1.   1  1) Algae have the formula C6H15O6N.  In the presence of oxygen and microbes:

a C6H15O6N + b O2 à c CO2 + d H2O + e NH3

NH3 + 2 O2 à 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)

 

 

1.     2) 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)

 

3)     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.)

 

 

4)     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?

d.     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.