Sources of Energy in the World

Lessons

Senior 4 · Physics

Sources of Energy in the World

4 lessons

Lesson 2 of 4

Characteristics of Energy Sources

25 min

Same rating, different output

Two power plants on Rwanda's grid have almost the same rating: the KivuWatt plant on Lake Kivu, 26.19 MW, burns methane drawn from deep in the lake; Nyabarongo I, 28 MW, runs on falling river water. Yet in REG's figures KivuWatt delivers about a quarter more energy in a year, and the largest solar plant on REG's list, the 8.5 MW Gigawatt Global plant, delivers less than a fifth of what its rating suggests. The rating of a source tells only part of the story. This lesson builds the measures that engineers use to compare sources: how much energy a kilogram of fuel stores, how much power a square metre of land can give, how much of the time a plant really runs, and whether it can run when it is needed.

Energy density: how much a fuel stores

The specific energy of a fuel is the energy released per kilogram when it burns completely (or, for nuclear fuel, when it undergoes fission). It decides how much fuel must be mined, carried and stored. Table 2.1 gives typical values.

Store of energy	Typical specific energy ( $\mathrm{MJ\,kg^{-1}}$ )
water held 100 m above a turbine ( $g h$ )	0.000 98
lithium-ion battery	0.5–0.9
dry firewood	16
coal	24–30
charcoal	29
diesel, petrol	45
methane (natural gas)	50
hydrogen	120
uranium-235, complete fission	$8 \times 10^{7}$

Table 2.1: Typical specific energies, rounded. Fuels are measured by the heat released on burning (lower heating value).

The table explains a great deal. Water behind a dam stores very little energy per kilogram, so hydroelectric power needs enormous volumes of water. Liquid fuels pack 50 to 90 times more energy into each kilogram than batteries, which is why lorries and aircraft still burn them. A kilogram of uranium-235 releases as much energy as about 1.8 million kilograms of diesel.

Definition

The specific energy of a fuel is the energy it releases per unit mass, in $\mathrm{J\,kg^{-1}}$ (often $\mathrm{MJ\,kg^{-1}}$ ). The energy density is the energy released per unit volume, in $\mathrm{J\,m^{-3}}$ . Example: 2.0 kg of charcoal at $29\,\mathrm{MJ\,kg^{-1}}$ releases $58\,\mathrm{MJ}$ when it burns completely.

Power from moving air and falling water

Wind. In a time $t$ the air that passes through the circle swept by a turbine's blades fills a cylinder of cross-section $A$ and length $v t$ (Figure 2.1). Its mass is $m = \rho A v t$ , where $\rho$ is the density of air, and its kinetic energy is

E_k = \tfrac{1}{2} m v^{2} = \tfrac{1}{2} \rho A v^{3} t \quad\Rightarrow\quad P_{\text{wind}} = \frac{E_k}{t} = \tfrac{1}{2} \rho A v^{3}

Figure 2.1: The air that passes through the swept area A in time t fills a cylinder of length vt.

The power in the wind grows as the cube of the wind speed: double the speed and the power rises eight times. A turbine cannot take all of it, because the air must keep moving to leave the blades. The fraction it extracts is its power coefficient $C_{p}$ , and no turbine can exceed the Betz limit, $C_p = \frac{16}{27} \approx 0.59$ ; good modern turbines reach about 0.45.

Water. If a volume $Q$ of water flows through the turbines each second (the flow rate, in $\mathrm{m^3\,s^{-1}}$ ) and falls through a vertical height $h$ (the head, Figure 2.2), the mass per second is $\rho Q$ and the gravitational potential energy released per second is $\rho Q g h$ . With an overall efficiency $\eta$ ,

P = \eta\, \rho\, g\, Q\, h

Figure 2.2: A hydroelectric scheme: water falls through the head h at flow rate Q.

Sunlight. Near the equator, sunlight on a clear day at noon delivers about $1000\,\mathrm{W\,m^{-2}}$ to a surface facing the Sun, and a sunny tropical site receives about $5\,\mathrm{kWh}$ per square metre per day on average. A panel of area $A$ and efficiency $\eta$ gives $P = \eta I A$ , where $I$ is the intensity of the sunlight. Because the output follows the Sun, it is zero at night and small on cloudy days.

Worked example 1: a wind turbine

A turbine has blades 40 m long and a power coefficient of 0.40. Find its output in a wind of 8.0 m/s, and in a wind of 4.0 m/s. Take $\rho = 1.2\,\mathrm{kg\,m^{-3}}$ .

A = \pi r^{2} = \pi \times 40^{2} = 5.03 \times 10^{3}\,\mathrm{m^2}

P_{\text{wind}} = \tfrac{1}{2} \times 1.2 \times 5.03 \times 10^{3} \times 8.0^{3} = 1.54 \times 10^{6}\,\mathrm{W}

P_{\text{out}} = C_p P_{\text{wind}} = 0.40 \times 1.54\,\mathrm{MW} = 0.62\,\mathrm{MW}

At 4.0 m/s the speed is halved, so the power falls by $2^{3} = 8$ : $P_{\text{out}} = \frac{0.62}{8} = 0.077\,\mathrm{MW}$ . A site with steady strong winds is worth far more than one with gentle winds.

