Converting and Transmitting Energy – Sources of Energy in the World (Senior 4 Physics) | Ganzaa

Sources of Energy in the World

Lessons

Senior 4 · Physics

Sources of Energy in the World

4 lessons

Lesson 3 of 4

Converting and Transmitting Energy

25 min

From a swamp to a socket

Rwanda Energy Group (REG) lists the Hakan peat-to-power plant, with a nameplate rating of 70 MW, as the largest power station in Rwanda (REG, 2025). Peat dug from the swamps of the south is burned to boil water; the steam spins a turbine; the turbine turns a generator; and the electricity travels along high-voltage lines, then local lines, to homes and factories. At every step some energy leaves as heat that can no longer do anything useful. In S3 you met transformers and saw why power lines carry high voltages. In this lesson you put numbers on both halves of the journey: how much of the fuel's energy becomes electricity, and how much of that electricity is lost on the way to the user.

Energy changes in a power station

No device creates energy; each one converts it from one form to another, and every real conversion also produces some internal energy (heat) that spreads into the surroundings. Table 3.1 lists the conversions you meet between a fuel, or a river, and a working appliance.

Device	Energy in	Useful energy out	Typical efficiency
boiler (furnace)	chemical energy of the fuel	internal energy of high-pressure steam	85–90%
steam turbine with condenser (the steam cycle)	internal energy of the steam	kinetic energy of the spinning shaft	about 40%
generator	kinetic energy of the shaft	electrical energy	98–99%
transformer	electrical energy at one voltage	electrical energy at another voltage	98–99.5%
water turbine and generator	gravitational potential energy of water	electrical energy	85–90%
solar (photovoltaic) panel	energy of sunlight	electrical energy	15–22%
electric motor	electrical energy	kinetic energy	85–95%

Table 3.1: Energy conversions and typical efficiencies of large modern equipment, rounded.

One row stands out. The steam cycle turns only about 40% of the steam's energy into work, and most of the rest is handed to cooling water in the condenser, where the used steam is turned back into water. This is not careless engineering: the second law of thermodynamics, which you meet in Unit 9, forbids any engine from turning all of the heat it takes in into work.

Definition

The efficiency $\eta$ of a device is the fraction of the energy put into it that comes out in the useful form:

\eta = \frac{\text{useful energy output}}{\text{total energy input}} = \frac{\text{useful power output}}{\text{total power input}}

It has no unit and is often given as a percentage. For every real device $\eta \lt 1$ ; the rest is wasted energy, usually internal energy of the surroundings. Example: a motor that takes 2.0 kW from the mains and delivers 1.7 kW of mechanical power has $\eta = \frac{1.7}{2.0} = 0.85$ , or 85%; the other 0.3 kW warms the motor and the air around it.

Chains of conversions and Sankey diagrams

In a power station the devices work in series: the useful output of the boiler is the input of the turbine, whose output is the input of the generator. If the stages have efficiencies $\eta_1, \eta_2, \eta_3$ , then $E_1 = \eta_1 E_{\text{in}}$ , $E_2 = \eta_2 E_1$ and $E_{\text{out}} = \eta_3 E_2$ , so

E_{\text{out}} = \eta_1 \eta_2 \eta_3\, E_{\text{in}} \quad\Rightarrow\quad \eta_{\text{total}} = \eta_1 \times \eta_2 \times \eta_3

Efficiencies in a chain multiply; they never add. A chain is always less efficient than its worst link.

A Sankey diagram shows the same idea as a picture. A band whose width is proportional to the energy flows from left to right; at each stage a branch peels off for the energy wasted there, and what is left flows on. Because energy is conserved, the widths of all the branches and the final band add up to the width of the input band. Figure 3.1 shows a typical fuel-burning power station.

Figure 3.1: A Sankey diagram of a steam power station, for every 100 MJ of chemical energy in the fuel. The values are typical, not those of a particular plant.

