Paul Mac Berthouex; Linfield C. Brown
Solved Material Balance
Problems
Pollution Prevention and Control

PAUL MAC BERTHOUEX,
LINFIELD C. BROWN
SOLVED MATERIAL
BALANCE PROBLEMS
POLLUTION PREVENTION
AND CONTROL
2
Solved Material Balance Problems: Pollution Prevention and Control
1st edition
© 2019 Paul Mac Berthouex, Linfield C. Brown & bookboon.com
ISBN 978-87-403-2843-1
3
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Contents
CONTENTS
Preface
12
1
The Fundamentals of Design
15
1.1
Block diagrams
15
1.2
Process synthesis
15
1.3
Block diagram - sorting municipal refuse
16
1.4
Block diagram - high quality process water
16
1.5
Process material balance
18
1.6
Air emissions
18
1.7
Conventional water treatment
19
1.8
Membrane water treatment
21
1.9
Separations – hot chocolate
22
1.10
Process invention - 1
22
1.11
Process invention - 2
23
1.12
Neutralization process
23
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Contents
2
Pollutants
25
2.1
Definitions
25
2.2
Lumped measurements
25
2.3
Solids in wastewater
26
2.4
Turbidity
27
2.5
Color
27
2.6
Wastewater solids
27
2.7
Carbon
28
2.8
BOD and COD - 1
28
2.9
Measuring BOD
29
2.10
Biodegradable organics in wastewater
30
2.11
BOD and COD - 2
30
2.12
Nitrogen compounds
31
2.13
Nitrogen as an air pollutant
31
2.14
Phosphorus
31
2.15
Sulfur
31
2.16
Particulates in air
32
2.17
Particle size distribution
33
2.18
Inertial impact collecter
33
2.19
Toxic metals
35
2.20
Toxic organic compounds
35
3
Quantifying pollutants
37
3.1
Sludge volume and mass
37
3.2
Cucumbers
37
3.3
Superfund creosote site
37
3.4
Sludge solids
38
3.5
Primary sludge
39
3.6
Slurry density
39
3.7
Salt by evaporation
39
3.8
Electrostatic precipitator
40
3.9
Solids measurement
42
3.10
Population equivalent - 1
43
3.11
Industrial equivalent
43
3.12
Population equivalent - 2
44
3.13
Sour mash stillage
45
3.14
Clarified effluent
46
3.15
Cadmium in sludge
47
3.16
Metals in solid waste
47
3.17
Dust fall
48
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Contents
3.18
Lead in baghouse dust
49
3.19
Printing plant air emissions
50
3.20
Gas tank capacity
51
3.21
Gas volumes
51
3.22
Volume concentration
52
3.23
Mixture of ideal gases
53
3.24
Gas adjusted for moisture
53
3.25
Dust pollution
53
3.26
Waste gas composition
54
3.27
Toluene emissions
55
3.28
Cadmium in wastewater
56
3.29
Liquid alum
56
3.30
Ion exchange water softening
57
3.31
Mass flow rate calculation
59
3.32
Soil contamination in nigerian landfill
60
3.33
Heavy metals in compost
61
3.34
Shredded municipal bulky waste
63
3.35
Mixed solid waste
64
3.36
Air pollution unit conversions
65
3.37
Coal-fired boiler exhaust gas
66
4
Conservation of mass
69
4.1
Solvent emissions
69
4.2
Mixing liquids
70
4.3
Mixing gases
70
4.4
Industries join municipality
70
4.5
Primary settling tank
71
4.6
Stack gas flow
72
4.7
Waste discharge to a stream
73
4.8
Boiler blowdown
74
4.9
Dye tracer to measure flow
76
4.10
Demineralized water
76
4.11
Well water contamination
78
4.12
Refuse derived fuel processing
79
4.13
Oil recovery with membranes
81
4.14
Water conservation for a refinery
83
4.15
Sludge thickening and disposal
86
4.16
Thickener recycle
88
4.17
Wet scrubber water use
90
4.18
Solvent removal from air
91
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Contents
4.19
Gas preheating
92
4.20
Biological treatment
93
4.21
Step aeration
94
4.22
Single-stage rinsing
95
4.23
Two-stage crossflow rinse
96
4.24
Reverse osmosis circuit
97
4.25
Smelter dust beneficiation
99
4.26
Drying solids
100
4.27
Anchovy processing
102
4.28
Anaerobic sludge digestion
103
4.29
Solvent vapor emissions
105
4.30
Oil and grease removal
107
4.31
River pollution newspaper story
107
4.32
River dilution
109
4.33
Oil sands
110
4.34
Water softening sludge
112
4.35
Alum coagulation sludge
113
4.36
Four-stage evaporation
114
4.37
Bioconcentration
115
4.38
Henry’s law and volatilization
117
4.39
Partitioning of pyrene in soil
118
4.40
Partitioning 1,2-dichlorobenzene
119
4.41
Partitioning 2,3,7,8-tetrachlorodibenzo-dioxin
121
4.42
Partitioning DDT
122
5Material balance and pollution prevention
124
5.1
Metabolism of a factory
124
5.2
Zero discharge
126
5.3
Pollution prevention focus questions
127
5.4
Inefficient water use
127
5.5
Semiconductor manufacturing
128
5.6
Water for soft drink manufacturing
130
5.7
Hidden and avoided costs
131
5.8
Solvent recovery payback
131
5.9
Waste minimization survey
132
5.10
Vegetable processing industry water reuse
133
5.11
Polymer recycling
135
5.12
Coca-cola
136
7
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Contents
6Material balance with chemical reactions
137
6.1
Burning octane
138
6.2
Combustion of gasoline
138
6.3
Combustion - 1
139
6.4
Combustion - 2
139
6.5
Composting stoichiometry
140
6.6
Combustion of municipal refuse
141
6.7
NOx control
143
6.8
Saline effluent
143
6.9
Flue gas desulfurization
144
6.10
Water softening
146
6.11
Chemical treatment of wastewater
147
6.12
Phosphorus precipitation
149
6.13
Metal plating
152
6.14
Chemical and biological phosphorus removal
154
6.15
Ammonium sulfate recovery from flue gas
156
6.16
Suspended solids removal
158
6.17
Sludge digester gas (methane)
162
7Reaction rates and reactor design
165
7.1
165
Detention time
7.2Rate coefficient determines the exponential decrease
165
7.3
First order reaction
166
7.4
Bacterial die-off in a river
166
7.5
Half-life for a first-oder reaction
167
7.6
Half-life reactions
168
7.7
Pollutant decomposition
169
7.8
Rate coefficient for a first-order reaction
170
7.9
Dye effluent degradation
171
7.10
Simple reaction - 1
173
7.11
Simple reaction - 2
173
7.12
Gaseous reaction
174
7.13
Ozone kinetics
177
7.14
Temperature and the reaction rate
178
7.15
Degradation of atrazine
178
7.16
Second-order reaction
179
7.17
Rate of dinitrotoluene (DNT) removal
180
7.18
Dissolution of lead
182
7.19
Biotransformation
186
7.20
Pesticide degradation – laboratory experiment
187
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Contents
7.21
Chlorine decomposition
189
7.22
DDT persistance
192
7.23
Bioconcentration factor
193
7.24
Bioaccumulation & depuration
195
7.25
Metabolites of pesticides
196
7.26
Parallel reaction
197
7.27
Ozone decolorization of acid yellow dye
198
7.28
Pesticide degradation
199
7.29
Pollutant removal in a CSTR
202
7.30
Pollutant removal in two CSTRs in series
203
7.31
Mixed-order model
203
7.32
Detention time for a CSTR
204
7.33
Detention time for two CSTRs in series
205
7.34
Graphical solution for two CSTRs in series
206
7.35
Fitting models
208
8Material balance for biological processes
212
8.1
BOD - 1
212
8.2
BOD - 2
212
8.3
BOD of soybean oil
214
8.4
Kinetics of BOD removal
215
8.5
BOD removal in a lagoon
217
8.6
BOD removal and temperature
217
8.7
Biomass yield factor
219
8.8
Waste activated sludge
220
8.9
Sludge age for nitrification
222
8.10
Sludge age control
224
8.11
Recycle to primary settling tank
225
8.12
Balance on activated sludge solids
227
8.13
Activated sludge aeration basin as cstrs in series
228
8.14
Aerobic lagoon
232
8.15
Wastewater treatment using 4 lagoons in series
233
8.16
Wastewater treatment plant mass balance - 1
234
8.17
Wastewater treatment plant mass balance - 2
242
8.18
Sludge volume index
249
8.19
Oxygen supply
251
8.20
Oxygen requirement
252
8.21
Oxygen demand
253
8.22
Deer island digesters
254
8.23
Methane production
255
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Contents
8.24
Anaerobic treatment
256
8.25
Anaerobic sludge digestion
256
8.26
Food waste
257
8.27
Anaerobic biodegrdation index
258
8.28
Mouras automatic scavenger
260
8.29
Anaerobic sludge digestion
261
8.30
Two-staged anaerobic digesters
263
8.31Temperature-phased digester (thermophilic + mesophilic)
264
8.32
Upflow anaerobic sludge blanket reactor
265
8.33
Agricultural biogas gas
266
8.34
Dairy manure
267
8.35
Co-digesting food waste and sludge
269
8.36
Biokinetics
273
8.37
Haldane inhibition model
275
8.38
COD removal from refinery wastewater
278
8.39
Reactor detention time
281
8.40
Synthetic media biofilters
283
8.41
Trickling filter hydraulic loading
284
8.42
Biofilter for odor control
284
8.43
Cannery waste treatment
285
8.44
Landfill gas activity
285
8.45
Composting temperature
286
8.46
Composting sludge cake
287
8.47
Sludge-refuse mixture for composting
287
8.48
Casio city and the ATOZINC rayon company
290
9The unsteady-state material balance
295
9.1
Filtration
295
9.2
Wetlands water budget
296
9.3
BOD load equalization
297
9.4
Pumping station
299
9.5
Batch reactor
300
9.6
Smoothing concentrations
302
9.7
Filling a leaky tank
305
9.8
Dilution of a salt solution
308
9.9
CSTRs in series
310
9.10
Sludge digesters
313
9.11
Lake pollution
318
9.12
Chlorides in the great lakes
318
9.13
PCB in lake trout
319
10
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Contents
10Water conservation and reuse
321
10.1
Industrial water consumption
321
10.2
Cooling water
321
10.3
Water recycle saves money
321
10.4
Water flow impacts cash flow
322
10.5
Industrial water balance
323
10.6
Cooling tower
324
10.7
Cooling tower with recycle makeup water
325
10.8
Cooling tower water consumption
326
10.9
Water reuse for cooling systems
327
10.10
Water reuse
328
10.11
Recycle and reuse
330
10.12
Membrane process for regeneration and reuse
333
10.13
Nearest neighbor - 1
334
10.14
Nearest neighbor - 2
335
10.15
Mass transfer process
336
10.16
Two process system
337
10.17
Water reuse – minimum flowrate
338
10.18
Composite mass-concentration curve
342
10.19
Process water regeneration and recycle
348
10.20
Water reclamation system - 1
351
10.21
Water reclamation system - 2
352
11Appendix 1 – Atomic Mass of Selected Elements
354
12Appendix 2 - Conversion Factors
355
13Appendix 3 – Densities and Specific Weights
357
11
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Preface
PREFACE
Solved Material Balance Problems: Pollution Prevention and Control is collection of solved
problems for use with the textbook Pollution Prevention and Control: Material Balances
published by Bookboon.com in 2019. The goal is to build the problem-solving strategies
and skills that are widely useful in water pollution control, air pollution control, and solid
waste control.
These skills are developed in two ways. The first way is to formulate and solve many kinds
of problems. The second way is by reading problems to learn more about technology and
the range of problems that can be solved with a mastery of the material balance.
The book is designed as an engineering text, but the concepts and calculations are accessible
to students in non-engineering disciplines. All the problems can be solved using algebra,
elementary calculus, and iterative calculations.
All of the problems are steady-state material balances except those in Chapter 6. Steady
state means that flows rates and concentrations are constant. Chapter 6 is about problems
where flow, concentration, or both change over time. All the problems can be solved using
algebra and simple iterative solutions, including those in Chapter 6.
Pollution prevention and control is a big subject and you will learn more and more quickly
if you vary what you read and how you read. Reading a textbook is not learning. Practicing
–thinking about problems and solving them - is learning. So, here is the practice material.
Two kinds of practice are needed. One is directed toward mastering everyday calculations
and procedures, such as calculating mass from a volume and concentration, or converting
units from a volume basis to a mass basis, or from a wet basis to a dry basis. The second
is discussion and open-ended questions. There are exercises of both kinds about water,
wastewater, air and other gases, soil, and solid waste, and with varying degrees of difficulty.
To make the practice more interesting, many of the problems are given in an engineering
context. The titles guide the student and instructor if there is a special area of interest. In a
typical introductory course on pollution prevention and control, the more varied experience
should be preferred. We believe this is true for engineers, and especially so for non-engineers.
If you are doing self-study, start by reading the problems; read as you would read a textbook
to discover the context in which certain calculations are needed. Read them to discover
new vocabulary and learn about systems and ideas that are not in the text. Then select a
few problems to solve. Draw a diagram and show all the given information and only then
start with the equations and calculations.
12
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Preface
Solutions are not isolated in the back of the study guide; they appear with the questions.
Work on your solution and then use our solution. There are often different paths to a correct
answer so use ours as a guide and not a strict pattern to be followed.
The numbering of figures and tables is not sequential. It is keyed to the problems.
• Figure P6.14 and Table P6.14 are part of problem 6.14.
• Figure S6.14 and Table S6.14 are part of the solution of problem 6.14.
A Note to Instructors
Some problems are short and quick, and some take a lot of work, so solve or carefully read
the problems before making an assignment. Many of the problems should be enhanced by
a quick explanation when they are assigned.
• Build professional vocabulary, not by listing and defining terms, but by working in
context. Briefly describe the treatment process or system, and explain new vocabulary.
• Provide a context, “This problem is about heavy metals in compost that is made
from sewage sludge. Heavy metals are toxic and we must be careful about moving
them from a waste disposal site into someone’s garden or park or playground.”
• In short, try to make the problem interesting beyond the obvious calculations
(which are necessary, but not interesting).
• Use problems as opportunities to teach the why and how of pollution prevention
and control engineering.
• Tell students to be resourceful in finding information. Electronic dictionaries are
great and should be used frequently. Wikipedia is a wonderful resource for concise
explanations.
• Augment the problems and solutions with photos and pictures. It was our idea
to use many photographs, one per problem would have been ideal, but there are
difficulties with copyrights and permissions, so that idea was abandoned. If the
problem mentions an electrostatic precipitator you can go to Google images and
look at them. Having found a suitable photo, don’t get sidetracked. Look fast and
get back to work.
13
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Preface
Paul Mac Berthouex
Emeritus Professor, Department of Civil and Environmental Engineering
The University of Wisconsin-Madison
Linfield C. Brown
Emeritus Professor, Department of Civil and Environmental Engineering
Tufts University
October 2018
14
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
1
The Fundamentals of Design
THE FUNDAMENTALS OF DESIGN
Tutorial Note
This chapter deals with process synthesis and understanding how processing systems are put
together. The tools for doing process analysis have not been explored yet, so the problems in
this chapter, with a few exceptions, are for discussion with the instructor’s guidance. The goal is
to initiate thinking about pollution prevention and control as being done by integrated systems
of processes.
1.1
BLOCK DIAGRAMS
Assemble the four basic processing elements into a block diagram for an operation or process
in your home, school, or work place.
Mixer
Reactor
Separator
Splitter
Figure P1.1 Processing elements
Solution
This is an open-ended question that is meant to provoke discussion and research. There are many
solutions.
1.2
PROCESS SYNTHESIS
Use the processing elements in Figure P1.1 to construct a process that will separate a
mixture of two solid materials, A and B, and dissolve B in water. Identify each block as a
transformation (reactor), separator, mixer, or splitter.
15
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The Fundamentals of Design
Solution
A
A&B
B
Separator
Aqueous
solution of B
Mixer
Water
Figure S1.2
1.3
BLOCK DIAGRAM - SORTING MUNICIPAL REFUSE
Municipal refuse arrives at a sorting station by truck. It undergoes size reduction in a
hammermill. This is followed by magnetic separation of ferrous metals, a screen that
removes small particles of broken glass, an air classifier to separate plastic and paper, and a
float-sink process to separate desirable species of plastic for recycle. Draw a block diagram
of the sorting process.
Solution
Refuse
Delivered
by Truck
Hammermill
Size
Reduction
Shredded
Refuse
Magnetic
Separator
Fine
Screen
Air
Classifier
Paper
Mixed
Plastics
Ferrous
Metals
Glass
Particles
Float-Sink
Separator
Recyclable
Plastics
Waste
Figure S1.3
1.4
BLOCK DIAGRAM - HIGH QUALITY PROCESS WATER
City water contains iron, calcium and other minerals that make it unsuitable for cleaning
sensitive products like computer chips. The first step is aeration to oxidize the iron so it
can be removed as particles of iron oxide by filtration, Next is a membrane process called
ultrafiltration to remove large dissolved molecules and very small particles, followed by a
16
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The Fundamentals of Design
ion exchange that removes all dissolved minerals, and finally it is passed through activated
carbon adsorption beds to remove dissolved organic chemicals. Draw a block diagram for
the process.
Solution
Water
Calcium
Iron
Minerals
Aeration
Water
Calcium
Iron oxide
Minerals
Filtration
Ultrafiltration
Filter backwash
water &
Iron oxide
Large dissolved
molecules &
small particles
Ion
Exchange
High-quality
water
Brine solution
containing
calcium & other
minerals
Figure S1.4
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SOLVED MATERIAL BALANCE PROBLEMS:
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1.5
The Fundamentals of Design
PROCESS MATERIAL BALANCE
Figure P1.5, a diagram of the overall material balance of an industry, is shown at a meeting.
Most everyone is pleased by the small quantities of waste that are shown. One person is
skeptical. What is the problem?
Water
1.6 m3
(423 gal)
Energy
2,100 GJ
(2 x 109 Btu)
Raw materials
1,200 kg
Industrial Production Unit
Air emissions
8 kg
Solid waste
40 kg
Product
1,000 kg
Waterborne waste
8 kg solids
Figure P1.5 Production Unit Material Balance
Solution
The material balance is not obvious because not all quantities are in the same units. Water input is
measured in gallons but wastewater output is in pounds. The mass of water input is (8.34 lb/gal)(423
gal) = 3,528 lb or 1,600 kg and this mass must leave the process as part of the product or as waste.
Using a material balance, and assuming the other mass quantities are correct, the quantity shown as
‘waterborne waste’ should be
‘wastewater’ = 1,600 kg + 1,200 kg – 1,000 kg – 8 kg – 40 kg = 1,752 kg
for everything to balance.
This quantity of wastewater may contain 8 kg of ‘waterborne solids’.
1.6
AIR EMISSIONS
An organic solvent (acetone) is used to clean the grease off of metal parts before they are
plated with nickel (Figure P1.6). The solvent input is 100 liters per month. No solvent leaves
on the clean metal parts. Sixty liters per month of used solvent is collected for disposal. The
volume unaccounted for is lost as an air emission. The solvent density is 0.95 kg per liter.
By law, air emissions must be reduced to less than 0.1 L/month. (a) What mass of solvent
is lost to the atmosphere? (b) The price of acetone solvent is $3.50/L. What is the cost of
solvent lost to the air? (c) What is the cost of the spent solvent that is sent to ‘disposal’?
18
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The Fundamentals of Design
The disposal cost is $5.00/L. (d) The company is agreeable to buying a system to recover
100% of the air emissions of acetone and also a distillation unit to recover 90% of the
acetone in the spent solvent if this system will pay for itself in 2 years or less. How much
can be spent on the project?
Air emissions
40 L/month
Solvent
100 L/month
Greasy
metal parts
Solvent
degreasing
process
Cleaned
metal parts
Spent solvent
60 L/month
Figure P1.6 Solvent degreasing process
Solution
a) Input = waste solvent + air emission loss
Air emission loss = 100 L/month – 60 L/month = 40 L/month
Air emission mass = (0.95 kg/L)(40 L/month) = 38 kg/month
b) Cost of acetone lost as air emissions = ($3.50/L)(40 L/month) = $140/month
c) Cost of acetone lost as spent solvent = ($3.50)(60 L/month) = $210/month
d) Value of acetone recovered from air emissions = $140/month
Value of acetone recovered from spent solvents = (0.9)($210/month) = $189/month
Savings on disposal costs = ($5/L)(60 L/month) = $300/month
Total savings on solvent purchase = value of recovered solvent + disposal cost
= $140/month +$189/month + $300/month = $629/month
Savings over 2 years = (24 months)($629/month) = $15,096
1.7
CONVENTIONAL WATER TREATMENT
The treatment process is shown in Figure P1.7. The raw water is turbid river water and
it contains particles of clay and silt, colloidal particles, bacteria and algae, and dissolved
minerals. The treated water must be free of turbidity, algae, and bacteria. Identify the
separation processes and the chemical transformations. Explain the sequence of operations.
19
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The Fundamentals of Design
Coagulant
addition
Filtration
Sedimentation
Raw
water
Disinfection
(chlorine)
Treated
water to
user
Coagulation &
Flocculation
Sludge to
disposal
Figure P1.7 Conventional water treatment
Solution
Raw water from a lake or river contains bacteria and turbidity (finely divided particles) that must be
removed to make safe drinkable water. A coagulating agent, usually containing ferric iron or aluminum
ions, is mixed into the raw water. The coagulant causes the finely divided particles to agglomerate
into larger particles. Some of these agglomerated particles can be removed by gravity settling in
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The Fundamentals of Design
the sedimentation basin. Those that are too small to settle put are removed by filtration. Figure P1.7
shows a filter with a bed of sand that captures the particles. From time to time the filter is cleaned by
backwashing (flushing clean water up through the sand bed to remove collected material). Coagulation
and filtration are not sufficient to make the water bacterially safe so disinfection is the final step.
Chlorine, shown here, is a popular disinfectant; ozone and UV light are others.
1.8
MEMBRANE WATER TREATMENT
The treatment process is shown in Figure P1.8. The raw water is turbid river water and it
contains particles of clay and silt, colloidal particles, bacteria and algae, and dissolved minerals.
The particles, including bacteria and algae, must be removed. The dissolved minerals are
not a problem. Identify the separation processes and the chemical transformations. Explain
the sequence of operations.
Membrane separation
(ultrafiltration)
Disinfectant
(chlorine)
addition
Permeate
tank
Raw
water
Treated
water to
user
Holding
tank
Disinfection
Figure P1.8 Membrane water treatment
Solution
Ultrafiltration uses porous membranes to remove very small particles, including microorganisms. The
permeate is the liquid that passes through the membrane. A fraction of the feed does not permeate;
it leaves as a concentrated stream that is wasted or recycled (recycle is shown here). The permeate is
disinfected (chlorine is used here) before going to the user. The holding tank provides no treatment.
It mixes the recycled concentrate from the ultrafiltration.
21
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
1.9
The Fundamentals of Design
SEPARATIONS – HOT CHOCOLATE
Imagine adding instant hot chocolate and marshmallows to a cup of hot water. Some solids
are readily dissolved and some are not. Momentarily stir the imaginary contents, and then
let the cup stand. Now separate the substances you added from the water.
Solution
This is an open-ended question that is meant to provoke discussion. No solution is provided.
1.10 PROCESS INVENTION - 1
A mixture of dry materials, A, B and C, are to be separated. They differ in these ways
Particle size
A>BB=C
Solubility in water A and B are soluble; C is not
Invent a process to separate the materials and draw the block diagram.
Solution
A, B, & C
Separator
based on
particle size
A
Water
Separator
based on
solubility in
water
B&C
B (in solution)
C (wet solids)
Figure S1.10
22
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The Fundamentals of Design
1.11 PROCESS INVENTION - 2
A mixture of dry materials, A, B and C, are to be separated. They differ in these ways
Particle size
A<B<C
Solubility in water A and B are soluble; C is not
Color
A and C are green; B is clear
Magnetic attraction C is attracted to a magnet, A and B are not
There will be several ways the materials can be separated. Draw block diagrams for at least
3 processes.
Solution
It helps to organize the data graphically as an ordered property list
a
Small
A, B
Soluble
A, C
B,C
Large
C
Insoluble
B
Size
Solubility in water
Color
Green
Clear
C
A,B
Magnetic
Not
magnetic
Magnetic attraction
Figure S1.11 Ordered Property List
There are so many possibilities we will not provide diagrams.
a) Separate based on size, followed by solubility
b) Separate based on size, followed by color
c) Separate based on color, followed by separation based on size
d) Separate C by magnet, separate A and B by size and/or color.
And so on.
1.12 NEUTRALIZATION PROCESS
Explain the process shown in Figure P1.12. No chemical equations are needed; just explain
clearly in words. Identify any assumptions that you make.
23
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Lime
Ca(OH)2
Wastewater
+
Sulfuric acid
(H2SO4)
The Fundamentals of Design
Acid
(HCl)
Polymer
coagulant
Mixer
Reactor
Settling
Tank
Water
for reuse
Reactor
Splitter
Water to
discharge
Dilute slurry of
CaSO4 precipitate
Centrate recycle
Centrifuge
Concentrated slurry
of low-grade gypsum
(CaSO4)
Figure P1.12 Neutralization process
Solution
The pH of the influent wastewater is too low. It must be raised by neutralizing the sulfuric acid before
the wastewater can be treated.
No other pollutants or contaminants are in the wastewater.
The neutralizing reagent is lime, Ca(OH)2
Calcium from the lime reacts with sulfate from the sulfuric acid to form solid particles of calcium sulfate,
CaSO4, commonly known as gypsum.
The gypsum precipitate (solids) are removed by gravity settling.
A polymer coagulant is added to make the removal of precipitate solids more efficient (more solids
removed faster).
Adding enough lime to precipitate gypsum raises the pH above the allowable limit for discharge.
The pH is lowered by adding (hydrochloric) acid.
No solids are formed so solids removal steps are not needed.
Gypsum solids are removed from the settling tank as a dilute slurry.
The slurry is concentrated in a centrifuge to produce a concentrated slurry of low-grade gypsum.
24
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
2
Pollutants
POLLUTANTS
Tutorial Note
Many of these questions are definitions and simple explanations. The answers are in the text
“Pollution Prevention and Control: Material Balances” and most are not repeated here.
2.1
DEFINITIONS
Define: (a) Pollutant and (b) Pollution.
No solution given
2.2
LUMPED MEASUREMENTS
a) Explain why lumped measurements are necessary in quantifying pollutants.
b) Name some lumped measurements and describe what they measure.
Solution
a) Lumped, aggregated, or collective measurements are used to define a collection of
substances when it is difficult or unnecessary to distinguishing individual species.
b) Examples are
• Turbidity – measures all colloids and particulates that scatter the passage of
light through water.
• Suspended solids – all solids, whether organic or inorganic, that can be collected
on a filter with a specified pore size.
• Total dissolved solids - all solids, whether organic or inorganic, that will pass
through a filter with a specified pore size.
• Color – measures hue and intensity without regard to the compounds that cause
the color.
• Particulate matter in air – like suspended solids in water and wastewater, there
is no differentiation of the kinds of particles.
• BOD and COD measure organic chemicals in water and wastewater, but do not
identify individual compounds.
25
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
2.3
Pollutants
SOLIDS IN WASTEWATER
A sample of wastewater has a total solids concentration of 1,500 mg/L, a volatile solids
concentration of 600 mg/L, total dissolved solids of 800 mg/L, volatile suspended solids
of 100 mg/L, and volatile dissolved solids = 70% of total dissolved solids. Calculate the
concentration of the other kinds of solids.
Solution
Given:
Total solids (TS) = 1,500 mg/L
Total dissolved solids (TDS) = 800 mg/L
Volatile solids (VS) = 600 mg/L
360°
thinking
Volatile suspended solids (VSS) = 100 mg/L
.
Fixed Solids (FS) = Total solids (TS) – Volatile solids (VS)
= 1,500 mg/L – 600 mg/L = 900 mg/L
Total suspended solids (TSS) = Total solids (TS) – Total dissolved solids (TDS)
= 1,500 mg/L – 800 mg/L = 700 mg/L
360°
thinking
.
360°
thinking
.
Discover the truth at www.deloitte.ca/careers
© Deloitte & Touche LLP and affiliated entities.
Discover the truth at www.deloitte.ca/careers
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26
Dis
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Pollutants
Fixed suspended solids (FSS) = Total suspended solids – volatile suspended solids
= 700 mg/L – 100 mg/L = 600 mg/L
Volatile dissolved solids (VDS) = 0.7(Total dissolved solids)
= 0.7(800 mg/L) = 560 mg/L
Fixed dissolved solids (FDS) = Total dissolved solids (TDS) - Volatile dissolved solids (VDS)
= 800 mg/L – 560 mg/L = 240 mg/L
2.4
TURBIDITY
a) Define colloid and explain how colloids cause turbidity.
b) Explain how turbidity is measured.
c) Why can turbidity not be measured in mass/volume units?
No solution given
2.5
COLOR
a) How is the intensity of color measured?
b) Explain the difference between true color and apparent color.
c) How is true color measured?
No solution given
2.6
WASTEWATER SOLIDS
These data are known about the solids composition of a wastewater.
Total solids (TS) = 1,500 mg/L
Volatile solids (VS) = 1,125 mg/L, thus VS = 0.75 TS
Total suspended solids (TSS) = 825 mg/L
Volatile suspended solids (VSS) = 675 mg/L, thus VSS = 0.6 VS
Calculate Total dissolved solids (TDS), Fixed dissolved solids (FDS), and Volatile dissolved
solids (VSS).
27
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Pollutants
Solution
Measured values in Table S2.6 are in BOLD type.
Kind of
Measured
Solids
(mg/L)
Relation
TS
1,500
VS
1,125
VS = 0.75TS
FS
375
FS = TS – VS
TSS
825
VSS
675
VSS = 0.65VS
FSS
150
FSS = TSS – VSS
TDS
675
TDS = TS – TSS
VDS
450
VDS = VS – VSS
FDS
225
FDS = FS – FSS
Table S2.6
2.7
CARBON
a) Explain the difference between organic and inorganic carbon.
b) Why is organic carbon considered a water pollutant and inorganic carbon is not?
c) Why is inorganic carbon treated as an air pollutant?
No solution given
2.8
BOD AND COD - 1
BOD and COD measure the organic strength of wastewater, but they do not measure the
same thing.
a) Define each and explain the difference.
No solution given
28
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
2.9
Pollutants
MEASURING BOD
Four 300 mL BOD bottles are used to measure the 5-day BOD of a municipal wastewater.
The test conditions are given in Table P2.9. Complete the calculations and calculate the
best estimate of the 5-day BOD.
Final
Bottle
Volume of
Dilution
Initial
ID
Wastewater
Factor DF
DO
No.
(mL)
= 300/V
(mg/L)
1
10
30
9.1
7.5
2
15
20
9.1
6.4
3
20
15
8.9
5.4
4
25
12
9
5.1
DO
(day 5)
(mg/L)
Depletion
5-day BOD
∆DO
(DF)(∆DO)
(mg/L)
(mg/L)
Table P2.9
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Pollutants
Solution
Bottle
Volume of
Dilution
Initial
Final DO
Depletion
5-day BOD
ID
Wastewater
Factor DF
DO
(day 5)
∆DO
(DF)(∆DO)
No.
(mL)
= 300/V
(mg/L)
(mg/L)
(mg/L)
(mg/L)
1
10
30
9.1
7.5
1.6
48
2
15
20
9.1
6.4
2.7
54
3
20
15
8.9
5.4
3.5
52
4
25
12
9
5.1
3.9
47
Average = 50 mg/L
Table S2.9
2.10 BIODEGRADABLE ORGANICS IN WASTEWATER
The influent to biological treatment process contains 400 mg/L ultimate BOD and 600
mg/L COD. The process removes 390 mg/L of ultimate BOD.
a) How much COD is removed?
b) What is the effluent COD?
c) What percent of ultimate BOD and COD are removed?
Solution
a) Ultimate BOD removed = COD removed = 390 mg/L
b) Effluent COD = 600 mg/L – 390 mg/L = 210 mg/L
c) Percent removal: BOD = 100(390 mg/L)/(400 mg/L) = 97.5%
COD = 100(390 mg/L)/(600 mg/L) = 65%
2.11 BOD AND COD - 2
Explain why COD is always greater than BOD.
No solution given
30
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Pollutants
2.12 NITROGEN COMPOUNDS
Nitrogen exists in many forms. The most important in water and wastewater are ammonia,
nitrate, nitrite, and total organic nitrogen.
a) Define each of these.
b) What are the two species of ammonia in water?
c) Why is it convenient for laboratories to report the concentration of these species
in terms of the nitrogen content (for example Ammonia = 20 mg NH3-N instead
of 24.3 mg NH3/L)?
d) If a sample contains 28 mg NH3/L, what is the concentration as mg NH3-N?
Solution
d) 28 mg NH3/L = (28 mg NH3/L)(14 mg N/17 mg NH3) = 23 mg NH3-N/L
2.13 NITROGEN AS AN AIR POLLUTANT
Explain what forms of nitrogen are considered to be air pollutants and explain why.
No solution given
2.14 PHOSPHORUS
a) Name the three important forms of phosphorus in water and wastewater treatment.
b) Which form is most common in wastewater?
c) What is the effect of having too much phosphorus in a lake?
No solution given
2.15 SULFUR
a) Two common forms of sulfur in wastewater are sulfate and sulfide. How is one
transformed to the other?
b) Both are a nuisance in drinking water. Why?
31
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Pollutants
c) Hydrogen sulfide must be controlled in the workplace and in biogas that will be
use as fuel. Explain.
No solution given
2.16 PARTICULATES IN AIR
a) Define an aerosol.
b) What is meant by PM10 and PM2.5?
c) Why are fine particulates a serious public health risk?
No solution given
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Pollutants
2.17 PARTICLE SIZE DISTRIBUTION
Figure P2.17 is the size distribution of particles collected from ambient air in a city.
a)
b)
c)
d)
What
What
What
What
is the diameter that is being reported?
is the PM10 diameter, in µm?
is the median diameter?
percentage of particles are larger than 5 µm?
Cumulative
CumulativeMass,
mass,%
%<” dd
100
90
80
70
60
50
40
30
20
10
0
1
10
100
Particle
d (—m)
(—m)
Particlesize,
size, d
Figure P2.17 Ambient air particle size distribution
Solution
a) The diameter is the aerodynamic diameter, which is defined based on the movement
of particles in air. It would be the physical diameter of spherical particles, but this is
not true for non-spherical particles, which includes most air borne particulates.
b) PM10 is particles with an aerodynamic diameter of 10 µm or less. From Figure P2.17
this is 50% of the particles in the ambient air sample.
c) The median diameter is the size that splits the particle size distribution at the 50%
size. Fifty percent of particles are larger and fifty percent are smaller than the median
diameter. D50% = 10 µm. This happens to be the PM10 value for this sample, but this
is a coincidence.
d) From Figure P2.17, the percentage of particles larger than 5 µm is 85%. Fifteen percent
of particles are smaller than 5 µm.
2.18 INERTIAL IMPACT COLLECTER
Data from a 5-stage impact collector are given in Table P2.18. The total mass of particles
collected is 15.7 mg. Calculate the mass fraction of particles in each size range and the
cumulative particle size distribution.
33
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Pollutants
Size Range
Mass Collected
(µm)
(mg)
1
> 9.0
0.2
2
4.0 – 9.0
1.8
3
2.2 – 4.0
4.9
4
1.2 – 2.2
6.2
5
0.7 – 1.2
2.2
Backup filter
0 – 0.7
0.4
Stage
Table P2.18
Solution
Mass fraction of particles captured in each stage is the mass of particles captured in a stage divided
by the total mass of particles collected.
For stage 1:
Mass fraction = 0.2 mg/15.7 mg = 0.013
Cumulative mass fraction is the cumulative sum of the collected mass fractions from smallest to largest
particle size.
Stage
Size Range
(µm)
Mass
Collected
(mg)
Mass
Cumulative
fraction
mass fraction
1
> 9.0
0.2
0.013
1.000
2
4.0 – 9.0
1.8
0.115
0.987
3
2.2 – 4.0
4.9
0.312
0.873
4
1.2 – 2.2
6.2
0.395
0.561
5
0.7 – 1.2
2.2
0.140
0.166
Backup filter
0 – 0.7
0.4
0.025
0.025
15.7
1
Table S2.18
34
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Pollutants
2.19 TOXIC METALS
Arsenic, cadmium, mercury, and lead are toxic metals. Briefly explain how they cause damage
when ingested or inhaled by humans.
Solution
This is an open-ended question that is meant to provoke discussion and research. A good solution
may be several paragraphs. The possible effects are many and varied.
2.20 TOXIC ORGANIC COMPOUNDS
List six toxic organic compounds, explain how people might be exposed to them, and
explain their toxic effects on humans.
35
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Pollutants
Solution
This is an open-ended question that is meant to provoke discussion and research. A good solution
may be several paragraphs. The possible effects are many and varied.
36
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
3
3.1
Quantifying pollutants
QUANTIFYING POLLUTANTS
SLUDGE VOLUME AND MASS
An industry is holding 600 m3 of dense industrial sludge that has specific gravity 1.4.
Calculate the sludge mass.
Solution
Mass of 1 m3 of water = 1,000 kg
Mass of 1 m3 of sludge = 1.4(1,000 kg) = 1,400 kg
Sludge mass = (600 m3)(1400 kg/m3) = 840,000 kg = 840 metric tons
3.2
CUCUMBERS
Yesterday, in preparation for making pickles, I bought 100 kg of cucumbers that were 99%
water. Today the cucumbers are 98% water. Now I have only 50 kg of cucumbers. Can
this be right?
Solution
Yes, I have only 50 kg of cucumbers.
Basis = 100 kg cucumbers
Yesterday:
100 kg cucumbers = 99 kg water + 1 kg cucumber meat
Today:
1 kg cucumber meat + 49 kg water = 50 kg cucumbers at 98% moisture
3.3
SUPERFUND CREOSOTE SITE
An industrial site has 90,000 tons of soil containing an average of 400 mg/kg pentachlorophenol
(PCP). The maximum concentration allowed in the soil after cleanup is 10 ppm. What
mass of PCP must be removed from the soil?
37
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
Solution
(90x106 kg soil)(400 – 10 mg PCP/kg soil)(10–6 kg/mg) = 35,100 kg PCP
3.4
SLUDGE SOLIDS
Sludge is discharged from a settling tank at 80 m3/d, specific gravity 1.02, and 6% total
solids concentration. Calculate the mass flow rate of wet sludge and the dry solids.
Solution
1 m3 of water has a mass of 1,000 kg
1 m3 of sludge has a mass of 1.02(1,000) =1,020 kg
Mass flow rate of sludge pumped = (80 m3/d)(1,020 kg/m3) = 81,600 kg/d
Mass flow rate of dry sludge solids = (0.06)(81,600 kg/d) = 4,896 kg/d
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
3.5
Quantifying pollutants
PRIMARY SLUDGE
Sludge (a slurry of solids and water) from a wastewater primary sedimentation basin is
pumped at specific gravity 1.03 and 6% total solids concentration. The mass flow of solids
in the sludge is 70,000 kg/d. What is the volume flow rate?
Solution
(Q m3/d)(1,030 kg/m3)(6 kg solids/100 kg sludge) = 70,000 kg/day
Q = 1,133 m3/d
3.6
SLURRY DENSITY
One liter of slurry contains 200,000 mg of particles that have specific gravity = 2.5 in water
that has specific gravity = 1.0000. What is the specific gravity of the slurry?
Solution
Basis = 200,000 mg = 200 g solids
Sp. gravity = 2.5 = 2.5 g/mL
Define
MS = mass of solids (g)
MW = mass of water (g)
VS = volume of solids (mL) = MS/2.5
VW = volume of water
VT = VS + VW
MT = MS + MW
Then:
VS = MS/2.5 = (200 g)/(2.5 g/mL) = 40 mL
1 L solution = 40 mL solids + 960 mL water
Mass of solution = 2.5(40) + 1(960) = 1,060 g
Sp. gravity of solution = mass/volume = 1,060 g/1,000 mL = 1.06
3.7
SALT BY EVAPORATION
A salt solution (brine) is concentrated by evaporating the water in a salt pan, with a
condensing surface above it to gather the evaporated water. Suppose 1200 g of salt solution
is emptied into the pan. Once all the water is evaporated, 100 g of salt remains. (a) What
39
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
percent of the original solution was water? (b) Now suppose that 0.1 L of the evaporated
water was added back to the salt, to bring it to the desired concentration. How much water
remains to be used elsewhere?
Solution
Basis = 1,200 g salt solution
Tie variables = salt
a) Mass of salt in brine solution = 100 g
Mass of water in brine solution = 1,200 g – 100 g = 1,100 g
Brine solution is
100(100 g salt/1200 g solution) = 8.33% salt
100(1,100 g water/1,200 g solution) = 91.67% water
b) Mass of water added to dry salt = (0.1 L water)(1,000 g/L) = 100 g water
Useful water = 1,100 g – 100 g = 1,000 g = 1 L
3.8
ELECTROSTATIC PRECIPITATOR
Before the installation of an electrostatic precipitator (Figure P8.2) the stack gas of a power
generating station had a particulate solids concentration of 6 g/m3. The gas flow rate was
50 m3/s. The new precipitator removes 24,000 kg/d of particulates.
a) What is the emission rate of particulates, in kg/d, before and after initiating
pollution control?
b) What is the efficiency of the new precipitator?
c) Does the new system meet an emission standard of 0.7 g/m3?
40
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
High voltage
power supply
Rappers for
cleaning
Particulate­
laden flue
gas
Clean gas
to
smokestack
Electrostaticall
y charged
metal collection
plates
Dust
collection
hoppers
Figure P3.8 Electrostatic precipitator
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
Solution
a) Emission before installation
= (6 g/m3)(50 m3/s)(86,400 s/d)(10-3 kg/g) = 25,920 kg/d
Emission after installation = 25,920 kg/d – 24,000 kg/d = 1,920 kg/d
b) Efficiency = 100(24,000 kg/d)/(25,920 kg/d) = 92.6%
This is low for an electrostatic precipitator.
c) New emission rate
= (1,920 kg/d)/[50 m3/s)(86,400 s/d)(10-3 kg/g)] = 0.44 g/m3
Yes, it meets the emission standard.
3.9
SOLIDS MEASUREMENT
A slurry containing 50 kg of total solids (soluble + insoluble) and 100 kg of water is filtered
to remove 100% of the insoluble solids (Figure P3.9). The filtrate (the liquid that leaves
the filter) contains dissolved solids. The moist mass of solids captured by the filter is the
filter cake, or simply cake. Use the results in Table P3.9 (a) to verify that all material has
been accounted for and to calculate (b) the percent moisture in the filter cake and (c) the
solubility of the solid material in filtrate.
Figure P3.9
Material
Feed (kg)
Cake (kg)
Filtrate (kg)
Water
100
4.85
95.15
Solids
50
48.05
1.95
Table P3.9
42
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
Solution
a) Feed = Cake + Filtrate
100 kg water = 4.85 kg water + 95.15 kg water
50 kg solids = 48.05 kg solids + 1.95 kg solids
150 kg total = 52.9 kg cake + 97.1 kg filtrate
b) Cake moisture = Water/(Water + Solids) = 100(4.85 kg)/(52.9 kg) = 9.17%
c) Solubility = (1.95 kg dissolved solids)/(97.1 kg filtrate) = 0.0205 kg solids/kg filtrate
3.10 POPULATION EQUIVALENT - 1
The average per capita contribution of BOD5 to sewage is 0.08 kg/day. The average is 0.1
kg/capita-day for suspended solids and 400 L /capita-day for flow. Compute the average
concentration (mg/L) of BOD5 and suspended solids in municipal sewage.
Solution
For BOD: (80,000 mg BOD5/cap-d)/(400 L/cap-d) = 200 mg BOD5/L
For Suspended solids: (100,000 mg SS/cap-d)(400 L/cap-d) = 250 mg SS /L
3.11 INDUSTRIAL EQUIVALENT
An industry produces 16,000 m3/day of wastewater that has a BOD5 of 900 mg/L. Use
the values from the previous problem to express this BOD5 load as an equivalent number
of people. What is the equivalent in terms of flow?
Solution
BOD5 load = (16,000 m3/d)(0.9 kg/m3) = 17,778 kg/d of BOD5
The BOD5 load from this industry is equivalent to the BOD5 load from
(17,778 kg/d)/(0.08 kg/cap-d) = 222,222 people.
The flow from this industry is equivalent to
(16,000 m3/d)/(0.4 m3/cap-d) = 40,000 people
43
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
3.12 POPULATION EQUIVALENT - 2
Three wastewaters are to be combined for treatment:
City sewage: Ave. flow = 3,000,000 gal/d, BOD5 = 240 mg/L, SS = 190 mg/L
Poultry processing: Ave. kill = 20,000 birds/d, flow = 7000 gal/1,000 birds
BOD5 = 26 lb/1,000 birds, SS = 34 lb/1,000 birds
Dairy plant: Ave. flow = 100,000 gal/d, BOD5 = 600 mg/L, SS = 200 mg/L
a) Calculate the total load of BOD5 and SS, as pounds per day, and the average
wastewater concentrations for BOD and SS, as mg/L.
b) Calculate the population equivalent of the two industries. Population equivalent
values are 100 gal/cap-d, 0.17 lb BOD5/cap-d, and 0.21 lb SS/cap-d
Solution
Note: 1,000,000 gal/d = 1 mgd
a) Mass Loads
BOD5 = 8.34(3 mgd)(240 mg/L) + (20,000 birds/d)(26 lb/1,000 birds)
+ 8.34(0.1 mgd)(600 mg/L)
= 6,005 lb/d + 520 lb/d + 500 lb/d = 7,025 lb/d
.
44
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
SS = 8.34(3)(190) + (20,000)(34/1,000) + 8.34(0.1)(200)
= 4,753 lb/d + 680 lb/d + 167 lb/d = 5,802 lb/d
Flow from poultry processing = (20,000 birds/d)(7,000 gal/1,000 birds)
= 140,000 gal/d = 0.14 mgd
Total flow = 3 mgd + 0.14 mgd + 0.1 mgd = 3.24 mgd
Average concentrations
BOD5 = (7,025 lb/d)/[(8.34)(3.24 mgd)] = 260 mg/L
SS = (5,802 lb/d)/[(8.34)(3.24 mgd)] = 215 mg/L
b) Population equivalents
Sample calculation for city wastewater.
Flow: (3,000,000 gal/d)/(100 gal/cap-d) = 30,000 persons
BOD5: (6,005 lb/d)/(0.17 lb/cap-d) = 35,323 persons
SS: (4,753 lb/d)/(0.21 lb/cap-d) = 22,633 persons
Complete results are in Table S3.13.
PE (persons)
Flow
BOD5
SS
(gal/d)
(lb/d)
(lb/d)
Flow
BOD5
SS
3,000,000
6,005
4,753
30,000
35,323
22,633
Poultry plant
140,000
520
680
1,400
3,059
3,238
Dairy
100,000
500
167
1,000
2,941
795
32,400
41,322
26,666
City
Totals
Table S3.13
3.13 SOUR MASH STILLAGE
A whiskey making operation produces 220,000 gal/d of waste stillage that has a 6.3% solids
concentration by weight and a Biological Oxygen Demand (BOD) of 40,000 mg/L. (BOD
is one measure of the pollution potential of a waste. For reference, municipal sewage has
a BOD of about 200 mg/L.) The nutritionally rich stillage had traditionally been sold for
animal feed, but as production increased this became unworkable. The disposal was discussed
in Food Processing (June 1973).
They said, “If the combined effluent … were to be directly discharged to a conventional
waste system, the system would have to be approximately the size of that needed for a
community the size of Louisville (450,000 people). What assumptions are imbedded in this
statement? It certainly gives a vivid picture of the problem. Is it a fair and valid picture?
45
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
Solution
Population equivalents:
Flow = (220,000 gal/d)/(100 gal/capita-d) = 22,000 people.
BOD = 8.34(0.22 mgd)(40,000 ppm)/(0.17 lb/cap-d) = 432,000 people
Suspended solids = 8.34(0.22 mgd)(63,000 ppm)/(0.22 lb/cap-d) = 525,400 people
The statement is correct on the basis of BOD and suspended solids loads. It is too high in terms of the
flow equivalent. It is often true that the different ways of expressing the population equivalent (flow,
DO, SS, etc.) will give different impressions.
Would the treatment plant need to be as large as Louisville’s? No. The statement assumes that the size
of the treatment system is proportional to the BOD and solids loading and that the same treatment
technology would be used. Neither assumption is entirely correct. Some parts of the treatment system
are more dependent on the flow rate, which is not too large. Some will depend on how easy it is
to remove the solids, and whether BOD is removed along with the solids. The processes that would
remove the solids (screens and settling tanks) would be sized mainly on the basis of flow rate. Sludge
processing processes will be roughly as large as Louisville’s. The size of the biological treatment system
is indeterminate. It will depend on how much of the influent BOD is removed along with the solids in
primary settling tanks.
A note on reasonable approximations:
Note on BOD - The 40,000 mg/L BOD has been used as though it is 40,000 ppm by weight. This is
correct when the specific weight of water 8.34 lb/gal (sp. gr. = 1.00). This is probably wrong, because the
solids content of the wastewater is very high. Suppose the real value is 8.7 lb/gal (which is undoubtedly
far too high). The change in population equivalent would be about 424,000 people instead of 432,000.
This is not important since population equivalent is used only to make a rough comparison.
Note on solids:
The 6.3% solids concentration is by weight. 6.3% = 63,000 ppm. The correct calculation would use
the true weight of the wastewater, which probably will be slightly more than 8.34 lb/gal because of
the high solids content.
3.14 CLARIFIED EFFLUENT
The clarified effluent from a wastewater treatment plant has an average daily flow of 42
million gallons per day (mgd) with an average monthly BOD5 concentration of 1.5 mg/L,
suspended solids concentration of 2 mg/L, ammonia concentration of 0.1 mg/L, and
phosphorus concentration of 0.4 mg/L. Calculate the average daily mass discharge of each
constituent.
46
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
Solution
Basis = One day’s flow
Flow = (42,000,000 gal/d)(8.34 lb/gal) = 350.28x106 lb/d
Interpret concentrations as mg/L = ppm = (1 lb/1,000,000 lb)
BOD5 = (1.5 lb BOD5/1,000,000 lb wastewater)(350,280,000 lb wastewater)
= 525.4 lb/day
SHORT CUT CALCULATION:
lb/d = 8.35(mgd)(mg/L)
SS = 8.34(42 mgd)(2 mg/L) = 700.6 lb/d
Ammonia = 8.34(42 mgd)(0.1 mg/L) = 35.0 lb/d
Phosphorus = 8.34(42 mgd)(0.4 mg/L) =140.1 lb/d
3.15 CADMIUM IN SLUDGE
How much cadmium, Cd, is added to a farm field if liquid sludge that is 10% solids (by
weight) is incorporated into soil at a rate of 150 m3/y? The density of the liquid sludge is
1.20 kg/m3. The measured concentration of cadmium in the sludge is 8 ppm, based on
the dry sludge solids.
Solution
Basis = One year’s application of sludge
8 ppm = 8 mg Cd/1,000,000 mg dry sludge solids = 8 mg Cd/kg dry solids
Dry solids in the sludge
= (150 m3 sludge)(1,200 kg/m3)(0.10 kg dry solids/kg sludge)
= 18,000 kg dry solids
Cd in the dry solids = (18,000 kg solids)(8 mg Cd/kg solids)
= 144,000 mg Cd = 0.144 kg Cd
3.16 METALS IN SOLID WASTE
A solid waste that is 80% organic matter is incinerated at a rate of 500 dry T/d. The waste
contains 430 ppm lead on a dry weight basis, 20% of the ash residue is collected as fly
ash, and 80% leaves the incinerator as bottom ash. The lead concentration in the fly ash is
2,000 ppm. What is the amount and concentration of lead in the incinerator bottom ash?
47
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
Solution
Incinerator input = 500 T dry solids per day
Ash input = (0.2)(500) = 100 T ash/d
Lead input = (500 T/d)(1,000 kg/T)(430 kg Pb/106 kg ash) = 215 kg Pb /d
Outputs:
Organic matter destroyed: (0.8)(500 T/d) = 400 T/d destroyed
Bottom ash: (0.8)(100 T ash/d) = 80 T bottom ash/d
Fly ash: (0.2)(100 T ash/d) = 20 T fly ash/d
Lead content of fly ash: 2,000 ppm = 0.2% by weight.
(0.002)(20 T Pb/d) = 0.04 T Pb/d = 40 kg Pb/d
Lead in bottom ash: 215 kg Pb/d– 40 kg Pb/d = 175 kg Pb/d
Concentration in bottom ash: 100(175 kg Pb/d)/(80,000 kg ash/d)
= 0.22% = 2,200 ppm by weight.
3.17 DUST FALL
The measurements given in Table P3.17 were obtained by placing wide-mouth glass jars
containing some water at five locations for a two-week sampling period. The jars had an
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
8 cm diameter mouth opening. At the end of the two weeks, the dust-water mixture was
filtered and the filter with the collected dust was dried. The dry weight of dust collected
can be calculated from the data. For the sampled area, what is the average dust fall in g/
m2 per month?
Weight (g)
Location
Dry filter +
Dry filter
Dry dust
Dry dust
1
2.1952
2.2146
0.0194
2
1.8335
1.8555
0.0220
3
1.9824
2.0052
0.0228
4
2.7495
2.7710
0.0215
Table P3.17
Solution
Total mass of dust collected at the four locations = 0.0857 g
Average mass of dust collected = (0.0857 g)/4 = 0.0214 g
Area of collection jar = πD2/4 = 3.1416(0.08 m)2/4 = 0.00503 m2
Average dust fall in 2 weeks: (0.0214 g)/(0.00503 m2) = 4.254 g/m2
Average dust fall per month: 2(4.254 g/m2) = 8.508 g/m2
3.18 LEAD IN BAGHOUSE DUST
The 800,000 m3/d gaseous emission from a secondary brass (zinc-alloyed copper) and
bronze (tin-alloyed copper) processes contains 50,000 µg/m3 of dust that is from 1% to
12% lead by weight. The dust will be removed in a bag house filter (Figure P3.18) and it
is proposed that the collected dust should be processed to recover the lead. Assuming 100%
dust removal, what amount of dust will be collected and what amount of lead is available
for reclamation?
49
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
Clean air
Compressed air or
mechanical shaker
Fabric
filter
(to clean filter)
Dust-laden
air in
Separated dust
out
Figure P3.18 Baghouse filter
Solution
Basis = One day’s gas emission
Mass of dust = (800,000 m3/d)(50,000 µg/m3)(10–9 µg/kg) = 40 kg/d
Mass of lead ranges from (0.01)(40 kg/d) = 0.4 kg/d to (0.12)(40 kg/d) = 4.8 kg/d
3.19 PRINTING PLANT AIR EMISSIONS
Volatile organic chemical (VOC) emissions from an offset printing plant are carried in
182,000 m3/h of air that has a VOC concentration of 25.9 mg/m3. The plant operates 10
hours per day, 250 days per year. What is the annual VOC emission?
Solution
(182,000 m3/h)(25.9 mg/m3)(10 h/d)(250 d/y)(10–6 mg/kg) = 11,800 kg/y
50
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
3.20 GAS TANK CAPACITY
A 25,000 L tank contains gas at 27°C and 50 atmospheres. The molar mass (MM) of the
gas is 79.9 g/mol. What is the mass of gas in the tank?
Solution
From the ideal gas law: PV = nRT
T = 273°C + 27°C = 300K
(50 atm)(25,000 L) = n (0.08205 L atm/mole K)(300K)
n = 50,782 moles
Molar mass = 79.9 g/mol
Mass of gas = (79.9 g/mol)(50,782 mol)/(1,000 g/kg) = 4,057 kg
3.21 GAS VOLUMES
Use the data in Figure P3.21 to calculate the actual volume of gas held in each tank.
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Oxygen
Amount
Molar mass
Temperature
Pressure
Quantifying pollutants
Combustion
gas
1 g-mol
32 g
50°C
1 atm
Chlorine
gas
1 g-mol
28 g
60°C
1.2 atm
1 g-mol
71 g
20°C
0.9 atm
Figure P3.21 Gas volumes
Solution
The volume of 1 g mole of gas is 22.41 L at standard conditions 0°C and 1 atm.
The volume change with change in temperature and pressure is determined using the Ideal Gas Law.
PV
T
PSVS
TS
V
or
VS
PST
PTS
where the subscript S indicates standard conditions
Oxygen
VO2 = (22.41 L)
(1 atm)(273°C + 50°C)
= 26.51 L
(1 atm)(273°C)
Combustion gas
VCG= (22.41 L)
(1 atm)(273°C + 60°C)
= 22.78 L
(1.2 atm)(273°C)
Chlorine VCl = (22.41 L)
(1 atm)(273°C + 20°C)
= 26.72 L
(0.9 atm)(273°C)
2
3.22 VOLUME CONCENTRATION
The concentration of a gaseous pollutant in air is 80 ppmv. The molecular mass of the
pollutant is 16 g/mole. The ambient air standard is 55 µg/L (at standard conditions). Is
the standard being satisfied?
Solution
Convert 80 ppmv to mg/m3
mg/m3 =
(ppmv)(MM)
(80 ppmv)(16 g/mol)
=
= 57.1 mg/m3
22.41 L/mol
22.41 L/mol
The standard is violated.
52
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
3.23 MIXTURE OF IDEAL GASES
The contents of four rigid flasks of one-liter volume are to be mixed together to prepare
a calibration gas mixture. The flasks contain sulfur dioxide (SO2) at 20 mm Hg pressure,
nitrogen (N2) at 180 mm Hg, methane (CH4) at 500 mm Hg, and carbon monoxide (CO)
at 60 mm Hg. What will be the final pressure after these four flasks are combined in a
one-liter flask?
Solution
The partial pressures of ideal gases add to give the total pressure.
PT
PSO2 PN2 PCH4 PCO
PT = 20 + 180 + 500 + 60 = 760 mm Hg
3.24 GAS ADJUSTED FOR MOISTURE
A gaseous emission has a pollutant concentration of Cwet basis = 25 ppmv and 15 volume percent
of water vapor. Adjust the measured ‘wet basis’ concentration to a ‘dry basis’ concentration.
Solution
CDry Basis
CWet Basis
0.25
=
= 29.4 ppmv
1w
1 - 0.15
where w = 0.15 = fraction, by volume, of water vapor in the emitted gas
3.25 DUST POLLUTION
An air pollution guideline stipulates not more than 20 tons of particulate fallout per square
mile per month (tons/mi2-month) for any area. An industrial area has a dustfall of 165 mg/
m2-day. Is the location in compliance?
Solution
Conversion factor: 1 ton/mi2-month = 11.68 mg/m2-d
(165 mg/m2-d)(1 ton/mi2-month/11.68 mg/m2-d) = 14.13 ton/mi2-month
Yes, the location is in compliance.
53
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
3.26 WASTE GAS COMPOSITION
A gaseous mixture, with total mass of 1,000 kg, is 5% benzene, 71% nitrogen, and 24%
oxygen. The ambient temperature and pressure are 25°C and 760 mm Hg. The molecular
mass of benzene (C6H6) is 78 g/g-mol. To reduce the benzene concentration to 1 mg/m3,
how much air, in m3, must be added to every m3 of the original mixture?
Solution
Separate gas masses can be calculated from their individual mass fractions.
Gas component = (mass fraction)(total gas mass)
Benzene = 0.05(1,000,000 g) = 50,000 g
Nitrogen = 0.71(1,000,000 g) = 710,000 g
Oxygen = 0.24(1,000,000 g) = 240,000 g
Molecular masses
C6H6 = 78 g/g mol
N2 = 28 g/g mol
O2 = 32 g/g mol
54
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
The number of gram moles of each gas is
C6H6 = (50,000 g)/(78 g/g-mol) = 641 g-mol
N2 = (710,000 g)/(28 g/g mol) = 25,357 g-mol
O2 = (240,000 g)/(32 g/g mol) = 7,500 g-mol
Total = 33,498 g-mol
Total volume of gas is calculated using 1 g-mol = 24.465 L at 25°C and 760 mm Hg.
Volume = (24.465 L/g-mol)(33,498 g-mol) = 819,529 L = 819.5 m3
Concentrations = (mass of gas)/(total volume of gas)
C6H6 = (50,000 g)/(819.5 m3) = 61.0 g/m3
N2 = (710,000 g)/(819.5 m3) = 866.4 g/m3
O2 = (240,000 g)/(819.5 m3) = 292.8 g/m3
The required volume of dilution air will be very close to 61 m3.
Material balance on benzene at the mixing chamber.
Initial benzene concentration is = 61.0 g/m3
Contaminated gas volume = 819.5 m3
Let V = volume of clean dilution air
(819.5 m3)(61.0 g/m3) + (V)(0 g/m3) = (819.5 m3 + V)(1 g/m3)
Q = 49,989.5 m3 – 819.5 m3 = 49,170 m3
Dilution ratio = (49,170 m3)/(819.5 m3) = 60
3.27 TOLUENE EMISSIONS
A printing company is limited to emitting 3 kg/h of toluene. The airflow that carries the
toluene is 20,000 m3/h at STP. The density of toluene is 4.12 kg/m3. What is the allowable
concentration (ppmv) of toluene in the emitted air?
Solution
Define C = allowable concentration of toluene (ppmv or m3/106 m3)
Allowed volume of toluene = (3 kg/h)/4.12 kg/m3) = 0.728 m3/h
(20,000 m3/h)(C m3/1,000,000 m3) = 0.728 m3/h.