Worked example 2: a hydroelectric station

Water flows through a station's turbines at $25\,\mathrm{m^3\,s^{-1}}$ with a head of 120 m. The overall efficiency is 88% (illustrative figures). Find the electrical power output.

P = \eta \rho g Q h = 0.88 \times 1000 \times 9.81 \times 25 \times 120 = 2.59 \times 10^{7}\,\mathrm{W} \approx 26\,\mathrm{MW}

Each kilogram of water gives only $gh = 1.18\,\mathrm{kJ}$ , so a 26 MW station needs 25 tonnes of water every second.

Capacity factor, availability and cost

A plant's rating is the most power it can deliver. Over a year it delivers less, because the wind drops, the Sun sets, rivers fall in the dry season, fuel runs short, demand is low or the plant is shut for maintenance. The capacity factor compares the energy actually delivered with the energy the plant would deliver at full power all the time:

\text{capacity factor} = \frac{E_{\text{delivered}}}{P_{\text{rated}} \times t}

Figure 2.3 shows the capacity factors REG uses for six of Rwanda's plants. The methane and peat plants that burn fuel can run day and night; the solar plants are limited by the hours of strong sunshine.

Figure 2.3: Capacity factors of six plants on Rwanda's grid. Source: REG, Least Cost Power Development Plan, 2024 Review (September 2025), Table 3.

Three further characteristics complete the picture.

Dispatchability. A dispatchable source can be turned up or down when the grid operator asks: fuel-burning plants and hydroelectric plants with reservoirs. A variable (intermittent) source, such as solar or wind, delivers power only when nature allows, so it needs storage (batteries, pumped water) or other plants beside it.
Cost. Some sources cost a lot to build but little to run (hydro, solar, wind: the "fuel" is free); others are cheaper to build but pay for fuel for their whole life (diesel). The diesel units at Jabana are used for emergencies and peak hours because of their high running cost (REG, 2025).
Location and lifetime. Hydroelectric power needs a river with a large head or flow; geothermal power needs hot rock near the surface; methane can only be taken from Lake Kivu. A dam may run for a century, a solar panel for 25–30 years.

Definition

The capacity factor of a power plant is the ratio of the energy it delivers in a period to the energy it would deliver running at its rated power for the whole period. It has no unit and is often given as a percentage. A dispatchable source is one whose output can be controlled to follow demand; a variable source depends on the weather or the time of day. Example: a 10 MW plant that delivers 35 GWh in a year has a capacity factor of $\frac{35\,000\,\mathrm{MWh}}{10\,\mathrm{MW} \times 8760\,\mathrm{h}} = 0.40$ .

Worked example 3: solar against methane

Use Figure 2.3. (a) Find the energy delivered in a year by the 8.5 MW Gigawatt Global solar plant and by the 26.19 MW KivuWatt plant. (b) What rating of solar plant, with the same capacity factor as Gigawatt Global, would deliver as much energy as KivuWatt?

E_{\text{solar}} = 8.5\,\mathrm{MW} \times 8760\,\mathrm{h} \times 0.1708 = 1.27 \times 10^{4}\,\mathrm{MWh} = 12.7\,\mathrm{GWh}

E_{\text{methane}} = 26.19 \times 8760 \times 0.9103 = 2.09 \times 10^{5}\,\mathrm{MWh} = 209\,\mathrm{GWh}

P_{\text{solar}} = \frac{2.09 \times 10^{5}\,\mathrm{MWh}}{8760\,\mathrm{h} \times 0.1708} = 140\,\mathrm{MW}

About 140 MW of solar panels would be needed, more than five times KivuWatt's rating, and the energy would still arrive only in daylight. A fair comparison of sources always uses energy delivered, not rating.

Warning

Common mistakes. (1) Using $v^{2}$ instead of $v^{3}$ for wind power: the mass of air arriving each second also grows with $v$ . (2) Using the length of the dam or the depth of the reservoir as the head: the head is the vertical drop between the two water surfaces. (3) Comparing sources by rated power (MW) alone: compare the energy they deliver (MWh) and when they deliver it. (4) Forgetting that a capacity factor needs the same units on top and bottom: MWh against MW × h.

Tip

Five questions for any source: How much energy per kilogram or per square metre? How much of the year does it run? Can it be switched on when needed? What does it cost to build and to run? Where can it be built?

How much land does it take to power the world? (TED-Ed). Watch for how the power that each square metre of land can give differs between solar, wind, hydroelectric and fossil-fuel sources: the idea of power density used in this lesson.

