A technician puts 5 mL of sewage in a bottle that holds 300 mL, and fills the rest of the bottle…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A technician puts mL of sewage in a bottle that holds mL, and fills the rest of the bottle with clean water that is full of oxygen. The oxygen in the bottle is mg per litre at the start and mg per litre after five days. A treated effluent has a BOD of mg per litre.
- What is the dilution factor?
- What is the BOD of the sewage?
- How many times is the BOD of the sewage greater than that of the treated effluent?
Need a reminder? The key ideaShow
Secondary treatment is biological. Aerobic microbes grow as flocs in the aeration tank and use up the organic matter. BOD is the oxygen consumed in oxidising the organic matter of one litre of water. In a BOD test with a diluted sample, BOD (fall of oxygen) (volume of the bottle) (volume of the sample). The sediment of the flocs is the activated sludge: a small part is the inoculum, and the rest goes to the anaerobic sludge digesters.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Given. Sample: mL. Bottle: mL. Oxygen: to mg per litre.
Plan. Find the dilution factor and the fall of oxygen, and scale up.
Step 1. The dilution factor. .
Step 2. The BOD. The fall of oxygen is mg per litre. The BOD is mg per litre.
Step 3. The ratio. .
Answer: ; mg per litre; times.
Check. Raw sewage has a BOD of a few hundred mg per litre, so is believable.
Watch out. A BOD test needs some oxygen left at the end. If all the oxygen is used, the sample must be diluted more.
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