Power and Efficiency

Site: Young Education
Cours: Work Energy Power
Livre: Power and Efficiency
Imprimé par: Guest user
Date: lundi, 9 novembre 2026, 16:40

1. Power

Learning outcomes
  • I can define power as the rate at which work is done or energy is transferred.
  • I can explain the difference between energy and power.
  • I can recall and use the equation  \( P = \frac{E}{t} \)
  • I can calculate power from energy transfers and time intervals.
  • I can compare the power outputs of different devices and systems.

Power

Power is the rate at which work is done or energy is transferred.

Power tells us how quickly energy is transferred.

Two machines can transfer exactly the same amount of energy but have very different power outputs if one transfers the energy in less time.

For example:

  • two people climb the same staircase, but one runs while the other walks
  • two kettles heat the same amount of water, but one heats it faster
  • two motors lift the same load, but one completes the lift more quickly
  • two lamps transfer energy continuously but at different rates

The faster energy is transferred, the greater the power.

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7

Energy and Power Are Different

Energy and power are closely related, but they are not the same quantity.

Energy describes how much energy is transferred or stored.

Power describes how quickly that energy is transferred.

For example, suppose two motors each transfer 10,000 J of energy.

Motor A takes:

20 s

Motor B takes:

5 s

They transfer the same amount of energy, but Motor B transfers it much faster.

Therefore, Motor B has the greater power.


The Power Equation

Power can be calculated using:

P = E/t

where:

  • P = power in watts (W)
  • E = energy transferred in joules (J)
  • t = time in seconds (s)

Power can also be expressed as:

P = W/t

where W represents work done, not watts.

This means:

power = energy transferred ÷ time

or:

power = work done ÷ time


The Watt

The SI unit of power is the watt (W).

One watt means that one joule of energy is transferred every second.

Therefore:

1 W = 1 J/s

A device operating at:

10 W

transfers:

10 J every second

A device operating at:

1000 W

transfers:

1000 J every second


Kilowatts and Megawatts

Large power values are often expressed using larger units.

1 kW = 1000 W

1 MW = 1,000,000 W

1 GW = 1,000,000,000 W

For example:

2000 W = 2 kW

5,000,000 W = 5 MW

Small electrical devices may use only a few watts, while power stations may produce hundreds or thousands of megawatts.


Worked Example: Basic Power Calculation

A machine transfers 600 J of energy in 3 s.

Calculate its power.

Use:

P = E/t

Substitute:

P = 600/3

P = 200 W

The machine has a power output of 200 W.

This means it transfers energy at a rate of:

200 J/s


Worked Example: Electric Device

An electrical device transfers 12,000 J of energy in 60 s.

P = E/t

P = 12,000/60

P = 200 W

The device operates at 200 W.


Worked Example: Larger Machine

A machine transfers 2.4 MJ of energy in 40 s.

First convert megajoules to joules:

2.4 MJ = 2,400,000 J

Then:

P = 2,400,000/40

P = 60,000 W

or:

P = 60 kW


Comparing Power

Suppose two machines perform the same task.

Machine A

Energy transferred = 5000 J
Time = 10 s

P = 5000/10

P = 500 W

Machine B

Energy transferred = 5000 J
Time = 5 s

P = 5000/5

P = 1000 W

Machine B has twice the power because it transfers the same energy in half the time.


Power and Time

For a fixed amount of energy:

shorter time → greater power

longer time → smaller power

For example:

Energy Transferred Time Power
1000 J 1 s 1000 W
1000 J 2 s 500 W
1000 J 5 s 200 W
1000 J 10 s 100 W

The total energy is identical in every case.

Only the rate of energy transfer changes.


Power and Energy

If the time remains constant:

more energy transferred → greater power

For example:

Energy Time Power
100 J 10 s 10 W
200 J 10 s 20 W
500 J 10 s 50 W
1000 J 10 s 100 W

Power is directly proportional to energy transferred when time remains constant.


Rearranging the Power Equation

Starting with:

P = E/t

we can rearrange the equation to calculate energy or time.

Finding Energy

Multiply both sides by t:

E = Pt

Finding Time

Rearrange:

t = E/P

These three forms are useful:

P = E/t

E = Pt

t = E/P


Worked Example: Finding Energy

A 500 W motor operates for 20 s.

Calculate the energy transferred.

Use:

E = Pt

E = 500 × 20

E = 10,000 J

The motor transfers 10 kJ of energy.


Worked Example: Finding Time

A machine must transfer 24,000 J of energy.

Its power is:

800 W

How long does it take?

Use:

t = E/P

t = 24,000/800

t = 30 s


Power and Work Done

Because work done represents energy transferred:

P = W/t

where:

  • P = power
  • W = work done
  • t = time

Recall:

work done = force × distance

For a constant force acting in the direction of motion:

W = Fd

Therefore:

P = Fd/t

Since:

v = d/t

we can also write:

P = Fv

for a constant force acting in the direction of motion.


Worked Example: Lifting a Load

A motor does 15,000 J of work lifting a load in 5 s.

Calculate its power.

P = W/t

P = 15,000/5

P = 3000 W

or:

P = 3.0 kW


Power and Gravitational Potential Energy

When an object is lifted:

ΔEₚ = mgΔh

If the lift takes time t:

P = mgΔh/t

This allows us to calculate the power required to raise an object.

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5

Worked Example: Crane

A crane lifts a 500 kg load vertically through 8 m in 10 s.

Use:

g = 9.8 N/kg

First calculate the change in gravitational potential energy:

ΔEₚ = mgΔh

ΔEₚ = 500 × 9.8 × 8

ΔEₚ = 39,200 J

Now calculate power:

P = E/t

P = 39,200/10

P = 3920 W

or:

P = 3.92 kW


Worked Example: Climbing Stairs

A 60 kg student climbs a staircase with a vertical height of 5 m in 6 s.

Use:

g = 9.8 N/kg

GPE gained:

ΔEₚ = 60 × 9.8 × 5

ΔEₚ = 2940 J

Power:

P = 2940/6

P = 490 W

The student's average mechanical power while climbing is approximately 490 W.


Comparing Two Students

Suppose two 60 kg students climb the same 5 m staircase.

Student A takes:

10 s

Student B takes:

5 s

Both gain the same GPE:

2940 J

Student A:

P = 2940/10 = 294 W

Student B:

P = 2940/5 = 588 W

Both perform the same amount of mechanical work, but Student B has twice the average power because the work is completed in half the time.


Power and Kinetic Energy

Power can also describe how quickly kinetic energy changes.

Recall:

Eₖ = ½mv²

If a motor increases an object's kinetic energy, the average power associated with that transfer is:

P = ΔEₖ/t

For an object accelerating from rest:

P = ½mv²/t

if we are calculating the average power over that time interval.


Worked Example: Accelerating a Car

A 1000 kg car accelerates from rest to 20 m/s in 10 s.

Ignore energy losses.

Final kinetic energy:

Eₖ = ½mv²

Eₖ = 0.5 × 1000 × 20²

Eₖ = 200,000 J

Average power:

P = E/t

P = 200,000/10

P = 20,000 W

or:

P = 20 kW

This is the average mechanical power transferred into kinetic energy during the acceleration.

A real engine or motor would need to provide more energy because of friction, air resistance, and other losses.


Average Power and Instantaneous Power

Power can change during a process.

The equation:

P = ΔE/Δt

usually gives average power over a time interval.

For example, a car's engine may produce different power at different moments during acceleration.

Instantaneous power describes the rate of energy transfer at one particular moment.

For many introductory problems, we calculate average power.


Electrical Devices

Electrical appliances are commonly labelled with their power.

For example, a device rated at:

1000 W

transfers approximately:

1000 J every second

while operating at that power.

A 2000 W device transfers energy twice as quickly as a 1000 W device.

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4

Example: Electric Kettle

A kettle operates at:

2000 W

How much energy does it transfer in 3 minutes?

First convert time:

3 min = 180 s

Use:

E = Pt

E = 2000 × 180

E = 360,000 J

or:

E = 360 kJ


Example: Light Bulb

A lamp operates at 10 W for 60 s.

E = Pt

E = 10 × 60

E = 600 J

The lamp transfers 600 J of energy during that minute.

Not all of this necessarily becomes visible light. Some energy is transferred thermally.


Power Does Not Mean Total Energy

A high-power device does not necessarily transfer more total energy than a low-power device.

It depends on how long each device operates.

For example:

A 1000 W device operating for 10 s transfers:

E = 1000 × 10

E = 10,000 J

A 100 W device operating for 200 s transfers:

E = 100 × 200

E = 20,000 J

The lower-power device transferred more total energy because it operated for much longer.

This is an important distinction:

power = rate

energy = total amount transferred


Power in Vehicles

Vehicle engines and motors are often compared by their power output.

Greater power allows energy to be transferred more rapidly.

This can affect a vehicle's ability to:

  • accelerate
  • climb hills
  • pull heavy loads
  • maintain speed against resistive forces

However, power alone does not determine vehicle performance. Mass, gearing, traction, drag, efficiency, and other factors also matter.

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4

Power in Sports

Power is important in many sports.

Activities requiring high power include:

  • sprinting
  • jumping
  • weightlifting
  • cycling
  • rowing
  • climbing quickly

Two athletes may perform the same amount of work, but the athlete who completes it faster produces greater average mechanical power.


Worked Example: Weightlifting

An athlete lifts a 100 kg mass vertically through 2 m in 1.5 s.

Use:

g = 9.8 N/kg

Work done against gravity:

W = mgΔh

W = 100 × 9.8 × 2

W = 1960 J

Power:

P = 1960/1.5

P ≈ 1307 W

or approximately:

1.31 kW


Power Stations

Power stations are designed to transfer enormous amounts of energy every second.

For example, a power station producing:

500 MW

has a power output of:

500,000,000 W

This means it transfers approximately:

500,000,000 J of energy every second

when operating at that output.

This demonstrates why watts are a rate, not an amount of energy.


Power and Efficiency

A device may receive energy at one power but provide useful energy at a smaller output power.

For example:

Input power = 1000 W

Useful output power = 800 W

Efficiency:

efficiency = useful output power / input power

efficiency = 800/1000

efficiency = 0.80

or:

80%

The remaining power is being transferred into other forms, often thermal energy and sound.


Worked Example: Motor Efficiency

An electric motor receives:

2500 W

of electrical power.

Its efficiency is:

80%

Useful mechanical power:

useful power = efficiency × input power

useful power = 0.80 × 2500

useful power = 2000 W

Therefore:

500 W

is being transferred into other forms while the motor operates under those conditions.


Energy Flow and Power

Energy flow diagrams can also include power.

For example:

2000 W electrical input → 1600 W useful mechanical output + 400 W thermal/sound

This means that every second:

2000 J enters

1600 J becomes useful mechanical energy

400 J is transferred elsewhere

The energy still balances.


Power and Energy Conservation

Power does not change the law of conservation of energy.

A powerful machine does not create more energy.

It simply transfers energy more rapidly.

For example, two motors may each transfer:

100 kJ

Motor A might take:

100 s

Motor B might take:

20 s

Motor A:

P = 100,000/100 = 1000 W

Motor B:

P = 100,000/20 = 5000 W

Both transfer the same energy.

Motor B simply does so five times faster.


Real-World Comparison: Kettles

Suppose two idealized kettles must transfer 300 kJ to water.

Kettle A

Power = 1000 W

t = E/P

t = 300,000/1000

t = 300 s

Kettle B

Power = 2000 W

t = 300,000/2000

t = 150 s

The higher-power kettle can transfer the required energy in half the time, assuming equal efficiency and identical conditions.


Real-World Comparison: Elevators

Consider two elevators lifting identical loads through the same height.

Both perform the same amount of work.

If Elevator A completes the lift in 20 s and Elevator B completes it in 10 s:

Elevator B has twice the average mechanical power.

This is because:

P = E/t

Halving the time while transferring the same energy doubles the power.


Power and Time Graphs

If a device operates at constant power, energy transferred increases steadily with time.

Since:

E = Pt

a graph of energy against time for constant power is a straight line.

The gradient is:

gradient = ΔE/Δt

Therefore:

gradient = power

A steeper energy-time graph represents a more powerful device.


Variable Power

Not every device operates at constant power.

Suppose a machine's power changes over time.

The total energy transferred depends on the area under a power-time graph.

For constant power:

E = Pt

which is simply:

area = height × width

For changing power, the area under the graph can still represent the total energy transferred.

This becomes useful when analyzing engines, motors, electrical systems, and changing mechanical processes.


Common Mistakes

Confusing Power with Energy

Energy is measured in:

joules (J)

Power is measured in:

watts (W)

Power describes the rate of energy transfer.


