Tank Water at Bikenibeu: How Long Can the School Stay Open?
A junior secondary school at Bikenibeu depends on one 10 000-litre rainwater tank. Rain is collected from the assembly hall roof, a single rectangle 24 m long and 7.5 m wide; the gutters deliver water to the tank, but the caretaker estimates that 12% of the rain falling on the roof is lost to splashing, first-flush diversion and inlet overflow. At the start of a wet week the tank holds 2 400 litres. Over that week the rain gauge records 62 mm. Daily use is recorded as: drinking water for 320 students at 3 litres each, 450 litres for the kitchen, and 200 litres for washing floors and toilets. At the end of the week the head teacher must decide whether to keep the school open for the following week or send students home, and whether to order a delivery of water at $25 per 200-litre drum. No other source is available and the next rain is not expected for ten days.
1. Calculate how much water the roof actually delivered into the tank during the week, and how many full days of normal use the tank can then support. Show every step, including any water lost because the tank was already full.
Read the worked answer
Step 1 - catchment area. 24 m × 7.5 m = 180 m².
Step 2 - gross catch. 1 mm of rain on 1 m² gives 1 litre, so 180 × 62 = 11 160 L falls on the roof.
Step 3 - losses before the tank. 12% is lost: 0.12 × 11 160 = 1 339.2 L, leaving 9 820.8 L arriving at the inlet.
Step 4 - storage limit. Free space in the tank is 10 000 − 2 400 = 7 600 L. Only 7 600 L can be kept, so 9 820.8 − 7 600 = 2 220.8 L overflows and is wasted. The tank ends the week full at 10 000 L. Total loss against rainfall = 1 339.2 + 2 220.8 = 3 560 L, about 32%.
Step 5 - daily demand. 320 × 3 = 960 L for drinking; + 450 L kitchen; + 200 L washing = 1 610 L per day.
Step 6 - days of supply. 10 000 ÷ 1 610 = 6.21, so 6 full days, with 10 000 − (6 × 1 610) = 340 L left on day 7.
The key judgement: a week of rain does not mean ten days of water. Because the tank was nearly a third empty, almost a quarter of the rain was lost, and demand of 1 610 L/day cannot be met for the full ten days without another source.
2. Compare three options for the ten rainless days: (i) buy all the shortfall, (ii) cut kitchen use by 25% and washing by 50% and buy the rest, (iii) buy a second tank. Use figures to recommend one course of action.
Read the worked answer
Option (i) - buy everything. Ten days need 10 × 1 610 = 16 100 L against 10 000 L in store, a shortfall of 6 100 L. Drums needed = 6 100 ÷ 200 = 30.5, so 31 drums × $25 = $775.
Option (ii) - reduce, then buy the rest. Kitchen saving = 0.25 × 450 = 112.5 L; washing saving = 0.5 × 200 = 100 L. New daily demand = 1 610 − 212.5 = 1 397.5 L. Ten days need 13 975 L, shortfall 3 975 L = 19.9 drums, so 20 drums = $500. Reduction alone still gives only 10 000 ÷ 1 397.5 = 7.15 days, so buying cannot be avoided - but it is cut by $275.
Option (iii) - second tank. The 2 220.8 L that overflowed would have added 2 220.8 ÷ 1 610 = 1.4 days. It does not solve this ten-day gap by itself, but it is the only option that pays back every wet season; the school lost 32% of its rain.
Recommendation: take option (ii) now, because $500 is affordable and keeps the school open, and budget for extra storage next. Buying water for the full demand wastes money on 200 L drums when a fixed tank would have stored the same rain free. Also check the 12% loss figure - repairing the first-flush diverter is cheaper than 20 drums.