe(OH)3 precipitate formed in mg/L and (c) calculate the
amount of alkalinity consumed in meq/L if the alkalinity consumed is equal to the sulfate
(SO24 ) generated:
Fe2 (SO4 )3 9H2O
Fe(OH)3
H
SO42
H2O
Instructor’s Notes: This problem is similar to Example 4-2.
Solution:
a.
Balance the reaction.
Balance iron on the right with iron on the left. There are 2 Fe on the left, so
increase to 2 on the right. There are 3 SO4 on the left, so increase to 3 on the
right. Lastly, there are 18 H and 9 O on the left so be sure the right adds up to
the same.
Fe2 (SO 4 )3 9H2O
b.
2Fe(OH)3
H
3SO24
3H2O
Calculate the amount of Fe(OH)3 precipitate formed in mg/L:
50 mg / L Fe 2 (SO 4 )3 9H2O
1 mmol Fe 2 (SO 4 )3 9H2O
561.7 mg Fe2 (SO 4 )3 9H2 O
2 mmol Fe(OH)3
106.85 mg Fe(OH)3
1 mmol Fe2 (SO 4 )3 9H2 O
1 mmol Fe(OH)3
19.0226 mg / L Fe(OH)3
c.
Compute the amount of alkalinity consumed if the alkalinity consumed is equal to
the sulfate (SO24 ) generated.
i.
Write the chemical equation and note the molecular weight of the reactants
and products involved in the reaction.
Fe2 (SO 4 )3 9H2O
561.7
ii.
2Fe(OH)3
2 x 162
H
6x1
3SO 24
3 x 96
3H2O
3 x 18
Determine the molar relationship for the disappearance of Fe2(SO4)3 9H2O
and formation of SO24 .
Homework Solution Manual
Principles of Water Treatment
Chapter 4 – Fundamental Principles of Environmental Engineering
Page 2 of 39
Version 1
50 mg / L Fe2 SO 4
=
3
9H2O
1 mmol
561.7 mg
3 mmol SO24
1 mmol ferric
2 meq SO24
mmol
0.5341 meq/L
PROBLEM 4-2
Problem Statement: Using information obtained from your local water utility, compute
the ionic strength of your drinking water. In addition, estimate the TDS concentration
and electrical conductivity (EC) of the water. If available, measure the TDS and / or EC
of the water and compare to computed values.
Instructor’s Notes: The solution obtained will depend on the local water quality. A
sample solution is provided using information from the City of Davis, CA, Public Works
Department.
Solution:
1.
Convert the concentration of each ion to molar concentration.
For example:
Ca2
34 mg / L
8.48 10
2.
4
1 g / 1000 mg
1 mole / 40 g
mole / L
Compute the ionic strength using Eq. 4-16 and tabulate the results.
I
1
2
Ions
Calcium
Magnesium
Sodium
Arsenic
Barium
Chromium
Nickel
Selenium
Aluminum
Iron
Manganese
Boron
Potassium
Sulfate
Ci Zi2
i
Ci (mole/L)
8.48×10-4
2.26×10-3
3.70×10-3
6.54×10-8
7.79×10-7
3.65×10-7
1.36×10-8
9.37×10-8
2.89×10-7
3.76×10-8
9.10×10-7
7.35×10-5
1.23×10-5
6.35×10-4
Homework Solution Manual
Principles of Water Treatment
Chapter 4 – Fundamental Principles of Environmental Engineering
Zi
2
2
1
3
2
3
2
4
3
2
2
3
1
-2
CiZi2
3.39×10-3
9.05×10-3
3.70×10-3
5.89×10-7
3.12×10-6
3.29×10-6
5.45×10-8
1.50×10-6
2.60×10-6
1.50×10-7
3.64×10-6
6.61×10-4
1.23×10-5
2.54×10-3
Page 3 of 39
Version 1
1.27×10-3
1.95×10-4
6.84×10-6
6.88×10-3
Chloride
Nitrate
Fluoride
Bicarbonate
I
Ionic strength (I)
3.