Thinking a More Powerful Device Always Uses More Energy

Not necessarily.

Energy depends on both:

power and operating time

because:

E = Pt


Forgetting to Convert Minutes to Seconds

For standard SI calculations:

1 min = 60 s

Therefore:

5 min = 300 s


Confusing Watts and Joules

Watt = power

Joule = energy

Remember:

1 W = 1 J/s


Forgetting to Convert Kilowatts

1 kW = 1000 W

Therefore:

2.5 kW = 2500 W


Using the Wrong Time

Power calculations require the time during which the specified energy transfer occurs.


Assuming Power Is the Same as Force

Power is not force.

Force is measured in:

newtons

Power is measured in:

watts


Forgetting That Power Can Change

Many calculations give average power over a time interval.

A machine's instantaneous power may vary during that interval.


Check Your Understanding

  1. Define power.
  2. Explain the difference between energy and power.
  3. State the equation for power.
  4. What is the SI unit of power?
  5. What does 1 W mean?
  6. A machine transfers 1200 J in 4 s. Calculate its power.
  7. A device transfers 15,000 J in 30 s. Calculate its power.
  8. A 2 kW machine operates for 10 s. Calculate the energy transferred.
  9. A 500 W device transfers 20,000 J. How long does it operate?
  10. Convert 3.5 kW to watts.
  11. Convert 2 MW to watts.
  12. Two machines each transfer 5000 J. Machine A takes 20 s and Machine B takes 5 s. Compare their power outputs.
  13. Explain why completing the same work in less time requires greater average power.
  14. A 50 kg student climbs vertically 4 m in 5 s. Calculate the student's average mechanical power using g = 9.8 N/kg.
  15. A crane lifts a 1000 kg load through 6 m in 20 s. Calculate its useful mechanical power.
  16. A 1500 W kettle operates for 3 minutes. Calculate the energy transferred.
  17. A 1000 kg car accelerates from rest to 30 m/s in 15 s. Calculate the increase in kinetic energy.
  18. Calculate the average power transferred into kinetic energy in Question 17.
  19. Why would the actual engine or motor need to transfer more energy than calculated in Question 17?
  20. A machine has an input power of 5000 W and a useful output power of 4000 W. Calculate its efficiency.
  21. A motor is 75% efficient and receives 8 kW. Calculate its useful output power.
  22. Explain how a high-power motor differs from a low-power motor performing the same task.
  23. A power station produces 300 MW. How many joules does it transfer each second?
  24. Explain why a 100 W device can transfer more total energy than a 1000 W device.
  25. Describe how power could be measured experimentally by having students climb a staircase.

Key Terms

Power — Rate at which work is done or energy is transferred.

Watt (W) — SI unit of power; one joule transferred per second.

Joule (J) — SI unit of energy.

Energy — Quantity that can be stored and transferred within physical systems.

Work done — Energy transferred when a force causes displacement.

Rate — Change in a quantity per unit time.

Kilowatt (kW) — Unit of power equal to 1000 W.

Megawatt (MW) — Unit of power equal to 1,000,000 W.

Gigawatt (GW) — Unit of power equal to 1,000,000,000 W.

Average power — Total energy transferred divided by the time interval over which the transfer occurs.

Instantaneous power — Rate of energy transfer at a particular moment.

Input power — Rate at which energy enters a device or system.

Useful output power — Rate at which energy is transferred into the desired output.

Efficiency — Fraction of input energy or power transferred into a useful output.

Energy-time graph — Graph showing how transferred energy changes with time; its gradient represents power.

Power-time graph — Graph showing how power varies with time; its area represents energy transferred.


Key Takeaways

  • Power describes how quickly energy is transferred.
  • Power can also be described as the rate at which work is done.
  • The main equation is P = E/t.
  • Power is measured in watts (W).
  • One watt equals one joule per second.
  • Energy and power are different quantities.
  • Energy measures the amount transferred; power measures the rate of transfer.
  • Two devices can transfer the same amount of energy while having different power outputs.
  • For the same energy transfer, completing the process in less time requires greater average power.
  • The equation can be rearranged to E = Pt and t = E/P.
  • Kilowatts, megawatts, and gigawatts are commonly used for larger power values.
  • Power can be combined with kinetic, gravitational potential, and elastic potential energy calculations.
  • A motor lifting an object can be analyzed using P = mgΔh/t.
  • Average power during acceleration can be found from P = ΔEₖ/t.
  • Higher power does not automatically mean greater total energy consumption; operating time also matters.
  • Efficiency can be calculated by comparing useful output power with input power.
  • Real machines often have a lower useful output power than input power because some energy is dissipated.
  • The gradient of an energy-time graph represents power.
  • The area under a power-time graph represents energy transferred.
  • Power is one of the most useful quantities for comparing how quickly different machines, devices, and people can transfer energy.

The central relationship is:

P = E/t

Remember:

same energy + less time = greater power

and:

1 W = 1 J/s

A powerful system does not necessarily transfer more total energy. It transfers energy at a faster rate.

2. Mechanical Power

Learning outcomes
  • I can explain how power relates to force, displacement, and velocity.
  • I can recall and use the equation \( P = \frac{W}{t} \)
  • I can recall and use the equation P = Fv for constant velocity motion.
  • I can calculate the mechanical power required in practical situations.
  • I can analyze how force, speed, and power are related in machines and transportation systems.

Mechanical Power

Mechanical power describes how quickly mechanical work is done or how quickly mechanical energy is transferred.

A machine that performs the same amount of work in less time has a greater power output.

Mechanical power is especially useful when studying:

  • engines and motors
  • vehicles
  • cranes and elevators
  • conveyor systems
  • pumps
  • bicycles
  • athletes
  • industrial machinery

Two machines may perform exactly the same amount of work, but the more powerful machine can perform that work more quickly.

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5

Mechanical Work and Power

Recall that work is done when a force causes an object to move through a displacement.

When the force acts in the same direction as the displacement:

W = Fd

where:

  • W = work done (J)
  • F = force (N)
  • d = displacement (m)

Mechanical power describes how quickly this work is done:

P = W/t

where:

  • P = power (W)
  • W = work done (J)
  • t = time (s)

Since:

1 J = 1 N·m

and:

1 W = 1 J/s

we can also think of mechanical power as the rate at which a force transfers energy through motion.


The Watt

The SI unit of power is the watt (W).

1 W = 1 J/s

Therefore, a machine producing 500 W of mechanical power transfers:

500 J of mechanical energy every second

A 10 kW motor transfers mechanical energy at a rate of:

10,000 J/s


Worked Example: Mechanical Work

A motor performs 12,000 J of work in 6 s.

Calculate its mechanical power.

Use:

P = W/t

P = 12,000/6

P = 2000 W

or:

P = 2.0 kW

The motor transfers mechanical energy at a rate of 2000 J/s.


Worked Example: Comparing Machines

Two motors each perform 30,000 J of work.

Motor A takes 15 s.

Motor B takes 5 s.

For Motor A:

P = 30,000/15

P = 2000 W

For Motor B:

P = 30,000/5

P = 6000 W

Motor B is three times as powerful.

Both motors perform the same amount of work, but Motor B performs it three times faster.


Connecting Force, Distance, and Power

We know:

P = W/t

and:

W = Fd

Substitute Fd for W:

P = Fd/t

We also know:

v = d/t

Therefore:

P = Fv

This is one of the most useful equations for mechanical power.


Mechanical Power Equation

For a constant force acting in the direction of motion:

P = Fv

where:

  • P = power (W)
  • F = force (N)
  • v = velocity or speed in the direction of the force (m/s)

This equation shows an important relationship:

power depends on both force and speed

A machine requires more power if it must:

  • produce a greater force at the same speed
  • move at a greater speed while producing the same force

Why P = Fv Works

Start with:

P = W/t

For a constant force parallel to the displacement:

W = Fd

Therefore:

P = Fd/t

Since:

v = d/t

we obtain:

P = Fv

So:

P = W/t = Fd/t = Fv

This connects work, force, displacement, velocity, and power.


Worked Example: Moving a Load

A machine pushes a load with a force of 500 N at a constant speed of 4 m/s.

Calculate the mechanical power.

Use:

P = Fv

P = 500 × 4

P = 2000 W

or:

P = 2.0 kW


Worked Example: Conveyor Belt

A conveyor system moves material while exerting a force of 800 N.

The belt moves at:

2.5 m/s

Calculate the mechanical power.

P = Fv

P = 800 × 2.5

P = 2000 W

Therefore:

P = 2.0 kW

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6

Rearranging P = Fv

The equation:

P = Fv

can be rearranged.

To find force:

F = P/v

To find speed:

v = P/F

These forms are especially useful when analyzing vehicles and machines.


Worked Example: Finding Force

A motor provides:

12,000 W

of mechanical power while moving an object at:

6 m/s

Calculate the force.

Use:

F = P/v

F = 12,000/6

F = 2000 N


Worked Example: Finding Speed

A machine provides 15 kW of mechanical power while exerting a force of 3000 N.

First convert:

15 kW = 15,000 W

Then:

v = P/F

v = 15,000/3000

v = 5 m/s


Constant Velocity

The equation P = Fv is particularly useful for objects moving at constant velocity.

At constant velocity:

resultant force = 0

This does not mean the engine or motor produces no force.

Instead, the driving force balances the resistive forces.

For a vehicle:

driving force = total resistive force

These resistive forces may include:

  • air resistance
  • rolling resistance
  • friction
  • mechanical resistance

The engine must continuously transfer energy to overcome these forces.


Worked Example: Car at Constant Speed

A car travels at a constant speed of:

20 m/s

The total resistive force is:

1200 N

Because the car moves at constant velocity:

driving force = 1200 N

Therefore:

P = Fv

P = 1200 × 20

P = 24,000 W

or:

P = 24 kW

The engine must provide at least 24 kW of useful mechanical power to maintain this speed under these simplified conditions.

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6

Why Power Is Needed at Constant Speed

Students sometimes think:

"If the car is not accelerating, why does the engine need power?"

The answer is that resistive forces are continually doing negative work on the vehicle.

The engine must continually transfer energy to overcome:

  • air resistance
  • tyre deformation
  • friction in moving components

Without engine power, these forces would cause the vehicle to slow down.

So at constant speed:

driving force = resistive force

but energy is still being transferred.


Force, Speed, and Power

From:

P = Fv

we can see several important relationships.

Constant Force

If F remains constant:

P ∝ v

Doubling speed doubles power.

Constant Speed

If v remains constant:

P ∝ F

Doubling force doubles power.

Constant Power

If P remains constant:

F = P/v

Therefore:

F ∝ 1/v

As speed increases, the available force decreases if power remains constant.

This relationship is extremely important in machines and transportation.


Constant Power Example

A motor provides constant mechanical power of:

20 kW

At 2 m/s:

F = 20,000/2

F = 10,000 N

At 5 m/s:

F = 20,000/5

F = 4000 N

At 10 m/s:

F = 20,000/10

F = 2000 N

Speed Force at 20 kW
2 m/s 10,000 N
5 m/s 4000 N
10 m/s 2000 N

At constant power:

higher speed → lower available force


Mechanical Power in Vehicles

Vehicle performance provides a useful application of mechanical power.

An engine or motor must transfer energy to:

  • accelerate the vehicle
  • overcome air resistance
  • overcome rolling resistance
  • climb hills
  • carry or tow loads

Greater power allows mechanical energy to be transferred more quickly.

However, the relationship between speed and power is not simple because resistive forces can change with speed.


Air Resistance and Speed

Air resistance generally increases as vehicle speed increases.

At higher speeds:

  • air resistance becomes greater
  • more driving force is required
  • more power is required to maintain speed

Because:

P = Fv

increasing both the required force and the speed can cause the required power to rise rapidly.

This is one reason high-speed travel can require very large power outputs.


Worked Example: Comparing Vehicle Speeds

At 10 m/s, suppose a vehicle experiences a total resistance of:

500 N

Required power:

P = 500 × 10

P = 5000 W

At 20 m/s, suppose resistance has increased to:

900 N

Required power:

P = 900 × 20

P = 18,000 W

The speed doubled, but the required power increased by much more than a factor of two because the resistive force also increased.


Power and Acceleration

When a vehicle accelerates, some of the engine's mechanical power increases the vehicle's kinetic energy.

Recall:

Eₖ = ½mv²

Average mechanical power used to increase kinetic energy can be calculated using:

P = ΔEₖ/t

If resistive forces are also present, the engine must provide additional power to overcome them.

So in a real accelerating vehicle:

engine power = power increasing KE + power overcoming resistance


Worked Example: Accelerating Vehicle

A 1200 kg car accelerates from rest to 20 m/s in 8 s.

