1. Heat Engines

Heat Engines

A heat engine is a device that converts part of the thermal energy transferred from a high-temperature source into useful mechanical work.

All heat engines operate using the same basic energy pathway:

hot reservoir → heat engine → useful work + cold reservoir

A heat engine cannot convert all of the thermal energy it receives into useful work. Some energy must be transferred to a lower-temperature region.

This limitation is a consequence of the Second Law of Thermodynamics.

 

 

 

7

The Basic Parts of a Heat Engine

A heat engine requires three main components:

• hot reservoir
• engine or working substance
• cold reservoir

The engine operates repeatedly in a cycle, taking energy from the hot reservoir, converting some of it into work, and rejecting the remainder to the cold reservoir.

The Hot Reservoir

The hot reservoir is the high-temperature energy source.

It supplies thermal energy to the engine.

Examples include:

• burning fuel
• hot combustion gases
• steam produced in a boiler
• a nuclear reactor
• geothermal heat
• concentrated solar thermal energy

Its temperature is usually represented by:

T_h

The energy transferred from it is often represented by:

Q_h

The Cold Reservoir

The cold reservoir receives the thermal energy that the engine cannot convert into useful work.

Its temperature is represented by:

T_c

and the rejected thermal energy is:

Q_c

Possible cold reservoirs include:

• atmosphere
• cooling water
• rivers
• lakes
• oceans
• cooling towers

The cold reservoir must have:

T_c < T_h

Energy Flow Through a Heat Engine

The basic process is:

Q_h enters engine

Some becomes:

W

and the remainder leaves as:

Q_c

Therefore:

Q_h = W + Q_c

This is an application of the:

First Law of Thermodynamics

Energy is conserved.

 

 

 

7

Heat Engine Energy Equation

The energy relationship is:

Q_h = W + Q_c

Rearranging:

W = Q_h − Q_c

where:

• Q_h = thermal energy absorbed from hot reservoir
• Q_c = thermal energy rejected to cold reservoir
• W = useful work produced

All quantities can be measured in:

joules (J)

Worked Example: Energy Flow

A heat engine absorbs:

Q_h = 2000 J

and transfers:

Q_c = 1300 J

to the cold reservoir.

Find the work output.

Use:

W = Q_h − Q_c

W = 2000 − 1300

W = 700 J

Therefore, the engine produces:

700 J of useful work

What Happens to the Energy?

For the previous example:

2000 J enters

Then:

700 J → useful work

1300 J → cold reservoir

Check:

700 + 1300 = 2000

Therefore:

energy in = energy out

The First Law is satisfied.

Heat Engine Efficiency

The efficiency of a heat engine describes the fraction of input thermal energy converted into useful work.

Efficiency is represented by:

η

The equation is:

η = W/Q_h

where:

• η = efficiency
• W = useful work output
• Q_h = thermal energy supplied by the hot reservoir

Efficiency has no unit.

Percentage Efficiency

Efficiency is often expressed as a percentage:

η = (W/Q_h) × 100%

For example, if an engine receives 1000 J and produces 300 J of useful work:

η = (300/1000) × 100%

η = 30%

This means 30% of the input thermal energy became useful work.

Worked Example: Engine Efficiency

An engine receives:

Q_h = 2500 J

and produces:

W = 900 J

Efficiency:

η = W/Q_h

η = 900/2500

η = 0.36

Therefore:

η = 36%

The remaining 64% is transferred elsewhere, primarily as rejected thermal energy.

Calculating Rejected Energy

An engine receives:

5000 J

and has an efficiency of:

40%

First find useful work:

W = ηQ_h

W = 0.40 × 5000

W = 2000 J

Now:

Q_c = Q_h − W

Q_c = 5000 − 2000

Q_c = 3000 J

Therefore:

• input = 5000 J
• useful work = 2000 J
• rejected heat = 3000 J

Efficiency Using Q_h and Q_c

Since:

W = Q_h − Q_c

and:

η = W/Q_h

we can substitute:

η = (Q_h − Q_c)/Q_h

Therefore:

η = 1 − Q_c/Q_h

This provides another useful efficiency equation.

Worked Example

A heat engine absorbs:

Q_h = 4000 J

and rejects:

Q_c = 2800 J

Efficiency:

η = 1 − Q_c/Q_h

η = 1 − 2800/4000

η = 1 − 0.70

η = 0.30

Therefore:

efficiency = 30%

Why Heat Engines Need Two Reservoirs

A heat engine requires a temperature difference.

Energy enters from:

high temperature

and some energy is ultimately transferred toward:

lower temperature

This temperature difference allows the engine to produce work.

Without a temperature difference:

T_h = T_c

there would be no spontaneous net heat flow available to drive the engine.

The Second Law

The First Law tells us:

Q_h = W + Q_c

But the First Law alone does not prevent:

Q_c = 0

which would imply:

W = Q_h

and therefore:

η = 100%

The Second Law of Thermodynamics tells us why this cannot happen for a cyclic heat engine operating between finite-temperature reservoirs.

Some thermal energy must be rejected to the colder reservoir.

Therefore:

Q_c > 0

and:

η < 100%

Why 100% Efficiency Is Impossible

A heat engine cannot completely convert thermal energy from a single reservoir into work during a cycle.

Some energy must be transferred to the colder surroundings.

This is sometimes expressed through the Kelvin-Planck statement of the Second Law.

In simple terms:

no heat engine can be 100% efficient

This is a fundamental physical limitation, not merely a failure of engineering.