Worked example 1: the chain in a peat-fired station

A peat-fired station has a boiler of efficiency 85%, a steam cycle of efficiency 40% and a generator of efficiency 98%. (a) Find its overall efficiency. (b) It delivers 70 MW of electrical power. At what rate must energy be supplied in the fuel, and at what rate is energy wasted? (c) Taking the specific energy of the peat as $15\,\mathrm{MJ\,kg^{-1}}$ (an assumed value), find the mass of peat burned each hour.

(a) $\eta = 0.85 \times 0.40 \times 0.98 = 0.333$ , or 33%.

(b) $P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{70\,\mathrm{MW}}{0.333} = 210\,\mathrm{MW}$ ; wasted: $210 - 70 = 140\,\mathrm{MW}$ .

(c) Mass per second: $\frac{2.10 \times 10^{8}\,\mathrm{J\,s^{-1}}}{1.5 \times 10^{7}\,\mathrm{J\,kg^{-1}}} = 14.0\,\mathrm{kg\,s^{-1}}$ ; per hour: $14.0 \times 3600 = 5.04 \times 10^{4}\,\mathrm{kg} \approx 50$ tonnes.

Two-thirds of the energy dug out of the swamp leaves as heat. For every unit of electricity, the station must also get rid of two units of heat, which is why steam stations need a large supply of cooling water or cooling towers.

Worked example 2: reading a Sankey diagram

Use Figure 3.1. (a) Find the efficiency of the station. (b) A station with this diagram delivers 50 MW to the grid. At what rate does it give heat to its cooling water? (c) Which single change would save the most energy, and why is it hard to make?

(a) $\eta = \frac{32}{100} = 0.32$ , or 32%.

(b) The branches are in proportion, so $P_{\text{condenser}} = \frac{52}{32} \times 50\,\mathrm{MW} = 81\,\mathrm{MW}$ .

(c) The condenser branch (52 out of 100) is by far the widest. It can be narrowed only a little by running the boiler hotter, which is limited by what the steel can withstand. Its heat can, however, be used, for example to heat buildings or dry crops, which is what "combined heat and power" stations do.

Sending the electricity: why the grid uses high voltages

A transmission line is a long conductor with resistance $R$ . When a power station sends power $P$ at voltage $V$ , the current in the line is $I = \frac{P}{V}$ , and the line turns electrical energy into internal energy at the rate

P_{\text{loss}} = I^{2} R = \left(\frac{P}{V}\right)^{2} R = \frac{P^{2} R}{V^{2}} \qquad\Rightarrow\qquad \frac{P_{\text{loss}}}{P} = \frac{P R}{V^{2}}

For a given power and line, the fraction lost falls as the square of the voltage: double the voltage, and the loss falls to a quarter. Transformers make this possible: a step-up transformer at the power station raises the voltage and lowers the current by the same factor, and step-down transformers near the users bring the voltage back to a safe level. Transformers work only with a changing current, which is one main reason why grids use alternating current.

Figure 3.2 shows the journey in Rwanda. REG's transmission network uses 220 kV and 110 kV lines, 1158 km of them at the end of June 2023 (REG). Medium-voltage lines run at 30 kV in rural areas and 15 kV in some urban areas, and transformers on poles step down to 400 V and 230 V for users (REG Reticulation Standard, 2020).

Figure 3.2: From power station to home: the voltages used on Rwanda's grid (REG). The current, and so the I²R loss, is smallest where the lines are longest.

Why not use even higher voltages? Each step up needs taller towers, longer insulators, wider cleared strips of land and costlier transformers, and above a few hundred kilovolts the air around a thin wire begins to conduct (a corona discharge, which wastes energy and makes a crackling sound). Engineers choose the voltage that gives the lowest total cost of losses plus equipment for the distance and power involved.

Definition

The transmission loss of a line is the rate at which it converts electrical energy into internal energy because of its resistance: $P_{\text{loss}} = I^{2} R$ , where $I$ is the current in the line and $R$ its total resistance. For power $P$ sent at voltage $V$ , the fraction lost is $\frac{PR}{V^{2}}$ . Example: 10 MW sent at 110 kV through 5.0 Ω loses $\frac{(1.0 \times 10^{7})(5.0)}{(1.1 \times 10^{5})^{2}} = 0.0041$ , or 0.41%.