C =36.4 ppmv
55
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
Alternate solution
mg/m3 =
(ppmv)(MM)
22.41 L/mol
or re-arranging
ppmv =
(mg/m3 )(22.41 L/mol)
MM
Allowable concentration of toluene
= (3 kg/h)/(20,000 m3/h) = 0.00015 kg/m3 = 150 mg/m3
Molar mass of toluene = 92.14 g/mol
ppmv = (150 mg/m3)(22.41 L/mol)/(92.14 g/mol) = 36.5 ppmv
3.28 CADMIUM IN WASTEWATER
A municipal wastewater treatment plant receives cadmium (Cd) in these amounts:
Municipal wastewater = 10,000 m3/d containing 0.5 µg/L Cd
Industrial wastewater = 2,000 m3/d containing 2 µg/L Cd
The plant will remove 75% of influent cadmium. The allowed concentration in the stream that
receives the effluent is 0.025 µg/L Cd. The flow in the receiving stream used for calculating
compliance with this concentration is 80,000 m3/d. The concentration of Cd upstream of the
discharge is zero. (a) What is the amount of Cd in the municipal wastewater? (b) What is
the amount of Cd in the industrial wastewater? (c) What is the concentration in the stream?
Solution
a) Municipal Cd = (10,000 m3/d)(1,000 L/m3)(0.5 µg/L) = 5,000,000 µg/d = 0.005 kg/d
b) Industrial Cd = (2,000 m3/d)(1,000 L/m3)(2 µg/L) = 4,000,000 µg/d = 0.004 kg/d
c) Cd total into treatment plant = 9,000,000 µg/d = 0.009 kg/d
Cd pass-through to stream = 0.25(9,000,000 µg/L)
= 2,250,000 µg/d = 0.00225 kg/d
After mixing in stream
Stream + Effluent flow = 80,000,000 L/d + 10,000,000 L/d + 2,000,000 L/d
= 92,000,000 L/d
Mass of Cd added to stream = 2,250,000 µg/d
Concentration of Cd in stream = (2,250,000 µg/d)/(92,000,000 L/d) = 0.0245 µg/L
This is acceptable.
3.29 LIQUID ALUM
A solution of aluminum sulfate (alum) is added to a flow of 1,900 m3/d to obtain a
concentration of 18 mg/L. The liquid alum is 0.05% alum by mass. (a) Calculate how
much alum per day is added to the water? (b) Calculate how much alum solution per day
is added to the water. (c) Calculate the volume of water used to make the solution.
56
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
Solution
Define
MW = Mass of water (kg/d)
MA = Mass of alum (kg/d)
MS = Mass of alum solution (kg/d)
a) MA = (1,900 m3/d)(18 mg/L)(1,000 L/m3)/(1,000,000 mg/kg) = 34.2 kg/d
b) MA = 0.05 MS = 34.2 kg/d
MS = 684 kg/d
c) MW = MS – MA = 684 kg/d - 34.2 kg/d = 650 kg
VW = (650 kg/d)/(1,000 kg/m3) = 0.65 m3/d
3.30 ION EXCHANGE WATER SOFTENING
Hardness in water is caused by calcium (and magnesium). Removing calcium is “softening”
the water. An ion exchange water softener, Figure P3.31, works by exchanging sodium ions
from a resin with calcium ions in the feed water. The exchange is 2 ions of sodium (Na+)
for one ion of calcium (Ca2+). The exchange is 2 plus charges from 2 sodium ions for 2
plus charges from one calcium ion. The water to be treated has a total dissolved solids
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
concentration of 600 mg/L and a calcium concentration of 150 mg/L. Assume that all the
calcium will be removed. Base your calculation on any convenient volume of water, say
1,000 L. What will be the total dissolved solids concentration of the water that has been
softened by this process?
Hard water
Ca2+, Mg2+
Waste brine
NaCl, CaCl2, MgCl2
Upflow of brine
expands the resin
bed and creates
turbulence
Softened water
Na+
Service cycle
NaCl brine
Regeneration cycle
Figure P3.30 Ion exchange softening
Solution
TDS = 600 mg/L
Ca2+ = 150 mg/L
Define: TDSS = total dissolved solids concentration of softened water; i.e. water that contains no calcium.
With the Ca2+ removed, the TDSS = 600 mg/L – 150 mg/L = 450 mg/L
Replace Ca2+ with Na+ at the ratio of 2 ions of Na+ for 1 ion of Ca2+
Notice that the number of positive charges balance. Two are removed with Ca2+ and 2 are replaced
with Na+.
Molar masses: Na+ = 23 g/mol and Ca2+ = 40 g/mol
2 ions of Na+ = 46 and 1 ion of Ca2+ = 40.
Sodium added to replace the Ca2+ = (150 mg/L)(46/40) = 172 mg/L
Softened water TDS = TDSS + 172 mg Na/L = 450 mg/L + 172 mg/L = 622 mg/L TDS
The water is ‘soft’ because the calcium is gone, but the total mineral content has increased.
Alternate process – Hydrogen Exchange (not alternate solution)
1 ion of Ca2+ can be replaced with 2 ions of hydrogen (H+)
Molar masses: H = 1 g/mol and Ca = 40 g/mol
58
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
or, 2 ions of H+ = 2 and 1 ion of Ca2+ = 40
Hydrogen added to replace the Ca2+ = (150 mg/L)(2/40) = 7.5 mg/L
Softened water TDS = 450 mg/L TDSS + 7.5 mg/L Na+ = 452.5 mg/L TDS
Actually the hydrogen exchange scheme will remove all the cations (ions with positive charges). The
clever engineer will say, ‘Why not replace all the negative ions (the anions) with OH–. Good idea! The
added H+ and OH– combine to make water. All the minerals are removed and what we added to remove
them is, after all is said and done, more water. This process is called demineralization.
3.31 MASS FLOW RATE CALCULATION
You have been asked to calculate the mass discharge from an industrial operation that is
losing a valuable material (identified only as X) to the plant drainage system. Measurements
over a critical 4-hour period reveal this pattern of waste flow and X concentration in Figure
P3.21.
-E
········ ....··......· ..··........· ..
20 ------··· ···
_ 400 ··············--- ............. ............. ..
vi'
2
3:
0
u....
1 5 .............. ............ · •,------,.............. ·
___.
§ 200
10 ........................................... ,___
5 ..............1------i.............. ..............
300
",i:,
�
+-'
C
QJ
u
C
0
.
0 '----..i....___....____,___.....i.
4
2
1
3
0
Time (hours)
u
·············· ............................ ------.
_
........... . .. ............. ,.,___
___...............
.
100-·
-· ·
·-· ··
··-·
··· ............................ .............. .
0 ,.____......___......____...___......_
4
2
1
3
0
Time (hours)
Figure P3.31 Flow and concentration patterns
You calculate:
Average flow = (20 L/s+5 L/s +15 L/s+10 L/s)/4 = 50 L/s/4 = 12.5 L/s.
Average concentration = (100 mg/L+400 mg/L+200 mg/L+300 mg/L)
= 1,000 mg/L/(4) = 250 mg/L
Estimated mass discharge = (12.5 L/s)(250 mg/L) = 3,125 mg/s
Just as you finish you notice that (10 L/s)(300 mg/L)
= 3,000 mg/s was discharged in period 4 alone.
And, the same is true in period 3. What is wrong?
Solution
The calculation must be done period-by-period to give proper weight to the hourly concentrations.
This is called making a flow weighted concentration and is commonly used in sampling.
59
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
Estimated mass average concentration is a flow weighted concentration
= Σ(flow)(concentration)/Σflows
= [(20 L/s)(100 mg/L) + (5)(400) + (15)(200) + (10)(300)]/(20 L/s + 5 + 15 +10)
= 200 mg/L
The Estimated Mass Flow = True Mass Flow = (12.5 L/s)(200 mg/L) = 2,500 mg/s
Check: Calculate the mass flow period by period
True Mass Flow = [(20 L/s)(100 mg/L) + (5)(400) + (15)(200) + (10)(300)]/4
= 2,500 mg/s
Composite sample with equal volumes per period
Making a composite sample with equal volumes, V, at each sampling interval is equivalent to making
the calculation in the problem statement. When the laboratory’s measurement of the concentration
of the composite sample is multiplied times the average flow rate, the estimated mass discharge rate
will be wrong.
The laboratory will measure the composite sample concentration as
[(100 mg/L) V + 400 V + 200 V + 300 V)]/(4 V) = 250 mg/L
Mass discharge calculated using this concentration and the average flow rate is
= (250 mg/L)(12.5 L/s) = 3,125 g/h, which is too high.
3.32 SOIL CONTAMINATION IN NIGERIAN LANDFILL
Soil samples were analyzed from four locations in a Nigerian landfill and from one control
location (believed to be unaffected by the landfill). Four samples were analyzed at each
location. The results (averages) given in Table P3.32 are physicochemical properties of soil
from a dumpsite in Kana, Nigeria. The measurements are on a dry mass basis. (a) Calculate
the Cd concentration in the natural (i.e., moist) soil for the North sample. (b) Write a short
report (not more than 1 page) about this study. You might, for example, have an opinion
about whether this is a good location for a dumpsite, based on the soil quality. You might
want to discuss the implications for groundwater contamination and for human health.
60
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Characteristic
Quantifying pollutants
North
South
East
West
Control
Moisture (%)
14.7
25.2
21.5
60.5
30.2
CEC (mmol/kg)
0.51
1.20
1.34
2.4
0.61
pH
7.3
7.6
7.6
7.2
7.8
Organic carbon (%)
1.73
1.49
1.79
1.38
0.39
Cd (mg/kg)
1.37
1.14
1.34
1.71
ND
Cr (mg/kg)
6.94
5.76
9.30
6.81
4.06
Ni (mg/kg)
0.39
0.58
0.99
1.24
0.06
Pb (mg/kg)
246
54.9
284
91.9
18.3
Sand (%)
81.8
80.4
78.2
79.4
82.0
Silt (%)
13.4
13.2
14.4
14.2
12.8
Clay (%)
4.80
6.4
7.4
6.4
5.2
CEC = Cation exchange capacity, a measure a soil’s capacity to adsorb cations.
ND = not detected
Source: Anake, WU, et al 2009, Bull. Chem. Soc. Ethiopia, vol. 23, pp281-289.
Table P3.32 Physicochemical properties of soil from a dumpsite in Kana, Nigeria
Solution
a) At 14.7% moisture, there is 0.853 kg dry soil per kg of natural soil
For Cd: (1.37 mg Cd/kg dry soil)(0.853 kg dry/1 kg natural soil)
= 1.17 mg Cd/kg natural soil
b) This is an open-ended discussion question that is meant to provoke discussion. No
solution is provided.
3.33 HEAVY METALS IN COMPOST
A European study provides these data (Table P3.33) for heavy metals in compost that was
made using municipal solid waste and sewage sludge. The compost is 60% moisture. What
are the concentrations of copper (Cu) and lead (Pb) in compost from Germany and France
per ton (1,000 kg) of moist compost?
61
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Country
Quantifying pollutants
Ranges and mean values [mg/kg dry matter]
Cd
Cr
Cu
Hg
Ni
Pb
Zn
0.4–3
16-27
39-64
0.3-3
9-80
13-221
142–2,000
Germany
1.4
46
274
1
23
63
809
Switzerland
1.7
74.1
341.1
1.7
31.9
94.5
929.3
Spain
4
162
258
-
43
212
955
France
2.9
58.8
309
3
31.9
106.7
754.2
Austria
1.3
40
200
0.8
25
55
900
90th percentile
2.2
80
320
2.9
45
110
1,400
EU range of means
Source: Amlinger, F, et.al 2004, Heavy Metals and Organic Compounds from Wastes Used as
Organic Fertilizers, Final Report, Technical Office for Agriculture, Vienna, Australia
Table P3.33 Heavy metals in European compost
Solution
Basis: 1,000 kg moist compost
Tie variable = dry solids
Mass dry solids = 0.4(1,000 kg) = 400 kg
Mass water = 1,000 kg – 400 kg = 600 kg
German compost:
Cu = (274 mg Cu/kg dry compost)(400 kg dry compost/T moist compost)
= 109,600 mg Cu/T moist compost = 0.110 kg Cu/T moist compost
Pb = (63 mg Pb/kg dry compost)(400 kg dry compost/T moist compost)
= 25,200 mg Pb/T moist compost = 0.0252 kg Pb/T moist compost
French compost:
Cu = (309 mg Cu/kg dry compost)(400 kg dry compost/T moist compost)
= 123,600 mg Cu/T moist compost = 0.124 kg Cu/T moist compost
Pb = (106.7 mg Pb/kg dry compost)(400 kg dry compost/T moist compost)
= 42,680 mg Pb/T moist compost = 0.0427 kg Pb/T moist compost
62
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
3.34 SHREDDED MUNICIPAL BULKY WASTE
In many Japanese cities, bulky waste is shredded and separated into burnable and inert
fractions. One way to do this is air classification – drop the shredded material into a rising
column of air and blow the lighter particles out the top while heavy particles settle to the
bottom. Experiments on air classification gave the data in Table P3.34. (The settling velocity
in air is proportional to density but it also depends on particle size and shape. That is why
the settling velocity of metal is not ten times more than the settling velocity of foam plastics,
even though the density is ten times greater.) Calculate the wet and dry composition (mass
fraction) of the mixture after foam plastic, glass and metal have been removed.
Material
Foam
plastics
Paper
Wood
Plastics
Rubber
Glass
Metals
Density (kg/m3)
0.5
0.15
0.42
1.11
1.15
2.50
5.03
Settling velocity
in air (m/s)
2.3
2.4
4.8
7.7
9.4
13.9
14.6
Wet mass (kg)
2
50
10
15
3
10
10
Moisture
content (%)
30
16
10
3
0
3
3
Table P3.34 Air classification data for shredded bulky waste
Solution
For convenience, start with 100 kg of mixed wet material.
After air classification, the wet mass of sorted bulky waste (foam plastic, glass and metal removed) is
100 kg – 2 kg – 10kg – 10 kg = 78 kg
Dry mass without foam plastic, glass and metals is
50 kg (1 -0.16) +10 kg (1-0.1) + 15 kg (1 – 0.03) + 3 kg (1 – 0)
= 42 kg + 9 kg + 14.55 kg + 3 kg = 68.55 kg
Calculations for paper
Wet mass fraction = 50 kg wet paper/78 kg wet material
= 0.64 kg wet paper/kg wet material
Mass of dry material = ((1 – 0.16) kg dry paper/kg wet paper))(50 kg wet paper)
= 42 kg dry paper
Dry mass fraction = 42 kg dry paper/68.55 kg dry material
= 0.612 dry paper/kg dry material
63
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
Calculations for paper, wood, plastics, and rubber are summarized in Table S3.34.
Material
Foam
plastics
Paper
Wood
Plastics
Rubber
Glass
Metals
Density (kg/m3)
0.5
0.15
0.42
1.11
1.15
2.50
5.03
Settling velocity in
air (m/s)
2.3
2.4
4.8
7.7
9.4
13.9
14.6
Wet mass (kg)
2
50
10
15
3
10
10
Moisture content (%)
30
16
10
3
0
3
3
Wet mass = 78 kg
0
50
10
15
3
0
0
Dry mass = 68.55 kg
0
42
9
14.55
3
0
0
Wet mass fractions
0
0.64
0.13
0.19
0.04
0
0
Dry mass fractions
0
0.61
0.13
0.21
0.05
0
0
Sorted bulky waste
Table S3.34
3.35 MIXED SOLID WASTE
The composition of a mixed solids waste is given in Table P3.35. What mass of waste is
combustible? What is the combustible fraction of each component and of the total mixed
waste?
Wet mass
Moisture
Ash
(kg)
(%)
(kg)
Paper
40
6
2.5
Cardboard
10
5
0.5
Plastics
15
1
1.5
Wood
10
20
0.15
Material
Table P3.35 Composition of a mixed solid waste
64
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
Solution
The table below gives the complete solution.
These are the definitions and calculations for paper:
Dry solids = (wet mass)(1 – moisture fraction)
Dry paper = (40 kg)(1 – 0.06) = 37.6 kg
Combustible mass = (dry mass)(1 – ash fraction)
Combustible paper = (37.6 kg)(1 – 0.066) = 35.1 kg
Combustible paper fraction = (Combustible paper solids)/(Total combustible solids)
= 35.1 kg/65.3 kg = 0.54
Wet
Moisture
mass
content
(kg)
(%)
Paper
40
6
2.5
37.6
Cardboard
10
5
0.5
Plastics
15
1
Wood
10
20
Total
75
Material
Ash
Ash
Dry
(% of
Combustible
Combustible
dry
solids (kg)
Fraction (%)
6.6
35.1
54
9.5
5.3
9.0
14
1.5
14.85
10.1
13.4
20
0.15
8.0
1.9
7.9
12
(kg)
solids
(kg)
solids)
69.95
65.3
Table S3.35
3.36 AIR POLLUTION UNIT CONVERSIONS
Table P3.36 gives concentrations for SO2, NO2 and CO. Convert the concentrations from
mass/volume (µg/m3 or mg/m3) to volume/volume (ppmv), or vice versa. Note: 1 mg =
1,000 µg; T = ambient temperature (K), MM = molar mass (g/mole), P = pressure (atm)
Pollutant
MM
T (K)
P (atm)
Concentration
SO2
64
298
1.0
0.350 µg/m3
NO2
46
293
1.5
0.650 µg/m3
CO
28
298
1.0
0.15 ppmv
Table P3.36
65
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
Solution
For standard conditions
mg pollutant
§ MM ·
= ppmv ¨
¸
m3 mixture
© 22.41 ¹
Concentration in ppmv is the same for all P and T.
The mass concentration (mg/m3) must be adjusted for different temperature and pressure because the
volume changes while the mass does not.
§ mg ·
¨ 3¸
© m ¹TP
§ mg ·
¨ 3¸
© m ¹STP
§ PTS ·
¨¨ P T ¸¸
© S ¹
and
§ mg ·
¨ 3¸
© m ¹STP
§ mg ·
¨ 3¸
© m ¹TP
§ PST ·
¨¨ PT ¸¸
© S¹
§ μg ·
¨ 3¸
© m ¹TP
§ μg ·
¨ 3¸
© m ¹STP
§ PTS
¨¨
© PST
and
§ μg ·
¨ 3¸
© m ¹STP
§ μg ·
¨ 3¸
© m ¹TP
§ PST
¨¨
© PTS
·
¸¸
¹
·
¸¸
¹
SO2: Adjust to T = 298 K, then convert to ppmv
μg · ª (1 atm)(298 K) º
§
3
¨ 0.350
¸«
» = 382 μg/m
m3 ¹ ¬ (1 atm)(273 K) ¼
©
ppmv = 22.41(382.0 mg/m3)/64 = 133.8 ppmv SO2
NO2: Adjust to P = 1.5 atm and T = 293 K, then convert to ppmv
μg · ª (1 atm)(293 K) º
§
3
¨ 0.650
¸«
» = 0.465 μg/m
m3 ¹ ¬ (1.5 atm)(273 K) ¼
©
ppmv = 22.41(465.1 mg/m3)/46 = 226.6 ppmv NO2
CO: Convert to mg/m3, then adjust to T = 298K and P = 1 atm
mg CO/m3 = (0.15 ppmv)(28)/22.41 = 0.1874 mg CO/m3
(0.15 ppmv)(28 g/mol)
= 0.1874 mg CO/m3
22.41 L/m3
§ mg ·
¨ 3¸
© m ¹TP
§ mg ·
¨ 3¸
© m ¹STP
§ PTS
¨¨ P T
© S
at 0°C and 1 atm
·
mg CO · ª (1 atm)(298 K) º
§
3
¸¸ = ©¨ 0.1874 m3 ¹¸ « (1 atm)(273 K) » = 0.205 mg CO/m
¬
¼
¹
3.37 COAL-FIRED BOILER EXHAUST GAS
The composition of wet exhaust from a coal-fired boiler is 70% nitrogen, 10% oxygen,
10% carbon dioxide and 10 % water vapor. These are volume percentages. The volume
of exhaust gas is 323 Nm3/GJ of fuel burned. (N indicates Normal conditions = 0°C and
1 atm. For convenience we will omit the N and understand m3 is a Normal conditions.)
66
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
a)
b)
c)
d)
Quantifying pollutants
What are the mole fractions for each gas?
How many cubic meters each gas result from burning 50,000 GJ/d of fuel?
How many kilograms of each gas result from burning 50,000 GJ/d of fuel?
What is the volume percent of CO2 in dry exhaust gas?
Solution
a) Mole fraction = volume fraction; so, N2 = 0.7, O2 = 0.1, CO2 = 0.1, H2O = 0.1
b) Burn 50,000 GJ/d of fuel = (323 Nm3/GJ)(50,000 GJ) = 16,150,000 m3/d
Volume of N2 = 0.7(16,150,000 m3) = 11,305,000 m3/d
Volume of O2 = 0.1(16,150,000 m3) = 1,615,000 m3/d
Volume of CO2 = 0.1(16,150,000 m3) = 1,615,000 m3d
Volume O2 in H2O = 0.1(16,150,000 m3) = 1,615,000 m3/d
c) Mass = (number of moles)(molar mass)
Volume of 1 mole of gas at 1 atm and 0°C = 22.41 L/g-mol = 22.41 m3/kg-mol
Molar masses:
N2 = 28 kg/kg-mol, O2 = 32 kg/kg-mol
CO2 = 44 kg/kg-mol, H2O = 18 kg/kg-mol
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67
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Quantifying pollutants
Mass of N2 =
(11,305,000 m3 /d)(28 kg/kg-mol)
= 14,125,000 kg/d
(22.41 m3 /kg-mol)
Mass of O2 =
(1,615,000 m3 /d)(32 kg/kg-mol)
= 2,306,000 kg/d
(22.41 m3 /kg-mol)
Mass of CO2 =
(1,615,000 m3 /d)(44 kg/kg-mol)
= 3,171,000 kg/d
(22.41 m3 /kg-mol)
Mass of H2O =
(1,615,000 m3 /d)(18 kg/kg-mol)
= 1,297,000 kg/d
(22.41 m3 /kg-mol)
d) Remove H2O vapor.
Volume of dry gas = 16,150,000 m3/d – 1,615,000 m3/d = 14,535,000 m3/d
Volume percent CO2 = 100(1,615,000 m3/d)/(14,535,000 m3/d) = 11.1%
68
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
4
4.1
Conservation of mass
CONSERVATION OF MASS
SOLVENT EMISSIONS
An organic solvent (e.g. acetone) is used to clean the grease from metal parts before nickel
plating. The solvent input is 100 liters per month. No solvent leaves on the clean metal
parts. Sixty liters per month of used solvent are collected for disposal, which costs $5/L.
Forty liters are lost as air emissions. The solvent density is 0.95 kg per liter. (a) What mass
of solvent is being lost to the atmosphere? (b) The cost of acetone solvent is $4.50/L. What
is the cost of solvent lost to the air? (c) What is the cost of the spent solvent that is sent to
‘disposal’? (d) The company is agreeable to buy a unit to recover the solvent air emissions
and buy solvent recovery distillation unit if the payback time for the project is 2 years or
less. The distillation unit can recover 90% of the spent solvent. How much can be spent
on the project?
Air emissions
40 L/month
Solvent
100 L/month
Greasy
metal parts
Solvent
degreasing
process
Cleaned
metal parts
Spent solvent
60 L/month
Figure P4.1 Solvent emissions
Solution
a) Input = waste solvent + air emission loss
Air emission loss = 100 L/month – 60 L/month = 40 L/month
Air emission mass = (0.95 kg/L)(40 L/month) = 38 kg/month
b) Cost of acetone lost as air emissions = ($4.50/L)(40 L/month) = $180/month
c) Cost of acetone lost as spent solvent = ($4.50)(60 L/month) = $270/month
d) Value of acetone recovered from air emissions = $180/month
Value of acetone recovered from spent solvents = (0.9)($270/month) = $243/month
Savings on disposal costs = ($5/L)(60 L/month) = $300/month
Total savings on solvent purchase =value of recovered solvent + disposal cost
= $180/month +$243/month + $300/month = $723/month
Savings over 2 years = (24 months)($723/month) = $17,352
69
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
4.2
Conservation of mass
MIXING LIQUIDS
A mixing process has two input streams, F1 and F2. Stream F1 is 4.8% A by mass and
95.2% B by mass. Input stream F2 is 100% B. The mixture of F1 and F2 leaving the mixer
is 0.6% A by mass. How much of stream F2 is mixed with stream F1?
Solution
Material balance on A
4.8F1 = 0.6(F1 + F2)
F2= 4.2/0.6 F1
7 kg F2 per 1 kg F1
4.3
MIXING GASES
A flow of air, F1, that is 21% oxygen by volume, is mixed with a stream, F2 of pure oxygen
(100% O2 by volume). The mixture is 50% oxygen. What is the ratio at which air is mixed
with pure oxygen?
Solution
Material balance on oxygen
0.21 F1 + 1 F2 = 0.5(F1 + F2) = 0.5 F1 + 0.5 F2
0.5 F2 = 0.29 F1
F2/F1 = 0.58
4.4
INDUSTRIES JOIN MUNICIPALITY
Two industrial wastes are to be mixed with a municipal wastewater for treatment. Calculate
the total suspended solids (TSS mg/L) of the mixed waste. Flow is million gallons per day
(mgd)
Slaughterhouse
QS = 1.5 mgd
TSSS = 650 mg/L
Dairy
QD = 1 mgd
TSSD = 240 mg/L
Municipal sewage QM = 8 mgd
TSSM = 190 mg/L
70
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Solution
QSTSSS + QDTSSD + QMTSSM = (QS + QD + QM)TSSMixed
TSSMixed =
(1.5 mgd)(650 mg/L) + (1 mgd)(240 mg/L) + (8 mgd)(190 mg/L)
1.5 mgd + 1 mgd + 8 mgd
= 260 mg TSS/L
The units of each term are not conventional units, like lb/day, but each term does have the same units
(mgd-mg/L) and that is the only requirement.
4.5
PRIMARY SETTLING TANK
A primary sedimentation basin, Figure P4.5, is designed for 3,000 m3/d flow and an inlet
concentration of 240 mg/L suspended solids. Suspended solids are the solids captured by on
a laboratory from a known volume of wastewater; some are organic, some are inorganic.
Some will settle within a reasonable time and some will not. The average suspended solids
removal efficiency of the sedimentation process is 60 percent. The slurry of solids and water
removed from the bottom of the tank (primary sludge) has an average solids concentration
of 4% and specific gravity of 1.023. (a) Calculate the mass of solids in the slurry. (b)
Calculate the mass and volume of slurry produced per day.
71
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Influent
Effluent
3,000 m3/d
240 mg/L
Primary sludge (solids & water)
4% solids by mass
sp. gr. = 1.023
Figure P4.5 Primary settling tank
Solution
a) Solids in slurry
240 mg/L = 0.24 kg/m3
Mass solids removed = (0.6)(3,000 m3/d)(0.240 kg/m3) = 432 kg/d
b) Mass of slurry = (432 kg/d)/(0.04) = 10,800 kg/d water plus solids
Volume of slurry = (10,800 kg/d)/(1.023 kg/m3) = 10,557 m3/d
4.6
STACK GAS FLOW
The flow rate of gases from a stack, Figure P4.6, is determined by adding pure CO2 at the
rate of 0.5 kg/min between the firebox and the stack. The wastes coming out of the firebox
contain 5.1% CO2 by weight. Gas analysis at the top of the stack gives 6.1% CO2 by weight.
Make a flow diagram for this problem and calculate the waste gas flowrate from the firebox.
Solution
Flow diagram:
t
Gas from firebox
Flow = MGas CO2
= 0.051 MGas
F::I �
Firebox
I \/ ,
CO2= 0.5 kg/min
Figure S4.6
72
Exhaust gas
Flow= M Gas + 0.5 kg/min
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Total gas flow
MGas = mass of gas out of the firebox, kg/min
Mass CO2 added = 0.5 kg/min
Total mass of gas out stack = MGas + 0.5 kg/min
CO2
CO2 out of the firebox = 0.051 MGas
Added CO2 = 0.5 kg/min
Total CO2 out stack = 0.051 MGas + 0.5 kg/min
Stack emission is 6.1% CO2
Material balance on CO2
0.061(MGas + 0.5 kg/min) = 0.5 kg/min + 0.051 MGas
0.061 MGas + 0.0305 kg/min = 0.5 kg/min + 0.051 MGas
0.01 MGas = 0.4695 kg/min
MGas = 46.95 kg/min ≈ 47 kg/min
4.7
WASTE DISCHARGE TO A STREAM
A river-bank discharge of Q2 = 40 m3/s and pollutant concentration C2 = 10 mg/L mixes
with a river flow of Q1 = 90 m3/s and C1= 3 mg/L, as shown in Figure P4.7. Calculate the
downstream concentration C3.
Figure P4.7
Solution
Pollutant Material Balance
Q1C1 + Q2C2 = Q3C3
(90 m3/s)(3 mg/L) + (40 m3/s)(10 mg/L) = (90 m3/s + 40 m3/s)C3
270 mg/L + 400 mg/L = 130 C3
C3 = (670 mg/L)/130 = 5.2 mg/L
73
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
4.8
Conservation of mass
BOILER BLOWDOWN
The total dissolved solids (TDS) in the circulating boiler water will increase with every cycle
because of steam consumption in the manufacturing process. The steam is pure water and
the salts that were in the water used to make the steam are left behind. In order to maintain
the salts at a level that does not interfere with boiler operation or cause damage, some salty
water must be removed and replaced with fresh water. The salty water that is removed is the
blowdown. The boiler feed water flow equals the steam consumption plus the blowdown.
Condensate
Feed water
QF (kg/h)
CF (mg/L)
Steam consumption
QS (kg/h), CS = 0
Boiler
Boiler blowdown (BD)
QBD (kg/h), CBD (mg/L)
Figure P4.8
74
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
The steam consumption is QS, = 800 kg/h and the blowdown concentration is CBD = 4,000
mg/L. Make the mass balance for the boiler system shown in Figure P4.8. Calculate and
compare the feed water and blowdown flow rates for the following three levels of feed
water quality.
1) Feed water from the city water supply has CF = 400 mg/L.
2) Recycled in-plant water can be used to dilute the city water to CF = 280 mg/L.
3) City water can be treated to have CF = 100.
Solution
Material balance equations
Water QF = QBD + QS = QBD + 800 kg/h
Salts QF CF = QBD CBD = (4,000 mg/L) QBD
Combine equations
QF CF = (4,000 mg/L)(QF – 800 kg/h))
QF = (3,200,000 mg/L kg/h)/(4,000 mg/L - CF)
1) Feed water from the city water supply @ CF = 400 mg/L
QF
3,200,000 mg/L kg/h
3,200,000 kg/h
3,200,000 kg/h
=
=
= 889 kg/h
4,000 mg/L - CF
4,000 - 400
3,600
QBD
QF - 800 kg/h = 889 kg/h - 800 kg/h = 89 kg/h
The solutions are given in Table S4.8
Source
CF (mg/L)
QF (kg/h)
QBD (kg/h)
City water supply
400
889
89
Recycled plant water
280
860
60
City water, treated
100
821
21
Table S4.8
75
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
4.9
Conservation of mass
DYE TRACER TO MEASURE FLOW
Flow will be measured by adding a known mass of inert organic dye and measuring the
concentration after it has mixed with the flow in the pipe, as shown in Figure P4.9. There
is no dye above the injection point. Dye solution at a concentration of 8 mg/L is added
at a rate of 1 L/min. The concentration measured downstream is 0.16 mg/L. There is no
accumulation of dye or water within the mixing zone. Find the flow rate, W.
Figure P4.9
Solution
Basis = 1 min
The tie component is the tracer dye.
Boundary = mixing zone
Material balance on dye:
(1 L/min)(8 mg/L) = (W + 1 L/min)(0.16 mg/L)
dye added
dye after mixing
8 mg/min = (0.16mg/L)W + 0.16 mg/min
W = (7.84 mg/min)/(0.16 mg/L) = 49 L/min
4.10 DEMINERALIZED WATER
A reverse osmosis system, Figure P4.10, removes salt from 1,000 lb/h seawater with a 3.1%
salt content to produce demineralized water at 500 ppm dissolved salts and brine waste
at 5.25 % dissolved salts. (a) How much brine and demineralized water are produced? (b)
The yield calculated in part (a) was not very good. Redo the calculation assuming you can
somehow get the brine waste concentration up to 12%.
76
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Brine recycle
Seawater
1,000 lb/h
3.1% TDS
Reverse
osmosis
Brine waste
5.2 % TDS
Demineralized water
500 ppm TDS
Figure P4.10
Solution
Basis = 1,000 lb/h
B = Brine, lb/h
D = Demineralized water, lb/h
Concentrations: 3.1% = 3.1 lb/100 lb = 31 lb/1,000 lb
5.25% = 5.25 lb/100 lb = 52.5 lb/1,000 lb
500 ppm = 500 lb/1,000,000 lb = 0.5 lb/1,000 lb
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77
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
a) Overall material balance: Recycle is irrelevant
1,000 lb/h = B + D
Salt balance: (31 lb/1,000 lb)(1,000 lb/h) = (0.5 lb/1,000 lb) D + (52.5 lb/1,000lb) B
Water balance:
31,000 lb/h = 0.5 D + 52.5 B
(1.0 – 0.031) = D + B(1 – 0.0525) Treating D as pure water
969 lb/h = D + 0.9475 B
One of these equations is redundant. Use the overall balance and the salt balance.
Substitute 31,000 lb/h = 0.5 D + 52.5 (1,000 lb/h – D)
31,000 lb/h = 0.5 D + 52,500 lb/h – 52.5 D
D = 413.5 lb/h
B = 586.5 lb/h
4.11 WELL WATER CONTAMINATION
You have been asked to give an opinion on whether or not a domestic well water supply
is being contaminated by leakage from a submerged irrigation system. You know that the
quality of local groundwater has been stable historically, and you are able to obtain data
on quality of the municipal well and the suspect irrigation water. The data, measured in
mg/L (ppm) are in Table P4.11.
Component
TDS
Calcium
Magnesium
Sulfate
Chloride
Nitrate
Ca
Mg
SO4
Cl
NO3
Municipal well (MW)
245
245
29
15
13
3.8
Irrigation water (IW)
55
55
6
3
2
2
Local groundwater (GW)
236
236
22
16
14
3.5
Table P4.11 Well water contamination data in mg/L (ppm)
If there is no gross contamination of the well, its quality should be the same as the local
groundwater. At a glance they compare closely, but the problem needs more analysis.
Presume that the well water is a mixture of one unit of groundwater and x units of irrigation
water. A material balance for each of the six components gives six estimates of x and each
provides some information on the true value of x. If there is no contamination the x’s are
the result of random measurement error. Does it appear that the well is contaminated?
78
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Solution
CGW, CIW and CMW = concentrations of groundwater, irrigation water, and municipal well water.
Material balance:
1 unit of groundwater + x units of irrigation water = 1 + x units of municipal well water
1 CGW + x CIW = (1 + x) CMW
x
CGW CMW
CMW CIW
Component
TDS
Ca
Mg
SO4
Cl
NO3
MW
245
29
15
13
3.8
4
IW
55
6
3
2
2
0.7
GW
236
22
16
14
3.5
3.9
x=
-0.05
-0.09
0.08
0.09
-0.167
-0.03
Table S4.11
The values of x are all small. Some are positive and some are negative. Obviously there cannot physically
be a negative x. These results are what one expects for quantities (the x’s) that are essentially the same,
but are affected by random measurement error. The precision of measurement of these chemicals is
probably something like ±5%. Chloride has the largest value of x of irrigation water, at – 0.167, but
the negative sign indicates no contamination. We conclude, based on these data, that there is no
contamination. It is possible that data on specific chemicals, such as pesticides might tell a different story.
4.12 REFUSE DERIVED FUEL PROCESSING
A plant to produce refuse derived fuel (RDF) will take in 1,000 T/d of municipal refuse of
the composition shown in Figure P4.12. The waste is 5% ferrous metals. The air classifier
feed contains 7% dust that is carried away in the air exhaust. Also 5% of the feed moisture
is lost in the exhaust. The heavy solids that drop out of the air classifier is 6% of the mass
fed to the unit. The trommel screen puts out three size fractions. The smallest size fraction
(≤ 25 mm) is 30% of the trommel feed and this material is waste. The larger particles (>
25 mm) can be used as RDF. Estimate the mass of dry RDF that can be produced.
79
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Figure P4.12
Solution
Ferrous metals removed = (0.05)1,000 T/d = 50 T/d
Mass to air classifier = 1,000 T/d – 50 T/d = 950 T/d
5% moisture loss = 0.05(950 T/d) = 47.5 T/d
‘Lights’ removed = 0.07(950 T/d) = 66.5 T/d
‘Heavies’ removed = 0.06(950 T/d) = 57 T/d
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Mass out of air classifier = (950 – 47.5 – 66.5 – 57)T/d = 779 T/d
Particles < 25 mm are rejected from the Trommel screens
Trommel reject = 0.3(779 T/d) = 233.7 T/d
Refuse derived fuel = 779 T/d– 233.7 T/d = 545.3 T/d
4.13 OIL RECOVERY WITH MEMBRANES
A waste stream (5.8 T/d) that contains 1% oil and 99% water (mass percent) will be
concentrated by a two-stage membrane process, Figure P4.13. A membrane filter (MF)
yields a retentate (concentrated stream) that is 10% oil. The second stage is an ultrafilter
(UF) has a retentate that is 35% oil. The MF and UF permeates contain less than 35 ppm
oil. This can be ignored when making the material balance. The ultrafilter concentrate goes
to an evaporator where 0.098 T/d of water is removed, leaving a material that is 85% oil
and can be burned as a fuel. The combined permeate can be recycled to the manufacturing
process. Calculate the mass of oil available for combustion, the water removed by evaporation,
and the mass of recycled water. Also calculate the composition of each retentate (R) and
permeate (P) stream.
RuF = 35% Oil
RMF = 10% Oil
Feed= 5.8 T/d
�
1 % oil (by mass)
PMF
Negligible Oil
Water
Evaporation
P uF
Negligible Oil
I
Ill----'-i ----�
Recycled
water ◄
•
Figure P4.13
Solution
Material balances on Membrane Filter (MF)
Feed to MF (total mass) = 5.8 T/d
Oil at 1% = 0.01(5.8) = 0.058 T/d
81
•
85%0il
Combustion
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Water = 5.8 – 0.058 = 5.742 T/d
Assume there is no oil in the MF permeate.
Retentate = 0.058 T/d Oil at 10% mass fraction
Total mass of MF retentate = 0.058/0.1 = 0.58 T/d
Mass of Water in MF retentate = Total mass – Oil
= 0.58 T/d – 0.058 T/d = 0.522 T/d
Permeate volume = Water in feed – Water in retentate
= 5.742 – 0.522 = 5.220 T/d
Material balance on the Ultrafilter (UF)
UF Feed = retentate from the MF = 0.58 T/d at 10% Oil
Oil in feed = 0.058 T/d Oil
Water in feed = 0.522 T/d water
Assume there is no oil in the UF permeate
UF Retentate at 35% oil (total mass) = (0.058 T/d)/0.35 = 0.166 T/d
Water in UF retentate = Total mass – Oil = 0.166 T/d – 0.058 T/d = 0.108 T/d
UF permeate = Water in feed – Water in retentate
= 0.522 T/d – 0.108 T/d = 0.414 T/d
Evaporator feed = UF retentate
0.058 T/d Oil
0.108 T/d water
Water evaporated = 0.098 T/d (given)
Water not evaporated = 0.108 T/d – 0.098 T/d = 0.01 T/d
Total mass output = 0.058 T/d Oil + 0.01 T/d water
Recycle water = MF Permeate + UF Permeate
= 5.220 + 0.414 = 5.614 T/d
The final quantities are shown in Figure S4.13
Figure S4.13
82
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
4.14 WATER CONSERVATION FOR A REFINERY
The process units in an oil refinery (stills, condensers, etc.) demand a high flow of water for
steam and cooing water. Steam is generated in boilers. After use the condensate returns as
clean condensate and can be used to make more steam. Cooling water is recycled through
cooling towers. Used process water, cooling tower blowdown, and boiler blowdown are
discharged as plant waste. Values shown on the diagram, Figure P4.14, are gallons per unit
of production. Complete the material balance on water.
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Cooling tower
makeup
(46 gal)
Conservation of mass
Evaporation
(31 gal)
Warm water from process
(1,660 gal)
Cooling tower blowdown
(15 gal)
Cool water
return
Total water
(100 gal)
Process water
(20 gal)
Process water to waste
Condensate to waste
(8 gal)
Condensate
return
(10 gal)
Steam
(18 gal)
Steam lost
Boiler blowdown
(2 gal)
Boiler makeup
Figure P4.14
Extra credit: Suggest ways that water use might be modified to reduce water intake or
waste output. For example, steam losses might be reduced by 50% by better condensation,
and the condensate could be used as process water. Modify the water budget diagram and
recalculate the material balance to illustrate the benefits of your suggested changes.
Solution
Basis: 1 unit of production
All values are gallons per unit of production (gal/unit)
For simplicity use gal instead of gal/unit
Check the overall material balance
Inputs = Outputs
100 gal = 46 gal + 20 + 34 = 31 gal + 15 + 20 + 8 + 24 + 2
Material balance on cooling towers
Makeup = Evaporation + Blowdown = 46 gal = 31 gal + 15 gal
Therefore, Cooled water return = Cooling water from process
Or, in a more complete form:
ªCooling tower º ªCooling water º
ªCooling tower º ªCooled water º
«
»+«
» = ¬ªEvaporation¼º + «
»+«
»
makeup
from
process
return
¬
¼ ¬
¼
¬ blowdown ¼ ¬
¼
ªCooled water º
«
» = 46 gal + 1,660 gal - 31 gal - 15gal = 1,660 gal
return
¬
¼
84
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Material balance on process units
ªCooledº
ª Process º + ªSteamº + « water » = ªCondensateº + ªCondensateº + ªProcess water º + ªCooling water º
¬
¼
¬«water in¼»
¬« return ¼» ¬« to waste ¼» ¬« to waste ¼» ¬« from process ¼»
« return »
¬
¼
20 gal + 18 gal + 1,660 gal = 10 gal + 8 gal + ª«Process water º» + 1,660 gal
¬ to waste ¼
ªProcess water º = 20 gal
«¬ to waste »¼
Material balance on feed water input split
ª Total water º
ªCooling water º
ªProcess º
«
» = «
» + «
» +
input
makeup
¬
¼
¬
¼
¬ water ¼
ª Boiler º
«
» = 100 gal - 46 gal - 20 gal = 34 gal
¬makeup¼
ª Boiler º
«
»
¬makeup¼
Material balance on boiler
ª Boiler º
ªCondensateº
ªSteamº
ª Boiler º
«
» + «
» = «
» + ª¬Steamº¼ + «
»
¬makeup¼
¬ return
¼
¬ lost ¼
¬blowdown¼
ªSteamº
«
» = 34 gal + 10 gal - 18 gal - 2 gal = 24gal
¬ lost ¼
The final quantities are shown in Figure S4.14.
Cooling tower
makeup
(46 gal)
Evaporation
(31 gal)
Warm water from process
(1,660 gal)
Cooling tower
blowdown
(15 gal)
Cool water
return
Total water
(100 gal)
Process water
(20 gal)
Condensate
return
(10 gal)
Process water to waste
(20 gal)
Condensate to waste
(8 gal)
Steam
(18 gal)
Steam lost
(24 gal)
Boiler blowdown
(2 gal)
Boiler makeup
(34 gal)
Figure S4.14
85
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
4.15 SLUDGE THICKENING AND DISPOSAL
Make a material balance for a primary settling and sludge processing system to handle
a design flow of 34,000 m3/d and influent suspended solids concentration of 300 mg/L.
Assume this includes all recycle flows. The system, shown in Figure P4.15, includes primary
sedimentation, gravity thickening of primary sludge, sludge dewatering by rotary vacuum
filtration, and disposal of sludge cake onto farmland.
Wastewater
Primary Settling
To Activated
Sludge Treatment
Supernatant
(recycle to influent)
Primary sludge
2.5% solids
Thickener
Filter cake
28% solids
Thickened sludge
6% solids
Vacuum Filter
Filtrate
(recycle to influent)
Figure P4.15
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
a) Suspended solids removal in primary settling = 70%. Primary sludge is to be
withdrawn from the primary settling tank continuously at a concentration of 2.5%
solids and specific gravity = 1.02. It is fed to a gravity thickener that operates
continuously to produce a thickened sludge at 6% solids by mass. Assume that
the solids capture efficiency of the thickener is 100%. The thickened sludge is
dewatered in a vacuum filter that operates 6 hours per day. Calculate the material
balance for the settling and thickening processes. How much sludge storage (m3)
is needed for the thickened primary sludge?
b) Lime (CaO), as a 10% solution, is added to the 6% thickened sludge to condition
it for dewatering by rotary vacuum filtration. The water in the lime solution can
be neglected. The lime dose is 28% CaO by weight, based on dry sludge solids
mass. For purpose of calculating the material balance on the solids assume that all
the lime becomes incorporated into the sludge cake. The sludge filter cake is 28%
solids by mass. The density of the sludge cake is 1,280 kg/m3. Assume that 100%
of the feed solids are captured in the filter.
c) The sludge cake is spread on the surface of farmland and then plowed into the soil.
Sludge application is limited to six months of the year, and sludge cake must be
stored for the other six months. How much storage capacity is required?
Solution
Basis = 1 day of operation = 34,000 m3 of wastewater influent
a) Primary sedimentation tank material balance
Raw wastewater = 34,000 m3
Raw wastewater solids = (0.3 kg/m3)(34,000 m3) = 10,200 kg
70% solids removal by sedimentation = 0.7(10,200) = 7,140 kg
Sludge = 2.5% solids
From definition of mass fraction:
Water in sludge = (7,140 kg)(1 – 0.025)/0.025 = 278,460 kg
Assume density = 1,000 kg/m3
Volume of sludge = (278,460 kg)/(1,000 kg/m3) = 278.5 m3
Gravity thickener material balance
100% solids capture = 7,140 kg
Thickened sludge = 6% solids
From definition of mass fraction:
Water in sludge = (7,140 kg)(1 – 0.06)/0.06 = 111,860 kg
Mass of sludge = solids + water = 7140 kg + 111,860 kg = 119,000 kg
Assume density = approx. 1,000 kg/m3
Sludge volume = 119 m3
87
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
b) Vacuum filter material balance
Lime added = 28% of sludge solids = 0.28(7,140 kg) = 1,999 kg
Feed to filter = solids + lime = 7,140 kg + 1,999 kg = 9,139 kg
Filter cake = 28% solids
From definition of mass fraction:
Water in cake = (9,139 kg)(1 – 0.28)/0.28 = 23,500 kg
Mass of sludge cake = 23,500 kg water + 9,139 kg solids = 32,639 kg
Sludge cake Density = 1,280 kg/m3
Sludge cake Volume = (32,639 kg)/(1,280 kg/m3) = 25.5 m3
c) Storage for 6 months
= (180 days)(25.5 m3/d) = 4,590 m3
4.16 THICKENER RECYCLE
A flow of 100,000 m3/d is processed in primary sedimentation, as shown in Figure P4.16
to remove 60% of the 0.25 kg/m3 influent suspended solids. Sludge is removed from the
sedimentation basin at a concentration of 2% solids and specific gravity of 1.02. This sludge
is thickened in a gravity thickener to a concentration of 6% and specific gravity 1.03. The
thickener captures 95% of the influent solids and the supernatant is returned to the inlet
of the primary sedimentation basin.
The recycled supernatant from the thickener will contain some solids. Assume this amount
is negligible (i.e. zero) and make the material balance on the system to estimate primary
sludge volume flow rate, mass of solids conveyed in the primary sludge, mass of solids in
the thickened sludge, volume flow of the thickened sludge, and the recycled supernatant
flow and solids load. Improved estimates can be obtained by adding the recycled amounts
to the inflow wastewater loads and making a second iteration on the material balance. The
second iteration is optional.
Influent
100,000 m3/d
0.25 kg/m3 TSS
Supernatant
recycle
Settling
Effluent
Sludge from settling tank
2% solids by mass
sp. gr. = 1.02
Thickener
Thickened sludge
6.5% solids by mass
sp. gr. = 1.03
Figure P4.16
88
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Solution
a) Solids removal in settling tank
Influent solids load = (100,000 m3/d)(0.25 kg/m3) = 25,000 kg/d
Solids removed in sludge = 0.6(25,000 kg/d) = 15,000 kg/d
Flow rate of sludge leaving settling tank: (1,020 kg/m3)(Qs)(0.02) = 15,000 kg/d
QS = 735 m3/d
b) Solids removal in thickener
Influent solids load = 15,000 kg/d
Solids captured in thickener underflow (95%) = 0.95(15,000 kg/d) = 14,250 kg/d
Flow rate of thickened sludge leaving the thickener
= (1,030 kg/m3)(QU)(0.06) = 14,250 kg/d
QU = 231 m3/d
c) Recycle flow stream = QR
Mass balance on thickener flow
QR = QS – QU
360°
thinking
QR = 735 m3/d – 231 m3/d = 524 m3/d
Mass balance on thickener solids
.
Recycled Solids = 15,000 kg/d – 14,250 kg/d = 750 kg/d
d) Revised estimate of loadings to the settling tank
Flow = 100,000 m /d + 524 m /d = 100,524 m /d
3
360°
thinking
.
3
3
360°
thinking
.
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Dis
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Solids = 15,000 kg/d + 750 kg/d = 15,750 kg/d
The additional flow is negligible.
The added solids load is about 5% of the raw wastewater loading.
4.17 WET SCRUBBER WATER USE
Figure P4.17 shows a wet scrubber that removes pollutants from the inlet gas. Assume that
dry inlet gas stream leaves the scrubber with 4.5 kg/min of evaporated water. The water
that evaporates, plus water that is withdrawn as blowdown, must be replaced with fresh
makeup water. The recycle is 80 L/min back to the scrubber. The blowdown is 8 L/min of
liquid waste that is withdrawn for treatment and disposal. Calculate the flow rates for the
scrubber water and the makeup water.
(3) Outlet Gas
4.5 kg/min
(1) Inlet Gas
Scrubber
(2) Recycle
80 m3/h
(4) Scrubber Water
(5) Makeup water
(6) Blowdown
8 L/min
Figure P4.17
Solution
Let M = mass flow rate (kg/min) and Q = volumetric flow rate (L/min)
Put outlet gas stream into consistent units:
M3 = 4.5 kg/min
Q3 = (4.5 kg/min)(1 L/kg) = 4.5 L/min
Table S4.17 summarizes what is known.
Process Stream
(1)
Inlet gas
(dry)
(2)
Recycle
liquid
(3)
Outlet
gas
(4)
Scrubber
water
(5)
Makeup
water
(6)
Blowdown
M = Mass (kg/min)
0
80
4.5
M4
M5
8
Q =Flow (L/min)
0
80
4.5
Q4
Q5
8
Unit of flow
Table S4.17
90
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Overall material balance for water on the scrubber system
Q1 + Q5 = Q3 + Q6
0 + Q5 = 4.5 L/min + 8 L/min
Q5 = 12.5 L/min
M5 = 12.5 kg/min Makeup water
Material balance around the scrubber water and makeup water mixing point
Q4 + Q5 = Q2 + Q6
Q4 + 12.5 L/min = 80 L/min + 8 L/min
Q4 = 88 L/min – 12.5 L/min = 75.5 L/min
M4 = 75.5 kg/min Scrubber water
4.18 SOLVENT REMOVAL FROM AIR
The air stream in Figure P4.18 is flowing at a rate of 5 kg/min contains 4% solvent (mass
basis). The air is contacted with an oil spray and the oil absorbs almost all of the solvent.
The exiting air is 0.01% solvent and 99.99% air. The exiting oil is 1% solvent and 99%
oil. All percentages are by weight (mass). Calculate the flow rate of oil entering the absorber
and the mass flow rate of the air/solvent mixture leaving the absorber.
Oil In = X kg/min
Air In (4% solvent)
Flow = 5 kg/min
Solvent = 0.2 kg/min
Air = 4.8 kg/min
Air Out (0.01% solvent)
Air = 4.8 kg/min air
Solvent = Y kg/min
Oil out = X kg/min
Solvent = (0.2 – Y) kg/min
Figure P4.18
Solution
Composition of air output (0.01% solvent):
Y/(4.8 kg/min + Y) = 0.0001
Y = 0.00048 kg solvent/min
Composition of oil output (1% solvent):
Solvent mass flow = (0.2 kg/min – Y) = 0.2 kg/min - 0.00048 kg/min
= 0.1995 kg solvent/min
(0.1995 kg/min)/(X + 0.1995 kg/min) = 0.01
X = 19.75 kg oil/min
91
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
4.19 GAS PREHEATING
Methane (CH4) and oxygen (O2) are mixed and heated before being sent to a burner, as
shown in Figure P4.19. What are the volumetric flow rate, the mass flow rate, and the mass
fraction of methane in the mixture? Assume the gases obey the ideal gas law.
CH4
100 kg-mol/min
75°C and 10 atm
Preheater
O2
200 kg-mol/min
25°C and 10 atm
Mixed gas
200°C
10 atm
Figure P4.19
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Solution
Basis: 1 minute of operation = 100 kg-mol of CH4
Standard conditions 0°C and 1 atm
Mass flows and mass fractions are unaffected by temperature and pressure; volumetric flows are.
Mass flow of inputs
CH4 in = (100 kg-mol/min)(16 kg/kg-mol) = 1,600 kg/min at 75°C and 10 atm
O2 in = (200 kg-mol/min)(32 kg/kg-mol) = 6,400 kg/min at 25°C and 10 atm
Gas flow in = 300 kg-mol/min = 8,000 kg/min
Mass flow of outputs
Mixed gas out = Gas flow in = 300 kg-mol/min
= 1,600 kg/min + 6,400 kg/min = 8,000 kg/min
Mass fraction of methane = 1,600 kg/8,000 kg = 0.2
Volumetric flow rate of methane
At STP:
Volumetric flow = (300 kg-mol/min)(22.41 m3/kg-mol) = 6,723 m3/min
At 200°C and 10 atm:
Volumetric Flow
§ T · § 1 atm ·
§ 473 K · § 1 atm ·
3
3
VSTP ¨
¸¨
¸ = (6,723 m /min) ¨
¸¨
¸ = 1,165 m /min
© 273 K ¹ © P ¹
© 273 K ¹ © 10 atm ¹
4.20 BIOLOGICAL TREATMENT
Figure P4.20 shows is a simplified experimental biological sludge process that was used to
measure the conversion of COD to suspended solids. This conversion happens as bacteria
consume the biodegradable organic compounds for energy and the production of new
bacterial cells. The mass of bacteria produced is proportional to the VSS produced. A fullscale activated sludge process will recycle most of the underflow from the settling tank (final
clarifier) to the bioreactor (aeration tank), but there is no sludge recycle in this problem.
(a) Calculate the mass of COD removed. (b) Calculate the mass of solids produced in the
bioreactor. (c) Calculate the mass of solids removed as underflow sludge.
Influent
QInf = 6 L/h
CODInf = 500 mg/L
TSSInf = 0 mg/L
Bioreactor
V = 240 L
TSS = 400 mg/L
VSS = 250 mg/L
Final
clarifier
Effluent
QEff = 6 L/h
CODEff = 75 mg/L
TSSEff = 40 mg/L
Waste sludge
QWaste = ?
TSSWaste= 8,000 mg/L
Figure P4.20
93
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Solution
a) Mass of COD removed is calculated from a material balance around the entire process.
= (6 L/h)(500 mg COD/L – 75 mg COD/L) = 2,550 mg COD/h
b) Mass of total suspended solids produced in the bioreactor
= (6 L/m)(400 mg TSS/L – 0 mg TSS/L) = 2,400 mg TSS/h
c) Mass of total suspended solids removed in the settling tank
= (6 L/m)(400 mg TSS/L – 40 mg TSS/L) = 2,160 mg TSS/h
Comments: In a full-scale activated sludge most of the sludge from the settling tank is recycled to
the aeration tank. What is not recycled is wasted, i.e. removed from the activated sludge process and
sent to some kind of sludge treatment. The concentration of the recycled sludge is typically 8,000 –
12,000 mg TSS/L.
4.21 STEP AERATION
Wastewater influent is divided and fed to the three aeration basins in series, as shown in
Figure P4.21. Calculate the mixed liquor suspended solids (MLSS) concentration in each
aeration basin. X = suspended solids concentration (mg/L) and V = volume (m3). The aeration
basin volumes are equal (V1 = V2 = V3 = 75 m3). Assume that the wastewater influent has X
= 0. The return activated sludge flow is 4 m3/s at a concentration of XRAS = 10,000 mg/L.
Solids added by bacterial growth will be ignored.
Wastewater
Influent
Q = 6 m3/s
Q2 = 2 m3/s
Q1 = 3 m3/s
Aeration 1
V1 , X1
Aeration 2
V2 , X2
Q3 = 1 m3/s
Aeration 3
V3 , X3
Return activated sludge (RAS)
QRAS = 4 m3/s,
XRAS = 10,000 mg/L
Figure P4.21
Solution
The flows out of each aeration basin are
Q1 = 3 m3/s + 4 m3/s = 7 m3/s
Q2 = Q1 + 2 m3/s = 9 m3/s
Q3 = Q2 + 1 m3/s = 10 m3/s
Make a material balance on X for each aeration basin.
94
Clarifier
Final
Effluent
Waste activated
sludge (WAS)
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Basin 1: (3 L/s)(0 mg/L) + (4 L/s)(10,000 mg/L) = (7 L/s)(X1)
X1 = 5,714 mg/L = 5.714 kg/m3
Basin 2: (7 L/s)(5,714 mg/L) = (9 m3/s)(X2)
X2 = 4,444 mg/L = 4.444 kg/m3
Basin 3: (9 m3/s)(4,444 mg/L) = (10 m3/s)(X3)
X3 = 4,000 mg/L = 4.00 kg/m3
Mass of suspended solids in Basin 1 = (75 m3)(5.714 kg/m3) = 429 kg
Masses in Basin 2 and Basin 3 are 333 kg and 300 kg, respectively.
4.22 SINGLE-STAGE RINSING
A single-stage rinse shown in Figure P4.22 is being redesigned. The current operation uses
QR = 10,000 L/h of fresh water to clean parts that have dragout of D = 1.0 L/h at CD
= 100,000 mg/L contaminant concentration. The rinse reduces the residual contaminant
concentration to C1 = 10 mg/L. Two changes are proposed.
1) Reduce the dragout from 1.0 L/h to 0.5 L/h, and
2) Allow the residual concentration to rise to 20 mg/l.
Calculate the new operating conditions and draw the new process flow diagram.
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Dragout
D = 1 L/h
CD = 100,000 mg/L
Clean Water
QR = 10,000 L/h
CR = 0 mg/L
Rinse
Tank
Conservation of mass
Dragout
D = 1 L/h
C1 = 10 mg/L
Dragout
D = 0.5 L/h
CD = 100,000 mg/L
Clean Water
QR = ?
CR = 0 mg/L
Rinse
Tank
Dragout
D = 0.5 L/h
C1 = 20 mg/L
Rinse Water
QR = ?
C1 =20 mg/L
Rinse Water
QR = 10,000 L/h
C1 = 10 mg/L
Proposed Rinse System
Existing Rinse System
Figure P4.22
Solution
Basis: 1 hour = 0.5 L of dragout
Material balance on the rinse tank
(0.5 L/h)(100,000 mg/L) + QR (0 mg/L) = (0.5 L/h)(20 mg/L) + QR (20 mg/L)
50,000 L/h – 10 L/h = 20 QR
QR = 2,500 L/h
Water savings = 10,000 L/h – 2,500 L/h = 7,500 L/h
4.23 TWO-STAGE CROSSFLOW RINSE
Figure P4.23 shows a two-stage crossflow rinsing system. The dragout rate from the plating
bath is D = 1 L/h and the dragout liquid contains CD = 60 g/L contaminant. The drag-out
and rinse water leaving rinse bath 2 must have concentration C2 = 20 mg/L. Calculate the
rinse water flow rates and effluent concentrations, assuming equal flow to each tank.
Clean Water = 2Q L/h
Plating
Bath
D = 1 L/h
CD = 60 g/L
D, C1
Rinse
Tank 1
Rinse
Tank 2
D = 1 L/h
C2 = 20 mg/L
Contaminated rinse water
Q L/h
Q L/h
C2 = 20 mg/L
C1 mg/L
Figure P4.23
96
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Solution
Q = rinse water flow rate to each stage; total rinse water = 2Q.
C1 = concentration of the dragout and the rinse water leaving rinse tank 1.
The incoming rinse water is contaminant free.
Material balance equations:
Rinse 1
(1 L/h)(60,000 mg/L) = (1 L/h) C1 + Q C1
Rinse 2
(1 L/h)C1 = (1 L/h)(20 mg/L) + Q(20 mg/L)
Solving these two equations gives
Rinse 1
60,000 mg/L = C1(1L/h + Q)
Rinse 2
(1 L/h + Q) = (C1)/(20 mg/L)
Combining
60,000 mg/L = (C1)2/(20 mg/L)
C1 = 1,095 mg/L
Substituting into Rinse 2 equation
(1 L/h + Q) = (1,095 mg/L)/(20 mg/L)
Q = 53.8 L/h
Total rinse flow needed for the two tanks is 2Q = 107.6 L/h
4.24 REVERSE OSMOSIS CIRCUIT
The feed to the two-stage reverse osmosis plant shown in Figure P4.24 has a total dissolved
solids (TDS) content of 4,200 mg/L; the Stage 2 permeate flows at 5 m3/h and 5 mg/L TDS.
The Stage 2 reject is recycled and mixed with the feed to Stage 1. The recovery fractions
are 66% for Stage 1 and 80% for Stage 2. The Stage 2 salt rejection is 98%. Complete
the material balance for the RO system, to include (a) flow rates for the feed, permeate
of stage 1, and the reject streams and (b) the salt concentration (TDS) for all the streams
where this is not specified
Feed
F=?
CF = 4,200 mg/L
Stage 1
P1 = 0.66 (F + R2) Stage 2
C1 = ?
R1 = (F + R2) – P1
C1 = ?
Recycle
R 2 = P 1 - P2
C2 = ?
