How Much Electricity Can a Small Stream Make?: Going Further 9–12

Power: energy every second

In Grade 9 you met power, the energy changed from one form to another in each second, measured in watts: 1 W = 1 J/s Grade 9: Electric power. A 10 W LED bulb turns 10 joules of electrical energy into light and heat every second.

A stream’s power comes from water losing height. When a mass m of water falls a height h, it loses gravitational potential energy mgh (Gravitational potential energy). A hydropower plant catches part of that energy before the water reaches the bottom. What matters is how much energy is released every second, and that depends on two things you can measure:

The head (H): the height the water falls, in meters, from the water’s surface where it is taken in to the turbine.
The flow (Q): how much water passes each second, in liters per second (L/s) or cubic meters per second (m³/s; 1 m³/s = 1,000 L/s). One liter of water has a mass of 1 kg.

If Q liters pass each second, then Q kilograms of water fall the height H each second, and the energy they release each second is QgH. So the power in the water is

$P = QgH$

with Q in kg/s (the same number as L/s), g = 9.8 m/s² and H in meters, giving watts. A quick way to remember it: one liter a second, falling one meter, gives about 10 W.

Set the stream’s head and flow. The shaded rectangle’s width is the flow and its height is the head, so its area shows the power. Every point on a curved line has the same power.

Head: 40 m Flow: 50 L/s

Power in the water: 19.6 kW
Electricity, at 50%: 9.8 kW
10 W LED bulbs it could light: 980

At about 50%, this stream would make a pico hydro plant (up to 10 kW of electricity): enough for a few houses or a small village.

An engineer finds a way to double the head of a scheme, by taking the water in higher up the stream, and the flow stays the same. The power in the water

Pick an answer to see why.

From water to wire

No plant turns all of the water’s power into electricity. You met efficiency in Grade 9: the useful energy out ÷ the energy put in, always less than 100% Grade 9: Efficiency. Follow the energy of a small hydropower plant from the stream to a village house, and at every step some of it is lost, mostly as heat:

Step	What is lost	Fraction passed on (rough values for a small plant)
the pipe (penstock)	friction between the water and the pipe’s wall	0.90
the turbine	splashing, swirling water, water leaving with speed it still has	0.70
the belt from turbine to generator	friction and bending of the belt	0.95
the generator	heating of its coils and core, friction in its bearings	0.85

Each step passes on a fraction of what it gets, so the overall efficiency is the fractions multiplied together:

$0.90 \times 0.70 \times 0.95 \times 0.85 = 0.51$

About half. Designers of small plants often plan on an overall efficiency of about 50% (some guides allow 60% or more for a well-built plant), and a small turbine on its own typically turns about 65% to 80% of the power it receives into work, depending on its kind and how well it is made. So the electrical power of a small plant is roughly

$P_\text{electric} \approx \eta\,QgH \approx 0.5 \times QgH$

where η (eta) is the overall efficiency. And the line that carries the electricity to the houses loses some more.

A small plant has 20 kW of power in its water and gives 9 kW of electricity. Where did most of the other 11 kW go?

Pick an answer to see why.

Measuring a real stream

To know how much a stream can give, you need its head and its flow. Neither needs special instruments.

Figure 13.1 A small hydropower scheme of the kind built in Nepal’s hills, drawn as a side view along the valley, to scale. A low weir sends part of the stream into a canal that runs almost level along the hillside, while the stream falls steeply away below it. From a small tank (the forebay) the penstock drops 47 m to the powerhouse beside the stream, and a short channel, the tailrace, takes the water from the turbine back into the stream. At this scale the buildings are as small as they really are.

Measuring the head

The head is the height from the water’s surface in the forebay to the turbine. On a steep hillside you can measure it with a long, clear plastic hose full of water, because water in a hose always settles at the same level at both ends, however the hose runs between them. One person holds one end with its water level at a mark on the ground; a second person, further down, holds the other end against a measuring stick and raises it until the water stops moving. The height of the water up the stick is the drop between the two places. Then the first person moves down to the second one’s place, and you do it again, step by step, down the hill. The head is the sum of the steps.

