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2020 · Section 1 · Multiple choice

Question 6

[1 mark] Simple familiar
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The question

An irrigation system uses a 7450 W electric motor to drive a pump that delivers 10 000 L of water per hour over a distance of 100 m. How efficient is the irrigation system? Assume that the system is without friction and that 1 L of water has a mass of 1 kg.

Not quite. 45% would need 3353 W to reach the water. Lifting 10 000 kg through 100 m in an hour takes only 2722 W, so the efficiency is lower than this.

That is it. Step through the hints below to see why.

Not quite. 27% would mean only 2012 W reaches the water. With no friction, all 2722 J the water gains each second is useful output, so the efficiency is higher than this.

Not quite. 10% would mean only 745 W reaches the water, far less than the 2722 W it takes to lift 10 000 kg through 100 m in an hour.

motor pump 10 000 L per hour 100 m lift input output 7450 W Where your working lands (percent) 0 100

Energy the water gains each hour
One hour is
Efficiency

Each hour the pump lifts 10 000 kg of water. The stem says “over a distance of 100 m”, and that has to be a height. With no friction, moving water sideways takes no energy at all.

$$E_{out}=mgh=10\,000\times9.8\times100$$
$$=9\,800\,000\text{ J}$$

This is the useful output. The water gains it as potential energy.

The motor is rated in watts, which are joules per second. Turn the hourly energy into a power. An hour is 3600 s.

$$P_{out}=\frac{9\,800\,000}{3600}=2722.22\text{ W}$$

You could instead turn the motor into an hourly energy, $7450\times3600=26\,820\,000$ J. Either way, top and bottom must be the same kind of quantity.

Efficiency is the useful output divided by the input.

$$\begin{aligned}\eta&=\frac{P_{out}}{P_{in}}\times100\%\\&=\frac{2722.22}{7450}\times100\%\end{aligned}$$
$$=36.54\%\approx37\%$$

The 2026 formula book prints this power form. The 2019 book that 2020 students had printed only $\eta=\text{MA}/\text{VR}$, so they had to know it.

$$\eta=\frac{mgh\div t}{P_{in}}\times100\%$$
$$=\frac{10\,000\times9.8\times100\div3600}{7450}\times100\%$$
$$\approx37\%$$

The useful output is the potential energy the water gains each second. Divide it by the motor’s 7450 W.

(B)
QCAA did not publish how students answered the 2020 multiple choice questions.

That means there is no record of the most common wrong answer, and QCAA did not say how it built these distractors. The usual slips land on none of them: leaving out $g$ gives 3.7%, and dividing the wrong way gives 274%. If your answer is not on the list, recheck it before you pick the nearest option.

Only mark this done when you could do it without help. Reading the solution does not count.

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Question wording and any figures are from the 2020 QCAA Engineering external assessment, © State of Queensland (QCAA) 2020, 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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