ENVIRONMENTAL SCIENCE AND ENGINEERING
ASSIGNMENT NO. 4: WATER QUALITY
Name: Manglo, John Andrae B.
Prepare clean and clear solutions to the following problems.
A new wastewater treatment plant proposes a discharge of 5 m3/s of treated waste to a river. Regulations prohibit
discharges that would raise the ultimate BOD of the river above 10 mg/L. The river has a flow of 5 m 3/s and an ultimate
BOD of 2 mg/L. Calculate the maximum 5-day BOD that can be discharged without violating the regulations. Assume a
decay coefficient of 0.1 /d for both the river and the proposed treatment plant.
Treatment Plant:
river:
Q= 5 m3/s
L= 2 mg/L
L5D= 10 mg/L
Q= 5 m3/L
K= 0.1/d
Solution:
L5D= L0e-0.1/d(5.79)
BOD-5
X= 7.02 mg/L
10 mg/L = L0 e-0.1/d(5.79)
X= L(1-e(-0.1/d)(5)
L0 = 17.84 mg/L
X= 17.84 mg/L(1-e(-0.1/d)(5))
1.
2.
Below a discharge from wastewater treatment plant, an 8.6-km stream has a reoxygenation constant of 0.4 /d, a velocity
of 0.15 m/s, a dissolved oxygen concentration of 6 mg/L and an ultimate oxygen demand of 25 mg/L. The stream is at 15
C. The deoxygenation constant is estimated at 0.25 /d. Will there be fish in this stream?
t = (8.6-km)/( 0.15 m/s) * (1000m/1km) * (1h/3600s) * (1d/24h) = 0.66d
Def= [(.25/d)(25mg/L)/(0.4-0.25)/d](e-.25(0.66) - e-.4(0.66))=3.33 mg/L
Def= s – c
s= Def + c = (3.33 + 6)= 9.33 mg/L , since 9.33>5, it can be assumed that there are fishes in this stream.
3.
A laboratory runs a solids test. The weight of the crucible is 48.6212 g. A 100-mL sample is placed in the crucible and the
water is evaporated. The weight of the crucible and dry solids is 48.6432 g. The crucible is then placed in a 600 C furnace
for 24 hours and cooled in a dessicator. The weight of the cooled crucible and residue is 48.6300 g. Find the
concentration (in ppm) of the different types of solids.
W1= 48.6212 g
Total sample:
W1+ drysolids = 48.6432 g
Wr= 48.6300 g
Solution:
Total solids= [(48.6432- 48.6212)/100 mL)]= 2.2*10-4 g/mL or 220 mg/L
Fixed solids= [(48.6300- 48.6212)/100 mL)]= 8.8*10-5 g/mL or 88 mg/L
Total Volatile Solids= TS-FS= 220 mg/L-88 mg/L= 132 mg/L
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ENVIRONMENTAL SCIENCE AND ENGINEERING
4.
A toxic contaminant is released 1 km upstream of a drinking water supply well. The relevant aquifer properties are as
follows: hydraulic conductivity 10-5 m/s; porosity 0.3; hydraulic gradient 10-2. Determine how long it will take for the
contaminant to reach the drinking water supply well.
V= [(10-5)(10-2)/0.3]= 3.33*10-7
t= [(1km(1000m/1km))/ 3.33*10-7]=3.003*109 s
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