Figure P4.24
97
Permeate 2
P2 = 0.8 P1 = 5 m3/h
C2 = 5 mg/L
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Solution
Basis: 1 hour of operation
Tie variable = salts (TDS)
Overall material balance: F = R1 + P2
Stage 1 material balance:
F + R2 = R1 + P1
Substitute overall balance
R1 + P2 + R2 = R1 + P1
Input to Stage 2
P1 = P2 + R2
Stage 2:P2 = 5 m3/h
Specified recovery = 80%
P1 = P2/0.8 = (5 m3/h)/0.8 = 6.25 m3/h
Material balance
R2 = P1 – P2 = 6.25 m3/h – 5.0 m3/h = 1.25 m3/h
Specified recovery = 0.66
P1/(F + R2) = 0.66
Stage 1:
P1 = 0.66(F + R2)
6.25 m3/h = 0.66(F + 1.25 m3/h)
F = 8.22 m3/h
Material balance
R1 = F – P2 = 8.22 m3/h – 5.0 m3/h = 3.22 m3/h
98
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Material Balance on TDS
Define CR1 = concentration in reject 1
CP1 = concentration in permeate 1, etc.
Overall TDS balance: FCF = R1 CR1 + P2 CP2
(8.22 m3/h)(4,200 mg/L) = 93.22 m3/h)CR1 + (5 m3/h)(5 mg/L)
CR1 = 10,714 mg/L
98% rejection of TDS in Stage 2
CP1 = (5 m3/h)(5 mg/L)/(0.02)(6.25 m3/h) = 200 mg/L
Material balance Stage 2 P1 CP1 = P2 CP2 + R2 CP2
(1.25m3/h)CR2 = (6.25 m3/h)(200 mg/L) + (5m3/h)(5 mg/L)
CR2 = 1,020 mg/L
Material balance at mixing point:
(8.22m3/h)(4,200 mg/L) + (1.25m3/h)(1,020 mg/L) = (9.47m3/h)CMix
CMix = 35,799 mg/L/9.47 = 3,780 mg/L
4.25 SMELTER DUST BENEFICIATION
The larger particles of dust emitted from a smelting process have a metal content high
enough to justify recovery and recycling. The smaller particles have less value. They will need
to be removed to meet the air pollution standards, but that can be done in a secondary air
pollution control system. The dust flow from the smelter is 1 kg/h, with the size distribution
shown in Table P4.25. Seventy percent of the particles are larger than 10 µm and they
have the value. A simple cyclone dust collector can remove the particles at the efficiencies
shown for each class of particle size. The efficiency of removal increases as the particle size
increases. Complete Table P4.25 to calculate the mass removed (kg/h) for each size class.
Calculate the mass of dust that passes through the cyclone and needs to be captured with
a secondary pollution control device.
Particle size (µm)
0–5µm
5–10µm
10–20µm
20–44µm
>44µm
Mass percent (%)
20
10
15
20
35
Mass silica dust input (kg/h)
0.2
0.1
0.15
0.20
0.35
Removal efficiency (%)
12
33
57
82
91
Table P4.25
99
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Solution
Example calculation
Mass removed = (mass input)(removal efficiency)
For 0–5µm particles: mass removed = (0.2 kg/h)(0.12) = 0.024 kg/h
The cyclone removes 0.624 kg dust/h. The input dust load was 1 kg/h. This is a removal efficiency of
62.4%. Table S4.25 shows the results.
Particle size range (µm)
0–5 µm
5–10 µm
10–20 µm
20–44 µm
>44 µm
Mass percent (%)
20
10
15
20
35
Mass silica dust input (kg/h)
0.2
0.1
0.15
0.20
0.35
Removal efficiency (%)
12
33
57
82
91
Amount removed (kg/h)
0.024
0.033
0.085
0.164
0.318
Table S4.25
Particles larger than 20 µm are removed with greater than 80% efficiency
0.55 kg/h (55%) of the dust that enters the cyclone is larger than 20 µm
0.482 kg/h of this 0.55 kg/h input is removed
About 50% of particles in the 10–20 µm size class are removed.
0.7 kg/h (70%) of the dust that enters the cyclone is larger than 10 µm
0.567 kg/h of this 0.7 kg/h input is removed
About 80% of particles smaller than 10 µm pass through the cyclone.
0.3 kg/h (30%) of the input is less than 10 µm.
0.057 kg/h of this fine dust is removed
0.243 kg/h of this fine dust is emitted
Note: ‘fine’ is a descriptive term, not a technical specification.
4.26 DRYING SOLIDS
Figure P4.26 shows an air dryer. For a basis of 1,000 kg/h of wet solids (15% moisture),
calculate the mass of air required to dry the material to 7% moisture. The inlet air has a
moisture content of 0.01 kg H2O/kg air. The outlet air has 0.1 kg H2O/kg air.
100
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Fresh air input
0.01 kg H2O/kg air
Dried solids out
7% H2O
Conservation of mass
Moist air out
0.1 kg H2O/kg air
Dryer
Wet solids Input
1,000 kg/h, 15% H2O
Figure P4.26
Solution
Basis: 1,000 kg/h Wet solids
Solids:
0.85(1,000 kg/h) = 850 kg/h
Water:
0.15(1,000 kg/h) = 150 kg/h
Mass balance on dry solids
Solids In = Solids out = 850 kg/h
Dried (but not bone dry) solids out
Specification = 7% moisture
W = water in dried (but not bone dry) solids, kg/h
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101
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
W/(850 kg/h + W) = 0.07
W = (850 kg/h)(0.07)/(1.0 - 0.07) = 64 kg/h
Mass balance on water
In with solids + In with fresh air = Out with solids + Out with moist air
Let A = mass of fresh air input, kg/h
150 kg/h + (A)(0.01 kg/kg) = 64 kg/h + (A)(0.1 kg/kg)
0.09 A = 150 kg/h – 64 kg/h
A = 956 kg/h fresh air
4.27 ANCHOVY PROCESSING
Anchovies are converted into fishmeal and fish oil. According to the process shown in
Figure P4.27, the fish are cooked and the liquor is pressed out. The liquor and the press
cake are then processed separately. The liquor is decanted in another kind of press and
more solids are recovered. The decanter liquor is processed to remove oil and the residual
sludge is combined with the other solids. The liquor from the oil separator, called stickwater,
is evaporated and the residual solids are combined with the press cake and sludge. The
combined solids stream is dried to produce fishmeal, which is 8% moisture and 92% fish
solids. The diagram contains all the information you need to fill in the missing values for
moisture, solids, and fat. Complete the material balance.
1,000 kg Whole Fish
710 kg Moisture
200 kg Solids
90 kg Fat
Cook &
Press
Liquor
Press Cake = 308 kg
156 kg Moisture
141 kg Solids
11 kg Fat
Dryer
Exhaust
Water
Dryer
Feed
Dryer
Fishmeal
8 % moisture
15 kg Fat
Decanter
Press Cake = 34 kg
19 kg Moisture
14 kg Solids
1 kg Fat
Liquor
Sludge = 8 kg
7 kg Moisture
1 kg Solids
0 kg Fat
Oil
Separator
Evaporator
Concentrate
Water = 461 kg
Figure P4.27 Anchovy processing
102
Stickwater = 575 kg
528 kg Moisture
45 kg Solids
3 kg Fat
Fish Oil
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Solution
Example calculation
Material balance around the Cook and Press operation for Liquor composition
Total mass = 1,000 kg – 308 kg = 692 kg
Moisture = 710 kg – 156 kg = 554 kg
Solids = 200 kg – 141 kg = 59 kg
Fat = 90 kg – 11 kg = 79 kg
The remaining calculations will not be shown. They are simple arithmetic to add and subtract the
quantities that are given. The material balance is in Figure S4.27
1,000 kg Whole Fish
710 kg Moisture
200 kg Solids
90 kg Fat
Dryer Feed = 464 kg
249 kg Moisture
200 kg Solids
15 kg Fat
Dryer Exhaust
230 kg Moisture
Dryer
Fish Meal = 234 kg
19 kg Moisture
200 kg Solids
15 kg Fat
Cook &
Press
Liquor = 692 kg
554 kg Moisture
59 kg Solids
79 kg Fat
Press Cake =
308 kg
156 kg Moisture
141 kg Solids
11 kg Fat
Decanter
Liquor = 658 kg
535 kg Moisture
45 kg Solids
78 kg Fat
Press Cake = 34 kg
19 kg Moisture
14 kg Solids
1 kg Fat
Sludge = 8 kg
7 kg Moisture
1 kg Solids
0 kg Fat
Concentrate = 114 kg
67 kg Moisture
44 kg Solids
3 kg Fat
Oil
Separator
Evaporator
Fish Oil
75 kg Fat
Stickwater = 575 kg
528 kg Moisture
44 kg Solids
3 kg Fat
Water = 461 kg
Figure S4.27
4.28 ANAEROBIC SLUDGE DIGESTION
Sludge with 4% total solids (mass percent) is feed to an anaerobic sludge digester, Figure
P4.28. The solids are 75% volatile solids (VS). Volatile solids destruction in the digester
is 45% to 55%, depending on the composition of the feed material. The diagram shows
biogas as 70% methane (CH4), but this is also variable and can range from 60% to 70%.
Assume the gas yield = 1 m3/kg VS destroyed. Using an input of 1,000 kg total solids and
the range of VSS destruction, calculate ranges for the outputs, to include TS, FS, VS, water,
methane and carbon dioxide.
103
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Biogas
70% CH4
Feed sludge
1,000 kg TS
750 kg VS
250 kg FS
Anaerobic sludge
digester
Digested
sludge
Figure P4.28
Solution
Basis = 1,000 kg of total solids
Total sludge mass, at 4% solids = (1,000 kg)/0.04 = 25,000 kg
Volatile solids = 0.75(1,000 kg TS) = 750 kg
Fixed solids = Total solids – Volatile solids = 1,000 kg – 750 kg = 250 kg
Water = Total sludge mass – mass of solids = 25,000 kg – 1,000 kg = 24,000 kg
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104
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Range of VS destroyed and VS in output digested sludge
At 45% VS destruction: 0.45(750 kg) = 337.5 kg destroyed
412.5 kg output
At 55% VS destruction: 0.55(750 kg) = 412.5 kg destroyed
337.5 kg output
Range of biogas, methane and carbon dioxide production
At 45% VS destruction
Minimum biogas = (337.5 kg VS destroyed)(1 m3/kg VS) = 337.5 m3
Minimum methane = 0.6(337.5 m3) = 202.5 m3
Carbon dioxide = 337.5 m3 – 202.5 m3 = 135 m3
At 55% VS destruction
Maximum = (412.5 kg VS destroyed)(1 m3/kg VS) = 412.5 m3
Maximum methane = 0.7(412.5 m3) = 288.8 m3
Carbon dioxide = 412.5 m3 – 288.8 m3 = 123.7 m3
The results are summarized in Table S4.28
At 45% VS destruction
Input
Output
At 55% VS destruction
Input
Output
Total sludge (kg)
25,000
24,662
25,000
24,588
Total solids (kg)
1,000
662.5
1,000
587.5
Volatile solids (kg)
750
412.5
750
337.5
Fixed solids (kg)
250
250
250
250
24,000
24,000
24,000
24,000
Water (kg)
Biogas (m3)
337.5
412.5
Methane (m3)
202.5
288.8
135
123.7
CO2 (m3)
Table S4.28
4.29 SOLVENT VAPOR EMISSIONS
Figure P4.29 shows how grease and dirt are removed from metal parts by ‘dipping’ them
in a tank filled with solvent vapor. The mass of solvent in the degreasing tank is unknown
but 800 kg of new solvent must be added each month to make up for solvent emissions to
the ventilation system and solvent that leaves on the cleaned parts (i.e., the dragout). The
dragout from the degreasing tank is solvent. The dragout from the rinse tank is a mixture
105
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
of solvent and water. The residual solvent is rinsed off the parts with clean water. The
concentration of solvent is the same in the contaminated rinse water and in the dragout
from the rinse. (a) Calculate the mass of solvent in the dragout + contaminated rinse water.
(b) Calculate the solvent concentration, in convenient units, (kg solvent/kg water or mg
solvent/kg water), in the dragout and the contaminated rinse water.
800 kg new
solvent
Dirty
parts
8Kg solvent
emission
1 kg solvent
emissions
Degreasing
tank
Parts + 9 kg
solvent dragout
1,000 kg clean
rinse water
Rinse
tank
Clean parts dragout
s (kg solvent/kg
water)
Solvent contaminated
rinse water
S (kg solvent/kg water)
Figure P4.29 Solvent vapor emissions
Solution
Basis = 800 kg new solvent and 1,000 kg of clean rinse water.
Define S = concentration of solvent in the rinse tank dragout and rinse water
= kg solvent/kg water
a) Solvent mass balance on the rinse tank
ª 9 kg solvent
«
«dragout from
«¬ degreasing
º
ª solvent in º
ª8 kg solvent º
ª solvent in º
»
«
»
» = « emissions » + «clean parts» + «rinse water »
¬
¼
¬
¼
»¼
«¬ dragout »¼
Solvent in dragout + Solvent in rinse water = 1 kg
b) Solvent concentration, S (kg solvent/kg water)
Assume the density of the solvent is the same as water.
There is 1 kg of solvent in the 1,000 kg of liquid leaving the process.
The mass of solvent in the dragout is very much less than the amount in the rinse water.
As a good approximation, assume solvent in the dragout = 0
Solvent concentration in rinse water
= 1 kg solvent/1,000 kg water= 106 mg solvent/1,000 L = 1,000 mg/L
106
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
4.30 OIL AND GREASE REMOVAL
Oil and grease (O&G) are removed in two stages: (1) API Separator that removes free
oil, and (2) dissolved air flotation that removes emulsified oil after emulsion breaking by
coagulation. The flow is 160 m3/d. The process and some performance data are shown in
Figure P4.30. (a) What mass and volume of O&G are removed in the API separator? The
mass density of the oil is 900 kg/m3. (b)What mass of COD, BOD and sulfide (S2-) are
removed by dissolved air flotation?
Oil & Grease
(O & G)
= 7,720 mg/L
API
Separator
O & G = 550 mg/L
COD = 1,460 mg/L
BOD5 = 550 mg/L
Sulfide = 8 mg/L
Coagulation
Flocculation
Dissolved
Air
Flotation
Concentrated
pollutants
O & G = 145 mg/L
COD = 970 mg/L
BOD5 = 340 mg/L
Sulfide = 2.7 mg/L
Concentrated
pollutants
Figure P4.30
Solution
a) API Separator
Mass of oil (O & G) removed = (160 m3/d)(7,220 mg/L – 550 mg/L)(103 L/m3)(kg/106 mg)
= 1,067 kg/d
Volume of oil (O & G) = (1,067 kg/d)/(900 kg/m3) = 1.2 m3/d
b) Dissolved air flotation
Influent flow = 160 m3/d (Neglect 1.2 m3 withdrawn in the API separator.)
O&G
(160 m3/d)(550 mg/L – 145 mg/L)/1,000 = 64.8 kg/d
COD (160 m3/d)(1,460 mg/L – 970 mg/L)/1,000 = 78.4 kg/d
BOD
(160 m3/d)(550 mg/L – 340 mg/L)/1,000 = 33.6 kg/d
Sulfide
(160 m3/d)(8 mg/L – 2.7 mg/L)/1,000 = 0.85 kg/d
4.31 RIVER POLLUTION NEWSPAPER STORY
A city of 200,000 population discharged sewage that had been treated to remove 90% of the
influent suspended solids into a small stream. Above the effluent discharge the stream flow
was 10,000 m3/d with zero suspended solids. A newspaper wrote that the city’s discharge
was equivalent to the untreated sewage of a population of 20,000 people. The actual and
implied situations are shown in the accompanying sketch. The mass of suspended solids
discharged is the same in each case. Compare the suspended solids concentration downstream
and comment on whether the newspaper’s analogy is fair. Make the material balance to see
the true picture.
107
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Sewage effluent
80,000 m3/d
2,000 kg SS/d
Sewage effluent
8,000 m3/d
2,000 kg SS/d
Actual Situation
Stated Equivalent
Situation
Figure P4.31
Solution
Assumptions:
1. With an average per capita flow of 0.4 m3/d, a city of 200,000 has a flow of 80,000
m3/d.
2. With an average per capita suspended solids (SS) load of 0.1 kg/d, the treatment plant
influent has 20,000 kg SS/d and the effluent, after 90% removal, has 2,000 kg SS/d.
Effluent SS concentration = (2,000 kg SS/d)/(80,000 m3/d)
= 0.25 kg SS/m3 = 250 mg SS/L
3. A city of 20,000 would have
Flow = (20,000 persons)(0.4 m3/cap-d) = 8,000 m3/d
Suspended solids load = (0.1 kg SS/cap-d)(20,000 persons) = 2,000 kg SS/d
SS concentration = (2,000 kg SS/d)/(8,000 m3/d) = 0.25 kg SS/m3 = 250 mg SS/L
In terms of suspended solids discharged the newspaper analogy is correct. The mass of solids discharged
is the same. However, this is misleading because it distorts the impact of the waste input on the river
quality.
Make the material balance to compute the downstream suspend solids concentrations.
Upstream solids + Input solids = Downstream solids
Upstream conditions:
Flow = 10,000 m3/d
Assume Upstream solids = 0
Actual situation:
2,000 kg/d = (10,000 m3/d + 80,000 m3/d)(C)
C = 0.022 kg SS/m3 = 22 mg SS/L
Stated equivalent situation:
2,000 kg/d = (10,000 m3/d + 8,000 m3/d)(C)
C = 0.11 kg SS/m3 = 110 mg SS/L
The actual downstream concentration is one-fifth that implied in the news story.
108
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
4.32 RIVER DILUTION
River water quality, measured by ultimate BOD, ammonia nitrogen (NH3 as N) and
dissolved oxygen (DO), is known at two locations as shown in Figure P4.32. The only water
flowing into the river between these two locations is clean groundwater; ‘clean’ meaning
that it contains negligible amount of ammonia and BOD, and no dissolved oxygen. Use
the given data to estimate the mass of BOD, ammonia nitrogen, and dissolved oxygen that
have been lost between the sampling stations due to reactions in the river.
Station 1
BODUlt = 10 mg/L
NH4-N = 8 mg/L as N
DO = 9 mg/L
Station 2
BODUlt = 2 mg/L
NH4-N = 5 mg/L as N
DO = 6 mg/L
Q = 5 m3/s
Q = 6 m3/s
Clean groundwater
Figure P4.32
Solution
Material balance around river section from Station 1 to Station 2:
Mass lost = Input upstream (Station 1) – Output downstream (Station 2)
Sample calculation for BOD:
Upstream BOD = (10 mg/L)(5 m3/s)(1,000 L/m3/(106mg/kg) = 0.050 kg/s
Downstream BOD = (2 mg/L)(6 m3/s)(1,000 L/m3)/(106mg/kg) = 0.012 kg/s
Difference (lost) = 0.050 kg/s – 0.012 kg/s = 0.038 kg/s
Table S4.32 shows the results for all three quality constituents.
Parameter
Upstream
Downstream
Difference
Flow
5 m3/s
6 m3/s
BOD
10 mg/L
0.050 kg/s
2 mg/L
0.0120 kg/s
0.0380 kg/s
Ammonia
8 mg/L
0.040 kg/s
5 mg/L
0.0300 kg/s
0.0100 kg/s
DO
9 mg/L
0.045 kg/s
6 mg/L
0.0360 kg/s
0.0090 kg/s
Table S4.32
109
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
4.33 OIL SANDS
The oil sands near Fort McMurray, AB, are a major contributor to Canada’s oil production.
The oil sands are 15% oil and 85% sand. In the initial stages of processing, the oil sands
are mixed with warm water and sent to a settling tank. As shown in Figure P4.33 three
layers form in the settling tank: a bottom layer containing 95% sand and 5% water, a
middle layer of 5 % oil and 95% water, and a top layer of 70% oil and 30% water. All
percentages are on a mass basis. The top layer is skimmed off and refined to recover 90%
of the oil that was in the oil sand input to the process. The middle and bottom layers are
sent for disposal. How much warm water is added to 100 tonnes of oil sands in this process
to recover the oil?
Top = 70% oil + 30% water
Mid = 5% oil + 95% water
Bottom= 5% water + 95% sand
Three layers form in the settling tank
Figure P4.33
.
110
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Solution
Basis: 100 T (100,000 kg) tar sand soil = 15 T oil + 85 T sand
Define
M = mass of oil
MTotal = total mass of oil input = 15 T
MT = mass of oil in top layer of the settling tank
MM = mass of oil in the middle layer of the settling tank
MB = mass of oil in the bottom layer of the settling tank = 0 T
WTotal = total mass of water added to the process
WT = mass of water in top layer of the settling tank
WM = mass of water in middle layer of the settling tank
WB = mass of water in bottom layer of the settling tank
S = mass of sand input = 85 T
Top Layer: 70% oil + 30% water
90% of oil of the 15 T oil input is recovered in the top layer
MT = 0.9 MTotal = 0.9 (15 T) = 13.5 T
MT = 0.7(MT + WT)
13.5 T = 0.7(13.5 T + WT)
WT = 0.3(13.5 T)/(0.7) = 5.8 T
Middle Layer: 5% oil + 95% water
MM = MTotal – MT = 15 T – 13.5 T= 1.5 T
MM = 0.05(MM + WM)
1.5 = 0.05(1.5 T + WM)
WM = (1.5 T)(0.95)/0.05 = 28.5 T
Bottom Layer: 95% sand + 5% water
S = 100 T soil – 15 T oil = 85 T
WB = 0.05(S + WB)
WB = (85 T)(0.05)/0.95 = 4.5 T
Overall Material Balance
Oil: MT + MM + MB = 13.5 T + 1.5 T + 0 T = 15 T
Water: WT + WM + WB = 5.8 T + 28.5T + 4.5 T = 38.8 T
Sand: S = 85 T
Summary
38.8 T of warm water added to 100 T of oil sands to yield 15 T of oil.
Middle layer of waste to disposal = 28.5 T + 1.5 T = 30 T
Bottom layer of waste to disposal = 85 T + 4.5 T = 89.5 T
111
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
4.34 WATER SOFTENING SLUDGE
Softening water by precipitation produces 200,000 kg/d of sludge, which is 15% solids by
weight and specific gravity 1.05. The sludge solids are calcium and magnesium precipitates,
CaCO3 and Mg(OH2). The calcium carbonate, CaCO3, can be converted to calcium oxide,
CaO, for reuse by burning (calcining). Before entering the calcining furnace, the sludge
is thickened with a centrifuge that captures 95% of the feed solids. The sludge that leaves
the centrifuge and enters the furnace is 65% solids by weight. The process flow diagram
is shown in Figure P4.34. How many kilograms of solids and kilograms of water are fed
to the furnace?
Soft water
To consumers
Hard water
Water
softening
process
Sludge
15% solids
Centrifuge
Non-recoverable solids
and carriage water
Thickened sludge
65% solids
Recovered
calcium oxide
Calcining
furnace
Ash & slag
Figure P4.34
Solution
Basis: 1 day = 200,000 kg sludge
Tie variable: Solids
Sludge to centrifuge
Mass total = 200,000 kg
Mass solids @ 15% solids = (0.15)(200,000 kg) = 30,000 kg
Mass water = Mass total – Mass solids = 200,000 kg – 30,000 kg = 170,000 kg
Material balance on centrifuge solids
Input: 30,000 kg
Thickened: 0.95(30,000 kg) = 28,500 kg
Centrate: 0.05(30,000 kg) = 1,500 kg
Material balances on centrifuge water
Input: 170,000 kg
Thickened:
0.65(Water + Thickened solids) = (Thickened solids)
0.65(Water + 28,500 kg) = 28,500 kg
Water = 0.35(28,500 kg)/0.65 = 15,350 kg
Centrate: 170,000kg – 15,350 kg = 154,650 kg
112
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Summary of material flows
Thickened solids to calcining furnace= 28,500 kg solids + 15,350 kg water
= 43,850 kg
Centrate output = 1,500 kg solids + 154,650 kg water = 156,150 kg
4.35 ALUM COAGULATION SLUDGE
Alum sludge from coagulation-sedimentation-filtration water treatment process has a gelatinous
appearance and it is difficult to thicken and dewater. Gravity thickening will increase the
solids concentration from 1% to 2% (dry mass basis). Dewatering by a pressure filter can
produce 18% solids (dry mass). Make the material balance for 40,000 L/day of 1% alum
sludge treated by (a) gravity thickening and (b) pressure filtration.
Solution
Basis: 40,000 L of alum sludge (1 day’s flow)
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113
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Sludge Feed
Volume of alum sludge = 40,000 L
Assume density of 1% alum sludge = 1 kg/L
Mass of alum sludge in = (1 kg/L)(40,000 L) = 40,000 kg
Mass of solids in sludge = 0.01(40,000 kg) = 400 kg
Mass of water in sludge = 40,000 kg – 400 kg = 39,600 kg
a) Gravity Sludge Thickener
Thickened sludge = 2% solids
Mass of solids in = 400 kg
Mass of solids out in thickened sludge = 400 kg
Sludge = 2% solids
From definition of mass fraction:
Water in sludge = (400 kg)(1 – 0.02)/0.02 = 19,600 kg
Mass of thickened sludge = 19,600 kg water + 400 kg solids = 20,000 kg
Thickener supernatant = Mass in – Mass of thickened sludge
= 40,000 kg – 20,000 kg = 20,000 kg
Flow rate of thickened sludge (assuming 1 kg/L)
= (20,000 kg/d)/(1 kg/L) = 20,000 L/d
Flow rate of thickener supernatant = 40,000 L/d – 20,000 L/d = 20,000 L/d
Note: Doubling the solids concentrations reduces the volume by half. The thickened sludge
flow rate and the thickener supernatant flow rate are equal.
b) Pressure filter = 18% solids
Mass of solids into filter = 400 kg
Mass of solids out in filter cake = 400 kg (ignore solids in the filtrate)
Sludge = 18% solids
From definition of mass fraction:
Water in sludge = (400 kg)(1 – 0.18)/0.18 = 1,822 kg
Mass of filter cake = 1,822 kg water + 400 kg solids = 2,222 kg
4.36 FOUR-STAGE EVAPORATION
A four-stage evaporator produces pure water as shown in Figure P4.36. Calculate the
composition of the concentrate stream that leaves stage 4.
114
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Feed
1,000 kg/h
1% TDS
Conservation of mass
250 kg/h water
0 TDS
Evaporator
230 kg/h water
0 TDS
Evaporator
Evaporator
170 kg/h water
0 TDS
Evaporator
130 kg/h water
0 TDS
Concentrate
Figure P4.36
Solution
Basis: 1 hour of operation = 1,000 kg feed
Overall material balance on water:
Water in feed = 0.99(1,000 kg) = 990 kg
Water in concentrate = 990 kg – 260 kg – 230 kg – 170 kg – 130 kg = 200 kg
Overall material balance on salt: salt in concentrate
= (1,000 kg)(0.01) – 0 – 0 – 0 – 0 = 10 kg
Percent salt in concentrate = 100[10 kg/(10 kg + 200 kg)] = 4.76%
4.37 BIOCONCENTRATION
The bioconcentration factor is the equilibrium ratio of the chemical concentration in the
fish flesh to the concentration in the water. The data in Figure P4.37 were obtained by
keeping fish in water that had a constant concentration of 10 µg/L of Chemical A and
then putting the fish into clean water. Fish were sacrificed every few days to measure the
chemical content of their flesh. The graph shows that the uptake of A was rapid and the fish
are near equilibrium with the water after 4 days. The equilibrium fish flesh concentration
at the end of the uptake period is Cf = 20 µg/kg. Calculate the bioconcentration factor.
115
Fish flesh conc. (—g/kg)
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
20
15
Uptake
10
Depuration
(cleansing)
5
0
0
1
2
3
4
5
6
7
8
9
Time of exposure (days)
Figure P4.37
Solution
The bioconcentration factor is
Kb = Cf/Cw = (20 µg/kg)/(10 µg/L) = 2 (L/kg)
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116
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
4.38 HENRY’S LAW AND VOLATILIZATION
Chemicals that are not highly soluble will move toward equilibrium between a water phase
and a gaseous phase (air in this case). Assume that a chemical, call it A, has been put into
water at time = 0 and that after some time it has come to equilibrium with the air that
is above the water. The conditions are shown in Figure P4.38. We will assume that the
chemical in the air compartment is not being swept away and the chemical in the water is
not disappearing due to biodegradation or hydrolysis or some other mechanism.
•
Air
• •
•
•• •
•
Water • • • •
• •••
• •• • •
Time
Time = 0
•
•
•
• •
•
•• •
• • ••
•
• • • ••
At Equilibrium
Figure P4.38
The relative concentrations at equilibrium are defined by Henry’s Law. Henry’s constant is
found with many different units and this can be very confusing. Use the dimensionless form,
which can be either mole/mole or ppm/ppm. Estimate the dimensionless Henry’s constant
based on the diagram, in which one dot equals 0.001 mole of chemical. The volume of the
water compartment is 2 L; the volume of the air compartment is 1 L.
Solution
Initial water phase concentration = (20 dots)(0.001 mol/dot) = 0.02 mol
Equilibrium water phase concentration = (16 dots)(0.001 mol/dot) = 0.016 mol
Equilibrium air phase concentration = (4 dots)(0.001 mol) = 0.004 mol
Henry’s constant = H = Caq/Cgas with both concentrations in mol/L
Caq = (0.016 mol)/(2 L) = 0.008 mol/L
Cgas = (0.004 mol)/(1 L) = 0.004 mol/L
H = (0.008 mol/L)/(0.004 moles/L) = 2 (dimensionless)
Optional: Redo the problem for the case where each dot represents 1 ppm mass in water and 1 ppmv
in air.
117
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
4.39 PARTITIONING OF PYRENE IN SOIL
An accident has spilled 12 kg of pyrene onto vacant land next to a factory. The chemical has
seeped into the soil and moved into the groundwater. For simplicity, assume that a volume
of soil that is 100 m long, 15 m wide, and 4 m deep is affected and that the concentration
of pyrene is uniform throughout this volume. Also, assume the system is at equilibrium
and that there is no movement of the groundwater or the chemical. The porosity of the
soil is 0.45 (45% voids and 55% solids). The bulk density of the dry soil is 1,300 kg/m3.
Estimate the mass of pyrene adsorbed onto the soil and dissolved in the groundwater, and
calculate the concentration in the water.
The solubility of pyrene in water is 0.135 mg/L. The soil-water partition coefficients is
Soil-Water
K d = 84
mg/kg soil
= 84 L/kg
mg/L water
Solution
Bulk volume of the affected soil = (100 m)(15 m)(4 m) = 6,000 m3
Mass of bulk soil = (1,300 kg/m3)(6,000 m3) = 7.8x106 kg (bulk soil)
Volume of groundwater in the pores = 0.45(6,000 m3) = 2,700 m3
Volume of soil in the affected volume = 3,300 m3
The equilibrium concentrations of pyrene for each compartment are:
Water solubility
Soil
Cs
Cw = 0.0135 mg/L
Kd Cw = 84
mg/kg soil
Cw
mg/L water
The material balance equation is:
ªmass of pyreneº
ªmass of pyreneº
ªmass of pyreneº
«
» = «
» + «
»
spilled
in soil
¬
¼
¬ in water
¼
¬
¼
Mass of pyrene adsorbed to the soil = (Cs)(Mass of bulk soil) = (Kd)(Cw)(Mass of bulk soil)
= (84 L/kg)(Cw)(7.8x106 kg)/(106 mg/kg) = (1,814 kg-L/mg)(Cw)
Mass of pyrene in the groundwater = Cw(Volume of groundwater)
= (Cw)(2,700 m3)(1,000 L/m3)/(106 mg/kg) = (2.7 kg-L/mg)(Cw)
This shows that nearly all the pyrene is adsorbed to the soil. There is little need to finish the calculation,
but for completeness the concentrations and masses are:
118
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
Mass balance
12 kg = (1,814 kg-L/mg) Cw + (2.7 kg-L/mg) Cw
Cw = (12 kg)/(1,816.7 kg-L/mg) = 0.0066 mg pyrene/L in water
and
Cs = Kd Cw = (84 L/kg)(0.0066 mg/L) = 0.55 mg pyrene/kg soil
Mass in soil = (1,814 kg-L/mg)(Cw) = (1,814 kg-L/mg)(0.0066 mg/L) = 11.98 kg
Mass in water = (2.7 kg-L/mg)(Cw) = (2.7 kg-L/mg)(0.0066 mg/L) = 0.018 kg
Check solubility: The concentration of pyrene in the ground water is Cw = 0.0066 mg/L, and is less
than the solubility of 0.135 mg/L, indicating that all the spilled pyrene is either dissolved in the water
or adsorbed to the soil.
4.40 PARTITIONING 1,2-DICHLOROBENZENE
The aquatic system described in Table P4.40 has been contaminated by a spill of 1400 kg
1,2-dichlorobenzene (1,2-DCB). Assume that by the time equilibrium has been reached 18%
of the chemical has disappeared because of biological or chemical reactions. Estimate the
concentration of 1,2-DCB in the water, sediment, and biota when the system has reached
equilibrium (you don’t need to know how long that takes). The organic carbon content of
the sediment is 5.5%. Bulk density of sediments = 1,500 kg/m3.
119
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Compartment
Water
Sediment
Biota
Conservation of mass
Volume
Density
(m )
(kg/m3)
1,000,000
1,000
30,000
1,500
100
1,100
3
Table P4.40
Solution
Aquatic ecosystem:
Mass water = (106 m3)(1,000 kg/m3) = 109 kg
Mass sediment = (30,000 m3)(1,500 kg/m3) = 4.5x107 kg
Mass biota = (100 m3)(1,100 kg/m3) = 1.1x105 kg
Chemical in ecosystem = (1 – 0.18)(1,400 kg) = 1,148 kg
1,2-DCB partitioning characteristics:
Koc = 1,700(mg/kg)/(mg/L) = 1700 L/kg
Kd = KOC fOC = (1,700 L/kg)(0.055) = 93.5 L/kg
Kb = (56 mg/kg)/(mg/L) = 56 L/kg
The material balance equation is:
ª mass of º
ª mass of º
ª mass of º
ª mass of º
«
»
«
»
«
»
«
»
«1,2-DCB» = «1,2-DCB» + « 1,2-DCB » + «1,2-DCB»
«¬ spilled »¼
«¬in water »¼
«¬in sediment »¼
«¬ in biota »¼
Mass of 1,2-DCB in the groundwater = (Cw)(Volume of groundwater)
= (Cw)(106 m3)(1,000 L/m3) = (109 L)(Cw)
Mass of 1,2-DCB adsorbed to the soil = (Cs)(Mass of sediment)
= (Kd)(Cw)(Mass of sediment)
= (93.5 L/kg)(Cw)(4.7x107 kg) = (4.21x109 L)(Cw)
Mass of 1,2-DCB in the biota = (Cs)(Mass of biota) = (Kb)(Cw)(Mass of biota)
= (56 L/kg)(Cw)(1.1x105 kg) = (0.0062x109 L)(Cw)
Mass balance
1,148 kg = (109 L)(Cw) + (4.21x109 L)(Cw) + (0.0062x109 L)(Cw)
120
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
= (5.216x109 L)(Cw)
Cw = (1,148 kg)(106 mg/kg)/(5.216x109 L) = 0.22 mg/L
and
Cs = Kd Cw = (93.5 L/kg)(0.22 mg/L) = 20.6 mg/kg
Cb = Kb Cw = (56 L/kg)(0.22 mg/L) = 12.3 mg/kg
Mass in water = (109 L)(Cw) = (109 L)(0.22 mg/L)/(106 mg/kg) = 220 kg
Mass in sediment = (4.21x109 L)(Cw) = (4.21x109 L)(0.22 mg/L)/(106 mg/kg)
= 926 kg
Mass in biota = (0.0062x109 L)(Cw) = (0.0062x109 L)(0.22 mg/L)/(106 mg/kg) = 1.4 kg
4.41 PARTITIONING 2,3,7,8-TETRACHLORODIBENZO-DIOXIN
A small marshy area described in Table P4.41 has been contaminated by 2,3,7,8-tetrachlorodibenzodioxin (2,3,7,8-TCDD). The chemical is a carcinogen so there is great interest its fate as a
cleanup plan is formulated. The amount that entered the ecosystem is unknown, but the
measured concentration in the water was 2 mg/L when the problem was discovered. The
scientists believe that the system can be considered at equilibrium. There is virtually no
flow of water through the marsh at this time of year. The chemical is highly soluble, which
means there will be virtually no loss to the atmosphere. The organic carbon content of the
sediment is 15%. Estimate the concentration of 2,3,7,8-TCDD in the water, sediment, and
biota when the system has reached equilibrium.
Volume
Density
(m )
(kg/m3)
1,000
1,000
Sediment
30
1,500
Biota
4
1,100
Compartment
Water
Air
3
10,000
Table P4.41
Solution
Salty marsh ecosystem:
Mass water = (103 m3)(1,000 kg/m3) = 106 kg
Mass sediment = (30 m3)(1,500 kg/m3) = 4.5x104 kg
Mass biota = (4 m3)(1,100 kg/m3) = 4.4x103 kg
Chemical concentration in water = 2 mg/L
121
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Conservation of mass
2,3,7,8-TCDD partitioning characteristics:
Koc = 3.3x106 (mg/kg)/(mg/L) = 3.3x106 L/kg
Kd = KOC fOC = (3.3x106 L/kg)(0.15) = 4.95x105 L/kg
Kb =5,000 (mg/kg)/(mg/L) = 5,000 L/kg
Concentrations
Cw = 2 mg/L
Cs = Kd Cw = (4.95x105 L/kg)(2 mg/L) = 990,000 mg/kg
Cb = Kb Cw = (5,000 L/kg)(2 mg/L) = 10,000 mg/kg
Mass of 2,3,7,8-TCDD in the water = (Cw)(Volume of water)
= (2 mg/L)(1,000 m3)(1,000 L/m3)/(106 mg/kg) = 2 kg
Mass of 2,3,7,8-TCDD adsorbed to the sediment = (Cs)(Mass of sediment)
= (990,000 mg/kg)(4.5x104 kg)/(106 mg/kg) = 44,600 kg
Mass of 2,3,7,8-TCDD in the biota = (Cs)(Mass of biota)
= (10,000)(4.4x103 kg)/(106 mg/kg) = 44 kg
Mass balance
ª mass of
º
ª mass of º ª mass of º
ªtotal mass of º
«
»
«
» «
»
=
2,3,7,8-TCDD
+
«
»
«
»
«2,3,7,8-TCDD» «2,3,7,8-TCDD»
¬2,3,7,8-TCDD¼
«¬ in water »¼
«¬ in sediment »¼ «¬ in biota »¼
2 kg
44,600 kg
44 kg
= 44,646 kg ≈ 45,000 kg
4.42 PARTITIONING DDT
DDT has leaked into an embayment described in Table P4.42 for many years. The leakage
has been stopped, but there is crystalline DDT on the bottom in some places, so the
water concentration should be at the solubility of DDT, which is 0.0055 mg/L. The water,
sediment, and biota can be assumed at equilibrium. The organic carbon content of the
sediment is 6%. The flow of water through the embayment will carry a very small amount
of DDT (at 0.0055 mg/L) and it will be ignored for purpose of making estimates of the
concentrations and masses of DDT in the sediment and biota.
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Compartment
Water
Sediment
Conservation of mass
Volume
Density
(m )
(kg/m3)
1,000,000
1,000
30,000
1,500
500
1,100
3
Biota
Table P4.42
Solution
Embayment ecosystem:
Mass water = (106 m3)(1,000 kg/m3) = 109 kg
Mass sediment = (3x104 m3)(1,500 kg/m3) = 4.5x107 kg
Mass biota = (500 m3)(1,100 kg/m3) = 5.5x105 kg
Chemical concentration in water = 0.0055 mg/L
DDT partitioning characteristics:
Koc = 243,000 (mg/kg)/(mg/L) = 243,000 L/kg
Kd = KOC fOC = (243,000 L/kg)(0.06) = 14,600 L/kg
Kb =54,000 (mg/kg)/(mg/L) = 54,000 L/kg
Concentrations
Cw = 0.0055 mg/L
Cs = Kd Cw = (14,600 L/kg)(0.0055 mg/L) = 80.3 mg/kg
Cb = Kb Cw = (54,000 L/kg)(0.0055 mg/L) = 297 mg/kg
Mass of DDT in the water = (Cw)(Volume of water)
= (0.0055 mg/L)(106 m3)(1,000 L/m3)/(106 mg/kg) = 5.5 kg
Mass of DDT adsorbed to the sediment = (Cs)(Mass of sediment)
= (80.3 mg/kg)(4.5x107 kg)/(106 mg/kg) = 3614 kg ≈ 3600 kg
Mass of DDT in the biota = (Cs)(Mass of biota)
= (297 mg/kg)(5.5x105 kg)/(106 mg/kg) = 163 kg ≈ 160 kg
Mass balance
ªtotal mass of º
ªmass of DDT º
ªmass of DDT º ªmass of DDT º
«
» = «
» + «
»«
»
DDT
¬
¼
¬ in water ¼
¬ in sediment ¼ ¬ in biota ¼
5.5 kg
3,614 kg
= 3,782.5 kg ≈ 3,800 kg
123
163 kg
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance and pollution prevention
5MATERIAL BALANCE AND
POLLUTION PREVENTION
Tutorial Note
At this stage in the book the student has only a basic repertoire of analytical and problem solving
tools. Because of this some instructors may postpone study of this chapter, or they may discuss
the text material but elect not to assign problems. Those are valid instructional options. But it
is possible to generate a lively flow of ideas and discussion without doing detailed calculations,
and this can allow some important ideas to be assimilated without the distraction of worry
about units and doing the algebra. There is even benefit to be gained by simply reading some
of the problems.
Some of the problems will require research on treatment technology, or some brief instruction
about a few separation processes, but most of them ask for ‘discussion’ or ‘explanation’ or ‘a
proposal for making things better’. This is very much in the spirit of the work at the outset of many
pollution prevention projects – that is, asking questions and making suggestions. Most of the
problems have no single correct answer. It should follow that they have no answers, suggestions,
or discussion that is judged to be silly or wrong. Feedback should, generally, be encouraging.
At the same time, everyone understands that there are ideas that need improvement, ideas
that are good, and some ideas that are better.
Some people call this ‘brain storming’, a term we do not much like because ‘storming’ sounds
unfocused and random. Engineering design, even in the cloudy preliminary steps, is not that
way. Perhaps those who favor the term, imagine that creativity happens by chance. It does
not. ‘Facilitated discussion’ is just as badly flawed because it suggests that someone with no
particular expertise can extract essential information and valuable ideas.
We have no particular name for this activity. We think of it as ‘engineering’ in the sense of the
German word ingenieur, which comes from ingenuity and ingenious. What we seek is ingenuity
and creativity. That requires practice, collaboration, and communication. We hope these exercises
will lead to thinking of this kind.
5.1
METABOLISM OF A FACTORY
A factory, like a city or a living organism, has a metabolism that must be fed and cleansed
as shown in Figure P5.1. Raw materials enter. Products, by-products, waste, solid waste, and
gaseous emissions leave. ‘Metabolism’ suggests that the system reacts to changes. A change
in raw material will change the product and the wastes. A change in a manufacturing step
will activate changes. A change in water availability (water is usually a critical raw material)
may propagate through the process and cause many kinds of changes. Have a discussion
124
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance and pollution prevention
with fellow students about industrial metabolism. What kinds of changes in inputs might
occur, by intent or by accident? What kinds of changes in outputs could occur, by intent
or by accident? Can the metabolism adapt to material reuse or recycle. Can it adapt to cuts
in key materials, or in processing conditions (e.g. temperature)?
Gaseous
Emissions
Byproducts
Product
Wastewater
Raw materials
Solid and liquid
wastes
Figure P5.1
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125
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance and pollution prevention
Solution
This is a problem for discussion. Figure S5.1 suggests some ideas for discussion.
Raw materials
Energy
Energy
Manufacturing
Product Use
Waste
Disposal to
landfill
Waste
(a) Original Process
Raw materials
Energy
Energy
Manufacturing
Recycle
Product Use
Disposal to
landfill
Waste
Waste
Reuse
Remanufacture, Recycle
(b) Possibilities for process redesign
Figure S5.1
5.2
ZERO DISCHARGE
The ideal process would create no waste and operate at a profit. All raw materials would
be converted into usable products, and all processing aids (catalysts, solvents, water, etc.)
would be fully regenerated to their original quality to be reused. However, the perfect or
ideal process does not exist. A small fraction of the raw material is lost as waste. Consumed
or degraded process inputs, including water, become part of the waste stream. The challenge
is to design processes that emphasize waste minimization. Propose some practical definitions
or short descriptions of ‘zero discharge’.
Solution
The following practical definitions of “zero” accept that a zero mass of all liquid, gaseous, and solid
outputs emissions is impossible. A practical definition could be made for zero discharge of dangerous
materials.
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SOLVED MATERIAL BALANCE PROBLEMS:
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Material balance and pollution prevention
1. Eliminate priority pollutants or toxic substances from the wastewater effluent.
2. Eliminate toxic air pollutants from gaseous emissions.
3. Eliminate RCRA controlled materials from solid wastes.
4. Discharge no water effluent stream from the processing site.
5. All wastewater, after treatment, is recycled and reused.
6. Slurries and sludge are converted to a solid waste by evaporation and the solid waste
is shipped to a landfill.
7. If a discharge is unavoidable, it should not create any harm in the receiving environment.
5.3
POLLUTION PREVENTION FOCUS QUESTIONS
Useful questions about pollution prevention design are stimulated by thinking in terms of
‘zero discharge’ of materials that are toxic or valuable. What are some questions that could
stimulate the creative design process?
Solution
Some questions are:
• What waste streams are generated from the plant?
• Which material handling or manufacturing operations generate these waste streams?
• Which wastes are hazardous and which are not?
• What raw materials are used in the waste generating processes?
• How much of each raw material leaves as useful products and how much is lost as
waste?
• Are wastes generated by mixing otherwise recyclable material with a waste material?
• Can the process be changed so that a troublesome material is not needed or not
generated?
• Can material be handled differently to reduce losses?
• Can the material be handled at a temperature more conducive to fume or dust
suppression?
• What process controls can be used to improve process efficiency?
• What housekeeping practices can be used to limit waste production?
5.4
INEFFICIENT WATER USE
Figure P5.4 shows an industry that has a once-through water use system. There is no recycle
or reuse. The effluent from the water softener is high in solids (dissolved or particulate,
depending on the method of softening that is used). The effluents from the boiler and the
127
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance and pollution prevention
cooling tower are blowdown, an intentional removal of salt-laden water that is necessary
to control scaling and corrosion in the equipment. The sanitary sewage contains toilet and
kitchen waste and must be handled in a way that protects humans from waterborne disease.
Do you see any opportunities to reuse or recycle water? This is difficult to answer precisely
without information about the water quality, but you may see opportunities that are worth
deeper investigation.
Figure P5.4
Solution
This is an open-ended discussion question that is meant to provoke discussion. No solution is provided.
5.5
SEMICONDUCTOR MANUFACTURING
A valuable raw material, Gallium arsenide (GaAs), is lost in process wastewater, as shown
in Figure P5.5. The rough process diagram shows three processing areas, the water input
to each, and the waste that leaves. All water that enters leaves as wastewater. Hydrofluoric
acid waste is mixed with the acidic process wastewater prior to neutralization with hydrated
lime, Ca(OH)2. The calcium (Ca) in the lime reacts with the fluoride (F) in the acid to
form a CaF2 precipitate. The CaF2 could be sold if it were pure, but it is contaminated by
GaAs. The GaAs also makes the sludge cake a hazardous waste that is an expensive disposal
problem.
128
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance and pollution prevention
You should be able to think of a number of pollution prevention opportunities. Do not
worry about which ones are the best, or even whether a proposal is technically easy or
difficult or infeasible. Generate ideas, supported with process flow diagrams that make your
ideas clear. (Verify the material balance.)
Figure P5.5
Solution
This is an open-ended discussion question that is meant to provoke discussion. No solution is provided.
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129
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
5.6
Material balance and pollution prevention
WATER FOR SOFT DRINK MANUFACTURING
An Australian company produces about 250 million liters of soft drinks per year and
it produces 1.5 million liters of process water daily using conventional flocculation and
filtration. The treatment plant can operate at 200,000 L/h. Three types of filters are used:
sand filters, carbon filters, and polishing filters. The company cleans the sand and carbon
filters daily by forcing water backwards through the filter, a process known as backwashing.
The daily volume of 200,000 L/d backwash water mixes with other factory wastewaters and
is discharged to the city sewer. The city is paid a discharge fee based on the concentration of
the effluent. The backwash water is more dilute that the other wastes. A cleaner production
initiative proposes recycling some of the backwash water to the inlet of the water treatment
plant. This can be done safely in the ratio of 1 part recycle to 4 parts non-recycled water.
This will have a capital cost of $150,000 but will save $100,000 per year on the purchase
of water. The city wants $15,000 more per year in wastewater discharge fees after recycling
is implemented.
a) After recycle is implemented, what will be the discharge to the sewer?
b) Is backwash recycle a good investment under these terms? Explain.
c) Is charging based on pollutant concentration fair? What other basis for discharge fees
might be proposed and how would the total cost be changed if it were implemented?
Solution
a) Discharge to the sewer.
Treatment plant influent of 1,500,000 L/d is sufficient to recycle all 200,000 L/d of filter
backwash water.
Effluent is reduced from 1,500,000 L/d to 1,300,000 L/d (a 13% reduction). The effluent
concentration will increase after the dilute backwash water is removed from the discharge.
b) This is a good investment because the cost of purchasing city water is reduced by
$100,000.
The payback time for the $150,000 capital investment is only 1.5 years.
c) The industry should negotiate new effluent discharge fees. Most plans charge on the
basis of flow and mass loadings; mass loading = (flow)(concentration). If the flow is
reduced by 13% the mass loading will be reduced as well, probably not by 13%, but
enough to reduce the load on the city wastewater treatment plant.
130
SOLVED MATERIAL BALANCE PROBLEMS:
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5.7
Material balance and pollution prevention
HIDDEN AND AVOIDED COSTS
Managers often have difficulty justifying investments because the costs of hazardous chemicals
and wastes are often allocated to overhead, and are therefore hidden. Consider the hidden
and direct costs of hazardous materials use and disposal, and the indirect benefits and cost
avoidance gained through using pollution prevention. Facilities that account for the costs
of using, managing, and disposing of toxic substances separately have an average of three
times as many pollution prevention projects as facilities that don’t!
a) Two hidden costs are permits and spill reporting. Name others.
b) List at least five benefits of pollution prevention.
Solution
a) Hidden Costs: Permits, monitoring and inspections, reporting and record keeping,
safety and health concerns, protective equipment, material and disposal costs, spill
reporting, public image, potential fines and penalties, long-term liability.
b) Benefits of Pollution Prevention: Reduced material costs, Accident and injury prevention,
Community relations, Reduced regulatory costs, Private property protection, Reduced
liability, Reduced insurance costs, and Decreased disposal costs.
5.8
SOLVENT RECOVERY PAYBACK
A solvent recovery/activated carbon system has solved the White Rubber Company’s (Ohio)
need for solvent recovery and the prevention of costly air pollution problems. A solvent
recovery/activated carbon system that can recover over 1,000 gallons of solvent per day has
resulted in a 100% payback and savings for White Rubber of more than $1.3 million. The
total cost of recovering a gallon of solvent is $0.35, which includes the cost of water, steam,
electricity and maintenance. Since the current price of a gallon of solvent is $1.40, the
company saves $1.05 on each gallon that is recovered. In addition to providing more than
95% recovery of solvent, the system also helps to prevent air pollution problems. Influent
to the recovery system contains solvent vapors ranging from 300 to 2,000 ppm. Stack
emissions after carbon treatment never exceed allowable limits. Create a dialog between the
purchasing department, the solvent process manager, and the air pollution control engineer
that might have led to the design and installation of this highly effective system.
131
SOLVED MATERIAL BALANCE PROBLEMS:
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Material balance and pollution prevention
Solution
This is an open-ended discussion question that is meant to provoke discussion. No solution is provided.
5.9
WASTE MINIMIZATION SURVEY
The first step in implementing waste minimization is (1) to eliminate or minimize the waste
volume and (2) to eliminate/minimize the pollutants that are toxic, hazardous, or difficult
to treat. (a) Outline the general steps to characterize the manufacturing process. (b) Outline
a general plan for identifying waste minimization options.
Solution
a) A new or existing process can be characterized by taking the following steps:
1. List all feed materials reacting to make salable products, any intermediates, and
all salable products. Call this List 1.
2. List all other materials in the process, such as non-salable byproducts, solvents,
water, air, nitrogen, acids, bases, and so on. Call this List 2.
3. For each compound in List 2, ask “How can a material from List 1 be used to
do the same function as the compound in List 2?” or “How can the process be
modified to eliminate the need for the material in List 2?”
4. For those materials in List 2 that are the result of producing non-salable products
(i.e., waste byproducts), ask, “How can the chemistry or process be modified
to minimize or eliminate wastes (for example, 100% reaction selectivity to a
desired product)?”
b) To uncover the best options, each waste stream should be analyzed as follows:
1. List all components in the waste stream, along with any key parameters. For
instance, for a wastewater stream these could be water, organic compounds,
inorganic compounds (both dissolved and suspended), pH, etc.
2. Identify the components triggering concern, e.g., hazardous air pollutants (HAPs),
carcinogenic compounds, wastes regulated under the Resource Conservation and
Recovery Act (RCRA), etc. These need to be reduced or eliminated. Determine
the sources of these components within the process.
3. Identify the highest volume materials - often these are diluents, such as water,
air, a carrier gas, or a solvent. These materials frequently control the investment
and operating costs associated with end-of-pipe treatment of the waste streams
and have a significant impact on the process cost of manufacture. Determine
the sources of these high-volume materials within the process.
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SOLVED MATERIAL BALANCE PROBLEMS:
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Material balance and pollution prevention
4. If the components identified in Step 2 are successfully minimized or eliminated,
identify the next set of components that have a large impact on investment and
operating cost (or both) for end-of-pipe treatment. For example, if the aqueous
waste stream was originally a hazardous waste and it was incinerated, eliminating
the hazardous compound(s) may permit the stream to be sent to the wastewater
treatment facility. However, this may overload the biochemical oxygen demand
(BOD) capacity of the existing wastewater treatment facility, making it necessary
to identify options to reduce organic load in the aqueous waste stream.
5.10 VEGETABLE PROCESSING INDUSTRY WATER REUSE
A food industry processing line comprises six steps, as shown in Table P5.10. Washing cleans
field dirt off the vegetables. Trimming removes stems, leaves, peels, bad spots, etc. Water is
used mainly to flush away large pieces of solid material. Finishers are the final cleaning step
before cooking, canning and pasteurizing. The cooker and cook room have high loads due
to spillage and cleanup. The pasteurizer uses a lot of water for cooling. One sewer collects
the wastewater from the entire factory. The flow and waste load (BOD and SS) in this
sewer has been estimated by summing the quantities from each of the six processing areas
and by measuring the full flow in the sewer. These quantities are in good agreement. Water
reuse/recycling may be possible, but water from the upstream processes cannot be used in
downstream because of public health regulations. Waste segregation is another strategy that
may be useful. Propose changes and improvements to the wastewater management system.
Flow
BOD
SS
Process
gpm
mgd
mg/L
lb/d
mg/L
lb/d
1
Washers
500
0.720
165
992
270
1,623
2
Trimming
100
0.144
450
541
3,000
3,067
3
Finishers
150
0.216
230
415
170
307
4
Cooker
10
0.014
3,500
421
7,600
914
5
Cook room
20
0.029
4,400
1,058
8,900
2,140
6
Pasteurizer
300
0.432
30
108
20
72
Calculated Total
1,080
1.555
273
3,535
668
8,663
Measured Total
1,120
1.613
220
2,959
660
8,877
Table P5.10
133
SOLVED MATERIAL BALANCE PROBLEMS:
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Material balance and pollution prevention
Solution
1) Pasteurizer has large flow but low contamination. The water has been used for cooling
and should be suitable for reuse in the Niagara washers or the trimming tables.
2) Small flow, but high BOD and SS from cook room and source cooker. Segregate this
flow for separate treatment to remove solids.
3) Trimming has a large flow, with high SS and low BOD loads. The water is used mainly
for flushing away solids. Consider dry clean up or a different means of transporting
the solids from the trimming tables.
Figure S5.10a shows the original process and Figure S5.10b shows one possible redesign to segregate
waste from the cook and cook room, and to recycle from the pasteurizer to the Niagara washers, and
from the Finishers to the trimming tables. The water use at the trimming station is arbitrarily reduced,
assuming dry conveyance of solids is at least partially possible. The revised BOD and SS loads cannot
be calculated with any precision for the trimming operation but there should be reductions.
Figure S5.10a Original process
Figure S5.10b Redesigned process
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SOLVED MATERIAL BALANCE PROBLEMS:
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Material balance and pollution prevention
5.11 POLYMER RECYCLING
The cartoon in Figure P5.11 shows the operator of a polymer recycling plant taking Euros
out of the process. The idea is that once-used plastic bottles and containers can be chipped,
extruded and granulated for reuse. Of course this consumes energy, water, and perhaps other
chemicals (e.g., detergent). The recycled polymer may have degraded properties compared to
the incoming material. There will be wastes produced. So, the cartoon asks, “Is it worth it?
Do you save money? Do you save resources? Do you help the environment?” These important
questions are not easy to answer. What information is needed to make the analyses? How
do you measure ‘help the environment’? How is the cost or cost savings calculated?
WHAT YOU NEED
Clean water
Energy for heating
detergent
POTENTIAL
PROBLEMS
EXTRUDER-GRANULATOR
Detergent
DETERGENT
effluent
needs
EFFLUENT
treatment
NEEDS
TREATMENT
Multicolored
polymer
Energy to melt the
polymer.
Possible
degradation of
properties
Energy to transport the
polymer.
Energy to granulate the
polymer.
IS IT WORTH IT?
Do you save money?
Do you save resources?
Do you help the environment?
Figure P5.11 Polymer recycling
Solution
This is an open-ended discussion question that is meant to provoke discussion. No solution is provided.
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Material balance and pollution prevention
5.12 COCA-COLA
In Coca-Cola’s 10-K SEC filing submitted in February 2010, the “Raw Materials” section
begins this way: “Water is a main ingredient in substantially all our products... our Company
recognizes water availability, quality, and sustainability ... as one of the key challenges facing
our business.” Coca-Cola bottles Dasani and Vitaminwater as well as soft drinks. (a) The
company needs 333 ounces of water to generate $1 of revenue. Every liter of beverage it
manufactures and sells requires 2.43 liters of water. Shortly after one liter of beverage is
consumed it is returned to the water cycle as urine. What happens to the other 1.43 liters?
(b) This volume of water use is almost 9% lower than ten years ago. That 9% over 10
years is 8,000,000,000 gallons of water saved. Learn how the volume of water needed for
manufacturing was reduced.
Solution
This is an open-ended discussion question that is meant to provoke discussion. No solution is provided.
Suggested reading
Fishman, C 2011, ‘Why GE, Coca-Cola and IBM are getting into the water business.
Fishman, C 2011. The Big Thirst: The Secret Life and Turbulent Future of Water, Simon & Schuster.
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
6MATERIAL BALANCE WITH
CHEMICAL REACTIONS
Tutorial Note
There is an art to making good approximations in chemical process calculations. ‘Good’ means
that the integrity of the answer is maintained while the calculations are simplified. Approximations
that are made in chemical process synthesis fall into 3 classes (Murphy, R2007. Introduction to
Chemical Processes: Principles, Analysis, Synthesis, McGraw-Hill, New York, pp 124-125)
Stream composition approximations
The raw material is pure
The product is pure
Air contains only oxygen and nitrogen
Reactants are fed at stoichiometric rates
System performance approximations
The reactants are completely consumed by reaction
No unwanted side reactions take place in the reactor
The separator separates all components into pure streams
Physical property approximations
Gases behave as ideal gases
Liquids behave as ideal solutions
Solid density is independent of temperature.
The convenience of making these approximations is offset by eliminating the process imperfections
that create waste and pollution. Raw materials are rarely pure. Exhaust air contains more than
oxygen and nitrogen. Pure products are created by removing impurities, which show up in
waste streams and emissions. There are side reactions, and reactants are not always completely
consumed. Separations do not cleanly split a mixture of two materials into two pure materials.
Nevertheless, these approximations are useful, especially in the early stages of design when
synthesis and analysis happen iteratively as the design details gradually emerge. Use them, but
be aware when you are doing so, and acknowledge them so a more refined analysis can verify
the validity or adjust the solution to a more precise conclusion.
137
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
6.1
Material balance with chemical reactions
BURNING OCTANE
Write a balanced stoichiometric equation for burning octane (C8H18) with oxygen. Interpret
the reaction by writing down the number of molecules, moles, and grams of each chemical
compound that appears in the equation.
C8H18 + a O2 → b CO2 + c H2O
Solution
Balanced reaction
C8H18 + 12.5 O2
→ 8 CO2 + 9 H2O
Conversion from moles to mass uses the molar mass of each compound.
Atomic masses:
C = 12, H = 1, and O = 16.
→
C8H18
+
12.5 O2
1 molecule of octane
reacts
12.5 molecules
with
of oxygen
+
12.5 g-mol O2
1 g-mol C8H18
8 CO2
+
9 H2O
to
8 molecules of
and
9 molecules
give
carbon dioxide
→
8 g-mol CO2
of water
+
9 g-mol H2O
Molar masses (g/g-mol)
114
32
44
18
12.5(2)(16) =
8(44) = 352
9(18) = 162
Reacting masses (g)
1(114) = 114
400
6.2
COMBUSTION OF GASOLINE
Gasoline is a mixture of hydrocarbons plus impurities. Let us suppose an empirical composition
of C8H16. (a) Calculate the amount of oxygen required for stoichiometric combustion. (b)
Air is 23.2% oxygen by mass. How much air is required per 1 kg of gasoline? (c) Calculate
the air to fuel ratio, as kg air/kg gasoline.
Solution
Balanced reaction
C8H16
+
12 O2
→
8 CO2
+
8 H2O
Molar masses (kg/kg-mol)
112
32
44
18
1(112) = 112
12(32) = 384
8(44) = 352
8(18) = 144
Reacting masses (kg)
138
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
a) Oxygen required per kg C8H16 = 384/112 = 3.43 kg O2/kg C8H16
b) Air required per kg C8H16 = (384 kg O2/kg C8H16)/(0.232 kg O2/kg air)
= 1,655 kg air/kg C8H16
c) Air/Fuel ratio = (1,655 kg air)/(112 kg C8H16) = 14.78 kg air/kg C8H16
6.3
COMBUSTION - 1
What is the minimum mass of oxygen, in grams, needed to burn 25.1g of C5H10O?
C5H10O + 7O2 → 5CO2 + 5H2O
Solution
Basis = 25.1 g of C5H10O
Molar masses: C5H10O = 86 g/mol
O2 = 32 g/mol
Moles of C5H10O burned = (25.1 g C5H10O)/(86 g C5H10O/mol C5H10O) = 0.291 mol C5H10O
Moles of O2 required = (0.291 mol C5H10O)(7 mol O2/mol C5H10O) = 2.04 mol O2
Mass of O2 required = (2.04 mol O2)(32 g O2/mol O2) = 65.3 g O2
6.4
COMBUSTION - 2
A 10.00 g mass of unknown substance is burned to yield 11.53 g H2O and 28.16 g CO2.
What is the empirical formula of substance?
Solution
Molar masses: C = 12 g/g-mol
CO2 = 44 g/g-mol
H = 1 g/g-mol
All C in the substance becomes CO2
Mass of CO2 = 28.16 g
Moles of CO2 = 28.16 g CO2/44 g/g-mol = 0.64 mol CO2
1 mole of C in CO2
Moles of C in the compound = 0.64 mol C
Mass of C = (0.64 mol C)(12 g/g-mol) = 7.68 g C
139
H2O = 18 g/g-mol
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
All H in the substance becomes H2O
Mass of H2O = 11.53 g
Moles of H2O = 11.53 g H2O/18 g/g-mol = 0.64 mol H2O
2 moles of H in H2O
Moles of H in the compound = 1.28 moles
Mass of H = (1.28 mol H)(1 g/g-mol) = 1.28 g H
Mass of C + H = 7.68 g C + 1.28 g H = 8.96 g
Mass of sample = 10.0 g
Mass of O by difference = 1.04 g O2
Moles of O = (1.04 g O2)/(16 g/g-mol) = 0.065 moles
Molar ratio of C:H:O = 0.64 : 1.28 : 0.065
Smallest number 0.065
Divide the ratio by the smallest number we get
Molar ratio of C:H:O = 10 : 20 : 1
Empirical formula is C10H20O
6.5
COMPOSTING STOICHIOMETRY
Composting is not combustion, but is a process that uses oxygen to convert organic
compounds into carbon dioxide and water. The empirical composition of the dry raw refuse
material is C32H50O26N. The decomposition products are an empirical residue with the
formula C11H14O4N plus carbon dioxide, ammonia and water. When the compost process
is stable (changes are too small to be of practical importance), 50% of the carbon in the
refuse (C32H50O26N) has been converted to CO2. The reaction is
C32H50O26N + x O2 → a C11H14O4 + b CO2 + c NH3 + d H2O
a) Derive the stoichiometric coefficients for oxygen and the products that balance
the reaction.
b) How much compost is obtained from 1,000 kg of original dry material?
c) How much oxygen is consumed per ton of original dry material?
Solution
50% conversion means that 50% of the 32 C atoms in the raw material are converted to CO2 and 50%
become part of the compost product.
140
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
a) Derive the stoichiometric coefficients
Balance on C: b = 32/2 = 16
32 = 11a + b = 11a + 16
a = 16/11 = 1.45
Balance on N: 1 = c
Balance on H: 50 = 14a + 3c + 2d = 14(1.45) + 3(1) + 2d
d = 26.7/2 = 13.35
Balance on O: 26 + 2x = 4a + 2b + d
2x = 4(1.45) + 2(16) + 13.35 – 26 = 25.15
x = 25.15/2 = 12.58
Balanced stoichiometric equation
C32H50O26N + 12.58 O2 → 1.45 C11H14O4 + 16 CO2 + NH3 + 13.35 H2O
b) Mass of compost from 1,000 kg refuse
Molecular masses
Refuse
C32H50O26N
864 kg/kg-mol
Compost
C11H14O4
210 kg/kg-mol
864 kg refuse yields 1.45(210 kg compost) = 304.5 kg compost
1,000 kg refuse yields (1,000 kg refuse)(304.5 kg compost)/(864 kg refuse)
= 352 kg compost
c) Oxygen consumption per 1,000 kg refuse
Oxygen consumption = 12.58 moles O2 per mole of refuse
(1,000 kg refuse)/(864 kg/kg-mol) = 1.157 kg-mol refuse
Oxygen = (12.58 kg-mol O2/kg-mol refuse)(1.157 kg-mol refuse)(32 kg O2/ mol O2)
= 465.8 kg O2
6.6
COMBUSTION OF MUNICIPAL REFUSE
A dry municipal refuse has the following empirical chemical composition, C59H93O37N. (a)
Write a balanced stoichiometric equation for the complete combustion of this material. (b)
Find the mass of oxygen consumed in the combustion of 1,000 kg/h of this material. How
much air will be required to sustain this combustion?
141
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
Solution
a) Stoichiometric coefficients that balance the reaction equation
C59H93O37N + x O2 → a CO2 + b H2O + c NO2
where a = 59
b = 93/2 = 46.5
c=1
2x + 37 = 2a + b + 2c = 2(59) + 46.5 + 2 = 166.5
x = (166.5 – 37)/2 = 129.5/2 = 64.75
Balanced reaction:
C59H93O37N
+
64.75 O2
→ 59 CO2 + 46.5 H2O + NO2
Molar mass (kg/kg-mol)
1,407
32
Reacting masses (kg)
1,407
32(64.75) = 2,072
b) Mass of oxygen to completely combust 1,000 kg refuse/h.
Stoichiometric oxygen requirement = (2,072 kg O2)/(1,407 kg dry refuse)
= 1.47 kg O2/kg dry refuse
142
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
Mass O2 per 1,000 kg refuse = (1,000 kg refuse/h)(2,072 kg O2)/(1,407 kg refuse)
= 1,470 kg O2/h
Mass of air (air = 23.2% O2) = (1,470 kg O2/h)/(0.232 kg O2/kg air) = 6,336 kg air/h
Excess air will be needed for complete combustion.
6.7
NOX CONTROL
Urea (CO(NH2)2) and nitric oxide (NO) are needed to evaluate a non-catalytic reduction
system for NOx control. How much urea is needed to react with nitric oxide being emitted
at a rate of 12 kg/min? Use the following reaction to calculate the urea requirement.
2 NO + CO(NH2)2 + 0.5 O2 →
2 N2
+ CO2 + 2 H2O
urea
Solution
Molar mass of reactants
NO = 14 + 16 = 30 g NO/g-mol = 30 kg NO/kg-mol
CO(NH2)2 =12 + 16 + 2(14 + 2) = 60 g/g-mol = 60 kg/kg-mol
Ratio of reactants
2 moles of NO = 60 kg
1 mole of urea = 60 kg
1 kg urea is needed to react with 1 kg of NO
12 kg/h of urea reacts with 12 kg/h of NO
6.8
SALINE EFFLUENT
An industrial effluent contains 253 mg/L sodium (Na+) and 40 mg/L calcium (Ca2+); other
minerals can be neglected. According an International Standard (IS 2490) this effluent would
be suitable for irrigation if the proportions of sodium and calcium are
Na ≤ 0.6(Na + 0.5 Ca)
where Na and Ca are measured in mg/L. Waste gypsum (CaSO4) is available to dissolve in
the effluent to change the Na/(Na + 0.5 Ca) ratio. (a) How much gypsum must be used
to bring the ratio down to 0.6? (b) What will be the sulfate (SO42-) concentration in the
effluent? (c) Instead of gypsum, quicklime (CaO) is to be used to lower the ratio. How
much quicklime must be used to lower the ratio to 0.6?
143
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
Solution
Basis = 1 L of effluent
Existing ratio: (253 mg Na)/(253 mg Na + 0.5(40 mg Ca)) = 0.927
Define:
X = mass of calcium to be added to 1 L of effluent
Desired:
253 mg ≤ 0.6([253 mg + 0.5(40 mg +X)]
253 mg ≤ 151.8 mg + 12 mg + 0.3X
89.2 mg ≤ 0.3 X
X ≥ 297 mg Ca
a) Adding Gypsum:
Molar mass of CaSO4 = 40 + 32 + 4(16) = 136 g/g-mol
Molar mass of SO4 = 40 + 4(16) = 96 g/g-mol
Molar mass of Ca = 40 g/g-mol
Adding 297 mg of Ca requires
(297 mg Ca/L)(136 mg CaSO4/40 mg Ca) = 1,010 mg CaSO4
Final mixture: 253 mg/L Na, 40 mg/L original Ca, plus 297 mg/L added Ca
Check Ratio = 253/(253+ 0.5(40 + 297)) = 253/(253 + 168.5) = 0.60
b) SO4 concentration = (1,010 mg CaSO4/L)(96 mg SO4/136 mg CaSO4) = 734 mg SO4/L
c) Add Ca in the form of CaO (quicklime) instead of CaSO4
Molar mass of CaO = 40 g Ca/g-mol + 16 g O/g-mol = 56 g CaO/g-mol
Adding 297 mg of Ca requires
(297 mg Ca)(56 mg CaO/40 mg Ca) = 416 mg CaO
6.9
FLUE GAS DESULFURIZATION
Flue gas contains 4,400 ppmv of SO2, which is dissolved in water in a reactor zone that
is continuously fed air and limestone. The dissolved SO2 reacts with oxygen in the air to
form sulfuric acid that reacts with limestone to form solid gypsum which is removed in
crystalline form.
SO2 + O2 + H2O → H2SO4 + H2
H2SO4 + CaCO3 → CaSO4 + H+ + HCO3–
The limestone used is 57% calcium carbonate and 43% non-reactive minerals. Assume that
95% of the SO2 reacts with limestone according to the ideal stoichiometry shown. (a) How
much limestone must be added, (b) How much gypsum will be formed, and (c) How much
solid material will be removed from the reactor for every 1,000 m3 of flue gas treated?
144
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
Solution
Basis: 1,000 m3 of flue gas
a) Limestone requirement
Volume of SO2 = (4,400 m3 SO2/106 m3 flue gas) = 4.4 m3 SO2
This is correct for any temperature and pressure of the flue gas.
Convert to mass of SO2
22.41 L/g-mol SO2 = 22.41 m3/kg-mol SO2
(4.4 m3 SO2)/(22.41 m3 SO2/kg-mol SO2) = 0.1963 kg-mol SO2
SO2 reacting = 0.95(0.1963 kg-mol SO2) = 0.1865 kg-mol SO2
H2SO4 produced = 1 mole per mole of SO2 reacting = 0.1865 kg-mol
CaCO3 required = 1 mole per mole of H2SO4 produced = 0.1865 kg-mol
Molar mass of CaCO3= 100 kg/kg-mol
Mass of CaCO3 required = (100 kg/kg-mol)(0.1865 kg-mol) = 186.5 kg CaCO3
Limestone is 57% CaCO3
= (186.5 kg CaCO3)/(0.57) = 327.2 kg limestone to produce 186.5 kg CaCO3
b) Gypsum formed = 1 mole per mole of H2SO4 produced = 0.1865 kg-mol
145
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
Molar mass of gypsum (CaSO4) = 40 kg Ca/kg-mol + 32 kg S/kg-mol + 4(16 kg O/kg-mol)
= 136 kg CaSO4/kg-mol
Gypsum formed = (0.1865 kg-mol CaSO4)(136 kg CaSO4/ kg-mol CaSO4)
= 25.4 kg CaSO4
c) Mass of solids produced = Mass of gypsum + mass of unreacted limestone
= 25.4 kg CaSO4 + (0.43)(327.2 kg limestone) = 25.4 kg + 140.7 kg = 166.1 kg
6.10 WATER SOFTENING
A city removes hardness from its water by lime precipitation. The hardness removed is 200
mg/L, measured as CaCO3. The finished water has a hardness of 50 mg/L CaCO3. The
average quantity treated is 8,000 m3/d. Operating experience shows a satisfactory degree of
softening for a dose of 0.36 kg/m3 lime, Ca(OH)2. An empirical estimate of the precipitate
solids production, which is mainly calcium carbonate (CaCO3), is 2.6 kg per kg of lime
added.
According to operator records the average lime dose was 365 mg/L, as CaCO3, and the amount
of lime sludge produced was 980 m3 per week. The sludge is 6% solids by weight with a
density of 1,020 kg/m3. The calcium concentration in the sludge is 2% by weight. Are the
operators’ records consistent with the estimated amounts using the empirical stoichiometry?
Solution
Empirical estimates
Lime added = (0.36 kg/m3)(8,000 m3/d) = 2,880 kg/d
Precipitate solids produced = (2.6 kg solids/kg lime)(2,880 kg lime/d) = 7,488 kg solids/d
At 6% solids by weight 7,488 kg solids/d is
(7,488 kg dry solids/d)/0.06 = 124,800 kg wet sludge/d
At 1,020 kg/m3, this is
(124,800 kg/d)/(1,020 kg/m3) = 122.4 m3 wet sludge/d
Calcium content:
CaCO3 solids are 40/100 = 40% calcium by weight
0.4(7,488 kg/d solids) = 2,995 kg/d Ca in sludge solids
Operator’s records
146
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
Lime added = (8,000 m3/d)(0.365 kg/m3) = 2,920 kg/d as CaCO3
According to the empirical stoichiometry the solids production should be
(2.6 kg solids/kg lime)(2,920 kg lime/d) = 7,592 kg solids/d
This agrees nicely with the earlier estimate of 7,488 kg/d.
Check the volume of sludge
Operator’s report = 980 m3/week of wet lime sludge = 140 m3/d
At 1,020 kg/m3 the mass of wet sludge
= (140 m3/d)(1,020 kg/m3) = 142,800 kg wet sludge /d
Dry solids, at 6% concentration, = 0.06(142,800 kg/d) = 8568 kg dry solids/d
This is slightly higher than the empirical estimate of 7,592 kg/d
Check Calcium content:
CaCO3 solids are 40/100 = 40% calcium by weight
Operator added (40/100)(2,920 kg/d)
= 1,168 kg Ca in the 2,920 kg of lime (as CaCO3)
Calcium in sludge – operator’s estimated quantity from solids production
0.4(7,592 kg solids/d) = 2,995 kg Ca/d in sludge solids
Calcium in sludge – operator’s estimated from sludge composition (2% Ca)
(0.02 kg Ca/kg wet sludge)(142,800 kg wet sludge/d) = 2856 kg Ca/d
The amount in the sludge agrees almost exactly with the amount added as lime.
Conclusion: Sludge volumes and flows are difficult to measure, so we would conclude that the operator’s
measurements are reliable. Empirical estimates are useful as quick estimates when more exact local
values are unknown. They are usually based on averages taken over several plants and the treatment
process and water chemistry will differ from plant to plant. Nevertheless, the empirical estimate serves
as an easy and useful check on the operator’s records.
6.11 CHEMICAL TREATMENT OF WASTEWATER
A chemical coagulant will be added to raw wastewater as it enters a primary settling tank.
The models for estimating sludge production are
SAl = 21.34 + 2.77(PIn - POut ) + 3.07 Al3+
Added + (TSSIn - TSSOut )
SFe = 27.41 + 2.65(PIn - POut ) + 1.78 Fe3+
Added + (TSSIn - TSSOut )
SAl and SFe are total sludge solids formed from iron or aluminum-based coagulants. Terms
in the model; P, AlAdded, FeAdded, TSSIn and TSSOut; are measured as mg/L. (Source of the
equations: Snurer, H 2008, ‘Sludge Production from chemical precipitation’, Lund Tekniska
Hogskola, Sweden)
147
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
The wastewater flow is 20,000 m3/d with a suspended solids concentration of TSS = 350
mg/L. The TSS removal will be 85% if the coagulant is FeCl3. The removal will be 80%
if the coagulant is Al2(SO4)3. The influent phosphate (PO43-) concentration, mg P/L, is 8
mg/L. The effluent concentration will be 1 mg/L, whichever coagulant is used. Assume ideal
stoichiometry for phosphate precipitation to calculate the amount of aluminum and ferric
iron added – that is, 1 mole of PO43+ reacts with 1 mole of Fe3+ or 1 mole of Al3+; also 1
mole of P reacts with 1 mole of Fe3+ or 1 mole of Al3+.
Calculate the masses and volumes of primary sludge if the solids concentrations are (a) 2%
for aluminum coagulation and (b) 3% for iron coagulation.
Solution
Chemical precipitation of phosphate
Phosphate is measured as mg/L of P (not mg/L of PO43+)
1 mole of P = 1 mole of PO43+
Metal required: 1 mole of metal per mole of PO43+
P removed = 8 mg P/L – 1 mg/P/L = 7 mg P/L
Molar masses: P = 30.97 g/g-mol
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
Al3+ = 26.98 g/g-mol
Fe3+ = 55.85 g/g-mol
AlAdded = (7 mg P/L)(26.98 g Al/g-mol)/(30.97 g P/g-mol) = 6.1 mg Al/L
FeAdded = (7 mg P/L)(55.85 g Fe/g-mol)/(30.97 g P/g-mol) = 12.6 mg Fe/L
Sludge production from Aluminum coagulation (SAlum)
SAl = 21.34 + 2.77 (PIn - POut ) + 3.07 Al3+
Added + (TSSIn - TSSOut )
(TSSIn - TSSOut) =0.8 TSSIn = 0.8(350 mg/L) = 280 mg/L
SAlum = 21.34 + 2.77(7 mg/L) + 3.07(6.1 mg/L) + 280 mg/L = 339 mg/L
Mass of sludge solids = (20,000 m3/d)(0.339 kg/m3) = 6,780 kg/d
Mass of sludge = (6,780 kg/d)/0.02 = 339,000 kg/d
Sludge production from Iron coagulation (SFe)
SFe = 27.41 + 2.65(PIn - POut ) + 1.78 Fe3+
Added + (TSSIn - TSSOut )
(TSSIn - TSSOut) =0.85 TSSIn = 0.85(350 mg/L) = 298 mg/L
SFe = 27.41 + 2.65(7 mg/L) + 1.78(12.6 mg/L) + 298 mg/L = 366 mg/L
Mass of sludge solids = (20,000 m3/d)(0.366 kg/m3) = 7,320 kg/d
Mass of sludge = (7,320 kg/d)/0.03 = 244,000 kg/d
6.12 PHOSPHORUS PRECIPITATION
A treatment plant that serves 145,000 people is ordered to reduce effluent phosphorus (P)
from 5 mg/L to 0.5 mg/L. The per capita contributions to the treatment plant are 400
L/d and 2.6 g P/d. There are a number of ways to remove phosphorus. One possibility is
precipitation with ferric iron (Fe3+). Pickle liquor is hydrochloric acid that contains ferric
iron in the form of ferric chloride (FeCl3). A nearby industry has large quantities of waste
pickle liquor that can be obtained for the price of hauling.
The phosphorus is in the form of phosphate (PO43-). The chemistry seems simple – one
molecule of ferric iron (Fe3+) reacts with one molecule of phosphate (PO43-) to form FePO4,
which is an insoluble particle.
Fe3+
Molar masses (g)
55.85
+ PO43- → FePO4
95
151
By this stoichiometry, removing one kg of P generates (151/30.97) = 4.87 kg of iron
phosphate (FePO4).
It seems simple. But – and it’s a big but – excess iron is required to make the reaction
go to completion. It takes more than one molecule of iron to precipitate one molecule of
149
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
phosphate. Some reports say it is as much as 2 Fe to 1 P. The excess iron reacts with non-P
compounds to form non-P precipitates. And the iron phosphate probably is not pure FePO4.
It could be something like Fe1.5(PO4)2•Gunk. And, all the iron in the pickle liquid may not
be in the ferric form (Fe3+). Ferrous iron (Fe2+) will form a precipitate with phosphorus, but
the stoichiometry is different.
Work out an estimate of the sludge production related to phosphorus removal using ferric
chloride. Unless you want to learn a lot more about the chemistry, this is a time to use
empirical estimates. The two graphs in Figure P6.12 should be useful. They show data for
the addition of ferric chloride to the aeration tank of an activated sludge process. They
are different studies so be careful with your interpretation. You may want to search for
additional data.
(a) P removed vs. Iron added
(b) Sludge yield vs. Effluent TP
Figure P6.12 Phosphorus precipitation
Solution
Wastewater volume = (145,000 persons)(400 L/d) = 58,000,000 L/d
Influent P load = (145,000 persons)(2.6 g P/person-d) = 377,000 g P/day = 377 kg P/d
Influent P concentration = (377,000 g/day)(1,000 mg/g)/(58,000,000 L/d) = 6.5 mg/L
Existing influent concentration = 6.5 mg P/L
Existing effluent concentration = 5 mg P/L
Required effluent concentration = 0.5 mg/L
P removal required = 6.5 – 0.5 = 6.0 mg P/L
Interpreting the graphs.
150
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
Graph (a) Experiments were done with a controlled effluent pH = 7.2.
This is a typical pH for domestic wastewater.
Influent P is not reported. A representative value for municipal wastewater is about
7 mg/L, or somewhat less.
P removal per unit of Fe added decreases at higher levels of removal (higher
Fe doses)
Graph (b) Sludge yield increases sharply when effluent P is forced below about 1 mg/L.
This is also shown by the flattening of the curve in graph (a) at higher Fe doses.
Formation of FePO4 is stoichiometric (linear) down to about 1 mg P/L.
At lower effluent TP levels FeCl3 is being used to depress the pH.
FeCl3 + 3 H2O → Fe(OH)3 + 3 H+ + 3 ClExcess sludge production at effluent TP below 1 mg/L is due to formation of non-FePO4 solids.
Estimation of sludge production
Based on Graph (b) and TP in and TP out (with no adjustment for removal with no Fe added)
Required P removal = (377 kg P/d)[(6 mg TP removed/L)/6.5 mg TP influent/L)]
= 348 kg P/d
Sludge yield coefficient at effluent TP = 0.5 mg/L = 1.5 kg TSS/kg TP removed
Sludge production = (348 kg P/d)(1.5 kg TSS/kg P) = 522 kg TSS/d
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
Note: If you want to learn more consult these references.
Jenkins, D, Ferguson, JF & Menar, AB 1971, ‘Chemical Processes for Phosphate Removal,’ Water
Research, vol. 5, pp 369–389.
Jenkins, D & Hermanowicz, SW 1991, Principles of Chemical Phosphate Removal, 2nd ed., Lewis
Publisher, pp. 91-110.
Snurer, H 2008, ‘Sludge Production from Chemical Precipitation’, Lund Tekniska Hogskola, Sweden
6.13 METAL PLATING
Acidic wastewater (pH 2) from a metal fabricating shop contains 200 mg/L Nickel (Ni), 800
mg/L Zinc (Zn), and 50 mg/L Lead (Pb). These are total concentrations and the metals are
100% soluble at the acidic feed conditions. The wastewater flow is 1,000 m3/d. The metals
can be precipitated as sulfides (NiS, ZnS, and PbS) by adding a sodium sulfide (Na2S). The
process is shown in Figure P6.13. The sulfur is soluble and free to react with the metals.
The reactor will operate at pH 8 and sodium hydroxide (NaOH) will be added to neutralize
the acidic influent. A polymer is also added to coagulate the fine sulfide precipitates and
facilitate separation by settling and filtration.
Polymer coagulant
Sodium hydroxide NaOH
Sodium sulfide (Na2S)
Acidic Influent
1,000 m3/d
200 mg Ni2+ L
800 mg Zn2+/L
50 mg Pb2+/L
Settling
Tank
Mixed
Reactor
Mixer
2.5% solids by weight
Sludge density = 1,020 kg/m3
Figure P6.13 Metal Plating
The chemistry of the sulfide reactions is:
→ NiS (s)
S2Ni2+ +
58.7 kg
32.1 kg
90.8 kg
→
S2Zn2+ +
65.8 kg
32.1 kg
ZnS (s)
97.9 kg
→ PbS (s)
S2Pb2+ +
207.2 kg
32.1 kg
239.3 kg
152
Filter
Effluent
at pH 8
contains traces of
Ni2+, Zn2+ and Pb2+
Sludge to
treatment and
disposal
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
One mole of metal ion combines with one mole of sulfide. (The molar masses have been
rounded to one decimal place.)
Some empirical stoichiometry is needed to account for precipitates, such as calcium sulfate
and iron hydroxides, that will form simultaneously with the sulfides, Experiments indicate that
2 kg of non-sulfide precipitates will be formed for each cubic meter of wastewater treated.
Ninety percent of the metal removal is due to settling and 10% is due to filtration. Assume
that metals in the effluent are in the form of very fine particles and that soluble metals are
negligible. The mass of metals in the filter backwash water can be neglected when making
the material balance. The sludge removed from the settling tank is 2.5% solids (mass basis)
with a density of 1,020 kg/m3.
Solution
Basis = 1,000 m3 of wastewater
Assumptions:
1) Estimated volume and mass of sludge are nearly the same whether the final effluent
concentration of zinc is 0.01mg/L or 0.1 mg/L. Therefore, sludge quantities will be
calculated as though all the metals have been removed.
2) Mass of solids removed in the filter is negligible compared with the amount removed
in the settler. Therefore, sludge quantities are calculated assuming 100% of the solids
are in the settler sludge.
Mass of metals in the wastewater feed
Mass Ni = (1,000 m3)(0.2 kg/m3) = 200 kg
Mass Zn = (1,000 m3)(0.8 kg/m3) = 800 kg
Mass Pb = (1,000 m3)(0.05 kg/m3) = 50 kg
Formation of particulate metal sulfides
Nickel sulfide
90.8 kg NiS/58.7 kg Ni = 1.547 kg NiS/kg Ni
(200 kg Ni)(1.547 kg NiS/kg Ni) = 309.4 kg NiS
Zinc sulfide
97.9 kg ZnS/65.8 kg Zn = 1.488 kg ZnS/kg Zn
(800 kg Zn)(1.488 kg ZnS/kg Zn) = 1,190.3 kg ZnS
Lead sulfide:
239.3 kg PbS/207.2 kg Pb = 1.155 kg PbS/kg Pb
(50 kg Pb)(1.155 kg PbS/kg Pb) = 57.7 kg PbS
MS = Metal sulfide precipitates = 309.4 kg + 1,190.3 kg + 57.7 kg = 1,557 kg (rounded)
Mother = Mass of other precipitates = (2 kg/m3)(1,000 m3) = 2,000 kg
153
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
Sludge mass and volume
Total mass of solids in sludge = MS + Mother = 3,557 kg
Mass of sludge, at 2.5% solids = 3,557/0.025 = 142,280 kg
Sludge density = 1,020 kg/m3
Volume of sludge = 142,280 kg/1020 kg/m3 = 139.5 m3 (rounded)
About 14% of the influent leaves the process as metal-contaminated sludge, as shown in Figure S6.13.
This is a toxic waste and it must be handled of according to the regulations for hazardous waste
disposal. The first step should be to thicken the sludge. Doubling the solids concentration to 5% will
reduce the volume by 50%, from 139.5 m3 to 69.7 m3. Another doubling, to 10% solids, will halve the
volume again to 34.9 m3. Reducing the volume to less than 10 m3 should be possible.
Sodium hydroxide (NaOH)
Sodium sulfide (Na2S)
Polymer coagulant
QIn = 1,000 m3
Ni = 200 mg/L
Zn = 800 mg/L
Pb = 50 mg/L
Precipitation &
Settling
Ni2+ + S2- o NiS
Zn2+ + S2- o ZnS
Pb2+ + S2- o PbS
Zn = ~ 0 kg
Ni = 0.01 mg/L
Pb = 0.02 kg
Metal-laden sludge
Mass = 142,280 kg
Volume = 139.5 m3
Figure S6.13
6.14 CHEMICAL AND BIOLOGICAL PHOSPHORUS REMOVAL
Figure P6.14a shows a chemical precipitation process for removing phosphorus that was
much used in the past but has been largely replaced today by biological phosphorus removal,
which is shown in Figure P6.14b. The chemical process shown adds ferric chloride (FeCl3)
to the influent and a phosphorus precipitate (FePO4) is removed with the influent settleable
suspended solids. (Other points of chemical addition have been used with success.) The
activated sludge with chemical precipitation produces 43,000 kg/d of sludge solids (dry
basis) that contains 952 kg/d of total phosphorus. The biological phosphorus removal process
produces 38,050 kg/d of sludge solids (dry basis) that contains 952 kg/d of phosphorus.
(a) Calculate the empirical solids yield coefficient for the chemically assisted primary settling
process. (b) Calculate the solids yield factor for the aerobic treatment stage, using the mass
of solids in the waste activated sludge. (c) Calculate the overall solids yield coefficient for
the chemically assisted treatment plant (d) Make the same calculations for the biological
154
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
phosphorus removal process. (Note: The two processes are contrived to remove the same
mass of phosphorus and to produce the same mass of waste solids.)
FeCl3
12,000 kg/d
Influent
190,000 m3/d
6.0 mg/L P
1,140 kg/d P
Activated Sludge
Aeration Tank
Primary
Settling
Primary Sludge
28,500 kg/d solids
662 kg/d P
Effluent
188,000 m3/d
1.0 mg/L P
188 kg/d P
Final
clarifier
Waste Activated Sludge
14,500 kg/d solids
290 kg/d P
Activated Sludge Recycle
(a) Chemical Precipitation of Phosphorus
Influent
190,000 m3/d
6.0 mg/L P
1,140 kg/d P
Primary
Settling
Anaerobic
Stage
Primary Sludge
20,400 kg/d solids
72 kg/d P
Anoxic
Stage
Aerobic
Stage
Activated Sludge Recycle
Effluent
188,000 m3/d
1.0 mg/L P
188 kg/d P
Final
clarifier
Waste Activated Sludge
17,650 kg/d solids
880 kg/d P
(b) Biological Phosphorus Removal
Figure P6.14 Chemical and Biological Phosphorus Removal
Solution
The solids yield factor might be expressed in terms of mass of chemical added, mass of P removed,
or volume of wastewater treated.
a) Empirical Solids Yield for Chemical Precipitation of Phosphorus
In the primary sludge
(662 kg P)/(12,000 kg FeCl3) = 0.055 kg P removed/kg FeCl3 added
(28,500 kg solids)/(12,000 kg FeCl3) = 2.4 kg solids produced/kg FeCl3 added
(28,500 kg solids)/(662 kg P removed) = 43.1 kg solids produced/kg P removed
(28,500 kg solids)/(190,000 m3) = 0.15 kg solids produced/m3 wastewater treated
Theoretical FePO4 produced, for 662 kg P removed in the primary settling tank (ignoring
precipitate carried in or formed in the aeration tank).
Fe3+
Molar masses (g)
55.85
+ PO43- → FePO4
95
151
By this stoichiometry
(1 kg P) yields (1 kg P)(151/30.97) = 4.87 kg of iron phosphate (FePO4).
Mass of P precipitated = 662 kg P/d
Mass of FePO4 formed = (4.87 kg FePO4/kg P)(662 kg P/d) = 3,224 kg FePO4/d
155
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
b) Solids Yield from Aerobic (Biological) Treatment Step
In the waste activated sludge
(14,500 kg solids)/(290 kg P removed) = 50 kg solids/kg P removed
(14,500 kg solids)/(190,000 m3) = 0.076 kg solids/m3 wastewater treated
c) Overall Treatment Plant Solids Yield
Waste solids streams = Primary Sludge + Waste Activated Sludge
Solids produced = 28,500 + 14.500 = 43,000 kg solids/d
Total Phosphorus removed = 662 + 290 = 952 kg P/d
(43,000 kg solids)/(12,000 kg FeCl3) = 3.6 kg solids/kg FeCl3 added
(43,000 kg solids)/(952 kg P removed) = 45.2 kg solids/kg P removed
(43,000 kg solids)/(190,000 m3) = 0.23 kg solids/m3 wastewater treated
d) Biological Removal of Phosphorus
In the primary sludge
(20,400 kg solids)/(72 kg P removed) = 283 kg solids/kg P removed
(20,400 kg solids)/(190,000 m3) = 0.11 kg solids/m3 wastewater treated
Biological Process - Waste activated sludge
(17,650 kg solids)/(880 kg P removed) = 20.0 kg solids/kg P removed
(17,650 kg solids)/(190,000 m3) = 0.09 kg solids/m3 wastewater treated
Overall treatment plant
(38,050 kg solids)/(952 kg P removed) = 40.0 kg solids/kg P removed
(38,050 kg solids)/(190,000 m3) = 0.20 kg solids/m3 wastewater treated
Ferric Chloride addition produces
Extra primary sludge solids = 28,500 – 20,400 = 8,100 kg solids/d
= 8,100 kg/12,000 kg FeCl3 = 0.675 kg solids/kg FeCl3
6.15 AMMONIUM SULFATE RECOVERY FROM FLUE GAS
Figure P6.15 shows a proposed system for converting sulfur dioxide (SO2) in flue gas (or
coke oven gas) into ammonium sulfate, (NH4)2SO4. Many details have been omitted but
the flow chart shows the main elements needed to make a material balance. Assume 100
% sulfur capture in the scrubber and ideal stoichiometry. Estimate the net cost of pollution
control based on these costs.
Flue gas is produced by burning fuel oil with a 2.5% sulfur content
Ammonia (anhydrous) = $1/kg
Ammonium sulfate = $0.2/kg
156
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
These are the reactions.
Flue gas: Combustion of
S + O2 → SO2
sulfur in fuel
(gas)
Scrubber: Absorbs
SO2 into liquid SO2 + H2O → H2SO3
(aqueous solution)
SO2 + H2O → H2SO3
(aqueous solution)
H2SO3 + 2NH3 → (NH4)2 SO3
(aqueous solution)
Oxidation
(NH4)2 SO3+ 0.5 O2 → (NH4)2 SO4
(aqueous solution)
Crystallizer
No chemical reaction.
Mixing tank
Water is removed to form crystals of (NH4)2 SO4
Overall reaction
SO2 + H2O + 2NH3 + 0.5O2 → (NH4)2SO4
Makeup
water
Exhaust
gas
Exhaust air to
dust removal
Flue gas
SO2
Aqueous
SO2
Air
Crystallizer
Mixing
tank
Oxidation
Scrubber
NH3
Hot Air
Product
(NH4)2SO4
Figure P6.15 Ammonium sulfate recovery from flue gas
Solution
Basis = 1,000 kg fuel oil, which contains 25 kg S
Molar masses
S = 32 kg/kg-mol
SO2 = 64 kg/kg-mol
NH3 = 17 kg/kg-mol
(NH4)2 SO4 = 132 kg/kg-mol
Flue gas combustion
Sulfur dioxide produced = (25 kg S)/(32 kg S/mol) = 0.78 mol S = 0.78 mol SO2
From the reaction stoichiometry
157
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
Ammonia required = (2 mol NH3/mol SO2)(0.78 mol SO2) = 2(0.78) = 1.56 mol NH3
Mass ammonia required = (17 kg NH3/mol)(1.56 mol) = 26.5 kg NH3
(NH4)2 SO4 produced = (1 mol (NH4)2 SO4 /mol SO2)(0.78 mol SO2)
= 0.78 mol (NH4)2 SO4
Mass (NH4)2 SO4 produced = (132 kg (NH4)2 SO4/mol)(0.78 mol) = 103 kg (NH4)2 SO4
Costs:
Ammonia = (26.5 kg)($1/kg) = $26.50
Ammonium sulfate = (103 kg)($0.2/kg) = $206
Net value = $206 - $26.5 = $179.50/1,000 kg fuel oil burned
This is not a profit of $179.50 because we have not estimated the cost of building the process or the
cost of operation and maintenance. It is revenue that would offset a portion of those costs.
6.16 SUSPENDED SOLIDS REMOVAL
The wastewater feed to an industrial treatment system is 200 L/min with a total suspended
solids concentration of TSS = 5,000 mg/L. The dissolved solids (TDS) are low. It increases
slightly through the process due to the addition of coagulants, lime (CaO) for sludge
conditioning, and acid for neutralization. Figure P6.16 shows the treatment process. Settleable
solids are removed in a tube settler that produces an effluent with 5 mg/L suspended solids.
The solids are non-toxic and can be discharged to the city sewer. The settled solids go to a
filter press. The filtrate suspended solids concentration is 100 mg/L, which can be neglected
in the material balance calculations. Calculate the unknown values Table P6.16.
158
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
40 mg/L acid
40 mg/L
coagulant
(2)
Tube settler
Flash mixer
(3)
(5)
(4) Sludge
ld
2.5% solids
(7) Lime addition (CaO)
0.1 kg/kg solids
(6) Effluent to sewer
pH control
Filtrate
(1) Feed = 200 L/min
TSS = 5,000 mg/L
TDS = 250 mg/L
(9)
Filter press
(8) Filter cake to disposal
35% solids
Figure P6.16 Suspended solids removal
Stream
1
Feed
Coagulants added (40 mg/L)
Flow (L/min)
TSS (ppm or %)
TDS (ppm)
pH
200
5,000
250
7
negligible
0
Nil
2
Tube clarifier influent
200
5,000
250
6.8
3
Tube clarifier effluent
200 - sludge
5
250
6.8
4
Sludge
unknown
2.5%
250
6.8
7
Lime addition
(0.1 kg/kg solids)
negligible
New SS = 1.5x
Lime added
New TDS =
unknown
10.5
7
Filter feed
unknown
unknown
unknown
10.5
8
Filter cake
35% mass
irrelevant
10.5
9
Filtrate
unknown
100
300
10.5
5
pH control influent
200 + Filtrate
5
300+ 50
= 350
≈ 9.5
6
Effluent to sewer
≈ 200
5
≈ 330
≈ 7.5
Table P6.16 Suspended solids removal
Solution
Basis = 1 hour of operation = (200 L/min)(60 min)(1 m3/1,000 L)(1,000 kg/m3)
= 12,000 kg (12 m3)
Tie variable = suspended solids. (Dissolved solids are a negligible amount of the total solids in the
sludge.)
159
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
Notation
W = water mass (kg)
S = solids mass TSS (kg)
M = W + S = total mass (kg)
Feed
W1 = 12,000 kg (12 m3)
S1 = (12 m3)(0.5 kg TSS/m3) = 60 kg
Tube Settler
S2 = S1 = 60 kg
Solids in added coagulant are negligible.
S4 = S2 = 60 kg
Solids capture in tube settler is 100%
M4 = 2.5% solids
S4/(W4 + S4) = 0.025
W4 = (60 kg)(1-0.025)/0.025 = 2,340 kg (2.34 m3)
S3 = 0
W3 = W1 – W4 = 12,000 kg – 2,340 kg = 9,660 kg (9.66 m3)
Lime Addition
Lime dose = 0.1(60 kg) = 6 kg
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160
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
The calcium in the lime that is added as a sludge conditioner will precipitate and become part of the
sludge. Assume that lime solids are 1.5 times the mass of added lime.
Lime solids produced = 1.5(6 kg) = 9 kg
Filter Press
S7 = S4 + 9 kg = 60 kg + 9 kg = 69 kg
W7 = W4 = 2,340 kg
S8 = S7 = 69 kg
Filter press captures 100% of applied solids
S9 = 0
M8 = 35% solids
S8/(S8 + W8) = 0.35
W8 = (69 kg)(1 – 0.35)/0.65 = 128 kg (0.128 m3)
W9 = W7 – W8 = 2,340 kg – 128 kg = 2,212 kg (2.212 m3)
pH Control
Mass of added acid is negligible
W5 = W3 + W9 = 9,660 kg + 2,212 kg = 11,872 kg (11.87 m3)
W6 = W5 = 11,872 kg (11.87 m3)
S5 = S6 = 0
Table S6.16 summarizes the calculations
Stream
1
Feed
Coagulants added (40 mg/L)
Water (W)
TSS (S)
TDS
3
(m /h)
(kg/h)
(ppm)
12
60
250
negligible
0
40
pH
7
2
Tube clarifier influent
12
60
250
6.8
3
Tube clarifier effluent
9.66
negligible
250
6.8
4
Sludge
2.34
60
250
6.8
negligible
9
unknown
10.5
Lime addition (0.1 kg/kg TSS)
7
Filter feed
2.34
69
unknown
10.5
8
Filter cake
0.13
69
irrelevant
10.5
9
Filtrate
2.21
negligible
300
10.5
5
pH control influent
11.87
negligible
340
≈ 9.5
6
Effluent to sewer
11.87
negligible
340
≈ 7.5
Table S6.16
161
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
Total dissolved solids are not important in this problem so they have been approximated. The TDS in
the water fraction of the sludge is assumed to be the same as the clarifier influent. Acid added for pH
control will increase the TDS by approximately the acid dose.
The polymer coagulant becomes part of the sludge so the clarifier effluent TDS is assumed to be the
same as the influent TDS.
The filtrate TDS may be higher than found in the clarifier sludge underflow, but a value cannot be
calculated. It is assumed to be a small and unimportant amount. An arbitrary increase of 50 ppm is
added.
6.17 SLUDGE DIGESTER GAS (METHANE)
The raw sludge input to an anaerobic digester is 7.5% total solids (mass percent). The solids
are 75% volatile solids (VS) and 25% fixed solids (FS). The digestion process converts 55%
of the VS to biogas at a yield is 0.75 – 1.1 m3/kg VSS destroyed. The volume fractions
are 0.6 for methane (CH4) and 0.4 for carbon dioxide (CO2). The gas is used as fuel in a
boiler. Air is supplied for combustion with 5% excess above the stoichiometric requirement.
Calculate the material balance on the digester and on the gas as it is burned in the boiler.
Use 1,000 kg total solids input as the basis.
Solution
Basis = 1,000 kg total solids
Inputs
Inert solids input = 250 kg
Inert solids output = 250 kg
Volatile solids input = 750 kg
Volatile solids destroyed = 0.55(750 kg) = 412 kg
Volatile solids output = 0.45(750 kg) = 338 kg
Total mass of sludge feed = (1,000 kg/)/0.075 = 13,333 kg
Mass of water fed with 1,000 kg total solids = 13,333 kg – 1,000 kg = 12,333 kg
Digested Sludge
Inert solids = 250 kg
Volatile solids = 0.45(750 kg) = 338 kg
Total solids = 250 kg + 338 kg = 588 kg
162
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
Gas Production in Digester
Assume median gas yield = 0.9 m3/kg VSS destroyed
Gas produced = (0.9 m3/kg)(412 kg) = 371 m3
Gas composition – 60% CH4 and 40% CO2
CH4 volume = 0.6(371 m3) = 223 m3
CO2 volume = 371 m3 – 223 m3 = 148 m3
Stoichiometric combustion of methane (CH4)
CH4
+ 2 O2
→
CO2
+
2 H2O
Molar masses (kg/kg-mol)
16
32
44
18
Reacting masses (kg)
16
64
44
36
Assume methane (CH4) conversion to CO2 in boiler is 100%
Gas densities (at STP)
Methane (CH4)
(16 kg/kg-mol)/(22.4 m3/kg-mole) = 0.714 kg/m3
Carbon dioxide (CO2) (44 kg/kg-mol)/(22.4 m3/kg-mole) = 1.964 kg/m3
Oxygen (O2)
(32 kg/kg-mol)/(22.4 m3/kg-mole) = 1.429 kg/m3
Gas inputs to boiler
Mass of Methane (CH4) (0.714 kg/m3)(223 m3) = 159 kg CH4
Mass of CO2
(1.964 kg/m3)(148 m3) = 291 kg CO2
Oxygen consumed: 16 kg methane burned consumes 64 kg oxygen
O2 consumed = (159 kg CH4)(64 kg O2/16 kg CH4) = 636 kg O2
Carbon Dioxide (CO2) produced:
=16 kg of methane burned produces 44 kg of carbon dioxide.
CO2 produced = (159 kg CH4)(44 kg O2/16 kg CH4) = 437 kg CO2
Material balance on carbon dioxide is
291 kg +
Gas In
437 kg
Produced
= 728 kg CO2
Output
Excess air. Five percent oxygen in excess of the stoichiometric requirement for 636 kg O2 is provided
to insure complete combustion.
Total required oxygen input = 1.05(636 kg) = 668 kg
Unconsumed O2 in exhaust gas = 668 kg – 636 kg = 32 kg
Air = 23.2% oxygen
Air required = (Oxygen required)/0.232 = 668 kg/0.232 = 2,880 kg
Nitrogen (N2) is an inert gas that enters with the supplied air and exits, unreacted, in the exhaust gases.
163
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance with chemical reactions
Air is 76.8% nitrogen and 23.2% oxygen by weight so each unit of oxygen brings along 76.8/23.2 =
3.31 units of nitrogen. Nitrogen carried into the burner with the 2,880 kg of air will be
0.768(2,880 kg) = 2,212 kg nitrogen.
The same mass of nitrogen leaves in the exhaust gas.
Water. The digester gas and the incoming air will contain some water vapor. The mass is unspecified
and will be ignored.
Mass of water produced by combustion
= (159 kg CH4)(36 kg H2O produced/16 kg CH4 combusted) = 358 kg.
Table S6.17 is a summary (assuming dry digester gas and input air) and Figure S7.2 shows the complete
flow diagram.
Input
Produced
Output
Destroyed
(kg)
(kg)
(kg)
(kg)
Methane (CH4)
159
+
0
=
0
+
159
Carbon dioxide (CO2)
291
+
437
=
728
+
0
Oxygen (O2)
668
+
0
=
32
+
636
Nitrogen (N2)
2,212
+
0
=
2,212
+
0
0
+
358
=
358
+
0
Water vapor (H2O)
Table S6.17
Digester gas
371 m3 = 223 m3 CH4 + 148 m3 CO2
450 kg = 159 kg CH4 + 291 kg CO2
Feed sludge
Total solids =1,000 kg
Inert solids = 250 kg
Volatile solids = 750 kg
Anaerobic Digester
55% VS destroyed
= 412 kg
Boiler
Exhaust gas
728 kg CO2
32 kg excess O2
2,212 kg N2
Water vapor
Combustion air (5% excess)
2,880 kg air = 668 kg O2 + 2,212 kg N2
Oxygen required = 636 kg O2
Digested sludge
Total Solids = 588 kg
Inert solids= 250 kg
Volatile solids = (0.45)(750 kg) = 338 kg
Figure S6.17
164
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
7REACTION RATES AND
REACTOR DESIGN
7.1
DETENTION TIME
a) Five hundred cubic meters per hour of wastewater is treated in a reactor of 650
cubic meters. What is the detention time?”
b) What reactor volume is required to process 0.064 m3/s with a detention time of
8 hours?
c) Solve part (a) for a packed-bed reactor of 650 m3 and a porosity of φ = 0.45.
Solution
The calculations use V = θQ
a) θ = V/Q so θ = (650 m3)/(500 m3/h) = 1.3 h
b) V = θQ so V = (0.064 m3/s)(3600 s/h)(8 h) = 1,843 m3
c) The packing in the reactor occupies space and reduces the effective volume
θ = ϕV/Q so θ = (0.45)(650 m3)/(500 m3/h) = 0.585 h
7.2RATE COEFFICIENT DETERMINES THE
EXPONENTIAL DECREASE
The model for a first-order reaction is ln(C/C0) = –kt. Figure P7.2 shows this reaction for the
same at four different conditions. What conditions could be changing to cause this behavior?
1
k = 0.5
0.1
k=1
C/C0
k=2
0.01
k=5
0.001
0
1
2
Time
Figure P7.2
165
3
4
5
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Solution
• Temperature – Generally the reaction rate increases when temperature increases.
• Pressure – In gas phase reaction a pressure increase will increase the reaction rate.
• Biomass concentration – In a biological process the reaction rate tends to be proportional
to the concentration of active bacteria, so long as there is sufficient food supply.
• pH – Many reactions are pH sensitive. Extreme pH will inhibit biological reactions.
pH will control the ionization of organic and inorganic chemicals, and this can change
the reaction rate.
Any combination of these:
• Higher temperature + higher pressure
• Higher temperature + higher biomass
7.3
FIRST ORDER REACTION
A first order reaction is 30% complete after 10 minutes.
a) Estimate the reaction rate coefficient.
b) How long will it take before the reaction is 90% complete?
Solution
a) First order rate model: ln(C/C0) = -kt
C/C0 = 1.0 – 0.3 = 0.7, at t = 10 min
ln(0.7) = -0.357 = -k (10 min)
k = 0.0357/min
b) Time for 90% completion
C/C0 = 1.0 – 0.9 = 0.1, at t = t90%
ln(0.1) = -2.303 = -(0.0357/min)(t90%)
t90% = 64.5 min
7.4
BACTERIAL DIE-OFF IN A RIVER
Studies of coliform bacteria die-off in a stream showed that 35% of the coliforms had died
at a location 10 hours travel time below a sewage discharge point. At a point 20 hours
below the discharge 60% had died. How far below the discharge will the reduction be
90%. Assume that the addition of coliforms along the river is negligible (not realistic, but
let’s keep it simple).
166
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Solution
The usual model for die-off of bacteria is an exponential decay, or first-order model.
Ct = C0 exp(-kt) or Ct/C0 = exp(-kt).
We do not know C0 so use the second form.
At 10 hours the loss is 35% so Ct/C0 = 0.65 and
0.65 = exp[-k (10 h)]
ln(0.65) = -k(10 h) giving k = 0.043/h
At 20 hours the loss is 60% so Ct/C0 = 0.4
0.4 = exp[-k (20 h)] and k = 0.046/h
We will now assume that k is not changing along the stream (also not realistic, but OK), and use an
average of the 10 h and 20 h estimates = 0.0445/h.
For 90% removal Ct/C0 = 0.1
0.1 = exp[(-0.0445/h)(t)] and t = 51.8 h
7.5
HALF-LIFE FOR A FIRST-ODER REACTION
Half-life is a self-descriptive name – the time for the concentration of a chemical or pollutant
to decrease by half. For a first order reaction, the half-life can be applied sequentially as
shown in Figure P7.7. At 2 half-lives, the concentration is one-quarter of its initial value
and at 3 half-lives it is one-eighth the initial value. (a) What is the concentration after four
half-lives? (b) Write a general expression for the concentration after n half-lives. (c) What
is the concentration after 1.5 half-lives? (d) How many half-lives are required to achieve
95% reduction from the initial concentration, C0?
t=0
Concentration, C
C0
t = t1/2
t = 2t1/2
C0/2
t1/2
t = 3t1/2
C0/4
t1/2
C0/8
t1/2
0
1
2
3
Time (number of half-lives)
Figure P7.5
167
4
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Solution
a) At four half-lives, C = half the value at three half-lives = (C0/8)/2 = C0/16
b) Note that the divisor for the half-life calculation is a geometric progression, 2n, where
n is the number of half-lives.
For example, for n = 3: C = C0/23 = C0/8
The general expression for the concentration after n half-lives is C = C0/2n
c) For n = 1.5
C = C0/21.5 = C0/2.83
d) For 95% reduction, C/C0 = 0.05
Rearranging the general expression gives C/C0 = 1/(2n) = 0.05
Solving for n
2n = 1/0.05 = 20
n ln(2) =ln(20)
n = ln(20)/ln(2) = 2.996/0.693 = 4.32
7.6
HALF-LIFE REACTIONS
Derive an equation for the half-life of a zero-order reaction, a first-order reaction, and a
second-order reaction.
Solution
The initial concentration is C0 and at t1/2 C =C0/2
a) Zero order reaction
C = C0 – kt
Half-life calculation
C0
2
C0 kt1/2
t1/2
and
C0
2k
b) First order reaction
C = C0 e-kt
Half-life calculation
C0
2
kt1/2
C0e
kt1/2
Ÿ
-ln(1/2) = 0.693
1
2
e
kt1/2
and
t1/2
168
0.693
k
SOLVED MATERIAL BALANCE PROBLEMS:
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c) Second order reaction
1/C = 1/C0 + kt
Half-life calculation
1
C0 /2
1
kt1/2
C0
Ÿ
kt1/2
2
1
1
=
C0 C0
C0
and
t1/2
1
kC0
Note all the equations are a function of the rate coefficient. The zero-order and second-order equations
are also a function of the initial concentration, C0. Only the first-order expression is independent of C0.
7.7
POLLUTANT DECOMPOSITION
Table P7.7 gives measurements for the decomposition of a pollutant. Find (a) The rate
model, (b) the rate coefficient, and (c) the half-life.
Pollutant (mol/L)
Time (min)
2.33
1.91
1.36
1.11
0.72
0.55
0
5.3
14.5
20.0
30.7
36.0
Table P7.7
Solution
a) Try a first-order model and estimate k by fitting the log-transformed data, shown in
Table S7.7 and Figure S7.8.
Time
C
(min)
(mol/L)
0
2.33
0.8459
5.3
1.91
0.6471
14.5
1.36
0.3075
20
1.11
0.1044
30.7
0.72
-0.3285
36
0.55
-0.5978
Table S7.7
169
ln(C)
SOLVED MATERIAL BALANCE PROBLEMS:
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Reaction rates and reactor design
The model and the transformed models are
C = C0 exp(-kt)
ln(C) = ln(C0) – kt
The fitted model is
ln(C) = -0.8648 – 0.0395 t
C0 = exp(0.8648) = 2.375 mol/L and k = 0.0395/min
C = (2.375 mol/L)e-(0.0395/min)t
1.0
0.8
ln(C) = 0.8648 -0.0395 t
0.6
ln (C)
0.4
0.2
0.0
-0.2
-0.4
-0.6
-0.8
0
5
10
15
20
25
30
35
40
Time (min)
Figure S7.7
b) The estimated rate coefficient is 0.0395/min
c) The half-life for a first order reaction is
t1/2 = 0.693/k = 0.693/(0.0395/min) = 17.5 min
7.8
RATE COEFFICIENT FOR A FIRST-ORDER REACTION
Determine the reaction rate coefficient for the data in Table P7.8.
Time (min)
0
10
20
30
40
50
60
100
Conc. (mg/L)
290
220
180
135
98
75
55
25
Table P7.8
170
SOLVED MATERIAL BALANCE PROBLEMS:
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Reaction rates and reactor design
Solution
Plot the data. The semi-log plot shows a straight line, which indicates a first-order reaction. Plotting
ln(C) vs. time (Figure S7.8) gives a regression line with slope k = 0.0252/min. This is the estimated
350
Concentration (mg/L)
Concentration (mg/L)
reaction rate coefficient.
300
250
200
150
100
50
0
0
20
40
60
80
500
200
100
50
20
10
0
100
20
40
60
80
100
Time (min)
Time (min)
ln(Concentration)
6
ln(C) = 5.637 - 0.0252t
5
4
3
2
0
20
40
60
80
100
Time (min)
Figure S7.8
Another way to estimate the reaction rate coefficient is from the formula for half-life. A graphical
estimate of the half-life is t1/2 = 27 minutes.
t1/2 =
0.693
k
or
k =
0.693
0.693
=
= 0.0257/min
t1/2
27 min
The two estimates, k = 0.0257/min and k = 0.0252/min are in good agreement.
7.9
DYE EFFLUENT DEGRADATION
Over 700,000 T per year of synthetic dyes are produced for dyeing and printing. The
family of dyes in use includes a broad spectrum of organic structures, including substituted
aromatic and heterocyclic groups. Many of these are suspected carcinogens. A very small
amount of dye in water is visible. Dyeing is inefficient and 10-15% of unused dyestuff
enters the wastewater.
This problem is based on dye house effluent in India. Dye destruction is accomplished with
a special bacterial culture of Pseudomonas stutzeri. The effluent COD = 3200 mg/L, the
BOD5 = 840 mg/L, and the pH = 8.1-8.3. The data from one test are in Table P7.9. The
bacterial growth was measured, but we omit these data.
171
SOLVED MATERIAL BALANCE PROBLEMS:
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Reaction rates and reactor design
Time (h)
0
6
12
18
24
36
48
COD (mg/L)
3,200
2,900
2,200
1,600
1,350
900
500
Table P7.9
Solution
A plot (ln(C) vs. t) shows that this is a first-order reaction.
The estimated reaction rate coefficient is k = 0.039/h
8.5
4,000
8.0
3,000
2,500
ln (COD)
COD (mg/L)
3,500
2,000
1,500
1,000
ln(COD) = 8.137 - 0.039 t
7.5
7.0
6.5
6.0
500
0
0
10
20
30
40
5.5
50
0
10
Time (h)
20
30
40
50
Time (h)
Figure S7.9
From the fitted model
C = C0 exp(-kt)
ln(C) = ln(C0) – kt
→
ln(C) = 8.137 - 0.039 t
C = (3,419 mg/L) exp[-(0.039/h)(t)]
The reaction rate coefficient can also be estimated directly from the data, using the initial concentration
C0 = 3,200 mg/L at t = 0 and C = 500 mg/L at t = 48 h
k =
ln(3,200) - ln(500)
8.0709 - 6.2146
=
= 0.0387/h
48 h
48 h
The model would be
C = (3,200 mg/L) exp[-(0.0387/h)(t)]
This estimate is not as precise as the one obtained from fitting the data. The measurement error of
the concentrations is small so the estimates are in good agreement.
Notice that fitting the model by regression treats the initial concentration like all the others. It is
assumed to have some random measurement error and the C is estimated along with k.
172
SOLVED MATERIAL BALANCE PROBLEMS:
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7.10 SIMPLE REACTION - 1
A certain reaction A → products is second order in A. If this reaction is 10% complete
after 20 minutes, how long would it take for the reaction to be 90% complete?
Solution
The model for a second-order reaction is
1
C
1
k2t
C0
Solve for the second order rate coefficient k2 to get k2
1/C - 1/C0
t
For convenience, say the initial concentration of A is C0 = 1.0 mg/L
Substitute known values to get the value of k2
1/(0.9 mg/L) - 1/(1.0 mg/L)
(1.111 - 1.0) L/mg
=
= 0.011 L/mg-min
10 min
10 min
k2
For 90% conversion C = 0.1 mg/L, and solving for t gives
t
1/C 1/C0
1/(0.1 mg/L) - 1/(1.0 mg/L)
9 min
=
=
= 818 min
k2
0.011 L/mg-min
0.011
7.11 SIMPLE REACTION - 2
The graphs in Figure P7.11 all refer to the same reaction, A → products, where CA is the
concentration of A. What is the order of this reaction?
0.7
3
20
0.6
2.5
0.5
0.3
0.2
ln(CA)
15
0.4
CA
1/CA
3.5
25
0.8
10
1
5
0.1
0.5
0
0
0
10
20
30
40
Time
50
60
70
2
1.5
0
10
20
30
40
Time
Figure P7.11
173
50
60
70
0
0
10
20
30
40
Time
50
60
70
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Solution
Zero-order. The rate of change of concentration is constant over time, that is, independent of the
concentration of A. This is shown by the middle plot.
7.12 GASEOUS REACTION
In a dark room, a balloon is filled with a gas A up to a volume of 4.8 liters to give a
pressure in the balloon of 2 atmospheres. Subsequently, the balloon is placed under a UVlight source. This causes the conversion of A to B according to:
A→3B
a) Formulate the mass balance of A and B for the UV reaction in the balloon and
plot the mass of A, B, and A + B as a function of time. (b) Optional: Determine
the volume of the balloon as a function of time. Assume that the:
• gases behave ideally.
• temperature in the balloon is constant at 20°C.
• balloon is an ideal batch reactor, in which the volume is directly proportional
to the pressure.
• reaction is a first order in A.
• reaction rate constant is k = 0.15/s
Solution
a) Mass balance for A and B
1 mole of A produces 3 moles of B
The initial amount of A can be found from the ideal gas law
n = A0 =
PV
(2 atm)(4.8 L)
=
= 0.40 g-mol
RT
(0.082 L-atm/mol-K)(293 K)
There is no B in the balloon initially, so B0 = 0
After all A is converted to B, the balloon contains 1.2 g-mol of B
The material balance for A is given by the first order decay rate expression
dA/dt = -kA or A = A0 e-kt
The material balance for B is the initial amount (B0 = 0) plus the moles of A converted to B
The moles of A converted to B at time t is (A0 – A), so that
B = B0 +3 (A0 – A) = 3 A0(1-e-kt)
where A and B are g-mol of gas.
174
SOLVED MATERIAL BALANCE PROBLEMS:
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Reaction rates and reactor design
Total g-mol of gas = A + B = A0 e-kt +3 A0(1-e-kt) = 3A0 -2A0e-kt
For k = 0.15/s, the progression of the reaction as g-mol of A, B, and A+B are shown in Figure
S7.12a
1.2
Mass (g-mol)
1
A+B
0.8
B
0.6
0.4
A
0.2
0
0
5
10
15
20
25
30
Time (s)
Figure S7.12a
b) Volume change over time.
This is a more complicated problem. The volume of gas in the balloon is proportional to the
moles of gas, which is known from part (a). Gas volume will change with temperature and
pressure according to the ideal gas law. Temperature is constant, so pressure is the important
variable and it depends on the elasticity of the balloon, which is not specified.
If we assume that the balloon expands during the reaction so the pressure remains at 2 atm
(i.e., constant pressure), the volume when the reaction is complete will be
V = 3(4.8 L) = 14.4 L
If the balloon cannot expand, the volume will remain constant at 4.8 L, but the pressure will
increase to 3(2 atm) = 6 atm (we hope the balloon will not burst at this pressure).
The reality is somewhere between these two extremes. The experiment is analogous to
blowing up a balloon. As air enters the balloon, the pressure, volume, and moles of gas in
the balloon increase; pressure and volume due to the elasticity of the balloon, and moles
due to the UV reaction. We know how the moles of gas are changing from part (a). What we
need is a relation between pressure and volume in order to solve for volume (or pressure)
changes with time.
175
SOLVED MATERIAL BALANCE PROBLEMS:
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Reaction rates and reactor design
To demonstrate how this might work, we hypothetically assume that volume increases linearly
with pressure, as
P = keV
where ke is an expansion coefficient. Estimate ke from the conditions at the beginning of the
reaction
ke = P/V = (2 atm)/(4.8 L) = 0.417 atm/L
Substituting P = keV into the ideal gas law gives
PV = nRT → (keV)(V) = keV2 =nRT
and V = (nRT/ke)0.5
Substituting for n (the moles of A+B) gives
V = ((3A0 -2A0e-kt)RT/ke)0.5
For k = 0.15/s and ke = 0.417 atm/L, the volume change is shown in Figure S7.12b.
