Raw sewage has a BOD of 260 mg per litre…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Raw sewage has a BOD of mg per litre. Primary treatment removes of the BOD, and secondary treatment removes of what is left. The plant treats m³ of sewage in a day.
- What is the BOD after primary treatment?
- What is the BOD after secondary treatment?
- What per cent of the BOD is removed in all?
- How many kilograms of oxygen demand are removed each day?
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. Raw BOD: mg per litre. Primary: . Secondary: of what is left. Flow: m³ a day.
Plan. Take each removal in turn.
Step 1. After primary. mg per litre.
Step 2. After secondary. mg per litre.
Step 3. Removed in all. .
Step 4. The oxygen demand. The fall is mg per litre, which is g per m³. For m³ this is g, which is kg each day.
Answer: mg per litre; mg per litre; ; kg each day.
Check. The two stages remove and , but the total is not . Each share applies to what is left.
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