1
2
Ci Zi 2
0.014
i
Estimate TDS using Eq. 4-17.
I
2.5 10
TDS
4.
1.27×10-3
1.95×10-4
6.84×10-6
6.88×10-3
-1
-1
-1
-1
5
TDS
I
2.5 10
5
0.014
2.5 10
5
554.53 mg / L
554.53 ppm
Estimate EC.
This drinking water can be treated as a dilute solution because ionic strength is
not high. Therefore, the following relationship between TDS and EC can be used
to estimate EC.
TDS = 0.5 (EC)
EC = TDS/0.5 = 554.53/0.5 = 1109 mho/cm
5.
Compare the computed values to measured data.
The computed TDS is 9 percent higher than the measured TDS. The estimation
is reasonable comparable to measured data. The computed EC is 32.5 percent
higher than measured EC, which occurs because a higher ratio of TDS to EC
exists in practice than was used in the estimation. The ratio of TDS to EC for the
measured data is approximately 0.6, which falls between 0.5 and 0.9. Therefore,
each water sample should be characterized separately.
PROBLEM 4-3
Problem Statement: Plot the activity coefficients of Na+, Ca2+, and Al3+ for ionic
strengths from 0.001 M (very fresh water) to 0.5 M (seawater). Determine the ionic
strength and TDS at which the activity coefficient corrections become important (activity
coefficient less than 0.95) for monovalent, divalent, and trivalent ions.
Solution:
1.
Calculate the activity coefficient using the Davies Equation (Eq. 4-19). At I =
0.001 M:
a. For Na+:
log10
Na
0.50(1)2
0.001
1
0.001
0.3(0.001)
Homework Solution Manual
Principles of Water Treatment
Chapter 4 – Fundamental Principles of Environmental Engineering
0.152
Na
= 0.97
Page 4 of 39
Version 1
b. For Ca2+:
log10
Ca 2
0.50(2)2
0.001
1
0.001
0.3(0.001)
0.0607
0.3(0.001)
0.14
Ca 2
= 0.87
c. For Al3+:
log10
2.
Al3
0.001
1
0.001
Al3
= 0.73
Tabulate and plot the activity coefficients.
Ionic Strength
0.001
0.0025
0.01
0.025
0.05
0.075
0.1
0.25
0.5
3.
0.50(3)2
Na+
0.97
0.95
0.90
0.86
0.82
0.80
0.79
0.74
0.74
Ca2+
0.87
0.81
0.67
0.55
0.46
0.41
0.38
0.30
0.30
Al3+
0.73
0.62
0.40
0.26
0.18
0.14
0.11
0.07
0.06
Calculate the ionic strength when the activity coefficient is 0.95.
a.
For monovalent ions:
Homework Solution Manual
Principles of Water Treatment
Chapter 4 – Fundamental Principles of Environmental Engineering
Page 5 of 39
Version 1
log10 (0.95)
b.
0.50(1)2
I
1
I
0.3(I)
I = 2.17 10 -3 M
0.3(I)
I = 1.27 10-4 M
0.3(I)
I = 2.45 10-5 M
For divalent ions:
log10 (0.95)
c.
0.50(2)2
I
1
I
For trivalent ions:
log10 (0.95)
0.50(3)2
I
1
I
PROBLEM 4-4
Problem Statement: Un-ionized ammonia (NH3) is toxic to fish at low concentrations.
The dissociation of ammonia in water has an equilibrium constant of pKa = 9.25 and is
described by the reaction
NH4
NH3 H
Calculate and plot the concentrations of NH3 and NH4 at pH values between 6 and 10 if
the total ammonia concentration (NH3 + NH4 ) is 1 mg/L as N.