Reversible Heat Engines

A perfectly reversible heat engine is an idealized engine with no entropy-producing processes such as:

• friction
• turbulence
• uncontrolled expansion
• heat transfer across finite temperature differences

Such an engine would have the:

maximum possible efficiency

between its two operating temperatures.

The Carnot Engine

The Carnot engine is an ideal reversible heat engine.

Its maximum theoretical efficiency depends only on the temperatures of its hot and cold reservoirs.

The Carnot efficiency is:

η_max = 1 − T_c/T_h

where:

• T_h = hot reservoir temperature
• T_c = cold reservoir temperature

Both temperatures must be measured in:

kelvin

 

 

 

8

Worked Example: Carnot Efficiency

An ideal heat engine operates between:

T_h = 600 K

and:

T_c = 300 K

Use:

η_max = 1 − T_c/T_h

η_max = 1 − 300/600

η_max = 1 − 0.50

η_max = 0.50

Therefore:

maximum efficiency = 50%

No heat engine operating between these two temperatures can exceed 50% efficiency.

Another Carnot Example

Suppose:

T_h = 800 K

T_c = 320 K

η_max = 1 − 320/800

η_max = 1 − 0.40

η_max = 0.60

Therefore:

maximum theoretical efficiency = 60%

A real engine would have:

η_real < 60%

because real engines contain irreversible processes.

Temperature Difference and Efficiency

From:

η_max = 1 − T_c/T_h

we can see that maximum efficiency increases when:

T_h increases

or:

T_c decreases

Therefore, a larger difference between hot and cold reservoir temperatures can allow greater theoretical efficiency.

Why Kelvin Must Be Used

Suppose:

T_h = 500°C

and:

T_c = 100°C

Do not calculate:

1 − 100/500

because Carnot efficiency requires absolute temperatures.

Convert:

T_h = 500 + 273

T_h = 773 K

T_c = 100 + 273

T_c = 373 K

Then:

η_max = 1 − 373/773

η_max ≈ 0.517

or approximately:

51.7%

Real Engines

Real heat engines have efficiencies lower than the Carnot limit because they contain:

• friction
• turbulence
• heat leakage
• electrical resistance
• finite temperature differences
• finite pressure differences
• incomplete combustion
• mechanical losses

These effects are:

irreversible

and produce entropy.

Therefore:

η_real < η_max

Internal Combustion Engines

Gasoline and diesel engines are examples of:

internal combustion engines

Fuel burns inside the engine.

 

The combustion produces hot gases.

These gases expand and push pistons.

The piston motion turns a:

crankshaft

which can ultimately drive the wheels.

Four-Stroke Gasoline Engine

A common gasoline engine uses four strokes:

Intake

Fuel-air mixture enters the cylinder.

Compression

The piston compresses the mixture.

Power

The mixture ignites.

Hot expanding gases push the piston downward.

Exhaust

Combustion products leave the cylinder.

The cycle then repeats.

Energy Flow in a Car Engine

A simplified pathway is:

chemical energy in fuel

↓

thermal energy from combustion

↓

expanding gases

↓

mechanical work on pistons

↓

rotational mechanical energy

↓

vehicle motion

However, substantial energy is also transferred through:

• hot exhaust
• engine cooling
• friction
• sound

Therefore only part of the fuel's energy becomes useful mechanical work.

Diesel Engines

Diesel engines are also internal combustion heat engines.

Unlike conventional gasoline engines, they ignite fuel primarily through:

high compression and resulting high temperature

rather than a spark plug.

Diesel engines are widely used in:

• trucks
• buses
• ships
• generators
• heavy machinery

Steam Engines

A steam engine is an example of an:

external combustion heat engine

Fuel burns outside the engine's working cylinder.

Thermal energy heats water to produce steam.

 

 

 

8

The steam expands and performs mechanical work.

Historically, steam engines were central to:

the Industrial Revolution

Steam Turbines

Modern thermal power stations generally use:

steam turbines

rather than traditional piston steam engines.

Water is heated to produce high-pressure steam.

The steam flows through turbine blades, causing the turbine to rotate.

The turbine drives an:

electrical generator

Thermal Power Stations

A simplified power-station energy pathway is:

fuel or nuclear energy

↓

thermal energy

↓

steam

↓

turbine rotation

↓

generator

↓

electrical energy

 

 

 

8

The steam must eventually release energy to a colder reservoir.

This commonly occurs in a:

condenser

The Condenser

After passing through a turbine, steam enters a condenser.

The condenser removes thermal energy from the steam.

The steam:

condenses back into liquid water

The water can then return to the boiler.

The condenser therefore helps establish the engine's:

cold side

Cooling Towers

Cooling towers are often associated with thermal power stations.

They help transfer waste thermal energy to the atmosphere.

The large white clouds commonly visible above cooling towers are mainly:

condensed water droplets

rather than smoke.

Cooling systems are necessary partly because a heat engine must reject thermal energy to a colder reservoir.

Fossil-Fuel Power Stations

In a fossil-fuel power station:

chemical energy → thermal energy → mechanical energy → electrical energy

Coal, natural gas or another fuel supplies the original energy.

Combustion provides the high-temperature reservoir.

Cooling systems provide the low-temperature reservoir.

Nuclear Power Stations

A nuclear power station is also based on a heat-engine cycle.

The major difference is the source of thermal energy.

Instead of combustion:

nuclear fission

produces the energy.

That energy is used to produce steam and operate turbines.