Ignore resistance.

Calculate the average mechanical power transferred into kinetic energy.

Final KE:

Eₖ = ½mv²

Eₖ = 0.5 × 1200 × 20²

Eₖ = 240,000 J

Initial KE:

0 J

Therefore:

ΔEₖ = 240,000 J

Average power:

P = ΔE/t

P = 240,000/8

P = 30,000 W

or:

P = 30 kW


Adding Resistive Forces

Suppose the previous car also experiences an average resistive force of 500 N while traveling an average distance of 80 m during the acceleration.

Work done against resistance:

W = Fd

W = 500 × 80

W = 40,000 J

Total mechanical energy supplied:

240,000 + 40,000 = 280,000 J

Average mechanical power:

P = 280,000/8

P = 35,000 W

or:

35 kW

This is more realistic because some engine energy must overcome resistance rather than increase the car's kinetic energy.


Mechanical Power in Lifting

Machines such as cranes and elevators perform mechanical work against gravity.

The work done in lifting an object is equal to its increase in gravitational potential energy:

W = mgΔh

Therefore:

P = mgΔh/t

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6

Worked Example: Crane

A crane lifts a 2000 kg load vertically through 12 m in 30 s.

Use:

g = 9.8 N/kg

Work done:

W = mgΔh

W = 2000 × 9.8 × 12

W = 235,200 J

Power:

P = W/t

P = 235,200/30

P = 7840 W

or:

P = 7.84 kW


Solving the Same Crane Problem Using P = Fv

The lifting force at constant speed is:

F = mg

F = 2000 × 9.8

F = 19,600 N

The vertical speed is:

v = d/t

v = 12/30

v = 0.40 m/s

Now:

P = Fv

P = 19,600 × 0.40

P = 7840 W

We obtain exactly the same answer.

This demonstrates the connection between:

P = W/t

and:

P = Fv


Worked Example: Elevator

An elevator has a total mass of 1500 kg and rises at a constant speed of 2 m/s.

Ignore friction.

At constant speed, the upward motor force equals the elevator's weight:

F = mg

F = 1500 × 9.8

F = 14,700 N

Power:

P = Fv

P = 14,700 × 2

P = 29,400 W

or:

P = 29.4 kW


Mechanical Power in Cycling

Cyclists produce mechanical power by applying forces to pedals.

Energy is transferred from chemical energy in the cyclist's body into mechanical energy.

Some mechanical power is used to:

  • increase kinetic energy
  • climb hills
  • overcome air resistance
  • overcome rolling resistance
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6

Worked Example: Cyclist

A cyclist travels at a constant:

8 m/s

against a total resistive force of:

35 N

Mechanical power required:

P = Fv

P = 35 × 8

P = 280 W

The cyclist must produce approximately 280 W of useful mechanical power to maintain this speed.


Cyclist Climbing a Hill

When climbing, the cyclist must also work against gravity.

Suppose the combined mass of cyclist and bicycle is:

80 kg

and the cyclist gains vertical height at:

0.50 m/s

The force associated with the weight is:

F = mg

F = 80 × 9.8

F = 784 N

Power required to gain GPE:

P = Fv

P = 784 × 0.50

P = 392 W

This is the mechanical power required just for the increase in gravitational potential energy.

Additional power would be required to overcome air resistance and rolling resistance.


Machines and Mechanical Advantage

Machines can trade force for speed or distance.

For an ideal machine:

input power = output power

Therefore:

Finput × vinput = Foutput × voutput

If a machine increases force, the output generally moves more slowly.

For example, an ideal machine might produce:

twice the force

but move the load at:

half the speed

The power remains the same.

Real machines also dissipate energy, so:

useful output power < input power


Worked Example: Ideal Machine

A machine receives an input force of:

200 N

moving at:

4 m/s

Input power:

P = Fv

P = 200 × 4

P = 800 W

The machine produces an output force of:

800 N

For an ideal machine:

800 = 800 × v

Therefore:

v = 1 m/s

The machine multiplies the force by four but reduces the speed by four.


Mechanical Power and Efficiency

Real machines are not perfectly efficient.

Efficiency can be calculated using:

efficiency = useful output power / input power

or:

efficiency (%) = useful output power / input power × 100


Worked Example: Motor Efficiency

An electric motor receives:

5000 W

of electrical power.

It provides:

4000 W

of mechanical power.

Efficiency:

efficiency = 4000/5000 × 100

efficiency = 80%

The remaining:

1000 W

is transferred into other forms, such as thermal energy and sound.


Worked Example: Finding Input Power

A crane must provide:

24 kW

of useful mechanical power.

Its motor system is:

80% efficient

Use:

efficiency = useful output power / input power

0.80 = 24/input power

Therefore:

input power = 24/0.80

input power = 30 kW

The system requires 30 kW of input power to provide 24 kW of useful mechanical power.


Transportation and Power

Power strongly affects transportation systems.

A vehicle needs power to:

  • overcome resistive forces
  • accelerate
  • climb slopes
  • tow loads
  • maintain high speeds

For constant-speed motion:

P = Fv

This means that high-speed transportation can require enormous power even when the vehicle is not accelerating.

Examples include:

  • cars
  • trucks
  • trains
  • aircraft
  • ships

Heavy Vehicles

A truck moving at the same speed as a smaller vehicle may require more power because it can experience greater:

  • rolling resistance
  • mechanical resistance
  • gravitational forces when climbing
  • drag, depending on its shape and size

When climbing a slope, a heavy vehicle may require particularly large power because it is increasing its gravitational potential energy.


Worked Example: Truck

A truck moves at a constant 25 m/s against a total resistive force of 6000 N.

Calculate the mechanical power required.

P = Fv

P = 6000 × 25

P = 150,000 W

or:

P = 150 kW


Worked Example: Train

A train moves at a constant speed of:

30 m/s

The locomotive provides a driving force of:

100,000 N

Calculate its mechanical power.

P = Fv

P = 100,000 × 30

P = 3,000,000 W

or:

P = 3 MW

Large transportation systems can therefore require power outputs measured in megawatts.


Force at Different Speeds

Suppose an engine can deliver a constant mechanical power of:

60 kW

At 5 m/s:

F = 60,000/5 = 12,000 N

At 10 m/s:

F = 60,000/10 = 6000 N

At 20 m/s:

F = 60,000/20 = 3000 N

At 30 m/s:

F = 60,000/30 = 2000 N

This illustrates the inverse relationship between force and speed at constant power.


Why Vehicles Use Gears

Gears help vehicles and machines trade force for speed.

At low speed, a vehicle may need a large driving force to:

  • start moving
  • accelerate
  • climb steep slopes
  • move heavy loads

At higher speeds, less driving force may be available for a given engine power.

Gear systems allow the motor or engine to operate effectively while changing the relationship between:

force and speed

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6

Power, Force, and Velocity Graphs

The equation:

P = Fv

allows several graphical relationships to be predicted.

Constant Force

If force remains constant:

P ∝ v

A graph of power against velocity is a straight line through the origin.

Constant Velocity

If velocity remains constant:

P ∝ F

A graph of power against force is a straight line through the origin.

Constant Power

If power remains constant:

F = P/v

Force decreases as velocity increases.

The force-velocity relationship is therefore an inverse relationship.


Direction Matters

The simple equation:

P = Fv

assumes the force and velocity are in the same direction.

More generally, mechanical power depends on the component of force in the direction of motion.

If force and motion are perpendicular, the force does no mechanical work on the object at that instant.

For the problems in this topic, unless stated otherwise, assume the force acts in the direction of motion.


Common Mistakes

Confusing Work and Power

Work measures energy transferred.

Power measures how quickly that energy is transferred.

W = Fd

but:

P = W/t


Using P = Fv During the Wrong Situation

The simple form:

P = Fv

is used when the force acts in the direction of motion.

For many introductory problems, it is especially useful for constant-speed motion.


Thinking Constant Velocity Means No Forces

Constant velocity means:

resultant force = 0

It does not mean:

all forces = 0

A vehicle can have a large driving force balanced by an equally large resistive force.


Thinking Constant Velocity Means Zero Power

A car moving at constant velocity against resistance still requires power.

The engine continuously transfers energy to overcome resistive forces.


Forgetting Unit Conversions

Remember:

1 kW = 1000 W

1 MW = 1,000,000 W


Using Distance Instead of Velocity

In:

P = Fv

v must be measured in:

m/s

not metres.


Forgetting That Power Depends on Both Force and Speed

A large force does not necessarily mean large power.

If the object moves very slowly, the power may still be small.


Assuming More Power Means More Force

Not necessarily.

From:

F = P/v

the force produced by a given power depends on speed.


Ignoring Energy Losses

Real machines dissipate energy through:

  • friction
  • thermal energy
  • sound
  • deformation

Therefore:

useful mechanical output power < input power


Check Your Understanding

  1. Define mechanical power.
  2. State the equation relating power, work, and time.
  3. State the equation relating power, force, and velocity.
  4. Explain how P = Fv can be derived from P = W/t.
  5. A machine performs 20,000 J of work in 10 s. Calculate its power.
  6. A motor exerts a force of 600 N while moving an object at 5 m/s. Calculate the mechanical power.
  7. A vehicle experiences a resistance of 1500 N while traveling at a constant 20 m/s. Calculate the required mechanical power.
  8. Explain why the vehicle in Question 7 needs power even though it is not accelerating.
  9. A 25 kW motor moves a load at 5 m/s. Calculate the force.
  10. A machine provides 40 kW while exerting a force of 8000 N. Calculate its speed.
  11. A crane lifts a 1500 kg load through 10 m in 20 s. Calculate the work done.
  12. Calculate the mechanical power of the crane in Question 11.
  13. Solve Question 12 again using P = Fv.
  14. A cyclist travels at 10 m/s against a resistance of 30 N. Calculate the mechanical power required.
  15. Explain why a cyclist generally needs more power to travel faster.
  16. A truck travels at 20 m/s against a resistance of 5000 N. Calculate its mechanical power.
  17. Convert your answer from Question 16 into kilowatts.
  18. A train produces a driving force of 80,000 N while traveling at 25 m/s. Calculate its mechanical power in MW.
  19. A motor provides a constant 50 kW. Calculate the available force at 5 m/s.
  20. Calculate the available force from the same motor at 10 m/s.
  21. What happens to available force when speed doubles at constant power?
  22. Explain why gears are useful in vehicles.
  23. A motor receives 10 kW of electrical power and produces 8 kW of useful mechanical power. Calculate its efficiency.
  24. A machine must provide 12 kW of useful mechanical power and is 75% efficient. Calculate its required input power.
  25. Explain how force, velocity, and power are related in a transportation system.
  26. Why does maintaining a high vehicle speed usually require substantial power?
  27. A 1000 kg car accelerates from rest to 20 m/s in 5 s. Ignoring resistance, calculate its average mechanical power.
  28. If the car in Question 27 also transfers 50,000 J to the surroundings during the acceleration, calculate the average engine power required.
  29. Explain the difference between input power and useful mechanical output power.
  30. Give one practical example where high force is more important than high speed and explain how power relates to the situation.

Key Terms

Mechanical power — Rate at which mechanical work is done or mechanical energy is transferred.

Power — Rate of energy transfer.

Work done — Energy transferred when a force causes displacement.

Watt (W) — SI unit of power, equal to one joule per second.

Force — Push or pull capable of changing an object's motion.

Displacement — Change in position in a specified direction.

Velocity — Rate of change of displacement, including direction.

Driving force — Force that propels a vehicle or machine.

Resistive force — Force opposing motion, such as friction or air resistance.

Constant velocity — Motion at constant speed and direction, meaning resultant force is zero.

Useful mechanical power — Rate at which energy is transferred into the intended mechanical output.

Input power — Rate at which energy is supplied to a machine.

Efficiency — Fraction of input power converted into useful output power.

Mechanical advantage — Increase in output force produced by a machine compared with its input force.

Air resistance — Resistive force caused by motion through air.

Rolling resistance — Resistive force associated with rolling objects such as tyres or wheels.

Gear — Mechanical device used to alter the relationship between force and speed.