Figure 13.2 Measuring the head with a water-filled hose, to scale. At each step the water stands at the same level at both ends of the hose, so the height of the water up the stick at the lower end is the drop from the upper mark. Here three steps of 1.50 m, 1.35 m and 1.45 m add up to 4.30 m.

A phone’s GPS height isn’t good enough for this: it is often wrong by several meters or more, as much as a whole small head.

Measuring the flow

For a very small stream, catch all of it in a bucket and time how long it takes to fill: a 15 L bucket that fills in 5 s means a flow of 15 ÷ 5 = 3 L/s.

For a bigger stream, use the float method. Choose a straight stretch where the water flows smoothly. Mark a start and a finish 10 m apart, drop a float (a small stick or an orange) into the middle above the start, and time it to the finish. That gives the speed of the water at the surface. Water near the bed and the banks is slowed by friction, so the average speed is less: about 0.8 to 0.9 times the surface speed, nearer 0.8 for a rough, rocky bed. Measure the depth at evenly spaced points across, to get the area of the stream’s cross-section: each depth stands for a strip of the stream as wide as the spacing, so the area is about the spacing × the sum of the depths. Then

$Q = \text{area of the cross-section} \times \text{average speed}$

Figure 13.3 The float method. Above, the stretch seen from above, to scale: the float is timed between two marks 10 m apart. Below, a close-up of the stream’s cross-section, drawn five times bigger, with the depth measured at five points 0.2 m apart.

Worked example

Measuring a stream

A stream is 1.2 m wide. Depths measured at five points across it, 0.2 m apart, are 0.10 m, 0.18 m, 0.22 m, 0.16 m and 0.09 m. A float takes 16 s to travel 10 m down the middle. Taking the average speed as 0.8 times the surface speed, what is the flow in liters per second?

Answer

Each depth stands for a strip of the stream 0.2 m wide (the water in the thin strips at the two edges is very shallow, and can be left out).
Area of the cross-section ≈ 0.2 m × (0.10 + 0.18 + 0.22 + 0.16 + 0.09) m = 0.2 m × 0.75 m = 0.15 m².
(Width × the average of the five depths, 1.2 m × 0.15 m = 0.18 m², would be too big: it treats the stream as 0.15 m deep right up to its banks, where it is really almost nothing.)
Surface speed = 10 m ÷ 16 s = 0.625 m/s; average speed ≈ 0.8 × 0.625 = 0.50 m/s.
$Q = 0.15\ \text{m}^2 \times 0.50\ \text{m/s} = 0.075\ \text{m}^3\text{/s} = \mathbf{75\ L/s}$

When to measure, and how much to take

Most small schemes in Nepal are run-of-river: they have no reservoir, and use the water as it comes. A stream that roars in the monsoon may be a trickle in the dry months before it, so engineers measure the flow in the dry season and design for that, or the lights go dim every spring.

A scheme also must not take the whole stream. Fish, the fields and the ghattas below need water too. Nepal’s Hydropower Development Policy (2001) asks every hydropower project to leave at least 10% of the river’s lowest monthly average flow in the river, or more where a study of the river’s life says it needs more. The water that has passed through the turbine goes back into the stream, so the only stretch that runs low is between the weir and the powerhouse.

Activity

Measure a stream’s flow

You need: an adult with you, a measuring tape, a stopwatch (a phone), a stick to measure depths with, a small float (a stick, or an orange, which floats low and isn’t blown by the wind), and a notebook.