Therefore:

nuclear reactor → heat source

steam turbine → heat engine

cooling system → cold reservoir

Gas Turbines

A gas turbine is another type of heat engine.

Air is compressed.

Fuel is burned.

The resulting hot gases expand through turbine blades.

Part of the turbine's output drives the compressor, while the remainder can provide useful mechanical or electrical power.

Gas turbines are used in:

• power stations
• aircraft
• ships

Jet Engines

Jet engines are heat engines.

A simplified sequence is:

air intake → compression → combustion → expansion → exhaust

 

 

 

6

Fuel combustion produces high-temperature gases.

These gases expand and ultimately produce thrust.

Jet engines demonstrate the conversion:

chemical energy → thermal energy → mechanical energy

Combined-Cycle Power Plants

A gas turbine produces extremely hot exhaust gases.

Instead of immediately releasing this energy, a combined-cycle power plant uses the hot exhaust to generate steam.

The steam then powers a second turbine.

Therefore:

gas turbine + steam turbine

extract more useful energy from the same fuel.

This improves overall efficiency.

Heat Engines in Nature

Not every heat engine needs to be a manufactured piston or turbine.

Any system that converts part of an energy flow driven by temperature differences into mechanical work can have heat-engine-like characteristics.

Atmospheric circulation, for example, is driven partly by unequal solar heating.

Temperature differences help produce:

• winds
• convection
• storms

The Earth's atmosphere can therefore be studied using thermodynamic concepts related to heat engines.

Human Body: Not a Simple Heat Engine

The human body converts chemical energy into mechanical work and thermal energy.

However, it is not accurately modeled as a conventional heat engine operating between hot and cold reservoirs.

Muscles obtain useful work primarily through:

chemical free-energy transformations

rather than by operating a thermal cycle between two temperatures.

This distinction is important when identifying genuine heat engines.

Heat Engine vs Electric Motor

An electric motor is not normally classified as a heat engine.

An electric motor converts:

electrical energy → mechanical energy

A heat engine converts energy supplied through a high-temperature thermal process into:

mechanical work

Examples of heat engines include:

• gasoline engine
• diesel engine
• steam engine
• steam turbine
• gas turbine
• jet engine

Heat Engine vs Refrigerator

A heat engine and refrigerator are related but operate differently.

Heat engine

Natural heat flow helps produce work:

hot → engine → cold

with:

work output

Refrigerator

Work is supplied to move thermal energy against its spontaneous direction:

cold → hot

with:

work input

Comparing the Two

Efficiency and Energy Conservation

Suppose an engine is:

35% efficient

This does not mean 65% of the energy disappears.

It means:

35% → useful work

and approximately:

65% → other energy transfers

such as rejected thermal energy.

Energy is always conserved.

Useful vs Rejected Energy

Imagine:

Q_h = 10,000 J

and:

η = 30%

Then:

W = 0.30 × 10,000

W = 3000 J

Therefore:

Q_c = 10,000 − 3000

Q_c = 7000 J

Energy accounting:

10,000 J = 3000 J + 7000 J

Nothing has disappeared.

Efficiency Cannot Exceed 1

Because:

η = W/Q_h

and:

W < Q_h

then:

η < 1

or:

η < 100%

If a calculation gives:

η > 100%

there must be an error in:

• calculation
• measurements
• interpretation of the energy transfers

Improving Engine Efficiency

Engineers improve real heat engines by reducing unnecessary irreversibility.

Strategies include:

• reducing friction
• improving lubrication
• improving combustion
• reducing heat leakage
• recovering exhaust energy
• improving turbine design
• reducing turbulence
• operating at higher hot-reservoir temperatures where materials permit
• improving cooling systems
• using combined cycles

However:

no engineering improvement can violate the Second Law

Waste Heat Recovery

Many engines release hot exhaust.

Instead of simply allowing this energy to enter the environment, some systems use:

waste heat recovery

For example:

hot exhaust → heat exchanger → additional useful energy

Combined-cycle power plants use this principle.

Recovering waste heat improves overall efficiency but still cannot produce 100% conversion.

Why Power Stations Need Cooling

A thermal power station cannot simply convert all boiler energy into electricity.

Some energy must be rejected to a colder reservoir.

Cooling systems provide this function.

Possible cold reservoirs include:

• cooling towers
• rivers
• lakes
• oceans
• atmospheric air

This requirement comes directly from the Second Law.

Heat Engines and Entropy

Heat engines operate because energy moves from:

higher temperature → lower temperature

The engine intercepts part of this energy flow and converts some of it into work.

However, entropy considerations require some energy to continue toward the cold reservoir.

Therefore heat-engine operation connects:

• First Law
• Second Law
• entropy
• temperature
• energy conservation
• efficiency

First Law and Heat Engines

The First Law gives:

Q_h = W + Q_c

It tells us:

energy is conserved

Second Law and Heat Engines

The Second Law tells us:

Q_c cannot simply be eliminated

for a cyclic heat engine operating between finite-temperature reservoirs.

Therefore:

η < 100%

Together the laws explain both:

where the energy goes

and:

why there is a limit to how much becomes useful work

Worked Example: Complete Engine Problem

A heat engine receives:

Q_h = 6000 J

and rejects:

Q_c = 3900 J

Find the work.

W = Q_h − Q_c

W = 6000 − 3900

W = 2100 J

Find the efficiency.

η = W/Q_h

η = 2100/6000

η = 0.35

η = 35%

Therefore:

• input energy = 6000 J
• work output = 2100 J
• rejected energy = 3900 J
• efficiency = 35%

Worked Example: Finding Input Energy

An engine is:

25% efficient

and produces:

500 J

of work.

Use:

η = W/Q_h

Rearrange:

Q_h = W/η

Q_h = 500/0.25

Q_h = 2000 J

Therefore the engine must absorb:

2000 J

from the hot reservoir.

Worked Example: Finding Rejected Energy

From the previous example:

Q_h = 2000 J

W = 500 J

Use:

Q_c = Q_h − W

Q_c = 2000 − 500

Q_c = 1500 J

Therefore 1500 J is transferred to the cold reservoir.

Worked Example: Comparing Engines

Engine A:

Q_h = 5000 J

W = 1500 J

η_A = 1500/5000

η_A = 30%

Engine B:

Q_h = 4000 J

W = 1600 J

η_B = 1600/4000

η_B = 40%

Therefore:

Engine B is more efficient

even though it receives less total energy.

Efficiency compares the fraction converted into useful work.

Worked Example: Real vs Maximum Efficiency

A real engine operates between:

T_h = 900 K

and:

T_c = 300 K

Maximum theoretical efficiency:

η_max = 1 − 300/900

η_max = 1 − 0.333

η_max ≈ 66.7%

Suppose the actual engine is:

40% efficient

This is physically reasonable because:

40% < 66.7%

If someone claimed the engine were 75% efficient, that would exceed the Carnot limit and would therefore be impossible between those reservoir temperatures.

Common Mistakes

Mistake: Saying a heat engine converts all heat into work.
Some energy must be rejected to a cold reservoir.

Mistake: Saying rejected energy disappears.
Energy is conserved and transferred to the surroundings.

Mistake: Confusing Q_h and Q_c.
Q_h enters from the hot reservoir; Q_c leaves toward the cold reservoir.

Mistake: Using Q_c as the input in η = W/Q_h.
Efficiency compares useful work with energy received from the hot reservoir.

Mistake: Forgetting to convert decimal efficiency to a percentage.
0.35 = 35%.

Mistake: Writing 35% as 35 in calculations.
Use 0.35.

Mistake: Saying a reversible engine is 100% efficient.
It has the maximum possible efficiency between its reservoir temperatures, not necessarily 100%.

Mistake: Using Celsius in η_max = 1 − T_c/T_h.
Use kelvin.

Mistake: Assuming an electric motor is a heat engine.
An electric motor primarily converts electrical energy directly into mechanical work.

Mistake: Saying power-station cooling towers are simply wasting energy because of poor engineering.
Real systems have avoidable losses, but the Second Law also imposes a fundamental requirement to reject thermal energy.

Key Terms

Heat engine — device operating cyclically that converts part of an energy transfer from a high-temperature source into useful work while rejecting energy to a colder reservoir.

Hot reservoir — high-temperature source supplying thermal energy to an engine.

Cold reservoir — lower-temperature region receiving rejected thermal energy.

Working substance — material, such as gas or steam, that undergoes thermodynamic changes inside an engine.

Q_h — thermal energy absorbed from the hot reservoir.

Q_c — thermal energy rejected to the cold reservoir.

Work (W) — useful mechanical energy transferred by the engine.

Efficiency (η) — fraction of input thermal energy converted into useful work.

Carnot engine — ideal reversible heat engine used to establish the maximum theoretical efficiency.

Carnot efficiency — maximum theoretical efficiency between two temperatures: η_max = 1 − T_c/T_h.

Internal combustion engine — engine in which fuel burns within the engine itself.

External combustion engine — engine in which heating occurs outside the working cylinder or turbine system.

Steam turbine — device in which expanding steam produces rotational mechanical work.

Gas turbine — heat engine using expanding hot gases to rotate turbine blades.

Condenser — component that removes thermal energy from vapor, causing it to condense.

Waste heat — thermal energy transferred away rather than converted into the desired useful output.

Combined-cycle power plant — plant that uses more than one thermodynamic cycle to recover additional useful energy.

Key Takeaways

• A heat engine converts part of an energy flow from a high-temperature source into useful work.
• Every heat engine requires a hot reservoir and a colder reservoir.
• Energy from the hot reservoir is represented by Q_h.
• Rejected energy is represented by Q_c.
• Useful work is represented by W.
• Energy conservation gives Q_h = W + Q_c.
• Therefore W = Q_h − Q_c.
• Heat-engine efficiency is η = W/Q_h.
• Percentage efficiency is η = (W/Q_h) × 100%.
• Efficiency can also be written as η = 1 − Q_c/Q_h.
• A heat engine cannot convert all its input thermal energy into work.
• Some energy must be rejected to a colder reservoir.
• Therefore no cyclic heat engine operating between finite-temperature reservoirs can be 100% efficient.
• The First Law explains energy conservation within the engine.
• The Second Law explains the fundamental efficiency limit.
• A reversible heat engine provides the maximum possible efficiency between two temperatures.
• The Carnot efficiency is η_max = 1 − T_c/T_h.
• Carnot temperatures must be measured in kelvin.
• Increasing T_h or decreasing T_c can increase the maximum theoretical efficiency.
• Real engines operate below the Carnot limit because of irreversibility.
• Friction, turbulence, heat leakage and finite temperature differences reduce real engine efficiency.
• Internal combustion engines include gasoline and diesel engines.
• Steam engines and steam turbines are examples of external-heating heat-engine systems.
• Gas turbines and jet engines are also heat engines.
• Thermal power stations use heat engines to convert thermal energy into mechanical and then electrical energy.
• Nuclear power plants also operate thermal heat-engine cycles; nuclear fission supplies the thermal energy.
• Cooling systems are essential because heat engines must reject energy to a colder reservoir.
• Combined-cycle systems improve efficiency by recovering energy that would otherwise leave in hot exhaust.
• An electric motor is not normally classified as a heat engine because its primary conversion is electrical energy to mechanical energy.
• Energy rejected by an engine is not destroyed; it is transferred to the surroundings.
• Heat-engine theory provides a practical connection between energy conservation, entropy, the Second Law and engineering efficiency.

