Math Problem Statement
In a BOD test, 5 mL of wastewater (DO=0.0) is mixed with 295 mL of diluting water (DO= 9.0 mg/L). After 5 days’ inoculation of the mixture, DO = 5.2 mg/L. What is the BOD5of the wastewater? Assume that all oxygen used represents the oxygen demand of the wastewater.
Solution
The Biochemical Oxygen Demand (BOD) is a measure of the amount of oxygen consumed by microorganisms while decomposing organic matter in water. In a BOD test, we can calculate the 5-day BOD (BOD₅) using the dilution method.
Given:
- Volume of wastewater = 5 mL
- Volume of diluting water = 295 mL
- Initial dissolved oxygen (DO) of wastewater = 0.0 mg/L
- Initial DO of diluting water = 9.0 mg/L
- DO after 5 days = 5.2 mg/L
To calculate the BOD₅ of the wastewater, we will use the formula:
Where:
- = Volume of wastewater (mL)
- = Volume of diluting water (mL)
- = Initial dissolved oxygen (mg/L)
- = Final dissolved oxygen after 5 days (mg/L)
Step 1: Calculate total volume of the mixture
Step 2: Calculate the BOD₅
Conclusion:
The 5-day Biochemical Oxygen Demand (BOD₅) of the wastewater is 228 mg/L.
Would you like more details on any of the steps, or do you have any questions?
Here are 5 related questions you might find interesting:
- How does BOD help in assessing water pollution?
- What is the significance of the 5-day incubation period in the BOD test?
- How would the BOD calculation change if the volume of wastewater increased?
- What other factors besides organic matter could affect BOD levels?
- How can BOD tests be used to improve wastewater treatment processes?
Tip: Always ensure the sample bottles for BOD tests are airtight to avoid the absorption of oxygen from the atmosphere during the incubation period.
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Math Problem Analysis
Mathematical Concepts
Biochemical Oxygen Demand
Proportions
Dilution Method
Formulas
BOD5 = (Vw + Vd) / Vw * (DOi - DOf)
Theorems
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Suitable Grade Level
University-level Environmental Science or Chemistry