Key Takeaways

  • Mechanical power describes how quickly mechanical work is done.
  • Mechanical power can be calculated using P = W/t.
  • When a constant force acts in the direction of motion, W = Fd.
  • Combining P = W/t, W = Fd, and v = d/t gives P = Fv.
  • Power is measured in watts.
  • 1 W = 1 J/s.
  • Greater force requires greater power when speed remains constant.
  • Greater speed requires greater power when force remains constant.
  • At constant power, force decreases as speed increases.
  • An object moving at constant velocity can still require substantial mechanical power.
  • Constant velocity means the driving and resistive forces balance; it does not mean that no forces act.
  • Vehicles need continuous power at constant speed because energy is continually transferred against resistive forces.
  • Air resistance generally increases with speed, so high-speed transportation can require very large power outputs.
  • Mechanical power can be calculated for cranes, elevators, vehicles, bicycles, trains, and industrial machines.
  • Lifting power can be calculated using P = mgΔh/t.
  • Power associated with acceleration can be found from the rate of increase of kinetic energy.
  • Real machines have lower useful output power than input power because some energy is dissipated.
  • Gears allow machines to trade force for speed.
  • In an ideal machine, increasing output force generally decreases output speed while power remains constant.
  • The equations P = W/t and P = Fv describe the same fundamental idea: the rate at which mechanical energy is transferred.

The central relationships are:

P = W/t

and:

P = Fv

So for machines and vehicles:

more force at the same speed → more power

more speed at the same force → more power

and, for a fixed power output:

more speed → less available force.

 
 
 

3. Electrical Power

Learning outcomes
  • I can describe electrical power as the rate of electrical energy transfer.
  • I can recall and use the equation P = IV
  • I can calculate electrical power in simple circuits.
  • I can determine energy consumption from power ratings and operating times.
  • I can interpret power ratings on electrical devices and explain their significance.

Electrical Power

Electrical power is the rate at which electrical energy is transferred by a device or circuit.

Every electrical device transfers energy. For example:

  • a lamp transfers electrical energy into light and thermal energy
  • a heater transfers electrical energy into thermal energy
  • a motor transfers electrical energy into mechanical energy
  • a speaker transfers electrical energy into sound
  • a charger transfers electrical energy into energy stored in a battery

The power rating tells us how quickly this energy is transferred.

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6

Electrical Power

Power is the rate of energy transfer:

P = E/t

where:

  • P = power in watts (W)
  • E = energy transferred in joules (J)
  • t = time in seconds (s)

For electrical circuits, power can also be calculated using:

P = IV

where:

  • P = electrical power (W)
  • I = current (A)
  • V = potential difference or voltage (V)

This is one of the most important equations for electrical circuits.


The Watt

Electrical power is measured in watts (W).

1 W = 1 J/s

Therefore, a 60 W device transfers:

60 J of energy every second

A 2000 W heater transfers:

2000 J every second

A higher power rating means energy is transferred at a greater rate.


Current, Voltage, and Power

The equation:

P = IV

shows that electrical power depends on both:

  • current
  • voltage

If voltage remains constant:

greater current → greater power

If current remains constant:

greater voltage → greater power

For example:

Current = 2 A
Voltage = 12 V

P = IV

P = 2 × 12

P = 24 W

The device transfers electrical energy at a rate of 24 J/s.


Worked Example: Simple Circuit

A lamp operates at:

V = 12 V

and draws:

I = 3 A

Calculate its power.

P = IV

P = 3 × 12

P = 36 W

The lamp transfers 36 J of energy each second.


Worked Example: Electric Motor

A motor operates at:

24 V

and draws:

5 A

Calculate the electrical power supplied to the motor.

P = IV

P = 5 × 24

P = 120 W

The motor receives electrical energy at a rate of:

120 J/s

Not all of this necessarily becomes useful mechanical power because the motor is not perfectly efficient.


Rearranging P = IV

Starting with:

P = IV

we can rearrange the equation.

To calculate current:

I = P/V

To calculate voltage:

V = P/I

So we have:

P = IV

I = P/V

V = P/I


Worked Example: Finding Current

An electrical heater has a power rating of:

1200 W

and operates at:

240 V

Calculate the current.

I = P/V

I = 1200/240

I = 5 A


Worked Example: Finding Voltage

A device operates at:

60 W

and draws:

2.5 A

Calculate the voltage.

V = P/I

V = 60/2.5

V = 24 V


Why P = IV?

Recall that voltage represents energy transferred per unit charge:

V = E/Q

Current represents charge flowing per unit time:

I = Q/t

Power is:

P = E/t

Since:

E = VQ

then:

P = VQ/t

But:

Q/t = I

Therefore:

P = IV

This connects electrical power directly to the movement of charge through a potential difference.


Power Ratings

Most electrical devices have a power rating.

The rating tells us the rate at which the device transfers energy when operating under its rated conditions.

For example:

Device Example Power
LED lamp 10 W
Laptop 60 W
Television 100 W
Refrigerator 150 W
Microwave oven 1200 W
Hair dryer 1500 W
Electric kettle 2000 W

These are illustrative values; actual devices vary considerably.

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7

What Does a 2000 W Rating Mean?

Suppose a kettle is rated:

2000 W

This means that while operating at its rated power, it transfers energy at:

2000 J/s

In 1 second:

2000 J

In 10 seconds:

20,000 J

In 60 seconds:

120,000 J

Therefore, power tells us how quickly a device transfers energy.


Electrical Energy Consumption

Because:

P = E/t

we can rearrange:

E = Pt

This allows us to calculate the energy transferred by an electrical device.

where:

  • E = energy (J)
  • P = power (W)
  • t = time (s)

Worked Example: Electric Heater

A 1500 W heater operates for 5 minutes.

First convert time:

5 min = 300 s

Then:

E = Pt

E = 1500 × 300

E = 450,000 J

or:

E = 450 kJ


Worked Example: Lamp

A 20 W lamp operates for 3 hours.

Convert time:

3 h = 10,800 s

Then:

E = Pt

E = 20 × 10,800

E = 216,000 J

or:

E = 216 kJ


Energy Depends on Power and Time

A device with a high power rating transfers energy quickly.

However, total energy consumption depends on both power and operating time.

Consider two devices.

Device A

Power = 2000 W
Time = 60 s

E = 2000 × 60

E = 120,000 J

Device B

Power = 100 W
Time = 3600 s

E = 100 × 3600

E = 360,000 J

Even though Device B has much lower power, it transfers more total energy because it operates for much longer.

Therefore:

high power does not automatically mean greater total energy consumption

Operating time matters.


Kilowatts

Household electrical devices are often rated in watts or kilowatts.

1 kW = 1000 W

Examples:

500 W = 0.5 kW

1500 W = 1.5 kW

2000 W = 2 kW

3500 W = 3.5 kW


Kilowatt-Hours

Electricity use in homes is commonly measured in kilowatt-hours (kWh) rather than joules.

A kilowatt-hour is a unit of energy, not power.

1 kWh is the energy transferred by a 1 kW device operating for 1 hour.

Therefore:

energy (kWh) = power (kW) × time cœur


Understanding 1 kWh

Suppose a 1 kW heater operates for 1 hour.

Energy:

E = Pt

E = 1 kW × 1 h

E = 1 kWh

The same energy could be transferred by:

  • a 2 kW device for 0.5 h
  • a 0.5 kW device for 2 h
  • a 0.25 kW device for 4 h

Each transfers:

1 kWh


Converting kWh to Joules

We know:

1 kW = 1000 W

and:

1 h = 3600 s

Therefore:

1 kWh = 1000 × 3600 J

1 kWh = 3,600,000 J

or:

1 kWh = 3.6 MJ

This is a useful conversion.


Worked Example: Kettle in kWh

A 2 kW kettle operates for 6 minutes.

Convert time to hours:

6 min = 0.1 h

Then:

E = Pt

E = 2 × 0.1

E = 0.2 kWh

In joules:

0.2 × 3.6 MJ = 0.72 MJ

or:

720,000 J


Worked Example: Air Conditioner

An air conditioner operates at an average electrical power of 1.5 kW for 8 hours.

Calculate its energy consumption.

E = Pt

E = 1.5 × 8

E = 12 kWh

The air conditioner uses 12 kWh of electrical energy over that period.


Electrical Energy Cost

If electricity costs a certain amount per kWh:

cost = energy used × price per kWh

For example, suppose a device uses:

5 kWh

and electricity costs:

$0.30 per kWh

Then:

cost = 5 × 0.30

cost = $1.50

This is why both power rating and operating time affect electricity bills.


Comparing Electrical Devices

Suppose two heaters operate for the same time.

Heater A:

1000 W

Heater B:

2000 W

Heater B transfers energy twice as quickly.

If both operate for 10 minutes:

Heater A:

E = 1000 × 600

E = 600,000 J

Heater B:

E = 2000 × 600

E = 1,200,000 J

The 2000 W heater transfers twice as much energy during the same period.


Higher Power Can Mean Shorter Operating Time

However, a higher-power appliance may complete a task more quickly.

Suppose two ideal heaters must transfer:

600,000 J

Heater A:

P = 1000 W

t = E/P

t = 600,000/1000

t = 600 s

Heater B:

P = 2000 W

t = 600,000/2000

t = 300 s

The more powerful heater completes the energy transfer in half the time.

This demonstrates why power and total energy consumption should not be confused.


Electrical Power in Heating Devices

Devices designed to produce heat often have relatively high power ratings.

Examples include:

  • kettles
  • ovens
  • heaters
  • toasters
  • hair dryers
  • irons

They need to transfer large amounts of thermal energy in relatively short periods.

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4

Electrical Power in Motors

Electric motors transfer:

electrical energy → mechanical energy

Examples include motors in:

  • fans
  • elevators
  • electric vehicles
  • washing machines
  • pumps
  • power tools

A motor's electrical input power can be calculated using:

Pinput = IV

Its useful mechanical output power may be calculated using:

Poutput = W/t

or:

Poutput = Fv

The output is normally less than the input because real motors are not perfectly efficient.


Worked Example: Electric Motor

A motor operates at:

120 V

and draws:

10 A

Electrical input power:

P = IV

P = 10 × 120

P = 1200 W

Suppose its useful mechanical output power is:

960 W

Efficiency:

efficiency = useful output power / input power × 100

efficiency = 960/1200 × 100

efficiency = 80%

The remaining power is transferred mainly into thermal energy and sound.


Power in Simple Circuits

Suppose a resistor has:

V = 6 V

across it and:

I = 0.5 A

through it.

Power:

P = IV

P = 0.5 × 6

P = 3 W

The resistor transfers electrical energy at a rate of:

3 J/s

Much of this energy becomes thermal energy.


Electrical Power and Resistance

Using Ohm's law:

V = IR

we can combine this with:

P = IV

Substituting V = IR:

P = I × IR

Therefore:

P = I²R

Alternatively:

I = V/R

Substitute into P = IV:

P = V × V/R

Therefore:

P = V²/R

So three useful electrical power equations are:

P = IV

P = I²R

P = V²/R

For this topic, P = IV is the central equation, but the other forms can be useful when resistance is known.


Worked Example: Power and Resistance

A 10 Ω resistor carries a current of:

2 A

Use:

P = I²R

P = 2² × 10

P = 40 W

The resistor transfers electrical energy at a rate of 40 J/s.


Worked Example: Voltage and Resistance

A 12 Ω resistor is connected across:

24 V

Use:

P = V²/R

P = 24²/12

P = 576/12

P = 48 W


Power and Current in Household Devices

At a fixed supply voltage:

I = P/V

This means higher-power appliances generally draw larger currents.

For example, compare two devices operating at 230 V.

230 W Device

I = 230/230

I = 1 A

2300 W Device

I = 2300/230

I = 10 A

The higher-power device requires a much greater current.

This is one reason electrical wiring and protective devices must be appropriate for the current a circuit carries.


Power Ratings and Safe Operation

Electrical devices are designed to operate under particular electrical conditions.

A rating label may include:

  • voltage
  • current
  • power
  • frequency
  • energy consumption information

For example, a label might indicate:

230 V

1500 W

This tells us that when operating as designed at the rated voltage, the appliance uses electrical power at approximately 1500 W.

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6

Why Device Ratings Matter

Power ratings help users and engineers determine:

  • how quickly a device transfers energy
  • how much electrical energy it may consume
  • how much current it may draw
  • appropriate electrical connections
  • likely operating costs
  • energy-efficiency considerations

A power rating therefore provides important information about both performance and energy use.


Power and Efficiency

Electrical devices rarely convert all input energy into the desired output.

For example, an electric motor might receive:

1000 W electrical power

but provide:

850 W mechanical power

The remaining:

150 W

may be transferred into:

  • thermal energy
  • sound
  • vibration

Efficiency:

efficiency = 850/1000 × 100

efficiency = 85%


Energy Flow Example

Consider a lamp receiving:

10 W electrical power

Suppose:

8 W → light

and:

2 W → thermal energy

Then:

10 W input = 8 W useful + 2 W other transfers

Energy is conserved.

Over 10 seconds:

Input energy:

E = Pt

E = 10 × 10

E = 100 J

Useful energy:

8 × 10 = 80 J

Thermal energy:

2 × 10 = 20 J

Therefore:

100 J = 80 J + 20 J


Choosing the Correct Equation

Use:

P = IV

when you know current and voltage.

