Appendix A
Answers to Chapter
Review Questions
CHAPTER 1 ANSWERS
CHAPTER 3 ANSWERS
1.1
3.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
A pattern or point of view that determines what is
seen as reality.
A change in the way things are understood and
done.
(1) Assessing and protecting drinking water
sources
(2) Optimizing treatment processes
(3) Ensuring the integrity of distribution systems
(4) Effecting correct cross-connection control
procedures
(5) Continuous monitoring and testing of the
water before it reaches the tap
Water/wastewater operations are usually low-profile activities and much of water/wastewater infrastructure is buried underground.
Secondary
Privatization means allowing private enterprise to
compete with government in providing public services, such as water and wastewater operations.
Reengineering is the systematic transformation of
an existing system into a new form to realize quality improvements in operations, systems capability, functionality, and performance at lower cost,
improved schedule, and less risk to the customer.
A process for rigorously measuring performance
vs. “best-in-class” operations and using the analysis to meet and exceed the best in class.
Planning, research, observation, analysis, adaptation
CHAPTER 2 ANSWERS
2.1
2.2
2.3
2.4
2.5
Operators are exposed to the full range of hazards
and work under all weather conditions.
Plants are upgrading to computerized operations.
Computerized maintenance management system
HAZMAT emergency response technician 24-hour
certification
Safe Drinking Water Act
Answers will vary
CHAPTER 4 ANSWERS
4.1
Matching answers
1.
o
14.
2.
c
15.
3.
t
16.
4.
j
17.
5.
s
18.
6.
p
19.
7.
d
20.
8.
i
21.
9.
e
22.
10. q
23.
11. u
24.
12. k
25.
13. a
26.
9219 lb/day
× 100 = 5.5%
169,011 lb/day
(3250 lb/day solids)(0.65) = 2113 lb/day VS
(4120 gpd)(8.34 lb/gal)(0.07)(0.70) = 1684 lb/day VS
98 ft – 91 ft = 7 ft drawdown
125 ft – 110 ft = 15 ft drawdown
161 ft – 144 ft = 17 ft drawdown
=
5.169 (3.7 psi)(2.31 ft/psi) = 8.5 ft sounding linne water depth
112 ft – 8.5 ft = 103.5 ft
1033.5 ft – 86 ft = 17.5 ft
5.170 (4.6 psi)(2.31 ft/psi) = 10.6 ft sounding line water depth
150 ft – 10.6 ft = 139.4 ft
171 ft – 139.4 ft = 31.4 ft drawdown
5.171 300 ÷ 20 = 15 gpm per ft of drawdown
5.172 420 gal ÷ 5 min. = 84 gpm
5.173 810 gal ÷ 5 min. = 162 gpm
5.174
856 gal
= 171 gpm
5 min
(171 gpm)(60 min/hr) = 10,260 gph
Fillet weld
Square butt weld
Single hem
Single flange
Location of weld
Steel section
Bevel weld
V weld
J-groove weld
4 times true size
Drawing number
180
Break line
Object line
3-D pictorial
Shape; complexity
Limits
Center line; finished surface
Volute
Hem
Seam
Relief valve
Gas; liquid
Butt
Architectural
Plot plan
Line
26 ft
77 ft
Eccentric, segmental
Flow nozzle
Ultrasonic flowmeter
4937 gal
4.57
213,904 ft3
103 ft
8064 lb
Always constant
Pressure due to the depth of water
The line that connects the piezometric surface
along a pipeline
0.28 ft
254.1 ft
6.2 × 10–8
0.86 ft
Pressure energy due to the velocity of the water
A pumping condition where the size of the impeller of the pump and above the surface of the water
from which the pump is running
The slope of the specific energy line
Alternator
The effect that causes current flow in a conductor
moving across magnetic lines of force
Mechanical, electrical
Increases, decreases, decreases, increases
To protect an electrical circuit and load
0.2 ohms
Orbits or shells
Protons and neutrons
The value of the resistor, the length of the conductors, and the diameter of the conductors
Direct current flow does not change direction,
whereas alternating current periodically changes
direction.
The magnetic poles
The flux lines, or magnetic flux along which a
magnetic force acts
Natural magnet, permanent magnets, and electromagnets
Chemistry
Battery, two
A series circuit has only one path for current
flow, whereas a parallel circuit has more than
one path.
Source voltage
Voltage drop
Counterclockwise
2 amps
12 volts
16 watts
80 watts
Less, more
Resistivity
Circular mil
Circular mil
Conductivity
Smaller
Doubles
It will withstand high voltages.
The two are directly proportional. As flux density
increases, field strength also increases.
The type of material and the flux density
North pole
Increases
141.4 volts
A voltage is induced in the conductor
AC, cut, counter
Counter
Current has an associated magnetic field
Increase
Increase
Positive-displacement
High-viscosity
Positive-displacement
High
High
Eye
Static, dynamic
Shut off
V2/2g
Total head
Head capacity, efficiency, horsepower demand
Water
Suction lift
Elevation head
Water hp and pump efficiency
Centrifugal force
Stuffing box
Impeller
Rings, impeller
Casing
A flexible piping component that absorbs thermal
and/or terminal movement
Fluid
Fluid
Connected
Flow
Pressure loss
Increases
Automatically
Insulation
Leakage
Four times
Routine preventive maintenance
12
Schedule, thickness
Increases
Ferrous
Increases
Iron oxide
Cast iron
Iron
Corrosion
Decreases
Clay, concrete, plastic, glass, or wood
Corrosion-proof
Cement
Pressed
Turbulent, lower
Steel
Na
H2SO4
7
Base
Changes
Solid, liquid, gas
Element
Compound
Periodic
Solvent, solute
An atom or group of atoms that carries a positive
or negative electrical charge as a result of having
lost or gained one or more electrons
Colloid
Turbidity
Result of dissolved chemicals
Toxicity
Organic
0; 14
Ability of water to neutralize an acid
Calcium and magnesium
Base
12.5
12.6
12.7
12.8
12.9
12.10
Spheres, rods, spirals
Typhoid, cholera, gastroenteritis
Amoebic dysentery, giardiasis
Cyst
Host
Plug screens, machinery; cause taste and odor
problems
12.11 No; bacteria is Machiavellian—it is a survivor.
Secondary maximum contaminant levels
Transpiration
Surface water
Agriculture, municipal wastewater plants, habitat
and hydrologic modifications, resource extraction,
and urban runoff and storm sewers
14.5 Solids content
14.6 Turbidity
14.7 Universal solvent
14.8 Alkalinity
14.9 Neutral state
14.10 Lead
CHAPTER 15 ANSWERS
15.1
15.2
15.3
15.4
15.5
CHAPTER 12 ANSWERS
12.1
12.2
12.3
12.4
Bacteria, viruses, protozoa
During rain storms
No
Binary fission
Muffle furnace, ceramic dishes, furnace tongs, and
insulated gloves
15 minutes
A sample collected all at one time; representative
of the conditions only at the time taken
For pH, dissolved oxygen, total residual chlorine,
fecal coliform, and any test by NPDES permit for
grab sample
A series of samples collected over a specified
period of time in proportion to flow
Collect from well-mixed location; clearly mark
sampling points; easy location to read; no large or
unusual particles; no deposits, growths, or floating
materials; corrosion-resistant containers; follow
safety procedures; test samples as soon as possible.
787
15.7
15.8
Refrigerate at 4°C.
Absorption of water during cooling, contaminants,
fingerprints, etc.
CHAPTER 16 ANSWERS
16.1
16.2
16.3
16.4
16.5
Cone of depression
12 in.
Concrete
Surface water, groundwater, GUDISW
Groundwater under the direct influence of surface
water
16.6 Easily located; softer than groundwater
16.7 The study of the properties of water and its distribution and behavior
16.8 Zone of influence
16.9 GUDISW
16.10 Prevent large material from entering the intake
CHAPTER 17 ANSWERS
17.1
A potential reserve area, usually distinct from the
treatment plant, where natural or artificial lakes are
used for water storage, natural sedimentation, and
seasonal pretreatment with or without disinfection
17.2 Collection area into which water drains
17.3 Either of two choices in water utility management—keep it out of the watershed or take it out
during treatment
17.4 Control algae and in turn decrease taste and odor
problems
17.5 Best management practices
17.6 True
17.7 True
17.8 True
17.9 False
17.10 False
A description of the soil encountered during well
construction, water quantity, well casing information, and well development and testing
Dug well
Disinfection residual, turbidity, coliform analysis
National Sanitation Foundation (NSF)
Fit for human consumption
First, determine the required chlorine feed rate:
Feed rate (lb/day) = Dose (mg/L) × flow (MGD)
× 8.34 = 0.6 mg/L × 1 MGD × 8.34 = 5.0 lb/day
If we require 5 lb/day of chlorine, we will require
more pounds of hypochlorite because it is not
100% chlorine; 68% of the hypochlorite is available chlorine, and 68% = 68/100 = 0.68. Next:
(Cl2 fraction)(hypochlorite) = Available chlorine
(0.68)(x lb/day hypochlorite) = 5.00 lb/day Cl2
x lb/day hypochlorite = 5.00/0.68
x = 7.36 lb/day hypochlorite
Public
The transport of water from one location to another
Acute
Reduction of pathogens to safe levels
Hypochlorites
Reduce the number of pathogens to safe levels in
water before the contact time is completed.
