7.34 A combined sewer overflow (CSO) discharges an effluent containing 0 mg/L of
dissolved oxygen to a stream that has a dissolved-oxygen concentration of 7 mg/L
upstream of the discharge. Calculate the dissolved-oxygen deficit at the mixing basin if
the saturation dissolved oxygen for the river is 9 mg/L. Assume that the CSO flowrate is
one half of the stream flowrate.
Solution:
7.35 A river traveling at a velocity of 10 km/day has a dissolved-oxygen content of 5
mg/L and an ultimate CBOD of 25 mg/L at distance x = 0 km, that is, immediately
downstream of a waste discharge. The waste has a CBOD decay coefficient k1 of 0.2/day.
The stream has a reaeration rate coefficient k2 of 0.4/day and a saturation dissolved-
oxygen concentration of 9 mg/L. (a) What is the initial dissolved-oxygen deficit? (b)
What is the location of the critical point, in time and distance? (c) What is the dissolved–
oxygen deficit at the critical point? (d) What is the dissolved-oxygen concentration at the
critical point?
a)
095 4
sat act L
mg
D DO DO ⇒= − −=
b) Use Equation 8-17 to determine the critical time and knowledge of the river’s velocity
to determine the critical distance:
( )
0 21
2
21 1 10
1ln 1
crit
D kk
k
tk k k kL
×−
= × ×−
−×
( )
4 0.4 / 0.2 /
1 0.4 /
ln 1
0.4 / 0.2 / 0.2 / 0.2 / 25
crit
mg
L
mg
L
day day
day
tday day day day
×−
= × ×−
−×
2.6
2.6 10 / 26
crit
crit
t day
x day km day km
=
=×=
c) Use Equation 7-16 to determine the oxygen deficit and
( )
( )
12 2
10
0
21
kt kt kt
t
kL
D e e De
kk
−× −× −×
×
= × − +×
−
( )
( )
0.2/ 2.6 0.4/ 2.6 0.4/ 2.6
0.2 / 25
4
0.4 / 0.2 /
day day day day day day
t
mg
mg
L
L
day
D ee e
day day
−× −× −×
×
= × − +×
−
7.4
t
mg
DL
=
This value is the deficit, not the concentration. The concentration of dissolved oxygen in
the stream is then 9 mg/L – 7.4 mg/L = 1.6 mg/L.
7
7.36 The wastewater treatment plant for Pine City discharges 1 × 105 m3/day of treated
waste to the Pine River. Immediately upstream of the treatment plant, the Pine River has
an ultimate CBOD of 2 mg/L and a flow of 9 × 105 m3/day. At a distance of 20 km
downstream of the treatment plant, the Pine River has an ultimate CBOD of 10 mg/L.
The state’s Department of Environmental Quality (DEQ) has set an ultimate CBOD
discharge limit for the treatment plant of 2,000 kg/day. The river has a velocity of 20
km/day. The CBOD decay coefficient is 0.1/day. Is the plant in violation of the DEQ
discharge limit?
Solution:
7.37 An industry discharges 0.5 m3/s of a waste with a 5-day CBOD of 500 mg/L to a
river with a flow of 2 m3/s and a 5–day CBOD of 2 mg/L. Calculate the 5–day CBOD of
the river after mixing with the waste.
Solution:
7.38 A high strength waste having an ultimate CBOD of 1,000 mg/L is discharged to a
river at a rate of 2 m3/s. The river has an ultimate CBOD of 10 mg/L and is flowing at a
rate of 8 m3/s. Assuming a reaction rate coefficient of 0.1/day, calculate the ultimate and
5-day CBOD of the waste at the point of discharge (0 km) and 20 km downstream. The
river is flowing at a velocity of 10 km/day.
Solution:
7.39 A new wastewater treatment plant proposes a discharge of 5 m3/s of treated waste to
a river. State regulations prohibit discharges that would raise the ultimate CBOD of the
river above 10 mg/L. The river has a flow of 5 m3/s and an ultimate CBOD of 2 mg/L.
Calculate the maximum 5-day CBOD that can be discharged without violating state
regulations. Assume a CBOD decay coefficient of 0.1/day for both the river and the
proposed treatment plant.
Solution:
7.40 A river flowing with a velocity of 20 km/day has an ultimate CBOD of 20 mg/L. If
the organic matter has a decay coefficient of 0.2/day, what is the ultimate CBOD 40 km
downstream?
Solution:
7.41 A river traveling at a velocity of 10 km/day has an initial oxygen deficit of 4 mg/L
and an ultimate CBOD of 10 mg/L. The CBOD has a decay coefficient of 0.2/day, and
the stream’s reaeration coefficient is 0.4/day. What is the location of the critical point: (a)
in time; (b) in distance?
Solution:
7.42 A paper mill discharges its waste (kL = 0.05/day) to a river flowing with a velocity
of 20 km/day. After mixing with the waste, the river has an ultimate carbonaceous BOD
of 50 mg/L. Calculate the 5-day carbonaceous BOD at that location and the ultimate
carbonaceous BOD remaining 10 km downstream.
Solution:
0.05/ 5
mg mg
day day
−×
7.43 For each of the following cases, assuming all other things unchanged, describe the
effect of the following parameter variations on the magnitude of the maximum oxygen
deficit in a river. Use the following symbols to indicate your answers: increase (+),
decrease (–), or remain the same (=).
Parameter
Magnitude of the Deficit
Increased initial deficit
——————
Increased ultimate CBOD @ x =0
——————
Increased deoxygenation rate
——————
Increased reaeration rate
——————
Increased ThOD @ x = 0
Solution:
7.44 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:
7.45 Assume 66% of phosphorus in human excrement in found in urine (the remaining
34% 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
capita per day in a single individual apartment, what is concentration (in mg P/L) in the
wastewater that is discharged from the apartment unit? Account for phosphorus in urine
and feces.
Solution:
7.46 Contact your local wastewater treatment plant to find out the average daily treated
flow rate and average concentration of phosphorus in the untreated influent and treated
effluent. Use census data to determine the current population of your area and assuming
a growth rate of 3%, the population in 2025 and 2050. (a) If nothing is done in how the
plant treats phosphorus and how each human discharges phosphorus, what is the current
and future P loading (kg P/day) to the local surface water that takes the plant effluent? (b)
Identify one technical and two nontechnical solutions to reduce future phosphorus
loading to the wastewater treatment plant. (c) If 50 percent of the treated wastewater is
reclaimed and applied to land for residential and agricultural purposes, how would the
current loading of phosphorus to local water change? (assume all the reclaimed water
infiltrates to groundwater)
Solution:
Students’ responses will vary.