environmental

1) If the 5-day BOD of an undiluted sample is 5 ppm and the DO
(or [O2]) after 5 days is: (Assume the initial DO is 9.1 ppm.)

5 ppm

0 ppm

9.1 ppm

4.1 ppm

2) The BOD5 of a sample is 33 ppm. The BOD7 is most probably:

> 33 ppm

= 33 ppm

< 33 ppm

3) The BOD5 of your sample is 1000 mg/L. You add 10 mL of the
sample to 290 mL of water (for a total volume of 300 mL). The
initial DO is 8.3 ppm. What is the DO after 5 days

0

8 ppm

5 ppm

3.5 ppm

4) You want to test disinfected (e.g. all the microorganisms
have been killed) wastewater for BOD5. You should use a:

Unseeded BOD test

Seeded BOD test

5) In a seeded BOD test the “seed” refers to:

The oxygen

The microorganisms

The BOD

The nutrients

6) A BOD5 bottle would best be represented as a:

CSTR

Batch reactor

PFR

7) If the temperature of a sample increases, the BOD5 will:

Increase

Stay the same

Decrease

8) An increase in temperature causes the DOs (saturation
concentration of DO) to:

Stay the same

Decrease

Increase

9) As you move down the stream in a DO sag curve problem,
which of the following changes (assuming constant temperature)

a) DO

b) k

c) L0

d)DOs

10) What best describes the environmental impact of releasing
wastewater with high level of BOD into a river

a) The oxygen levels in the river will decrease and that will
damage aquatic life.

b) The pH will increase and that will damage aquatic life.

c) Solid material will accumulate on the river floor.

d) Toxins will be released that will damage aquatic life.

11) Algae have the formula C6H15O6N. In the presence of oxygen
and microbes:

a C6H15O6N + b O2à c CO2 + d H2O + e NH3

NH3 + 2 O2à NO3- + H+ + H2O

a. Balance the chemical equation.

b. Calculate the ultimate CBOD for a 100 mg/L solution of
algae. (97.5 ppm)

c. Calculate the NBOD for a 100 mg/L solution of algae. (32.5
ppm)

Calculate 5-day BOD of a 100 mg/L solution of algae with a k =
0.2 day-1. (61.6ppm)

d. If you were measuring the 5-day BOD in the lab, how many
mL of this solution would you add to a 300 mL BOD bottle to ensure
that the final DO doesn’t drop below 1 ppm Assume that the initial
DO is 8.26 ppm. (35.4 mL)

12) A mixture consisting of 30 mL of waste and 270 mL of seeded
dilution water has an initial DO of 8.55 mg/L; after five days, it
has a final DO of 2.40 mg/L.

a. Graph the DO vs. time and the BOD vs. time in the mixture
bottle. Assume the rate constant is 0.2 day-1. (hint: first you
will need to calculate BOD5 and Lo in the bottle. Also – the time
should go from 0 days to 12 days. Please include the excel graph in
your solutions.

b. At what time will the oxygen be zero in the mixture bottle
(10.5 days)

c. This sample requires a seeded BOD test. Given an example
of the kind of samples that required a seeded BOD test.

d. The seeded bottle contains just the seeded dilution water
and has an initial DO of 8.75 mg/L and a final DO of 8.53 mg/L.
Find the five-day BOD of the waste. (59.5 ppm)

13) The Metropolitan Wastewater Treatment plant in St. Paul (180
MGD; BOD5 = 220 mg/l, DO = 3 ppm) discharges into the Mississippi
River (flow = 18,000 ft3/s; mean depth = 14 ft; mean width = 0.3
miles). The dissolved oxygen in the river is at the saturation
concentration of 9.0 ppm and has an ultimate BOD of 2.0 mg/L. The
deoxygenation rate for BOD is 0.23 day-1 and the re-aeration rate
is 0.15 day-1.

a. Calculate the ultimate BOD in the Wastewater discharge.
(322 ppm)

b. Calculate the ultimate BOD in the river at the point where
the discharge and the river meet. What is the value of the initial
deficit to be used in the DO sag curve (0.09 ppm)

c. Graph DO (in ppm) as a function of distance in miles.
Print out the graph and attach it to the homework.

d. What is the distance where the concentration of DO would
be a minimum (feel free to just label the graph) (70 miles)

e. Calculate the minimum DO at the critical time (feel free
to label the graph) (5.87 ppm)

f. What is the minimum DO concentration if kr is 0.1 day-1
instead of 0.15

day-1 (To answer you can either print out a new graph or solve
the equation to find the answer.) (5.33 ppm)

g. Name one assumption that you made to solve this problem.
What would happen if this assumption is not true (note: You can use
the givens as assumptions since in the real world you would need to
question these givens.

14) Wastewater from a paper mill contains many organic
compounds including lignin C31H34O11.

a. Calculate the ultimate BOD resulting from 150 ppm lignin.

b. Assume that you using an unseeded BOD5 test. You know the
ultimate BOD is 200 ppm. What volume of sample would you add to a
300 mL 5-day BOD bottle if you want the final DO after 5 days to be
2 ppm Assume the initial DO is 9 ppm and the k = 0.15 day-1.

c. Now assume that the discharge has a 5-day of 180 ppm and a
DO of 2.5 ppm. This water is discharged at a rate of 0.2 MGD into a
river with DO = DOs of 9.09 ppm and no BOD. The river has a flow of
1 million gallons per day before the discharge. kd = 0.18 day-1 and
kr = 0.2 day-1. The area of the river is 100 ft2.At what distance
downstream will the DO be a minimum

Imagine that initially the temperature is 25 °C and then it
changes to 20 °C. Explain in words what would happen to the DO when
the temperature changes.

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