✦

Practice

How much energy is released when 3.5 kg of dry firewood burns completely? How much charcoal would release the same energy?
A small turbine has blades 1.5 m long. Find the power in a wind of 6.0 m/s, and its output if $C_{p} = 0.35$ . Take $\rho = 1.2\,\mathrm{kg\,m^{-3}}$ .
By what factor does a wind turbine's output change when the wind speed rises from 5.0 m/s to 7.5 m/s?
A micro-hydro scheme on a hillside stream has a head of 45 m and a flow rate of $0.060\,\mathrm{m^3\,s^{-1}}$ . Its efficiency is 70%. Find its output, and how many 10 W LED bulbs it could light.
Use Figure 2.3 to find the energy delivered in a year by the 28 MW Nyabarongo I plant.
A school needs 15 kWh of electricity a day. Its site receives 5.0 kWh of sunlight per square metre per day and its panels are 18% efficient. Find the area of panels needed, and explain why the school also needs batteries.
Explain, using the idea of dispatchability, why a grid with a large share of solar power still needs hydroelectric or fuel-burning plants.
A delivery company is choosing between diesel lorries and battery lorries. Using Table 2.1, explain the main physical disadvantage of the battery lorry, and one way in which electric motors partly make up for it.

Show answersHide answers

$3.5 \times 16 = 56\,\mathrm{MJ}$ ; charcoal: $\frac{56}{29} = 1.9\,\mathrm{kg}$ .
$A = \pi \times 1.5^{2} = 7.07\,\mathrm{m^2}$ ; $P_{\text{wind}} = \tfrac{1}{2} \times 1.2 \times 7.07 \times 6.0^{3} = 916\,\mathrm{W} \approx 0.92\,\mathrm{kW}$ ; output $0.35 \times 916 = 321\,\mathrm{W} \approx 0.32\,\mathrm{kW}$ .
$\left(\frac{7.5}{5.0}\right)^{3} = 1.5^{3} = 3.4$ : it rises about 3.4 times.
$P = 0.70 \times 1000 \times 9.81 \times 0.060 \times 45 = 1.85 \times 10^{4}\,\mathrm{W} \approx 19\,\mathrm{kW}$ : enough for about 1850 bulbs.
$28 \times 8760 \times 0.6854 = 1.68 \times 10^{5}\,\mathrm{MWh} = 168\,\mathrm{GWh}$ .
$A = \frac{15}{0.18 \times 5.0} = 17\,\mathrm{m^2}$ . Panels give power only in daylight, mostly around noon, so batteries must store energy for the evening and for cloudy days.
Solar output cannot be raised when demand rises, and it falls to zero every evening, which is when Rwanda's daily peak load occurs (REG, 2025). Dispatchable plants (hydro with a reservoir, methane, peat, diesel) or storage must cover the evening peak and cloudy days.
Batteries store only about 0.5–0.9 MJ per kg against 45 MJ per kg for diesel, so the batteries for the same range are far heavier and take load space. Electric motors waste much less energy than diesel engines (roughly 90% against 40% efficient), so the battery needs to store only about half the energy the diesel tank holds; the gap narrows but does not close.

Quick check

The wind speed at a turbine doubles. By what factor does the power in the wind change?

Key points

Specific energy ( $\mathrm{J\,kg^{-1}}$ ) ranges from about 1 kJ per kg for water 100 m up to $8 \times 10^{13}\,\mathrm{J\,kg^{-1}}$ for uranium-235.
Wind: $P = \tfrac{1}{2} \rho A v^{3}$ , of which at most $\frac{16}{27}$ can be extracted. Water: $P = \eta \rho g Q h$ . Sunlight: $P = \eta I A$ .
Capacity factor $= \frac{E_{\text{delivered}}}{P_{\text{rated}}\,t}$ : about 91% for KivuWatt but 17% for the 8.5 MW Gigawatt Global solar plant (REG, 2025).
Dispatchable sources follow demand; variable sources need storage or back-up. Cost, location and lifetime also matter.

Sources and credits

Rwandan facts: Rwanda Energy Group, Rwanda Least Cost Power Development Plan, 2024 Review (September 2025), Table 3 and section 4; REG, Methane gas, read October 2026.
Photo: Wind turbines near Fairfield, Montana, May 2022 (cropped) by the U.S. Department of Agriculture, Wikimedia Commons, public domain.
Photo: Mead, Nebraska solar farm aerial 2026 (cropped) by AllTheUsernamesAreInUse, Wikimedia Commons, released under CC0 1.0.
Video: How much land does it take to power the world?, TED-Ed, embedded from YouTube.

Written by Ganzaa: the lesson text, derivations, diagrams, chart, table, worked examples and practice exercises.

1Same rating, different output

2Energy density: how much a fuel stores

3Power from moving air and falling water

4Capacity factor, availability and cost

Same rating, different output

Energy density: how much a fuel stores

Power from moving air and falling water

Capacity factor, availability and cost