Solution:
1. The concentration of each species can be calculated using -notation where the
concentration of each species is represented with Eqs. 4-40 and 4-41:
[NH3 ]
0
[NH4 ]
1
CT
CT
2. Calculate and plot the concentrations of [NH3] and [NH4+] at pH values between
6 and 10. The values can be calculated with Eqs. 4-46 and 4-47, where Ka =
10-9.25 (see Eq. 4-21). For pH = 6:
0
10 6
10 6 10
[NH3 ]
1
10
[NH4 ]
9.25
0.9994
(0.9994)(1 mg/L)
10
9.25
6
10
9.25
0.9994 mg/L
0.000562
(0.000562)(1 mg/L)
0.000562 mg/L
Homework Solution Manual
Principles of Water Treatment
Chapter 4 – Fundamental Principles of Environmental Engineering
Page 6 of 39
Version 1
3. Set up a spreadsheet to calculate [NH3] and [NH4+] at additional values of pH and
plot the result. The graph is shown below.
PROBLEM 4-5
Problem Statement: A scrubber is used to remove sulfur dioxide (SO2) from the flue
gas from a coal-fired power plant. The scrubber works by spraying high-pH water
downward through a tower while the flue gas passes upward, transferring the SO2 from
the gas to the water. The influent flue gas enters the tower at a rate of 50,000 m3/h and
contains 645 mg/m3 of SO2. The scrubber must reduce the SO2 in the exhaust flue gas
by 90 percent to meet emission requirements. The maximum possible concentration of
SO2 in the water is 820 mg/L. Calculate the required water flow rate to meet emission
requirements. Assume there is no SO2 in the influent water and the air and water flow
rates do not change in the tower.
Instructor’s Notes: This problem is similar to Example 4-5.
Solution:
QL
CLI
QG
CGO
QG
CGI
QL
CLO
1. Draw a diagram of the system.
Homework Solution Manual
Principles of Water Treatment
Chapter 4 – Fundamental Principles of Environmental Engineering
Page 7 of 39
Version 1
2. Write the general mass balance equation:
[accum]
[in] [out] [rxn]
3. State the assumptions:
Steady state, so [accum] = 0
No reactions, so [rxn] = 0
4. The mass balance then becomes:
0
[in]
QL CLI
QGCGI
in
out
0
[out]
QLCLO
QGCGO
5. Rearrange the mass balance equation and solve the flow rate, QL. Note CLI = 0.
QL
QL
QG CGI - CGO
CLO
(50000m3 h) (0.9)(645mg m3 )
3
(820mg L)(1000L m )
35.39m3 h
PROBLEM 4-6
Problem Statement: A rancher needs to provide water for his cattle but the only water
source is a brackish well that has a total dissolved solids concentration (TDS) of 4,800
mg/L. The cattle need 400 L/d of water with TDS < 1,600 mg/L. The rancher has
purchased a solar still that operates at 37 percent recovery of water (distillate) and 96
percent removal of dissolved solids. The rancher wants to recycle the blowdown from
the still to a 20-m3 feed tank to maximize his fresh water recovery and minimize the
waste that has to be hauled off, but the still cannot operate effectively above 52,000
mg/L TDS in the blowdown because of scaling problems. The system will operate as
shown in the diagram below.
Homework Solution Manual
Principles of Water Treatment
Chapter 4 – Fundamental Principles of Environmental Engineering
Page 8 of 39
Version 1
QS = ?
CS = ?
Still feed
5,000 gal
feed tank
QI = ?
CI = 4800 mg/L
Well
QB=?
CB < 52,000
mg/L
Distillate
Solar
still
QC = 400 L/d
CC< 1600 mg/L
Blowdown
QW = ? Waste
CW = ?
Stock pond
a. Prepare a table showing the flow rate and concentration of TDS in the (i) well, (ii) still
feed, (iii) distillate, (iv) blowdown, and (v) waste. Explain all assumptions you make.
b. Propose a modification (i.e., using the existing equipment) that would decrease the
waste that has to be hauled off, and determine how much reduction in waste flow
this modification would achieve.
Instructor’s Notes: This problem is similar to Example 4-5.
Solution:
1. State the assumptions.
Steady state, so [accum] = 0
Feed tank is CMFR
No water loss in process
Constant feed tank volume
No reactions, so [rxn] = 0
2. There are 4 unknown flow rates (QI, QS, QB, and QW) and 4 unknown concentrations
(CS, CB, CC, and CW). We can assume that the operation of the still will be limited by
either the maximum distillate concentration (CC = 1600 mg/L) or the maximum
blowdown concentration (CB = 52,000 mg/L). Although we do not know initially
which one controls the system, ultimately one of them will be a known value, leaving
7 unknowns. Thus, 7 equations are needed.
Two equations are given by the operation of the still.
(Eqn. 1)
(Eqn. 2)
QC = 0.37 QS
CC = (1 - 0.96)CS
One equation is given by virtue of the feed tank being a CMFR
(Eqn. 3)
CW = CS
Homework Solution Manual
Principles of Water Treatment
Chapter 4 – Fundamental Principles of Environmental Engineering
Page 9 of 39
Version 1
The remaining 4 equations can be formed by performing 2 mass balances. The
choices for mass balances are (1) around the entire system (1 input, 2 outputs),
(2) around the feed tank (2 inputs, 2 outputs), and (3) around the still (1 input, 2
outputs). Since the feed tank has 2 inputs, the other two are slightly easier and
will be used.
3. Mass balance around the still
a.
(Eqn. 4) Flow balance:
QS = QC + QB
Qs =
QC 400 L/d
=
= 1081.08 L/d
0.37
0.37
QB = QS - QC = 1081.08 L/d - 400 L/d = 681.08 L/d
b.
(Eqn. 5) Mass balance:
Q S CS
QCCC + QBCB
QSCS = QC (1-0.96)CS + QBCB
QSCS - 0.04QCCS = QBCB
CS (QS - 0.04QC ) = QBCB
CS =
c.
Assume that CB = 52,000 mg/L
CS =
d.
QBCB
QS - 0.04QC
681.08 L/d 52,000 mg/L
1081.08 L/d - 0.04 400 L/d
= 33,252 mg/L
Now that we know Cs we can solve for CC in Eqn 2:
CC = (1 - 0.96)CS=(1 - 0.96)(33,252 mg/L) = 1330.08 mg/L
e.
Since CC < 1600 mg/L, both CB and CC satisfy the given design criteria. If
CC was above 1600 mg/L, it would be necessary to solve Eqn 5 in terms of
CC, substitute CC = 1600 mg/L and solve for the other concentrations
around the still.
4. Using Eqn 3: CW = 33,252 mg/L
5. Mass balance around the overall system
a.
(Eqn 6) Overall flow balance: QI = QC + QW
b.
(Eqn 7) Overall mass balance: QICI = QCCC + QWCW
Homework Solution Manual
Principles of Water Treatment
Chapter 4 – Fundamental Principles of Environmental Engineering
Page 10 of 39
Version 1
c.
Rearranging algebraically:
QC + Q W CI = QCCC + Q W CW
QCCI + QW CI = QCCC + QW CW
Q W CI - Q W CW = QCCC - QCCI
Q W CI - CW = QCCC - QCCI
QW =
400 L/d 1330.08 mg/L -400 L/d 4800 mg/L
QCCC - QCCI
=
CI - CW
4800 mg/L - 33,252 mg/L
Q W = 48.78 L/d
d.