Worked example 3: the same power at three voltages

A power station sends 50 MW along a line of total resistance 4.0 Ω. Find the current, the power lost and the percentage lost if the power is sent at (a) 30 kV, (b) 110 kV, (c) 220 kV.

Voltage	$I = \frac{P}{V}$	$P_{\text{loss}} = I^{2}R$	Percentage lost
30 kV	$\frac{5.0 \times 10^{7}}{3.0 \times 10^{4}} = 1.67 \times 10^{3}\,\mathrm{A}$	$(1667)^{2} \times 4.0 = 11.1\,\mathrm{MW}$	22%
110 kV	$455\,\mathrm{A}$	$(455)^{2} \times 4.0 = 0.83\,\mathrm{MW}$	1.7%
220 kV	$227\,\mathrm{A}$	$(227)^{2} \times 4.0 = 0.21\,\mathrm{MW}$	0.41%

Raising the voltage from 110 kV to 220 kV halves the current and quarters the loss. At 30 kV more than a fifth of the power would warm the wires, which is why medium-voltage lines are kept for shorter distances.

Worked example 4: losses on Rwanda's grid

REG's plan for 2025 expects a total supply of 1619.5 GWh, with 50.0 GWh lost in transmission and 202.0 GWh lost in distribution (REG, 2025, Table 9). Find each loss as a percentage of the supply, and explain why one is much larger.

\text{transmission: } \frac{50.0}{1619.5} \times 100\% = 3.1\%, \qquad \text{distribution: } \frac{202.0}{1619.5} \times 100\% = 12.5\%

Together about 15.6% of the supply, roughly one unit in every six, never reaches a customer. Distribution loses four times as much because its lines run at 30 kV, 15 kV and 400 V: for the same power the current is many times larger than at 110 kV or 220 kV, and there are many thousands of kilometres of these lines and many transformers. (Distribution figures also include energy that is used but never billed, for example through illegal connections.)

Warning

Common mistakes. (1) Adding efficiencies in a chain: 85% and 40% give $0.85 \times 0.40 = 34\%$ , not 125% or 62.5%. (2) Using $\frac{V^{2}}{R}$ with the supply voltage to find the line loss: the supply voltage is not across the line. Use $P_{\text{loss}} = I^{2}R$ with $I = \frac{P}{V}$ . (3) Thinking high voltage lowers the power sent: the power $P = V I$ is the same; it is the current that falls. (4) Drawing a Sankey diagram whose branches do not add up to the input.

Tip

Memory hook: "Loss goes as one over V squared." Same power, twice the voltage, half the current, a quarter of the loss. For a chain of devices, multiply the efficiencies, and remember the answer can never be larger than the smallest one.

Why do Power Lines use High Voltage? (Physics Ninja). Watch how the power lost in the line, I²R, is compared for the same power sent at a low and at a high voltage, and check the result against Worked example 3.

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Practice

An electric motor takes 1.5 kW from the mains and delivers 1.2 kW of mechanical power. Find its efficiency and the energy it wastes in 10 minutes.
A hydroelectric station has a turbine of efficiency 92%, a generator of 97% and a step-up transformer of 99%. Find its overall efficiency.
An LED lamp takes 9.0 W and gives out 3.6 W as light. Find its efficiency and sketch a Sankey diagram for it, drawn to scale.
A power station sends 120 MW at 220 kV along lines of total resistance 6.0 Ω. Find (a) the current, (b) the power lost, (c) the percentage lost.
The station in question 4 sends the same power at 110 kV. Find the new loss, and explain why it is four times as large.
Show that the fraction of power lost in a line is $\frac{PR}{V^{2}}$ . Find the smallest voltage that keeps the loss in question 4 below 1.0%.
REG lists the available capacity of the Hakan plant as 26.6 MW (REG, 2025). Using the chain efficiency of Worked example 1 and peat of $15\,\mathrm{MJ\,kg^{-1}}$ , estimate the mass of peat burned in a day at this output.
A classmate suggests sending all electricity at 1000 kV, "because then the losses would be almost zero". Evaluate this suggestion.

Show answersHide answers

$\eta = \frac{1.2}{1.5} = 0.80$ , or 80%; wasted $0.3\,\mathrm{kW} \times 600\,\mathrm{s} = 1.8 \times 10^{5}\,\mathrm{J}$ .
$0.92 \times 0.97 \times 0.99 = 0.88$ , or 88%.
$\eta = \frac{3.6}{9.0} = 0.40$ , or 40%. The diagram: an input band of 9.0 W splitting into 3.6 W of light and 5.4 W of heat, widths in the ratio 2 : 3.
(a) $I = \frac{1.2 \times 10^{8}}{2.2 \times 10^{5}} = 545\,\mathrm{A}$ ; (b) $545^{2} \times 6.0 = 1.79 \times 10^{6}\,\mathrm{W} = 1.8\,\mathrm{MW}$ ; (c) $\frac{1.79}{120} = 1.5\%$ .
$I = 1091\,\mathrm{A}$ , $P_{\text{loss}} = 7.1\,\mathrm{MW}$ (6.0%). Halving $V$ doubles $I$ , and $P_{\text{loss}} = I^{2}R$ goes up by $2^{2} = 4$ .
$P_{\text{loss}} = \left(\frac{P}{V}\right)^{2}R$ , so $\frac{P_{\text{loss}}}{P} = \frac{PR}{V^{2}}$ . Need $\frac{PR}{V^{2}} \le 0.010$ : $V \ge \sqrt{\frac{1.2 \times 10^{8} \times 6.0}{0.010}} = 2.7 \times 10^{5}\,\mathrm{V}$ , about 270 kV.
$P_{\text{in}} = \frac{26.6}{0.333} = 79.8\,\mathrm{MW}$ ; in a day $79.8 \times 10^{6} \times 86\,400 = 6.9 \times 10^{12}\,\mathrm{J}$ ; mass $\frac{6.9 \times 10^{12}}{1.5 \times 10^{7}} = 4.6 \times 10^{5}\,\mathrm{kg}$ , about 460 tonnes.
Losses would indeed be tiny, but 1000 kV needs very tall towers, very long insulators, wide cleared corridors, costly transformers at both ends, and suffers corona losses; for Rwanda's distances and powers (a few hundred MW over tens to hundreds of km) 110 kV and 220 kV already keep transmission losses near 3%. The best voltage balances the cost of losses against the cost of equipment.

Quick check

A station sends the same power along the same line, but at twice the voltage. The power lost in the line becomes:

Key points

Efficiency $\eta = \frac{\text{useful output}}{\text{total input}} \lt 1$ ; in a chain the efficiencies multiply.
A Sankey diagram shows energy flows to scale; its branches add up to the input.
A steam station turns only about a third of its fuel's energy into electricity; most of the rest leaves in the cooling water.
Line loss $P_{\text{loss}} = I^{2}R = \frac{P^{2}R}{V^{2}}$ : high voltage means small current and small loss. Rwanda transmits at 220 kV and 110 kV and distributes at 30 kV, 15 kV, 400 V and 230 V (REG).

Sources and credits

Rwandan facts: Rwanda Energy Group, Rwanda Least Cost Power Development Plan, 2024 Review (September 2025), Table 3, Table 9 and section 6.1.1; REG, Transmission, read October 2026; REG, Reticulation Standards for Electricity Distribution Planning, Construction and Maintenance, version 6 (August 2020).
Photo: Low-pressure turbine rotor from the Greifswald Nuclear Power Plant by Siarhei Besarab, Wikimedia Commons, licensed under CC BY-SA 4.0.
Photo: High voltage transmission towers and lines by Stefan Andrej Shambora, Wikimedia Commons, licensed under CC BY 2.0.
Video: Why do Power Lines use High Voltage?, Physics Ninja, embedded from YouTube.

Written by Ganzaa: the lesson text, derivations, Sankey diagram, grid diagram, tables, worked examples and practice exercises. This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1From a swamp to a socket

2Energy changes in a power station

3Chains of conversions and Sankey diagrams

4Sending the electricity: why the grid uses high voltages

From a swamp to a socket

Energy changes in a power station

Chains of conversions and Sankey diagrams

Sending the electricity: why the grid uses high voltages