Use:

P = E/t

when you know energy and time.

Use:

E = Pt

when you know power and operating time.

Use:

E(kWh) = P(kW) × tcœur

when calculating household electricity consumption.

Use:

P = I²R

when current and resistance are known.

Use:

P = V²/R

when voltage and resistance are known.


Multi-Step Example

A heater operates at:

230 V

and draws:

8 A

It operates for:

15 minutes

Calculate Power

P = IV

P = 8 × 230

P = 1840 W

Convert Time

15 min = 900 s

Calculate Energy

E = Pt

E = 1840 × 900

E = 1,656,000 J

or:

E = 1.656 MJ

Alternatively:

P = 1.84 kW

t = 0.25 h

Therefore:

E = 1.84 × 0.25

E = 0.46 kWh

Both calculations describe the same energy transfer.


Common Mistakes

Confusing Power and Energy

Power is measured in:

watts

Energy is measured in:

joules or kilowatt-hours.


Thinking kWh Is a Unit of Power

It is not.

kW = power

kWh = energy


Forgetting Time Conversion

For joule calculations:

1 min = 60 s

1 h = 3600 s

For kWh calculations, time should normally be in hours.


Forgetting to Convert Watts to Kilowatts

For kWh:

1000 W = 1 kW

Therefore:

2500 W = 2.5 kW


Mixing Units

Do not calculate:

2000 W × 3 h

and call the answer joules.

Use compatible units:

W × s → J

or:

kW × h → kWh


Assuming High Power Always Means High Energy Consumption

A high-power device operating briefly may use less energy than a low-power device operating for many hours.


Confusing Current and Voltage

Current measures the rate of charge flow.

Voltage measures energy transferred per unit charge.

Power depends on both:

P = IV


Ignoring Efficiency

Electrical input power is not always equal to useful output power.

Real devices transfer some energy into other forms.


Check Your Understanding

  1. Define electrical power.
  2. State the SI unit of electrical power.
  3. What does a power of 1 W mean?
  4. State the equation relating power, current, and voltage.
  5. A device operates at 12 V and draws 4 A. Calculate its power.
  6. A motor operates at 24 V and draws 8 A. Calculate its electrical input power.
  7. A 1500 W heater operates at 250 V. Calculate the current.
  8. A 60 W device draws 2 A. Calculate its operating voltage.
  9. Explain what a 2000 W power rating means.
  10. A 500 W device operates for 20 s. Calculate the energy transferred.
  11. A 2 kW heater operates for 5 minutes. Calculate the energy transferred in joules.
  12. Calculate the answer to Question 11 in kWh.
  13. What is the difference between kW and kWh?
  14. Convert 1 kWh into joules.
  15. A 1.5 kW air conditioner operates for 6 h. Calculate its energy consumption in kWh.
  16. A 100 W device operates for 10 h. How much energy does it consume in kWh?
  17. Explain why a lower-power device can sometimes consume more energy than a higher-power device.
  18. A 10 Ω resistor carries 3 A. Calculate its power.
  19. A 20 Ω resistor has 40 V across it. Calculate its power.
  20. A motor receives 2000 W of electrical power and produces 1600 W of useful mechanical power. Calculate its efficiency.
  21. How much power is transferred into other forms in Question 20?
  22. A heater operates at 240 V and 10 A for 30 minutes. Calculate its power.
  23. Calculate the energy consumed by the heater in Question 22 in kWh.
  24. If electricity costs $0.30 per kWh, calculate the operating cost for Question 23.
  25. Explain why power ratings are useful when comparing electrical devices.

Key Terms

Electrical power — Rate at which electrical energy is transferred.

Power rating — Power a device is designed to use or transfer under specified operating conditions.

Watt (W) — SI unit of power; one joule per second.

Kilowatt (kW) — Unit of power equal to 1000 W.

Energy — Quantity that can be stored and transferred.

Joule (J) — SI unit of energy.

Kilowatt-hour (kWh) — Energy transferred by 1 kW of power operating for 1 hour.

Current (I) — Rate of flow of electric charge, measured in amperes.

Potential difference (V) — Energy transferred per unit charge, measured in volts.

Resistance (R) — Opposition to current in a circuit, measured in ohms.

Electrical energy — Energy transferred through electrical processes.

Input power — Rate at which energy enters a device.

Useful output power — Rate at which energy is transferred into the intended output.

Efficiency — Fraction of input power or energy transferred usefully.

Power consumption — Rate at which a device receives or transfers electrical energy.

Energy consumption — Total electrical energy transferred during a period of operation.


Key Takeaways

  • Electrical power is the rate of electrical energy transfer.
  • Electrical power is measured in watts.
  • 1 W = 1 J/s.
  • The main electrical power equation is P = IV.
  • Power increases if current increases while voltage remains constant.
  • Power increases if voltage increases while current remains constant.
  • The equation can be rearranged to I = P/V and V = P/I.
  • A device's power rating tells us how quickly it transfers energy under its rated conditions.
  • Energy consumption depends on both power and operating time.
  • Energy can be calculated using E = Pt.
  • For household electricity calculations, E(kWh) = P(kW) × tcœur.
  • kW is a unit of power, while kWh is a unit of energy.
  • 1 kWh = 3.6 MJ.
  • High-power appliances transfer energy rapidly, but they do not necessarily consume more total energy if they operate for shorter periods.
  • Combining P = IV with Ohm's law gives P = I²R and P = V²/R.
  • Higher-power devices generally draw greater current when operated at the same voltage.
  • Electrical input power can be compared with useful output power to determine efficiency.
  • Real electrical devices transfer some energy into unwanted forms, commonly thermal energy.
  • Power ratings are useful for comparing device performance, estimating energy consumption, and understanding electrical requirements.

The central relationship is:

P = IV

and for energy consumption:

E = Pt

So remember:

power tells us how fast energy is transferred

while:

energy tells us how much has been transferred in total.

4. Efficiency

Learning outcomes
  • I can explain why no real system is 100% efficient.
  • I can distinguish between useful and wasted energy transfers.
  • I can calculate efficiency using energy or power values.
  • I can analyze the causes of energy dissipation in mechanical and electrical systems.
  • I can compare the efficiencies of different devices and processes.

Efficiency

Efficiency describes how much of the energy transferred into a system becomes a useful energy output.

Whenever a machine or device operates, energy is transferred. However, not all of the input energy is transferred in the form we actually want.

For example:

  • a motor transfers electrical energy into useful mechanical energy, but also becomes warm
  • a lamp transfers electrical energy into light, but also transfers thermal energy to its surroundings
  • a car engine transfers chemical energy into mechanical energy, but large amounts are transferred as thermal energy and sound
  • a generator transfers mechanical energy into electrical energy, but some energy is dissipated through friction and heating

Real systems are therefore less than 100% efficient.

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5

Useful and Wasted Energy Transfers

When analyzing efficiency, energy transfers can be divided into two broad categories.

Useful Energy

Useful energy is energy transferred in the form required for the intended purpose of the device.

Examples:

  • mechanical energy from a motor
  • light from a lamp
  • thermal energy from a heater
  • electrical energy from a generator
  • kinetic energy from a vehicle engine

Wasted Energy

Wasted energy is energy transferred in forms that are not useful for the intended purpose.

Common examples include:

  • unwanted thermal energy
  • sound
  • vibration
  • deformation

"Wasted" does not mean that the energy has disappeared.

Energy is always conserved.

A better physical description is that the energy has been dissipated or transferred into less useful stores.


Energy Is Conserved

Consider a motor receiving:

1000 J

of electrical energy.

Suppose it produces:

800 J useful mechanical energy

and transfers:

200 J into thermal energy and sound

The energy balance is:

1000 J input = 800 J useful + 200 J wasted

Nothing has disappeared.

The total energy remains:

1000 J

Energy conservation still applies even though the motor is not 100% efficient.


Calculating Efficiency

Efficiency compares the useful output with the total input.

Efficiency can be written as a decimal or percentage.

For example:

efficiency = 0.80

is the same as:

efficiency = 80%


Worked Example: Electric Motor

A motor receives:

500 J

of electrical energy.

It produces:

400 J

of useful mechanical energy.

Efficiency:

efficiency = 400/500

efficiency = 0.80

As a percentage:

0.80 × 100 = 80%

The motor is 80% efficient.


Finding Wasted Energy

Because energy is conserved:

input energy = useful energy + wasted energy

Therefore:

wasted energy = input energy − useful energy

For the previous motor:

wasted energy = 500 − 400

wasted energy = 100 J

So:

500 J input → 400 J useful + 100 J wasted


Worked Example: Lamp

A lamp receives:

200 J

of electrical energy.

It transfers:

60 J

as useful light.

Efficiency:

efficiency = 60/200 × 100

efficiency = 30%

Wasted energy:

200 − 60 = 140 J

Therefore:

200 J electrical → 60 J useful light + 140 J thermal energy

The lamp is 30% efficient at producing the desired light output.


Efficiency Using Power

Efficiency can also be calculated using power.

When input and output are measured over the same time interval:

efficiency = useful output power / total input power

As a percentage:

efficiency (%) = useful output power / total input power × 100

This is especially useful for machines operating continuously.


Worked Example: Motor Power

An electric motor receives:

2500 W

of electrical power.

It provides:

2000 W

of useful mechanical power.

Efficiency:

efficiency = 2000/2500 × 100

efficiency = 80%

Power dissipated:

2500 − 2000 = 500 W

This means energy is being dissipated at a rate of:

500 J/s


Why Energy and Power Give the Same Efficiency

Suppose the motor operates for 10 s.

Input energy:

E = Pt

E = 2500 × 10

E = 25,000 J

Useful output energy:

E = 2000 × 10

E = 20,000 J

Efficiency:

efficiency = 20,000/25,000

efficiency = 0.80

or:

80%

The same result is obtained using either energy or power.


Rearranging the Efficiency Equation

If efficiency and input energy are known:

useful output = efficiency × input

Remember to convert a percentage to a decimal.

For example:

75% = 0.75

Therefore:

useful output = 0.75 × input

If useful output and efficiency are known:

input = useful output / efficiency


Worked Example: Finding Useful Energy

A machine receives:

2400 J

and is:

75% efficient

Convert:

75% = 0.75

Useful energy:

useful output = 0.75 × 2400

useful output = 1800 J

Wasted energy:

2400 − 1800 = 600 J


Worked Example: Finding Input Energy

A machine must produce:

3600 J

of useful energy.

It is:

80% efficient

Convert:

80% = 0.80

Input energy:

input = 3600/0.80

input = 4500 J

Therefore:

900 J

is transferred into unwanted forms.


Worked Example: Finding Useful Power

A motor receives:

10 kW

of electrical power.

It is:

85% efficient

Useful mechanical power:

Puseful = 0.85 × 10

Puseful = 8.5 kW

Wasted power:

10 − 8.5 = 1.5 kW


Why Real Systems Are Not 100% Efficient

In real systems, interactions between components cause some energy to be transferred into unwanted forms.

Common causes include:

  • friction
  • electrical resistance
  • air resistance
  • sound
  • vibration
  • deformation
  • turbulence
  • heating of components

These processes transfer energy into the surroundings.

Because these transfers cannot normally be eliminated completely:

real systems are not 100% efficient


Friction

Friction is one of the most common causes of energy dissipation in mechanical systems.

When surfaces move against one another, mechanical energy is transferred into thermal energy.

Examples include:

  • gears
  • bearings
  • tyres
  • engine components
  • conveyor systems
  • bicycle chains

You may notice machinery becoming warm during operation because of these transfers.

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6

Reducing Friction

Engineers can reduce friction using:

  • lubricants
  • ball bearings
  • smoother surfaces
  • appropriate materials
  • improved alignment

Reducing friction means less mechanical energy is transferred into unwanted thermal energy.

This can increase efficiency.


Electrical Resistance

Electrical devices also dissipate energy because conductors have resistance.

When current passes through resistance:

electrical energy → thermal energy

This is useful in devices designed for heating.

Examples include:

  • kettles
  • heaters
  • toasters

But it is usually unwanted in:

  • motors
  • power cables
  • computers
  • transformers
  • generators

Electrical components becoming warm is evidence that some energy is being transferred thermally.


Useful or Wasted Depends on Purpose

An important point is that whether an energy transfer is considered useful depends on the purpose of the device.

Consider an electric heater.

Electrical energy becoming thermal energy is:

useful

But in an electric motor, excessive heating is:

wasted

Similarly, light from a lamp is useful, while light from a heating element may not be.

Efficiency must therefore always be considered in relation to the intended output.


Sound and Vibration

Machines often produce sound and vibration.

