Question

De constant kiday"). 8. A standard BOD test is run using seeded dilution water. In one bottle, the water sample is mixed with seeded dilution water giving a dilution of 1:30. Another bottle, the blank, contains just seeded dilution water. Both bottles begin the test with DO at the saturation value of 9.2 mg/L. After five days, the bottle containing waste has DO equal to 2.5 mg/L while that containing just seeded dilution water has DO equal to 8 mg/L. Find the five-day BOD of the waste (BOD).

Answer 1

agero1 Answer: According to the given Pnformation, to find the five-day Bon of the neceste (BODs). Given, uring the equation, BoDs Do – 04 - Let - Bf] (e-P) —O Hel , Lo; is the initial dissolved oxygen. Cool of the dillested waste water sample. Dog. is the dissolved oxygen (no) at tie End of tests Boos Ps five day biochemical oxygen demand, By is the pritial Do Pn the seeded dillution water Be is the final do in the seeded dillution water. NOW) * calculate the chilufton facter by curing the equation :- het,

pageida Here, vw is volume of waste water in the mixture Vm is volume of the mixture. Vws 1 Vm = 30 Substitute tee values. P= 1/30 P: 0.033 Doi a a. 2 mg/h Dog a 2.5 mghh P . 0.033 BP = qua mogla Bf = 80mghh. Substitute the above values in Equation o BODS - DO? - 001 - (B; - Bf) (1-P)/P

- page 103 BODS - 962 mg la – das mgle - (9.angle - 8 mgle) (1-0.033) 0.033 = 6.4 mg/h -(1.2 mg/h) x 0,964 0.033 2 6it ngth - 1.1604 mall 0.033 5.5396 mild 0.033 a = 167. 86 mgla :: the five-day Bop of the waste (B00s) : 167086 mg/.