Now that you have completed this study session, you can assess how well you have achieved its Learning Outcomes by answering these questions.
Imagine you have to inform the local population of the new water treatment plant that is to be built in their town. Draw a simple flow diagram to show the different stages of treatment, and write one or two sentences to describe what happens and why at each of the stages.
Your diagram should look something like Figure 5.9.
Stage 1 Screening removes large floating and suspended solids, which can damage equipment in the plant, or block pipework.
Stage 2 Aeration expels acidic gases such as carbon dioxide and hydrogen sulphide, and gaseous organic compounds that can give an unpleasant taste to the water. Aeration also oxidises iron and manganese to their solid form so that they can be removed, thereby eliminating bad flavour and staining. If excess algae are present in the raw water, pre-chlorination is carried out. The chlorine also oxidises compounds that cause taste and odour.
Stage 3 Coagulation neutralises the negative electrical charge in particles in the water, which enables the particles to come together to form flocs. Flocculation results in large flocs forming.
Stage 4 The large flocs formed during flocculation settle out in the sedimentation stage, leaving largely clear water.
Stage 5 Any solids remaining in the water after sedimentation are removed during filtration.
Stage 6 The water is then chlorinated to kill any pathogenic micro-organisms present.
Stage 7 Supplementary treatment, such as the addition of fluoride, or the reduction of the fluoride level, then takes place.
Recalling your study of the wastes produced during water treatment, assign the different wastes to the management options shown below.
|Sent to landfill|
|Discharged to sewer|
|Taken to a sewage treatment plant|
You should have identified the following wastes for each option.
|Sent to landfill||Coarse screenings; fine screenings (if no sewer present); sludge from the sedimentation tank|
|Recycled||Plastic chemical drums|
|Discharged to sewer||Fine screenings; backwash from the rapid gravity sand filter|
|Reused||Wooden and cardboard packaging|
|Taken to a sewage treatment plant||Sludge from the sedimentation tank|
Which of the following statements do not contribute to the attainment of sustainability and resilience in a water treatment plant? Give reasons for your choice.
- using locally available materials
- making sure the plant is protected from natural hazards
- using a diesel generator for running the pumps and compressors
- using simple systems where possible
- using the cheapest equipment that is in the market.
3. Using a diesel generator for running the pumps and compressors will not contribute to sustainability and resilience. Diesel is a non-renewable source of energy and will run out in time. It is better to use a renewable source of energy, for example electricity generated by solar or wind power.
5. Using the cheapest equipment that is on the market will also not contribute to sustainability and resilience. The cheapest equipment is often the least robust and is likely to fail before long.
Gideon, a new urban WASH worker, needs guidance on how to calculate the water requirement and service reservoir size in a new area that is to be developed in a town that already has a population of 150,000. In this new area, the population will be 30,000, and there will be three new health centres. The three health centres together will treat 250 people a day. There will also be a day school for 1500 pupils.
Draw up the calculations to show Gideon how it’s done.
Using data from Tables 5.1 and 5.2, the water requirement each day for the extra population will be 80 litres x 30,000 = 2,400,000 litres, or 2400 m3.
The three health centres will need 135 litres x 250 = 33,750 litres a day, or 33.75 m3 a day.
The day school will need 18.5 litres x 1500 = 27,750 litres a day, or 27.75 m3 a day.
The total additional water requirement will be 2400 + 33.75 + 22.75 = 2461.5 m3.
With allowance for leakage, etc. the water requirement will be 2461.5 m3 x 1.15 = 2830.7 m3, say, 2831 m3.
The service reservoir has to hold 2831 m3 x 1.5 = 4246.5 m3, say, 4247 m3.
Thus, the water requirement in the new area would be 2831 m3 per day, and the size of service reservoir required would be 4247 m3.