These transfers usually represent energy that is not contributing to the intended output.

Examples include:

  • noisy engines
  • vibrating motors
  • rattling machinery
  • tyre noise
  • electrical transformers

Reducing unwanted vibration and sound can improve efficiency as well as make machinery quieter and more reliable.


Air Resistance

Moving objects transfer energy to the surrounding air.

This is particularly important for:

  • cars
  • trains
  • aircraft
  • bicycles

Air resistance generally increases as speed increases.

Therefore, high-speed vehicles may transfer large amounts of energy into the surrounding air.

Streamlining can reduce these unwanted energy transfers.

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4

Improving Mechanical Efficiency

Mechanical systems can often be made more efficient by:

  • reducing friction
  • using lubrication
  • using bearings
  • reducing unnecessary mass
  • improving aerodynamics
  • reducing vibration
  • improving gear design
  • using suitable materials

However, completely eliminating unwanted transfers is usually impossible.


Improving Electrical Efficiency

Electrical systems can be improved through:

  • lower-resistance conductors
  • efficient motors
  • improved insulation
  • better electronic control systems
  • reduced standby consumption
  • efficient lighting
  • improved cooling systems
  • appropriate operating voltages

The goal is to increase the fraction of input energy transferred into the intended output.


Comparing Devices

Suppose three motors receive the same input energy:

1000 J

Motor Useful Output Wasted Energy Efficiency
A 600 J 400 J 60%
B 750 J 250 J 75%
C 900 J 100 J 90%

Motor C is the most efficient because the greatest fraction of its input energy becomes useful output.

It also dissipates the least energy.


Comparing Devices with Different Inputs

Efficiency is a ratio, so devices can be compared even when their input energies are different.

Device A

Input = 500 J
Useful = 400 J

efficiency = 400/500 × 100 = 80%

Device B

Input = 2000 J
Useful = 1500 J

efficiency = 1500/2000 × 100 = 75%

Device B produces more useful energy overall, but Device A is more efficient.

This is an important distinction.


Efficiency Does Not Mean Power

A device can be highly efficient but have low power.

Another device can be less efficient but have much greater power.

For example:

Device A:

input power = 100 W

useful power = 90 W

Efficiency:

90%

Device B:

input power = 2000 W

useful power = 1600 W

Efficiency:

80%

Device B provides much more useful power, but Device A is more efficient.


Sankey Diagrams

A Sankey diagram visually represents energy transfers.

The width of each arrow represents the amount of energy being transferred.

For example:

100 J input

might divide into:

70 J useful

and:

30 J wasted

Because:

70 + 30 = 100

the total width of the output arrows equals the width of the input arrow.

Sankey diagrams make it easy to compare the useful and wasted energy transfers in different systems.


Reading a Sankey Diagram

Suppose a motor receives:

500 J

The diagram shows:

350 J useful mechanical energy

and:

150 J thermal and sound energy

Efficiency:

efficiency = 350/500 × 100

efficiency = 70%

The wasted fraction is:

150/500 × 100

= 30%

Therefore:

70% useful + 30% wasted = 100%


Light Bulbs

Different lighting technologies can provide the same useful function while transferring different amounts of energy into unwanted thermal energy.

An efficient lamp transfers a greater proportion of its electrical input into useful light and a smaller proportion into unwanted heating.

This means less electrical energy is required to produce a particular useful light output.

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5

Worked Comparison: Lamps

Suppose two lamps each receive:

100 J

of electrical energy.

Lamp A produces:

15 J useful light

Lamp B produces:

80 J useful light

Lamp A:

efficiency = 15/100 × 100

efficiency = 15%

Lamp B:

efficiency = 80/100 × 100

efficiency = 80%

Lamp B converts a much greater fraction of the electrical input into the desired output.


Electric Motors

Electric motors convert:

electrical energy → mechanical energy

Some energy is also transferred through:

  • heating of wires
  • friction in bearings
  • sound
  • vibration

A more efficient motor provides more mechanical output for the same electrical input.

This can reduce:

  • electricity consumption
  • operating costs
  • unwanted heating
  • environmental impacts associated with energy production

Vehicle Engines

Vehicle engines and motors convert stored energy into mechanical energy.

In combustion engines, significant energy can be transferred into:

  • hot exhaust gases
  • cooling systems
  • friction
  • sound

Only part of the input energy becomes useful mechanical output.

Electric motors generally have different energy-loss pathways, including electrical resistance, magnetic losses, friction, and heating.

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6

Worked Example: Vehicle

An engine receives:

50 MJ

of chemical energy.

It transfers:

18 MJ

into useful mechanical energy.

Efficiency:

efficiency = 18/50 × 100

efficiency = 36%

Energy dissipated:

50 − 18 = 32 MJ

The remaining energy has been transferred mainly into thermal energy and other unwanted outputs.


Power Station Efficiency

Efficiency is also important in electricity generation.

Suppose a power station receives:

1000 MW

of energy per second from its fuel source and produces:

400 MW

of electrical power.

Efficiency:

efficiency = 400/1000 × 100

efficiency = 40%

The remaining:

600 MW

is transferred into other forms, especially thermal energy.


Efficiency Through Several Stages

Many real systems involve several energy conversions.

For example:

fuel → electricity → motor → motion

Suppose:

Power station efficiency = 40%

Motor efficiency = 80%

Start with:

100 J

After the power station:

100 × 0.40 = 40 J

After the motor:

40 × 0.80 = 32 J

Overall useful output:

32 J

Therefore:

overall efficiency = 32%

For several stages, the overall efficiency can be found by multiplying the decimal efficiencies:

overall efficiency = efficiency₁ × efficiency₂ × efficiency₃ ...


Worked Example: Multi-Stage System

A system contains:

Generator efficiency = 90%

Motor efficiency = 80%

Machine efficiency = 75%

Convert to decimals:

0.90 × 0.80 × 0.75

= 0.54

Therefore:

overall efficiency = 54%

Only 54% of the original input reaches the final useful output.

This demonstrates how losses at several stages can significantly reduce overall efficiency.


Efficiency and Sustainability

Improving efficiency can reduce the amount of energy required to perform a task.

This can lead to:

  • lower energy consumption
  • lower operating costs
  • reduced fuel use
  • reduced waste heat
  • lower demand for energy resources
  • reduced environmental impacts

For example, if two motors perform the same useful task but one is more efficient, the more efficient motor requires less input energy.


Worked Example: Energy Savings

Machine A is 60% efficient.

Machine B is 80% efficient.

Both must provide:

2400 J

of useful energy.

Machine A

input = 2400/0.60

input = 4000 J

Machine B

input = 2400/0.80

input = 3000 J

Machine B requires:

1000 J less input energy

to perform the same useful task.


Can Efficiency Be 100%?

In an idealized model, we sometimes assume:

efficiency = 100%

This simplifies calculations.

Real machines, however, always have some unwanted energy transfers.

For example:

  • wires have resistance
  • surfaces experience friction
  • objects interact with air
  • machines vibrate
  • components become warm

Therefore, real energy-transfer systems do not achieve perfect efficiency.


Can Efficiency Be Greater Than 100%?

No.

An efficiency greater than 100% would mean:

useful output energy > input energy

That would require energy to be created.

This violates conservation of energy.

Therefore:

0% ≤ efficiency ≤ 100%

for a correctly defined system.

If a calculation gives more than 100%, check:

  • the data
  • the units
  • which quantity is input
  • which quantity is useful output

Efficiency Versus Energy Conservation

These ideas should not be confused.

Conservation of Energy

Total energy is always conserved.

input energy = total output energy

Efficiency

Efficiency asks what fraction of that output is useful.

For example:

100 J input → 70 J useful + 30 J thermal

Energy conservation:

100 J = 70 J + 30 J

Efficiency:

70%

The system is only 70% efficient, but 100% of the energy still exists.


Practical Example: Electric Elevator

An elevator motor receives:

50 kW

of electrical power.

The elevator receives:

42 kW

of useful mechanical power.

Efficiency:

efficiency = 42/50 × 100

efficiency = 84%

Power dissipated:

50 − 42 = 8 kW

This means:

8000 J

is transferred into other forms every second.


Practical Example: Pump

A water pump receives:

3000 W

of electrical power.

It transfers:

2250 W

usefully to the water.

Efficiency:

efficiency = 2250/3000 × 100

efficiency = 75%

Wasted power:

3000 − 2250 = 750 W


Improving Efficiency

Engineers try to increase efficiency by reducing unwanted energy transfers.

Possible approaches include:

Mechanical Systems

  • lubrication
  • bearings
  • aerodynamic design
  • reducing unnecessary vibration
  • improving alignment
  • reducing friction

Electrical Systems

  • low-resistance conductors
  • efficient motors
  • improved electronic components
  • appropriate cooling
  • improved circuit design

Buildings

  • insulation
  • efficient lighting
  • improved heating and cooling systems
  • reducing unwanted air leakage

Transportation

  • aerodynamic shapes
  • lightweight materials
  • efficient motors
  • regenerative braking
  • reduced rolling resistance

Efficiency improvements often involve engineering trade-offs involving cost, reliability, safety, materials, and performance.


Common Mistakes

Saying Energy Is "Used Up"

Energy is not destroyed.

It is transferred or dissipated.


Saying Wasted Energy Disappears

Wasted energy still exists.

It has been transferred into forms that are less useful for the intended purpose.


Dividing Input by Output

The correct relationship is:

useful output ÷ total input

not the other way around.


Forgetting to Multiply by 100

If the question asks for percentage efficiency:

0.75 = 75%


Using 80 Instead of 0.80

When using an efficiency in another calculation:

80% = 0.80

not 80.


Assuming the Largest Output Means the Highest Efficiency

Efficiency is a ratio.

A machine producing more useful energy may still be less efficient if it requires much more input energy.


Confusing Power with Energy

Efficiency can be calculated using either energy or power, but the quantities must be compared appropriately.

Use:

useful energy / input energy

or:

useful power / input power


Assuming Thermal Energy Is Always Wasted

Thermal energy from a heater is useful.

Thermal energy from a motor is generally unwanted.

The intended purpose of the device determines what counts as useful.


Assuming 100% Efficiency Is Realistic

Idealized calculations may assume 100%, but real systems experience energy dissipation.


Check Your Understanding

  1. Define efficiency.
  2. Explain the difference between useful and wasted energy transfers.
  3. Why is the term dissipated energy often more scientifically useful than saying energy is "lost"?
  4. A machine receives 500 J and produces 400 J of useful energy. Calculate its efficiency.
  5. How much energy is wasted in Question 4?
  6. A motor receives 2000 J and produces 1500 J of useful mechanical energy. Calculate its efficiency.
  7. A lamp receives 100 J and transfers 25 J as useful light. Calculate its efficiency.
  8. A device receives 500 W and provides 400 W of useful output power. Calculate its efficiency.
  9. How much power is dissipated in Question 8?
  10. A motor is 80% efficient and receives 5000 J. Calculate its useful output.
  11. Calculate the wasted energy in Question 10.
  12. A machine produces 3000 J of useful output and is 75% efficient. Calculate its required input energy.
  13. A motor receives 10 kW and is 85% efficient. Calculate its useful mechanical power.
  14. Explain why friction reduces mechanical efficiency.
  15. Explain why electrical resistance can reduce the efficiency of a motor.
  16. Give three examples of unwanted energy transfers in machines.
  17. Why can thermal energy be useful in one device but wasted in another?
  18. Machine A receives 1000 J and produces 700 J usefully. Machine B receives 500 J and produces 400 J usefully. Which is more efficient?
  19. Explain why the device producing the greatest useful output is not necessarily the most efficient.
  20. A power station is 40% efficient and receives 2000 MW. Calculate its electrical output power.
  21. A 90%-efficient motor receives 5 kW. Calculate its useful output power.
  22. A system contains two stages with efficiencies of 80% and 75%. Calculate the overall efficiency.
  23. A system contains three stages with efficiencies of 90%, 80%, and 70%. Calculate the overall efficiency.
  24. Explain why several efficient devices connected together can have a noticeably lower overall efficiency.
  25. Describe three ways engineers can improve the efficiency of a mechanical system.
  26. Describe two ways engineers can reduce unwanted electrical energy transfers.
  27. Explain why improving efficiency can reduce environmental impacts.
  28. A machine must produce 6000 J of useful output. Compare the required input energy if it is 60% efficient and if it is 80% efficient.
  29. Explain why an efficiency of 120% is impossible.
  30. Explain how a system can be only 60% efficient while still obeying the law of conservation of energy.

Key Terms

Efficiency — Fraction of the total input energy or power transferred into the intended useful output.

Useful energy — Energy transferred in the form required for the intended purpose.

