Homework Answers
Solution –
Population = 20,000
Daily average demand = 400 L/capita
= 400 * 20,000
= 8,000,000 L
= 8,000 m3.
Maximum Daily demand = 1.25 * 8000
= 10,000 m3.
- For a constant flow supply to the reservoir for 24 hours.
Volume in is the average of volume out i.e., 1250 m3.
We can formulate a table as show.
| Period of the day | %age of daily demand | Daily demand / Volume Out (m3) | Volume In (m3) | Accumulated volume | Overall cumulative volume (m3) |
| 12 AM to 3 AM | 5 | 500 | 1250 | 750 | 750 |
| 3 AM to 6 AM | 7 | 700 | 1250 | 550 | 1300 |
| 6 AM to 9 AM | 13 | 1300 | 1250 | -50 | 1250 |
| 9 AM to 12 PM | 20 | 2000 | 1250 | -750 | 500 |
| 12 PM to 3 PM | 20 | 2000 | 1250 | -750 | -250 |
| 3 PM to 6 PM | 15 | 1500 | 1250 | -250 | -500 |
| 6 PM to 9 PM | 10 | 1000 | 1250 | 250 | -250 |
| 9 PM to 12 AM | 10 | 1000 | 1250 | 250 | 0 |
So, the storage volume should be 1300 m3.
- For a constant flow of supply for a period of 12 hours only,
Volume in is the average of volume out over 12 hours i.e., 2500 m3.
We can formula a table as shown.
| Period of the day | %age of daily demand | Daily demand / Volume Out (m3) | Volume In (m3) | Accumulated volume | Overall cumulative volume (m3) |
| 12 AM to 3 AM | 5 | 500 | 0 | -500 | -500 |
| 3 AM to 6 AM | 7 | 700 | 0 | -700 | -1200 |
| 6 AM to 9 AM | 13 | 1300 | 2500 | 1200 | 0 |
| 9 AM to 12 PM | 20 | 2000 | 2500 | 500 | 500 |
| 12 PM to 3 PM | 20 | 2000 | 2500 | 500 | 1000 |
| 3 PM to 6 PM | 15 | 1500 | 2500 | 1000 | 2000 |
| 6 PM to 9 PM | 10 | 1000 | 0 | -1000 | 1000 |
| 9 PM to 12 AM | 10 | 1000 | 0 | -1000 | 0 |
So, the storage volume should be 2000 m3.
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