Feed rate (lb/day) = Dose (mg/L) × flow (MGD)
× 8.34
= 0.4 mg/L × 5/3 MGD × 8.34 = 17.68 lb/day Cl2
Residual = Dose – demand
= 10 (mg/L) – 2.6 (mg/L) = 7.5 mg/L
Turbidity can entrap or shield microorganisms
from the chlorine.
Feed rate (lb/day) = Dose (mg/L) × flow (MGD)
× 8.34
= 0.8 × 2.6 × 8.34 = 17.35 lb/day soda ash
Given:
Flow
0.75 MGD
Shape Circular
Size
Radius = 20 ft
Depth 10 ft
Find the detention time.
a. Find tank volume:
Volume = πr2h
Volume = π × (20 ft)2 × 10 ft = 12,560 ft3
b. Flow = 0.75 MGD × 1,000,000 = 750,000 gpd
Handbook of Water and Wastewater Treatment Plant Operations, Second Edition
18.27 Yes
18.28 Chlorine residual
18.29 A link that connects two systems and a force that
causes liquids in a system to move
18.30 Moderate
18.31 Negative; low
18.32 Peristaltic metering pump
18.33 Purchase of buffer zone around a reservoir; inspection of construction sites; public education
18.34 Given:
No. of filters 3
Size (each)
10 ft × 7 ft
Operating
1 out of service
Filtration rate 280 gal/min (total capacity for both
filters)
Find the filtration rate square foot of filter:
Area of each filter = 10 ft × 7 ft = 70 ft2
Total area of filters = 70 ft2 = 140 ft2 total
Filtration rate =
280 gal/min
= 2 gal/min/ft 2 of filter
140 ft 2
18.35 Given:
Filter area
300 ft2
Backwash rate 15 gal/ft2/min
Backwash time 8 min
Find the amount of water for backwash.
We have been given information on per foot of
filter but we want to find the total water required
to backwash the entire filter.
a. Find total filtration rate:
15 gal
= 4500 gal/min
ft 2 /min
b. Gallons per 8-minute backwash time
300 ft 2 ×
4500 gal
× 8 min = 36,000 gal used
min
Velocity = Distance traveled ÷ time
Velocity = 600 ft ÷ 5 min = 120 ft/min
Material Safety Data Sheets (MSDS)
Chlorination and filtration
Pump more than rated capacity
Hypochlorous acid
Protozoa
Removal/inactivation of most resistant pathogens
Corrosivity
Turbidity, paddles speed, pH
We want to find the velocity; therefore, we must
rearrange the general formula to solve for velocity:
V = Q/A
Given:
Q = Rate of flow = 11.2 cfs
A = Area in square feet
Width = 2.5 ft
Depth = 1.4 ft
Find the average velocity.
Step 1.
Area = Width × depth = 2.5 ft × 1.4 ft = 3.5 ft2
Step 2.
Velocity (ft/sec) = Flow rate (ft3/sec) ÷ area (ft2)
Velocity = 11.2 ft3/sec ÷ 3.5 ft2 = 32 ft/sec
Given:
Height
100 ft
Diameter 20 ft
Shape
Cylindrical
Find total gallons of water contained in the tank.
a. Find the volume in cubic feet:
Volume = 0.785 × (diameter)2 × height
= 0.785 × (20 ft)2 × 100 ft
= 0.785 × 400 ft2 × 100 ft = 31,400 ft3
b. Our problem asks how many gallons of water
will it contain:
31,400 ft3 × 7.48 gal/ft3 = 234,872 gal
Rapid mix, flocculation, sedimentation
Removal of color, suspended matter, and organics
Transform soluble ions to insoluble compounds
3 to 4 times the theoretical amount
5%
Given:
Distance 1500 ft
Time
4 min
a. Find the velocity in ft/min:
Velocity = 1500 ft ÷ 4 min = 375 ft/min
b. Convert minutes to seconds:
375 ft/min × 1 min/60 sec = 6.25 ft/sec
Gate
Achieve optimum corrosion control
50%
Sodium fluoride (NaF)
Mottled teeth enamel
0.75 mg/L
Amount of chlorine to add for breakpoint chlorination; correct amount of coagulant to use for
proper coagulation; length of flash mix; proper
amount of mixing and settling time
Corrosion control technology
Given:
Flow
350 GPM
Pipe size 6 in.
Find the velocity (ft/sec) = Distance ÷ time
a. Convert gallons to ft3:
350 gal/min ÷ 7.48 gal/ft3 = 46.8 ft3/min
b. Find cross-sectional area of pipe:
Area of circle = πr2
= 3.14 × (3 in. × 3 in.) = 28.26 in.2
c. Convert square inches to square feet:
28.26 in.2 ÷ 144 in.2/ft2 = 0.20 ft2
d. Find ft/min:
46.8 ft3/min ÷ 0.20 ft2 = 234 ft/min
e. Convert minutes to seconds:
234 ft/min × 1 min/60 sec = 3.9 ft/sec
Air, chlorine, or potassium permanganate
pH, alkalinity, hardness
Adsorption
Prior to the rapid mix basin
Before the backwash, water reaches the lip of the
wash water trough.