Check Your Understanding

1. Define a heat engine.
2. Identify the three main parts of a basic heat-engine system.
3. What is the role of the hot reservoir?
4. What is the role of the cold reservoir?
5. What do Q_h, Q_c and W represent?
6. Write the energy-conservation equation for a heat engine.
7. An engine absorbs 3000 J and rejects 1900 J. Calculate its work output.
8. Write the equation for heat-engine efficiency.
9. An engine absorbs 5000 J and produces 1750 J of work. Calculate its efficiency.
10. An engine is 30% efficient and receives 8000 J. Calculate its useful work output.
11. For the engine in Question 10, calculate the energy rejected to the cold reservoir.
12. Explain why a heat engine cannot be 100% efficient.
13. What is a Carnot engine?
14. Write the equation for maximum Carnot efficiency.
15. Why must temperatures in the Carnot equation be measured in kelvin?
16. An ideal engine operates between 750 K and 300 K. Calculate its maximum theoretical efficiency.
17. Identify three examples of real heat engines.
18. Describe the main energy transformations occurring in a thermal power station.
19. Explain why thermal power stations require cooling systems.
20. Explain how the First and Second Laws of Thermodynamics together determine the operation and limitations of a heat engine.

2. The Carnot Cycle

The Carnot Cycle

The Carnot cycle is an idealized thermodynamic cycle that describes the most efficient possible heat engine operating between two temperatures.

It was developed by French physicist Sadi Carnot in the 19th century.

A Carnot engine operates between:

• a hot reservoir at temperature T_h
• a cold reservoir at temperature T_c

and converts some of the thermal energy received from the hot reservoir into useful work.

Its importance comes from a fundamental result:

No heat engine operating between the same two temperatures can be more efficient than a Carnot engine.

 

 

 

9

Why the Carnot Cycle Matters

Real engines contain unavoidable sources of irreversibility such as:

• friction
• turbulence
• electrical resistance
• heat transfer across finite temperature differences
• rapid expansion and compression
• combustion

The Carnot cycle assumes that all processes are:

perfectly reversible

Therefore, it produces no entropy through irreversibility.

This makes it an ideal benchmark for determining:

the maximum possible efficiency of a heat engine

The Carnot Engine

A Carnot engine is an ideal heat engine containing a working substance, usually modeled as an ideal gas.

During one complete cycle, the gas:

1. expands isothermally
2. expands adiabatically
3. compresses isothermally
4. compresses adiabatically

After the fourth stage, the gas returns to:

its original thermodynamic state

The cycle can then repeat.

The Four Stages

The Carnot cycle consists of:

isothermal expansion

↓

adiabatic expansion

↓

isothermal compression

↓

adiabatic compression

↓

original state

Two processes occur at constant temperature.

Two processes occur without heat transfer.

Isothermal Processes

An isothermal process occurs at:

constant temperature

Therefore:

ΔT = 0

For an ideal gas, internal energy depends only on temperature.

So during an isothermal process:

ΔU = 0

From the First Law:

ΔU = Q − W

therefore:

Q = W

during an ideal-gas isothermal expansion.

Adiabatic Processes

An adiabatic process occurs with:

no heat transfer

Therefore:

Q = 0

From:

ΔU = Q − W

we obtain:

ΔU = −W

during expansion.

Thus the gas performs work using its internal energy and its temperature decreases.

During adiabatic compression, work is done on the gas and its temperature increases.

Stage 1: Isothermal Expansion

The gas begins in contact with the:

hot reservoir at T_h

It expands slowly while remaining at constant temperature.

 

 

 

5

During this stage:

• heat Q_h enters the gas
• gas expands
• gas performs work
• temperature remains T_h
• internal energy remains constant for an ideal gas

Energy relationship:

Q_h = W

for this stage.

The thermal energy entering the gas is converted into work while the temperature remains constant.

Why Temperature Remains Constant

As the gas expands, it performs work.

Normally, this would reduce its internal energy and temperature.

However, the hot reservoir continuously supplies energy.

Therefore:

energy entering as heat = energy leaving as work

and:

ΔU = 0

So the temperature remains constant.

Stage 2: Adiabatic Expansion

The gas is now thermally isolated.

It continues expanding.

During this stage:

Q = 0

 

 

 

6

The gas performs work on the surroundings.

Since no heat enters:

internal energy decreases

and therefore:

temperature decreases

The temperature falls from:

T_h → T_c

Energy During Adiabatic Expansion

Since:

Q = 0

the First Law becomes:

ΔU = −W

Because the gas performs positive work:

W > 0

therefore:

ΔU < 0

The gas loses internal energy and cools.

Stage 3: Isothermal Compression

The gas is placed in thermal contact with the:

cold reservoir at T_c

The surroundings slowly compress the gas.

 

 

 

6

During this stage:

• work is done on the gas
• gas volume decreases
• thermal energy Q_c leaves the gas
• temperature remains constant at T_c

For an ideal gas:

ΔU = 0

Therefore, the work done on the gas is transferred as heat to the cold reservoir.