Wasted energy — Energy transferred in forms that are not useful for the intended purpose.

Energy dissipation — Transfer and spreading of energy into less useful stores, often thermal energy in the surroundings.

Input energy — Total energy supplied to a system.

Output energy — Energy transferred from a system.

Useful output — Portion of the output that performs the intended function.

Input power — Rate at which energy is supplied to a system.

Useful output power — Rate at which useful energy is transferred.

Watt (W) — Unit of power equal to one joule per second.

Friction — Force opposing motion that can transfer mechanical energy into thermal energy.

Electrical resistance — Opposition to electric current that causes electrical energy to be transferred thermally.

Air resistance — Resistive force caused by motion through air.

Sankey diagram — Diagram showing energy transfers using arrows whose widths represent energy quantities.

Overall efficiency — Efficiency of a system containing several energy-transfer stages.

Conservation of energy — Principle stating that energy cannot be created or destroyed.


Key Takeaways

  • Efficiency measures the fraction of input energy transferred into the desired useful output.
  • Efficiency can be expressed as a decimal or percentage.
  • Efficiency can be calculated using either energy values or power values.
  • Useful output is determined by the intended purpose of the device.
  • Wasted energy has not disappeared; it has been transferred into less useful forms.
  • Total energy is always conserved.
  • Real systems are less than 100% efficient because some energy is dissipated.
  • Common causes of energy dissipation include friction, electrical resistance, air resistance, sound, vibration, and deformation.
  • Mechanical efficiency can often be improved by reducing friction and resistance.
  • Electrical efficiency can often be improved by reducing unwanted heating and resistance.
  • A more efficient device requires less input energy to produce the same useful output.
  • A device with a greater useful output is not necessarily more efficient.
  • Efficiency and power are different quantities.
  • Several energy conversions can reduce the overall efficiency of a system.
  • For multi-stage systems, decimal efficiencies can be multiplied to determine overall efficiency.
  • Sankey diagrams show how input energy divides into useful and wasted outputs.
  • Improving efficiency can reduce energy consumption, operating costs, resource use, and environmental impacts.
  • An efficiency greater than 100% is impossible because it would violate conservation of energy.

The central idea is:

Energy is never destroyed — but not all transferred energy remains useful.

A more efficient system transfers a greater fraction of its input energy into the output we actually want.

 
 
 

5. Energy Resources and Energy Density

Learning outcomes
  • I can define energy density as the amount of energy stored per unit mass or volume.
  • I can compare the energy densities of different fuels and energy storage systems.
  • I can explain why energy density is important in transportation and energy production.
  • I can evaluate the advantages and disadvantages of various energy resources using energy density data.
  • I can analyze how energy density influences technological and environmental decisions.

Energy Resources and Energy Density

Different energy resources store very different amounts of energy. A small mass of one fuel may contain far more usable energy than the same mass of another.

Energy density describes the amount of energy stored or available per unit mass or per unit volume.

Energy density is important because energy resources must often be:

  • stored
  • transported
  • carried by vehicles
  • delivered to power stations
  • contained safely
  • converted into useful energy

A resource with a high energy density can provide a large amount of energy without requiring as much mass or volume.

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6

Energy Density by Mass

Energy density by mass tells us how much energy is available from each unit of mass.

It can be calculated using:

energy density = energy stored / mass

Common units include:

J/kg

kJ/kg

MJ/kg

For fuels, MJ/kg is particularly common.

For example, if 2 kg of a fuel stores 80 MJ:

energy density = 80/2

energy density = 40 MJ/kg

This means each kilogram contains approximately 40 MJ of energy.


Energy Density by Volume

Sometimes volume is more important than mass.

Volumetric energy density can be calculated using:

energy density = energy stored / volume

Possible units include:

J/m³

MJ/m³

MJ/L

kWh/L

This is particularly important when storage space is limited.

For example, fuel tanks in:

  • cars
  • aircraft
  • ships
  • spacecraft

have limited volume.

A fuel with high energy per litre can therefore be extremely useful even if another resource has a similar energy density per kilogram.


Gravimetric and Volumetric Energy Density

Two terms are useful when comparing energy-storage systems.

Gravimetric energy density means energy stored per unit mass.

Volumetric energy density means energy stored per unit volume.

These can lead to different conclusions.

A resource may have:

  • high energy per kilogram
  • but low energy per litre

Hydrogen is an important example. Hydrogen has a very high energy content per kilogram, but hydrogen gas occupies a large volume unless it is compressed or liquefied.

Therefore, both mass and volume must sometimes be considered.


Comparing Energy Resources

Approximate energy densities by mass for some fuels are:

Energy Resource Approximate Energy Density
Wood 15–20 MJ/kg
Coal 20–30 MJ/kg
Ethanol 27 MJ/kg
Gasoline 44–46 MJ/kg
Diesel 45–48 MJ/kg
Natural gas 50–55 MJ/kg
Hydrogen about 120 MJ/kg

Values vary depending on composition and measurement conditions.

The important idea is the comparison, not memorizing every value.

Hydrogen has a very high energy density by mass, while gasoline and diesel also contain large amounts of energy for their mass.

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6

Worked Example: Comparing Fuels

Suppose Fuel A has an energy density of:

20 MJ/kg

and Fuel B has:

45 MJ/kg

For 10 kg of Fuel A:

E = 20 × 10

E = 200 MJ

For 10 kg of Fuel B:

E = 45 × 10

E = 450 MJ

The same mass of Fuel B stores more than twice as much energy.

This can be a major advantage when mass must be minimized.


Finding Energy from Energy Density

The relationship can be rearranged:

energy stored = energy density × mass

For example, gasoline has an approximate energy density of:

46 MJ/kg

If a vehicle carries:

30 kg

of gasoline:

E = 46 × 30

E = 1380 MJ

or:

1.38 GJ

This represents the approximate chemical energy stored in the fuel, not the useful mechanical energy ultimately delivered by the vehicle.


Energy Density and Efficiency

Energy density and efficiency are different concepts.

Energy density tells us how much energy a resource contains per unit mass or volume.

Efficiency tells us how much of that input energy becomes useful output.

A fuel may have a high energy density but be used in a relatively inefficient system.

For example, suppose:

Fuel energy = 1000 MJ

Engine efficiency = 30%

Useful mechanical energy:

useful energy = 0.30 × 1000

useful energy = 300 MJ

The fuel contained 1000 MJ, but only 300 MJ became the intended mechanical output.


Energy Density of Batteries

Batteries store energy chemically and release it electrically.

Battery energy density is often expressed as:

Wh/kg

or:

Wh/L

Recall:

1 Wh = 3600 J

Therefore:

1 kWh = 3.6 MJ

Modern rechargeable batteries store much less energy per kilogram than liquid fuels when comparing the raw stored-energy values.

However, electric motors can convert stored electrical energy into mechanical energy very efficiently.

This means energy density alone does not determine overall vehicle performance.

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6

Worked Example: Battery

A battery has an energy density of:

250 Wh/kg

and has a mass of:

400 kg

Stored energy:

E = 250 × 400

E = 100,000 Wh

Therefore:

E = 100 kWh

Convert to joules:

100 kWh × 3.6 MJ/kWh = 360 MJ

The battery stores approximately:

360 MJ

of energy.


Comparing Battery and Gasoline Energy

Suppose:

Battery system = 250 Wh/kg

Convert this to MJ/kg:

250 Wh/kg × 3600 J/Wh

= 900,000 J/kg

= 0.90 MJ/kg

Gasoline ≈ 46 MJ/kg

By raw stored energy per kilogram, gasoline contains far more energy.

However, this does not mean a gasoline vehicle is automatically more energy efficient.

Electric motors typically convert a greater fraction of their input energy into useful mechanical energy than combustion engines.

Therefore, meaningful comparisons must consider:

energy density + efficiency + system mass + storage requirements + environmental effects


Why Energy Density Matters in Transportation

Transportation systems must carry their energy source with them.

Every kilogram of fuel or battery adds mass.

Greater vehicle mass can require additional energy to:

  • accelerate
  • climb hills
  • support the vehicle
  • transport the energy-storage system itself

High energy density is therefore particularly valuable for:

  • aircraft
  • spacecraft
  • long-distance vehicles
  • ships
  • portable equipment
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4

Aviation and Energy Density

Aircraft provide a clear example of why energy density matters.

An aircraft must lift:

  • passengers
  • cargo
  • engines
  • structure
  • fuel or energy storage

If the energy-storage system becomes much heavier, additional lift and energy are required.

This makes energy per kilogram extremely important.

Liquid hydrocarbon fuels have historically been useful for aviation partly because they store large amounts of energy for their mass and volume.

Battery-powered aircraft face a major engineering challenge because batteries generally store considerably less energy per kilogram.


Volume Matters Too

Mass is not the only limitation.

Consider two fuels:

Fuel A:

40 MJ/kg

Fuel B:

50 MJ/kg

Fuel B appears better by mass.

But suppose Fuel B requires a very large storage tank.

A vehicle may not have enough physical space.

Engineers therefore consider:

energy per kilogram

and:

energy per litre

A practical energy source must often perform well in both categories.


Hydrogen

Hydrogen demonstrates why the distinction is important.

Hydrogen has a very high energy content by mass.

Approximately:

120 MJ/kg

This is much higher than gasoline.

However, hydrogen gas has a low density under ordinary conditions.

To store useful quantities, hydrogen may need to be:

  • compressed to high pressure
  • liquefied at extremely low temperature
  • stored chemically in another material

These storage methods require additional equipment and energy.

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6

Hydrogen Example

Suppose 5 kg of hydrogen stores approximately:

120 MJ/kg

Total energy:

E = 120 × 5

E = 600 MJ

That is a large amount of energy for only 5 kg of hydrogen.

However, the complete storage system includes:

  • pressure tanks or cryogenic tanks
  • valves
  • insulation
  • safety equipment
  • fuel delivery systems

Therefore, the energy density of the whole system is lower than that of the hydrogen alone.

This is an important engineering principle:

Always consider the complete system, not just the energy-storage material.


Nuclear Fuel and Energy Density

Nuclear fuels have vastly greater energy density than chemical fuels because nuclear reactions release energy from changes within atomic nuclei rather than ordinary chemical bonds.

A relatively small mass of nuclear fuel can therefore release enormous amounts of energy.

This reduces the quantity of fuel that must be transported to a power station.

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7

Nuclear Energy: Important Trade-Off

Very high energy density does not automatically make an energy resource the best choice.

Nuclear energy also requires consideration of:

  • radioactive materials
  • radioactive waste
  • reactor safety
  • construction costs
  • decommissioning
  • security
  • long-term waste management

Therefore:

energy density is only one criterion for evaluating an energy resource.


Fossil Fuels

Fossil fuels include:

  • coal
  • oil
  • natural gas

They have been widely used partly because they:

  • contain substantial energy
  • can be stored
  • can be transported
  • provide energy when needed
  • have established infrastructure

However, their combustion releases carbon dioxide.

Other environmental concerns can include:

  • air pollution
  • extraction impacts
  • habitat disruption
  • spills
  • methane leakage

Their energy density must therefore be considered alongside their environmental effects.


Renewable Energy Resources

Some renewable resources do not use a stored fuel in the same way.

Examples include:

  • solar
  • wind
  • hydroelectric
  • tidal
  • wave energy

For these systems, fuel energy density may not be the most useful comparison.

Instead, engineers may consider:

  • power per unit area
  • availability
  • storage requirements
  • capacity factor
  • location
  • infrastructure
  • environmental effects
  • cost
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4

Solar Energy

Solar energy does not require fuel to be transported to a solar panel.

This is a major advantage.

However, sunlight is spread over a large area.

Producing large amounts of power may therefore require:

  • large areas of solar panels
  • suitable locations
  • energy storage
  • electrical transmission infrastructure

Solar energy illustrates an important point:

low concentration of an energy resource does not necessarily make it unsuitable.

It simply creates different engineering requirements.


Wind Energy

Wind turbines extract kinetic energy from moving air.

There is no fuel tank and therefore no conventional fuel energy density.

Advantages include:

  • renewable resource
  • no fuel combustion during operation
  • no direct carbon dioxide emissions from fuel combustion

Challenges include:

  • variable wind
  • suitable locations
  • energy storage or grid balancing
  • land or offshore infrastructure

Again, energy-resource decisions require more than one measurement.


Hydroelectric Energy

Hydroelectric systems store energy as gravitational potential energy in elevated water.

Recall:

ΔEₚ = mgΔh

The amount of stored energy depends on:

  • mass of water
  • gravitational field strength
  • height difference

A reservoir may therefore require an enormous volume of water to store large amounts of energy.

