CEE 370 
November 16, 2006 
Closed Book, two sheets of notes allowed
Please answer any combination of the following 9 questions such that the total potential is 100%. Show all work. Be neat, and boxin your answer.
See book or class overheads. Models include: linear, exponential, logistic, Monod
Carrying capacity, which might include such factors as population density, presence of toxins, availability of food (electron donor, and carbon source), temperature, pH, light energy (for phototrophs), disease.
First determine the Darcy’s velocity between the wells
_{}
_{}
Then determine the true velocity between the wells
_{}
_{}
Now determine the time it takes for water to travel between the two wells
_{}
From Equation 55:
_{}
_{}
_{}
_{}
so:
_{}
This may be viewed as a pseudofirst order reaction:
_{}
where: k=k_{2}[Chlorine]=2.23x10^{3}M^{1}min^{1}(0.001M)=2.23min^{1}=3,210d^{1}
and now:
_{}
_{}
1. 
T 
PFRs are
always more efficient that CMFRs for 1^{st} order processes 
2. 
T 
Chemoheterotrophs
use inorganic chemicals for energy 
3. 
F 
Primary
producers include all of the higher animals 
4. 
F 
Fick’s
first law describes advective flow 
5. 
F 
The NOAEL
is the dose level that causes 100% mortality 
6. 
T 
Darcy’s law
describes flow in porous media 
7. 
F 
Evapotranspiration
is the process by which water moves from groundwater to a river 
8. 
F 
Rotifers
are a form of algae 
9. 
T 
Protozoa
can exist in a cyst form 
10. 
T 
Most nitrogenous
BOD is in the form of ammonia or organic amines 
A. A 5day BOD test was conducted at standard temperature (20 °C) on a treated wastewater. To run this test, the analyst used 40 mL of wastewater and 260 mL of dilution water. The initial and final (5 day) DO levels in the 300 mL BOD bottle were 8.0 mg/L and 4.0 mg/L respectively. What is the BOD_{5} of the (undiluted) treated wastewater (7 points)?
B. If both the firstorder BOD bottle constant (k or k_{b}) and the deoxygenation rate constant (k_{d} or k_{1}) at 20 °C are 0.38 day^{1}, what is the ultimate BOD of the treated wastewater (7 points)?
C. Assuming that the same concentration and characteristics of carbonaceous oxygen demanding waste is in the treated wastewater throughout the year, what is the BOD concentration (L) in the river in the winter at a downstream travel time of two days (13 points)?
D. What is the initial dissolved oxygen deficit in the river at the point of mixing in the summer (6 points)?
E. In the summer the stream velocity is 0.2 m/s. How far downstream (in m) does the minimum dissolved oxygen concentration occur in the summer? What is the value of the minimum dissolved oxygen concentration (17 points)?
No one selected this problem
For a PFR:
_{}
First let’s calculate the volume of the pipe:
_{}
Next let’s determine the pseudofirst order rate constant, k:
_{}
where:
_{}
now combining:
_{}
This is also equal to 11.4 ppb.
a. What is the steady state phosphorous concentration (in mg/L) leaving the lake if there are no inputs of P other than the inflow stream?
_{}
or
0=P_{in}Q_{in}k_{s}VP_{out}P_{out}Q_{out}
0=(0.030g/m^{3})(0.8m^{3}/s)5.79x10^{7}s^{1})(5.6x10^{6}m^{3})P_{out}(0.8m^{3}/s)P_{out}
P_{out} = 0.0059 mg/L
b. If the maximum desirable in lake phosphorous concentration is 0.015 mg/L, what is the maximum allowable rate of phosphorous input (S, in g/s) due to other sources such as direct runoff?
c. What fundamental chemistry principle is the basis for the “limiting nutrient” concept in water quality?
Liebig’s
law of the minimum states that the nutrient in the shortest supply will limit
overall growth. This is determined by
comparing the ratios of the aqueous concentrations with the ratios needed for
biological cells.
d. If the nitrogen concentration in the lake is 0.18 mg/L as N, would the increased P loading in part “b” above lead to increased algal growth in the lake? Prove your answer quantitatively.
The
Redfield ratio is 16 N atoms per 1 P atom.
Therefore when nitrogen is 0.18 mg/L (13 µM), the corresponding P
concentration should be onesixteenth if both are equally limiting. This would be 0.8 µM, or 0.025 mg/L as
P. Since the actual P concentration is
well below this, it is limiting growth and any increase in P will cause an
increase in growth.
A. The first WWTP discharges 0.5 m^{3}/s of flow with an ultimate BOD (BOD_{ult} or L) of 45 mg/L. The BOD bottle rate constant (k_{b} or k) is 0.4375 d^{1}. The dissolved oxygen of this waste flow is 1.5 mg/L. Calculate the BOD and dissolved oxygen level at a point 5 km downstream (which is immediately upstream of the second WWTP discharge (25 points).
First calculate the velocity after the initial point of mixing:
_{}
Next determine the deoxygenation rate constant:
_{}
And the reaeration rate:
_{}
Now determine the BODu and D at the point of mixing
_{}
And for BOD
_{}
Now for Dissolved oxygen
_{}
Which means the initial deficit is: 9.177.21 = 1.96 mg/L
The time of travel to 5 km is:
_{}
So the downstream ultimate BOD is:
_{}
Now the downstream deficit is determined by the Streeter Phelps equation:
_{}
Which gives a dissolved oxygen concentration of 9.173.40 = 5.77 mg/L
B. The second WWTP discharges 5 km downstream of the first one. Its wastewater flow is 0.3 m^{3}/s and it has an ultimate BOD (BOD_{ult} or L) of 55 mg/L. As with the first wastewater discharge, the BOD bottle rate constant (k_{b} or k) is 0.4375 d^{1}. The dissolved oxygen of this waste flow is 0.5 mg/L. Calculate the BOD and dissolved oxygen level at a point 5 km downstream of the second WWTP discharge (i.e., 10 km downstream of the first discharge) (25 points).
You need to reevaluate the “initial” DO and BODs for this second reach of the river
First calculate the velocity after the initial point of mixing:
_{}
Next determine the deoxygenation rate constant:
_{}
And the reaeration rate:
_{}
Now determine the BODu and D at the point of mixing
_{}
And for BOD
_{}
Now for Dissolved oxygen
_{}
Which means the initial deficit is: 9.175.16 = 4.01 mg/L
The time of travel to 5 km is:
_{}
So the downstream ultimate BOD is:
_{}
Now the downstream deficit is determined by the Streeter Phelps equation:
_{}
Which gives a dissolved oxygen concentration of 9.174.82 = 4.35 mg/L
Some
additional useful information:
O’Connor Dobbins equation:
_{}
Where: k_{r} (the reaeration coefficient; same as k_{2} or k_{a}) is in d^{1}, u (velocity) is in m/s and h (depth) is in m.
Selected Chemical Constants
Element 
Symbol 
Atomic # 
Atomic Wt. 