Why Heat Leaves During Compression

Compression tends to increase the gas's internal energy and temperature.

However, the gas must remain at:

T_c

Therefore the energy supplied by compression is transferred to the cold reservoir.

The gas remains at constant temperature.

Stage 4: Adiabatic Compression

The gas is thermally isolated again.

The surroundings continue compressing it.

During this stage:

Q = 0

 

 

 

6

Work is done on the gas.

Its internal energy increases.

Therefore its temperature rises:

T_c → T_h

At the end of the process, the gas returns to its original state.

The cycle is complete.

The Complete Carnot Cycle

The four stages can be summarized as:

The working gas then returns to its initial state.

Energy Flow Through the Carnot Engine

During the cycle:

hot reservoir

↓

Q_h

↓

Carnot engine

↙︎　　　　　　　　　↘︎

W　　　　　　　Q_c

　　　　　　　　　　　↓

　　　　　　　cold reservoir

Energy conservation requires:

Q_h = W + Q_c

Therefore:

W = Q_h − Q_c

The P–V Diagram

The Carnot cycle is often shown on a:

pressure-volume (P–V) diagram

 

 

 

6

The cycle forms a closed loop.

The two isothermal curves correspond to:

T_h

and:

T_c

The two adiabatic curves connect them.

Area Under a P–V Curve

For expansion or compression:

W = area under the P–V curve

During expansion:

work is done by the gas

During compression:

work is done on the gas

For a complete Carnot cycle:

net work = area enclosed by the cycle

This is an important graphical interpretation.

Why the Cycle Produces Net Work

During the high-temperature expansion, the gas performs relatively large work.

During the lower-temperature compression, less work must be supplied to compress it.

Therefore:

work during expansion > work required for compression

The difference becomes:

net useful work

Heat Engine Efficiency

Efficiency is:

η = W/Q_h

Since:

W = Q_h − Q_c

then:

η = (Q_h − Q_c)/Q_h

Therefore:

η = 1 − Q_c/Q_h

This equation applies to heat-engine efficiency in terms of heat transfers.

Carnot Efficiency

For a reversible Carnot engine:

Q_c/Q_h = T_c/T_h

Therefore:

η_Carnot = 1 − T_c/T_h

where:

• η_Carnot = maximum theoretical efficiency
• T_c = cold reservoir temperature
• T_h = hot reservoir temperature

Temperatures must be measured in:

kelvin

Worked Example: Carnot Efficiency

A Carnot engine operates between:

T_h = 600 K

and:

T_c = 300 K

Use:

η_Carnot = 1 − T_c/T_h

η = 1 − 300/600

η = 1 − 0.50

η = 0.50

Therefore:

η = 50%

The maximum possible efficiency is:

50%

Worked Example: Higher Hot Temperature

Suppose:

T_h = 900 K

T_c = 300 K

η = 1 − 300/900

η = 1 − 0.333

η ≈ 0.667

Therefore:

η ≈ 66.7%

Increasing the hot-reservoir temperature increased the maximum possible efficiency.

Worked Example: Lower Cold Temperature

Suppose:

T_h = 600 K

T_c = 200 K

η = 1 − 200/600

η = 1 − 0.333

η ≈ 66.7%

Lowering the cold-reservoir temperature also increases maximum efficiency.

Temperature Difference and Efficiency

From:

η_Carnot = 1 − T_c/T_h

we can see:

Increasing T_h tends to increase efficiency.

Decreasing T_c tends to increase efficiency.

Therefore:

larger temperature difference → greater possible efficiency

However, the ratio T_c/T_h is what directly determines Carnot efficiency.

Why Kelvin Must Be Used

Suppose:

T_h = 400°C

and:

T_c = 100°C

First convert:

T_h = 400 + 273

T_h = 673 K

T_c = 100 + 273

T_c = 373 K

Now:

η = 1 − 373/673

η ≈ 1 − 0.554

η ≈ 0.446

Therefore:

η ≈ 44.6%

Using Celsius directly would give an incorrect result.

Finding the Cold-Reservoir Temperature

Suppose a Carnot engine has:

η = 60%

and:

T_h = 750 K

Use:

η = 1 − T_c/T_h

0.60 = 1 − T_c/750

T_c/750 = 0.40

T_c = 0.40 × 750

T_c = 300 K

Finding the Hot-Reservoir Temperature

Suppose:

η = 40%

and:

T_c = 360 K

Use:

0.40 = 1 − 360/T_h

360/T_h = 0.60

T_h = 360/0.60

T_h = 600 K

Calculating Work

Suppose a Carnot engine:

• receives Q_h = 5000 J
• operates at η = 60%

Use:

η = W/Q_h

Therefore:

W = ηQ_h

W = 0.60 × 5000

W = 3000 J

The engine produces:

3000 J of work

Calculating Rejected Heat

From:

Q_h = W + Q_c

we have:

Q_c = Q_h − W

Q_c = 5000 − 3000

Q_c = 2000 J

Therefore:

• Q_h = 5000 J
• W = 3000 J
• Q_c = 2000 J

Checking with Reservoir Temperatures

Suppose:

T_h = 750 K

T_c = 300 K

Then:

Q_c/Q_h = T_c/T_h

Q_c/5000 = 300/750

Q_c/5000 = 0.40

Q_c = 2000 J

This agrees with our previous calculation.

Entropy During the Carnot Cycle

The Carnot cycle is completely reversible.

During isothermal expansion at T_h:

ΔS_hot = −Q_h/T_h

The working substance gains the corresponding entropy.

During isothermal compression at T_c, entropy is transferred to the cold reservoir.

For the complete reversible cycle:

ΔS_total = 0

No entropy is produced.

The Working Substance Returns to Its Starting State

Entropy is a:

state function

After one complete cycle, the working substance returns to exactly the same state.

Therefore:

ΔS_working substance = 0

over the complete cycle.

Similarly:

ΔU_cycle = 0

because internal energy is also a state function.

This does not mean no energy has been transferred.

The engine has:

• absorbed Q_h
• rejected Q_c
• produced W

Why the Carnot Engine Has Maximum Efficiency

Any real source of irreversibility produces entropy.

Examples include:

• friction
• turbulence
• heat transfer across finite temperature differences
• rapid expansion
• rapid compression
• electrical resistance

Entropy production reduces the maximum amount of energy that can be converted into useful work.

The Carnot engine avoids all these idealized losses.

Therefore:

no engine between the same T_h and T_c can be more efficient

Carnot's Theorem

Carnot's theorem states that:

No heat engine operating between two reservoirs can be more efficient than a reversible engine operating between those same reservoirs.

It also implies that:

all reversible engines operating between the same two temperatures have the same efficiency

regardless of their particular design or working substance.

The Working Substance Does Not Determine Maximum Efficiency

Imagine two ideal reversible engines.

Engine A uses:

helium

Engine B uses:

another ideal gas

If both operate between:

T_h = 700 K

and:

T_c = 350 K

then both have:

η = 1 − 350/700

η = 50%

Their maximum efficiency is determined by reservoir temperatures, not by the choice of working substance.

Real Engines

Real engines cannot follow the Carnot cycle perfectly.

They contain:

• friction
• finite temperature differences
• rapid gas flow
• turbulence
• combustion
• heat leakage
• mechanical resistance

Therefore:

real engines are irreversible

and:

η_real < η_Carnot

Comparing Carnot and Real Engines

Example: Real vs Carnot Efficiency

A real engine operates between:

T_h = 800 K

and:

T_c = 300 K

Carnot limit:

η_Carnot = 1 − 300/800

η_Carnot = 1 − 0.375

η_Carnot = 62.5%

Suppose the real engine operates at:

40% efficiency

This is possible because:

40% < 62.5%

But an engine claiming:

70% efficiency

between the same temperatures would be impossible because:

70% > 62.5%

Why Real Engines Cannot Be Carnot Engines

A perfectly reversible Carnot process must proceed through infinitesimally small differences in:

• temperature
• pressure

This means the processes must occur:

extremely slowly

But a practical engine must produce:

useful power

Power is:

P = W/t

If an engine took an infinitely long time to complete each cycle:

t → ∞

then its practical power output would approach:

P → 0

Thus perfect Carnot operation is not useful as a practical engine.

Efficiency vs Power

This creates an important engineering trade-off.

To approach maximum reversible efficiency:

operate increasingly slowly

But to produce useful power:

operate at a finite rate

Real engineers therefore balance:

• efficiency
• power
• cost
• size
• reliability
• materials
• operating temperature

The most efficient theoretical engine is not necessarily the most practical engine.

Carnot Cycle on a P–V Diagram

The Carnot P–V diagram contains four curves.

A → B

Isothermal expansion at T_h

Heat enters.

B → C

Adiabatic expansion

Temperature falls from T_h to T_c.

C → D

Isothermal compression at T_c

Heat leaves.

D → A

Adiabatic compression

Temperature rises from T_c to T_h.

 

 

 

7

The area enclosed by the loop represents:

net work produced per cycle

Carnot Cycle on a Temperature-Entropy Diagram

The Carnot cycle can also be represented using a:

temperature-entropy (T–S) diagram

 

 

 

7

For an ideal Carnot cycle:

• isothermal processes appear horizontally
• reversible adiabatic processes occur at constant entropy and appear vertically

The result is a rectangular cycle on an idealized T–S diagram.

Reversible Adiabatic Means Isentropic

During a reversible adiabatic process:

Q = 0

and no entropy is produced.

Therefore:

ΔS = 0

Such a process is called:

isentropic

Thus Stages 2 and 4 of the ideal Carnot cycle are:

reversible adiabatic = isentropic

The Four Stages in Energy Terms

Stage 1

Isothermal expansion

Q_h enters.

Gas performs work.

Stage 2

Adiabatic expansion

No heat transfer.

Gas performs work and cools.

Stage 3

Isothermal compression

Work is done on gas.

Q_c leaves.

Stage 4

Adiabatic compression

No heat transfer.

Work is done on gas and it warms.

Then the cycle repeats.

Why Heat Is Absorbed at T_h

During Stage 1, the working gas remains in thermal contact with the hot reservoir.

To make this transfer perfectly reversible, the temperature difference between reservoir and gas would need to be:

infinitesimally small

This allows heat transfer without entropy production in the ideal limit.

Why Heat Is Rejected at T_c

During Stage 3, the working substance is in thermal contact with the cold reservoir.

Again, the temperature difference must approach:

zero

for perfectly reversible heat transfer.

The gas releases Q_c while remaining at T_c.

Why 100% Carnot Efficiency Is Impossible

For:

η_Carnot = 1

we would require:

T_c/T_h = 0

Therefore either:

T_c = 0 K

or an unboundedly large T_h.

Neither provides a realizable finite-temperature heat engine.