However, hydroelectric systems can provide substantial power and can be highly useful for grid-scale energy storage and electricity production where geography allows.


Pumped Hydroelectric Storage

Pumped-storage systems use excess electrical energy to pump water uphill.

The energy transformation is approximately:

electrical → mechanical → gravitational potential

When electricity is needed:

gravitational potential → kinetic → mechanical → electrical

This system may require a very large volume of water, but it can store enormous total quantities of energy.

This shows that low energy density does not necessarily mean low total energy storage.

Large-scale systems can compensate by using large masses or volumes.


Energy Density and Portable Technology

Energy density is also important in:

  • smartphones
  • laptops
  • cameras
  • drones
  • medical devices
  • power tools
  • wearable electronics

A battery with greater energy per kilogram can provide longer operating time without making the device excessively heavy.

A battery with greater energy per litre can store more energy without making the device excessively large.

For portable devices, engineers often want both:

high gravimetric energy density

and:

high volumetric energy density


Energy Density Versus Power Density

Energy density should not be confused with power density.

Energy density tells us:

how much energy can be stored

Power density tells us:

how quickly energy can be delivered relative to mass or volume

A storage system might hold a large amount of energy but be unable to release it very quickly.

Another might store less energy but deliver extremely high power for a short period.

This distinction matters for:

  • vehicles
  • power tools
  • grid storage
  • electronics
  • acceleration systems

Worked Comparison

Suppose Battery A stores:

200 Wh/kg

Battery B stores:

300 Wh/kg

A device requires:

60,000 Wh

Battery A mass:

mass = energy / energy density

mass = 60,000/200

mass = 300 kg

Battery B:

mass = 60,000/300

mass = 200 kg

Using Battery B reduces the required battery mass by:

100 kg

This could have a major effect on vehicle performance.


Comparing Fuels Using Required Mass

Suppose a system needs:

900 MJ

of stored chemical energy.

Fuel A:

30 MJ/kg

Required mass:

m = 900/30

m = 30 kg

Fuel B:

45 MJ/kg

Required mass:

m = 900/45

m = 20 kg

Fuel B requires 10 kg less fuel to store the same amount of energy.


Including Efficiency in the Comparison

Suppose a vehicle needs:

300 MJ

of useful mechanical energy.

System A

Fuel energy density:

45 MJ/kg

Efficiency:

30%

Required input energy:

300/0.30 = 1000 MJ

Fuel mass:

1000/45 ≈ 22.2 kg

System B

Energy storage density:

1 MJ/kg

Efficiency:

90%

Required input energy:

300/0.90 ≈ 333 MJ

Storage mass:

333/1 ≈ 333 kg

System B is much more efficient but requires a heavier energy-storage system in this simplified comparison.

This demonstrates why transportation engineering involves trade-offs.


Evaluating Energy Resources

Energy density is important, but it should never be the only factor considered.

A complete comparison might include:

Factor Question
Energy density How much energy is available per kg or litre?
Efficiency How much becomes useful output?
Power How quickly can energy be delivered?
Renewability Can the resource be replenished?
Emissions What pollutants or greenhouse gases are produced?
Storage How easily can the energy be stored?
Transport How easily can the resource be moved?
Safety What hazards are involved?
Reliability Can energy be supplied when required?
Cost What are the total economic costs?
Infrastructure What systems are required to use it?
Waste What waste products are produced?

Transportation Decisions

Consider choosing an energy system for a vehicle.

A designer might ask:

  • How far must the vehicle travel?
  • How heavy can the energy-storage system be?
  • How much space is available?
  • How quickly must the vehicle refuel or recharge?
  • How much power must be delivered?
  • What infrastructure exists?
  • What emissions are produced?
  • How efficient is the drivetrain?
  • How safe is the storage system?

Different vehicles can therefore favor different technologies.


Cars

Cars can potentially use:

  • gasoline
  • diesel
  • batteries
  • hydrogen
  • biofuels

Liquid fuels offer high energy density and rapid refueling.

Battery systems can offer high drivetrain efficiency and avoid direct exhaust emissions.

Hydrogen can offer high energy per kilogram of fuel but requires specialized storage.

The best comparison depends on the whole system and intended use.


Aircraft

Aircraft place especially strong limits on mass.

This makes gravimetric energy density extremely important.

An energy-storage system that is practical for a stationary building or road vehicle may be much more difficult to use in an aircraft because the aircraft must continually carry the storage mass.


Ships

Ships can carry much greater masses than aircraft.

Therefore, mass energy density may be less restrictive.

However, ships travel long distances and require very large total quantities of energy.

Designers must consider:

  • fuel volume
  • fuel mass
  • safety
  • emissions
  • cost
  • refueling infrastructure

Grid Energy Storage

Stationary energy storage has different priorities.

A grid battery or pumped-storage facility does not have to carry its own energy-storage system while moving.

Therefore, mass may be less important.

Engineers may prioritize:

  • cost
  • lifespan
  • efficiency
  • safety
  • response time
  • total storage capacity
  • environmental impact

This demonstrates why the "best" energy technology depends strongly on its application.


Technological Trade-Offs

Imagine two storage technologies.

Technology A

  • very high energy density
  • expensive
  • difficult to recycle
  • high power output

Technology B

  • lower energy density
  • inexpensive
  • long lifespan
  • easy to recycle

There is no automatically correct choice.

For an aircraft, Technology A's low mass may be extremely important.

For a stationary storage facility, Technology B may be preferable.

Engineering decisions involve balancing competing requirements.


Environmental Trade-Offs

Energy density can indirectly affect environmental impacts.

A higher-energy-density fuel may require:

  • less material extraction
  • less transport mass
  • smaller storage systems

But it might also:

  • produce greenhouse gases
  • create hazardous waste
  • require environmentally damaging extraction
  • create safety risks

Therefore, high energy density does not automatically mean environmentally preferable.


Life-Cycle Thinking

A complete environmental comparison considers the entire life cycle of an energy technology.

This may include:

raw materials → manufacturing → transportation → operation → maintenance → disposal/recycling

For example, evaluating a battery might include:

  • mineral extraction
  • battery manufacturing
  • electricity used for charging
  • operating lifetime
  • recycling
  • disposal

Similarly, evaluating gasoline should include:

  • oil extraction
  • refining
  • transportation
  • combustion
  • emissions

This produces a more meaningful comparison than considering energy density alone.


Energy Density and Sustainability

A sustainable energy system must balance many factors.

These include:

  • sufficient energy supply
  • acceptable environmental impact
  • affordability
  • reliability
  • resource availability
  • safety
  • efficiency
  • appropriate energy density

Energy density can influence which technologies are practical, but sustainability requires a broader evaluation.


Common Mistakes

Confusing Energy Density with Material Density

Material density is:

density = mass / volume

Energy density describes:

energy / mass

or:

energy / volume

They are different quantities.


Confusing Energy Density with Power

Energy density tells us how much energy is stored.

Power tells us how quickly energy is transferred.


Confusing Energy Density with Efficiency

A system can have:

high energy density + low efficiency

or:

low energy density + high efficiency

These quantities describe different characteristics.


Assuming the Highest Energy Density Is Always Best

Energy density is only one criterion.

Other factors include:

  • safety
  • emissions
  • cost
  • renewability
  • efficiency
  • storage
  • infrastructure

Comparing Different Units Directly

Do not directly compare:

MJ/kg

with:

MJ/L

One is energy per mass.

The other is energy per volume.

They answer different questions.


Ignoring the Storage System

The energy-storage material may require:

  • tanks
  • containers
  • cooling
  • insulation
  • electronics
  • structural support

These components add mass and volume.


Ignoring Efficiency

Stored energy is not necessarily equal to useful energy.

If a system is:

30% efficient

only 30% of the stored energy becomes the desired useful output.


Assuming Renewable Means High Energy Density

Renewability and energy density are separate characteristics.

Wind and sunlight are renewable but are not stored fuels like gasoline.


Check Your Understanding

  1. Define energy density.
  2. Explain the difference between gravimetric and volumetric energy density.
  3. State two common units for energy density.
  4. A fuel contains 100 MJ in 4 kg. Calculate its energy density.
  5. A fuel has an energy density of 40 MJ/kg. How much energy is stored in 15 kg?
  6. A system requires 600 MJ. Calculate the mass of a fuel with an energy density of 30 MJ/kg.
  7. Repeat Question 6 for a fuel with an energy density of 50 MJ/kg.
  8. Which fuel requires less mass?
  9. Explain why high energy density is particularly important for aircraft.
  10. Explain why energy density may be less important for stationary energy storage.
  11. What is the difference between energy density and power density?
  12. Explain why hydrogen can have a high gravimetric energy density but present volumetric storage challenges.
  13. Why must the mass of a hydrogen tank be considered when evaluating the complete system?
  14. A battery has an energy density of 200 Wh/kg and a mass of 300 kg. Calculate its stored energy in kWh.
  15. Convert the answer to Question 14 into MJ.
  16. Explain why gasoline's high energy density does not automatically make a gasoline vehicle more efficient than an electric vehicle.
  17. What is meant by the efficiency of an energy-conversion system?
  18. A fuel contains 1000 MJ and is used in a 35%-efficient engine. Calculate the useful mechanical energy.
  19. Explain why nuclear fuels are described as having extremely high energy density.
  20. Give two disadvantages that must still be considered when evaluating nuclear energy.
  21. Explain why solar and wind resources cannot be evaluated simply by comparing fuel energy density.
  22. Describe how pumped hydroelectric storage stores energy.
  23. Explain why a low-energy-density storage system can still store a very large total amount of energy.
  24. Why is volumetric energy density important for vehicles with limited storage space?
  25. Explain how energy density influences the design of portable electronic devices.
  26. List five factors other than energy density that should be considered when comparing energy resources.
  27. Explain how life-cycle analysis can improve comparisons between energy technologies.
  28. Why might the best energy-storage technology for an aircraft differ from the best technology for an electrical grid?
  29. Explain how energy density can influence environmental impacts.
  30. Evaluate the statement: "The energy source with the highest energy density is always the best energy source."

Key Terms

Energy density — Amount of energy stored or available per unit mass or volume.

Gravimetric energy density — Energy stored per unit mass.

Volumetric energy density — Energy stored per unit volume.

Energy resource — Source from which usable energy can be obtained.

Fuel — Substance that can release stored energy through a physical, chemical, or nuclear process.

Energy storage — Retaining energy so that it can be used later.

Battery — Device that stores chemical energy and can transfer it electrically.

Power density — Rate at which energy can be delivered relative to mass or volume.

Efficiency — Fraction of input energy transferred into useful output.

Renewable resource — Energy resource replenished naturally on a human timescale.

Non-renewable resource — Resource consumed faster than natural processes replace it.

Fossil fuel — Carbon-rich fuel such as coal, oil, or natural gas formed from ancient biological material.

Hydrogen — Energy carrier with high energy content per unit mass but challenging volumetric storage requirements.

Nuclear fuel — Material capable of releasing energy through nuclear reactions.

Pumped storage — Energy-storage system that stores gravitational potential energy by moving water to a higher elevation.

Life-cycle analysis — Evaluation of environmental impacts throughout production, use, and disposal.

Energy carrier — Substance or system used to transport or deliver energy from one location or process to another.


Key Takeaways

  • Energy density describes how much energy is stored per unit mass or volume.
  • Gravimetric energy density measures energy per unit mass.
  • Volumetric energy density measures energy per unit volume.
  • High energy density allows large amounts of energy to be carried using less mass or space.
  • Energy density is particularly important in transportation and portable technology.
  • Aircraft are especially sensitive to energy-storage mass.
  • Batteries generally have lower gravimetric energy density than liquid hydrocarbon fuels, although electric motors can be highly efficient.
  • Hydrogen has very high energy content per kilogram but presents significant storage challenges because of its low density.
  • The complete storage system must be considered, not just the fuel or battery material.
  • Nuclear fuels have extremely high energy densities compared with chemical fuels.
  • High energy density does not automatically make an energy resource safe, clean, economical, or sustainable.
  • Renewable resources such as wind and solar require different comparisons because they do not use conventional stored fuels.
  • Energy density is different from efficiency.
  • Energy density is different from power density.
  • Energy density by mass and energy density by volume answer different engineering questions.
  • Energy resources should be compared using multiple criteria, including efficiency, cost, emissions, storage, safety, reliability, infrastructure, renewability, and environmental impact.
  • Life-cycle analysis provides a broader way to compare energy technologies.
  • The best energy resource depends strongly on the intended application.

The central idea is:

energy density = energy stored / mass or volume

But choosing an energy technology requires a much broader question:

How much energy can it provide, how efficiently can we use it, what does it take to store and deliver it, and what are the consequences of doing so?