Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
One litre of effluent mixes with L of stream water. The stream water holds mg of oxygen in each litre, and it has no other organic matter. Take the oxygen in the effluent itself as zero. The raw sewage has a BOD of mg per litre and the treated effluent has a BOD of mg per litre.
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.
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. L of effluent and L of stream water. Stream oxygen: mg per litre.
Plan. Spread the BOD and the oxygen over the L, then compare.
Step 1. Raw sewage. The demand is mg per litre.
Step 2. The oxygen. mg per litre.
Step 3. Treated effluent. The demand is mg per litre.
Step 4. Compare. With raw sewage, , so the oxygen runs out and the fish die. With treated effluent, , so some oxygen is left.
Answer: mg per litre; mg per litre; mg per litre; with raw sewage.
Check. The treated effluent has of the demand of the raw sewage.
Watch out. A river also takes in oxygen from the air as the microbes use it, so a real river recovers a little. This model leaves that out.
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