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p. 1/5 ENV 4001: E NVIRONMENTAL S YSTEMS E NGINEERING Spring 2010 University of South Florida Homework #8 Civil & Environmental Eng. Due: Friday, April 2 Prof. J. A. Cunningham Each homework assignment will be graded on a 100-point scale. Point values for this assignment are indicated for each question. Background information: Recall from Homework #6 that a wastewater treatment plant in the city of Nastyville operates with only primary treatment, and they can’t meet the requirements of the Clean Water Act, so now they have to add secondary treatment to their facility. You work for an engineering firm that has been hired to design and build an activated sludge system for secondary treatment at the wastewater treatment plant. To design the activated sludge system, you must specify three things: the volume of the aeration basin , the oxygen input into the aeration basin , and the sludge return (recycle) rate . You also have to find the solids retention time and indicate how much waste sludge they will have to handle as a result of your design. (Another engineer in your firm will design a sludge-handling system based on what you tell him/her.) NOTE: In 2010 we will skip the oxygen input; see me in office hours if you are interested in learning about this. The design flow rate is 150 million gallons per day (6.6 m 3 /sec) of primary-settled wastewater. Under current operation, the BOD 5 of the primary-settled wastewater is usually no greater than 165 mg/L. Of that, 85 mg/L of BOD 5 come from soluble BOD 5 , and 80 mg/L of BOD 5 come from solids. (You might recognize the flow rate and the BOD 5 from homework #6.) To meet the requirements of the Clean Water Act, the plant’s treated secondary effluent (which they will continue to discharge into the river) must contain no more than 20 mg/L of BOD 5 (total, including both soluble and suspended forms) and no more than 15 mg/L suspended solids. Based on the information above, we have the following schematic diagram.
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p. 2/5 1. (28 pts) In problem 1, you will find the required volume of the aeration basin. a. (4 pts) Assume that the BOD 5 of the treated effluent comes from both the suspended solids and the soluble substrate (dissolved organic carbon). That is, (BOD 5 ) total = (BOD 5 ) soluble + (BOD 5 ) suspended The BOD 5 of the suspended solids is equal to 60% of the suspended solids concentration. [NOTE: in my experience, that statement gives students a lot of trouble each year. It is actually quite simple: (BOD 5 ) suspended = 0.60* X .] Assume that the suspended solids concentration in the treated effluent ( X E ) is at its maximum allowable concentration of 15 mg/L, as shown in the figure on the previous page.
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