Chlorine
True
True
79,269 gal
Powdered activated carbon
Iron and manganese
Copper
Soluble polyvalent cations
Gains an electron in going from the +2 oxidation
state to the +3 form
Bicarbonate
Negative head
Gravity
Influent
Uniform
Maximize the conversion of organic carbon from
the dissolved phase to the particulate phase; the
removal of natural organic material; optimize the
removal of DHP precursor material.
30 hours
Phenyl arsine oxide
Given:
Surface area of pond = 20 ac
Height of water collected = 2 in.
Find the number of gallons collected in the reservoir after the storm.
a. Convert acres to ft2:
20 ac × 43,560 ft2/ac = 871,200 ft2
b. Convert inches to feet:
2 in. × 1 ft/12 in. = 0.167 ft
c. Calculate volume of water collected:
Area × height = 871,200 ft2 × 0.167 ft = 145,490 ft3
d. Convert ft3 to gallons:
145,590 ft3 × 7.48 gal/ft3 = 1,089,013 gal
20.0 lb/day Cl2
70 ft × 0.4 = 28 ft
28 ft × 0.433 = 12.1 psi
Groundwater
Aeration, boiling, adsorption
Addition of powdered activated carbon
Permeability
Water table
Waterborne
The licensed operator and the responsible official
The amount of organic material in a sample that
can be oxidized by a strong oxidizing agent
Prevent disease, protect aquatic organisms, protect
water quality
Dissolved and suspended
Organic indicates matter that is made up mainly
of carbon, hydrogen, and oxygen and will decompose into mainly carbon dioxide and water at
550°C; inorganic materials, such as salt, ferric
chloride, iron, sand, gravel, etc.
Algae, bacteria, protozoa, rotifers, virus
Carbon dioxide, water, more organics, stable solids
Toxic matter, inorganic dissolved solids, pathogenic organisms
Raw effluent
From body wastes of humans who have disease
Disease-causing
Domestic waste
Industrial waste
4.4%
2.3 ft
5250 gal × 8.34 lb/gal = 43,785 lb
14,362 gal
850.7 lb/day
686 kg/day
0.121 MGD
8477 people
9.41 lb/gal
Cutter may be sharpened or replaced when needed.
Cutter alignment must be adjusted as needed
Grit is heavy inorganic matter; sand, gravel, metal
filings, egg shells, coffee grounds, etc.
0.7 fps
A large amount of organic matter is present in the
gut. The aeration rate must be increased to prevent
settling of the organic solids.
To remove settleable and flotable solids
To remove the settleable solids formed by the biological activity
7962 gpd/ft
Stabilization pond, oxidation pond, polishing pond
Settling, anaerobic digestion of settled solids, aerobic/anaerobic decomposition of dissolved and
Handbook of Water and Wastewater Treatment Plant Operations, Second Edition
colloidal organic solids by bacteria producing stable solids and carbon dioxide, photosynthesis
Products of oxygen by algae; summer effluent is
high in solids (algae) and low in BOD; winter
effluent is low in solids and high in BOD.
Eliminates wide diurnal and seasonal variation in
pond dissolved oxygen
Standard, high rate, roughing
Increase waste rate
Decrease, decrease, decrease, increase, increase
10 containers
88 days
103 cylinders; $2823.49
4716 lb/day
21.5 lb
27 days
64.1%
National Pollutant Discharge Elimination System
By increasing the primary sludge pumping rate or
by adding dilution water
7.0 pH
Because the microorganisms have been killed or
they are absent
The time to do the test, 3 hours vs. 5 days
Dark, greasy
Increases
Temperature, pH, toxicity, waste rate, aeration tank
configuration
Can function with or without dissolved oxygen;
prefer dissolved oxygen but can use chemically
combined oxygen such as sulfate or nitrate
Organic
Living organisms
Final
Colloidal
Not possible
Aerobic, facultative
Different
Reduced
Temperature
BOD
F/M
Secondary clarifier weirs
To separate and return biosolids to the aeration
tank
Declining
1.5 and 2.5 mg/L
Increased MLVSS concentration
Decreased waste rate
Decreased MCRT
Concentration of aeration influent solids
Complete mix is more resistant to shock loads
Decrease the grit channel aeration rate
Increase
Floor level
$22.77
Anoxic
C:N:P
Secondary
Are not
2 ft/sec
Lower
Chlorine residual
2 hr
0.1
800 gpd/ft2
Monochloramine
0.2 to 0.5
Nitrogen
Decrease explosive hazard, decrease odor release,
maintain temperature, collect gas
Algae
Dissolved solids
0.0005 ppm