Question

A test bottle containing only seeded dilution water has it DO final 1mg/l in 5 days incubation. A 300ml bottle filled with 10 ml of waste water and rest seeded dilution water experience of initial DO 6.2mg/l in the same time period. What should be the 5 days BOD of waste water

Answer 1

Explanation:

MODULE 1

SOLVED NUMERICAL PROBLEMS

Problem 1: Find the TON when odor is just barely detected in a flask containing 40 ml of

sample water.

Solution:

5

40

200

Sample volume (ml)

200 ml      A

A B TON

Problem 2: The BOD6 of a wastewater is determined to be 400 mg/L at 20o

C. The k value at

20o

C is known to be 0.23 per day. What would be BOD8 value if tests were run at 15o

C?

Answer: Solved using MATHCAD, however, can be solved otherwise also.

Problem 3: 6 ml of wastewater is diluted to 300 ml distilled water in standard BOD bottle. Initial

DO in the bottle is determined to be 8.5 mg/l. DO after 5 days at 20 C is found to be 5 mg/l.

Determine BOD5 of wastewater and compute the ultimate BOD.

Soln. We know,

0 5

5 wd

w

(DO - DO ) (8.5-5) BOD = ×(V +V ) = ×(300) 175 mg/l V 6 

Since BODt = BODu (1- -kt e ) at any particular temperature

BODu = BOD5 /(1- -5k e ) = 175 / (1- -5 0.23 e 

) = 256 mg/l.

BOD BOD     k t t T  u   T , 1 exp

    

  20

20

20

20 1.047     T T

T k k  k

Given that: BOD6 20   400 mg

L

k20  0.23 d

 1

BODu

BOD6 20 

  1  exp k    206

  534.458 mg

L

k15 k20 1.047( ) 15 20     0.183 d

 1

BOD8 15   BODu  1  exp k    158  410.643 mg

L