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9
Volume (L)
8
7
6
5
4
3
0
5
10
15
20
25
30
Time (s)
Figure S7.12b
7.13 OZONE KINETICS
Ozone disappears and oxygen appears according to the reaction
2 O3 → 3 O2
a) At a particulate instant, the rate of oxygen appearance is 3x10-5 mol/s, what is the
rate of ozone disappearance at that same instant?
b) If, at a particulate instant, the rate of ozone disappearance is 6x10-5 mol/s, what is
the rate of oxygen appearance at that same instant?
Solution
Ozone disappears and oxygen appears according to the reaction
2 O3 → 3 O2
The relative rates of change of [moles] of each gas are:
-
1 d[O3 ]
1 d[O2 ]
=
2 dt
3 dt
or
-
d[O3 ]
2 d[O2 ]
=
dt
3 dt
a) Ozone disappearance for oxygen appearance = 3x10-5 mol/s
-
d[O3 ]
2 d[O2 ]
=
dt
3 dt
2
(3x10-5mol/s) = 2x10-5mol/s
3
b) Oxygen appearance for ozone disappearance is 6x10-5 mol/s
d[O2 ]
3 d[O3 ]
=
dt
2 dt
-
3
(-6x10-5mol/s) = 9x10-5mol/s
2
177
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
7.14 TEMPERATURE AND THE REACTION RATE
A first order reaction has a rate coefficient k = 0.22 min-1 at 20°C. Calculate the rate
coefficient at 25°C and 30°C for θ = 1.05.
Solution
k25°C = (k20°C)1.05(25 - 20) = (0.22 min-1)(1.055) =0.281 min-1
k30°C = (k20°C)1.05(30 - 20) = (0.22 min-1)(1.0510) =0.358 min-1
7.15 DEGRADATION OF ATRAZINE
The chlorinated compound, atrazine, can be degraded by biological oxidation. The data in
Table P7.15 are from a kinetic experiment
Time (min.)
0
5
12
22
31
40
50
60
Atrazine (µg/L)
18
15
11
6.8
4.2
2.4
1.5
0.8
Table P7.15
Fit a first order reaction model to the data.
Solution
First-order reaction, where [A] represents the concentration of atrazine
[A] = [A]0e-kt
Regression on ln[A] vs. time (Figure S7.15) gives
ln[A] = 2.985 - 0.5217 t
which translates to
[A]0 = e2.985 = 19.8 μg/L
k = 0.5217/min (round to k = 0.52/min)
and
[A] = (19.8 μg/L)e-(0.52/min)t
178
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20
3.5
15
2.5
ln[Atrazine (μg/L)]
Atrazine (μg/L)
3
10
5
ln[Atrazine (μg/L)] = 2.985 - 0.05217 t
2
1.5
1
0.5
0
0
0
10
20
30
40
50
-0.5
60
0
10
Time (min)
20
30
40
50
60
Time (min)
Figure S7.15
7.16 SECOND-ORDER REACTION
Does a second-order model of the form d[A]/dt = -k[A]2, where [A] is the concentration
of substance A, describe the data in Table P7.16?
Time (min)
0
10
[A] (mol/L)
8
3.1
360°
thinking
40
60
80
100
1.9
1.1
0.8
0.6
0.5
Table P7.16
360°
thinking
.
.
20
360°
thinking
.
Discover the truth at www.deloitte.ca/careers
© Deloitte & Touche LLP and affiliated entities.
Discover the truth at www.deloitte.ca/careers
© Deloitte & Touche LLP and affiliated entities.
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179
Dis
SOLVED MATERIAL BALANCE PROBLEMS:
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Solution
The solution for the second-order model is
1
[A]
1
kt
[A]0
Time (min)
0
10
20
40
60
80
100
[A] (mol/L)
8
3.1
1.9
1.1
0.8
0.6
0.5
1/[A] (L/mol)]
0.125
0.323
0.526
0.909
1.250
1.667
2.000
Table S7.16
The reciprocal of [A] will plot as a straight line if this is the correct model. The straight shows a very
good fit (Figure S7.16). The rate coefficient is the slope of the line: k = 0.0188 L/mol-min.
The complete model is: 1 = 0.139 L/mol + (0.188L/mol-min)t
[A]
2.5
1/[A] (L/mol)
2.0
1/[A] = 0.139 + 0.0188 t
1.5
1.0
0.5
0.0
0
20
40
60
80
100
Time (min)
Figure S7.16
7.17 RATE OF DINITROTOLUENE (DNT) REMOVAL
Experiments on the oxidation of dinitrotoluene (DNT) and its intermediate by-products
with hydrogen peroxide (H2O2) produced the data in Table P7.17. The conditions were
15 moles of H2O2 per mole of DNT. Determine the reaction rate model and the reaction
rate coefficient.
180
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Time (min)
0
15
30
45
60
120
DNT (mol/L)
0.2020
0.0820
0.0550
0.0450
0.0230
0.0020
Table P7.17
Source: Nihar R. et al. 1993. ‘Oxidation of 2,4-dinitrotoluene using Fenton’s
reagent: reaction mechanisms and their practical applications’, Civil and
Environ. Engr. Faculty Publications, Northeastern University.
Solution
Plot the data as in Figure S7.17a. The concentration appears to decay exponentially, so try a first
order model.
0.25
DNT (mol/L)
0.2
0.15
0.1
0.05
0
0
20
40
60
80
100
120
Time (min)
Figure S7.17a
First-order reaction, where C represents the concentration of DNT
C = C0e-kt
Time (min)
0
15
30
45
60
120
DNT (mol/L)
0.2020
0.0820
0.0550
0.0450
0.0230
0.0020
ln(DNT)
-1.5995
-2.5010
-2.9004
-3.1011
-3.7723
-6.2146
Table S7.17a
Regression on ln(C) vs. time (Figure S7.17b) gives
ln(C) = -1.687 - 0.0369 t
181
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
which translates to
C0 = e-1.687 = 0.185 mol/L
k = 0.0369/min
The concentration of DNT vs. time and ln(DNT) vs. time are plotted in Figure S7.17b. The fitted firstorder model is
C = (0.185 mol/L) e-(0.0369/min)t
The solid lines are the concentration predicted from the fitted model.
-1.0
0.25
-2.0
ln [(DNT (mol/L)]
DNT (mol/L)
0.2
0.15
0.1
0.05
ln[DNT (mol/L)] = -1.687 - 0.0369
-3.0
-4.0
-5.0
-6.0
0
-7.0
0
20
40
60
80
100
0
120
Time (min)
20
40
60
80
100
120
Time (min)
Figure S7.17b
7.18 DISSOLUTION OF LEAD
More than seventy percent of the lead (Pb) used in manufacturing is to make lead-acid
batteries. Effective recycling and recovery of waste is an urgent need for environmental and
economic reasons. More than ninety percent of lead recovery technologies involve melting
the battery lead paste in furnaces at high temperatures exceeding 1,000°C. The paste contains
a high percentage of sulfur, in the form of lead sulfate (PbSO4). A serious problem is SO2
emissions resulting from decomposition of PbSO4 at high temperatures exceeding 1,000°C
and emissions of Pb by evaporation at these high temperatures.
A proposed hydrometallurgical process for the recovery of lead from batteries dissolves the lead
compounds (PbSO4, PbO and PbO2) in dilute sulfuric acid and calcium chloride solution.
The calcium and the sulfate react to form gypsum (CaSO4), which can be recovered in
almost pure form. The lead forms the hydroxide, Pb(OH)2. This can be dissolved in acetic
acid, CH3COOH, to form lead acetate, Pb(CH3COO)2. The dissolution is shown by the
data in Table P7.18 and the graph in Figure P7.18. The concentration of lead in solution
approaches an asymptote, which we shall call α and interpret as the saturation limit for lead
solubility under the test conditions. A plausible model to describe the change in soluble
lead concentration might be that the rate of dissolution is proportional to the difference
182
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
in the solution concentration [Pb] and the asymptote, α. Fit a model to the [Pb] data. As
a starting point consider
k1 (D [Pb])
d [Pb]
dt
or perhaps
kn (D [Pb]) n
Pb(CH3COO)2
CH3COOH
(mol/L)
(mol/L)
0
0
0.33
1.8
4.95
0.01
0.31
3.6
9.52
0.02
0.29
7.2
17.74
0.04
0.25
14.4
30.93
0.07
0.18
21.6
40.72
0.10
0.14
28.8
47.98
0.12
0.10
43.2
57.40
0.14
0.06
57.6
62.60
0.15
0.03
68.4
64.89
0.16
0.02
72.0
65.52
0.16
0.02
Time (s x10-3)
Pb (g/L)
0
Table P7.18
70
Soluble Lead [Pb] (g/L)
d [Pb]
dt
60
50
40
30
20
10
0
0
10
20
30
40
50
Time (s x 10-3)
Figure P7.18
183
60
70
80
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Solution
The plot in Figure P7.18, appears exponential that approaches an asymptotic soluble lead concentration
[Pb] around 70 g/L.
So as an initial fit, try a first order model, n = 1.
The model is
d[Pb]
dt
k1(D [Pb])
The solution is
[Pb]= α(1 – exp(-k1t)) = α – α exp(-k1t)
Rearrange and take logarithms
α - [Pb] = α exp(-k1t) ln(α - [Pb]) = ln α - k1t
This is the model for a straight line when plotting ln(α - [Pb]) vs. t. The slope is k1 and intercept is ln
α. The calculations and plot are shown in Table S7.18 and Figure S7.18, for a trial guess of α =70 g/L.
[Pb]
α – [Pb]
(g/L)
(α = 70 )
0
0
70.00
4.248
1.8
4.95
65.05
4.175
3.6
9.52
60.48
4.102
7.2
17.74
52.26
3.956
14.4
30.93
39.07
3.665
21.6
40.72
29.28
3.377
28.8
47.98
22.02
3.092
43.2
57.40
12.60
2.534
57.6
62.60
7.40
2.001
68.4
64.89
5.11
1.631
72
65.52
4.48
1.500
Time (s x 10-3)
Table S7.18
184
ln(α – [Pb])
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
ln ([Pb]-70 g/L)
4.5
4.0
ln([Pb] - 70 g/L) = 4.226 - 0.0383 t
3.5
3.0
2.5
2.0
1.5
1.0
0
10
20
30
40
Time (s x
50
60
70
80
10-3)
Figure S7.18
The fitted model looks pretty good, and gives
α = e4.226 = 68.45 g/L and k1 = -0.0383/(s x 10-3).
The estimated value of α is a bit smaller than the assumed value of 70 mg/L, so we might want to try
again with a smaller value, say 69 g/L. Try several values of α.
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POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
The plot of ln([Pb] – α) is linear, so our initial guess of a first order reaction (n = 1) is good.
Using the solver function in a spreadsheet program for the linear plot will yield the best values of α
(69.06 g/L) and k1 (0.0412/(s x 10-3).
7.19 BIOTRANSFORMATION
The data in Table P7.19 are for biotransformation of 1,1,1-trichloroethane. The proposed
kinetic model is second order. Fit a second-order model to the data, and compare that with
a first-order model.
Time (h)
0
2
5
10
24
48
Conc. (mg/L)
0.5
0.48
0.45
0.41
0.30
0.18
Table P7.19
Solution
a) Second-order model is
dC
dt
kC 2
Ÿ
1
C
1
kt
C0
where C is the concentration of 1,1,1-trichloroethane
A plot of 1/C vs. t (Figure S7.19a) should be a straight line.
The slope of the line is k. The estimated value is k = 0.0739 L/mg-h
The estimated C0 = 1/(1.843 L/mg) = 0.543 mg/L
0.6
6
0.5
5
0.4
4
1/C (L/mg)
C (mg/L)
The fitted model is: 1/C = 0.543 mg/L + (0.0739 L/mg-h) t
0.3
0.2
0.1
1/C = 1.843 + 0.0739 t
3
2
1
0
0
0
10
20
30
40
50
0
Time(h)
10
20
30
Time (h)
Figure S7.19a
186
40
50
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
The plotted data (1/C) show some convex curvature, so perhaps the second order model
isn’t correct, and a better model can be found.
b) Try a first-order model: C = C0e-kt
A plot of ln(C) vs. t (Figure S7.19b) should be a straight line.
The slope of the line is k. The estimated value is k = 0.0213/h
The estimated C0 = e-0.6894 = 0.502 mg/L
The fitted model is: C = (0.502 mg/L)e-(0.0213/h) t
-0.4
ln(C) = -0.6894 - 0.0213 t
ln [C (mg/L)]
-0.8
-1.2
-1.6
-2.0
0
10
20
30
40
50
Time (h)
Figure S7.19b
The first-order model is the better description of the data.
Notice that fitting the model by regression, as done here, treats the all the concentrations the same,
that is, as though they all have measurement error. If the test was arranged so the initial concentration
is known, the curve should be forced to go through the given value.
7.20 PESTICIDE DEGRADATION – LABORATORY EXPERIMENT
Table P7.20 gives pesticide degradation data from a laboratory experiment. Note that the
measurements were done in duplicate. Derive and fit a model to describe the data.
187
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Time
Concentration
(d)
(µg/L)
0
88.3
91.4
1
85.6
84.5
2
78.9
77.6
3
72.0
71.9
5
50.3
59.4
7
47.0
45.1
14
27.7
27.3
21
10.0
10.4
30
2.9
4.0
Table P7.20
Solution
Plot the data, Figure S7.20.
The duplicates are so nearly the same that many plotted values are obscured.
A first-order model seems justified. Propose
100
5.0
80
4.0
60
3.0
ln (C)
C (μg/L)
dC/dt = -kC and C = C0 exp(-kt)
40
20
0
ln(C) = 4.582 - 0.108 t
2.0
1.0
0
5
10
15
20
25
0.0
30
Time (d)
0
5
10
15
20
Time (d)
Figure S7.20
Replot as ln(C) vs. time and use the slope of the regression line as the estimate of k.
ln(C) = 4.582 – 0.108t
The (negative) slope is an estimate of the reaction rate coefficient: k = 0.108/d
188
25
30
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
The intercept is an estimate of ln(C0): ln(C0) = 4.582 and C0 = e4.582 = 97.7 μg/L
The fitted model is: C = (97.7μg/L)e-(0.108/d)t
The discrepancy between the estimated (97.7 μg/L) and the observed (88.3 and 91.4 μg/L) value of C0
is quite large in view of the precision of the measured values. Perhaps a different model will provide
a better fit to the data.
7.21 CHLORINE DECOMPOSITION
A dose of 6 mg/L chlorine was added to water from Lake Mendota, Madison, WI. The
water contains a variety of compounds that will react with chlorine. Some chlorine will
be oxidized to chloride and some will become chloramines which form when chlorine
reacts with ammonia. The measured response is free chlorine, which is HOCl. The pH
and temperature may be assumed constant throughout the experiment (pH = 8.25 and T
= 19.8°C). Use the data in Table P7.21 to model the disappearance of free chlorine (data
are from a UW-Madison class experiment).
Jar 1
Elapsed
Time (h)
Jar 2
Free
Elapsed
Chlorine
Time (h)
(mg/L)
Free
Chlorine
(mg/L)
0.05
1.50
0.04
1.55
0.37
1.15
0.37
1.15
0.70
1.00
0.71
0.95
1.04
0.90
1.03
0.90
1.53
0.70
1.55
0.80
2.05
0.65
2.04
0.73
2.53
0.69
2.53
0.67
Table P7.21
189
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Solution
Temperature and pH do affect the ionization of chlorine and the fraction of total chlorine that will be free
chlorine. We shall ignore that since the pH and temperature change very little during the experiment.
First plot the data in a variety of ways, as in the draft plots in Figure S7.21a, [the four plots are for a
zero order (C vs. t), first order (ln(C) vs. t, second order (1/C vs. t), and power function [ln(C) vs. ln(t)]
to see whether any appear linear.
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0.5
0.4
0.3
0.2
0.1
0
-0.1
-0.2
-0.3
-0.4
-0.5
C vs. t
0
0.5
1
1.5
2
2.5
3
0
0.5
0.4
0.3
0.2
0.1
0
-0.1
-0.2
-0.3
-0.4
-0.5
1.8
1.6
1/C vs. t
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.5
1
ln(C) vs. t
1.5
2
2.5
1
1.5
2
2.5
3
-1
0
1
2
ln(C) vs. ln(t)
-4
3
0.5
-3
-2
Figure S7.21a
None of the plots appear particularly linear, so we have to dig a bit deeper. Note that the plot of C
vs. t appears to be an exponential decay, but the asymptote is probably not zero. Rather the data
appear to level off at about 0.6 to 0.65 mg/L. So let’s try a first order model with an asymptote of,
say, α =0.63 mg/L. (This is a similar model to the one in Problem P7.18).
The model has the form (C – α) = (C0 – α)exp(-kt)
The calculations using an initial guess of α = 0.63 mg/L, are in Table S7.21, and the relevant plot [ln(C
– 0.63 mg/L) vs. t] is in Figure S7.21b
190
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Jar 1
Jar 2
Reaction rates and reactor design
Elapsed
Free Chlorine
C-α
Time
(C)
α = 0.63
(h)
(mg/L)
(mg/L)
0.05
1.5
0.87
-0.139
0.37
1.15
0.52
-0.654
0.7
1
0.37
-0.994
1.04
0.9
0.27
-1.309
1.53
0.7
0.07
-2.659
2.05
0.65
0.02
-3.912
2.53
0.69
0.06
-2.813
0.04
1.55
0.92
-0.083
0.37
1.15
0.52
-0.654
0.71
0.95
0.32
-1.139
1.03
0.9
0.27
-1.309
1.55
0.8
0.17
-1.772
2.04
0.73
0.1
-2.303
2.53
0.67
0.04
-3.219
Table S7.21
191
ln (C - α)
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
0.0
ln(C - 0.63 mg/L) = -0.1454 - 1.265 t
ln(C - 0.63 mg/L)
-1.0
-2.0
-3.0
-4.0
-5.0
0
0.5
1
1.5
2
2.5
3
Time (h)
Figure S7.21b
Figure S7.21b shows the data are well described by a straight line. The estimate of the rate coefficient
is k = 1.265/h. The intercept gives the estimate of C0.
ln(C0 – 0.63 mg/L) = -0.1454
C0 = e-0.1454 + 0.63 mg/L = 0.865 mg/L + 0.63 mg/L = 1.495 mg/L
The complete model is C - 0.63 mg/L = (1.495 mg/L – 0.63 mg/L)e-(1.265/h)t
or C = 0.63 mg/L – (0.865 mg/L)e-(1.265/h)t
Our guess that α = 0.63 mg/L may not be the best one. Try other values.
Note that the data become more dispersed after t = 1 h. This is an indication that a non-linear regression
method may better describe the data and give better parameter estimates. That subject is beyond
the scope of this problem, and one that is very useful.
7.22 DDT PERSISTANCE
DDT (dichloro diphenyl trichloroethane) was for many years considered a wonder chemical
and because of this it was widely used, especially for mosquito control to fight malaria. It is
a persistent chemical, almost insoluble in water but soluble in fat, and it accumulates in fish
and birds, with the concentration increasing in animals higher up the food chain (algae <
minnows < trout < eagles). The half-life of DDT in an embayment below a manufacturing
plant (which was closed many years ago) is estimated to be 15 years. The data are in Table
P7.23. Assume an in initial mass of 100 kg of DDT in 1945, almost all in crystalline form
in or on the sediments in the embayment. Estimate the mass remaining in 2020.
192
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Year
1945
1960
1975
1990
2005
Mass (kg) remaining
100
50
25
12.5
6.25
Table P7.22
Solution
Half of the mass is lost every 15 years.
There is no need to make an exact calculation. The table shows sufficient detail.
2020 is one half life beyond 2005, and the mass remaining should be about
6.25 kg/2 = 3.12 kg
7.23 BIOCONCENTRATION FACTOR
An aquatic organism that lives in contaminated water may accumulate heavy metals or toxic
organic chemicals. The same is true for insects that live in contaminated soil. While the
animals take in the pollutant they also rid themselves of it, in a process called depuration (see
Figure P7.23). Assume the pollutant concentration in water is CW and the concentration in
the animal is C. The rate coefficient for uptake is ku and the rate coefficient for elimination
is ke. Assume uptake and elimination are first-order processes.
a) Assume a fish is moved from clean water into polluted water. Derive the model
for accumulating pollutant concentration in the fish.
b) What is the steady-state pollutant concentration CSS in the fish?
c) What is the bioconcentration factor, defined as BCF = CSS/CW
d) Assume the fish is removed from the polluted water, after the concentration C has
reached steady-state, and put into clean water. Derive the model for depurating
pollutant concentration in the fish.
CW
C
ku
Figure P7.23
193
ke
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Solution
a) This is a single compartment model. The compartment is the fish.
The material balance for the accumulation of C is
dC
dt
kuCW keC
and the solution is
C
§ CW ku
¨¨
© ke
·
k t
¸¸ (1 e e )
¹
b) Steady-state concentration, CSS, at large t is
§ CW ku
¨¨ k
© e
C
·
ke t
¸¸ (1 e )
¹
Ÿ
CSS
§ CW ku ·
¨¨ k ¸¸
© e ¹
c) Bioconcentration factor, at steady-state
BCF
CSS
CW
ku
ke
d) In clean water, CW = 0
The material balance for the depuration of C is
dC
dt
C
kuCW keC
CSS e-ket
Ÿ
dC
dt
Ÿ
§C k
C = ¨¨ W u
© ke
keC
· ket
¸¸ e
¹
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POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
7.24 BIOACCUMULATION & DEPURATION
Modify the uptake/depuration model in the previous problem (Problem 7.23) to account
for uptake and depuration through the gills (k1 and k2), and metabolic biotransformation,
km, as shown in Figure P7.24. Neglect elimination by growth dilution.
Metabolic
biotransformation, km
Gill uptake, k1
Dietary
uptake, ku
Fecal egestion, ke
Gill elimination, k2
Figure 7.24
Solution
a) Accumulation material balance. C is the concentration in the fish
dC
dt
kuCW k1CW k2C keC kmC
(ku k1 )CW (k2 + k e + km )C
the solution is
C
§ CW (ku k1 ) ·
( k k k )t
¨¨
¸¸ (1 e 2 e m )
© k2 ke km ¹
b) Steady state fish concentration, CSS
CSS
CW (ku k1 )
k2 ke km
c) Bioconcentration factor, at steady-state
BCF
CSS
CW
(ku k1 )
(k2 ke km )
d) In clean water, CW = 0
The material balance for the depuration of C is
195
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
dC
dt
Reaction rates and reactor design
ku CW k1CW k2C keC kmC
(k2 ke km )C
the solution is
C
CW (ku k1 ) (k2 ke km )t
e
(k2 ke km )
CSS e (k2 ke km )t
7.25 METABOLITES OF PESTICIDES
% of Applied Radioactivity
Some pesticides create a metabolite when they decompose in soil or sediment. Figure P7.25
shows such a case. The parent compound was manufactured to contain radioactive carbon
(C14) and the vertical axis shows the percentage of applied radioactivity that remains days
after the application. Formulate a model that is a plausible description of the pattern shown
in Figure P7.25.
100
Parent compound
80
60
Metabolite
40
20
0
0
20
40
80
60
100
120
140
160
Days after application
Figure P7.25
Solution
First-order models very often apply to pesticide degradation so the simplest model will be first-order
reactions in series.
Define
P = concentration of the parent compound (e.g. pesticide)
M = concentration of the metabolite
Assume 100% formation of the metabolite.
For the Parent material dP/dt = -kPP
For the Metabolite
dM/dt = kPP – kMM
196
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
where P and M are the concentrations, and kP and kM are the rate coefficients of the parent and
metabolite compounds, respectively.
The solutions with M0 = 0, are
P
M
P0e kP t
kP P0
ªe kP t e kM t º
¼
kM kP ¬
7.26 PARALLEL REACTION
Compound A decomposes to form compounds B, C, and D as shown in Figure P7.26.
Assume all reactions are first order with respect to A. The rate coefficients are k1 = 0.01/h,
k2 = 0.02/h, and k3 = 0.03/h. (a) Write a model for the concentration of A (CA). (b) If CA0
= 100 mg/L, what is the concentration of A after 1 day?
197
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
B
k1
k2
A
C
k3
D
Figure P7.26
Solution
a) First order model for parallel reactions
dC A
dt
k1C A k2C A k3C A
(k1 k2 k3 )C A
with solution
CA = CA0 exp[-(k1 + k2 + k3)t]
b) Concentration after 1 d, with CA0 = 100 mg/L
C A = C A0exp ª¬-(k1 k2 k3 )t º¼ = C A0exp ª¬-(0.01/h + 0.02/h + 0.03/h)t º¼
= (100 mg/L) exp -(0.06/h)t
= (100 mg/L) exp -(0.06/h)(24 h)
= (100 mg/L)(0.237) = 23.7 mg/L
7.27 OZONE DECOLORIZATION OF ACID YELLOW DYE
Acid yellow dye (C32H22N2OS2CoNa) was decolorized by oxidation with ozone. It is resistant
to biodegradation (the BOD is near zero) and chemical oxidation is the alternative treatment.
The postulate reaction pathway goes through two intermediate products and then to a
colorless product. The reaction rate coefficients are shown above the arrows. The units are
h-1. The initial concentration of the acid yellow dye is 5.75x10-3 mol/m3. The intermediate
and final products have an initial concentration of zero.
k3 = 0.01/h
k1 = 0.07/h
k2 = 0.033/h
Dye 
o Intermediate 
o Color Product 
o Clear Product
Construct a model using first-order reactions. (Source: Mackowska, e, et al. 2003, Polish J.
Envir. Studies, vol. 4, pp 425-429).
198
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Solution
The material balance equations for Dye, Intermediate (Int), Color product Color), and Clear product
(Clear) are
dCDye
k1CDye
dt
dCInt
dt
(0.07/h)CDye
k1CDye k2CInt
(0.07/h)CDye (0.033/h)CInt
dCColor
dt
k2CInt k3CColor
dCClear
dt
k3CColor
(0.033/h)CInt (0.01/h)CColor
(0.10/h)CColor
The solutions to these equations are complex and will not be provided here. However, they are shown
graphically in Figure S7.27. Note that it takes over 4 days for the reaction to approach completion.
Concentration (mol/m3 x 10-3)
6
5
Acid yellow dye
4
Clear Product
Intermediate
3
2
Color Product
1
0
0
20
40
60
80
100
Time (h)
Figure S7.27
7.28 PESTICIDE DEGRADATION
Table P7.28 and Figure P7.28 show field data on degradation of a pesticide. The pesticide
(P) degrades to a metabolite (M). The metabolite decays to minerals. Is the degradation of
the parent compound first-order? If so, what is the degradation rate coefficient? Modeling
the metabolite is optional.
Compound
Time (days)
0
1
3
7
14
28
59
91
Pesticide
6.8
6.5
6.3
5
4.0
3.0
0.6
0.3
Metabolite
0.69
1.11
0.97
2.36
3.05
1.94
0.42
0.3
Table P7.28
199
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Concentration (mg/L)
8
7
6
Parent Compound
5
4
3
2
1
Metabolite
0
0
20
40
60
80
100
Time (d)
Figure P7.28
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200
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Solution
Define
P = concentration of the parent compound (e.g. pesticide)
M = concentration of the metabolite
Is the degradation of the parent compound first-order?
Yes. A plot of ln(P) vs. time, Figure S7.28a is a straight line.
2.5
2.0
ln[P (mg/L)] = 1.910 - 0.0358 t
ln [P (mg/L)]
1.5
1.0
0.5
0.0
-0.5
-1.0
-1.5
-2.0
0
20
40
60
80
100
Time (d)
Figure S7.28a
The slope estimates the reaction rate coefficient: kP = 0.036/d
The intercept estimates ln(P0) = 1.910, and P0 = e1.910 = 6.75 mg/L
The complete model is
dP
dt
kP P
and
P
P0ekPt
(6.75 mg/L)e-(0.036/d)t
Optional: To model the metabolite, postulate two first-order models in series.
dP
dt
kP P
and
dM
dt
kP P kM M
The solutions are
P
M
P0e kP t
kP P0
ªe kP t e kM t º M0e kM t
¼
kM kP ¬
This is a multi-response estimation problem and can be done with an advanced regression program,
or with the ‘solver’ function in a spreadsheet program. The fitted models are shown in Figure S7.28b,
with parameter estimates P0 = 6.93 mg/L, kP= 0.0391/d, M0 = 0.889 mg/L, and kM = 0.0601/d.
201
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Concentration (mg/L)
8
7
6
Parent Compound
5
4
3
2
Metabolite
1
0
0
20
40
60
80
100
Time (d)
Figure S7.28b
7.29 POLLUTANT REMOVAL IN A CSTR
A pollutant is destroyed in a CSTR reactor with a detention time 2 hours. The initial
concentration is 800 mg/L and the effluent concentration is 40 mg/L. The reaction is firstorder and the process operates at steady state. (a) What is the pollutant concentration in the
reactor? (b) Calculate the reaction rate coefficient. (c) What will the effluent concentration
be if the flow rate increases by 40%?
Solution
a) Concentration in the reactor is the same as the effluent concentration, i.e. 40 mg/L.
b) The material balance for steady-state operation is
QC0 kCV
40 mg/L =
QC
Ÿ
800 mg/L
1+k(2 h)
C
C0
1 kV /Q
Ÿ
C0
1 kT
k = 9.5/h
c) Increasing the flow by 40% reduces the detention time by 40%, giving
θ = (2 h)(1 – 0.4) = 1.2 h
C
C0
800 mg/L
800 mg/L
=
=
= 64.5 mg/L
1 kT
1 + (9.5/h)(1.2 h)
12.4
202
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
7.30 POLLUTANT REMOVAL IN TWO CSTRS IN SERIES
Figure P7.30 shows two CSTRs in series. The process is steady state and the reactions are
first order. The reaction rate coefficients are the same in reactors 1 and 2. What are the
detention times? (The diagram shows equal sizes, but this is misleading.)
C0 = 250 mg/L
C1 = 125 mg/L
Reactor 1
Reactor 2
C2 = 20 mg/L
θ2 V2
θ1 V1
C1 = 125 mg/L
C2 = 20 mg/L
Figure P7.30
Solution
The material balances are
QC0 kC1V1
C1
QC1
and
C0
1 kT1
and
QC1 kC2V2
C2
QC2
C1
1 kT2
Because the flow rate, rate coefficient, and detention times are unknown, we can only solve for the
ratio of the detention times.
Reactor 1
kT1
Reactor 2
C0
250 mg/L
-1=1
1 =
C1
125 mg/L
and
kT2
C1
125 mg/L
-1=4
1 =
C2
20 mg/L
The ratio of the two material balances gives
kT2
T
4
= 2 =
=4
1
kT1
T1
and
V2
=4
V1
7.31 MIXED-ORDER MODEL
A well-mixed reactor was operated at steady-state at a high concentration and it was
determined that the maximum rate of removal was k = 20 mg/L-min. It was also determined
that the decomposition was 10 mg/L-min when the concentration was C = 15 mg/L. The
kinetic model is known to be
dC
dt
kC
K C
What will be the rate of decomposition when the concentration is C = 5 mg/L?
203
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Solution
The maximum rate of removal is k and that value is given: k = 20 mg/L-min.
The coefficient K is the concentration when the rate is half the maximum, which is K = 15 mg/L at the
rate of 10 mg/L-min. Thus, the model is
dC
dt
kC
K C
(20 mg/L-min) C
15 mg/L + C
At C = 5 mg/L the removal rate is
dC
kC
(20 mg/L-min) (5 mg/L)
=
=
= 5 mg/L-min
dt
K C
15 mg/L + 5 mg/L
7.32 DETENTION TIME FOR A CSTR
A CSTR has a feed pollutant concentration C0 = 250 mg/L and produces an effluent with
20 mg/L. Calculate the detention time if the CSTR operates with the kinetic model
(600 mg/L-d)C
dC
=
dt
40 mg/L + C
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Solution
The concentration in the CSTR is the same as the effluent, that is, 20 mg/L.
The reaction rate in the CSTR is constant and is the rate when C = 20 mg/L.
(600 mg/L-d)C
(600 mg/L-d)(20 mg/L)
ª dC º
=
=
= 200 mg/L-d
« dt »
40
mg/L
+
40 mg/L + 20 mg/L
C
¬
¼C = 20mg/L
The pollutant removed is (250 mg/L – 20 mg/L) = 230 mg/L
Detention time =
T =
Pollutant removed
Rate of reaction
(250 mg/L - 20 mg/L)
= 1.15 d
200 mg/L-d
7.33 DETENTION TIME FOR TWO CSTRs IN SERIES
Calculate the detention times and reactor volumes for the two CSTRs shown in Figure
P7.30. Both reactors have k = 600 mg/L-d and K = 40 mg/L, as in the previous problem
(P7.32). The pollutant concentration in CSTR 1 is 125 mg/L and the concentration in
CSTR 2 is 20 mg/L. Reactor 1 removes (250 – 125) mg/L = 125 mg/L. Reactor 2 removes
(125 – 20) mg/L = 105 mg/L.
Solution
For the conditions shown in Figure P7.32, the pollutant removal rates are
dC1
dt
kC1
K C1
(600 mg/L-d)(125 mg/L)
= 455 mg/L-d
40 mg/L + 125 mg/L
dC2
dt
kC2
K C2
(600 mg/L-d)(20 mg/L)
= 200 mg/L-d
40 mg/L + 20 mg/L
For a general rate expression, dC/dt, the general material balance on a CSTR is
ª dC º
QC0 V «
»
¬ dt ¼C
QC
Noting V/Q = θ, and solving for θ gives T
C0 C
V
Q
¬ªdC /dt ¼º C
The detention time in reactor 1, using ¬ªdC /dt ¼ºC = 455 mg/L-d, is
1
T1 =
C0 C1
V1
(250 mg/L - 125 mg/L)
=
=
= 0.275 d
Q
455 mg/L-d
ª¬dC /dt º¼C
1
Reactor 2 reduces the reactor 1 effluent (125 mg/L) to 20 mg/L. The detention time, using ª¬dC /dt ¼ºC
= 200 mg/L-d, is
2
205
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
T2 =
Reaction rates and reactor design
V2
C1 C2
(125 mg/L - 20 mg/L)
=
=
= 0.525 d
Q
200 mg/L-d
ª¬dC /dt º¼C
2
The total hydraulic detention time for the system is
qTotal = q1 + q2 = 0.275 d + 0.525 d = 0.8 d
This is much less than θ = 1.15 d that was calculated for the single reactor in Problem 7.33. The reason
is that half the pollutant removal is done at a greater reaction rate because the pollutant concentration
in reactor 1 is higher than in reactor 2.
NOTE: Imagine that a series of five or six reactors in series were used. The total detention time would
be even less. Adding more reactors in series makes the system more like an ideal a plug flow reactor.
7.34 GRAPHICAL SOLUTION FOR TWO CSTRs IN SERIES
Construct a graphical solution to the 2 CSTRs-in-series as described in Problem 7.33. The
reaction rate model is
(600 mg/L-d)C
dC
=
dt
40 mg/L + C
Solution
The solution for the detention time in CSTR 1 was
T1 =
C0 C1
V1
(250 mg/L - 125 mg/L)
=
=
= 0.275 d
Q
455 mg/L-d
ª¬dC /dt º¼C
1
The blue line in Figure S7.34a is the rate curve for the reaction. Draw a vertical line from C1 = 125
mg/L on the concentration axis to intercept the rate curve. Then draw a line from that intersection
point to C0 = 250 mg/L on the concentration axis. The negative reciprocal of the slope of this line is
the value of θ1 that was calculated above. Repeat the procedure to get the detention time for CSTR 2.
206
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Reaction rate, dC/dt (mg/L-d)
600
dC
(600 mg/L-d)(C )
=
dt
40 mg/L + C
500
400
T1 = -1/Slope
= - (250 – 125)/(0 – 450)
= 0.275 d
300
T2 = -1/Slope
200
= - (125 – 20)/(0 - 200)
= 0.525 d
100
0
0
50
100
150
200
250
300
C0 = 250 mg/L
C1 = 125 mg/L
C2 = 20 mg/L
Concentration, C (mg/L)
Figure S7.34a Graphical solution for two CSTRs in series for k = 600 mg/L-d and K = 40 mg/L.
Figure S7.34b shows the solution for 3 CSTRs in series, where the effluent concentrations from the
three reactors are C1 = 125 mg/L, C2 = 50 mg/L and C3 = 20 mg/L. The total detention time is reduced
to 0.65 d, compared with 1.15 d for 1 CSTR and 0.8 d for 2 CSTRs.
Reaction rate, dC/dt (mg/L-d)
600
dC
(600 mg/L-d)(C )
=
dt
40 mg/L + C
500
TTotal = 0.275 + 0.225 + 0.15
400
= 0.65 d
300
T1 = 0.275 d
T2 = 0.225 d
200
T3 =
0.15 d
100
0
0
50
C3 = 20 C2 = 50
100
150
200
C1 = 125
250
300
C0 = 250
Concentration, C (mg/L)
Figure S7.34b Graphical solution for three CSTRs in series for k = 600 mg/L-d and K = 40 mg/L.
207
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
7.35 FITTING MODELS
Using the data in Table P7.35, determine the least squares estimates of β and θ by plotting
the sum of squares for these models:
a) y1, Calc = β x12
b) y2, Calc = 1 – exp(-θ x2)
where yCalc represents the calculated value of y for an assumed value of β, and yObs represents
the observed value of y.
x1
y1, Obs.
x2
y2, Obs.
2
2.8
2
0.44
4
6.2
4
0.71
6
10.4
6
0.81
8
17.7
8
0.93
Table P7.35
Solution
a) Model: y1, Calc = βx12
The calculations are summarized in Table S7.35a.
The observed data (x1 and y1, Obs) are listed in the first two columns.
The top row of the table is the trial values of β.
The top section of the table has the values of y1, Calc for the given x1 and the trial β.
The middle section has the values of the residual errors; (y1, Obs – y1, Calc)
The bottom section is the residual errors squared; (y1, Obs – y1, Calc)2.
The bottom line is the residual sum of squares for the trial β.
SSQ = (2.8 - E 22 )2 + (6.2 - E 42 )2 + (10.4 - E 62 )2 + (17.7 -E 82 )2
The minimum sum of squares, SSQmin = 5.86, is for β = 0.29.
Figure S7.35a is a plot of SSQ vs. β and shows β = 0.285 is has marginally smaller SSQ.
208
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Observed
beta
Data
x1
y1,
0.24
0.25
0.26
0.27
0.28
0.29
0.30
0.31
Obs.
y1, Calc
2
2.8
0.96
1
1.04
1.08
1.12
1.16
1.2
1.24
4
6.2
3.84
4
4.16
4.32
4.48
4.64
4.8
4.96
6
10.4
8.64
9
9.36
9.72
10.08
10.44
10.8
11.16
8
17.7
15.36
16
16.64
17.28
17.92
18.56
19.2
19.84
Residuals
2
2.8
1.84
1.8
1.76
1.72
1.68
1.64
1.6
1.56
4
6.2
2.36
2.2
2.04
1.88
1.72
1.56
1.4
1.24
6
10.4
1.76
1.4
1.04
0.68
0.32
-0.04
-0.4
-0.76
8
17.7
2.34
1.7
1.06
0.42
-0.22
-0.86
-1.5
-2.14
Squared Residuals
2
2.8
3.39
3.24
3.10
2.96
2.82
2.69
2.56
2.43
4
6.2
5.57
4.84
4.16
3.53
2.96
2.43
1.96
1.54
6
10.4
3.10
1.96
1.08
0.46
0.10
0.00
0.16
0.58
8
17.7
5.48
2.89
1.12
0.18
0.05
0.74
2.25
4.58
Sum
17.53
12.93
9.46
7.13
5.93
5.86
6.93
9.13
Table S7.35a
Residual sum of squares
20
18
16
14
12
Minimum
sum of squares
10
8
6
4
0.23
0.24
0.25
0.26
0.27
0.28
Beta
Figure S7.35a
209
0.29
0.3
0.31
0.32
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
b) Model: y2, Calc = 1 – exp(-θ x2)
The calculations are summarized in Table S7.35b, in a manner similar to Table S7.35a
The minimum sum of squares, SSQmin = 0.0013, is for θ = 0.30.
Figure S7.35b is a plot of SSQ vs. θ and shows θ = 0.298 has marginally smaller SSQ.
.
210
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Reaction rates and reactor design
Observed
theta
Data
x2
y2,
0.26
0.27
0.28
0.29
0.30
0.31
0.32
0.33
Obs.
y2, Calc
2
0.44
0.4055
0.4173
0.4288
0.4401
0.4512
0.4621
0.4727
0.4831
4
0.71
0.6465
0.6604
0.6737
0.6865
0.6988
0.7106
0.7220
0.7329
6
0.81
0.7899
0.8021
0.8136
0.8245
0.8347
0.8443
0.8534
0.8619
8
0.93
0.8751
0.8847
0.8935
0.9017
0.9093
0.9163
0.9227
0.9286
Residuals
2
0.44
0.0345
0.0227
0.0112
-0.0001
-0.0112
-0.0221
-0.0327
-0.0431
4
0.71
0.0635
0.0496
0.0363
0.0235
0.0112
-0.0006
-0.0120
-0.0229
6
0.81
0.0201
0.0079
-0.0036
-0.0145
-0.0247
-0.0343
-0.0434
-0.0519
8
0.93
0.0549
0.0453
0.0365
0.0283
0.0207
0.0137
0.0073
0.0014
Squared Residuals
2
0.44
0.0012
0.0005
0.0001
0.0000
0.0001
0.0005
0.0011
0.0019
4
0.71
0.0040
0.0025
0.0013
0.0006
0.0001
0.0000
0.0001
0.0005
6
0.81
0.0004
0.0001
0.0000
0.0002
0.0006
0.0012
0.0019
0.0027
8
0.93
0.0030
0.0021
0.0013
0.0008
0.0004
0.0002
0.0001
0.0000
Sum
0.0086
0.0051
0.0028
0.0016
0.0013
0.0019
0.0031
0.0051
Table S7.35b
211
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
8MATERIAL BALANCE FOR
BIOLOGICAL PROCESSES
8.1
BOD - 1
Three wastewater samples had a measured 5-day, 20°C BOD of 250 mg/L. The k coefficients
for the three samples were 0.25 d-1, 0.35 d-, and 0.45 d-1. Calculate the ultimate BOD for
each sample.
Solution
BODt = BODUlt (1 - e-kt )
Ÿ
BODUlt =
BODt
(1 - e-kt )
for k = 0.25/d
BODUlt
250 mg/L
250 mg/L
250 mg/L
=
=
= 350 mg/L
1 - 0.2865
0.7135
(1 - e-(0.25/d)(5d) )
For k = 0.35/d: BODUlt = 303 mg/L
For k = 0.45/d: BODUlt = 279 mg/L
8.2
BOD - 2
Use the long term BOD data in the table to determine the ultimate carbonaceous BOD
(CBODUlt) and the ultimate nitrogenous demand (NODUlt).
Time (d)
BOD
(mg/L)
Time (d)
BOD
(mg/L)
0
1
2
3
4
5
6
7
8
9
10
0
10
18
23
26
29
31
32
33
46
56
11
12
13
14
16
18
20
25
30
40
63
69
74
77
82
85
87
89
90
90
Table P8.2
212
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Solution
Plot the data (Figure S8.2)
A graphical estimate of CBODUlt and NODUlt can be made from the asymptotes of the CBOD portion
of the data (days 0 – 8) and the NOD portion (days 9 – 40)
100
BOD (mg/L)
80
NODUlt = 90 mg/L - 33 mg/L
= 57 mg/L
60
40
CBODUlt = 33 mg/L
20
0
0
5
10
15
20
25
30
35
40
Time (d)
Figure S8.2
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SOLVED MATERIAL BALANCE PROBLEMS:
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8.3
Material balance for biological processes
BOD OF SOYBEAN OIL
The data in Table P8.3 are from a long-term BOD test on soybean oil. BOD is reported as
g BOD per g of soybean oil, and t = time in days. There are four replicates at each time.
Fit the first-order BOD model
BOD T1 1 eT2t
where
θ1 = ultimate BOD (g/g soybean oil)
θ2 = reaction rate coefficient (d-1)
a) Estimate the ultimate BOD θ1
b) Estimate the rate coefficient θ2.
Time (d)
1
2
BOD replicate
3
4
7
12
20
BOD (g/g soybean oil)
1
0.40
0.99
0.95
1.53
1.45
2.12
2.42
2
0.55
0.95
1.00
1.47
1.35
2.21
2.28
3
0.61
0.98
1.05
1.75
1.90
2.34
1.96
4
0.66
0.95
1.20
1.60
1.95
1.95
1.92
Table P8.3
Solution
Plot the data – Figure S8.3 and draw free hand smooth curve.
An approximate solution would be to use the graphical estimate of the ultimate BOD and estimate
the one remaining parameter, θ2.
The approximate ultimate BOD can be estimated graphically.
θ1 = 2.2 g BOD/g soybean oil
Estimates of the rate coefficient θ2 can be calculated at a few times along the curve and averaged.
For t = 2 d, BOD2 = 1.0 g/g
BOD2
1.0 g/g = T1 1 eT2t
eT2 (2d) = 1 -
= (2.2 g/g)(1 - eT2 (2d) )
1.0 g/g
= 0.5455
2.2 g/g
Ÿ
For t = 4 d, BOD4 = 1.5 g/g
214
T2
0.303/d
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
BOD4
1.5 g/g = T1 1 eT2t
eT2 (4d) = 1 -
Material balance for biological processes
= (2.2 g/g)(1 - eT2 (4d) )
1.5 g/g
= 0.3182
2.2 g/g
T2
Ÿ
0.286/d
For t = 7 d, BOD7 = 1.85 g/g
BOD7
1.85 g/g = T1 1 eT2t
eT2 (7d) = 1 -
= (2.2 g/g)(1 - eT2 (7d) )
1.85 g/g
= 0.1591
2.2 g/g
Ÿ
T2
0.263/d
Averaging the estimated values from days 2, 4, and 7 gives θ2 = 0.284/d
Approx. BODult = 2.2 g/g
BOD (g/g soybean oil)
2.5
2
1.5
1
0.5
0
0
5
10
15
20
Time (d)
Figure S8.3 (Curve drawn free-hand)
Estimating θ2 at these various times gives different values. This is because the graphical estimate of
θ1 is probably not the best value. Using a solver or regression function in a spreadsheet program to
simultaneously estimate θ1 and θ2 will give the best values, which are: θ1 = 2.157 g/g and θ2 = 0.270/d.
Given the scatter in the replicate data, the graphical estimates are pretty good.
8.4
KINETICS OF BOD REMOVAL
Figure P8.4 shows the progress of BOD removal over time for two levels of mixed liquor
suspended solids (MLSS). The initial BOD5 is 300 mg/L. The tests were done in a batch
reactor. The MLSS will increase as BOD5 is removed, but that can be ignored. The BOD5
removal rate is expected to be proportional to the MLSS concentration. Does this seem to
be the case?
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SOLVED MATERIAL BALANCE PROBLEMS:
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Material balance for biological processes
500
BOD5 (mg/L)
200
100
MLSS =
1000 mg/L
50
MLSS =
2000 mg/L
20
10
0
2
4
6
8
10
Aeration Time (h)
Figure P8.4
Solution
Start by calculating the BOD removal rate coefficients for the data. The initial BOD is 300 mg/L.
The straight line on a plot of log(BOD) vs. time indicates a first-order reaction.
The tests were done in a batch reactor so the model is
BODt = BOD0 e-Kt
where K, the effective BOD removal rate coefficient, has units of h-1. K does depend on the MLSS
concentration.
For MLSS = 1,000 mg/L, BODt = 25 mg/L at t = 8 h
20 mg/L = (300 mg/L) e-K(8 h)
and
ln(20/300) = -K(8 h) and K = 0.338/h
For MLSS = 2,000 mg/L, BODt = 20 mg/L at t = 4 h
20 mg/L = (300 mg/L) e-K(4 h)
and
ln(20/300) = -K(4 h) and K = 0.677/h
Doubling the MLSS concentration did double the BOD5 removal rate.
Defining K = kX, where X = MLSS concentration, we can calculate k, the rate of BOD removal per unit
mass of MLSS, for both conditions (which should be the same).
k = K/X
For 1,000 mg MLSS/L k = (0.338/h)/(1,000 mg MLSS/L) = 0.000338 L/mg MLSS-h
For 2,000 mg MLSS/L k = (0.667/h)/(2,000 mg MLSS/L) = 0.000338 L/mg MLSS-h
216
SOLVED MATERIAL BALANCE PROBLEMS:
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8.5
Material balance for biological processes
BOD REMOVAL IN A LAGOON
A batch treatability test to support the design of a wastewater treatment lagoon will treat
an influent with BOD5 = 350 mg/L to produce, after 20 days, an effluent with BOD5 =
20 mg/L. Assume a first-order removal rate model and calculate the BOD removal rate
coefficient k.
Solution
The model for first-order BOD removal in a batch process is
BODEff
(k )(20 d)
k
8.6
BODInf e kt
Ÿ
20 mg/L = (350 mg/L)e(k )(20 d)
ln[(20 mg/L)/(350 mg/L)] = ln(0.0571) =-2.862
2.862/20 d = 0.143 d1
BOD REMOVAL AND TEMPERATURE
Activated sludge reactors (CSTRs) were operated at six temperatures to evaluate waste
treatment in a climate with cold winters and hot summers. The reactors were completely
mixed and the volume was 25 L. The results are given in Table P8.6.
a) Calculate the COD removal rate coefficient for each condition. Assume first-order
reactions
b) Derive an equation for k, the COD removal rate coefficient, as a function of
temperature.
Temp.
Flow
Inf. COD
MLVSS
Eff. COD
(°C)
(L/d)
(mg/L)
(mg/L)
(mg/L)
10
20
800
1,640
27
12
20
800
1,800
22
15
20
800
2,150
16
18
20
800
1,700
20
22
20
800
2,000
15
25
20
800
1,920
12
Table P8.6
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SOLVED MATERIAL BALANCE PROBLEMS:
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Material balance for biological processes
Solution
a) COD removal rate coefficients
The material balance model for a CSTR with first order reaction is
QS0 kSXV
QS
Ÿ
k
Q(S0 S)
SXV
where Q = flow, S0 = influent COD, S = effluent COD
X = mixed liquor volatile suspended solids (MLVSS)
Example calculation for reactor at 10 °C:
k10°C
(20 L/d)(800 mg COD/L - 27 mg COD/L)
(27 mg COD/L)(1,640 mg MVSS/L)(25 L)
0.00140 L/mg MLVSS-d
Calculations for each reactor are summarized in Table S8.6
Temp.
Flow
Inf. COD
MLVSS
Eff. COD
k
(°C)
(L/d)
(mg/L)
(mg/L)
(mg/L)
(L/mg MLVSS-d)
10
20
800
1,640
27
0.0140
12
20
800
1,800
22
0.0157
15
20
800
2,150
16
0.0182
18
20
800
1,700
20
0.0184
22
20
800
2,000
15
0.0209
25
20
800
1,920
12
0.0274
Table S8.6
b) The temperature correction model for activated sludge reactions is
kT = k20qC T T 20qC
This is referenced to 20°C because the operating temperature is almost always between
10°C and 30°C
Linearize the model by taking natural logarithms
ln(kT ) = ln(k20°C ) + (T - 20°C) ln(T )
Plot the data as ln(kT) vs (T – 20°C) to get Figure S8.6 and calculate θ from the slope of the
regression model.
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SOLVED MATERIAL BALANCE PROBLEMS:
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Material balance for biological processes
-1.55
ln(kT) = ln(k20qC) + ln(θ) (T - 20qC)
-1.60
= -1.6785 + 0.0169(T - 20qC)
ln(kT)
-1.65
-1.70
-1.75
-1.80
-1.85
-1.90
-15
-10
-5
0
5
10
T - 20qC
Figure S8.6
The regression model is
ln(kT) = ln(k20°C) + ln(θ)(T - 20°C) = -1.6785 + 0.0169(T - 20°C)
From the slope:
ln(θ) = 0.0169, giving θ = 1.017
From the intercept:
ln(k20°C) = -1.6785, giving k20°C = 0.187 L/mg MLVSS-d
The complete model is
8.7
kT = (0.187 L/mg MLVSS-d)(1.017)(T -20°C)
BIOMASS YIELD FACTOR
Figure P8.7 shows the flow rates, total suspended solids (TSS) and the influent and effluent
BOD (5-day) for an activated sludge process. These values can be used to calculate the
biomass yield factor (Y) for the process, in kg TSS produced/kg BOD removed. WAS is
waste activated sludge, which is intentionally removed from the activated sludge process to
balance the new biomass that is produced by the treatment process.
219
SOLVED MATERIAL BALANCE PROBLEMS:
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Influent
QInf = 1,000 m3/d
TSSInf = 80 mg/L
BODInf = 200 mg/L
Material balance for biological processes
Final Clarifier
Aeration Tank
QRAS = 25,000 m3/d
TSSRAS = 12,000 mg/L
Effluent
QEff = 984 m3/d
TSSEff = 15 mg/L
BODEff = 15 mg/L
WAS
QWAS = 16 m3/d
TSSWAS = 12,000 mg/L
Figure P8.7 Biomass yield factor
Solution
Overall Material Balance on Flow
Influent = Effluent + Waste Activated Sludge (WAS)
QInf = QEff + QWAS
Overall Material Balance on Solids
Solids in influent + Solids yield from BOD removal = Solids in effluent + Solids in WAS
QInf TSSInf + Solids yield from BOD removal = QEff TSSEff + QWASTSSWAS
Solids in influent = QInf TSSInf = (1,000 m3/d)(0.08 kg TSS/m3) = 80 kg TSS/d
Solids in effluent = QEff TSSEff = (984 m3/d)(0.015 kg TSS/m3) = 14.8 kg TSS/d
Solids in WAS = QWASTSSWAS = (16 m3/d)(12 kg TSS/m3) = 192 kg TSS/d
Total solids out = TSSWAS + TSSEff = 192 kg/d + 14.8 kg/d = 207 kg TSS/d
Solids yield from BOD removal = Total solids out – Solids in influent
= 207 kg/d – 80 kg/d = 127 kg TSS produced/d
BOD removed = (1,000 m /d)(0.2 – 0.015) kg BOD/m3) = 185 kg BOD removed/d
3
Biomass yield factor = (127 kg TSS produced/d)/(185 kg BOD removed/d)
= 0.69 kg TSS produced/kg BOD removed
8.8
WASTE ACTIVATED SLUDGE
Figure P8.8 shows an activated sludge process that will treat 24,000 m3/d with a COD =
0.5 kg/m3 (500 mg/L). This is a daily load of 12,000 kg/d. Ninety percent of the influent
COD is removed, so the effluent contains 0.05 kg/m3 (50 mg/L), or 1,200 kg/d. The
effluent contains 0.01 kg/m3 of total suspended solids, or 10 mg/L TSS. The total suspended
solids concentration of the return activated sludge (RAS) is 1% solids by weight, which is
220
SOLVED MATERIAL BALANCE PROBLEMS:
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Material balance for biological processes
10 kg/m3 = 10,000 mg/L. The solids are 75% volatile and 25% fixed, or 7,500 mg/L volatile
solids and 2,500 mg/L fixed solids. Assume the density of the waste activated sludge is
1,000 kg/m3. Assume that the dissolved solids are negligible in comparison to the suspended
solids (roughly 200 mg/L dissolved and 10,000 mg/L suspended solids). The empirical
stoichiometry is given in the diagram. Complete the material balance by determining the
WAS and effluent flows.
EMPIRICAL STOICHIOMETRY
1 kg COD removed Æ 0.5 kg TSS produced
Influent
Q = 24,000 m3/d
COD = 500 mg/L
TSS = negligible
WAS to primary settling
TSSWAS = 10,000 mg/L
QWAS =?
Aerobic
Reactor
Final Clarifier
Effluent
QE = ?
CODE = 50 mg/L
TSSE = 10 mg/L
VSSE = 4 mg/L
Return activated sludge (RAS)
TSSRAS = 10,000 mg/L
Figure P8.8 Waste activated sludge
Solution
Basis = 24,000 m3 of wastewater and 12,000 kg COD.
COD removed = (24,000 m3)(0.5 kg/m3 - 0.05 kg/m3) = 10,800 kg COD
Empirical stoichiometry for the conversion of COD to biomass
0.5 kg TSS produced per 1 kg COD removed
TSS produced = (0.5 kg/kg)(10,800 kg COD removed) = 5,400 kg TSS
VSS produced = 0.75 TSS produced = (0.75)(5,400 kg) = 4,050 kg VSS
Waste Activated Sludge Calculation.
Mass of total suspended solids (and volatile suspended solids) in the reactor remains constant.
All of solids produced are removed, either in the effluent or as waste activated sludge (WAS).
TSS produced = TSS removed as WAS + TSSEff
5,400 kg = (10 kg/m3)(QWAS) + (0.01 kg/m3)(24,000 m3 - QWAS)
5,400 kg = (10 kg/m3)QWAS + 240 kg – (0.01 kg/m3)QWAS
(9.99 kg/m3)QWAS = 5,160 kg
QWAS = 516.5 m3
221
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Waste activated sludge flow (WAS) is (516.5 m3)(100)/(24,000 m3) = 2.2% of the influent.
WAS flow and solids are added to the influent wastewater load of the primary settling tank.
Effluent flow = 24,000 m3 – 517 m3 = 23,500 m3
Note that a short-cut approximation can be made by assuming that the effluent suspended solids are
negligible in comparison to the WAS solids. The relative amounts calculated in this example were 240
kg effluent solids and 5,160 kg WAS solids. This assumption gives a mass balance of:
TSS produced = TSS removed as WAS
5,400 kg = (10 kg/m3)(QWAS)
QWAS = 5400 kg/(10 kg/m3) = 540 m3
The approximate WAS volume is 540 m3 instead of 516 m3.
Approximations based on reasonable assumptions are sometimes useful.
8.9
SLUDGE AGE FOR NITRIFICATION
The activated sludge system in Figure P8.9 is required to maintain a sludge age of 8 days in
the summer and 15 days in the winter in order to nitrify (remove ammonia). (a) Calculate
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222
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
the waste activated sludge flow rate for summer. (b) Will you be able to make the system
nitrify in the winter at MLSS = 2,500 mg/L and effluent solids = 25 mg/L?
Influent
QInf = 10,000 m3/d
TSSInf = 0 mg/L
Aeration Tank
V = 1,000 m3
MLSS = 2,500 mg/L
Final Clarifier
Effluent
QE = ??? m3/d
TSSE = 25 mg/L
QWAS = ??? m3/d
TSSWAS = 10,000 mg/L
Figure P8.9 Nitrification sludge age
Solution
a) Mass of solids in aeration basin = (1,000 m3)(2.5 kg/m3) = 2,500 kg
Mass of solids in effluent = (10,000 m3/d - QWAS)(0.025 kg/m3)
Mass of solids in WAS = (QWAS)(10 kg/m3)
Sludge age: TC =
XV
QWAS XWAS QE X E
XV
QWAS (XWAS X E ) QInf X E
For Sludge age = 8 d
XV /TC QInf X E
X WAS X E
QWAS
=
=
(2.5 kg/m3 )(1,000 m3 )/(8 d) - (10,000 m3 /d)(0.025 kg/m3 )
10 kg/m3 - 0.025 kg/m3
312.5 m3 /d - 250 m3 /d
= 6.27 m3 /d
9.975
b) If QWAS = 0 the sludge age will be 10 days.
TC =
XV
(2.5 kg/m3 )(1,000 m3 )
=
= 10 d
QE X E
(10,000 m3 /d)(0.025 kg/m3 )
To increase the sludge age you must reduce the effluent solids concentration. If effluent
solids could be reduced to 10 mg/L, the solids mass lost in the effluent would be 100 kg/d
instead of 250 kg/d
For sludge age = 15 d and XE = 10 mg/L
XV /TC QInf X E
X WAS X E
QWAS
=
(2.5 kg/m3 )(1,000 m3 )/(15 d) - (10,000 m3 /d)(0.010 kg/m3 )
10 kg/m3 - 0.010 kg/m3
=
166.7 m3 /d - 100 m3 /d
= 6.67 m3 /d
9.99
223
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Note: It is difficult to operate a system with a long sludge if the effluent solids are high (say
more than 10-20 mg/L) or if they are highly variable. Imagine a system operating at sludge
age 10 days with effluent solids 10 mg/L when suddenly the effluent solids rise to 30 mg/L.
Solids are being lost in the effluent at 3 times the rate and the sludge age will drop rapidly.
8.10 SLUDGE AGE CONTROL
Sludge age is a key design and operating parameter of the activated sludge process. The
process shown in Figure P8.10 is supposed to produce an effluent that is virtually free of
ammonia. To do this it needs to operate with a sludge age of about 8 days. Will the system
nitrify for the conditions shown? If the sludge age is too low, how will you increase it? If
the sludge age is higher than needed, how will you reduce it?
Influent
QInf = 10,000 m3/d
TSSInf = 0 mg/L
Aeration Tank
V = 1,000 m3
MLSS = 2,000 mg/L
Effluent
QE = ??? m3/d
TSSE = 25 mg/L
Final Clarifier
QRAS = 2,000
TSSRAS = 10,000 mg/L
m3/d
QWAS = 40 m3/d
TSSWAS = 10,000 mg/L
Figure P8.10
Solution
For the conditions shown the sludge age is too low.
TC =
XV
QWAS X WAS QE X E
XV
QWAS (X WAS X E ) QInf X E
=
(2 kg/m3 )(1,000 m3 )
(40 m3 /d)(10 kg/m3 - 0.025 kg/m3 ) + (10,000 m3 /d)(0.025 kg/m3 )
=
2,000 kg
= 3.08 d
399 kg/d + 250 kg/d
The only control available is to decrease the waste sludge rate. This will retain solids in the system
and the MLSS will increase.
Look at this more carefully. For example, suppose that the sludge wasting is reduced to zero for a
short time. At 2,000 mg/L the aeration basin contains (1,000 m3)(2 kg/m3) = 2,000 kg solids. The mass
of solids removed as WAS is (10 kg/m3)(40 m3/d) = 400 kg/d. If these are solids are not removed they
will remain in the aeration basin and in the final clarifier. Suppose they are distributed 300 kg to the
aeration basin and 100 kg to the final clarifier.
224
SOLVED MATERIAL BALANCE PROBLEMS:
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Material balance for biological processes
After one day the MLSS would be increased to 2,300 kg. At this MLSS concentration, what QWAS will
give a sludge age = 8 d?
XV /TC QInf X E
X WAS X E
QWAS
=
(2.3 kg/m3 )(1,000 m3 )/(8 d) - (10,000 m3 /d)(0.025 kg/m3 )
10 kg/m3 - 0.025 kg/m3
=
287.5 m3 /d - 250 m3 /d
= 3.76 m3 /d
9.975
And TSSWAS = 37.6 kg/d
If the solids are not wasted for 2 days, the MLSS concentration would increase to 2,600 mg/L. What
QWAS will give a sludge age = 8 d?
XV /TC QInf X E
X WAS X E
QWAS
=
(2.6 kg/m3 )(1,000 m3 )/(8 d) - (10,000 m3 /d)(0.025 kg/m3 )
10 kg/m3 - 0.025 kg/m3
=
325 m3 /d - 250 m3 /d
= 7.52 m3 /d
9.975
And TSSWAS = 75.2 kg/d
8.11 RECYCLE TO PRIMARY SETTLING TANK
A primary clarifier in a wastewater treatment plant receives raw sewage (RS), waste activated
sludge (WAS), and centrate (C) from a sludge dewatering centrifuge, as shown in Figure
P8.11. Waste activated sludge (WAS) is intentionally removed from the activated sludge
process to balance the new biomass that grows in the process. Centrate is the dilute effluent
from a centrifuge that is used to dewater digested sludge. Primary effluent (PE) and primary
sludge (PS) are removed from the clarifier. The diagram shows the known data. The primary
effluent flow actually was measured, but the value was accidentally omitted. Calculate the
value and also the primary sludge solids concentration.
225
SOLVED MATERIAL BALANCE PROBLEMS:
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Material balance for biological processes
Figure P8.11
Solution
Tie component = suspended solids
Note: Suspended solids in the primary clarifier do not react or dissolve. Mass of solids is conserved.
226
SOLVED MATERIAL BALANCE PROBLEMS:
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Material balance for biological processes
Primary effluent flow rate
QPE = QRS + QC + QWAS – QPS
= 290 L/s + 52 L/s + 4 L/s – 30 L/s = 316 L/s
Solids into primary clarifier = QRSCRS + QCCC + QWASCWAS
= (290 L/s)(350 mg/L) + (52 L/s)(12,000 mg/L) + (4 L/s)(2,500 mg/L)
= 735,500 mg/s
Solids balance on primary clarifier
Solids In = Solids Out = QPE CPE + QPS CPS
735,500 mg/s = (316 L/s)(70 mg/L) + (30 L/s)CPS
CPS = 23,780 mg/L
8.12 BALANCE ON ACTIVATED SLUDGE SOLIDS
In the activated sludge process shown in Figure P8.12, solids are removed from the process
in the waste activated sludge (WAS) that has flowrate QWAS and concentration CWAS. The
WAS and primary settling tank sludge go to a centrifuge for thickening before additional
sludge treatment is performed. The water removed by the centrifuge, the centrate, is recycled
to the primary settling tank. Calculate the primary sludge quantities (QPS and CPS) and the
dewatered centrifuge output quantities (QDW and CDW).
Figure P8.12
227
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Solution
Calculate Primary Sludge Quantities
Water balance on the primary settling tank
QRS + QC = QPE + QPS
290 L/s + 40 L/s = 322 L/s + QPS
QPS = 8 L/s
Solids balance on the primary settling tank
QRSCRS + QCCC = QPECPE + QPSCPS
(290 L/s)(350 mg/L) + (40 L/s)(1,800 mg/L) = (322 L/s)(70 mg/L) + (8 L/s) CPS
CPS = 18,870 mg/L (approximately 2% solids)
Water balance on centrifuge
QDW = QWAS + QPS - QC
QDW = 52 L/s + 8 L/s – 40 L/s = 20 L/s
Solids balance on centrifuge
QDWCDW = QWASCWAS + QPSCPS - QCCC
(20 L/s)CDW = (52 L/s)(12,000 mg/L) + (8 L/s)(18,870 mg/L) – (40 L/s)(1800 mg/L)
CDW = 35,150 mg/L (approximately 3.5% solids)
Figure S8.12
8.13 ACTIVATED SLUDGE AERATION BASIN AS CSTRS IN SERIES
An activated sludge process consists of a biological reactor, the aeration basin, and a final
clarifier (settling tank). A common aeration tank geometry is a long channel, or a long
channel that is folded to create several aeration bays. Figure P8.13 shows an activated sludge
aeration basin that is folded to create three independent compartments (there is no backflow
from compartment 2 to compartment 1).
228
SOLVED MATERIAL BALANCE PROBLEMS:
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Material balance for biological processes
Mixing is provided by compressed air that is introduced through submerged fine-bubble
diffusers. There are three constituents that might be modeled in a conventional activated
sludge process: (1) BOD or COD concentration, (2) microorganism concentration, and
(3) dissolved oxygen concentration. In a biological nutrient removal modification of the
process there would also be soluble phosphorus, ammonia nitrogen, and nitrate nitrogen. The
aeration basin might be homogeneous with respect to one, two, or all of the constituents.
For example, the COD and microorganism concentrations change rather slowly along the
axis of the aeration basin and the difference might not be enough to seriously invalidate
a CSTR approximation. Dissolved oxygen may be high at one end and low at the other,
depending on the design and control of the air supply.
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229
SOLVED MATERIAL BALANCE PROBLEMS:
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Material balance for biological processes
Figure P8.13 A three-bay folded activated sludge aeration basin (courtesy of Pixabay)
The process removes 95% of the influent biodegradable COD.
COD removal is described by a first-order reaction (r = – kC) with k = 0.5/h
Influent concentration is C0 = 600 mg/L
Effluent concentration is C = 30 mg/L
Average flow is Q = 300 m3/h
Detention time of the aeration basin at this flow is qT = 9 h
Aeration basin volume is VT = QqT = (300 m3/h)(9 h) = 2,700 m3
Volume of each reactor compartment is V = (2,700 m3)/3 = 900 m3
Detention time of one compartment is θ = qT /3 = 3 h
Find a reasonable reactor model for this process. One way to approximate this aeration basin
is as n CSTRs in series. The value of n is not known, but we can check a few possibilities.
Solution
Single CSTR Model – This cannot be correct. The concentrations certainly will not be the same in all
three bays. The estimated effluent concentration for θ = 9 h is 109 mg/L, which is too large.
C1
C0
600 mg/L
600 mg/L
=
=
= 109 mg/L
1 kT
1 + (0.5/h)(9 h)
5.5
Plug Flow Model - This predicts an effluent COD that is too small.
C = C0 exp(–kθ) = (600 mg/L) exp[(-0.5/h)(9 h)] = (600 mg/L)(0.0111) = 6.6 mg/L
230
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
3-CSTRs-in-Series Model – This should be better than a 1 CSTR model or a plug flow model.
For n CSTRs in series, the concentration of the effluent from the nth reactor is
Cn
C0
(1 kT )n
Using C0 = 600 mg/L, k = 0.5/h and qTotal = 9 h. For n = 3, the total detention time of 9 h is divided
by 3 to get the detention time for a single CSTR.
θ = 9 h/3 = 3 h
The model for 3 CSTRs in series is
C3
C0
(1 kT )
3
=
600 mg/L
600 mg/L
=
= 38.4 mg/L
3
[1 + (0.5/h)(3 h)]
2.53
This is reasonably close to the observed effluent COD concentration of 30 mg/L.
We conclude that 3 CSTRs in series is a useful description of the process.
The COD concentrations through the hypothetical three-stage process are shown in Figure S8.13.
C1 = C0/(1 + kθ) = (600mg/L)/[1 + (0.5/h)(3 h)] = (600 mg/L)/2.5 = 240 mg/L
The denominator of (1 + (0.5/h)(3h)) = 2.5 can be used to calculate the effluent from reactors 2 and 3.
C2 = C1/2.5 = 96 mg/L
C3 = C2/2.5 = 38.4 mg/L
The COD reduction is greatest in reactor 1 because it operates with r = – (0.5/h)(240 mg/L) = -120
mg/L-h. In contrast, reactor 3 operates with r = – (0.5/h)(38.4 mg/L) = -19.2 mg/L-h.
Influent
Q = 300 m3/d
C0 = 600 mg/L
C1
240 mg/L
C1
240 mg/L
C2
96 mg/L
C2
96 mg/L
C3
38.4 mg/L
Effluent
Q = 300 m3/d
C3 = 38.4 mg/L
Figure S8.13 Concentrations for the three hypothetical stages of the process. k= 0.5/d
The composition of wastewater is continually changing, and so composition of the biodegradable
organic compounds and the value of k are also changing. If we checked these models on several days
over several months a good deal might be learned.
The flow rate is also changing, but we will ignore this for now.
231
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
8.14 AEROBIC LAGOON
An aerobic lagoon with no recycle functions as a completely-mixed reactor (with respect to
COD) will be used to treat 400 m3/d of wastewater that has influent COD = 150 mg/L
to produce an effluent with 50 mg/L COD. Assume first-order kinetic removal of COD
with k = 0.2 d-1 and a net solids yield factor of 0.25 mg TSS/mg COD removed. Calculate:
a)
b)
c)
d)
The required hydraulic detention time and lagoon volume.
The mass of total suspended solids produced per day.
The total suspended solids concentration in the lagoon effluent
Will additional processing be required to remove solids from the effluent? If so,
how will this be managed?
Solution
a) First-order model for a completely mixed reactor
C
C0
1 kT
Ÿ
T
§ C0
·1
§ 150 mg COD/L
· 1
= ¨
- 1¸
= 10 d
1¸
¨
50
mg
COD/L
C
k
©
¹ 0.2/d
©
¹
V = Qθ = (400 m3/d)(10 d) = 4,000 m3
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232
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
b) COD removed = (400 m3/d)(0.15 kg COD/m3 – 0.05 kg COD/m3) = 40 kg COD/d
Solids yield = (0.25 kg TSS/kg COD removed)(40 kg COD/d) = 10 kg TSS/d
c) It is given that the lagoon is completely mixed (i.e. homogeneous) with respect to
COD. This does not mean it is homogeneous with respect to suspended solids. Most
lagoons are not. However, let us assume complete mixing for solids to estimate the
maximum possible effluent solids concentration.
Assume all solids produced leave the lagoon in the effluent (i.e. no solids are removed by
settling)
TSS concentration in effluent = (10 kg TSS/d)/(400 m3/d) = 0.025 kg/m3 = 25 mg/L
d) Probably not. The effluent TSS concentration will be 25 mg/L if all the solids produced
exit the lagoon. It is likely that some of these solids will settle (say in dead zones), so
the actual concentration will be lower. Settling can be promoted with a small unmixed
zone near the lagoon outflow, or by adding a settling pond to receive the lagoon
effluent prior to discharge.
8.15 WASTEWATER TREATMENT USING 4 LAGOONS IN SERIES
The wastewater treatment system shown in Figure P8.15 has four lagoons in series. The
first three lagoons are aerated with submerged fine bubble diffusers; there are 28 in lagoon
1, 7 in lagoon 2, and 6 in lagoon 3. We shall assume that the aeration provides sufficient
mixing to assume these lagoons will function as CSTRs. The first lagoon is approximately
14,000 m3, the second and third are approximately 8,800 m3, and the largest (the 4th) is
approximately 121,000 m3. Why is this arrangement beneficial?
Effluent
Influent
Figure P8.15 Wastewater treatment system with 4 lagoons in series to treat 1,020 m3/d (0.27
mgd). The first 3 lagoons are aerated with fine bubble diffusers. (Courtesy of Mars Hill Utility
District, Maine, USA)
233
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Solution
The pollutant concentration (BOD or COD) is highest in lagoon 1 and that is why more aeration is
needed. The series arrangement allows that aeration to be tapered as each lagoon operates at a lower
pollutant concentration. The large lagoon, which is not aerated, serves as a storage facility and a final
clarifier. The aerated lagoons prevent most solids from settling.
NOTE: The next two problems are long, with iterative calculations, but they build strength
in solving material balance problems and in understanding a large treatment system.
8.16 WASTEWATER TREATMENT PLANT MASS BALANCE - 1
The activated sludge wastewater treatment plant shown in Figure P8.16 has average design
flow of 1,000 m3/d. Influent waste strengths are: BOD = 300 mg/L and TSS = 400 mg/L.
Table P8.16a gives the plant’s performance data. The supernatant return flows do carry
suspended solids, and the data to calculate the solids load are given. Use these data to perform
a material balance on BOD and TSS by filling in the unshaded cells in Table P8.16b of
mass flows. To start, assume the recycle contributions to the influent can be ignored.
Influent
2
1
12
Primary
Settling
4
3
Activated Sludge
aeration tank
Final
clarifier
5
Effluent
6
Return activated sludge (RAS)
WAS Supernatant recycle
Waste activated
sludge (WAS)
Centrate
recycle
8
11
Digested
sludge
centrifuge
10
9
Digested
sludge
Anaerobic
digester
7
Thickened
WAS
WAS
Thickening
Thickened digested
sludge to disposal
Figure P8.16
234
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Process
Primary settling
Activated sludge process
Anaerobic digestion
WAS Thickener
Digested sludge centrifuge
Material balance for biological processes
Performance Characteristic
Value
BOD removal (%)
30
TSS removal (%)
65
% Solids in sludge
2
Effluent BOD (mg/L)
10
Effluent TSS (mg/L)
20
Solids yield (kg TSS/kg BOD)
0.77
VSS fraction
0.77
% Solids in WAS
1.5
Volatile SS fraction (%)
72
VSS destruction (%)
60
WAS solids capture (%)
60
Thickened sludge conc. (%)
4.5
Solids capture (%)
90
Thickened sludge conc. (%)
25
Table P8.16a
235
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Stream
Flow
(m /d)
3
Material balance for biological processes
Mass
Flow
(kg/d)
1,000
BOD
TSS
(kg/d)
(kg/d)
300
400
Notes
1
Raw influent
12
Recycle
2
Primary
influent
Sum of 1 + 12
BOD and TSS
Removed
30% BOD & 65% TSS
removal
4
Primary
sludge
2% solids
3
Primary
effluent
Difference of 4 - 2
5
Secondary
effluent
BOD = 10 mg/L; TSS =
20 mg/L
6
Waste
activated
sludge (WAS)
Biomass solids produced
= 0.77 kg TSS/kg BOD
removed
WAS = 1.5% solids
7
Thickened
WAS
4.5% solids; 60% solids
capture
8
WAS
centrifuge
recycle
Difference of 6 - 7
4+7
Digester feed
(TSS)
Thickened WAS + Primary
Sludge
Digester feed
(VSS)
72% VSS
VSS destroyed
60% destroyed
9
Digested
sludge to
centrifuge
TSS feed – VSS destroyed
10
Dewatered
sludge
90% capture; 25% solids
11
Digested
sludge
centrifuge
recycle
Difference of 9 - 10
12
Recycle
Sum of 8 + 11
0
0
0
Table P8.16b Mass flows to be computed
236
Given
Assume 0 to start
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Solution
Hint: (a) To start, assume the recycle contributions to the influent are zero. (b) To get a more exact
solution, add the recycle loads to the influent and repeat the calculation. The second iteration gives
a good solution and a third iteration gives an almost exact balance. (The return flows also carry BOD
but the data are not given so they can be ignored in this problem.)
Values are calculated in this order. Calculations are not shown in detail, because most are easy to follow.
Raw wastewater influent loads:
Mass flow = water plus solids = 1,000,000 kg/d + 400 kg/d = 1,000,400 kg/d
Primary influent loads (assuming recycle loads = zero)
Primary sludge solids produced: S4 = 65% of influent solids = 260 kg
Primary sludge flow = W4 = S4 (1 - 0.02)/0.02
= (260 kg)(0.98)/0.02 =12,740 kg = 12.7 m3
W4 = water in primary sludge (kg)
S4 = solids in primary sludge (kg)
Assume volume flow of sludge = volume of water in the sludge
Total mass flow = water + solids
= W4 + S4 = 12,740 kg + 260 kg = 13,000 kg
This calculation is used on the WAS thickener and the digested sludge centrifuge.
Waste activated sludge
Waste activated sludge mass (kg) = Mass of solids produced in the aeration tank
= (0.77 kg solids/kg BOD removed)(200 kg BOD removed) = 154 kg
60% of WAS solids are captured in the thickened sludge
40% of WAS solids return to the primary settling tank in the supernatant recycle
Digester
60% of volatile suspended solids fed to the digester are destroyed
90% of solids in the digested sludge are captured in the centrifuge discharge cake
10% of the centrate solids are returned to the primary settling tank in the supernatant recycle
The solution for the first iteration is presented in Table S8.16a below.
237
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Material Balance - Iteration 1
Stream
1
Raw influent
12
2
Flow
(m /d)
3
Mass
Flow
(kg/d)
1,000
1,000,400
Recycle
0
0
Primary
influent
1,000
1,000,400
BOD
TSS
(kg/d)
(kg/d)
300
400
0
BOD and TSS
Removed
Notes
Given
Assume 0 to start
300
400
Sum of 1 + 12
90
260
30% BOD & 65% TSS
removal
260
2% solids
4
Primary
sludge
12.7
13,000
3
Primary
effluent
987.3
987,400
210
140
Difference of 4 - 2
5
Secondary
effluent
977.1
977,13
10
20
BOD = 10 mg/L; TSS =
20 mg/L
6
Waste
activated
sludge (WAS)
10.1
10,267
154
Biomass solids produced
= 0.77 kg TSS/kg BOD
removed
WAS = 1.5% solids
7
Thickened
WAS
2.0
2,053
92.4
4.5% solids; 60% solids
capture
8
WAS
centrifuge
recycle
8.2
8,213
61.6
Difference of 6 - 7
4+7
Digester feed
(TSS)
14.7
15,053
352.4
Thickened WAS +
Primary Sludge
Digester feed
(VSS)
253.7
72% VSS
VSS
destroyed
152.2
60% destroyed
9
Digested
sludge
to
centrifuge
14.7
14,901
200.2
TSS feed – VSS destroyed
10
Dewatered
sludge
0.5
721
180.1
90% capture; 25% solids
11
Digested
sludge
centrifuge
recycle
14.2
14,081
20.0
Difference of 9 - 10
12
Recycle
22.3
22,394
81.6
Sum of 8 + 11
Table S8.16a
238
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Note that the calculated recycle flow of 22.3 m3/d does not agree with the assumed value of 0 m3/d.
For the second iteration, use 22.3 m3/d for the recycle flow and recalculate the values.
Tables S8.16b and S8.16c show the results of Iteration 2 and Iteration 3 of the material
balance calculation.
239
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Material Balance - Iteration 2
Stream
Flow
Mass Flow
BOD
TSS
3
(m /d)
(kg/d)
(kg/d)
(kg/d)
300
400
1
Raw influent
1,000
1,000,400
12
Recycle
22.3
22,394
2
Primary
influent
1,022.3
1,022,794
BOD and TSS
Removed
Notes
Given
81.6
From iteration 1
300
481.6
Sum of 1 + 12
90
313.0
30% BOD & 65% TSS
removal
313.0
2% solids
210
168.6
Difference of 4 - 2
10
20
BOD = 10 mg/L; TSS
= 20 mg/L
4
Primary
sludge
15.3
15,652.0
3
Primary
effluent
1,007.0
1,007,142
5
Secondary
effluent
6
Waste
activated
sludge (WAS)
10.1
10,266.7
154.0
Biomass solids
produced
= 0.77 kg TSS/kg
BOD removed
WAS = 1.5% solids
7
Thickened
WAS
2.0
2,053.3
92.4
4.5% solids; 60%
solids capture
8
WAS
centrifuge
recycle
8.2
8,213.3
61.6
Difference of 6 - 7
4+7
Digester
feed (TSS)
17.3
17,705.3
405.4
Thickened WAS
Primary Sludge
Digester
feed (VSS)
291.98
72% VSS
VSS
destroyed
175.2
60% destroyed
996.8
9
Digested
sludge to
centrifuge
17.3
17,530.2
230.3
TSS feed – VSS
destroyed
10
Dewatered
sludge
0.6
829.0
207.3
90% capture; 25%
solids
11
Digested
sludge
centrifuge
recycle
16.7
16,701.1
23.0
Difference of 9 - 10
12
Recycle
24.8
24.914.5
84.6
Sum of 8 + 11
Table S8.16b
240
+
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Material Balance - Iteration 3
Stream
Flow
Mass Flow
BOD
TSS
3
(m /d)
(kg/d)
(kg/d)
(kg/d)
300
400
Given
84.6
From iteration 2
300
484.6
Sum of 1 + 12
90
315.0
30% BOD & 65% TSS
removal
315.0
2% solids
210
169.6
Difference of 4 - 2
10
20
BOD = 10 mg/L; TSS
= 20 mg/L
1
Raw influent
1,000
1,000,400
12
Recycle
24.8
24,914.5
2
Primary
influent
1,024.8
BOD and TSS
Removed
Notes
4
Primary
sludge
15.4
15,750.4
3
Primary
effluent
1,009.4
1,009,546.0
5
Secondary
effluent
6
Waste
activated
sludge (WAS)
10.1
10,266.7
154
Biomass solids
produced
= 0.77 kg TSS/kg
BOD removed
WAS = 1.5% solids
7
Thickened
WAS
2.0
2,053.3
92.4
4.5% solids; 60%
solids capture
8
WAS
centrifuge
recycle
8.2
8,213.3
61.6
Difference of 6 - 7
4+7
Digester feed
(TSS)
17.4
17,803.8
407.4
Thickened WAS +
Primary Sludge
Digester feed
(VSS)
393.3
72% VSS
VSS destroyed
176.0
60% destroyed
999.3
9
Digested
sludge to
centrifuge
17.4
17,627.8
231.4
TSS feed – VSS
destroyed
10
Dewatered
sludge
0.6
833.1
208.3
90% capture; 25%
solids
11
Digested
sludge
centrifuge
recycle
16.8
16,794.7
23.1
Difference of 9 - 10
12
Recycle
24.9
25,008.0
84.7
Sum of 8 + 11
Table S8.16c
241
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
8.17 WASTEWATER TREATMENT PLANT MASS BALANCE - 2
A wastewater treatment plant, Figure P8.17, considers adding chemicals in the primary
sedimentation process. How would this change the mass balance of the plant? The average
design flow is 1,000 m3/d. Influent waste strengths are: BOD = 300 mg/L, TSS = 400
mg/L, and colloidal solids = 50 mg/L. The “with chemical” performance shown in Table
P8.17a is for the addition of 40 mg/L FeCl3 and 2 mg/L polymer in the primary settling
process. Use these data to perform a material balance on BOD and TSS by filling in the
unshaded cells in Table P8.17b of mass flows. To start, assume the recycle contributions to
the influent can be ignored.
Chemical
Addition
Influent
2
1
12
Primary
Settling
4
3
Activated Sludge
aeration tank
Final
clarifier
5
Effluent
6
Return activated sludge (RAS)
WAS Supernatant recycle
Waste activated
sludge (WAS)
Centrate
recycle
8
11
Digested
sludge
centrifuge
10
9
Digested
sludge
Anaerobic
digester
7
Thickened
WAS
WAS
Thickening
Thickened digested
sludge to disposal
Figure P8.17 Wastewater treatment plant mass balance
242
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Without
With
Chemicals
Chemicals
Chemical solids formed (kg/kg FeCl3)
---
0.6
Colloidal solids removed (%)
---
45
BOD removal (%)
30
40
TSS removal (%)
65
75
Primary sludge solids (%)
2.5
3.0
BOD removal (%)
90
95
Effluent TSS (mg/L)
20
10
Sludge yield (kg TSS/kg BOD)
0.77
0.77
VSS fraction
0.77
0.80
Volatile SS fraction (%)
72
75
VSS destruction (%)
60
65
WAS solids capture (%)
60
90
Thickened sludge conc. (%)
4.5
5
Solids capture (%)
90
95
Thickened sludge conc. (%)
25
25
Process
Chemical addition
Primary settling
Activated sludge process
Anaerobic digestion
WAS Thickener
Digested sludge
centrifuge
Material balance for biological processes
Performance Characteristic
Table P8.17a Process characteristics
243
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Stream
1
Raw influent
12
Recycle
2
Material balance for biological processes
Flow
Mass
BOD
TSS
(m /d)
(kg/d)
(kg/d)
(kg/d)
300
400
3
1,000
0
0
0
Notes
Given
Assume 0 to start
FeCl3 added
40
40 mg/L = 0.04 kg/m3
Polymer added
2
2 mg/L = 0.02 kg/m3
Primary influent
Sum of 4 rows above
Chemicals solids
produced
0.6 kg/kg chemical
added
Coagulated colloids
produced
45% of colloidal solids
Total primary solids
Sum of 3 rows above
Removals
45% BOD & 75% TSS
removal
4
Primary sludge
3% solids
3
Primary effluent
Difference of 2 - 4
Activated sludge
BOD removal
90%
overall
removal
Activated sludge
solids production
0.77kg/kg BOD
removed
6
WAS
1.5% solids
5
Secondary effluent
Effluent TSS = 10 mg/L
7
Thickened WAS
5% TSS; 90% solids
capture
8
WAS centrifuge
recycle
Difference of 6 - 7
Digester feed (TSS)
Thickened WAS +
Primary Sludge
Digester feed (VSS)
75% VSS
VSS destroyed
65% VSS Destruction
9
Digested sludge to
centrifuge
TSS feed – VSS destroyed
10
Dewatered sludge
95% capture; 25% solids
11
Digested sludge
centrifuge recycle
Difference of 9 - 10
12
Recycle
Sum of 8 and 11
Table 8.17b Mass flows to be computed
244
BOD
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Solution
The solution is presented in the tables below.
The calculations are not shown in detail, because most are easy to follow.
Values were calculated in this order
Raw wastewater influent loads
Mass flow = water plus solids = 1,000,000 kg/d + 400 kg/d = 1,000,400 kg/d
Primary influent loads (assuming recycle loads = zero)
Solids produced from chemical addition
FeCl3 = (0.040 kg/m3)(1,000 m3/d) = 40 kg/d
Polymer = (0.002 kg/m3)(1,000 m3/d) = 2 kg/d
Primary sludge produced = 75% of influent solids = 0.75(446.5 kg/d) = 334.9 kg/d
Primary sludge flow = W4 = S4(1-0.03)/0.03
= (334.9 kg/d)(0.97)/0.03 =10,828 kg/d = 10.8 m3/d
W4 = water in sludge (kg)
S4 = solids in sludge (kg)
Assume volume flow of sludge = volume of water in the sludge
Total mass flow = water + solids = W4 + S4
= 10,828 kg/d + 334.9 kg/d = 11,163 kg/d
This calculation is used on the WAS thickener and the digested sludge centrifuge.
Waste activated sludge mass (kg) = Mass of solids produced in the aeration tank
= (0.77 kg solids/kg BOD removed) 148.5 kg BOD removed) = 114.3 kg
90% of WAS solids are captured in the thickened sludge
10% of WAS solids return to the primary settling tank in the supernatant recycle
65% of volatile suspended solids fed to the digester are destroyed
95% of solids in the digested sludge are captured in the centrifuge discharge cake
5% of the centrate solids are returned to the primary settling tank in the supernatant recycle
The following three tables show Iterations 1, 2 and 3 of the material balance calculations.
245
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Material Balance - Iteration 1
Stream
1
Raw influent
12
Recycle
2
Flow
Mass
BOD
TSS
3
(m /d)
(kg/d)
(kg/d)
(kg/d)
1,000
1,000,400
300
400
0
0
0
Notes
Given
Assume 0 to start
FeCl3 added
40
40 mg/L = 0.04 kg/m3
Polymer added
2
2 mg/L = 0.02 kg/m3
Primary influent
1,000
1,000,442
300
400
Sum of 4 rows above
Chemicals solids
produced
24
0.6 kg/kg chemical
added
Coagulated
colloids produced
22.5
45% of colloidal solids
Total
solids
446.5
Sum of 3 rows above
135
334.9
45% BOD & 75% TSS
removal
primary
Removals
4
Primary sludge
10.8
11,163
NR
334.9
3% solids
3
Primary effluent
989.2
989,280
165
111.6
By Difference
Activated sludge
BOD removal
148.5
Activated sludge
solids production
6
WAS
5
Secondary
effluent
7
8
90% overall BOD
removal
114.3
0.77kg/kg BOD
removed
114.3
1.5% solids
7.5
7,623
981.7
981,657
Thickened WAS
2.0
2,058
102.9
5% TSS; 90% solids
capture
WAS centrifuge
recycle
5.6
5,565
11.4
Difference of 6 - 7
Digester feed
(TSS)
12.8
13,221
437.8
Thickened WAS
Primary Sludge
Digester feed
(VSS)
328.3
75% VSS
VSS destroyed
213.4
65% VSS Destruction
16.5
9.8
Effluent TSS = 10
mg/L
9
Digested sludge
to centrifuge
12.8
13,007
224.4
TSS feed – VSS
destroyed
10
Dewatered
sludge
0.6
853
213.1
95% capture; 25%
solids
11
Digested sludge
centrifuge
recycle
12.1
12,155
11.2
Difference of 9 - 10
12
Recycle
17.7
17,719
22.7
Sum of 8 and 11
Table S8.17a – Iteration 1
246
+
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Material Balance - Iteration 2
Stream
Flow
Mass
BOD
TSS
3
(m /d)
(kg/d)
(kg/d)
(kg/d)
Notes
1
Raw influent
1,000
1,000,400
300
400
Given
12
Recycle
17.7
17,719
0
22.7
From Iteration 1
2
FeCl3 added
40
40 mg/L = 0.04 kg/m3
Polymer added
2
2 mg/L = 0.02 kg/m3
Primary influent
1,018
1,018,161
300
Chemicals solids
produced
423
Sum of 4 rows above
24
0.6 kg/kg chemical
added
Coagulated
colloids
produced
22.5
45% of colloidal solids
Total primary
solids
469.2
Sum of 3 rows above
135
351.9
45% BOD & 75% TSS
removal
Removals
4
Primary sludge
11.4
11,729
NR
351.9
3% solids
3
Primary effluent
1,006.3
1,006,433
165
117.3
By Difference
Activated sludge
BOD removal
Activated sludge
solids production
6
WAS
5
Secondary
effluent
7
8
90% overall BOD
removal
148.5
7.5
7,623
998.8
998,810
Thickened WAS
2.0
WAS centrifuge
recycle
Digester feed
(TSS)
16.5
114.3
0.77kg/kg BOD
removed
114.3
1.5% solids
0.0
Effluent TSS = 10 mg/L
2,058
102.9
5% TSS; 90% solids
capture
5.6
5,565
11.4
Difference of 6 - 7
13.3
13,787
454.8
Thickened WAS +
Primary Sludge
Digester feed
(VSS)
341.1
75% VSS
VSS destroyed
221.7
65% VSS Destruction
9
Digested sludge
to centrifuge
13.3
13,565
233.1
TSS feed – VSS
destroyed
10
Dewatered
sludge
0.7
886
221.4
95% capture; 25%
solids
11
Digested sludge
centrifuge
recycle
12.7
12,680
11.7
Difference of 9 - 10
12
Recycle
18.2
18,244
23.1
Sum of 8 and 11
Table S8.17b – Iteration 2
247
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION
PREVENTION AND CONTROL
SOLVED MATERIAL BALANCE PROBLEMS:
Material balance for biological processes
POLLUTION PREVENTION AND CONTROL
Stream
Flow
2
4
3
6
5
7
8
9
10
11
12
Raw influent
Recycle
FeCl3 added
Polymer
added
Primary
influent
Chemicals
solids
produced
Coagulated
colloids
produced
Total primary
solids
Removals
Primary
sludge
Primary
effluent
Activated
sludge BOD
removal
Activated
sludge solids
production
WAS
Secondary
effluent
Thickened
WAS
WAS
centrifuge
recycle
Digester
feed (TSS)
Digester
feed (VSS)
VSS
destroyed
Digested
sludge
to
centrifuge
Dewatered
sludge
Digested
sludge
centrifuge
recycle
Recycle
Material Balance - Iteration 3
Mass
BOD
TSS
Notes
(m /d)
1,000
18.2
(kg/d)
1,000,400
18,244
40
2
(kg/d)
300
(kg/d)
400
23.1
1,018
1,018,686
300
423
Sum of 4 rows above
24
0.6 kg/kg chemical
added
3
1
12
Material balance for biological processes
Given
From Iteration 2
40 mg/L = 0.04 kg/m3
2 mg/L = 0.02 kg/m3
22.5
45% of colloidal solids
469.6
Sum of 3 rows above
135
352.2
11.4
11,740
NR
352.2
45% BOD & 75% TSS
removal
3% solids
1,006.8
1,006,947
165
117.4
By Difference
148.5
90% overall BOD
removal
114.3
0.77kg/kg BOD
removed
114.3
0.0
1.5% solids
Effluent TSS = 10 mg/L
5% TSS; 90% solids
capture
Difference of 6 - 7
7.5
999.3
7,623
999,324
2.0
2,058
102.9
5.6
5,565
11.4
13.3
13,798
455.1
16.5
341.3
Thickened WAS +
Primary Sludge
75% VSS
221.9
65% VSS Destruction
13.3
13,576
233.2
TSS feed – VSS
destroyed
0.7
886
221.6
95% capture; 25% solids
12.7
12,690
11.7
Difference of 9 - 10
18.2
18,255
23.1
Sum of 8 and 11
248
Table S8.17c – Iteration 3
Table
S8.17c – Iteration 3
248
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
8.18 SLUDGE VOLUME INDEX
The sludge volume index, SVI, is a measure of the settleability of activated sludge. It is
measured by letting a sample of the activated sludge mixed liquor (aeration tank contents)
settle quietly in a 1,000 mL graduated cylinder for 30 minutes. The SVI is the volume (in
mL) of the settled sludge, V, divided by the mass of solids (in grams) contained in that
volume. An SVI = 100 mL/g indicates a well settling sludge.
Figure P8.18 shows how we can imagine a small activated sludge treatment plant in which
a 1 L cylinder serves as the final settling tank and the recirculated sludge is withdrawn from
the bottom of the cylinder. In a well-operated stable process, when all the solids are settling
and not overflowing with the effluent, the SVI is a useful measure of process performance.
Derive each of these relations, where V = volume (mL) of the settled solids in the cylinder.
a) QR/(Q + QR) = V/1,000 mL
b) QR =VQ/(1,000 mL - V)
c) XR = 1,000 X/V = 1,000,000/SVI
Figure P8.18
Solution
Material balance around the cylinder, assuming XEff = 0
(Q + QR)X = QRXR
Material balance from the ratio of the column volume and the settled sludge volume
(1,000 ml)X = VXR
and
XR = (1,000 mL)(X/V)
a) Substituting for XR in the material balance gives
(Q + QR)X = QR(1,000 mL)(X/V) and
QR
Q QR
V
1,000 mL
249
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
b) Rearranging the solution in part a gives
QR(1,000 mL) = V(Q + QR)
QR(1,000 mL – V) = VQ and
QR
VQ
1,000 mL - V
c) Note: the units and conversion factors are a bit messy, but we carry them through
the derivation for completeness.
From the definition of SVI, with the mass of solids (grams) = X(1 L)/(1,000mg/g)
SVI
V
(1,000 mg/L-g) V
=
X
( X )(1 L)/(1,000 mg/g)
Substituting for V/X from the ratio of the column volume and the settled sludge volume
XR = 1,000 X/V and V/X = 1,000/XR gives
SVI =
(1,000 mg/L-g) V
X
(1,000 mg/L-g)(1,000 mL)
XR
mL-mg
L-g
SVI
1,000,000
X R=
Check units: with SVI in mL/g, the units of XR are mg/L. (OK)
250
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
8.19 OXYGEN SUPPLY
An activated sludge process removes 50,000 kg BOD5 per day. The aeration basin volume
is 10,000 m3 and the mixed liquor volatile suspended solids (MLVSS) concentration is
1,800 mg/L.
a) Calculate an empirical estimate of the oxygen required (kg/d) if 0.5 kg O2 is
required for each kg BOD5 removed and 0.05 kg O2 is required per day for each
kg VSS under aeration.
b) What volume of diffused air per day must be supplied, assuming a 10 percent
oxygen transfer rate at field operating conditions?
c) What power is required for mechanical aerators that have a standard oxygen transfer
efficiency of 2.4 kg O2/kWh. Technically this is called the Standard Aeration
Efficiency (SAE) in clean water, at 20°C, zero dissolved oxygen concentration, and
at sea level. Operating conditions will 25°C, 1.5 mg/L dissolved oxygen, at sea
level, in wastewater that has α = 0.75 and β = 0.95?
Solution
a) The empirical estimate of oxygen requirement is given by:
kg O2 required/d = a(kg BOD5 removed/d) + b(kg VSS under aeration)
where a = 0.5 kg O2/kg BOD5 removed
b = 0.05 kg O2/d-kg VSS under aeration
kg O2/d = (0.5 kg O2/kg BOD5)(50,000 kg BOD5/d)
+ (0.05 kg O2/d-kg VSS)(1.8 kg VSS/m3)(10,000 m3)
kg O2/d = 25,000 kg O2/d + 900 kg O2/d = 25,900 kg O2/d
b) Mass of diffused air required at 10% transfer.
Air is 23.13% oxygen by mass
Mass of air required = (25,900 kg O2/d)/0.2313 kg O2/kg air) = 112,000 kg air/d
Specific weight of dry air = 1.1293 kg/m3
Volume of air required = (112,000 kg/d)/(1.1293 kg/m3) = 99,200 m3/d
At 10% transfer the volume of air to be supplied is
(99,200 m3/d)/0.1 = 992,000 m3/d
c) Power requirement for mechanical aerators. Conversion of aeration efficiency from
standard to operating conditions
ª
º T -20°C
EC* - C
FAE = (SAE) D «
» T
9.18
mg/L
0
mg/L
¬
¼
251
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
where FAE = aeration efficiency at field operating conditions (kg O2/kWh)
SAE = aeration efficiency at standard conditions (kg O2/kWh)
C = DO concentration in the activated sludge aeration tank (mg/L)
C* =DO saturation concentration in water at operating temperature T (mg/L)
α = alpha factor (dimensionless)
βC* = saturation concentration in wastewater at operating temperature T (mg/L)
β = dimensionless correction factor for DO saturation concentration
θ = coefficient for temperature correction, typically 1.02
T = operating temperature (°C)
At T = 25°C, C* = 8.38 mg/L. Substituting the problem data gives
ª
º T -20qC
EC* - C
FAE = (SAE) D «
» T
9.18
mg/L
0
mg/L
¬
¼
§ 0.95(8.38) mg/L - 1.5 mg/L ·
25qC 20qC
= (2.4 kg O2 /kWh)(0.75) ¨
¸ 1.02
9.18 mg/L - 0 mg/L)
©
¹
= (2.4 kg O2 /kWh)(0.75)(0.704)(1.104) = 1.4 kg O2 /kWh
Power = Mass of oxygen required/FAE = (25,900 kg O2/d)/(1.4 kg O2/kWh)
= 18,500 kWh/d
= (18,500 kWh/d)(d/24h) = 771 kW
8.20 OXYGEN REQUIREMENT
An activated sludge process requires 3,700 lb/d of oxygen to remove 6,000 lb/d of BOD5.
This will be supplied as diffused air. The oxygen transfer efficiency is 6.2%. Air is 20.95%
oxygen by volume, 23.13% oxygen by weight, and the density of dry air at 1 atm and
50°F is 0.0778 lb/ft3.
a) What volume of air per day is needed?
b) How does the volume you calculated compare with the rule-of-thumb recommendation
of 500-700 ft3 air/lb of BOD5 removed?
Solution
a) Mass of oxygen supplied in diffused air = (3,700 lb O2/d)/0.062 = 59,680 lb O2/d
Mass of air supplied in diffused air = (59,680 lb O2/d)/(0.2313 lb O2/lb air)
= 258,000 lb air/d
Volume of air supplied = (258,000 lb air/d)/(0.0778 lb air/ft3) = 3,316,000 ft3 air/d
252
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
b) Rule of thumb comparison
(3,316,000 ft3/d)/(6,000 lb BOD5 removed/d) = 553 ft3 air/lb BOD5 removed.
Value is within the rule-of-thumb range.
8.21 OXYGEN DEMAND
An activated sludge process treats 5 million gallons per day at 300 mg/L BOD5 to give an
effluent with 20 mg/L BOD5. The net oxygen demand is 0.52 lb O2 per lb BOD5 removed.
The aerators have a standard oxygen transfer rate (manufacturer’s rating) of 5 lb O2/hph. Technically this is called the Standard Aeration Efficiency (SAE).The field operating
conditions are 28°C, dissolved oxygen concentration in the aeration tank = 1.2 mg/L, α =
0.6, β = 0.95. Dissolved oxygen saturation at 28°C is C* = 7.4 mg/L. What is the power
required to meet the oxygen demand of the process?
Solution
Use shortcut formula to get lb BOD5 removed/d
BOD5 removed (lb/d) = 8.34(5 mgd)(300 mg/L - 20 mg/L) = 11,700 lb BOD5/d
Oxygen demand (lb/d) = (0.52 lb O2/lb BOD5)(11,700 BOD5 lb/d) = 6,100 lb O2/d
Power requirement. Conversion of aeration efficiency from standard to operating conditions
ª
º T -20°C
EC* - C
FAE = (SAE) D «
» T
9.18
mg/L
0
mg/L
¬
¼
where FAE = aeration efficiency at field operating conditions (lb O2/hp-h)
SAE = aeration efficiency at standard conditions (lb O2/hp-h)
C = DO concentration in the activated sludge aeration tank (mg/L)
C* =DO saturation concentration in water at operating temperature T (mg/L)
α = alpha factor (dimensionless)
βC* = saturation concentration in wastewater at operating temperature T (mg/L)
β = dimensionless correction factor for DO saturation concentration
θ = coefficient for temperature correction, typically 1.02
T = operating temperature (°C)
Substituting the problem data gives
ª
º T -20qC
EC* - C
FAE = (SAE) D «
» T
¬ 9.18 mg/L - 0 mg/L ¼
§ 0.95(7.4) mg/L - 1.2 mg/L ·
28qC 20qC
= (5.0 lb O2 /hp-h)(0.6) ¨
¸ 1.02
9.18
mg/L
0
mg/L)
©
¹
= (5.0 lb O2 /hp-h)(0.6)(0.635)(1.172) = 2.23 lb O2 /hp-h
253
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Power = Mass of oxygen required/FAE = (6,100 lb O2/d)/(2.23 lb O2/hp-h)
= 2,740 hp-h/d
= (2,740 hp-h/d)(d/24h) = 114 hp
8.22 DEER ISLAND DIGESTERS
The egg-shaped sludge digesters at Boston’s Deer Island plant, shown in Figure P8.22, are
140 feet tall and hold 3 million gallons each. What volume of sludge can be treated per
day in one digester if the detention time is 15 days?
254
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Figure P8.22
Solution
Detention time = Volume/Flow
15 d = (3,000,000 gal)/Q
Q = (3,000,000 gal)/(15 d) = 200,000 gal/d
8.23 METHANE PRODUCTION
What is the volume of methane produced from the conversion of 1200 mg/L COD to
CH4 in a flow of 275 m3/day?
Solution
Basis: 1 day
1,200 mg COD/L = 1.2 kg COD/m3
COD destroyed = (1.2 kg COD/m3)(275 m3/d) = 330 kg COD/d
The empirical stoichiometry is 0.3 m3 -0.44 m3 CH4 produced/kg COD destroyed
Methane produced is between
(330 kg COD destroyed/d)(0.3 m3 CH4 produced/kg COD destroyed) = 99 m3/d
and
(330 kg COD destroyed/d)(0.44 m3 CH4 produced/kg COD destroyed) = 145 m3/d
255
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
8.24 ANAEROBIC TREATMENT
An anaerobic biological process treats a high-BOD5 soluble wastewater with a sludge age =
15 days. The volatile suspended solids concentration in the reactor is 3,000 mg/L. Several
weeks of operation at steady state conditions indicates that the net solids growth rate is
0.015 kg TSS/kg BOD5 removed. The process removes 15,000 kg BOD5 per day.
a) What is volume of the anaerobic reactor?
b) What is the waste sludge flow rate, in m3/day, if the waste sludge concentration is
1.5% solids by weight, and has density of 1020 kg/m3?
Solution
Basis: 1 day of operation
Sludge age = (Solids in reactor)/(Solids leaving the reactor)
a) Reactor Volume
Solids leaving the reactor = Solids produced in the reactor
Solids yield = (15,000 kg BOD5 removed/d)(0.015 kg TSS/kg BOD5 removed)
= 225 kg TSS/d
Mass of solids in the reactor = (3 kg TSS/m3)(Volume)
Sludge age = Mass of solids in reactor/mass flow of solids from reactor
15 d= ((3 kg TSS/m3)(Volume))/(225 kg TSS/d)
Volume = (15 d)(225 kg TSS/d)/(3 kg TSS/m3) = 1,125 m3
b) Waste sludge flow rate
Mass flow rate of sludge = (225 kg TSS/d)/(0.015 kg TSS/kg sludge)
= 15,000 kg sludge/d
Volumetric flow rate of sludge = (15,000 kg/d)/(1020 kg/m3) = 14.7 m3/d
8.25 ANAEROBIC SLUDGE DIGESTION
A plant is to treat 100 m3/d of sludge from wastewater treatment. The sludge is 5 percent
solids by weight, 70 percent of which are organic (volatile solids). A well-mixed digester
with no recycle operating with a sludge age of 15 days will give 50 percent destruction of
the volatile solids. Use the experienced-based criteria to estimate gas production.
256
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Solution
Assume sludge density is = 1020 kg/m3
Sludge feed = (100 m3/d)(1020 kg/m3) = 102,000 kg/d
Solids in feed = (102,000 kg sludge/d)(0.05 kg solids/kg sludge) = 5,100 kg solids/d
Volatile solids in feed = (0.7 kg VSS/kg solids)(5,100 kg solids/d) = 3,570 kg VSS/d
Volatile solids destroyed = (0.5 kg VSS destroyed/kg VSS)(3,570 kg VSS/d)
= 1,785 kg VSS destroyed/d
Typical gas production = 0.8-1.1 m3 gas/kg VSS destroyed
Using an average of 1.0 m3 gas/kg VSS destroyed, the estimated gas production is
(1.0 m3 gas/kg VSS destroyed)(1,785 kg VSS destroyed/d) = 1,785 m3 gas/d
8.26 FOOD WASTE
Wastewater from a food processing industry contains starches and sugars (carbohydrates)
and grease. It contains very little nitrogen or phosphorus so ammonia will be added as a
nitrogen source. The influent contains 1,000 kg/d of the organic compound and the process
can be operated so that 80% of it can be destroyed anaerobically to produce methane. Make
a preliminary assessment of this treatment. The reaction is
C9H18O2 + 0.25 NH4+ + 0.25 HCO3– + 2.5 H2O → 5.6 CH4 + 0.25 C5H7O2N + 2.4 CO2
a) How much methane (kg/d) will be produced?
b) How much ammonia (kg/d) needs to be added?
Solution
Basis: 1,000 kg/d of organic compound at 80% destruction = 800 kg/d
a) Methane production
Each mole of organic destroyed produces 5.6 moles of methane.
The molecular mass of the organic substance is 158 kg/kg-mol.
The molecular mass of methane is 16 kg/kg-mol.
Mass of organic destroyed
= (800 kg/d)/158kg/kg-mol = 5.06 kg-mol organic destroyed/d.
This produces
(5.06 kg-mol organic/d)(5.6 kg-mol methane/kg-mol organic)
= 28.3 kg-mol methane/d
Mass of methane produced
(16 kg methane/kg-mol)(28.3 kg-mol/d) = 453 kg CH4/d
257
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
b) Ammonia to be added
Each mole of organic destroyed requires 0.25 moles of ammonium.
The molecular mass of ammonium is 18 kg/kg-mol.
Moles of ammonium required
(5.06 kg-mol organic/d)(0.25 kg ammonium/kg-mol organic)
= 1.26 kg-mol ammonium/d
Mass of ammonium required
(18 kg ammonium/kg-mol)(1.26 kg-mol/d) = 22.7 kg ammonium/d
Alternatively, in terms of ammonia (NH3), this is equivalent to
(22.7 kg ammonium/d)(17 kg NH3/18 kg ammonium) = 21.4 kg NH3/d
8.27 ANAEROBIC BIODEGRDATION INDEX
A biomethane production (BMP) test is done using the liquid fraction of a centrifuged
sample from a healthy municipal anaerobic sludge digester. Biogas production, measured as
normal liters (NL) of biogas (1 atm, 0°C) produced per kg of total dry solids mass (DM)
is measured over a long period of incubation. The units of BMP are NL gas/kg DM. The
test reactors were operated at 37°C. The plots in Figure P8.27 are for an inoculated reactor
along with control reactor. No inoculum is added to the control, but some gas is produced
by organisms that are naturally in the sludge sample. The test was done in triplicate, that is,
three identical reactors of each kind. The curves are drawn free-hand through the average
of the three measured values. Estimate the biogas production, as m3/T of wet sludge, in a
reactor with 20 days detention time that is fed sludge with total solids concentration of 6%.
258
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Biogas Production
(NL/kg dry solids fed)
1000
Innoculated
Reactor
800
600
400
Control (no innoculum)
200
0
0
20
40
60
80
100
120
140
Time (days)
Figure P8.27
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259
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Solution
Basis = 1 T = 1,000 kg of wet sludge
Dry solids loading = (0.06 kg dry solids/kg wet sludge)(1,000 kg wet sludge)
= 60 kg dry solids
From the graph, for a detention time of 20 days, biogas production is 600 NL/kg dry solids
Biogas production = (600 NL/kg dry solids)(60 kg dry solids) = 36,000 NL
For wet sludge basis
Biogas production = (36,000 NL)/(1 T wet sludge) = 36 m3/T wet sludge
8.28 MOURAS AUTOMATIC SCAVENGER
According to an 1882 report,
“… this mysterious contrivance which has been used for 20 years, consists of a closed vault with
a water seal, which rapidly transforms all the excrementitious matter which it receives into
homogeneous fluid, only slightly turbid, and holding all the solid matters in suspension in the
form of scarcely visible filaments. The vault is self-emptying, and continuous in its working, and
the escaping liquid, while it contains all the organic and inorganic elements of the feces, is almost
devoid of small solid matter, and can be received into water carts for horticultural purposes, or
may pass away into the sewer for use in irrigation.”
As for the theory of action, it was said, “May not the unseen agents be those vibrions or
anaerobies which, according to Pasteur, are destroyed by hydrogen, and only manifest there
activity in vessels from which air is excluded.”
What are the unseen agents? Do they work only in vessels from which air is excluded? Is
air excluded from the Mouras Scavenger? Will there be some solid residue? Why would the
effluent be beneficial for horticultural purposes? Would using it for irrigation pose a threat
to human health? Do you believe it will be odor free?
Solution
This is an open ended problem for group or class discussion. No solution is provided.
260
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
8.29 ANAEROBIC SLUDGE DIGESTION
The feed to an anaerobic digester has a total solids (TS) concentration of 6% solids (dry
mass basis); 80% of the solids are volatile (VS) and 20% are inert (IS). The volatile solids
fraction is higher than typical municipal sewage sludge because some food industry waste
is mixed with the sludge. The volatile solids fraction of the digested sludge solids is 45%.
The gas yield is 1 m3 per kg of volatile solids destroyed. The gas is 75% methane. Ignore
the loss of water vapor in the gas. Calculate the material balance using a basis of 10,000
kg wet sludge per day.
Solution
Basis:10,000 kg sludge (wet basis)
80% of total dry solids in digested sludge are volatile.
20% of total dry solids are inert.
Tie component = inert solids
All solids masses are on a dry basis
Total solids (TS) in feed sludge = 6% sludge mass
TS = (0.06 kg TS/kg sludge)(10,000 kg sludge) = 600 kg TS
Water in feed sludge = 10,000 kg - 600 kg = 9,400 kg water
Inert solids in feed = IS = 20% of TS
= (0.20 kg IS/kg TS)(600 kg TS) = 120 kg IS
Inert solids in digested sludge = Inert solids in feed = 120 kg IS
Volatile solids (VS) in feed sludge = 80% of TS in feed
= (0.8 kg VS/kg TS)(600 kg TS) = 480 kg VS
VS of digested sludge: By definition
%VS =
VS =
100 VS
VS + IS
(%VS)(IS)
100% %VS
=
(45%)(120 kg)
= 98.2 kg
100% - 45%
Round 98.2 kg to 98 kg VS
VS destroyed = 480 kg – 98 kg = 382 kg
Note: Destroyed means converted to gas or solubilized
Digested sludge solids = 120 kg IS + 98 VS = 218 kg
261
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Total mass of digested sludge = 9,400 kg water + 120 kg IS + 98 kg VS = 9,618 kg
Digested sludge solids concentration = 100(120 kg IS + 98 kg VS)/(9,618 kg) = 2.3%
Gas produced at 1 m3/kg VS destroyed = (1m3/kg VS)(382 kg VS destroyed) = 382 m3
Gas is 75% methane
Methane produced = 0.75(382 m3) = 286 m3
The complete material balance is shown in Figure S8.29.
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Digested Gas (Biogas)
75% CH4 + 25%CO2
1 m3 gas/kg VS destroyed
382 kg/d VS destroyed
382 m3/d gas produced
286 m3/d CH4 produced
Feed Sludge
Mass = 10,000 kg/d
TS = 600 kg/d
VS = 480 kg/d
IS = 120 kg/d
Water = 9,400 kg/d
6% solids
Anaerobic
Sludge Digester
Digested Sludge
VS = 98 kg/d
IS = 120 kg/d
TS = 120 + 98 = 218 kg/d
Water = 9,400 kg/d
Mass = 9,618 kg/d
2.3% solids
Figure S8.29
8.30 TWO-STAGED ANAEROBIC DIGESTERS
The design loadings for a two-phase anaerobic digester shown in Figure P8.30 are given
in Table P8.30. The sludge ages are 2 days for the acid producing phase and 12 days for
the methane producing phase. Calculate the digester volumes. Assume the sludge density
is 1,000 kg/m3.
CH4 and CO2
Thickened
Raw Sludge
Acid
Phase
Digested Sludge
to Dewatering
and Disposal
Methane
Phase
Figure P8.30
Volume sludge to digester (m3/d)
216
Mass of sludge to digester (kg/d)
216,000
Total solids concentration (%)
2.4
TS load to digestion (kg/d)
5,184
VS load to digestion (kg/d)
3,900
Table P8.30 Design Loadings
263
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Solution
Sludge age = Hydraulic detention time
Sludge volume into acid phase digester = Sludge volume into methane phase digester
Acid phase
Sludge age = 2 d = V/Q = VAcid/(216 m3/d)
VAcid = (2 d)(216 m3/d) = 432 m3
Methane phase
Sludge age = 12 d = VMethane/(216 m3/d)
VMethane = (12 d)(216 m3/d) = 2,592 m3
8.31TEMPERATURE-PHASED DIGESTER
(THERMOPHILIC + MESOPHILIC)
A two-stage temperature-phase anaerobic digester (TPAD) shown in Figure P8.31 will be
designed for the loadings in Table P8.31. The thermophilic stage is designed for a volatile
solids loading of 5 kg VS/m3-d. The mesophilic stage is designed for a sludge age of 12
days. Volatile solids are 70% of total solids. Calculate the digester volumes. Assume the
sludge density is 1,000 kg/m3.
CH4 and CO2
Thickened
Raw Sludge
Thermophilic
Phase
Digested Sludge
to Dewatering
and Disposal
Mesophilic
Phase
Figure P8.31
Volume sludge to digester (m3/d)
216
Total solids concentration (%)
2.4
Table P8.31 Design Loadings
264
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Solution
Digester loadings
Mass of sludge to digester = (216 m3/d)(1,000 kg/m3) = 216,000 kg/d
Total solids to digester = (0.024)(216,000 kg/d) = 5,184 kg TS/d
Volatile solids to digester = (0.7)(5,184 kg/d) = 3,629 kg VS/d
Thermophilic phase operating at a loading of 5 kg VS/m3-d
Volume = (3,628 kg VS/d)/(5 kg VS/m3-d) = 726 m3
Mesophilic phase operates with a sludge age of 12 days.
Sludge age = Hydraulic detention time
12 d = V/Q = V/(216 m3/d)
V = (12 d)(216 m3/d) = 2,592 m3
8.32 UPFLOW ANAEROBIC SLUDGE BLANKET REACTOR
An upflow anaerobic sludge blanket (UASB) reactor is used to treat food processing wastewater
at 20oC. Note: The reactions in an upflow sludge blanket digester are the same as in other
anaerobic digesters. This is an interesting process that is much used for high-strength wastes
(e.g. food industry). For extra credit do some self-study about the process.
The flow rate is 2,000 m3/h with a mean soluble COD of 7,000 mg/L. (a) Using empirical
stoichiometric factors, calculate the theoretical maximum CH4 generation rate in m3/day.
(b) What is a more realistic estimate of methane production for 80% COD removal?
Solution
a) Complete degradation of organic matter in the waste will give the theoretical maximum
methane generation rate.
Total COD removed = (2,000 m3/h)(7 kg COD/m3) = 14,000 kg COD/h
Empirical stoichiometry:
1 kg COD destroyed produces 0.35 m3 CH4 at 20°C and 1 atm
For complete removal of COD the estimated methane production is
(0.35 m3 CH4/kg COD)(14,000 kg COD/h) = 4,900 m3 CH4/h at 20°C and 1 atm
265
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
b) More realistic estimate of methane production.
Not all COD (organic matter) is completely degraded and not all that is degraded is converted
to CH4. The fate of COD during anaerobic treatment process includes
Residual COD in effluent
COD converted to CH4 gas
COD diverted to biomass synthesis
COD utilized for sulfate reduction (if sulfate is present)
Assuming 80% COD removal for all processes
COD removed = 0.8(14,000 kg/h) = 11,200 kg COD/h
Assume that 10% of the 11,200 kg COD/h removed has been utilized for biomass synthesis.
Assume that 90% of the 11,200 kg COD/h removed has been converted to CH4 gas.
COD converted to biogas = 0.9(11,200 kg COD/h) = 10,080 kg COD/h
Empirical stoichiometry for gas production
1 kg COD produces 0.35 m3 CH4 at 20°C and 1 atm
(0.35 m3 CH4/kg)(10,080 kg COD/h) = 3,528 m3 CH4/h at 20°C and 1 atm
8.33 AGRICULTURAL BIOGAS GAS
An agricultural anaerobic digester produces 75,000 m3/d (at standard conditions (0°C, 1
atm) of biogas that is 50% methane (CH4), 35% carbon dioxide (CO2) 13% water vapor
(H2O) and 2% hydrogen sulfide (H2S) These are volume percentages. The CO2, H2O and
H2S are removed to increase the fuel value of the gas so it can be sold. (a) What is the
yield of CH4 by volume (m3/d) and by mass (kg/d)? (b) What mass of CO2, H2O, and
H2S are removed each day.
Solution
a) Methane (CH4) yield
Total = 75,000 m3/d
50% Methane (CH4) = 37,500 m3/d
35% Carbon dioxide (CO2) = 26,250 m3/d
13% Water vapor (H2O) = 9,750 m3/d
2% Hydrogen sulfide (H2S) = 1,500 m3/d
At 0°C and 1 atm the molar volume of a gas is 22.414 m3/kg-mol
Molar mass of CH4 = 16 kg/kg-mol
266
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Methane density = (16 kg/kg-mol)/(22.414 m3/kg-mol) = 0.714 kg/m3
Mass of CH4 = (37,500 m3/d)(0.714 kg/m3) = 26,800 kg/d
b) Mass of CO2, H2O, and H2S removed
Density of CO2, H2O and H2S
CO2 (44 kg/kg-mol)/(22.414 m3/kg-mol) = 1.96 kg/m3
H2O (18 kg/kg-mol)/(22.414 m3/kg-mol) = 0.803 kg/m3
H2S (34 kg/kg-mol)/(22.414 m3/kg-mol) = 1.52 kg/m3
Mass removed
Mass of CO2 removed = (26,250 m3/d)(1.96 kg/m3) = 51,500 kg/d
Mass of H2O removed = (9,750 m3/d)(0.803 kg/m3) = 7,830 kg/d
Mass of H2S removed = (1,500 m3/d)(1.52 kg/m3) = 2,280 kg/d
Note: the mass percentages of the digester gases are quite different from the volume
percentages.
The total mass of digester gas is
26,800 kg CH4/d + 51,500 kg CO2/d + 7,830 kg H2O/d + 2,280 kg H2S/d
= 88,410 kg/d
Mass% CH4 = 100(26,800 kg/d)/(88,410 kg/d) = 30.3%
Likewise CO2 = 58.2%, H2O = 8.9%, and H2S = 2.6%
8.34 DAIRY MANURE
A dairy farm has 600 cows with an average weight of 635 kg. Each cow produces 54.4 kg/d
of manure (wet mass) that contains 7.6 kg/d total solids and 6.4 kg/d volatile solids. Ten
percent of the manure is lost in handling and 90% goes to an anaerobic manure digester.
The digester is 20-45% efficient in the destruction of volatile solids. Biogas production is
0.5-1.1 m3 CH4/kg VS destroyed. The gas is 50-65% methane (CH4). (a) Calculate the
maximum and minimum quantities for total gas production and for methane production.
(b) Calculate the material balance for total solids, volatile solids, and inert (fixed) solids
for the digester.
267
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Solution
a) Gas Production
Mass of manure = (600 cows)(54.4 kg manure/cow) = 32,600 kg manure/d
Mass of TS in manure = (600 cows)(7.6 kg TS/cow) = 4,560 kg TS/d
Mass of VS in manure = (600 cows)(6.4 kg VS/cow) = 3,840 kg VS/d
VS destruction = 20% to 45%
Gas yield = 0.5 m3 CH4/kg VS destroyed to 1.1 m3 CH4/kg VS destroyed
Sample calculation: 20% VS destruction efficiency and gas yield 0.5 m3/kg VS destroyed
Gas yield = (3,840 kg VS/d)(0.2 kg VS destroyed/kg VS input)(0.5 m3/kg VS destroyed)
= 384 m3 CH4/d
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
b) The material balance for total solids, volatile solids, and inert (fixed) solids for the
digester are shown in Table S8.34.
Stream
Solids
Gas Yield
Gas Yield
(kg/d)
(@ 0.5 m /kg)
(@ 1.1 m3/kg)
Total Solids Input
4,560
Fixed Solids Input
720
Volatile Solids Input
3
3,840
20% destroyed
768
384 m3/d
845 m3/d
45% destroyed
1,728
864 m3/d
1,900 m3/d
Volatile Solids Output
20% destroyed
3,072
45% destroyed
2,112
Fixed Solids output
720
Total Solids Output
20% destroyed
3,792
45% destroyed
2,832
Table S8.34
8.35 CO-DIGESTING FOOD WASTE AND SLUDGE
About 1,200 m3/y (240,000 kg/y) of wet food waste was being used as a food source for
domestic animals, mainly pigs, until this use was banned. This caused an increase in organic
waste going to the city landfill and it was decided run a full-scale experiment of feeding
the food waste to the wastewater treatment anaerobic digesters in combination with the
wastewater sludge. The wastewater’s sludge volatile solids load is 60% from primary sludge
and 40% from waste activated sludge. Sludge is fed to the digesters semi-continuously (every
3 hours) from a sludge thickener. Sludge volatile solids concentrations (VS) range from 10
to 20 g/L, total solids (TS) range from 20 to 30 g/L, and the sludge COD ranges from 18
to 30 g/L. The sludge retention time in the digesters is 20 days.
269
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Table P8.35 shows the VS and COD loading rates (kg/d) to the digesters, which include the
organic waste from January 2004 to March 2005, and otherwise are wastewater sludge only.
Before and after the food waste experiment the average gas yield was 0.39 m3/kg VS fed and
the VS degradation was 72.3%. During the experiment, when food waste was added, the
gas yield was 0.60 m3/kg VS fed and the VS degradation was 83.6%. (Note: Normally gas
production is expressed in terms of VS destroyed and not VS fed). (a) Use the information
provided to calculate total gas production and VS remaining after digestion. (b) Plot these
calculated values, along with VS and COD feed (kg/d) as a function of time. (c) Describe
your findings.
Year
2003
Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec
VS
feed
(kg/d)
1,399 1,688 1,366
1,563
1,761 1,516 1,481 1,598 1,445
1,307
1,680 1,440
COD
feed
(kg/d)
1,588 1,892 1,550
1,774
1,998 1,720 1,680 1,855 1,640
1,483
1,989 1,919
Apr
May
Oct
Nov
Year
2004
Jan
Feb
Mar
Jun
Jul
Aug
Sep
Dec
VS
feed
(kg/d)
1,226 1,805 1,788
1,608
2,012 1,703 1,964 1,881 1,569
1,445
1,528 1,891
COD
feed
(kg/d)
2,124 2,157 2,273
2,257
2,574 2,873 3,350 2,842 2,271
2,157
2,370 2,687
Food
Waste
(kg/d)
Year
2005
187
Jan
173
Feb
277
Mar
228
150
Apr
May
212
Jun
254
Jul
253
Aug
401
Sep
310
Oct
VS
feed
(kg/d)
1,434 1,835 1,455
1,792
1,891 1,262
992
750 1,020
1,434
COD
feed
(kg/d)
2,362 3,491 1,742
2,726
2,682 1,553 1,184
868 1,233
1,597
Food
Waste
(kg/d)
185
343
185
Table P8.35
270
384
369
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Solution
a) Gas production and VS remaining after destruction
Sample Calculation for January 2003
Gas production = (0.39 m3/kg VS fed(1,399 kg VS fed/d) = 546 m3/d
VS output = (1 – 0.723)(1,399 kg VS fed/d) = 388 kg VS/d
Sample calculation for January 2004
Gas production = (0.60 m3/kg VS fed(1,226 kg VS fed/d) = 736 m3/d
VS output = (1 – 0.836)(1,226 kg VS fed/d) = 201 kg VS/d
The complete calculations are summarized in Table S8.36
271
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Year =
2003
VS feed
(kg/d)
Jan
Feb
Mar
Material balance for biological processes
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec
1,399 1,688 1,366 1,563 1,761 1,516 1,481 1,598 1,445 1,307 1,680 1,440
COD feed
1,588 1,892 1,550 1,774 1,998 1,720 1,680 1,855 1,640 1,483 1,989 1,919
(kg/d)
Food
Waste
(kg/d)
0
0
0
0
0
0
0
0
0
0
0
0
Gas
Produced
(m3/d)
546
658
533
610
687
591
578
623
564
510
655
562
VS
Output
(kg/d)
388
468
378
433
488
420
410
443
400
362
465
399
Year =
2004
VS feed
(kg/d)
Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec
1,226 1,805 1,788 1,608 2,012 1,703 1,964 1,881 1,569 1,445 1,528 1,891
COD feed
2,124 2,157 2,273 2,257 2,574 2,873 3,350 2,842 2,271 2,157 2,370 2,687
(kg/d)
Food
Waste
(kg/d)
187
173
277
Gas
Produced
(m3/d)
736 1,083
1073
VS
Output
(kg/d)
201
Year =
2005
VS feed
(kg/d)
Jan
296
Feb
293
Mar
228
253
401
310
384
965 1,207 1,022 1,178 1,129
941
867
917 1,135
264
257
237
251
Apr
150
212
330
May
279
Jun
254
322
Jul
1,434 1,835 1,455 1,792 1,891 1,262
308
Aug
Sep
Oct
992
750 1,020 1,434
COD feed
2,362 3,491 1,742 2,726 2,682 1,553 1,184
(kg/d)
868 1,233 1,597
Food
Waste
(kg/d)
185
343
185
0
0
0
0
0
0
0
Gas
Produced
(m3/d)
860 1,101
873
699
737
492
387
293
398
559
VS
Output
(kg/d)
235
239
496
524
350
275
208
283
397
301
Table S8.35
272
369
310
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
b) Data Plots
Figure S8.35
c) There is a noticeable boost in gas production when the food waste is added to the
normal sludge feed. The added mass of food waste volatile solids is barely evident
in the VS feed data, and one might overlook it if not for being told that the sludge
feed was being augmented. The food waste, while not high in volatile solids, is high
in readily bioavailable COD, and this is what drives the gas production.
8.36 BIOKINETICS
The mixed-order rate expression for a biological reaction, known as the Monod model, is
dS
dt
kS
K S
273
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Accounting for the active biomass this becomes
dS
dt
where
kXS
K S
dS/dt = the rate of substrate utilization in the reactor (mg COD removed/L-d)
S = substrate concentration (mg COD/L)
X = volatile suspended solids concentration (mg VSS/L)
K = Half-saturation coefficient (mg COD/L)
k = reaction rate coefficient (mg COD/mg VSS-d). k is also known as the maximum
substrate utilization rate
Plot curves for reaction rate as a function of substrate concentration (from 0 mg/L to 300
mg/L) that would be observed in completely mixed reactors that operate with X =1800
mg/L, 2,000 mg/L, and 2,200 mg/L. The kinetic parameters are k = 6 mg COD/mg VSS-d
and K = 50 mg COD/L.
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Solution
For the given kinetic parameters the model is
COD removal rate =
dS
(6 mg COD/mg VSS-d) X S
=
dt
(50 mg COD/L) + S
A spread sheet solution will give the model plots in Figure S8.36.
COD reaction rate, dS/dt
(mg COD/L-d)
14,000
X = 2,500 mg VSS/L
12,000
X = 2,000 mg VSS/L
10,000
X = 1,500 mg VSS/L
8,000
6,000
4,000
2,000
0
0
50
100
150
200
250
300
Substrate concentration, S (mg/L)
Figure S8.36
8.37 HALDANE INHIBITION MODEL
The Haldane model has been used to describe substrate inhibition kinetics for bacterial
growth.
k
kMax S
K S
where
S2
KI
k = operating specific degradation rate coefficient (h-1)
S = substrate concentration (mg/L)
kMax = maximum specific degradation rate coefficient (h-1)
K = substrate saturation coefficient (mg S/L)
KI = substrate inhibition coefficient (mg S/L).
275
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Note: The actual substrate removal rate can be determined from
dS
dt
where
X
k
Y
X = mixed liquor volatile solids concentration (mg VSS/L)
Y = yield coefficient (mg VSS produced/mg substrate removed)
Phenol can be a problem in wastewater from steel mills, refineries, and other industries.
It is inhibitory to the activated sludge process, but an acclimated population of can been
developed in the biological system. A laboratory study of phenol biodegradation (Dey and
Mukherjee 2010) provides these kinetic coefficients.
K = 51.8 mg/L
KI = 404 mg/L
kMax = 0.150 h-1
a) Plot the value of k for the Haldane inhibition model and compare it with the
Monod model over a range of phenol concentrations from 0 mg/L to 200 mg/L.
The Monod model kinetic parameters are kMax and K as given in Problem 8.36.
b) Is inhibition a problem if a phenolic wastewater were treated in a completely mixed
reactor with an effluent concentration of 5 mg/L?
c) Is a first-order model a good approximation for the treatment conditions given in
part (b)?
Solution
a) Sample calculation for Haldane model for S = 100 mg P/L. (P = phenol).
kMax S
k
K S S2 /K I
=
(0.150/h)(100 mg P/L)
= 0.0854/h
(51.8 mg P/L) + (100 mg P/L) + (100 mg P/L)2 /(404 mgP/L)
Sample calculation for Monod model for S = 100 mg P/L
k
kMax S
(0.150/h)(100 mg P/L)
=
= 0.0988/h
K S
(51.8 mg P/L) + (100 mg P/L)
Figure S8.37 compares the specific degradation rate coefficients with and without inhibition.
276
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Specific Degradation Rate
Coefficient (mg/L-h)
0.14
0.12
Monod Model
0.10
0.08
Haldane Model
0.06
0.04
0.02
0.00
0
40
360°
thinking
80
120
160
200
.
Phenol Concentration (mg/L)
Figure S8.37 Substrate inhibition of phenol biodegradation.
360°
thinking
.
360°
thinking
.
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277
Dis
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
b) There is no inhibition until about S = 30 mg P/L. Inhibition is fairly strong for S > 100
mg P/L. A completely mixed reactor with effluent of 5 mg P/L has a concentration of
5 mg/L in the reactor. There will be no inhibition.
c) At low concentrations, say below 25 mg/L phenol, the reaction can be modeled as
first-order because the rate is nearly linear in concentration.
8.38 COD REMOVAL FROM REFINERY WASTEWATER
Bacteria that have had prior exposure to petroleum derivatives are expected to have a higher
rate of COD degradation than un-exposed bacteria. This test used Micrococcus luteus. Use
the data in Table P8.38 to discover the reaction rate model and the rate coefficient(s).
Time (d)
COD
(mg/L)
0
2
4
6
8
10
12
14
16
18
600 550 496 440 388 338 286 256 2v32 215
20
22
24
26
28
30
200 190 180 175 171 168
Table P8.38
Solution
Plot the data as in Figure S8.38a. The COD does not drop to zero, as we tend to expect with first-order
reactions. This expectation is not valid for COD, because the COD test measures many compounds
that are not biodegradable. It appears that about one-third of the COD in this wastewater is not
biodegradable.
The arithmetic plot (left) looks like it could be first order, but we need to look at the semi-log plot
(right). This is not a straight line, so a simple first-order model is not a good choice.
700
600
500
COD (mg/L)
COD (mg/L)
600
400
300
200
400
300
200
100
0
0
5
10
15
Time (d)
20
25
100
30
0
5
10
15
Time (d)
20
25
30
Figure S8.38a
The Monod model is popular model for biological reactions. This model transitions from zero-order
278
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
high concentrations to first-order at low concentrations. The COD removal rate for the first 10-15 days
does look like a zero-order reaction. The rate is constant; it does not decrease as the concentration
decreases. From about 15 days on it appears that the reaction rate does decrease as concentration
decreases.
Look at the COD removal rates. Estimate the rates by taking the difference of subsequent COD
concentrations. For example, the estimated rate at time t = 0, with COD = 600 mg/L is
Rate = (600 mg/L – 550 mg/L)/(2 d) = 25 mg/L-d
Plotting the removal rates as a function of concentration shows, in Figure S8.38b, that from about
from 600 mg/L down to about 350 mg/L the rate is constant at r0 = 26 mg/L-d. Below 350 mg/L the
rate decreases in proportion to the concentrations – this is a first-order reaction. The slope of the
first-order portion of the data estimates the reaction rate coefficient.
r1
15 mg/L-d - 1.5 mg/L-d
= 0.12 d-1
286 mg/L - 171 mg/L
Reaction Rate (mg/L-d)
30
25
Zero order
20
15
10
First order
5
0
0
100
200
300
400
500
600
700
COD (mg/L)
Figure S8.38b
Graphical estimates of the Monod model are
r
where
k COD
(26 mg/L-d)(COD)
=
K + COD
260 mg/L + COD
k = maximum rate = 26 mg/L-d
K is the value of COD at 0.5k = 13 mg/L-d
K = 260 mg/L
More precise estimates could be obtained from a least squares (nonlinear regression) fit of the data.
This is not necessary here as we are most interested in identifying the model.
279
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Alternate solution
The plot of concentration vs. time in Figure S8.38a suggests a first order reaction with a non-zero
asymptote. This model is of the form used in Problem 7.31. The data appear to level off at about COD
= 150 mg/L. So let’s try a first order model with an asymptote of, say, α =150 mg/L.
The model has the form (C – α) = (C0 – α)exp(-kt)
where
C = COD concentration, mg/L
k = first order rate coefficient, d-1
The calculations using an initial guess of α = 150 mg/L, are in Table S8.38, and the plot [ln(C – 150
mg/L) vs. t] is in Figure S8.38c
C–α
Time
COD
(d)
(mg/L)
0
600
450
6.11
2
550
400
5.99
4
496
346
5.85
6
440
290
5.67
8
388
238
5.47
10
338
188
5.24
12
286
136
4.91
14
256
106
4.66
16
232
82
4.41
18
215
65
4.17
20
200
50
3.91
22
190
40
3.69
24
180
30
3.40
26
175
25
3.22
28
171
21
3.04
30
168
18
2.89
α = 150
ln(C - α)
(mg/L)
Table S8.38
280
SOLVED MATERIAL BALANCE PROBLEMS:
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Material balance for biological processes
ln(C - 150 mg/L)
7
ln(C – 150 mg/L) = 6.23 - 0.116 t
6
5
4
3
2
0
5
10
15
20
25
30
Time (d)
Figure S8.38c
Figure 8.38c shows the data are well described by a straight line. The estimate of the rate coefficient
is k = 0.116/d. The intercept gives the estimate of C0.
ln(C0 – 150 mg/L) = 6.23
C0 = e6.23 + 150 mg/L = 508 mg/L + 150 mg/L = 658 mg/L
The complete model is C – 150 mg/L= (658 mg/L – 150 mg/L)e-(0.116/d)t
or C = 150 mg/L + (508 mg/L)e-(0.116/d)t
Our guess that α = 150 mg/L may not be the best one. Try other values. Non-linear regression method
may better describe the data and give better parameter estimates. That subject is beyond the scope
of this problem, and one that is very useful.
8.39 REACTOR DETENTION TIME
Calculate the detention time for a CSTR to reduce an influent pollutant concentration of
250 mg/L to an effluent concentration of 20 mg/L. The Monod kinetic parameters are k
= 600 mg/L-d and K = 40 mg/L.
Solution
Figure S8.39 shows dC/dt as a function C for the given kinetic parameter values (k = 600 mg/L-d, K
= 40 mg/L)
dC
dt
kC
K C
Ÿ
dC
(600 mg/L-d)C
=
dt
40 mg/L + C
281
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Reaction Rate, dC/dt (mg/L-d)
600
500
400
300
dC
dt
200
100
C 20 mg/L
(600 mg/L-d)(20 mg/L)
= 200 mg/L-d
40 mg/L + 20 mg/L
C = 20 mg/L
0
0
50
100
150
200
250
300
Concentration, C (mg/L)
Figure S8.39
The CSTR will operate at one point on this curve. If C = 20 mg/L, then the rate removal is
dC
dt
kC
(600 mg/L-d)(20 mg/L)
1,200 mg/L-d
=
=
= 200 mg/L-d
K C
40 mg/L + 20 mg/L
60 mg/L
The material balance for a completely-mixed reactor operating at steady-state is
QC0 QC dC
dt
V
0
C
Solving the material balance for the hydraulic detention time, V/Q, gives
T
V
Q
C0 C
(dC /dt ) C
For an influent pollutant concentration of 250 mg/L and effluent concentration of 20 mg/L, the required
hydraulic detention time is
T
V
Q
(C0 C )
(dC /dt ) C=20mg/L
250 mg/L - 20 mg/L
= 1.15 d
200 mg/L-d
282
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
8.40 SYNTHETIC MEDIA BIOFILTERS
Figure 8.40 shows wastewater being applied to the top of an aerobic biofilter (trickling
filter) that is packed with rigid synthetic media on which the working microorganisms grow.
Name three beneficial characteristics of this kind of media.
Figure P8.40
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SOLVED MATERIAL BALANCE PROBLEMS:
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Material balance for biological processes
Solution
1) Strong and lightweight so the filter bed can be made tall.
2) Large surface area on which organisms can attach and grow.
3) High void volume which is good for air circulation and oxygen transfer.
4) Accepts high hydraulic loading rates which helps prevent clogging.
8.41 TRICKLING FILTER HYDRAULIC LOADING
The biofilter in Figure P8.44 shows a rotating distributor arm that sprays the influent
wastewater onto the top of the trickling filter. The distributor nozzles are not evenly spaced
along the arm. Why not?
Solution
The area swept by nozzles further out on the rotating arm cover a larger surface area than those close
to the center of rotation. If the nozzles each deliver the same volumetric flow rate (say m3/h), they
must be spaced closer together farther out on the rotating arm in order to achieve a uniform surface
hydraulic loading rate (say m3/m2-h) of the biofilter.
8.42 BIOFILTER FOR ODOR CONTROL
A yard waste compost bed is used as a biofilter to treat 50,000 scfm odorous air from a waste
treatment process. The dimensions of the compost bed are 140 ft by 70 ft by 3.5 ft deep.
The odorous air is saturated with moisture at 80°F. When it is compressed the temperature
increases to 87°F and the relative humidity is 75%. It exits a humidifier at 82°F and at
100% relative humidity and goes to the biofilter. A flow of 5,000 gpd of water at 80°F is
sprayed onto the biofilter to maintain the proper moisture content. What is the detention
time (empty bed) of the air in the biofilter?
Solution
Biofilter bed volume = (140 ft)(70 ft)(3.5 ft) = 34,300 ft3
Detention time (empty bed) = 34,300 ft3/50,000 scfm = 0.686 min (based on standard conditions).
284
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
8.43 CANNERY WASTE TREATMENT
A vegetable cannery operates only from July through September. The wastewater has BOD5
= 8,000 mg/L, TSS = 6,000 mg/L, pH = 10, NH3-N = 100 mg/L, and P = 250 mg/L.
Suggest some biological treatment alternatives and outline the constraints and advantages
associated with each.
Solution
High TSS - If the solids can be removed by settling, screening or primary sedimentation will be useful.
This may also remove some BOD5 - how much cannot be predicted from the information given.
BOD/N/P ratio should be about 100/20/1 for good biological treatment. This would apply to the
wastewater after primary settling, if any. Based on the raw wastewater: BOD5/N ratio = (8,000 mg/L)/(100
mg/L) = 80. This ratio should be 20/1 for good biological treatment. Nitrogen will need to be added.
BOD5/P ratio = (8,000 mg/L)/(250 mg/L) = 32. There is more than enough P to keep the bacteria
healthy. Phosphorus removal probably will be required to meet a standard of 1 mg/L.
pH is too high. Neutralization will be required.
A major difficulty with seasonal systems, such as canning, is that the waste load comes from zero
before the canning season to a very high amount immediately as the season starts. Municipal treatment
plants normally are not designed to accept such loads. Lagoons and irrigation systems are common
for canneries. Other systems are difficult to start up in a short time.
8.44 LANDFILL GAS ACTIVITY
Figure P8.44 shows the ratio of CH4/CO2 in the landfill gas across a landfill site. Red indicates
a higher methane content (>65% CH4), orange is a more typical 50% CH4 content. Using
the gas composition, can you explain the work history of the landfill?
285
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Figure P8.44
Solution
This is an open-ended discussion question that is meant to provoke discussion. No solution is provided.
8.45 COMPOSTING TEMPERATURE
The interior temperature of a compost windrow rises quickly and remains high through the
composting period. This is true in winter and summer. Explain this process.
Solution
Composting is an aerobic process. Compost piles must be turned or mixed from time to time to keep
them aerobic. Decomposition reactions by aerobic organisms are exothermic - they give off heat.
The heat is generated by the aerobic microorganisms in the compost heap as long as there is proper
moisture (about 50% moisture) and a suitable ratio of carbon and nitrogen (about 40 to 1).
286
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
8.46 COMPOSTING SLUDGE CAKE
Sludge cake in the amount of 625 wet tones per day at 16% total solids is mixed with
750 wet tones per day of cured compost that has 40% total solids. What is the moisture
content of the mixture?
Solution
Sludge cake = 625 wet T/d at 16% dry solids
Dry solids = 100 T/d dry solids
Water = 625 wet T/d – 100 T/d = 525 T/d
Cured compost = 750 wet T/d at 40% dry solids
Dry solids = 300 T/d
Water = 750 wet T/d – 300 T/d = 450 kg/d
Mixture
Dry solids = 100 T/d + 300 T/d = 400 Td
Water = 525 T/d + 450 T/d = 925 T/d
Total mass = 1,325 T/d
% moisture = 100(925 T/d water)/(1,325 T/d total mass) = 69.8%
8.47 SLUDGE-REFUSE MIXTURE FOR COMPOSTING
It is proposed that sludge and refuse are to be mixed for composting. The available sludge
is 7% solids by weight, nitrogen accounts for 2.5% of the solids, and the carbon/nitrogen
ratio is 10/1. The refuse is 25% moisture by weight and the solids contain 35% carbon
and a negligible amount of nitrogen. The mixture should have a moisture content of 50%
(between 45-55%) and C/N of 40/1 (between 30/1 and 50/1). These are the ideal conditions
for composting.
a) Design a mixture that is 50% moisture. Check the material balance on carbon and
nitrogen and the C/N ratio. Does this mixture meet the specifications? If not, what
is lacking? What might be done to remedy the deficiency?
b) Design a mixture that meets the C/N ratio specification of C/N = 40. Check the
material balance on carbon and nitrogen and the C/N ratio. Does this mixture
meet the specification for moisture? What improvements could be made?
287
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Solution
a) Basis: 100 kg of mixed sludge and refuse solids
Let X = kg of sludge input and Y = kg of refuse input.
Sludge has 25 % moisture, N = 2.5% of solids, C/N = 10/1, so C = 25% of solids.
Refuse has 28% moisture, C = 35% of solids, and N= 0%
Overall material balance: X + Y = 100 kg
Composition of sludge: solids = 0.07 X, water = 0.93 X
Composition of refuse: solids = 0.75 Y, water 0.25 Y
Mixture at 50% moisture
Material balance on water: 0.93 X + 0.25 Y = 0.5(100 kg) = 50 kg
Material balance on solids: 0.07 X + 0.75 Y = 0.5(100 kg) = 50 kg
Solving the solids material balances gives
0.07 X + 0.75(100 kg - X) = 50
X = 36.8 kg sludge and Y = 63.2 kg refuse
Proper moisture content: Mix approximately 1 part sludge and 2 parts refuse by weight
Check the C/N ratio for this mixture:
36.8 kg sludge contains:
Water = 0.93(36.8 kg) = 34.2 kg water
Solids = 0.07(36.8 kg) = 2.6 kg solids
N = 0.025(2.6 kg) = 0.0645 kg N
C = 0.25(2.6 kg) = 0.645 kg C
63.2 kg refuse contains:
Water = 0.25(63.2 kg) = 15.8 kg water
Solids = 0.75(63.2 kg) = 47.4 kg solids
N = 0 kg N
C = 0.35(47.4) = 16.59 kg C
Mixture contains:
50 kg water, 50 kg solids, 0.0645 kg N, 17.23 kg C
C/N = 17.23 kg C/0.0645 kg N = 267.2
This mixture is nitrogen deficient by at least a factor of 6. To get more nitrogen, add more
sludge or add nitrogen in another form. A sludge/refuse mixture that is 55% moisture, the
maximum allowable, will still be deficient in N. Perhaps the sludge could be thickened or
dewatered. If the sludge were dewatered to 25% solids, a 50-50 mixture by weight would
be 50% moisture. The C/N ratio would be greatly improved. Checking whether it meets
specification is left as an exercise.
288
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
b) Calculate mixture to meet C/N = 40/1
Overall material balance: X + Y = 100 kg
Composition of sludge: solids = 0.07 X, water = 0.93 X
Composition of refuse: solids = 0.75 Y, water 0.25 Y
Material balance on carbon: C = 0.25 (0.07 X) + 0.35 (0.75 Y)
Material balance on nitrogen: N = 0.025 (0.07 X) + zero in refuse
Design for C/N = 40/1
0.25 (0.07 X) + 0.35 (0.75 Y) = 40 [0.025 (0.07 X)]
0.0175 X + 0.2625 Y = 0.07 X
0.0175 X + 0.2625 (100 kg - X) = 0.07X
0.315 X = 26.25 kg
X = 83.3 kg sludge and Y = 16.7 kg refuse
Check moisture balance for this mixture:
83.3 kg sludge contains:
Water = 0.93(83.3 kg) = 77.5 kg water
Solids = 0.07(83.3 kg) =5.8 kg solids
N = 0.025(5.8 kg) = 0.146 kg N
C = 0.25(5.8 kg) = 1.46 kg C
16.7 kg refuse contains:
Water = 0.25(16.7 kg) = 4.2 kg water
Solids = 0.75(16.7 kg) = 12.5 kg solids
N = 0 kg N
C = 0.35(12.5 kg) = 4.38 kg C
Mixture contains: 81.7 kg water, 18.3 kg solids, 0.146 kg N, 5.84 kg C
Moisture content = 81.7%, C/N = 5.84 kg C/0.146 kg N = 40
Jumbo Problem
The following is a jumbo problem. That is a problem that takes some time to understand
and requires the student to present a solution as a well-written and detailed report. The time
allocated to solve the problem and prepare a report should be at least five days, and one week
is probably better.
289
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
8.48 CASIO CITY AND THE ATOZINC RAYON COMPANY
Casio City is in violation of its discharge permit. The State is threatening court action if a
plan for improving effluent quality is not submitted within three months and implemented
within one year. The city blames the ATOZINC Rayon Company for causing a variety of
problems at the Casio City Wastewater Treatment Plant. The problem is serious. Tempers
are hot.
ATOZINC has not accepted responsibility for any problems, but they are not uncooperative.
They intend to help solve problems that they cause. For now they want to turn the current
emotional arguments toward a rational scientific investigation of the problem. Your job is
to help ATOZINC accomplish this. To begin you need to identify any problems existing
at the city’s treatment plant, their causes, and to what extent the problems are related to
the wastewater coming from ATOZINC.
Prepare a report that can serve as the basis for objective discussions.
STATEMENT BY ATOZINC RAYON COMPANY
We use zinc (Zn) as a retardant in the critical reactions of the rayon manufacturing process.
The zinc does not enter directly into any of the reactions, but it must be present. It is lost
from the process in several ways; for example, (1) drag out from the reactor on the work
pieces, (2) leaks, and (3) equipment washing.
We have no wastewater treatment at our factory. Cooling water is discharged directly to
the river. We have a discharge permit to do this. All other wastewater (sanitary wastes from
showers, toilets, lunch room, etc.) goes into the city sewer. The average wastewater flow is
3,600 m3/d. The maximum is 4,500 m3/d; this cannot be exceeded because of production
limitations. The flow might be as low as 1,900 m3/d on some occasions. Some data on our
plant effluent, measured on Sept. 24, 2018, are: pH = 1.5 - 3.0, Zn = 80 mg/L, BOD5 =
24 mg/L, and COD = 53 mg/L
STATEMENT BY CASIO CITY
ATOZINC has no waste treatment so they dump their problems on us. Their discharge
is nearly one-half of the flow into our plant. They cause all the operating problems at our
plant, the most serious being with our activated sludge process.
We think they should treat their own waste. We may deny them permission to put anything
into the city sewers. If we are fined for being in violation of our discharge permit, we are
going to sue ATOZINC to recover damages.
290
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Our discharge permit requires that our monthly average effluent BOD5 and SS cannot
exceed 25 mg/L and the effluent must be free of chemicals in toxic amounts. The permit
also requires that the dissolved oxygen in the Three Sisters River be not less than 4 mg/L at
any point between our point of discharge and the next downstream wastewater discharge.
Treatment is by primary settling and conventional activated sludge with final settling. Waste
activated sludge is treated by anaerobic digestion and then spread on farmland.
Some performance data that represent recent conditions are:
Influent:
Flow average = 8,000 m3/d; range = 6,000-15,000 m3/d
BOD5 average = 300 mg/L; range = 200-500 mg/L
TSS average = 275 mg/L; range = 200-350 mg/L
pH = 5.5 - 6.5
Zinc = 20 - 40 mg Zn/L
Effluent:
BOD5 average = 50 mg/L; range = 30-150 mg/L
Ultimate BOD = 1.5 BOD5
TSS average = 60 mg/L; range = 20-100 mg/L
Zinc = 10 - 20 mg/L
Dissolved oxygen = 4.4 mg/L average
Activated sludge operating conditions:
Mixed liquor suspended solids (MLSS) = 1100 mg/L
Sludge volume index (SVI) = 250
DO conc. in aeration basin = 0.2 mg/L (all blowers at full capacity)
Aeration basin volume = 1200 m3
Waste activated sludge (WAS) rate = 40 m3/d at 8,000 mg/L solids
Water quality in the Three Sisters River is in Table P8.48
Water Quality
Above Casio
400 meters below
Indicator
City discharge
discharge
8.5 mg/L
7.6 mg/L
1 mg/L
unmeasured
≤ detectable limit
unmeasured
17°C
17°C
DO
BOD5
Heavy metals
Maximum water temp.
Table P8.48
291
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Dissolved oxygen model for the Three Sisters River
The low river flow (7Q10) to be used for design is 26,000 m3/d. Studies have shown that
a very simple model describes the dissolved oxygen concentration in the Three Sisters River
for at least 6 days flow below the treatment plant discharge. The model is
Dt
where
3.0L0 (e(0.3/d)t e(0.4/d)t ) D0e(0.4/d) t
Dt = dissolved oxygen deficit at t days flow below the discharge, mg/L
L0 = ultimate BOD at t = 0, mg/L
D0 = the dissolved oxygen deficit at t = 0, mg/L
t = time of travel below the discharge, d
Activated sludge efficiency tests
When our plant was expanded in 2010, the designers did some tests on the activated sludge
process. The results are given in Figure P8.48a. ATOZINC was discharging wastewater to
our sewer at the time of the study, but not at full capacity. No zinc measurements were
made at that time. We have not had much time to search for data on how zinc affects
the activated sludge process. One reference (sorry, but the citation has been lost) reports
that 0.3 to 5 mg Zn/L can inhibit the activated sludge process. It further reports that in
two plants, influent zinc was 75% insoluble and that 50% of the zinc was removed by the
treatment process.
500
BOD5 (mg/L)
200
100
MLSS =
1000 mg/L
50
MLSS =
2000 mg/L
20
10
0
2
4
6
8
10
Aeration Time (h)
Figure P8.48a ATOZINC activated sludge test data
Zinc Toxicity Information.
The federal water quality criteria for acute and chronic toxicity of zinc are
Acute toxicity criterion (μg Zn/L) = exp(0.8473 ln(H) + 0.8604)
Chronic toxicity criterion (μg Zn/L) = exp(0.8473 ln(H) + 0.7614)
where H = water hardness measured in mg/L as CaCO3.
292
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Figure P8.48b is from a report on the acute toxicity (as percent of exposed trout surviving
for 96 hours) of zinc to rainbow trout. The 96-hour LC50 is 4.8 mg/L. Fish can survive
longer at lower concentrations, and the death rate is higher at higher concentrations. Other
data may be available for trout, other fish species, and aquatic organisms.
% Surviving after 96 h
120
100
80
60
40
96-h LC50
= 4.8 mg/L
20
0 1
2
3
4
5
6
8
10
Zinc concentration (mg/L)
Figure P8.48b Acute bioassay data for zinc on rainbow trout
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SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Material balance for biological processes
Solution
This is an open-ended problem for class and group discussion. Prepare a report that can serve as the
basis for objective discussions. No solution given.
294
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
9THE UNSTEADY-STATE
MATERIAL BALANCE
9.1
FILTRATION
The pressure filter in Figure P9.1 treats 20,000 m3/h of turbid water to which chemicals
have been added to flocculate colloids and fine clay particles. The influent suspended solids
concentration is 8 mg/L and the effluent is, for all practical purposes, free of particulate
solids. The filter can operate for 12 hours, at which time it is shut down and cleaned by a
backwash with clean water. Write the unsteady-state material balance equation and calculate
each term in the balance. If the backwash water is 1% of the throughput, calculate the
volume of backwash water per cycle, the suspended solids concentration in the backwash
water, and the net production of clean water.
Raw water
Dirty
backwash
water to
disposal
Clean
filtered
water
Filtered
water
Backwash Mode
Filtration Mode
Figure P9.1
Solution
Material balance on solids
Solids Accumulation = Solids Mass In – Solids Mass Out
Influent solids concentration = 8 mg/L = 0.008 kg/m3
Solids Mass in = (20,000 m3/h)(12 h/cycle)(0.008 kg/m3) =1,920 kg/cycle
Solids Mass Out = 0 kg/cycle
Solids Accumulation = 1,920 kg/cycle – 0 kg/cycle = 1,920 kg/cycle
Backwash cleaning will remove 1,920 kg suspended solids
295
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
Assume backwash volume = 1% of throughput
Throughput = (20,000m3/h)(12h) = 240,000 m3/cycle
Backwash volume = 0.01(240,000 m3/cycle) = 2,400 m3/cycle
SS concentration in backwash = (1,920 kg/cycle)/(2,400 m3/cycle)
= 0.8 kg/m3 = 800 mg/L
Net production of clean water per cycle = 0.99(240,000 m3/cycle) = 237,600 m3/cycle
9.2
WETLANDS WATER BUDGET
A small community discharges effluent from an aerated lagoon into a 1,500 hectare peat
wetlands. Calculate the annual outflow from the wetlands using the data in Table P9.2. All
quantities are in the same units (cm).
Net precipitation = Precipitation – Evaporation – Transpiration.
Negative net precipitation indicates that more water was evaporated and transpired than
fell as rain or snow.
296
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
Net Precip.
Year
(cm)
Effluent
Watershed
Added
Runoff
(cm)
(cm)
1989
-153
672
132
1990
-43
622
0
1991
-100
724
73
1992
-250
719
0
Table P9.2
Solution
Annual outflow = Net Precipitation + Wastewater effluent added + Runoff
Example calculation for 1989:
Annual outflow = -153 cm + 672 cm + 132 cm = 651 cm
Year
Net Precip.
(cm)
Effluent
Watershed
Net Annual
Added
Runoff
Outflow
(cm)
(cm)
(cm)
1989
-153
672
132
651
1990
-43
622
0
579
1991
-100
724
73
697
1992
-250
719
0
469
Table S9.2
9.3
BOD LOAD EQUALIZATION
An industry is required to discharge wastewater to a municipal sewer such that the BOD
load, measured in kg/h, is constant over 24 hours. The BOD concentration is constant at
1200 mg/L, but the flow varies over a typical 24-h period as shown in Figure P9.3. Design
an equalization tank that will have a constant outflow over the 24-h work day. (a) What is
the discharge rate? (b) What is the storage volume of the tank?
297
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
Figure P9.3
Solution
Because the BOD concentration is constant (1,200 mg/L) the BOD load will be equalized if the flow
is equalized.
a) Calculate inflow volume over 24 h
Total inflow = (3 h)(5 m3/h + 20 m3/h) + (6 h)(30 m3/h + 15 m3/h + 5 m3/h)
= 75 m3 + 300 m3 = 375 m3
Average flow rate = 375 m3/24 h = 15.6 m3/h
Constant discharge rate = 15.6 m3/h
b) Graphical solution – Draw tangent lines with slope equal to the average outflow rate
(the equalized flow rate). This identifies the points in time when the tank changes status
from filling to emptying. The tank fills from 9 am until 6pm and empties between 6
pm and 6 am. The required storage is estimated graphically as 100 m3.
Cumulative inflow volume (m3)
400
350
300
Equalized outflow
= 375 m3/24 h
= 15.6 m3/h
250
200
Required storage
volume (appx)
= 250 m3 – 150 m3
= 100 m3
150
100
50
0
6AM
0
12N
6
6PM
12
Time of day
Figure S9.3
298
12M
18
6AM
24
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
9.4
The unsteady-state material balance
PUMPING STATION
Pumps draw their input from a reservoir (wet-well) that always contains a minimum depth
of water so the pump will not run dry. The inflow rate is variable so the wet-well inventory
increases and decreases. Figure P9.4 shows a cumulative inflow diagram for a wet well that
will equalize flow over short time periods. The goal is to pump from the reservoir at the
average rate of inflow over the 24-hour period. Use the graph to estimate the volume of
water that must be stored.
Figure P9.4
Solution
The average rate of inflow over 24 hours is 40,000 m3/24 h = 1,667 m3/h. A straight line from (0, 0)
to (24, 40,000) has slope = 1,667 m3/h and would show the average rate outflow. From 0 – 6 h, the
inflow and outflow are nearly equal. From 6-8 the inflow is near zero, but it sharply increases at 8 h and
water accumulates until about 16 h. A volume of 11,000 m3 must be stored between about 8 h and 16
h. From 15-24 h the wet-well empties. The maximum volume that must be stored is about 11,000 m3.
299
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
Figure S9.4
9.5
BATCH REACTOR
A quantity of reactive material is charged into a batch reactor. After allowing time for the
reaction, the material is removed to another processing step, and the cycle is repeated. The
batch reactor may be cleaned between loads. Batch smoothing and blending is done in
three steps, as shown in Figure P9.5. A feed of 10 m3 is dumped into a batch reactor that
contains 50 m3, the contents are heated and mixed for 2 hours to allow the reaction to
proceed. Then a 10 m3 portion is removed. The reactor contents prior to charging have a
concentration of 0.2 kg/m3 and the charged material has a concentration of 4 kg/m3. The
reaction runs according to an exponential decay model
Mt = M0 exp(-t)
where t = time (h)
M0 = mass at t = 0 (kg)
Mt = mass at time t (kg)
Calculate the mass and concentration of the material that is removed.
300
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Step 1 - Addition
The unsteady-state material balance
Step 2 - Blend
Step 3 -Removal
Figure P9.5
Solution
This is a mixing problem with reaction. Two materials are mixed, allowed to react, and the batch
removed has the same concentration as the mixture.
Step 1 Addition
Blender contents: Vblender = 50 m3
Mblender = (50 m3)(0.2 kg/m3) = 10 kg
Added material: Vadded = 10 m3
Madded = (10 m3)(4 kg/m3) = 40 kg
Step 2 Blending at t = 0
Vblender = 50 m3 + 10 m3 = 60 m3
Mblender = 10 kg + 40 kg = 50 kg
Step 2 Reaction for 2 hours
M0 = 50 kg
Mt=2 = (50 kg) exp(-2) = (50 kg)(0.135) = 6.8 kg
Step 3 Removal
Blender mass = 6.8 kg
Blender concentration C = (6.8 kg)/(60 m3) = 0.11 kg/m3
If a second batch is charged, the initial concentration will be calculated using 6.8 kg. Or, the reactor
might be cleaned between batches, in which case the initial concentration will be some small value.
301
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
9.6
The unsteady-state material balance
SMOOTHING CONCENTRATIONS
Table P9.6 gives the volume and concentration of 45 batches of wastewater. Batch volumes
are random values between 4 m3 and 12 m3. Concentrations vary randomly between 100
mg/L and 200 mg/L. Find an effective design for smoothing the concentrations. There is
no single correct answer, because there is no precise definition of ‘effective’, so you will have
to explain your recommendation.
Volume
Conc.
Volume
Conc.
Volume
Conc.
(m )
(mg/L)
(m )
(mg/L)
(m )
(mg/L)
1
12
104
16
7
184
31
5
147
2
10
140
17
8
197
32
8
163
3
12
160
18
12
200
33
9
151
4
7
141
19
7
132
34
12
105
5
5
100
20
12
102
35
7
155
6
6
168
21
12
128
36
9
144
7
4
170
22
8
140
37
5
198
8
6
143
23
10
107
38
8
108
9
9
142
24
6
118
39
11
136
10
6
104
25
4
189
40
9
139
11
9
200
26
9
107
41
6
145
12
12
106
27
5
146
42
8
168
13
12
117
28
12
124
43
6
122
14
8
138
29
12
197
44
9
112
15
11
167
30
5
154
45
9
192
Batch
3
Batch
3
Batch
3
Table P9.6 Wastewater batch volumes and concentrations
Solution
The average batch size put into the blender is 8.42 m3 and that is the volume that is withdrawn at the
end of each blending cycle.
Assumed initial values:
VBlender = 10 m3
CBlender = 150 mg/L
MBlender = (10 m3)(0.150 kg/m3) = 1.5 kg
302
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
Material balances
The volume in the blender after a new batch has been added, blended, and the output removed is
Vi = Vi -1 + BVi – VOut
where
Vi = Volume in blender after each batch has been processed
BVi = Volume of input batch
VOut = Volume of batch removed after blending = 8.42 m3
The mass of material in the blender is
Mi = Mi-1 + BMi – MOut
where
Mi = Mass of material in blender after each batch has been processed
BMi = Mass of material in input batch
MOut = Mass of material in batch removed after blending
The concentration of material during blending is
Ci = (Mi-1+BMi)/(Vi-1 + BVi)
A summary of the calculations are in Table S9.6a.
303
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Batch Characteristics
Batch
Volume
Conc.
Mass
The unsteady-state material balance
Contents During
Blending
Volume
Mass
Conc.
0
Contents After Removal
Volume
Mass
Conc.
10
1.5
150
1
12
104
1.25
22.00
2.75
124.9
13.58
1.70
124.91
2
10
140
1.40
23.58
3.10
131.3
15.16
1.99
131.3
3
12
160
1.92
27.16
3.91
144.0
18.73
2.70
144.0
4
7
141
0.99
25.73
3.68
143.2
17.31
2.48
143.2
5
5
100
0.50
22.31
2.98
133.5
13.89
1.85
133.5
…
…
…
…
…
…
…
…
…
…
10
6
104
0.62
11.20
1.37
122.4
2.78
0.34
122.4
…
…
…
…
…
…
…
…
…
…
20
12
102
1.22
24.98
3.43
137.4
16.56
2.27
137.4
…
…
…
…
…
…
…
…
…
…
30
5
154
0.77
23.76
3.73
157.0
15.33
2.41
157.0
…
…
…
…
…
…
…
…
…
…
45
9
192
1.73
18.42
2.92
158.4
10.00
1.58
158.4
…
…
…
…
…
…
…
…
…
…
8.42
144.67
1.20
21.11
3.00
142.7
12.69
1.80
142.7
Maximum
12
200
2.40
29.71
4.30
188.5
21.29
2.96
188.5
Minimum
4
100
0.50
11.20
1.37
119.0
2.78
0.34
119.0
Range
8
100
1.90
18.51
2.93
69.5
18.51
2.62
69.5
Average
Table S9.6a
We can investigate the benefit of using a larger blending tank. The concentration range without
blending is 100 mg/L. A 10 m3 tank reduces the range to 70 mg/L; a 50 m3 tank reduces it to 28 mg/L.
Blender volume (m3)
10
20
30
40
50
Minimum conc. (mg/L)
119
127
131
133
134
Maximum conc. (mg/L)
189
178
171
166
162
Range (mg/L)
70
51
40
33
28
Table S9.6b
There is no answer for the ‘correct’ tank size. Nor is there a clear answer of whether a blending tank
of any size is a good investment. This would depend on downstream processing requirements that
304
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
are unknown to us. Such questions are not unusual in preliminary design. Someone will say, ‘Make
some calculations to investigate equalization.’ and the calculations, once made, may never lead to
implementation. That is the exploratory aspect of process design.
9.7
FILLING A LEAKY TANK
A 2.0 m diameter storage tank is being filled at the rate of QIn = 2.0 m3/min. When the
height of the liquid in the tank is 2 m, the bottom of the tank springs a leak. The rate of
leaking is proportional to the depth of fluid, h, so that it is leaking at a rate of QOut = 0.4h
m3/min, where h is in m. (Note: for dimensional consistency, the coefficient 0.4 carries
units of m2/min). Calculate and plot the depth of liquid in the tank as a function of time.
Calculate the steady state height of the fluid in the tank.
Solution
The material balance for the change in stored volume, ∆V/∆t, over some small interval of time, ∆t is
∆V/∆t = QIn – QOut = 2 m3/min – (0.4 m2/min)h
At steady-state: ∆V/∆t = 0
The steady-state liquid depth is
h = (2 m3/min)/(0.4 m2/min) = 5 m
The rate of leakage increases as the liquid level rises from 2 m, the level when the leak started, until
a 5 m depth is reached. Until then the rate of inflow exceeds the rate of leakage.
Numerical solution for the change in h with time.
Write volume change in terms of change in depth, h.
'V
A'h
'V
't
S
S
D2 'h
4 't
D2
'h
4
QIn QOut
QIn 0.4h
Solving for ∆h
'h
ht 't ht
4
(QIn 0.4ht )'t
S D2
and
ht 't
ht 4
(QIn 0.4ht )'t
S D2
Adding known values D =2 m and QIn = 2 m3/min yields
305
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
ht 't
ht ht 't
ht 4
S (2 m)2
The unsteady-state material balance
(2 m3 /min - 0.4 m2 /min ht )'t
2 m/min (0.4 /min) ht
S
't
Starting at h0 = 2 m and using ∆t = 0.1 min to illustrate the calculation.
t = 0.1 min
h0.1
min
=2m+
2m/min - 0.4/min(2 m)
S
(0.1 min) = 2.0382 m
t = 0.2 min
h0.2
min
2.0382 m +
2 m/min - 0.4/min (2.0382 m)
S
(0.1min) = 2.0759 m
and so on…
The exact solution is
ht
ª - (0.4/min) t º
5 m - (3 m)exp «
»
S
¬
¼
The approximate and the exact solutions are compared in Table S9.7 and Figure S9.7. The agreement
is excellent. The approximate solution is slightly higher than the exact solution (by about 0.1%). This
error can be reduced by using a smaller Δt. Using Δt = 0.2 gives estimates that are about 0.2% too high.
306
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Interval
The unsteady-state material balance
t
Δh
h + Δh
Exact h
(min)
(m)
(m)
(m)
0
2
1
0.1
0.0382
2.0382
2.0379
2
0.2
0.0377
2.0759
2.0752
3
0.3
0.0372
2.1131
2.1121
4
0.4
0.0368
2.1499
2.1486
5
0.5
0.0363
2.1862
2.1846
…
…
…
…
…
10
1
0.0340
2.3608
2.3578
20
2
0.0299
2.6782
2.6729
40
4
0.0232
3.2031
3.1949
60
6
0.0179
3.6094
3.5998
80
8
0.0139
3.9237
3.9139
100
10
0.0107
4.1671
4.1575
150
15
0.0057
4.5611
4.5535
200
20
0.0030
4.7687
4.7634
300
30
0.0008
4.9358
4.9336
500
50
0.0001
4.9951
4.9948
Table S9.7
Figure S9.7
307
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
9.8
The unsteady-state material balance
DILUTION OF A SALT SOLUTION
A tank holds 500 L of a salt-water solution in which 4.0 kg of salt are dissolved. Pure water
runs into the tank at the rate of 25 L/min and salt solution overflows at the same rate.
Mixing in the tank is adequate to keep the concentration of salt in the tank uniform at
all times. The salt solution in the tank initially is (4kg)/(504 kg) = 0.8% salt. The density
of a 0.8% salt (NaCl) solution is 1.0043 kg/L, essentially the same as that of pure water.
For convenience, assume the salt solution and the water have the same density of 1 kg/L.
Plot the concentration of salt exiting the tank and calculate the mass of salt in the tank at
the end of 50 min.
Solution
The general unsteady state material balance on the mass of salt in the tank is
V
'C
't
QInCIn QOutCOut
where t = time (min)
V = volume of salt solution in the tank (L)
QIn and QOut = Volumetric flow rate of liquid into and out of the tank
CIn and COut = concentrations of NaCl flowing in and out of the tank
The inflow is clean water, so CIn = 0
'C
't
QOut COut
V
Because the tank is completely mixed, the concentration of salt in the tank, C, is the same as that in
the outflow.
C = COut
Iterative numerical solution for time varying salt concentration. Solve the material balance for ΔC
'C = Ct 't Ct = and
Ct 't = Ct QOut Ct
't
V
QOut Ct
Q 't º
ª
't = Ct «1 Out »
V
V ¼
¬
The mass of salt in the tank at any time is Mt = CtV
The exact solution for the time varying salt concentration is
Ct
ª Q
º
C0 exp « Out t »
V
¬
¼
308
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
A comparison of the iterative (∆t = 0.1 min) and exact solutions is in Table S9.8 and Figure S9.8. After
50 min, the concentration of salt in the tank is 0.653 g/L and the mass remaining in the tank is 0.326 kg.
Iterative Solution
Exact Solution
t
∆C
C
M
C
M
(min)
(g/L)
(g/L)
(kg)
(g/L)
(kg)
8
4
8
4
0
1
-0.0382
7.6089
3.8044
7.6098
3.8049
2
-0.0364
7.2369
3.6184
7.2387
3.6193
3
-0.0346
6.8831
3.4415
6.8857
3.4428
4
-0.0329
6.5466
3.2733
6.5498
3.2749
5
-0.0313
6.2265
3.1133
6.2304
3.1152
…
…
…
…
…
…
10
-0.0244
4.8462
2.4231
4.8522
2.4261
…
…
…
…
…
…
15
-0.0190
3.7718
1.8859
3.7789
1.8895
…
…
…
…
…
…
20
-0.0148
2.9357
1.4678
2.9430
1.4715
…
…
…
…
…
…
30
-0.0089
1.7783
0.8892
1.7850
0.8925
…
…
…
…
…
…
40
-0.0054
1.0773
0.5386
1.0827
0.5413
…
…
…
…
…
…
50
-0.0033
0.6526
0.3263
0.6567
0.3283
Table S9.8
309
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
Figure S9.8
9.9
CSTRS IN SERIES
Calculate and plot the tracer dispersion curve for two CSTR reactors in series with an
instantaneous step-change in the input concentration from C0 = 0 to C0 = 40 mg/L. The
volume of each tank is V= 6 m3 and the flow rate is Q = 3 m3/min.
C0
C1
Figure P9.9
Solution
Define
C0 = Influent to Tank 1
C1 = Effluent from Tank 1 = Influent to Tank 2
C2 = Effluent from Tank 2
The model for the first CSTR is
V
'C1
't
QC0 QC1
Q
1
C0 C1 't
C C1 't
V
T 0
C1 t 't
C1 t 'C1 t
'C1
310
2
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
The model for the second CSTR is
V
'C2
't
QC1 QC2
Q
1
C1 C2 't
C C2 't
V
T 1
C2 t 't
C2 t 'C2 t
'C2
The system has V = 12 m3, Q = 3 m3/min, θ = V/Q = 4 min.
Each tank has V = 6 m3, Q = 3 m3/min, θ = V/Q = 2 min.
These are the calculations for the first 3 time steps for Δt = 0.1 min.
ΔC1
0.1
ΔC2
0.1
ΔC1
0.2
ΔC2
0.2
ΔC1
0.3
ΔC2
0.3
=
=
=
=
=
=
1
40 mg/L - 0 mg/L 0.1 min = 2.0 mg/L
2 min
C1 0.1 = 0 mg/L + 2.0 mg/L = 2.0 mg/L
1
0 mg/L - 0 mg/L 0.1 min = 0 mg/L
2 min
C2 0.1 = 0 mg/L + 0 mg/L = 0 mg/L
1
40 mg/L - 2.0 mg/L 0.1 min = 1.9 mg/L
2 min
C1 0.2 = 2.0 mg/L + 1.9 mg/L = 3.9 mg/L
1
2.0 mg/L - 0 mg/L 0.1 min = 0.1 mg/L
2 min
C2 0.2 = 0 mg/L + 0.1 mg/L = 0.1 mg/L
1
40 mg/L - 3.9 mg/L 0.1 min = 1.805 mg/L
2 min
C1 0.3 = 3.9 mg/L + 1.805 mg/L = 5.705 mg/L
1
3.9 mg/L - 0.1 mg/L 0.1 min = 0.19 mg/L
2 min
C2 0.3 = 0.10 mg/L + 0.19 mg/L = 0.29 mg/L
A summary of the numerical solution is in Table S9.9 and Figure S9.9.
311
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
t
∆C1
C1
∆C2
C2
(min
(mg/L)
(mg/L)
(mg/L)
(mg/L)
0
0
0
0.1
2.000
2.000
0.000
0.000
0.2
1.900
3.900
0.100
0.100
0.3
1.805
5.705
0.190
0.290
0.4
1.715
7.420
0.271
0.561
0.5
1.629
9.049
0.343
0.904
0.6
1.548
10.596
0.407
1.311
0.7
1.470
12.067
0.464
1.775
0.8
1.397
13.463
0.515
2.290
0.9
1.327
14.790
0.559
2.848
1
1.260
16.051
0.597
3.446
…
…
…
…
…
1.5
0.975
21.468
0.719
6.838
…
…
…
…
…
2
0.755
25.661
0.755
10.566
…
…
…
…
…
3
0.452
31.414
0.690
17.858
…
…
…
…
…
4
0.271
34.860
0.555
24.037
…
…
…
…
5
0.162
36.922
0.418
28.823
…
…
…
…
…
7
0.058
38.897
0.211
34.832
…
…
…
…
…
10
0.012
39.763
0.065
38.517
Table S9.9
312
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
Figure S9.9
9.10 SLUDGE DIGESTERS
The egg-shaped anaerobic sludge digesters in Figure P9.10 use very efficient mixers to
eliminate the formation of supernatant or scum layers. All raw sludge that enters the
digesters leaves as digested sludge, except for the portion that is converted to gas, which is
withdrawn from the top of the tanks. The gas has a very large value economically, but its
mass is unimportant in the material balance calculations for this problem. This problem is
about the startup of the digesters. The gas-producing microorganisms grow slowly and they
are sensitive to overloading that will cause a decrease in the pH, which must be neutral. The
organic load must be increased gradually to establish a healthy population of gas producing
microorganisms. At the outset the digester has been filled with 4,000 m3 of digested sludge
from a neighboring treatment plant. This sludge is 2% total solids and 1% volatile solids.
Normal operation will be a feed rate of 200 m3/d at 6% total solids and 4% volatile solids.
The sludge removal rate will be 200 m3/d. Assume that 60% of the volatile solids input is
converted to gas every day.
Calculate, starting from the time normal feeding begins until steady state is reached, (a) the
mass (kg) of total solids and volatile solids in the digester and (b) the mass flow rate (kg/d)
of total solids and volatile solids out of the digester. Assume the sludge in the digester at
startup, the feed sludge, and the sludge withdrawn have the same density.
313
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
Figure P9.10
Solution
The digester is very well mixed; the contents are homogeneous.
The feed is continuous at 200 m3/d
Digester detention time = 4,000 m3/(200 m3/d) = 20 d
Assume all sludge flows have density = 1,000 kg/m3
Steady state is the condition when the reactor contents and the reactor outputs are not changing. A
mixed reactor will reach 95% of steady state in about 3 detention times. Calculation must cover 60+
days to reach steady state.
Initial conditions: Digester Contents at t = 0 d
Sludge volume = 4,000 m3
Sludge mass = (4,000 m3)(1,000 kg/m3) = 4,000,000 kg digested sludge
Total Solids (TS) = (0.02)(4,000,000 kg) = 80,000 kg TS
Volatile Solids (VS) = (0.01)(4,000,000 kg) = 40,000 kg VS
Inert solids (IS) = TS – VS = 40,000 kg
Basis for calculations:
Sludge and solids added and converted to gas each day
Sludge volume added = 200 m3
Sludge mass added = (200 m3)(1,000 kg/m3) = 200,000 kg
Mass TS added = (0.06)(200,000 kg) = 12,000 kg TS
Mass VS added = (0.04)(200,000 kg) = 8,000 kg VS
Mass IS added = 4,000 kg IS
Mass VS converted to gas = 0.6(Mass VS added)
= 0.6(8,000 kg) = 4,800 kg VS
314
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
Sludge and solids in outflow each day
Sludge volume of outflow = 200 m3
Sludge mass in outflow = (200 m3)(1,000 kg/m3) = 200,000 kg
Mass fraction of digester contents (IS, VS, TS) in outflow
= (200,000 kg)/(4,200,000 kg) = 1/21
Mass IS in outflow = (Mass IS in tank)/21
Mass IS in tank during day = Mass IS start + Mass IS added
Mass VS in outflow = (Mass VS in tank)/21
Mass VS in tank during day
= Mass VS start + Mass VS added – Mass VS converted to gas
Mass TS in outflow = Mass IS in outflow + Mass VS in outflow
Sludge and solids in tank at end of day
Mass IS in tank at end of day = Mass IS in tank during day - Mass IS in outflow
Mass VS in tank at end of day = Mass VS in tank during day - Mass VS in outflow
Mass TS in tank at end of day
= Mass IS in tank at end of day + Mass VS in tank at end of day
Summary of Mass Balances over a one day time interval
IS
end
= IS start + IS
IS start = IS
IS
outflow
VS
end
VS
to gas
VS
outflow
TS
end
TS
)/21 = (IS
in tank
= VS start + VS
VS start = VS
outflow
end
= (VS
end
= IS
added
start
+ IS
)/21
added
– VS to gas - VS
outflow
for prior day
= (0.6)(VS
= IS
outflow
for prior day
end
= (IS
– IS
added
)
added
)/21 = (VS
in tank
+ VS
outflow
start
+ VS
added
- VS
to gas
)/21
end
+ VS
outflow
Iterative Numerical Solution
Calculations for Day 1
Inert Solids:
IS start = 40,000 kg
IS added = 4,000 kg
IS in tank = 40,000 kg + 4,000 kg = 44,000 kg
IS outflow = (44,000 kg)/21 = 2095 kg
IS end = 40,000 kg + 4,000 kg – 2095 kg = 41,905 kg
Volatile Solids:
VS start = 40,000 kg
VS added = 8,000 kg
VS to gas = (0.6)(8,000 kg) = 4,800 kg
315
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
VS in tank = 40,000 kg + 8,000 kg – 4,800 kg = 43,200 kg
VS outflow = (43,200 kg)/21 = 2057 kg
VS end = 40,000 kg + 8,000 kg – 4,800 kg – 2057 kg = 41,143 kg
Total Solids
TS start = IS start + VS start = 40,000 kg + 40,000 kg = 80, 000 kg
TS outflow = IS outflow + VS outflow = 2095 kg + 2057 kg = 4142 kg
TS end = IS end + VS end = 41,905 kg + 41,143 kg = 83048 kg
Calculations for Day 2
Inert Solids:
IS start = 41,905 kg
IS added = 4,000 kg
IS in tank = 41,905 kg + 4,000 kg = 45,905 kg
IS outflow = (45,905 kg)/21 = 2186 kg
IS end = 41,905 kg + 4,000 kg – 2186 kg = 43,719 kg
Volatile Solids:
VS start = 41,143 kg
VS added = 8,000 kg
VS to gas = (0.6)(8,000 kg) = 4,800 kg
VS in tank = 41,143 kg + 8,000 kg – 4,800 kg = 44,343 kg
VS outflow = (44,343 kg)/21 = 2112 kg
VS end = 41,143 kg + 8,000 kg – 4,800 kg – 2112 kg = 42,231 kg
Total Solids
TS start = IS start + VS start = 41,905 kg + 41,143 kg = 83,048 kg
TS outflow = IS outflow + VS outflow = 2186 kg + 2112 kg = 4298 kg
TS end = IS end + VS end = 43,719 kg + 42,231 kg = 85,950 kg
The calculations for subsequent days are obtained by iterative solution of the mass balance equations
until steady state is reached. Summary results are in Table S9.10 and Figure S9.10.
316
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Inert Solids (kg)
In tank
In tank
Day Added during Outflow
at end
The unsteady-state material balance
Total Solids (kg)
Volatile Solids (kg)
In tank
In tank
In tank
Conv.
Added
during Outflow at end Outflow at end
to gas
of day
day
of day
40,000
40,000
0
80,000
8,000 4,800 43,200
2,057
41,143
4,152
83.048
8,000 4,800 44,343
2,112
42,231
4.298
85,950
0
1
2
day
40,000
4,000 44,000
4,000 45,905
2,095
2,186
of day
40,000
41,905
43,719
3
4
5
4,000
4,000
4,000
47,719
49,446
51,092
2,272
2,355
2,433
45,446
47,092
48,659
8,000
8,000
8,000
4,800
4,800
4,800
45,431
46,468
47,455
2,163
2,213
2,260
43,268
44,255
45,195
4,436
4,567
4,693
88,714
91,347
93,854
10
4,000
58,216
2,772
55,443
8,000
4,800
51,729
2,463
49,266
5,235
104,710
20
4,000
68,171
3,246
64,924
8,000
4,800
57,702
2,748
54,955
5,994
119,879
40
4,000
78,034
3,716
74,318
8,000
4,800
63,620
3,030
60,591
6,745
134,909
80
4,000
83,153
3,960
79,193
8,000
4,800
66,692
3,176
63,516
7,135
142,709
120
4,000
83,880
3,994
79,885
8,000
4,800
67,128
3,197
63,931
7,191
143,817
Table S9.10
One check on the steady state is the inert solids.
Inert solids input = 4,000 kg/d.
Inert solids output becomes 4,000 kg/d on day 170.
Inert solids reach 95% of steady state on day 66 – about 3 detention times.
and 99% of the steady state value on day 80 – about 4 detention times.
For practical purposes, steady state is reached at day 80.
