Chapter 9. Wastewater and Stormwater: Collection, Treatment, Resource
Recovery
9.1 Research an emerging chemical of concern that might be discharged to a wastewater
treatment plant or household septic systems. Examples would include pharmaceuticals, caffeine,
surfactants found in detergents, fragrances, and perfumes. Write an essay of up to three pages on
the concentration of this chemical found in wastewater influent. Determine whether the chemical
you are researching is treated in the plant, passes through untreated, or accumulates on the
sludge. Identify any adverse ecosystem or human health impacts that have been found for this
chemical.
Solution:
Students’ responses will vary.
9.2 Research whether there are state or regional pollution prevention programs to keep mercury
out of your local municipal wastewater treatment plant. This mercury might come from your
university laboratories or local dental offices and hospitals. What are some of the specifics of
these programs? How much mercury has been kept out of the environment since the program’s
inception?
Solution:
Students’ responses will vary.
9.3 A laboratory provides the following solids analysis for a wastewater sample: TS = 225
mg/L, TDS = 40 mg/L, FSS = 30 mg.L. (a) What is the total suspended solids concentration of
this sample? (b) Does this sample have appreciable organic matter? Why or why not?
Solution:
9.4 A 100 mL water sample is collected from the activated sludge process of municipal
wastewater treatment. The sample is placed in a drying dish (weight 5 0.5000 g before the
sample is added) and then placed in an oven at 104°C until all moisture is evaporated. The
weight of the dried dish is recorded as 0.5625 g. A similar 100 mL sample is filtered, and the 100
mL liquid sample that passes through the filter is collected and placed in another drying dish
(weight 0.5000 g before the sample is added). This sample is dried at 104°C, and the dried
dish’s weight is recorded as 0.5325 g. Determine the concentration (in mg/L) of (a) total solids,
(b) total suspended solids, (c) total dissolved solids, and (d) volatile suspended solids. (Assume
VSS=0.7TSS .).
Solution:
9.5 Obtain the World Health Organization (WHO) report on “Urine diversion: Hygienic risks
and microbial guidelines for reuse.” Review Figure 2 in Chapter 1 of this report (Introduction).
(a) How many grams of N, P, and K are excreted every day in a Swedish person’s urine?
Solution:
9.6 Humans produce 0.8 to 1.6 L of urine per day. The annual mass of phosphorus in this urine
on a per capita basis ranges from 0.2 to 0.4 kg P. a) What is the maximum concentration of
phosphorus in human urine in mg P/L? b) What is the concentration in moles P/L? c) Most of
this phosphorus is present as HPO42-. What is the concentration of phosphorus in mg HPO42–/L?
Solution:
9.7 Assume 50% of phosphorus in human excrement in found in urine (the remaining 50% is
found in feces). Assume humans produce 1 L of urine per day and the annual mass of
phosphorus in this urine is 0.3 kg P. If indoor water usage is 80 gallons per day in a single
individual apartment, what is the concentration (in mg P/L) in the wastewater that is discharged
from the apartment unit? Make sure you account for phosphorus found in urine and feces.
Solution:
9.8 The following equation shows the stoichiometry for recovery of phosphorus and nitrogen
from wastewater through preciptation of struvite.
Mg2+ + NH4+ + PO43- + 6H2O
↔
MgNH4PO4•6H2O(s)
If the composition of the wastewater under consideration for struvite recovery is 7 mg P/L, NH4+
is 25 mg NH4+ – N/L, and Mg is 40 mg Mg2+/L, is there sufficient Mg and NH4+ to precipitate all
the phosphorus, assuming it all exists as PO43-?
Solution:
9.9 Design an aerated grit chamber system to treat a 1 day sustained peak hourly flow of 1.6 m3/s
with an average flow of 0.65 m3/s. Determine: (a) the grit chamber volume (assuming two
chambers will be used); (b) the dimensions of the two grit chambers; (c) the average hydraulic
retention time in each grit chamber; (d) air requirements, assuming 0.20 m3 of air per m length of
tank per minute; and (e) the quantity of grit removed at peak flow, assuming a typical value of
0.20 m3 of grit per one thousand m3 of untreated wastewater.
Solution:
9.10 A wastewater treatment plant receives a flow of 35,000 m3/day. Calculate the required
volume (m3) for a 3 m deep horizontal flow grit chamber that will remove particles with a
specific gravity of more than 1.9 and a size greater than 0.2 mm diameter.
Solution:
hQ
9.11 A wastewater treatment plant will receive a flow of 35,000 m3/day. Calculate the surface
area (m2), diameter (m), volume (m3), and hydraulic retention time of a 3 m deep circular,
primary clarifier that would remove 50 percent of suspended solids. Assume the surface
overflow rate used for the design is 60 m3/m2–day.
