ENV4001_s11_hw8 - ENV 4001: ENVIRONMENTAL SYSTEMS...

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p. 1/4 ENV 4001: E NVIRONMENTAL S YSTEMS E NGINEERING Spring 2011 University of South Florida Homework #8 Civil & Environmental Eng. Due: Monday, April 4 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 2011 we will skip the oxygen input. 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.) Based on your findings in HW #6, the plant’s treated secondary effluent (which they will continue to discharge into the river) must contain no more than 10 mg/L of BOD 5 (total, including both soluble and suspended forms) and no more than 5 mg/L suspended solids. Based on the information above, we have the following schematic diagram. Q I = 6.6m 3 /sec S 0 = 85 mg/L from primary settling aeration basin V , X , S ( Q I + Q R ) X , S clarifier Q E S X E = 5 mg/L treated effluent to river Q W X W S waste sludge recycle stream (sludge return) Q R X R S
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p. 2/4 1. (20 pts) In problem 1, you will find the solids retention time required for the Nastyville plant. a. (5 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 about 60% of the suspended solids concentration. [NOTE: in my experience, that statement gives students a lot of trouble each
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This note was uploaded on 06/12/2011 for the course ENV 4001 taught by Professor Staff during the Fall '08 term at University of South Florida - Tampa.

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ENV4001_s11_hw8 - ENV 4001: ENVIRONMENTAL SYSTEMS...

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