Electronegativity 

Aluminum 
Al 
13 
26.98 
3 
1.47 

Boron 
B 
5 
10.81 
3 
2.01 

Calcium 
Ca 
20 
40.08 
2 
1.04 

Carbon 
C 
6 
12.01 
2,4 
2.50 

Cerium 
Ce 
58 
140.12 
3,4 
1.06 

Helium 
He 
2 
4.00 
0 


Holmiuum 
Ho 
67 
164.93 
3 
1.10 

Hydrogen 
H 
1 
1.01 
1 
2.20 

Magnesium 
Mg 
12 
24.31 
2 
1.23 

Manganese 
Mn 
25 
54.94 
2,3,4,6,7 
1.60 

Osmium 
Os 
76 
190.2 
2,3,4,8 
1.52 

Oxygen 
O 
8 
16.00 
2 
3.50 

Potassium 
K 
19 
39.10 
1 
0.91 

Sodium 
Na 
11 
22.99 
1 
1.01 

Sulfur 
S 
16 
32.06 
2,4,6 
2.44 

Selected Acidity Constants (Aqueous Solution, 25°C, I = 0)
NAME 
FORMULA 
pK_{a} 

Hydrochloric acid 
HCl = H^{+ }+ Cl^{} 
3 

Sulfuric acid 
H_{2}SO_{4}= H^{+ }+ HSO_{4}^{}_{} 
3 

Nitric acid 
HNO_{3 }= H^{+ }+ NO_{3}^{}_{} 
0 

Bisulfate ion 
HSO_{4}^{}_{ }= H^{+ }+ SO_{4}^{2} 
^{ }2 

Phosphoric acid 
H_{3}PO_{4 }= H^{+} + H_{2}PO_{4}^{}_{} 
_{ }2.15 

Hydrofluoric acid 
HF = H^{+} + F^{} 
^{ }3.2 

Nitrous acid 
HNO_{2 }= H^{+} + NO_{2}^{}_{} 
_{ }4.5 

Acetic acid 
CH_{3}COOH = H^{+} + CH_{3}COO^{} 
^{ }4.75 

Propionic acid 
C_{2}H_{5}COOH = H^{+} + C_{2}H_{5}COO^{} 
^{ }4.87 

Carbonic acid 
H_{2}CO_{3 }= H^{+} + HCO_{3}^{}_{} 
_{ }6.35 

Hydrogen sulfide 
H_{2}S = H^{+} + HS^{}_{} 
_{ }7.02 

Dihydrogen phosphate 
H_{2}PO_{4}^{}_{ }= H^{+} + HPO_{4}^{2} 
^{ }7.2 

Hypochlorous acid 
HOCl = H^{+} + OCl^{} 
^{ }7.5 

Ammonium ion 
NH_{4}^{+}_{ }= H^{+} + NH_{3} 
_{ }9.24 

Hydrocyanic acid 
HCN = H^{+} + CN^{} 
^{ }9.3 

Phenol 
C_{6}H_{5}OH = H^{+} + C_{6}H_{5}O^{} 
^{ }9.9 

Bicarbonate ion 
HCO_{3}^{}_{ }= H^{+} + CO_{3}^{2} 
10.33 

Monohydrogen phosphate 
HPO_{4}^{2} = H^{+} + PO_{4}^{3} 
12.3 

Bisulfide ion 
HS^{} = H^{+} + S^{2} 
13.9 

Temperature °C 
DO saturation (C_{s}) in mg/L 
0 
14.62 
1 
14.23 
2 
13.84 
3 
13.48 
4 
13.13 
5 
12.80 
6 
12.48 
7 
12.17 
8 
11.87 
9 
11.59 
10 
11.33 
11 
11.08 
12 
10.83 
13 
10.60 
14 
10.37 
15 
10.15 
16 
9.95 
17 
9.74 
18 
9.54 
19 
9.35 
20 
9.17 
21 
8.99 
22 
8.83 
23 
8.68 
24 
8.53 
25 
8.38 
26 
8.22 
27 
8.07 
28 
7.92 
29 
7.77 
30 
7.63 
31 
7.51 
32 
7.42 
33 
7.28 
34 
7.17 
35 
7.07 
36 
6.96 
37 
6.86 
38 
6.75 
Use a θ (theta) of 1.135 for adjusting deoxygenation coefficients
Conversion factors: 1 gal = 3.7854x10^{3} m^{3}, 1 ft = 0.3048 m
Other Constants of interest: