The Leaking Tank: Why Doubling the Pump Doesn't Halve the Time
An off-grid household stores water in a rectangular tank with internal dimensions 1,20 m long, 0,80 m wide and 1,50 m deep. A borehole pump feeds the tank at a steady 15 litres per minute. A leaking outlet valve loses water at a constant 3 litres per minute whenever the tank holds water. At 06:00 on Monday the tank holds 300 litres and the pump is switched on; it runs continuously. The householder is considering replacing the pump with one rated at 30 litres per minute and claims that this will halve the time needed to fill the tank. Water is billed at R24,50 per kilolitre, where 1 kL = 1 000 litres.
1. Calculate the capacity of the tank in litres and determine the clock time at which the tank becomes full on Monday.
Read the worked answer
Capacity V = 1,20 × 0,80 × 1,50 = 1,44 m³ 1 m³ = 1 000 L, so V = 1 440 L
Net inflow rate The leak removes water for as long as the tank holds water, so the net rate is 15 − 3 = 12 L/min
Volume still needed 1 440 − 300 = 1 140 L
Time 1 140 ÷ 12 = 95 minutes = 1 h 35 min
Clock time 06:00 + 1 h 35 min = 07:35
Note the leak is subtracted as a rate, not as a percentage: the 12 L/min is a constant while the tank is filling.
2. Test the householder's claim that a 30 L/min pump will halve the filling time. Support your answer with calculation.
Read the worked answer
New net rate 30 − 3 = 27 L/min (the leak does not double)
New time 1 140 ÷ 27 = 42,22 minutes ≈ 42 minutes 13 seconds
Compare with "half" Half of 95 minutes would be 47,5 minutes. The actual time is about 5 minutes faster than half, so the claim is wrong in direction of the reasoning even though the answer is close.
Why the claim fails in general Time = (volume needed) ÷ (pump rate − leak rate). Doubling the numerator's rate while subtracting a fixed leak does not halve the quotient unless the leak is zero. In symbols, with R replaced by 2R:
t(2R) = V / (2R − L) vs t(R)/2 = V / (2R − 2L)
The second is larger whenever L > 0, so the new pump always beats the naive "half the time" prediction.
A caution on the model: 42 minutes assumes the pump delivers 30 L/min at the actual head and that the leak rate stays at 3 L/min - a higher head may increase leakage, which would lengthen the time.
3. Assuming the tank always holds water, calculate what the leak costs per day at the stated tariff, and state one assumption that could make this figure misleading.
Read the worked answer
Water lost per day 3 L/min × 60 min × 24 h = 4 320 L = 4,32 kL per day
Cost 4,32 kL × R24,50/kL = R105,84 per day
(About R3 175 per 30-day month - far more than the cost of a new valve washer.)
Assumption that could mislead "The tank always holds water" is doing all the work. If the tank runs dry for part of the day, the leak stops and the true loss is lower. The figure also assumes the full R24,50/kL applies to this water - if the household is on a rising block tariff, water lost through the leak may sit in a higher block than average, making the true cost larger, not smaller. Conversely, if the borehole water is unmetered, the R105,84 is an opportunity cost of pump electricity and standing time rather than a cash bill.