Choose a small stream with a straight, smooth stretch at least 10 m long, which you can reach from the bank. Never do this in the monsoon, or in water deeper than your knees or flowing fast enough to push you.
Mark a start and a finish 10 m apart along the bank.
Measure the stream’s width, and its depth at five evenly spaced points across (for a stream 1.2 m wide, every 0.2 m).
Time the float from start to finish three times, and take the average.
Work out the area of the cross-section (the spacing × the sum of the depths) and the flow, using 0.8 times the surface speed as the average speed.
Estimate the head you could get: how far the stream falls over the next few hundred meters. Work out the power in the water, and the electricity a small plant there might give at 50%.

What you should see

A typical small hill stream in the dry season gives somewhere between a few liters and a few hundred liters a second. With a head of a few tens of meters, that means anything from about a kilowatt to tens of kilowatts in the water. Repeat your measurement after rain, and you will see how much the flow changes: one reason engineers measure over a whole year if they can.

In the float method, why is the average speed of the water taken as less than the speed of the float?

Pick an answer to see why.

Turbines: catching the energy

A turbine is a wheel that the water turns. Its job is to take as much as possible of the water’s energy, and leave the water with as little as possible.

Inside a small mud-walled room: a round millstone on a stone base, dusted white with flour, with a wooden hopper hanging on ropes above it — **Figure 13.4** Inside a ghatta (water mill) at Gadsera village in Doti. Grain trickles from the wooden hopper into the hole in the middle of the upper millstone, which the water wheel under the floor turns; the flour collects all round it. Photo: Janak Bhatta, CC BY 4.0

The ghatta’s wheel is one of the oldest turbines: a ring of flat wooden paddles on an upright shaft, turned by a jet of water from the chute, with the millstone fixed to the top of the same shaft. It is simple and cheap, but it wastes most of the water’s power. Can a better wheel catch more?

Turning the head into speed

Most small turbines in Nepal first turn the head into speed. The water is sent down the penstock and out of a narrow nozzle at the bottom as a fast jet. Its gravitational potential energy has become kinetic energy, just as for a falling ball (Swapping one for the other), so if nothing were lost

$\tfrac{1}{2}mv^2 = mgH \quad\Rightarrow\quad v = \sqrt{2gH}$

The best speed for the wheel

Now the jet hits the cups around the rim of a wheel. How fast should the wheel turn?

If the wheel is held still, the jet pushes hard, but nothing moves, so no work is done: no power.
If the cups move as fast as the jet, the water can’t catch up with them: no push, so again no power.

The best is in between. If the water simply pushes a flat paddle aside, the most a paddle can take is half the jet’s energy, when the paddle moves at half the jet’s speed: the water flies off sideways, still carrying the other half. But if the cup is shaped to turn the water right round and send it back the way it came, at half the jet’s speed backwards relative to a cup moving forwards at half the jet’s speed, the water ends up almost at rest. It has given the cup nearly all its kinetic energy.

Choose the shape of the blade, then set how fast the blades move compared with the jet. The arrows show the speeds to scale: the jet coming in, the blade moving, and the water leaving. The faster the water leaves, the more energy it carries away wasted. (An ideal wheel, without friction.)

Blade:

Blade speed: 0.50 × jet speed

Energy taken by the wheel: 98%
Water leaves at: 0.13 × jet speed
Energy the water takes away: 2%

At about half the jet’s speed, the cup turns the water back so that it drops off almost still: nearly all its energy went into the wheel.

A story from science

The wheel that slipped

Lester Pelton went to California in 1850, in the gold rush. The mines needed power for their machines, and many got it from water wheels. Pelton, a carpenter and millwright, spent years making better wheels for the small, steep streams near the mines. In the 1870s, so the story goes, he noticed a wheel whose key had slipped on its shaft, so that the jet struck the edge of the cups instead of their middle. The water was thrown back out of the cups, and the wheel ran faster. Pelton designed a cup in two halves with a sharp ridge between them, which splits the jet and turns each half back.

