Q1) 10,000 gallons per day of wastewater enters a 1500 gallon activated sludge tank. The raw wastewater has a SS = 200 ppm and a BOD = 230 ppm. The MLSS needs to be 2800 ppm and the effluent SS should be less than 10 ppm. The total wasted sludge has a Q of 5 GPD with a SS concentration of 6000 ppm.
 Calculate the recycle flow, Q_{r}, required. (8710 GPD)
 Calculate the F/M ratio in lbs BOD/lbs SS – day (0.55 lb BOD/lb SSday)
 Calculate the BOD Loading in lbs BOD/day – 1000 ft^{3}(95.6)
 Based on these calculations, is the activated sludge tank over designed or under designed? What could you do to remedy this situation?
Q2) Colloidal gold has been released into the water supply from an electronics manufacturing plant. These particles are 0.1 mm in size with a specific gravity of 19.3. Your water treatment plant has 3 clarifiers each with a volume of 0.3 million gallons and a height of 6 ft. The total flow going into all 3 clarifiers is 5 MGD.
 Will these particles settle in this clarifier or will we be drinking gold? Assume that they are not flocculated. Defend your answer quantitatively.
 What is the minimum detention time needed to settle these gold particles over a height of 6 ft? (214 days)
 If you cannot remove these gold particles, should you add a sand filter or a reverse osmosis unit? Give two reasons for your answer and state your assumptions.
 If you decide to flocculate these particles, what is the minimum diameter of a gold particle (with a specific gravity of 19.3) that can be removed in this clarifier? (3.4 mm)
Q3) You are given the following information:
Conc,
mg/L 
MW
g/mole 
EW
mg/meq 
Conc,
meq/L 
Conc
mg/L as CaCO_{3} 

Ca^{2+}  180  40  
Mg^{2+}  24.3  173  
Na^{+}  30  23  
CO_{3}^{2}–  60  169  
HCO_{3}^{–}  8  61  
SO_{4}^{2}  11  96  
OH^{–}  17 
 Calculate the hardness in meq/L. (12.5 meq/L)
Q4) An activated sludge tank treats 1 MGD from the primary clarifier and has an [SS], or X of 2200 ppm. The water effluent is 0.9 MGD. 1.0 MGD of the sludge is recycled back into the tank and has an [SS] of 3980 ppm. The concentration of BOD exiting the primary clarifier is 200 ppm.a. Calculate the minimum volume of the tank that will satisfy the following requirements: (Hint: Calculate the volume for each condition and then determine which volume satisfied all the requirements).
 BOD loading < 35 lb BOD/1000 ft^{3}– day
 Sludge age > 5 days
iii. Aeration period > 4 hours
Q5) 10,000 gallons per day of wastewater enters a 1500 gallon activated sludge tank. The raw wastewater has a SS = 200 ppm and a BOD = 230 ppm. The MLSS needs to be 2800 ppm and the effluent SS should be less than 10 ppm. The total wasted sludge has a Q of 5 GPD with a SS concentration of 6000 ppm.
 Calculate the recycle flow, Q_{r}, required. (8710 GPD)
 Calculate the F/M ratio in lbs BOD/lbs SS – day (0.55 lb BOD/lb SSday)
 Calculate the BOD Loading in lbs BOD/day – 1000 ft^{3}(95.6)
Based on these calculations, is the activated sludge tank over designed or under designed? What could you do to remedy this situation
Q6) Water flows out of the primary clarifier with a SS = 200 ppm and a BOD = 250 ppm. An activated sludge tank treats 5 MGD from the primary clarifier and has an [SS], or X of 2500 ppm. The water effluent is 95% of the influent and 50% of the sludge is recycled back into the activated sludge tank with an [SS] of 6800 ppm. What is Qe? in MGD
Q7) Water flows out of the primary clarifier with a SS = 200 ppm and a BOD = 250 ppm. An activated sludge tank treats 5 MGD from the primary clarifier and has an [SS], or X of 2500 ppm. The water effluent is 95% of the influent and 50% of the sludge is recycled back into the activated sludge tank with an [SS] of 6800 ppm. What is Qr? in MGD
Q8) Water flows out of the primary clarifier with a SS = 200 ppm and a BOD = 250 ppm. An activated sludge tank treats 5 MGD from the primary clarifier and has an [SS], or X of 2500 ppm. The water effluent is 95% of the influent and 50% of the sludge is recycled back into the activated sludge tank with an [SS] of 6800 ppm. What is Qc (the flow leaving the bottom of the secondary clarifier)? in MGD
Q9) Water flows out of the primary clarifier with a SS = 200 ppm and a BOD = 250 ppm. An activated sludge tank treats 5 MGD from the primary clarifier and has an [SS], or X of 2500 ppm. The water effluent is 95% of the influent and 50% of the sludge is recycled back into the activated sludge tank with an [SS] of 6800 ppm. What is Xw? in ppm
Q10)The BOD loading you calculate is 53 lb BOD/1000 ft3day. The BOD loading should be between 30 – 50 lb BOD/1000 ft3day. Is the tank over or under designed?
under designed 
over designed 
It is just fine 
Q11)BOD in wastewater can be represented in all the following ways EXCEPT:
SS 
BOD 
S 
F 
Q12)n an activated sludge tank it is best if the microorganisms have:
less food than they need so that all the food is gone. 
more food than they need so that the microorganisms continue to reproduce. 
no food so they all die. 