Figure S9.10
317
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
9.11 LAKE POLLUTION
A lake of volume, V = 180,000 m3 is fed by a river discharging QIn = 50,000 m3/y. Evaporation
from the lake is 10,000 m3/y and the lake’s outflow, QOut, is (50,000 m3/y – 10,000m3/y)
= 40,000 m3/y. The river carries in a pollutant at concentration CIn = 12 mg/L. We will
assume that the lake is nearly homogeneous, enough so that there will be no serious error
in assuming that the concentration of pollutant, C, is uniform throughout the lake. The
pollutant decays at a rate that is proportional to the concentration of the pollutant. The
proportionality constant is called a decay coefficient, and is denoted by k, with units of
1/y. Solve for the steady state pollutant concentration in the lake and in the river outflow
(because of the homogeneous conditions, these two concentrations are the same) using k
= 0.2/y.
Solution
The material balance model for a reactive pollutant in a well-mixed lake is
V
'C
't
QInCIn QOut COut kVCOut
At steady state, the pollutant concentration does not change with time, and ∆C/∆t = 0
Solving for COut
COut =
QInCIn
(50,000 m3 /y)(12 mg/L)
=
= 7.89 mg/L
QOut kV
(40,000 m3 /y) + (0.2/y)(10,000 m3 )
9.12 CHLORIDES IN THE GREAT LAKES
The diagrams in Figure P9.12 show the increase in chlorides in the five Great Lakes. Explain
the rapid increase from 1900 to about 1970. Explain why the concentrations decreased for
several years in Lakes Ontario, Erie and Huron, and then started to increase again.
318
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
5
30
Superior
Chloride Conc. (mg/L)
0
Michigan
10
5
0
The unsteady-state material balance
30
Erie
25
25
20
20
15
15
10
10
5
5
Huron
5
0
1800
0
0
1850
1900
1950
2000
Ontario
1800
1850
1900
1950
2000
1800
1850
1900
1950
2000
Year
Figure P9.12
Solution
This is an open-ended discussion question that is meant to provoke class or group discussion. No
solution is provided.
9.13 PCB IN LAKE TROUT
Figures P9.13a and 9.13b show the history of PCBs in Great Lakes lake trout. Figure P9.13a
compares the trends for Lakes Michigan and Superior. Figure P9.13b shows more detailed
data for PCBs in Lake Ontario lake trout The red values are Arochlor 1254, measured by
Environment Canada. The blue values are the sum of PCB congeners measured by the
U.S. EPA.
(a) Why is Lake Michigan the most heavily polluted? (b) What was the source of the PCBs
in Lake Michigan? (c) Why have the PCB concentrations in Lake Michigan dropped so
rapidly from 1974 to 1984? Why is the rate of decline so much slower after 1984? (d) Why
has the cleaning process of the lake has taken so many years?
You can have a discussion about these questions based on your general knowledge about
lakes, the Great Lakes in particular, and PCB. For extra credit read about the PCB problem
in the Great lakes. A good reference is McGoldrick, D, Clark, M & Murphy, E 2012.
‘Contaminants in Whole Fish’, State of the Great Lakes 2012.
319
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
The unsteady-state material balance
25
PCBs (ppm)
20
15
Lake Michigan
10
5
0
Superior
70 72 74 76
78
80 82 84
86
88 90 92
Year
94
96 98 00 02
Figure P9.13a PCB concentrations in lake trout in Lakes Superior and Michigan
Figure P9.13b PCB concentrations in lake trout in Lake Ontario
Solution
This is an open-ended discussion question that is meant to provoke class or group discussion. No
solution is provided.
320
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Water conservation and reuse
10WATER CONSERVATION
AND REUSE
10.1 INDUSTRIAL WATER CONSUMPTION
Look up the water consumption for a food industry, a refinery, a textile industry, and
phosphate fertilizer plant. How much water do they use per unit of product? Where is it
used: cooling, cleaning, added to product, etc.?
Solution
This is an open-ended question that is meant to encourage research on industrial water consumption.
There are many correct answers.
10.2 COOLING WATER
Water for cooling is one of the biggest water demands in industry, including many industries
outside the thermal power generation industry. (a) Look up the cooling water use in five
or six industries. (b) Discuss how reducing energy use in an industry will reduce water use.
Solution
This is an open-ended question that is meant to encourage research on industrial cooling water. No
solution is provided.
10.3 WATER RECYCLE SAVES MONEY
Some simple changes in an industry - installing a few pumps, some piping, and a clarifier saved 400,000 m3 of freshwater per year and reduced wastewater treatment demand by the
same amount. The amortized cost of the equipment is $175,000 per year (over 10 years)
and the annual operating costs for labor, power, chemicals and equipment maintenance
is $40,000 per year. The cost for water is $1/m3 and the cost for wastewater disposal is
$1.30/m3. Calculate the annual savings, the annual costs, and the net annual savings. The
payback time = (total annual cost, $/y)/(net annual savings, $). How long does it take for
the project to payback its cost?
321
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Water conservation and reuse
Solution
Water savings = (400,000 m3)($1/m3) = $400,000/y
Wastewater savings = (400,000 m3)($1.30/m3) = 520,000/y
Total annual savings = $920,000/y
Cost of equipment (installed) = $175,000/y (amortized over 10 years)
Cost of operation = $40,000/y
Annual cost = $215,000
Net annual savings = $920,000/y - $215,000/y = $705,000/y
Payback time = (total annual cost)/(net annual savings)
= ($215,000)/($705,000/y) = 0.3 y
The project pays for itself more than 3 times per year.
10.4 WATER FLOW IMPACTS CASH FLOW
The cost of water is €0.56/m3 and the sewer fee is €2.00/m3 (average costs in Denmark in
2000). The city has enacted a surcharge of 20% on customers who increase water use above
a base of 8,000 m3/y. This is a powerful incentive to decrease water use and not increase it.
Calculate (a) the savings by reducing water use by 10%, 25% and 50% and (b) the cost of
increasing water use by 10%, 25% and 50% and (Note: Danish water prices are used as a
reference but this problem is not based on a Danish water policy.)
Solution
Base water use = 8,000 m3/y
Water fees = (€ 0.56/m3)(8,000 m3/y) = € 4,480/y
Sewer fees = (€ 2.00/m3)(8,000 m3/y) = € 16,000/y
City surcharge fees = 0.2(Water fees above 8,000m3/y)
Total cost per year = Water fees + Sewer fees + City surcharge fees
Net economic impact = (Total cost for increased [decreased] use) – (Total cost for base use)
The calculations are summarized in Table S10.4.
322
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Cost Factor
Water conservation and reuse
Deceased Use
Base
50%
25%
10%
Water use (m3/y)
4,000
6,000
7,200
Water fees (€)
2,240
3,360
Sewer fees (€)
8,000
Total water & sewer
(€)
10,240
Increased Use
10%
25%
50%
8,000
8,800
10,000
12,000
4,032
4,480
4,928
5,600
6,720
12,000
14,400
16,000
17,600
20,000
24,000
15,360
18,432
20,480
22,528
25,600
30,720
986
1,120
1,344
23,514
26,720
32,064
3,034
6,240
11,584
City surcharge fees
(€)
Total cost per year
(€)
10,240
15,360
18,432
Net economic
impact (€)
-10,240
-5,120
- 2,048
20,480
Table S10.4
10.5 INDUSTRIAL WATER BALANCE
Figure P10.5 shows an industry that has no recycle or reuse in their water network.
Stormwater water runoff (not shown on the diagram) mixes with other wastewater streams.
Sanitary sewage includes wastewater from kitchens, toilets, showers, laundries, etc. Process
wastes are mixed with boiler blowdown and once-through cooling water. (a) Explain why
the water out is less than the water in. (b) Propose changes that will reduce the use of city
water and the volume of wastewater discharged to the Mississippi River.
Figure P10.5
323
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Water conservation and reuse
Solution
a) Water losses are due to evaporation from cooling towers, steam loss, and water
incorporated into the manufactured product.
b) Multiple possible solutions, but all should include:
Separation of stormwater runoff from the other waste streams.
Separation of process wastewater and sanitary sewage from boiler and cooling tower
blowdown, and from the water softener waste.
The boiler and cooling tower blowdown, and water softener waste may need to be treated,
but the technology will be different than what is used for sanitary sewage and process wastes.
Depending on the characteristics of the process waste, it might be combined with or
segregated from the sanitary sewage.
10.6 COOLING TOWER
A forced-draft cooling tower operates at 5 cycles of recirculation, with makeup water flow
of 100 m3/d. (a) Calculate the blowdown. (b) Calculate the evaporation. (c) Calculate the
ratio of TDS in the blowdown to the TDS in the makeup water. (d) Calculate how much
the blowdown could be reduced if the system could operate at TDSBD/TDSM = 8 cycles.
Solution
By definition
Cycles =
TDSBD
M
=
TDSM
BD
Drift is included in blowdown.
By material balance
M = BD + E
Combining with the definition of cycles
Cycles =
TDSBD
M
M
E
=
=
= 1+
TDSM
BD
M-E
BD
a) BD = M/Cycles = (100 m3/d)/5 = 20 m3/d
b) E = M – BD = 100 m3/d – 20 m3/d = 80 m3/d
c) By definition TDSBD/TDSM = 5
d) Changing the cycles of operation, blowdown, and makeup does not change the
evaporation, which is a function of temperature and remains: E = 80 m3/d
324
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
New Cycles =
Water conservation and reuse
TDSBD
M
M
=
=
= 8
TDSM
M - E
M - 80 m3 /d
8 M – 640 m3/d = M
M = (640 m3/d)/7 = 91.4 m3
10.7 COOLING TOWER WITH RECYCLE MAKEUP WATER
A cooling tower system uses 120 m3/d of makeup water that is purchased from the city. The
makeup has a total dissolved solids (TDS) concentration of 200 mg/L. There is a proposal
to change the makeup water to 50% high quality wastewater effluent plus 50% city water.
The reclaimed wastewater has a TDS = 800 mg/L. (a) Calculate the water balance with
100% city water as makeup, using E = 100 m3/d. (b) Calculate the cycles and TDS of the
blowdown for the 100% city water operation. (c) For the proposed water reuse scheme,
calculate the TDS of the blended water mixture, the cycles of operation, the blowdown,
and the makeup water requirement.
Solution
a) 100% city water operation:
The original water balance for city water makeup is
M = Makeup = 120 m3/d
E = Evaporation = 100 m3/d
BD = Blowdown = 20 m3/d (drift is included in the blowdown)
b) Cycles for 100% city water operation
Cycles = M/BD = (120 m3/d)/(20 m3/d) = 6
Cycles = TDSCity/TDSBD = 6
TDSBD = 6 TDSCity = 6(200 mg/L) = 1,200 mg/L
c) Operation with 50% reclaimed water (Evaporation remains 100 m3/d)
TDS of makeup water mixture = (200 mg/L + 800 mg/L)/2 = 500 mg/L
Cycles = TDSBD/TDSMix = (1,200 mg/L)/(500 mg/L) = 2.4
BD = E/(Cycles – 1) = (100 m3/d)/1.4 = 71.4 m3/d
M = E + BD = 100 m3/d+ 71.4 m3/d = 171 m3/d
The new operation uses 86 m3/d reclaimed water plus 86 m3/d city water. This is a reduction
in city water purchases of 34 m3/d, or 12,500 m3/y, a substantial savings.
325
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Water conservation and reuse
10.8 COOLING TOWER WATER CONSUMPTION
A cooling system with 300 tons of refrigeration capacity operates 250 days per year at 2.5
cycles of recirculation. Evaporation is 9 gal/min. The combined cost of for city water and
wastewater service is $10.90 per 1,000 gallons.
a) Calculate the blowdown, makeup, and total water use, in gal/day or operation and
gal/year.
b) The cycles have been increased from 2.5 to 5.0 by installing a conductivity controller,
and a pH controller. Calculate the blowdown, makeup, and total water use, in gal/
day of operation and gal/year.
c) Calculate the savings in water use, in water and sewer charges.
d) The capital cost (installed equipment, etc.) of the project is $25,000. Calculate
the payback period, in years. Payback period = (total annual cost)/(net savings/y)
Solution
a) Pre-conservation project
Evaporation = (9 gal/min)(1,440 min/d) = 12,960 gal/d
Blowdown = (Evaporation)/(Cycles – 1) = (12,960 gal/d)/(2.5 – 1) = 8,640 gal/d
Makeup = Evaporation + Blowdown = 12,960 gpd + 8,640 gpd = 21,600 gpd
Annual makeup water use = (21,600 gal/d)(250 d/y) = 5,400,000 gal/y
b) Post-conservation project
Evaporation does not change = 12,960 gal/d
Blowdown = (Evaporation)/(Cycles – 1) = (12,960 gal/d)/(5.0 – 1) = 3,240 gal/d
Makeup = Evaporation + Blowdown = 12,960 gal/d +3,240 gal/d = 16,200 gal/d
Annual makeup water use = (16,200 gal/d)(250 d/y) = 4,050,000 gal/y
c) Annual savings in cost of makeup water
Makeup water savings = 5,400,000 gal/y - 4,050,000 gal/y = 1,350,000 gal/y
City water & sewer charges = (1,350,000 gal/y)($10.90/1,000 gal) = $14,715/y
d) Payback period = (total annual cost)/(net savings/y)
Annual cost of makeup water = (4,050,000 gal)($10.90/1,000 gal) = $44,145
Payback period = (Annual cost of water)/(net savings/y)
= ($44,145)/($14,715/y) = 3 y
326
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Water conservation and reuse
10.9 WATER REUSE FOR COOLING SYSTEMS
A company buys 500 m3/d of city water for cooling. They wish to reduce this by blending
city water with recycled process water. The city water has a TDS concentration of 300
mg/L. The maximum operating concentration in the blowdown (and the circulating water)
is TDS= 1,800 mg/L. Higher concentrations cause serious problems with corrosion and
mineral scaling. For these conditions the system operates at 6 cycles of concentration. The
TDS discharge is 150 kg/d. This is the maximum allowed for blowdown discharge under
the current wastewater discharge permit. Process wastewater from the five sources listed in
Table P10.9 is available for reuse. Determine a plan minimizes city water use.
Wastewater
TDS
Available Flow
for recycle
(mg/L)
(m3/d)
Scrubber water
3,000
40
Process water A
460
24
Process water B
230
200
Process water C
570
80
RO reject
2,400
4
City water
300
500
Table P10.9
Solution
Start with some simple calculations and observations Table S10.9.
Convert mg/L to kg/m3 (1 mg/L = 1 ppm = 1,000 kg/m3; ppm/1,000 = kg/m3)
Wastewater
TDS
Available Flow
TDS Mass
(kg/m )
3
for recycle
(m /d)
(kg/d)
Scrubber water
3
40
120
Process water A
0.46
24
11
Process water B
0.23
200
46
Process water C
0.57
80
46
RO reject
2.4
4
10
348
232.2
500
150
3
Totals
City water
0.3
Table S10.9
327
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Water conservation and reuse
a) Calculate the blowdown and evaporation flow rates
The maximum TDS loading for the makeup water, at 300 mg/L TDS.
= (500 m3/d)(0.3kg/m3) = 150 kg/d
TDS loading in blowdown = 150 kg/d
At 6 cycles of operation this gives a blowdown concentration of
TDSBD = (Cycles)(TDSM) = 6(0.3 kg/m3) = 1.8 kg/m3
The blowdown flow rate is the mass flow divided by its concentration
BD = (150 kg/d)/(1.8 kg/m3) = 83.3 m3/d
Evaporation and drift is the difference between makeup and blowdown
E = M – BD-= 500 m3/d – 83.3 m3/d = 416.7 m3/d
b) Can the cooling system operate only on recycled process wastewater and stay at 6
cycles?
The total wastewater flow available for recycle = 348 m3/d
If all the process wastewater can be recycled, the city water use is reduced by 348 m3/d
Required city water purchase = 500 m3/d – 348 m3/d = 152 m3/d
Total TDS load of all 5 recycle streams is 232 kg/d
232 kg/d >150 kg/d TDS loading limit to operate at 6 cycles
No, the cooling system cannot operate using all recycled wastewater streams and stay
at 6 cycles.
c) Blend all recycle water with city water and calculate the new cycles.
TDS load from all wastewater streams = 232 kg/d
TDS from 152 m3/d city water = (152 m3/d)(0.3 kg/m3) = 46 kg/d
Total TDS load of blended recycled wastewater as makeup
= 232 kg/d + 46 kg/d = 278 kg/d
TDS concentration of makeup water = (278 kg/d)/(500 m3/d) = 0.556 kg/m3
To maintain the desired 1.8 kg/m3 TDS level in the system the cycles of concentration
must be reduced
New Cycles = TDSBD/TDSM = (1.8 kg/m3)/(0.556 kg/m3)
=
3.2
3.2 cycles is acceptable since it recycles the wastewater and saves 348 m3/d of city water.
10.10
WATER REUSE
Figure P10.10 shows four water-consuming operations with no recycle or reuse. Assuming
there are no restrictions because of water quality, propose 3 water reuse schemes. You may
not be able to reuse all the water.
328
SOLVED MATERIAL BALANCE PROBLEMS:
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Water conservation and reuse
'
SOT/h
F resh water
130 T/h
..
10 T/h evaporation
Operation
1
20T/h
.. Operation
2
SOT/h
.
Operation
40T/h
20T/h .
SOT/h
3
lOT/h . Operation
.
.
lOT/h
Wastewat er
120T/h
.
.
Figure P10.10
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SOLVED MATERIAL BALANCE PROBLEMS:
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Water conservation and reuse
Solution
There are many correct answers.
10.11
RECYCLE AND REUSE
A factory has three processes (Figure P10.11), each supplied with freshwater with a total
dissolved solids (TDS) concentration of C =10 ppm TDS (0.01 kg TDS/m3). The total
freshwater supply is F = 2,100 m3/d. (Hereafter understand that F is in m3/h and C is in
kg TDS/m3.) Process 1 discharges a concentrated dye waste (F1 = 100, C1 = 10), process 2
produces a high strength waste (F2 = 1,000, C2 = 3), and process 3 has a low-strength effluent
(F3 = 1,000, C3 = 0.5). Diagram a system with water recycle and show the material balance.
Figure P10.11
Consider two options: (a) Recycle but no regeneration and (b) Regeneration and recycle.
a) Diagram a system with water recycle and show the material balance. The mass of
total solids in the recycled wastewater will be added to the mass of pollutant that
exists when there is no recycle.
b) Evaluate a proposal to regenerate the low strength waste from Process 3 for reuse.
The regeneration will use a membrane process that yields 80% of the feed as a
permeate with CP = 10 ppm (0.01 kg TDS/m3), and 20% as a concentrated reject.
The permeate will be recycled and the reject will be combined with the other
wastewater for treatment. Make the material balance for the proposed system.
330
SOLVED MATERIAL BALANCE PROBLEMS:
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Water conservation and reuse
Solution
a) Recycle with no regeneration
The attractive recycle possibility is to use the 1,000 m3/h effluent from Process 3 as the feed
to Process 2 as shown in Figure S10.11a. The (0.5 kg/m3 )(1,000 m3/h) = 500 kg/h solids in the
recycle passes through Process 2 and does not change the 3,000 kg/h mass of solids added
by the process. The Process 2 output then carries 3,500 kg/h of solids in a flow of 1,000 m3/h
and has a concentration of 3.5 kg/m3. This mixes with the 100 m3/h effluent from Process 1.
The total mass of pollutant leaving the system is unchanged, and is (1,000 kg/h + 3,500
kg/h) = 4,500 kg/h. The required freshwater input is reduced from 2,100 m3/h to 1,100 m3/h.
The wastewater volume going to treatment is also reduced to 1,100 m3/h and the waste is
more concentrated, C = (4,500 kg/h)/(1,100 m3/h) = 4.091 kg/m3. This may be an advantage
(depending on the type of treatment that is used).
Figure S10.11a System with Process 3 output recycled as input to Process 2
Is there a chance this will not work? Of course, preliminary designs that exist only on paper
may fail when tested with more detailed information. This does not make them bad proposals,
but simply proposals that need more study or testing.
One reason it might not work is our use of a single pollutant, total solids, as a measure of
pollution. There may be other pollutants or substances in the recycle that will interfere with
Process 2. There are ways to test for this before going to full scale, and certainly this would
be done.
b) Regeneration and recycle
The membrane process (Figure S10.11b) splits the 1,000 m3/h waste stream from Process 3
into a high quality permeate, FP = 800 m3/h at 0.01 kg/m3, and a concentrated reject, FR =
331
SOLVED MATERIAL BALANCE PROBLEMS:
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Water conservation and reuse
200 m3/h at an as yet unknown concentration. The value is not needed because we know the
mass of pollutant leaving the system will be the same as in the original system. The volume
will be reduced and the concentration will be increased. Nevertheless, do the material balance
on the membrane process.
Mass in feed = Mass in permeate + Mass in reject
(1,000 m3/h)(0.5 kg/m3) = (800 m3/h)(0.01 kg/h) + (200 m3/h) CR
CR = 2.46 kg/m3
The concentration in the waste going to treatment is computed from a mass balance around
the mixing point
(100 m3/h)(10 kg/m3) + (1,000 m3/h)(3 kg/m3) + (200 m3/h)(2.46 kg/m3)
= (1300 m3/h)(C)
C = 3.46 kg/m3
This reduces the feed of city water from 2,100 m3/h to 1,300 m3/h. The flow to the wastewater
treatment plant is reduced by the same amount. Note: Recycling permeate to Process 2
gives an identical result.
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SOLVED MATERIAL BALANCE PROBLEMS:
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Water conservation and reuse
Figure S10.11b System with Process 3 output regenerated and recycled to replace some of the fresh water
input to Process 3
10.12
MEMBRANE PROCESS FOR REGENERATION AND REUSE
Redo Problem 10.11 with the output from all three processes being regenerated with a
membrane process. The membrane will deliver 80% of the input for recycle at a concentration
of 100 ppm. Calculate the required freshwater supply and the volume and concentration
of the wastewater mixture. Diagram the re-designed system.
Solution
Membrane system
Feed: F = 2,100 m3/h and C = 2.143 kg/m3; m = 4,500 kg/h,
Permeate: 80% of feed becomes permeate at CP = 0.01 kg/m3
FP = 0.8(F) = (0.8)(2,100 m3/h) = 1,680 m3/h
mP = FP CP = (1,680 m3/h)(0.01 kg/m3) = 16.8 kg/h
Reject: 20% of feed becomes a concentrated reject
FR = 0.2(F) = (0.2)(2,100 m3/h) = 420 m3/h
Material balance on membrane system
m = FPCP + FRCR
4,500 kg/h = (1,680 m3/h)(0.01 kg/m3) + (420 m3/h)CR
CR = (4,500 kg/h - 16.8 kg/h)/(420 m3/h) = 10.67 kg/m3
333
SOLVED MATERIAL BALANCE PROBLEMS:
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Figure S10.12
10.13
NEAREST NEIGHBOR - 1
For the process in Figure P10.13 find the best way to satisfy the demand for 80 T/h of water
that has a concentration not exceeding 40 ppm. The goal is to use effluent from sources
1 and 2 so far as possible, subject to the limitations that Q1 ≤ 60 T/h and Q2 ≤ 20 T/h.
The combined flow (20 + 60 = 80 = QD) can satisfy the quantity demand, but the quality
constraint, CD ≤ 40 ppm, must also be satisfied.
Source 1
Q1 ≤ 60 T/h
C1 = 50 ppm
Freshwater
QF = unlimited
CF = 0 ppm
Source 2
Q2 ≤ 20 T/h
C2 = 100 ppm
Demand
QD = 80 T/h
CD ≤ 40 ppm
Figure P10.13
Solution
Define 1 mass unit = (1 T/h)(1 ppm)
1 T/h of Freshwater contributes zero mass units
1 T/h from Source 1 contributes 50 mass units
1 T/h from Source 2 contributes 100 mass units
334
QD = 80 T/h
CD ≤ 40 ppm
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Water conservation and reuse
The constraint on mass units accepted by the Demand node is
(80 T/h)(40 ppm) = 3200 mass units
The nearest neighbor rule says to use first the source that is most like the required final product, and
to use as much of it as possible. This would be Source 1.
Use all of Source 1 = (60 T/h)(50 ppm) = 3,000 mass units
Add 2 T/h = 200 mass units from Source 2 to give a total of
62 T/h and 3,200 mass units
Add 80 T/h – 62 T/h = 18 T/h of fresh water.
Check the material balance
QF + Q1 + Q2 = 18 T/h + 60 T/h + 2 T/h = 80 T/h
QFCF + Q1C1 + Q2C2 = (40 ppm)(80 T/h) = 3,200 mass units
(18 T/h)(0 ppm) + (60 T/h)(50 ppm) + (2 T/h)(100 pm)
= 0 + 3,000 + 200 = 3,200 mass units
You can readily see another solution, starting from Source 2
Add 20 T/h from Source 2 = (20 T/h)(100 ppm) = 2,000 mass units
Add 24T/h from Source 1 = 1,200 mass units
Add (80 T/h - 20 T/h – 24 T/h) = 36 T/h of fresh water
The nearest neighbor rule did give the solution that uses the least amount of freshwater.
Can you prove that it always works?
10.14
NEAREST NEIGHBOR - 2
The demand for the process in Figure P10.14 is for 80 T/h of water with a concentration
of 100 ppm or less. Blend water from Sources A and B to minimize the use of freshwater.
Source A
QA ≤ 60 T/h
CA = 60 ppm
Fresh water
QF = unlimited
CF = 0 ppm
Source B
QB ≤ 40 T/h
CB = 200 ppm
Mixer
Figure P10.14
335
Demand
QD = 80 T/h
CD ≤ 100 ppm
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Water conservation and reuse
Solution
Define 1 mass unit = (1 T/h)(1 ppm)
1 T/h from Source A = 60 mass units
1 T/h from Source B = 200 mass units
Demand node limit = (80 T/h)(100 ppm) = 8,000 mass units
Start with Source A, according to the nearest neighbor rule
Use 60 T/h from Source A = 3,600 mass units
Demand node can accept another 4,400 mass units
4,400 mass units from Source B
4,400 mass units/200 ppm = 22 T/h
Constraint on QD = 80 T/h is exceeded (QA + QB) = (60 T/h + 22 T/h) > 80 T/h
Take 20 T/h from Source B = (20 T/h)(200 ppm) = 4,000 mass units
Mass units = 3,600 + 4,000 = 7600 ≤ 8,000, so constraint on CD is satisfied.
Flow to Demand = 60 T/h + 20 T/h = 80 T/h
All constraints are satisfied.
No fresh water needed QF = 0
Alternate solution
Dilute Source B with freshwater?
QF + QB = 80 T/h QF + 40 T/h = 80 T/h QF = 40 T/h
QBCB = (40 T/h)(200 ppm) = 8,000 mass units
This alternate uses QF = 40 T/h of freshwater.
The nearest neighbor rule gave a very good solution.
10.15
MASS TRANSFER PROCESS
Figure P10.15 shows a mass transfer process in terms of the inlet and outlet concentrations
of the water that is flowing at a rate of FW = 20 T/h. Only the water stream that gains in
pollutant load is shown. Construct the concentration (kg/m3) vs. mass load (kg/h) diagram
for the process.
FW = 20 T/h
CIn = 80 ppm
Mass Transfer
Process
Figure P10.15
336
FW = 20 T/h
COut = 200 ppm
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Water conservation and reuse
Solution
Mass (kg/h)
m = FC/1,000 when F is T/h and C is ppm
Mass exchanged
'm (kg/h) =
(F T/h)('C ppm)
1000
Concentration units conversion
C [kg/m3] = C [ppm] /1,000
mIn = (20 T/h)(80 ppm)/1,000 = 1.6 kg/h
mOut = (20 T/h)(200 ppm)/1,000 = 4.0 kg/h
Mass exchanged = Δm = 4.0 kg/h – 1.6 kg/h = 2.4 kg/h
The plotting points are:
In
CIn = 80 ppm/1,000 = 0.08 kg/m3 at mIn = 1.6 kg/h
Out
COut = 200 ppm/1,000 = 0.20 kg/m3 at mOut = 4.0 kg/h
Figure S10.15
10.16
TWO PROCESS SYSTEM
Figure P10.16 shows two processes that are operated with fresh water inputs. Process 1
gains 4 kg/h of pollutant mass and Process 2 gains 6 kg/h. Process 1 needs fresh water as
input, but Process 2 can operate with a feed of up to 50 ppm. Show how the use of fresh
water can be reduced.
'
'
Figure P10.16
337
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Water conservation and reuse
Solution
Maximum mass load to Process 2
Maximum concentration is 50 ppm = 0.00005 kg/kg = 0.05kg/T
Maximum mass load = (0.05 kg/T)(40 T/h) = 2 kg/h
Output of Process 1 carries 4 kg/h.
So we can mix half of process 1 effluent (10 T/h) with 30 T/h fresh water to feed Process 2, giving
C2, In = 50 ppm.
Process 2 output
The increase in pollutant mass through Process 2 remains at 6 kg/h. That is
m2, Out = m2, In + 6 kg/h
m2, In = maximum mass load = 2 kg/h
m2, Out = 2 kg/h + 6 kg/h = 8 kg/h
The pollutant concentration leaving Process 2 is
C2, Out = (m2, Out)/(F2, Out) = (8 kg/h)/[(40 T/h)(1,000kg/T)]
= 0.0002 kg/kg = 200 ppm
The process flow diagram with recycle is shown in Figure S10.16
'
'
Figure S10.16
This reduces the fresh water input by 10 T/h and it reduces the volume of wastewater going to
treatment by 10 T/h. The mass of pollutant to be treated has not changed – the total is 10 kg/h. The
concentration has increased.
COriginal = (10 kg/h)/[(60 T/h)(1,000 kg/T)] = 0.000133 kg/kg = 133 ppm
CRecycle = (10 kg/h)/[(50 T/h)(1,000 kg/T)] = 0.0002 kg/kg = 200 ppm
10.17
WATER REUSE – MINIMUM FLOWRATE
The mass-concentration diagram in Figure P10.17 plots the data for three processes. The
maximum input concentrations are 0 ppm for Process 1, 100 ppm for Process 2, and 200
338
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Water conservation and reuse
ppm for Process 3. The mass exchanged in each process is 4 kg/h, 6 kg/h, and 4 kg/h,
respectively. Draw the composite mass exchange diagram, find the graphical solution for the
pinch point, and calculate the minimum possible freshwater input flow (i.e. the limiting
flow). Draw the composite flow diagram for the combined processes.
Figure P10.17
Solution
First, determine the flow rates for each process, given the maximum input concentrations and mass
exchanged. For example, for Process 1
F1 = Dm1(1,000)/(C1, Out - C1, In, Max)
F1 = (4 kg/h)(1,000)/(400 ppm – 0 ppm) = 10 T/h
The flow value for each process are in Table S10.17a
F when CIn
Process
Dm
COut
CIn, Max
= CIn, Max
(kg/h)
(ppm)
(ppm)
F = 1,000Dm/DC
(T/h)
1
4
400
0
4,000/400 = 10
2
6
250
100
6,000/150 = 40
3
4
400
200
4,000/200 = 20
Table S10.17a
Composite mass exchange diagram
The composite mass exchange diagram is determined numerically by tabulating the mass exchanged by
each process and summing over all processes exchanging mass in each of the respective concentration
intervals. The cumulative mass exchanged gives the composite mass exchange diagram.
339
SOLVED MATERIAL BALANCE PROBLEMS:
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Water conservation and reuse
There are four concentration intervals over which mass is exchanged: 0 – 100 ppm; 100 – 200 ppm;
200 – 250 ppm; and 250 – 400ppm. For the concentration interval from 0 - 100 ppm, only process 1
is exchanging mass, and it contributes 1 kg/h. This gives the first point on the composite exchange
diagram [1 kg/h, 100 ppm].
In the concentration interval 100 to 200 ppm, process 1 exchanges 1 kg/h and process 2 exchanges
4 kg/h for a total of 5 kg/h. The cumulative mass exchanged from 0 ppm to 200 ppm is 6 kg/h. This
gives the second point on the diagram [6 kg/h, 200 ppm].
The calculations are repeated for each concentration interval. Table S10.17b has the complete mass
exchange calculations and Figure S10.17a is the composite mass diagram.
Mass Exchanged (kg/h)
Cumulative
Mass
Concentration
Process
Process
Process
Sum
Exchanged
Interval (ppm)
1
2
3
(kg/h)
(kg/h)
0
-
-
-
0
0
0 - 100
1
-
-
1
1
100 - 200
1
4
-
5
6
200 - 250
0.5
2
1
3.5
9.5
250 - 400
1.5
-
3
4.5
14
Table S10.17b
Figure S10.17a
340
SOLVED MATERIAL BALANCE PROBLEMS:
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Water conservation and reuse
Draw a straight line from the origin to locate the vertex (pinch point) with the largest mass flow that
can be intersected without going above the mass-concentration composite curve. This is the blue
dashed line.
The pinch point occurs at [9.5 kg/h, 250 ppm] and the limiting flow is
FLimiting = 1,000∆m/∆C = 1,000(9.5 kg/h)/(250 ppm) = 38 T/h.
Composite flow diagram
Process 1 requires fresh water input.
Process 1: F = 1,000∆m/∆C = 1,000(4 kg/h)/400ppm = 10 T/h
Process 3 (CIn, Max = 200 ppm) can use all the output from process 1 if diluted with 10 T/h fresh water.
Process 3: F = 10 T/h @ 400 ppm from Process 1 plus
10 T/h @ 0 ppm fresh water = 20 T/h @200 ppm
Process 2 (CIn, Max = 100 ppm) uses the remainder of the limiting freshwater flow not used for Processes
1 and 3 (38 T/h – 20 T/h) = 18 T/h as input and is mixed with a recycled portion of Process 3 output
to yield CIn, Max = 100 ppm.
The required recycle flow from process 3, R, is obtained from a material balance on Process 2 input
(18 T/h)(0 ppm) + R (400 ppm) = (18 T/h + R)(100 ppm) R = 6 T/h
The total flow into Process 2 is
F = 18 T/h + 6 T/h = 24 T/h
Process 2 output concentration is
∆C = 1,000∆m/F = 1,000(6 kg/h)/(24 T/h) = 250 ppm
COut = CIn +∆C = 100 ppm + 250 ppm = 350 ppm
The composite flow diagram is shown in Figure S10.17b.
'
'
'
Figure S10.17b Composite process flow diagram (Freshwater inputs are in blue)
341
SOLVED MATERIAL BALANCE PROBLEMS:
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10.18
Water conservation and reuse
COMPOSITE MASS-CONCENTRATION CURVE
A four-process system is described by the data in Table P10.18. (a) Construct a mass exchange
diagram for each process in the system. (b) Construct the composite mass-exchange diagram.
(c) Determine the limiting flow rate. (d) Construct the process flow diagram for the system
showing an efficient water recycle plan.
F when CIn = CIn, Max
Dm
COut
CIn, Max
(kg/h)
(ppm)
(ppm)
1
4
200
0
4,000/200 = 20
2
6
300
100
6,000/200 = 30
3
4
600
200
4,000/400 = 10
4
2
800
400
2,000/400 = 5
Process
F = 1,000Dm/DC
(T/h)
Table P10.18
Solution
a) Mass exchange diagram for each process is in Figure S10.18a.
Figure S10.18a
342
SOLVED MATERIAL BALANCE PROBLEMS:
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Water conservation and reuse
b) Composite mass-exchange diagram
The composite mass exchange diagram is determined numerically by tabulating the mass
exchanged by each process and summing over all processes exchanging mass in each of
the respective concentration intervals. The cumulative mass exchanged gives the composite
mass exchange diagram.
There are six concentration intervals over which mass is exchanged: 0 – 100 ppm; 100 – 200
ppm; 200 – 300 ppm; 300 – 400 ppm; 400 – 600 ppm; and 600 – 800 ppm. For the concentration
interval from 0 – 100 ppm, only process 1 is exchanging mass, and it contributes 2 kg/h. This
gives the first point on the composite exchange diagram [2 kg/h, 100 ppm].
In the concentration interval 100 to 200 ppm, process 1 exchanges 2 kg/h and process 2
exchanges 3 kg/h for a total of 5 kg/h. The cumulative mass exchanged from 0 ppm to 200
ppm is 7 kg/h. This gives the second point on the diagram [7 kg/h, 200 ppm].
The calculations are repeated for each concentration interval. Table S10.18 is the complete
mass exchange table and Figure S10.18b is the composite mass exchange diagram.
Mass Exchanged (kg/h)
Cumulative
Mass
Concentration
Process
Process
Process
Process
Sum
Exchanged
Interval (ppm)
1
2
3
4
(kg/h)
(kg/h)
0
-
-
-
-
0
0
0 – 100
2
-
-
-
2
2
100 – 200
2
3
-
-
5
7
200 – 300
-
3
1
-
4
11
300 – 400
-
-
1
-
1
12
400 – 600
-
-
2
1
3
15
600 – 800
-
-
-
1
1
16
Table S10.18
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SOLVED MATERIAL BALANCE PROBLEMS:
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Figure S10.18b
c) Limiting flow. Draw a straight line from the origin to locate the vertex (pinch point)
with the largest mass flow that can be intersected without going above the massconcentration composite curve. This is the blue dashed line.
The pinch point occurs at m = 11 kg/h and C = 300 ppm.
Limiting flow = F = 1,000 ∆m/∆C = 1,000(11 kg/h)/(300 ppm) = 36.67 T/h
d) Process flow diagram with efficient recycle.
Process 1 must use freshwater as input. This requires
F = 1,000 ∆m/∆C = 1,000(4 kg/h)/(200 ppm) = 20 T/h
This leaves 16.67 T/h of the limiting flow for the remaining processes.
Process 2 requires an input concentration of no more than 100 ppm. This can be achieved
by recycling a portion of the output from process 1 (at 200 ppm), and diluting it with the
remaining fresh water (16.67 T/h) not used in process 1. Let R be the flow to be recycled
from process 1 into process 2.
A material balance on the input is:
(16.67 T/h)(0 ppm) + R (200 ppm) = (16.67 T/h + R)(CIn)
The mass exchange for process 2 is
(R + 16.67 T/h) = 1,000(6 kg/h)/(600 ppm – CIn)
Solving simultaneously yields R = 10 T/h and CIn = 75 ppm
Process 3 can use the remainder of process 1 ouflow (10 T/h at C = 200 ppm) directly as input.
Process 4 can use a portion of the output from process 2 (300 ppm) because it is less than
the maximum input concentration (400 ppm) for process 4. The flow requirement is
F = 1,000 ∆m/∆C = 1,000(2 kg/h)/(800 ppm – 300 ppm) = 4 T/h
344
SOLVED MATERIAL BALANCE PROBLEMS:
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Water conservation and reuse
The complete process flow diagram is shown in Figure S10.18c.
'
'
'
'
Figure S10.18c Composite process flow diagram with recycle
Tutorial note
The next few problems are about using separation processes for reclamation of water for reuse.
There are four hypothetical separation processes and two pollutants, COD and TDS.
Separation processes do not destroy pollutants. They simply split an input into two streams,
A and B. One stream (B) will concentrate COD or TDS, or both, and one stream (A) will
be cleaner than the feed.
Separations can be used alone or in combination, for example two stages of Separation
1, or Separation 1 followed by Separation 2 and/or 3. Branching systems are allowed, for
example Separation 1 followed by Separation 2 on F1A and Separation 3 on F1B.
The four available separation processes are shown in Tutorial Figure 10.1, and perform as
follows:
1) Separation process 1 will concentrate 90% of the COD mass in 20% of the influent
volume. The TDS concentration is not affected by this separation process. That
is, the TDS in the concentrate stream will be the same as in the influent and the
effluent streams. This technology acts as a splitter for TDS.
2) Separation process 2 will concentrate 80% of the TDS mass in 5% of the influent
volume but it removes no COD. All influent COD appears in the dilute process
effluent.
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SOLVED MATERIAL BALANCE PROBLEMS:
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Water conservation and reuse
3) Separation process 3 will remove 98% of the TDS mass and 100% of the COD
mass from the influent and concentrate it in 10% of the influent volume. The feed
to Process 3 is constrained to have COD ≤ 4 kg/m3.
4) Separation process 4 will remove 50% of COD and 50% of TDS and concentrate
them in 20% of the influent volume.
Tutorial Figure 10.1 Four separation processes.
The Tutorial Table 10.1 summarizes the separation process performances.
.
346
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Component
Water conservation and reuse
Percent (%) of input mass leaving in output streams A and B
Process 1
Process 2
Process 3
Process 4
Flow A (volume/time)
80
95
90
80
COD A (mass/volume)
10
100
0
50
TDS A (mass/volume)
80
20
2
50
Flow B (volume/time)
20
5
10
20
COD B (mass/volume)
90
0
100
50
TDS B (mass/volume)
20
80
98
50
Tutorial Table 10.1 Performance of four separation processes.
The four separation processes do different amounts of purification and they have different
costs. The relative costs, in hypothetical units of thetas/m3 (Θ/m3) of feed to the process, are
Separation 2 = 3 Θ/m3
Separation 1 = 1 Θ/m3
Separation 3 = 4 Θ/m3
Separation 4 = 2 Θ/m3
Sample material balance calculation
Tutorial Figure 10.2 shows the material balance for Separation 1 for an influent flow of 100 m3/d, COD
= 50 kg/m3, and TDS = 10 kg/m3. Recall that separation 1 concentrates 90% of the COD mass in 20%
of the influent volume and does not affect the TDS concentration.
Tutorial Figure 10.2 Example separation result for Process 1
The cost for this processing is (1 Θ/m3)(100 m3/d) = 100
347
Θ/d.
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
10.19
Water conservation and reuse
PROCESS WATER REGENERATION AND RECYCLE
One section of a textile dying plant is summarized as two processes, a concentrated dye bath
and a strong effluent from washing the dyed textiles. These are shown in Figure P10.19.
Water quality is measured by COD (ppm) and TDS (ppm). Flow is in cubic meters per
day (m3/d). The material balance is given in the diagram.
There is an opportunity for regeneration and recycle. Process 1 can accept influent with TDS
≤ 2 kg/m3 (2,000 ppm) and COD ≤ 1 kg/m3 (1,000 ppm). Process 2 can accept influent
with TDS ≤ 0.5 kg/m3 (500 ppm) and COD ≤ 0.5 kg/m3 (500 ppm).
'
'
'
'
Figure P10.19 Textile dying wastewater
You may use the four separation processes shown in Tutorial Figure 10.1, alone or in
combination.
Propose at least two plausible arrangements for water recycle. Sketch the proposed systems
and show all flow rates, pollutant concentrations and pollutant mass flows.
Solution
We show one plausible recycle system.
Effluent material balance on the original system
Mass = (Flow)(Concentration)
For example, the COD mass flow from the concentrated dye bath is
COD = (100 m3/d)(50 kg/m3) =5,000 kg/d
The remaining calculations are in Table S10.19.
348
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Effluent
Water conservation and reuse
Mass
Flow
COD
TDS
(m /d)
(kg/m )
(kg/m )
3
3
3
COD
(kg/d)
Mass TDS
(kg/d)
Concentrated dye bath
100
50
10
5,000
1,000
Strong effluent
1,000
3
2
3,000
2,000
Combined effluent
1,100
7.272
2.727
8,000
3,000
Table S10.19
Proposed Regeneration System
Treat the combined effluent of 1,100 m3/d using Separation 1 and Separation 2 in series. Figure S10.19a
shows the material balance on the separations system. A flow of 836 m3/d has been regenerated,
with COD and TDS concentrations that need to be checked against the maximum acceptable input
process water concentrations.
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349
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Water conservation and reuse
Figure S10.19a Proposed regeneration using separations 1 and 2 in series
The recycle water is acceptable for direct use in the concentrated dye bath. Recycle to the washing
process must be diluted with freshwater. The acceptable ratio is 535 m3/d of recycle diluted with 455
m3/d of freshwater. Assume that using recycled water does not increase the concentrations of COD
and TDS in the concentrated dye bath and strong effluent process waste streams. The proposed
regeneration/reuse process is shown in Figure S10.19b.
The total wastewater produced is the 264 m3 concentrate from the separation processes plus 201 m3
of regenerated water that cannot be recycled.
'
'
'
'
Figure S10.19b Proposed reuse of regenerated wastewater
350
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
10.20
Water conservation and reuse
WATER RECLAMATION SYSTEM - 1
Wastewater in the amount of 100 m3/h with COD = 10 kg/m3 and TDS = 4 kg/m3 can
be recycled and reused if the COD ≤ 2 kg/m3 and TDS ≤ 1 kg/m3. Using the separation
processes in Tutorial Figure 10.1 design a separation system that will deliver the maximum
flow of suitable water for reuse. Calculate the flow rate (m3/h) and concentration (kg/m3)
of each stream. Compare the costs of the original and your designed system.
Solution
Separation 3 cannot be used directly because the input COD > 4 kg/m3
Using Separations 1 and 2 in series is an attractive design. Separation 1 has the lowest cost and it
will produce an output stream with 1.25 kg /m3 COD and 4 kg/m3 TDS. Separation 2 will recover the
greatest volume of water and it will produce TDS < 1 kg/m3.
Sample calculation for Separation 1
Recall that Separation 1 concentrates 90% of the COD mass in 20% of the influent volume and does
not affect the TDS concentration.
Influent stream:
F1 = 100 m3/h
COD1 = (10 kg/m3)(100 m3/h) = 1,000 kg/h
TDS1 = (4 kg/m3)(100 m3/h) = 400 kg/h
Concentrated stream:
F1B = (0.2)(100m3/h) = 20 m3/h
COD1B = (0.9)(1,000 kg/h) = 900 kg/h
TSS1B = (20 m3/h)(4 kg/h) = 80 kg/h
Dilute stream:
F1A = (0.8)(100 m3/h) = 80 m3/h
COD1A = (0.1)(1,000 kg/h) = 100 kg/h
= (100 kg/h)/(80 m3/h) = 1.25 kg/m3
TSS1A = (80 m3/h)(4 kg/m3) = 320 kg/h
The complete material balance is shown in Figure S10.20. The production of reclaimed water is 76
m3/h with 1.32 kg/m3 COD and 0.84 kg/m3 TDS.
351
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Water conservation and reuse
Figure S10.20
The reclamation cost = Cost of feed to Separation 1 + Cost of feed to Separation 2
= (1 Θ/m3)(F1) + (3 Θ/m3)(F1A)
= (1 Θ/m3)(100 m3/h) + (3 Θ/m3)(80 m3/h) = 340 Θ/h.
The two concentrated streams, F1B and F2B, go to waste and have a combined flow of 24 m3/h and mass
loads of 900 kg/h (37.5 kg/m3) COD and 336 kg/h (2.83 kg/m3) TDS. The reclaimed water reduces the
freshwater requirements by 76 m3/h to 24 m3/h.
Cost without reclaimed water
= purchase of 100 m3/h fresh water + disposal of 100 m3/h of wastewater
Cost with reclaimed water
= 340 Θ/h + purchase of 24 m3/h fresh water + disposal of 24 m3/h of wastewater
10.21
WATER RECLAMATION SYSTEM - 2
Wastewater in the amount of 100 m3/d with COD = 5 kg/m3 and TDS = 18 kg/m3 can
be recycled and reused if both the COD and TDS can be reduced to ≤ 0.5 kg/m3. Use the
separation processes in Tutorial Figure 10.1 to design a separation system that will deliver the
maximum flow of suitable water for reuse. Calculate the flow rate (m3/h) and concentration
(kg/m3) of each stream. Compare the costs of the original and your designed system.
Solution
Only Separation 3 can deliver TDS ≤ 0.5 kg/m3, but it cannot be used directly because the COD
concentration is too high. Consider a design that uses Separation 1 to reduce the COD, followed by
Separation 3.
352
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Water conservation and reuse
The material balance is shown in Figure S10.21. The production of reclaimed water is 72 m3/h with 0
kg/d (0 kg/m3) COD and 28.8 kg/d (0.4 kg/m3) TDS. This is very high quality water.
Figure S10.21
The reclamation cost = Cost of feed to Separation 1 + Cost of feed to Separation 3
= (1 Θ/m3)(F1) + (4 Θ/m3)(F1A)
= (1 Θ/m3)(100 m3/d) + (4 Θ/m3)(80 m3/d) = 420 Θ/d.
The two concentrated streams, F1B and F3B, go to waste and have a combined flow of 28 m3/d and
mass loads of 500 kg/d (17.9 kg/m3) COD and 1,771.2 kg/d (63.3 kg/m3) TDS. The reclaimed water
reduces the freshwater requirements by 72 m3/d to 28 m3/d.
Cost without reclaimed water
= purchase of 100 m3/d fresh water + disposal of 100 m3/d of wastewater
Cost with reclaimed water
= 420 Θ/d + purchase of 28 m3/d fresh water + disposal of 28 m3/d of wastewater
353
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Appendix 1 – Atomic Mass of Selected Elements
11APPENDIX 1 – ATOMIC MASS
OF SELECTED ELEMENTS
Symbol
At
Atomic
No.
Mass
Element
Symbol
At
Atomic
No.
Mass
Aluminum
Al
13
26.98154
Mercury
Hg
80
200.59
Antimony
Sb
51
121.75
Molybdenum
Mo
42
95.94
Argon
Ar
18
39.948
Neon
Ne
10
20.179
Arsenic
As
33
74.9216
Nickel
Ni
28
58.70
Barium
Ba
56
137.33
Nitrogen
N
7
14.0067
Beryllium
Be
4
9.01218
Oxygen
O
8
15.9994
Bismuth
Bi
83
208.9804
Phosphorus
P
15
30.97376
Boron
B
5
10.81
Platinum
Pt
78
195.09
Bromine
Br
35
79.904
Plutonium
Pu
94
(244)
Cadmium
Cd
48
112.41
Polonium
Po
84
(209)
Calcium
Ca
20
40.08
Potassium
K
19
39.0983
Carbon
C
6
12.011
Radium
Ra
88
226.0254
Chlorine
Cl
17
35.453
Radon
Rn
86
(222)
Chromium
Cr
24
51.966
Selenium
Se
34
78.96
Cobalt
Co
27
58.9332
Silicon
Si
14
28.0855
Copper
Cu
29
63.546
Silver
Ag
47
107.868
Fluorine
F
9
18.99840
Sodium
Na
11
22.98977
Gallium
Ga
31
69.72
Strontium
Sr
38
87.62
Gold
Au
79
196.9665
Sulfur
S
16
32.06
Helium
He
2
4.0026
Tin
Sn
50
118.69
Hydrogen
H
1
1.0079
Titanium
Ti
22
47.90
Iodine
I
53
126.9045
Tungsten
W
74
183.85
Iron
Fe
26
55.847
Uranium
U
92
238.029
Krypton
Kr
36
83.80
Vanadium
V
23
50.9414
Lead
Pb
82
207.2
Xenon
Xe
54
131.30
Lithium
Li
3
6.941
Zinc
Zn
30
65.38
Magnesium
Mg
12
24.305
Zirconium
Zr
40
91.22
Manganese
Mn
25
54.9380
354
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Appendix 2 - Conversion Factors
12APPENDIX 2 - CONVERSION
FACTORS
Linear Measure Equivalents
meter
foot
centimeter
inch
1
3.2808
100
39.37
0.2048
1
30.48
12.0
100
0.03281
1
0.3937
0.0254
0.0833
2.54
1
Area Equivalents
hectare
sq. meter
acre
sq. feet
1
10,000
2.471
107,639.1
0.001
1
0.000,247
10.764
0.4047
4,046.9
1
43,560
9.29x10-6
0.0929
0.000,023
1
Volume Equivalents (U.S. units)
cubic foot
U. S. gallon
acre-foot
barrel (U.S. petroleum)
1
7.48
0.000,023
0.1781
0.1337
1
0.000,003,1
43,560
325,851
1
65.615
42.0
---
355
1
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Appendix 2 - Conversion Factors
Volume Equivalents (Metric & U.S.)
liter
cubic meter
U.S. gallon
cubic foot
1
0.001
0.2642
0.0353
1,000
1
264.172
35.315
3.785
0.00378
1
0.1337
28.317
0.02832
7.48
1
Power Equivalents
horsepower
kilowatt
ft-lb/sec
Btu/sec
1
0.7457
550
0.7068
1.341
1
737.56
0.9478
0.001,818
0.001,356
1
0.001285
1.415
1.055
778.16
1
Heat, Energy, or Work Equivalents
joule = kg-m
ft-lb
kWh
hp-h
Liter-atm
Btu
1
7.233
2.724x10–6
3.653x10–6
0.0968
0.009,296
0.1383
1
3,766x10–7
5.050x10–7
0.0134
0.000,324
367,100
2,655,000
1
1.341
35,534
3412.8
273,750
1,980,000
0.7455
1
26,494
2,545
10.33
74.73
2.815x10–5
3.774x10–5
1
0.0242
426.7
3,086
0.001,162
0.001,558
41.29
1
356
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Appendix 3 – Densities and Specific Weights
13APPENDIX 3 – DENSITIES
AND SPECIFIC WEIGHTS
U. S. Units
Temperature
(°F)
Density
ρ
(lb/ft )
3
SI Units
Specific
Weight
Temperature
γ
(°C)
(lb/ft3)
Density
Specific Weight
ρ
γ
(kg/m )
(N/m3)
3
-40
0.09464
0.09464
-40
1.514
14.85
-20
0.09032
0.09032
-20
1.395
13.68
0
0.08639
0.08639
0
1.293
12.67
10
0.08456
0.08456
5
1.269
12.45
20
0.08279
0.08279
10
1.247
12.23
30
0.08111
0.08111
15
1.225
12.01
32
0.08063
0.08063
20
1.204
11.81
40
0.07950
0.07950
25
1.184
11.61
50
0.07792
0.07792
30
1.165
11.43
60
0.07641
0.07641
40
1.127
11.05
70
0.07499
0.07499
50
1.109
10.88
80
0.07361
0.07361
60
1.060
10.40
90
0.07226
0.07226
70
1.029
10.09
100
0.07097
0.07097
80
0.9996
9.803
120
0.06852
0.06852
90
0.9721
9.533
140
0.06624
0.06624
100
0.9461
9.278
160
0.06408
0.06408
200
0.7461
7.317
180
0.06208
0.06208
300
0.6159
6.040
200
0.06021
0.06021
400
0.5243
5.142
300
0.05229
0.05229
500
0.4565
4.477
400
0.04621
0.04621
1,000
0.2772
2.719
500
0.04138
0.04138
750
0.03284
0.03284
1,000
0.02721
0.02721
1,500
0.02025
0.02025
The correct U.S. unit for density is slugs/ft3; the more convenient and unit is lb/ft3
Specific weight = (density)(acceleration of gravity)
Table A3.1 Density and specific weight of air (at 1 atm)
357
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Gas
Appendix 3 – Densities and Specific Weights
Formula
Molar
Density
Density
Mass
lb/ft
kg/m3
3
Air
----
29
0.0808
1.2943
Ammonia
NH3
17.03
0.0482
0.7721
Carbon dioxide
CO2
44
0.1235
1.9783
Chlorine
Cl2
70.91
0.2011
3.2213
Hydrogen
H2
2.016
0.0056
0.0897
Methane
CH4
16.03
0.0448
0.7176
Nitrogen
N2
28.022
0.0782
1.2527
Oxygen
O2
32
0.0892
1.4289
Sulfur dioxide
SO2
64.06
0.1825
2.9234
Table A3.2 Densities of selected gases at 1 atm and 0°C
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358
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Gas
Acetylene
Formula
C2H2
Appendix 3 – Densities and Specific Weights
Density
Molar Mass
(g/g-mol)
(kg/m3)
(lb/ft3)
26.02
1.1708
0.0732
1.2928
0.0808
Air
Ammonia
NH4
17.03
0.7708
0.0482
Butane
C4H10
58.08
2.5985
0.1623
Carbon dioxide
CO2
44.00
1.9768
0.1235
Carbon monoxide
CO
28.00
1.2501
0.0781
Chlorine
Cl2
70.91
3.2204
0.2011
Cyanogen
C2N2
52.02
2.3348
0.1459
Ethane
C2H6
30.05
2.8700
0.1793
Ethylene
C2H4
28.03
1.2644
0.0783
Fluorine
F2
38.00
1.6354
0.1022
Hydrogen
H2
2.016
0.0898
0.0056
Hydrogen
chloride
HCl
36.47
1.6394
0.1024
Hydrogen sulfide
H2S
34.08
1.5992
0.0961
Methane
CH4
16.03
0.7167
0.0448
CH3Cl
50.48
2.3044
0.1440
19.5
0.7-0.9
0.0440.056
Methyl chloride
Natural gas
Nitrogen
N2
28.02
1.2507
0.0782
Oxygen
O2
32.00
1.4289
0.0892
Propane
C3H6
44.09
1.882
0.1175
Sulfur dioxide
SO2
64.06
2.9268
0.1828
Water vapor
(steam)
H 2O
18.016
0.804
0.048
Table A3.3 Density of gases at standard conditions (O°C and 1 atm)
359
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Mass Al2(SO4)3
(15°C)
%
1%
2%
4%
6%
8%
10 %
12 %
16 %
20 %
24 %
26 %
28 %
30 %
35 %
40 %
50 %
NH4NO3
(25°C)
CaCl2
(20°C)
1.0093
1.0195
1.0404
1.0011
1.0051
1.0132
1.0148
1.0316
1.0659
1.0837
1.0297
1.1015
1.1293
1.1770
1.2272
1.2803
1.3079
1.0464
1.0633
1.0806
1.0982
1.1386
1.1775
1.2284
Appendix 3 – Densities and Specific Weights
H2CrO4
(15°C)
FeCl3
(20°C)
1.006
1.014
1.0086 1.0085 1.0032 1.0086
1.0152
1.0082 1.0190
1.0324 1.0375 1.0181 1.0398
1.0279 1.0816
1.0669 1.0785 1.0376 1.1244
1.0474
1.104
1.122 1.0574 1.1244
1.000
1.1675 1.0878
1.182
1.2135 1.0980
1.1187
1.1290
1.1392
1.000
1.1493
1.353
1.4175
1.1980
1.551
1.045
1.076
1.127
1.163
1.220
1.1161
1.1252
1.1727
1.2229
1.2816
1.3373
1.3957
1.260
1.371
1.505
Table A3.4 Density of aqueous inorganic solutions
360
FeSO4
HCl NaCO3 NaCl H2SO4
(18°C) (20°C) (20°C) (20°C) (20°C)
1.0095 1.0051
1.0207 1.0104
1.0428 1.0250
1.0385
1.0869 1.0522
1.0661
1.1309 1.0802
1.1751 1.1094
1.2191 1.1394
1.2629 1.1704
1.1862
1.3064 1.2023
1.2185
1.2599
1.4300 1.3028
1.5253 1.3951
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Appendix 3 – Densities and Specific Weights
Concentration
Acetic
Methyl
Ethyl
(mass %)
Acid
Alcohol
Alcohol
0
0.9982
0.9982
0.99820
0.9982
1
0.9996
0.9965
0.99636
1.0006
2
1.0012
0.9948
0.99453
1.0030
3
1.0025
0.9931
0.99275
1.0054
4
1.0040
0.9914
0.99103
1.0078
5
1.0055
0.9896
0.98938
1.0102
10
1.0141
0.9815
0.98043
1.0221
15
1.0213
0.9740
0.97514
1.0345
20
1.0283
0.9666
0.97864
1.0469
25
1.0349
0.9592
0.96168
1.0598
30
1.0411
0.9515
0.95362
1.0727
40
1.0488
0.9345
0.93518
1.0993
50
1.0575
0.9156
0.91384
1.1263
60
1.0642
0.8946
0.89113
1.1538
70
1.0685
0.8715
0.86766
1.1812
80
1.7000
0.8469
0.84344
1.2085
90
1.0661
0.8202
0.81797
1.2351
100
1.0498
0.7917
0.78934
1.2611
Table A3.5 Density of aqueous organic solutions at 20°C
361
Glycerol
SOLVED MATERIAL BALANCE PROBLEMS:
POLLUTION PREVENTION AND CONTROL
Appendix 3 – Densities and Specific Weights
Ave.
Material
Sp. gr.
Ave.
Density
Material
Sp. gr.
(lb/ft )
Density
(lb/ft3)
3
Metals
Various liquids
Aluminum
2.55-2.8
165
Alcohol, ethyl
(100%)
0.79
49
Bronze
7.4-8.9
554
Alcohol, methyl
(100%)
0.80
50
7.03-7.10
442
Acid, nitric (91%)
1.50
94
5.2
325
Acid, sulfuric (87%)
1.80
112
4.9-5.2
315
Chloroform
1.50
95
11.34
710
Oils, vegetable
0.91-0.94
58
galena ore
7.3-7.6
465
Concrete masonry
Steel, cold-drawn
7.83
489
cement, stone,
sand
2.2-2.4
144
slag. etc.
1.9-2.3
130
cinder, etc.
1.5-1.7
100
1.00
63
1.76
110
1.20
76
Iron, gray cast
hematite ore
magnetite ore
Lead
Various solids
Cereals, corn
(bulk)
0.73
45
Cotton, flax,
hemp
1.47-1.50
93
Earth, etc.,
excavated
2.4-2.8
162
Clay, dry
Glass, plate
2.45-2.72
161
Glass, flint
3.2-4.7
247
Earth, dry loose
Leather
0.86-1.02
59
dry, packed
1.5 95
Paper
0.70-1.15
58
moist, loose
1.30
78
1.0-2.0
94
moist, packed
1.60
96
0.77
48
Bituminous
substances
1.93-2.07
125
Asphalt
1.11-1.5
81
Refined (kerosene
0.78-0.82
54
0.70-0.75
45
1.2
75
Glass, common
Rubber, goods
Salt,
granulated
(piled)
Sulfur
damp plastic
Timber
Fir, Douglas
0.48-0.55
32
Gasoline
Maple, white
0.53
33
Tar, bituminous
Oak, white
0.77
4
Coal and coke,
piled
Redwood,
California
0.42
26
anthracite
0.75-0.93
47-58
Teak, African
0.99
62
bituminous
0.64-0.87
40-54
charcoal
0.166-0.23
10-14
coke
0.37-0.51
23-34
Stone, quarried
& piled
Limestone,
marble, quartz
1.50
95
Sandstone
1.30
82
Table A3.6 Approximate specific gravities and densities of miscellaneous solids and liquids.
362