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Assign_3-1_BOD_Solutions - BOD Calculations Assumptions...

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Unformatted text preview: EVEG 31 10 BOD Calculations Assumptions: Lt=Loe'kt Yt= Lo(l- e'kt) Where k=first ordr decay constant decay constant (0.2 days'l) Lt: BOD remaining at time tin mg-Osz (Organic material remaining) LO=BOD remaining at time t=0 in mg-OZ/L (Organic material at start) t = time (days) Yt= BOD exerted in mg-OzlL (Organic matter already decayed) C5: Oxygen saturation level mg-Oz/L Ct= Oxygen level at time t Problem I: A 0.5 mgd waste flow with a £40ng of 100 and an oxygen concentration of 4.5 (C) is mixed with a 3.5 mgd river flow with a background BOD20 of 5 and DO level of 8.0. What is the BODZO and the dissolved oxygen at the point of mixing. ,4, #flrwreitrs) a t W— 449 [L 1‘ ZM7/L Problem 2: What is the BODgo one (1) mile down stream if the stream velocity is l ft/sec? V c / ff/flé {f f 49% KW? Was/:2 7 2 /3 Problem 3: You are asked to estimate the BOD20 in a stream seven miles downstream of a point of mixing. You are told the BOD5 in the river after mixing is 37 mg/L. What is the BODZO at the point of mixing. What is common between the two numbers? Why are the numbers different? XL: 1» ( / '6'“) :7? Z" flfléA/ufé flag-274’ flaw My L. W _ x t .’ pkg/5653A}? flW/wyg (3F EKPKETS/Oflf Let f 4r: Zfl 6522;” w: flew—if Problem 4: A 500 ml BOD bottle is dosed with 50 of sewage and aerated until the bottle’s oxygen is at 85% of saturation. The bottle is sealed. The oxygen level in the bottle is 2.5 mg/l after five days. What was the BOD20 of the undiluted wastewater sample? 65 “Wig-5M :y‘g'i i X ,zx) r f. ZZZ 8a : <9~Z)(0,M): xvii? ’02: 5’9,” 3: L f er“? .— j: Afibo) 925352;) ‘" ll ,. [a fl/éas [07 3/3 Problem 5: After you finish measuring the dissolved oxygen in the BOD bottle described in Problem 4, you re-aerate it and ciose it up again. How much dissolved oxygen will be left in the bottle afier five days? ...
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Assign_3-1_BOD_Solutions - BOD Calculations Assumptions...

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