Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
5.19 When food supplies have been exhausted, populations die away. This exponential decay is
described by a simple modification of the exponential-growth model. Engineers use this model to
calculate the length of time that a swimming beach must remain closed following pollution with
fecal material. For a population of bacteria with an initial biomass of 100 mg/L and a
d
0.4 / dayk=
,
calculate the time necessary to reduce the population size to 10 mg/L.
Solution:
Under conditions of exhausted food supplies (S=0), the overall population growth model is used
max 1d
s
dt K K S
+
However, because S equals zero, the above expression reduces to a first order decay
d
dX kX
dT =−×
The above expression can be solved
( )
d
kt
to
XXe
−×
= ×
And this expression can be solved for time, t. Then given values can be substituted as follows:
Xo = 100 mg/L, Xt = 10 mg/L, and kd= 0.4/day.
10 /
ln ln 100 / 5.8
0.4 /
t
o
d
Xmg L
Xmg L
t days
k day
−−
= = =
5.20 A population having a biomass of 2 mg/L at
0t=
days reaches a biomass of 139 at
10t=
days. Assuming exponential growth, calculate the value of the specific growth coefficient.
Solution:
5.21 Fecal bacteria occupy the guts of warm-blooded animals and do not grow in the natural
environment. Their population dynamics in lakes and rivers—that is, following a discharge of
untreated domestic wastewater—can be described as one of exponential decay or death. How
many days would it take for a bacteria concentration of 106 cell/mL to be reduced to the public
health standard of 102 cell/mL if the decay coefficient is 2/day?
Solution:
2
6
10 /
ln 10 / 4.6
2/
kt
to
X Xe
cells mL
cells mL days
day
−×
= ×
−
=
5.22 The 2012 Living Planet Report from the World Wildlife Fund (WWF, 2012) reported that
in the year 2008, the Earth’s total biocapacity was 12.0 billion global hectares, which equates to
1.8 global hectares per person. In contract, humanity’s ecological footprint was reported to be
18.2 billion global hectares, which equates to 2.7 global hectares per person. (a) Using these
values, how many years would it take the Earth to fully regenerate the renewable resources that
humanity consumed in one year. (b) “Ecological Overshoot” is a term that describes when the
global ecological footprint is larger than the Earth’s biocapacity. What was the “ecological
overshoot” in 2008 reported in global hectares per person? (c) Review Chapter 1 of the Living
Planet Report, what percent did the the global living planet index decline between 1970 and
2008? What percent did the freshwater index decline over the same time period?
Solution:
Solutions Manual prepared by: Colleen Naughton, Ziad Katirji, Heather E. Wright Wendel, and
James Mihelcic
Environmental Engineering: Fundamentals, Sustainability, Design, 2nd Edition
James R. Mihelcic and Julie Beth Zimmerman, John Wiley & Sons, New York, 2014.
c) The Global Living Planet Index is depicted on page 18 of the report. The percent decline of
the global living planet index was -28% from 1970 to 2008.
A graph of the freshwater living planet index over time is found on page 25 of the report. From
1970-2008, this index declined 37%.
5.23 According to the Web site maintained by Redefining Progress, its latest Footprint Analysis
indicates that humans are exceeding their ecological limits by 39 percent. Go to the following
Web site, and determine your own ecological footprint. Record your value, and compare it with
those for your country and the world. Identify some changes you can make in your current
lifestyle, and then rerun the footprint calculator to reflect those changes. Summarize the changes
you make and how they affect your ecological footprint. The Web site is
www.rprogress.org/ecological_footprint/about_ecological_footprint.htm.
Solution:
Students’ answers may vary
5.24 Remediation of toluene in a contaminated groundwater aquifer has been found to have the
following biokinetic coefficients for microbial growth. µmax = 1.2/day and Ks = 0.31 mg/L.
What is the growth rate of the microorganism (day-1) removing the toluene if the concentration
of the pollutant is 1 ppb and 1 ppm?
Solution:
5.25 A biological treatment process used to treat wastewater was found to have the following
biokinetic coefficients: yield coefficient = 0.52 mg VSS/mg COD, half saturation constant = 60
mg COD/L, and maximum specific growth rate = 0.96/day. What is the growth rate of the
organisms (units of day-1) if the organic matter in the reactor is: (a) a low strength wastewater
with 125 mg COD/L, (b) a high strength wastewater with COD = 325 mg/L? (c) if the
concentration of microorganisms in the biological reactor is 1,000 mg VSS/L, what is the rate of
COD utilization?
Solution:
5.26 What is the ThOD of the following chemicals? Show the balanced stoichiometric equation
with your work: (a) 5 mg/L C7H3; (b) 0.5 mg/L C6Cl5OH; (c) C12H10.
5.27 A waste contains 100 mg/L ethylene glycol (C2H6O2) and 50 mg/L NH3-N. Determine the
theoretical carbonaceous and the theoretical nitrogenous oxygen demand of the waste.
Solution:
5.28 Calculate the NBOD and ThOD of a waste containing 100 mg/L isopropanol (C3H7OH)
and 100 mg/L NH3-N.
Solution:
5.29 A waste contains 100 mg/L acetic acid (CH3COOH) and 50 mg/L NH3-N. Determine the
theoretical carbonaceous oxygen demand, the theoretical nitrogenous oxygen demand, and the
total theoretical oxygen demand of the waste.
Solution:
5.30 A waste has an ultimate CBOD of 1,000 mg/L and a kL of 0.1/day. What is its 5-day
CBOD?
Solution:
( )
( )
5
0.1/ 5
5
1
1,000 1 393
L
kt
o
day day
yL e
mg mg
ye
LL
−×
−×
= ×−
= ×− =
5.31 A new manufacturing facility is being located in your town. It plans to produce 2,000
m3/day of a wastewater that consists primarily of water and the chemical phenol dissolved in it at
a concentration of 5 mg/L. Phenol has a chemical formula of C6H5OH. The company has asked
the municipal wastewater treatment to consider treating this industrial waste. Your plant
currently treats 30,000 m3/day with an average influent of 350 mg COD/L. (a) Estimate the
increase in COD loading (kg COD/day) if you accept the industrial waste discharge? (b)
Estimate the additional amount of oxygen (in kg O2/day) needed to oxidize the phenol at the
treatment plant.
Solution:
5.32 Untreated municipal wastewater in Europe may average 600 mg/L for the carbonaceous 5-
day BOD while in the U.S. this average value can be as low as 200 mg/L. One reason for this is
because the U.S. has a greater water use per capita in the home than in Europe and also has
problems associated with the infiltration/inflow of water into their wastewater collection system.
(a) If the BOD rate constant for untreated wastewater is 0.35/day, calculate the ultimate BOD of
the untreated European and U.S. wastewaters. (b) Assume the dissolved oxygen concentration of
oxygen saturated dilution water used in the BOD test is 8 mg/L and you are using a 300-mL
BOD bottle. Estimate the volume of sample you would add to the BOD bottle to ensure
satisfactory test results for the European and U.S. samples (mL).
Solution:
5.33 (a) Calculate the ultimate BOD of a waste that has a measured 5-day BOD of 20 mg/L,
assuming a BOD rate coefficient of 0.15/day measured at 20oC. (b) Estimate the rate coefficient
and resulting BOD ultimate if the temperature of the waste is raised to 30oC.
Solution:
5.34 A 5 mL wastewater sample is placed in a standard 300 mL BOD bottle, and the bottle is
filled with dilution water. The bottle had an initial dissolved-oxygen concentration of 9 mg/L and
a final dissolved-oxygen concentration of 3.5 mg/L. A blank (a BOD bottle filled with dilution
water) run in parallel showed no change in dissolved oxygen over the 5-day incubation period.
The BOD reaction rate coefficient for the waste is 0.3/day. Calculate the 5-day (y5) and ultimate
(L0) BOD of the wastewater.
Solution:
