Question

7. A BOD test is run using 100 mL of treated wastewater effluent mixed with 200 mL of dilution water (containing no BOD). The initial DO of the mix is 9.0 mg/L After 5 days, the DO is 4.0 mg/L. After a long period of time, the DO is 2.0 mg/L, and it no longer seems to be dropping. Assume nitrification has been inhibited so that only carbonaceous BOD is measured (a) What is the 5-day BOD of the wastewater? (b) Assuming no nitrification effects, estimate the ultimate carbonaceous BOD. (c) What would be the remaining BOD after five days have elapsed? (d) Estimate the reaction rate constant k (day).

Answer 1

For this question,first we have to calculate the difference between initial and after 5 days DO values adjusted it due to wastewater. After that for ultimate carbonaceous BOD,take difference between inital and final DO values and adjusted it due to wastewater.Remaining BOD is difference between a and b. For reaction rate constant,we should use "first order model".Detailed solution is given below please see this for better understanding.....

Solution. Composition of mixture = 100 mL of treated Wastewater + 200mL of dilution water - (Containing no BOD) Therefore Wastewater comprised of test mixture a change in dissolved oxygen content in the test mixture during the first five days = (-4) ing/= 5mg/L S-day BoD of Wastewater (BODs) = 5mg/x3 - 15 mg/L Am (because wastewater comprises ord of mixture) b) ultimate Carbanaceous BOD (BODu) is the difforence between initial & final Do levels, atte ditute Therefore, BODU = (0-2) x 3 mg/L = 21 mg/L Aus e since we know that Ultimate BOD = 5-day BOD of Wastewater + BOD remaining after sdays 7 BOD remaining after 5 days = Ultimate BoD - 5day BOD 5 Wastewater = (21-15) m = 6 m Am a) constant Reaction rate the is estimated from first order model, 300 remaining after 5 days = Ultimate BOD & el-KBOD Solay 6m/= 21 mg, el-kBont Sdays) =) - KBOD 5 days = ln (1) - KIBOD = - 1 day in (62) = -5 days 1125) Therefore reaction rate Constant = kBOD= 0.25d Am Ans - ay 15 mg/L b) 21 mah c) Gmah d) 0.25d-