Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
360 views
in General by (113k points)
closed by
Say a raw wastewater sample from AA WWTP has 5-day BOD equals to 2000 mg/L (reaction constant k = 0.23/day at 20°C ). Calculate value of ultimate BOD?
1. 2826 mg/L
2. 2296 mg/L
3. 2000 mg/L
4. 2926 mg/L

1 Answer

0 votes
by (115k points)
selected by
 
Best answer
Correct Answer - Option 4 : 2926 mg/L

Concept:

Bio-chemical oxygen demand:

(i) The amount of O2 required to carry out the decomposition of bio-degradable organic matter present in the system is termed as BOD.

(ii) BOD during 5 days at 20°C is taken as standard BOD and is approximately 68% of Ultimate BOD.

BOD at any time is calculated by,

BODt = BODu (1 - e-kt) or BODu (1 - 10-kDt)

Where k and kD represent deoxygentation constant at base 'e' and at base '10' respectively that signifies the rate of BOD reaction without affecting the ultimate BOD (BODu)

Calculation:

Given,

BOD5 = 2000 mg/l, k = 0.23

∵ we know that, BODt = BODu (1 - e-kt

So BOD at 5 days, BOD5 = BODu (1 - e-kt)

2000 = BODu (1 - e-0.23 × 5)

⇒ 2000 = BODu × 0.6833

BODu = 2926 mg/l

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...