0. What is the BODs of the wastewater sample if the DO values for the blank and diluted samples afher 5 days are R1 mg/L and 3.5 mg/L, respectively, and the wastewater is diluted from 2 miL to 200 milL? (10 points) If the 20 day BOD (assume that this is the ultimate BOD) is 700.0 mg/L determine the BOD rate constant k (in base e). (10 points)
0. What is the BODs of the wastewater sample if the DO...
ml of river water with BODs of 1000 ppm is diluted to 1 L, aerated and seeded. The If 2 disolved oxygen content was 7.8 mg/L initially. After 20 days, the dissolved oxygen ssolv content had dropped to 5.3 mg/L. What is the ultimate BOD, and remaining BOD? Which of the two BODs give better prediction of BODL, and why?
ml of river water with BODs of 1000 ppm is diluted to 1 L, aerated and seeded. The If 2...
(20 pts) The ultimate BOD of a domestic wastewater is 275 mg/L and the e based rate constant k=0.24 d a. (4 pts) What will be the 5-day BOD (Sample-1)? b. (4 pts) What will be the 7-day BOD? (Sample-1)? c. (4 pts) A different sample has the same ultimate BOD given in the question, but the 20 °C k value (e based) is equal to 0.36 dl Determine the 5-day BOD of this sample (Sample-2). d. (8 pts) Please...
Problem Statement: A midnight dumper discharges a truck load of industrial waste in a gravel pit. Three days after the truck is spotted, a sample of the remaining pool of pure waste is collected and is sent to laboratory for forensic analyses. Test results indicate that sample has a BODs of 80 mg/L 02 with a rate constant of 0.1/day. Three nearby factories generate organic wastes: a winery (ultimate BOD = 275 mg/L Oz), a vinegar manufacturer (ultimate BOD =...
Determine the ultimate BOD (BODu) and 5-day BOD (BOD5) given that the BOD4 of a waste is 130 mg/L and K is 0.075 d-1 . Use yt=Lo(1-e-kt) and k=2.303(K).
ENV3006C HW#4 The following data provided to you. Calculate the BODS of the WW sample (mg/L) are Seeded wastewater (WW): Initial DO 8.2 mg/L; Volume of wastewater Final DO (after 5 days) 2.5 mg/L Total volume in BOD bottle = 300 mL 5 mL Seeded dilution water: Final DO (after 5 days) 7.3 mg/L Initial DO 8.3 mg/L; Volume of seeded dilution water 300 mL
Calculate the BOD Ultimate value given the following information BOD5= 316 mg/L k = 0.23 day-1 Temperature is 20 C for both BOD5 and BOD Ultimate (L)
1. Calculate (a) BODs, (b) Le or BOD and (c) plot the O2 consumed vs time for the DO data shown below in a BOD bottle. Additionally, clearly show BODs, L. or BODult, and organic matter remaining in the plot. Time (days) DO (mg/L) 7.5 6.6 5.2 4.8 3.8 Ans.: (a) 3.4 mg/L, (b) 6.2 mg/L, (e) Plot of O, consume vs. time
2. The 5-day, 20°C BOD of a wastewater (K 0.23/day, 0 1.047) is 185 mg/L. What is the corresponding 10-day demand and ultimate BOD (in mg/L)? If the bottle had been incubated at 33°C, what would the 5-day BOD value have been?
The Big Bear town council has asked that you determine whether the discharge of the town's wastewater into the Salmon River will reduce the DO below the state standard of 3.00 mg/L. The DO at saturation at 28 C is 7.81 mg/L. The other pertinent data follow. (16 pts) 2. Salmon River Big Bear Wastewater Parameter0,580 Flow, m/sec BODs, mg/L Ultimate BOD, mg/L DO, mg/L ka, d k d Speed, m/sec Temp, C 0.580 9 mg/L 7.00 6.70 0.322 0.852...