He patented it in 1880. In a test at a gold mine in 1883, his wheel turned about 90% of the jet’s energy into work, while the best of the other wheels managed less than 77%. Today the Pelton wheel turns the generators of high-head stations all over the world, from tiny village plants in Nepal to giant machines of over 400 MW in the Alps.

A large metal wheel standing on its edge, with rows of curved double cups all round its rim — **Figure 13.5** The wheel (runner) of a Pelton turbine, on display in Karlsruhe, Germany. Each cup is two scoops side by side with a sharp ridge between them, which splits the jet in two. Photo: Dietmar Rabich, CC BY-SA 4.0

Try it

The size of the wheel

For the 40 m head above, the jet moves at 28 m/s, so the cups should move at about 14 m/s. A generator must turn 1,500 times a minute (you will see why below). If the wheel drives the generator directly, how big across should the wheel be, measured to the middle of the cups?

Hint

In one turn, a cup goes once round a circle of circumference πD. 1,500 turns a minute is 25 turns a second.

Answer

The cups travel πD × 25 meters every second, and this must be 14 m/s:
$D = \frac{14\ \text{m/s}}{\pi \times 25\ \text{per second}} = 0.18\ \text{m}$
A wheel only about 18 cm across. With a higher head the jet is faster, so the wheel can be bigger or turn faster; with a lower head the wheel must be smaller or a belt must speed it up.

Different heads, different turbines

A Pelton wheel needs a fast jet, so it suits a big head. With a low head, the jet would be slow, and the wheel would turn too slowly to drive a generator. Other turbines take the water differently.

Figure 13.6 Three kinds of small water turbine, each drawn on its own scale. A Pelton wheel is turned by a fast jet; in a crossflow turbine the water passes through the blades twice, in at the top and out at the bottom; a propeller turbine is turned by a large flow passing through a tube.

A rough guide that micro-hydro engineers use:

Head	Turbines that suit it
high (over about 50 m)	Pelton, Turgo (a cousin of the Pelton)
medium (about 10 to 50 m)	crossflow, Pelton with several jets, Turgo, Francis
low (under about 10 m)	crossflow, propeller, Kaplan (a propeller whose blades can turn)

In Nepal the Pelton and the crossflow turbine are the most common in small plants, and most of them are made in Nepal’s own workshops. The crossflow suits a wide range of heads and flows, and it is simple to make from steel plate and pipe. For the smallest plants there is the peltric set: a small Pelton wheel on the same shaft as an ordinary electric motor, which works as a generator when it is turned. According to the Alternative Energy Promotion Centre (AEPC), peltric sets made in Nepal give up to 5 kW.

A village has a stream with a big flow but a drop of only 3 m near the village. Which turbine is the best choice?

Pick an answer to see why.

From turbine to generator

The turbine turns a generator, which you met in Grade 10: a coil turning in a magnetic field Grade 10: The a.c. generator. Two things make a small plant’s generator harder to run than the bicycle dynamo.

The speed must be right. Mains electricity in Nepal goes through 50 complete cycles a second (50 Hz), and lights and machines are made for that. A common generator has four magnetic poles and gives one complete cycle for every half turn, so it must turn 25 times a second, 1,500 times a minute. Most turbines turn more slowly, so a belt from a big pulley on the turbine to a small one on the generator speeds it up (a wheel and axle at work).

The speed must stay the same. At dusk, the villagers switch their lights on all at once. The generator then has more load, the turbine slows down, and the voltage and frequency drop; switch them off, and it races. A big station opens and closes a valve. A small plant uses something simpler: an electronic load controller, which keeps the total load the same. Whatever power the village isn’t using, it sends to a ballast, often a heater in a tank of water. Switch on 100 W of lights, and the controller sends 100 W less to the ballast. The turbine never notices.

Design, build, test, improve

Engineers don’t only work out a design on paper. They build it, test it by measuring how well it works, and improve it, again and again. Every design must meet some criteria (what it must do: give 6 kW, say) within some constraints (what it mustn’t break: cost, the materials to hand, the water that must be left in the stream).

Worked example

Designing a village scheme

A village of 60 houses wants about 100 W of electricity for each house. The stream’s flow in the dry season was measured as 40 L/s, and 10% of it must be left in the stream. Taking the overall efficiency as 50%, what head is needed?

Answer

Electrical power needed = 60 × 100 W = 6,000 W.
At 50% efficiency, the power in the water must be 6,000 ÷ 0.5 = 12,000 W.
Flow that may be used = 40 − 4 = 36 L/s, that is 36 kg/s.
From P = QgH:
$H = \frac{P}{Qg} = \frac{12{,}000\ \text{W}}{36\ \text{kg/s} \times 9.8\ \text{m/s}^2} = 34\ \text{m}$
So the engineers look for a place where they can get at least 34 m of head. Taking the water in further upstream gives more head, but needs a longer canal, which costs more and has more to go wrong. Finding the best balance is the design.

Now try it yourself, on a small scale. Your wheel won’t make electricity (unless you have a small motor from an old toy, which works as a generator when it is turned), but you can measure its output another way: by the weight it lifts. The energy you put in is the potential energy of the water you pour through it; the energy you get out is the potential energy of the load it lifts.

$\text{efficiency} = \frac{\text{output}}{\text{input}} = \frac{m_\text{load}\,g\,h_\text{lift}}{m_\text{water}\,g\,h_\text{drop}}$

Figure 13.7 A water wheel to build at home (a close-up), to scale. A cork with plastic spoons pushed into it turns on a bamboo skewer, held in two straws taped to a frame. Water from an upturned bottle, with a hole in its cap, strikes the spoons; the turning skewer winds up a thread and lifts a small bag of coins. The drop is measured from half-way up the water in the full bottle to the bottom of the wheel.

Activity

Build a water wheel and make it better

You need: a cork (or a thick piece of a bamboo stem), 6 to 8 plastic spoons, a bamboo skewer, two drinking straws, a frame to hold the axle (a shoebox with its ends cut away, or two pieces of wood), tape, a 2 L plastic bottle with a small hole made in its cap, thread, a small plastic bag of coins (weigh it on a kitchen scale), a ruler, and a large tray or basin. Do it outside, or in a bath or a sink.

Push the spoons into the cork all the way round, evenly spaced, all facing the same way. Push the skewer through the middle of the cork.
Tape a short piece of straw to each side of the frame, and rest the skewer’s ends in them, so the wheel turns freely over the basin.
Tie the thread to the skewer at one end, and the bag of coins to the thread, hanging down to the floor.
Fill the bottle with 2 L of water, hold it upside down at a fixed height above the wheel, and let the jet strike the spoons. Measure the drop: from half-way up the water in the full bottle to the bottom of the wheel. The water starts above the cap, and on average it falls from about half-way up the bottle.
When the bottle is empty, measure how high the bag has been lifted.
Work out the input energy, the output energy and the efficiency. Then change one thing (the blade shape, the number of blades, where the jet hits, the size of the hole, the bearings, the load), and test again. Keep everything else the same, so you know what made the difference.

Water that splashes past the spoons without pushing them still counts in the input; that is part of what the efficiency measures, and a better design wastes less of it. Measure the same way every time, so that your designs can be compared fairly.

What you should see

Expect only a few per cent at first, perhaps less. A home-made wheel loses most of the energy: water splashing past the blades, the jet hitting the blades at the wrong place, friction in the bearings. Cupped spoons should do better than flat blades, and a narrow jet aimed at the cups better than a wide spray, just as Pelton found. Too heavy a load stalls the wheel; too light a load lifts so little that the output is small. Somewhere in between is a best load.

Type in each trial of your own water wheel, and press Add to see its input, output and efficiency. The trials shown first are made-up examples, to show how it works.

#	Change	Input	Output	Efficiency
1	flat card blades	10.78 J	0.088 J	0.8%
2	spoons instead	10.78 J	0.167 J	1.5%
3	jet aimed at the cups	10.78 J	0.255 J	2.4%
4	heavier load, 100 g	10.78 J	0.402 J	3.7%
5
6
7
8

Input = water’s mass × g × drop. Output = load’s mass × g × height lifted. Efficiency = output ÷ input.

Small hydropower in Nepal

The Alternative Energy Promotion Centre (AEPC), the government’s agency for renewable energy, sorts small hydropower plants by their electrical power:

Kind	Power
pico hydro	up to 10 kW
micro hydro	over 10 kW, up to 100 kW
mini hydro	100 kW to 1 MW

As of July 2025, plants built with AEPC’s support give 40,655 kW together, about 41 MW, and 11,104 improved water mills have been installed with its support: ghattas given metal wheels in place of wooden ones (AEPC’s figures at its 29th anniversary, November 2025). An improved mill can grind faster than the old ghatta, and drive a small generator as well, giving up to 3 to 5 kW. There are still thought to be about 25,000 traditional ghattas in Nepal, perhaps more.

41 MW is small next to Nepal’s big stations: Upper Tamakoshi alone, at 456 MW, gives about eleven times as much Grade 10: Electricity in Nepal. But a micro-hydro plant reaches villages days’ walk from the nearest power line, where a big station’s electricity never could, and it is built, run and repaired by the village itself.

A narrow concrete-lined canal of rushing water running along a rocky hillside to a small tank, with a steep mountainside behind — **Figure 13.8** The canal of the micro-hydro plant at Ghunsa, in Taplejung, close to Kangchenjunga, running almost level along the hillside to the forebay tank at the back. Photo: Nirmaljoshi, CC BY-SA 4.0

A metal pipe on concrete supports leaving a concrete tank and running down a stony slope, with rocky mountains behind — **Figure 13.9** The penstock of the small hydropower plant at Kyanjin Gompa in the Langtang valley, Rasuwa, leaving the forebay tank on its way down to the turbine. Photo: Pustam.EGR, CC BY 4.0

In short

The power in falling water is P = QgH: the flow (Q, in kg/s, the same number as L/s) × g × the head (H, in m). One liter a second falling one meter gives about 10 W.
Every step from water to wire loses some energy, mostly as heat. The overall efficiency is the steps’ efficiencies multiplied; for a small plant it is about 50%.
The head can be measured with a water-filled hose, step by step; the flow with a bucket or by the float method, Q = area × average speed. Design for the dry season’s flow, and leave water in the stream.
A turbine should leave the water with as little speed as possible. A Pelton wheel’s cups, moving at half the jet’s speed, turn the water back and leave it almost still. Pelton wheels suit high heads; crossflow and propeller turbines suit lower ones.
A plant’s generator must turn at a steady speed: an electronic load controller sends unused power to a ballast heater.
Engineers design, build, test and improve, within constraints; you can do the same with a home-made wheel, measuring output for a given input.

Questions

A plant has a head of 25 m and a flow of 40 L/s. What power is in its water?

Pick an answer to see why.

A Pelton wheel’s cups move at the same speed as the jet. How much power does the wheel give?

Pick an answer to see why.

a.

A stream gives 120 L/s down a head of 15 m. a. What is the power in the water? b. If the overall efficiency is 55%, what electrical power will the plant give? c. Is this a pico, micro or mini hydro plant, by AEPC’s classes?

Answer

a.P = QgH = 120 × 9.8 × 15 = 17,640 W, about 17.6 kW.
b.0.55 × 17,640 = 9,702 W, about 9.7 kW.
c.Under 10 kW, so a pico hydro plant, only just.

b.

A small plant’s penstock passes on 92% of the water’s power, its turbine 75%, its belt drive 96% and its generator 88%. a. What is the overall efficiency? b. The engineers can afford to improve one part: a better turbine at 82%, or a better generator at 92%. Which improves the plant more? Show your working.