Solution: (SOR means surface overflow rate)
Q
SOR
=
9.12 Assume a plant flow of 12,000 m3/day. Determine the actual detention time observed in the
field of two circular settling tanks with depth of 3.5 m that were designed to have an overflow
rate not to exceed 60 m3/m2-day and a detention time of at least 2 hr.
Solution:
9.13 A wastewater treatment plant has a flow of 35,000 m3/day. Calculate the mass of sludge
wasted each day ( expressed in kg/day) for an activated–sludge system operated at a
solids retention time (SRT) of 5 days. Assume an aeration tank volume of 1,640 m3 and an
MLSS concentration of 2,000 mg/L.
Solution:
ww
,QX
9.14 You are provided with the following information about a municipal wastewater treatment
plant. This plant uses the traditional activated-sludge process. Assume the microorganisms are 55
percent efficient at converting food to biomass, the organisms have a first-order death rate
constant of 0.05/day, and the microbes reach half of their maximum growth rate when the BOD5
concentration is 10 mg/L. There are 150,000 people in the community (their wastewater
production is 225 L/day-capita, 0.1 kg BOD5/capita-day). The effluent standard is
and . Suspended solids were measured as 4,300 mg/L in a
wastewater sample obtained from the biological reactor, 15,000 mg/L in the secondary sludge,
200 mg/L in the plant influent, and 100 mg/L in the primary clarifier effluent. SRT is equal to 4
days. (a) What is the design volume of the aeration basin (m3)? (b) What is the plant’s aeration
period (days)? (c) How many kg of secondary dry solids need to be processed daily from the
treatment plants? (d) If the sludge wastage rate (Qw) is increased in the plant, will the solids
retention time go up, go down, or remain the same? (e) Determine the F/M ratio in units of kg
BOD5/kg MLVSS–day. (f) What is the mean cell residence time?
Solution:
5
BOD = 20 mg/L
TSS = 20 mg/L
( )
6
6
L g SS
33.75 10 0.55
day g BOD
1 mg mg 0.05
day 444 0.7 20
mg SS
4 L L day
4,300 L
Solve for 4.2 10 L
V
V
×
= −−
= ×
b) Aeration period is hydraulic detention time
6
6
4.2 10 L 0.12 day 2.9 h
L
33.75 10 day
V
Q
×
= = = =
×
c) Using the SRT equation again,
( )
6
9
6
mg SS
4.2 10 L 4,300 L
4
mg SS kg kg
4.515 10 4,515
day day
10 mg
ww
ww
ww
VX VX
SRT Q X
Q X SRT days
QX
×
=⇒==
=×=
d) If Qw is increased the SRT will decrease
e)
( )
( )
6
6
L mg
33.75 10 444 0.7
day L
mg SS MLVSS
4,300 0.6 4.2 10 L
L MLSS
lbs BOD
0.97 lbs MLVSS-day
oo
QS
FM XV
×
= = ×
=
f) Mean cell residence time equals sludge age equals solids retention time = 4 days
9.15 Using information provided in Example 9.4, determine the critical SRT value (sometimes
referred to as SRTmin). This term refers to the SRT where the cells in the activated–sludge process
would be washed out or removed from the system faster than they can reproduce.
Solution:
9.16 If the specific growth rate for a completed mixed activated sludge process equals 0.10 day-1,
what is the solids retention time for this system (units of days). (b) what is the mean cell
retention time for the same system (units of day)?
Solution
9.17 In the following sentences, circle the correct term in boldface. If the solids retention time
(SRT) is low (for example, 4 days), which conditions exist? (a) The F/M ratio is low/high. (b)
The power requirements for aeration will be less/greater. (c) The microorganisms will be
starved/saturated with food. (d) The mean cell retention time is low/high. (e) The sludge age is
low/high. (f) The sludge wastage rate may have been recently increased/decreased. (g) The
MLSS may have been increased/decreased.
Solution:
9.18 The suspended-solids concentration entering a treatment and resource recovery plant is 200
mg/L in the plant influent; 3,000 mg/L in the primary sludge; 12,500 mg/L in the secondary
sludge; and 3,500 mg/L exiting the aeration basin. The concentration of total dissolved solids in
the plant influent is 350 mg/L, and the concentration of total dissolved solids exiting the aeration
basins is 2,300 mg/L. The BOD5 is 100 mg/L measured after primary treatment and 3 mg/L
exiting the plant. Total nitrogen levels in the plant are approximately 35 mg N/L.
If the F/M ratio is 0.35 gram BOD5/ gram MLSS–day, estimate the hydraulic retention time of
the aeration basins if the daily plant flow is 15 million liters.
Solution:
9.19 Determine the sludge volume index (SVI) for a test where 3 g MLSS occupy a 450 mL
volume after 30 min settling.
Solution: