Question 21
A factory uses a two-pump system to vertically lift water into a main reservoir.
The first pump is 80% efficient and lifts 500 litres of water per minute into a holding tank 6 m above the ground. The second pump is 75% efficient and lifts 500 litres of water per minute from the holding tank into the main reservoir 10 m above the ground.
This current system is being replaced with a single-pump system that is 90% efficient and has the same power input as the two-pump system. Determine the rate at which the new system will lift water into the main reservoir. Answer to the nearest whole unit (litres per minute).
Note: 1 litre of water has a mass of 1 kg.
Each pump is worked on its own height and its own efficiency, and the two inputs add
Heights and power bars are to scale. Grey bars are power in, blue bars are useful power out. Watch how far each pump actually lifts the water.
Backwards for each old pump, divide its output by its efficiency. Forwards for the new pump, multiply its input by 0.9, then lift through the full 10 m.
Change the efficiencies and the tank height, then try the two traps
The question is 80%, 75% and 90% with the tank 6 m up. The two trap switches are the errors this question is built to catch.
This is the question. The new pump lifts 577.5 L a minute. That is exactly half-way, which is why the guide accepts 577 or 578.
Treating 80% as 100% plus a 20% loss and multiplying by 1.2 undercounts the input. The 2023 subject report names this exact error. 80% efficient means divide by 0.8.
Pump 2 starts at the holding tank, so it only lifts the water the remaining distance. Using the full 10 m counts the first 6 m twice and inflates the answer by more than half.
A less efficient old pump needs more input, and the new pump inherits all of it. Moving the tank shifts lift from one pump to the other. The new pump always lifts the full 10 m.
Work through the solution one mark at a time
Ten marks, one for every number in the chain. Try each step yourself before you reveal it.
500 L is 500 kg, lifted 6 m every minute. The force is the weight, so the work is $mgh$.
Marker: determines the work done by the first pump
That work is done every minute, which is 60 s.
$$P_{out}=\frac{W}{t}=\frac{29\,400}{60}=490\text{ W}$$
Marker: determines the power output of the first pump. “Allow FT error for incorrect work done when determining power output of the first pump.”
The output is known and the input is wanted, so the efficiency runs backwards. Divide by 0.8.
The 2026 formula book prints $\eta=\dfrac{P_{out}}{P_{in}}\times100\%$. The guide writes the same idea as useful output over input.
Marker: determines the power input of the first pump. “Allow FT error for incorrect power output when determining power input of the first pump.”
The second pump starts at the holding tank, 6 m up, and delivers to the reservoir, 10 m up. It lifts the water $10-6=4\text{ m}$.
$$\begin{aligned}W&=mgh=500\times9.8\times4\\&=19\,600\text{ J}\end{aligned}$$
Marker: determines the work done by the second pump
$$P_{out}=\frac{19\,600}{60}=326.67\text{ W}$$
Marker: determines the power output of the second pump. “Accept power output from 326.66–326.67 W inclusive.” “Allow FT error for incorrect work when determining power output of the second pump.”
Backwards again, with this pump’s own efficiency.
$$P_{in}=\frac{326.67}{0.75}=435.56\text{ W}$$
Marker: determines the power input of the second pump. “Accept power input from 435.54–435.56 W inclusive.” “Allow FT error for incorrect power when determining power input of the second pump.”
Each pump takes its own power, so the inputs add. Do not multiply the efficiencies here. That rule is for one machine driving another.
$$P_{in}=612.5+435.56=1048.06\text{ W}$$
Marker: determines the overall power input of the current two-stage system. “Accept overall power input from 1048–1048.06 W inclusive.” “Allow FT error/s for incorrect power input of either or both pumps when determining the overall power input of the two-stage system.”
The new pump has the same input. Now the input is known and the output is wanted, so the efficiency runs forwards. Multiply by 0.9.
$$P_{out}=0.9\times1048.06=943.25\text{ W}$$
Marker: determines the power output of the new single-pump system. “Accept power output of the new single-pump system from 943.20–943.26 W inclusive.” “Allow FT error for incorrect power input when determining power output of the new single-pump system.”
Turn the power into the work done in one minute.
$$\begin{aligned}W&=P\times t=943.25\times60\\&=56\,595\text{ J}\end{aligned}$$
Marker: determines the work done by the new single-pump system. “Accept work done from 56 592–56 596 J inclusive.” “Allow FT error for incorrect power output when determining work done by the new single-pump system.”
The new pump lifts from the ground to the reservoir, the full 10 m. Read $W=mgh$ in reverse for the mass lifted each minute.
$$m=\frac{W}{gh}=\frac{56\,595}{9.8\times10}=577.5\text{ kg}$$
577.5 is exact, so it sits on the half-way mark. Rounding half up gives 578, and the guide accepts 577 as well.
Marker: determines the answer in litres per minute to the nearest whole unit. “Accept answer from 577–578 litres per minute inclusive.” “Allow FT error for incorrect work done when determining the answer.”
Each phrase of the question gave you something
The 2023 subject report says the effective responses found the work, power and efficiency for each pump on the way to the two inputs, then the new pump’s output and the rate to the nearest whole unit. Its additional advice names the 1.2 error: 80% efficiency means divide by 0.8, never multiply by 100% plus the 20% loss.
QCAA did not publish a band for this question. It reads as complex unfamiliar because you have to design the chain yourself: two machines that each take their own power, a replacement that inherits their combined input, and efficiencies that run backwards for the old pumps and forwards for the new one. It is the ten-mark question that closes the 2023 paper.
If I had to guess where you might lose marks, it would be the height for the second pump. Using 10 m gives 938 L a minute. Follow-through limits the damage to the second pump’s work mark, but only if you carry the wrong number correctly through every later step. The other trap is the one QCAA reported, multiplying by 1.2 and 1.25 instead of dividing by 0.8 and 0.75, which gives 549 L a minute.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording, any figures and marking-guide steps are from the 2023 QCAA Engineering external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and have been adapted, except where a third-party source is named. Tangent Tuition is not affiliated with the QCAA.
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