Answer

a.0.92 × 0.75 × 0.96 × 0.88 = 0.58, about 58%.
b.With the better turbine: 0.92 × 0.82 × 0.96 × 0.88 = 0.637, about 64%. With the better generator: 0.92 × 0.75 × 0.96 × 0.92 = 0.609, about 61%. The turbine improves it more, because it was the part losing the most. (Engineers improve the weakest part first.)

c.

Explain why the float method gives a flow that is too big unless you multiply the float’s speed by a number less than 1.

Answer

The float moves with the water at the surface in the middle of the stream, which is the fastest-moving water. Near the bed and the banks, friction slows the water down. The flow is the area of the cross-section × the average speed of all the water in it, which is less than the surface speed in the middle, about 0.8 to 0.9 times it. Using the float’s speed alone would give a flow 10 to 25% too big.

d.

A team tests three designs of a home-made water wheel. Each time they pour 2.0 kg of water through a drop of 0.50 m, and measure how high it lifts a 40 g load. (These results are made up.)

Design	Lift
A: 6 flat blades	0.20 m
B: 6 spoon cups	0.45 m
C: 10 spoon cups	0.40 m

a. Work out the input energy and each design’s efficiency. b. Which design is best, and what might explain the result for C? c. Suggest the next change the team should test, and how to make the test fair.

Answer

a.Input = mgh = 2.0 × 9.8 × 0.50 = 9.8 J. Output = 0.040 × 9.8 × lift: A, 0.040 × 9.8 × 0.20 = 0.078 J, efficiency 0.078 ÷ 9.8 = 0.8%; B, 0.176 J, 1.8%; C, 0.157 J, 1.6%.
b.B. Cups turn the water back and take more of its energy than flat blades. With 10 cups (C), they are crowded: one cup may shield the next from the jet, or water from one cup may splash onto the back of the next, pushing it the wrong way; the extra weight may add friction too.
c.For example: a narrower jet aimed at the middle of the cups, or a lighter or heavier load to find the best load, or smoother bearings. Change only that one thing; keep the water, the drop, the load (unless the load is what is being tested) and the way of measuring the same; and repeat each test two or three times and take the average.

e.

A village has two possible sites for its plant. Site 1: a head of 60 m, with a 600 m canal. Site 2: a head of 30 m, with a 150 m canal. The usable flow is 30 L/s at both. a. Work out the electrical power at each site, at 50% efficiency. b. Give one reason for choosing each site.

Answer

a.Site 1: 0.5 × 30 × 9.8 × 60 = 8,820 W, about 8.8 kW. Site 2: 0.5 × 30 × 9.8 × 30 = 4,410 W, about 4.4 kW.
b.Site 1 gives twice the power, enough for more houses or a mill. Site 2 costs much less to build (a quarter of the canal), and a short canal has less to be damaged by landslides in the monsoon and is easier to look after. Which is better depends on how much electricity the village needs and how much it can spend: a criterion against a constraint.

f.

Explain why a small hydropower plant needs an electronic load controller, and what it does.

Answer

The generator must turn at a steady speed to give a steady voltage and 50 Hz. If the load changes (people switch lights on or off), a turbine with a fixed flow of water would slow down or speed up. The load controller keeps the total load on the generator the same: it sends whatever power the users aren’t taking to a ballast load, usually a heater in water. When users take more, the ballast gets less, and the other way round, so the turbine always has the same load and keeps the same speed.

Where to go next

Where Does the Energy Go When a Ball Bounces? The kinetic and gravitational potential energy behind this topic, counted in joules, and how the total always stays the same.
Why Does Heat Never Flow from Cold to Hot? Why the energy lost in every machine ends up as heat, and why engines that run on heat can never be as efficient as a water turbine.
How Do Magnets Push Without Touching? The energy stored in magnetic and electric fields, like the fields inside the generator.