We can now solve QI in Eqn 6:
QI = QC + QW = 400 L/d + 48.78 L/d = 448.78 L/d
6. Summarizing the results in a table:
(i) Well
Feed tank
(ii) Still feed
(iii) Distillate
(iv) Blowdown
(v) Waste
Flowrate (L/d)
448.8
-1,081
400
681
48.8
Concentration (mg/L)
4,800
33,252
33,252
1,330
52,000
33,252
Mass flow (mg/d)
2,154,156
35,948,250
532,034
35,416,216
1,622,122
Overall balance
Still balance
0
0
7. One modification would be to withdraw the waste stream from the blowdown line
instead of from the feed tank. This modification would increase the concentration of
the waste stream from 33,252 mg/L to 52,000 mg/L. An increase in the
concentration of the waste would result in a reduction of waste volume if the same
mass of salts had to be hauled off per day. A second modification would be to
provide a bypass of raw water around the still to increase the concentration of the
water in the stock tank to 1600 mg/L. Since the concentration of the distillate is
lower than required by the cattle, some of the required flowrate could be provided by
a bypass directly from the well, reducing the flow rate and therefore the waste flow
rate by the still.
PROBLEM 4-7
Problem Statement: The following time and concentration data were measured in a
batch reactor. For the specified data set (to be selected by instructor), determine the
reaction order that yields the best fit and estimate the rate constant for the reaction.
Homework Solution Manual
Principles of Water Treatment
Chapter 4 – Fundamental Principles of Environmental Engineering
Page 11 of 39
Version 1
Time, min
0
1
2
3
4
5
6
7
8
9
10
A
40.0
31.5
21.5
17.9
12.2
10.1
6.84
5.25
4.30
2.95
2.42
B
1.18
1.11
1.06
1.00
0.93
0.92
0.81
0.76
0.73
0.66
0.59
Concentration, mg/L
C
120.0
36.1
21.5
16.3
11.5
9.3
7.8
6.9
5.9
5.4
4.9
D
120.0
51.0
24.0
8.7
4.1
1.8
0.55
0.35
0.096
0.052
0.022
E
20.0
9.52
6.38
4.27
3.96
3.11
2.65
2.25
2.15
1.97
1.70
Instructor’s Notes: This problem is similar to Example 4-6. The solution is worked out
for Problem A and the answers for the remaining problems are included in a table at the
end of the solution.
Solution for Problem A:
1. Calculate ln(C) and 1/C for plotting as a function of time.
Time, min
0
1
2
3
4
5
6
7
8
9
10
C, mg/L
40.0
31.5
21.5
17.9
12.2
10.1
6.84
5.25
4.3
2.95
2.42
ln(C)
3.688879
3.449988
3.068053
2.884801
2.501436
2.312535
1.922788
1.658228
1.458615
1.081805
0.883768
1/C
0.025
0.031746
0.046512
0.055866
0.081967
0.099010
0.146199
0.190476
0.232558
0.338983
0.413223
2. Plot ln(C) versus time for a first-order reaction and 1/C versus time for a secondorder reaction
Homework Solution Manual
Principles of Water Treatment
Chapter 4 – Fundamental Principles of Environmental Engineering
Page 12 of 39
Version 1
3. The plot of ln(C) versus time yields the best fit; therefore, a first-order reaction is the
best fit for concentration A.
4. The reaction rate constant is determined by finding the slope of the best-fit line for
the data. The first-order reaction rate constant for concentration A is 0.2851 min-1.
Solution for all problems:
Reaction order and rate constants are shown below. The graphs for each problem are
shown after that.
A
B
C
D
E
Reaction order
1st
1st
2nd
1st
2nd
Rate constant
0.285 min-1
0.0668 min-1
0.0199 L/mg·min
0.869 min-1
0.0525 L/mg·min
Problem B:
Homework Solution Manual
Principles of Water Treatment
Chapter 4 – Fundamental Principles of Environmental Engineering
Page 13 of 39
Version 1
Problem C:
Problem D:
Problem E:
Homework Solution Manual
Principles of Water Treatment
Chapter 4 – Fundamental Principles of Environmental Engineering
Page 14 of 39
Version 1