Therefore:

η_Carnot < 100%

for engines operating between finite, physically realizable temperatures.

Example: Very Hot Engine

Suppose:

T_h = 1500 K

and:

T_c = 300 K

η = 1 − 300/1500

η = 1 − 0.20

η = 80%

Even with an extremely hot reservoir, the maximum theoretical efficiency is still below 100%.

A real engine would achieve less than 80%.

Engineering and High Temperatures

Increasing T_h can improve theoretical efficiency.

However, real materials place limits on operating temperatures.

At very high temperatures:

• metals weaken
• turbine blades can deform
• oxidation increases
• cooling becomes difficult
• components may melt or fail

Therefore engineers cannot simply increase T_h indefinitely.

Engineering and Low Temperatures

Lowering T_c can also increase efficiency.

However, cooling the cold reservoir itself may require:

additional energy

and environmental conditions impose limits.

Therefore practical engines cannot arbitrarily lower T_c either.

Steam Power Plants

Steam power plants do not operate on a perfect Carnot cycle.

They commonly use cycles related to the:

Rankine cycle

 

 

 

5

The system includes:

• boiler
• turbine
• condenser
• pump

The Carnot cycle remains useful because it establishes the maximum theoretical efficiency between given temperatures.

Internal Combustion Engines

Gasoline engines are commonly modeled using idealized cycles such as the:

Otto cycle

Diesel engines can be modeled using the:

Diesel cycle

Gas turbines are often modeled using the:

Brayton cycle

These are not Carnot cycles.

But Carnot efficiency still provides a fundamental upper benchmark based on operating temperatures.

Carnot Efficiency as a Benchmark

Suppose a real power plant has:

η_real = 45%

and its Carnot limit is:

η_Carnot = 60%

It may be useful to compare:

45/60 = 0.75

The real plant achieves about:

75% of the Carnot-limit efficiency

This does not mean the plant is 75% efficient overall.

Its actual efficiency remains:

45%

It means it achieves 75% of the maximum theoretical value available between those temperatures.

Worked Example: Full Carnot Problem

A Carnot engine operates between:

T_h = 1000 K

and:

T_c = 400 K

It absorbs:

Q_h = 5000 J

Find the efficiency

η = 1 − 400/1000

η = 0.60

η = 60%

Find the work output

W = ηQ_h

W = 0.60 × 5000

W = 3000 J

Find Q_c

Q_c = Q_h − W

Q_c = 5000 − 3000

Q_c = 2000 J

Therefore:

Q_h = 5000 J

W = 3000 J

Q_c = 2000 J

Worked Example: Testing an Engine Claim

An engineer claims an engine operating between:

700 K

and:

350 K

has an efficiency of:

60%

Find the Carnot limit:

η_Carnot = 1 − 350/700

η_Carnot = 0.50

η_Carnot = 50%

The claimed efficiency is:

60%

Since:

60% > 50%

the claim violates the Carnot limit.

Such an engine cannot operate as claimed between those two reservoirs.

Worked Example: Comparing Two Engines

Engine A:

T_h = 600 K

T_c = 300 K

η_A = 1 − 300/600

η_A = 50%

Engine B:

T_h = 900 K

T_c = 300 K

η_B = 1 − 300/900

η_B ≈ 66.7%

Engine B has the higher theoretical maximum efficiency because it operates with the larger temperature ratio difference.

Carnot Cycle and the Second Law

The Carnot cycle demonstrates several major consequences of the Second Law:

• heat engines require hot and cold reservoirs
• not all thermal energy can become work
• reversible processes establish maximum efficiency
• irreversible processes reduce efficiency
• entropy production reduces available useful work
• no engine can exceed the Carnot limit

The Carnot cycle therefore provides one of the clearest links between:

temperature, entropy, energy and efficiency

Common Mistakes

Mistake: Saying the Carnot engine is a real engine design.
It is an idealized thermodynamic model.

Mistake: Saying the Carnot engine is 100% efficient.
Its efficiency is η = 1 − T_c/T_h and is below 100% for finite reservoir temperatures.

Mistake: Using Celsius in the Carnot equation.
Always use kelvin.

Mistake: Confusing the hot and cold temperatures.
T_h is the higher temperature and T_c is the lower temperature.

Mistake: Saying all four stages transfer heat.
The two adiabatic stages have Q = 0.

Mistake: Saying temperature changes during the isothermal stages.
Isothermal means constant temperature.

Mistake: Saying temperature stays constant during adiabatic expansion.
The gas cools because it performs work without receiving heat.

Mistake: Saying temperature stays constant during adiabatic compression.
The gas warms because work is done on it without heat leaving.

Mistake: Saying real engines can reach Carnot efficiency if friction is reduced enough.
Perfect Carnot efficiency requires completely reversible processes, an unattainable idealization.

Mistake: Saying a larger temperature difference always means a practical engine will automatically be more efficient.
Carnot efficiency sets the theoretical limit; real efficiency also depends on design and irreversibilities.

Key Terms

Carnot cycle — ideal reversible thermodynamic cycle consisting of two isothermal and two adiabatic processes.

Carnot engine — ideal heat engine operating through a Carnot cycle.

Hot reservoir — high-temperature source supplying thermal energy Q_h.

Cold reservoir — lower-temperature region receiving Q_c.

Working substance — material undergoing expansion and compression during the cycle.

Isothermal process — process occurring at constant temperature.

Adiabatic process — process involving no heat transfer.

Isentropic process — process occurring at constant entropy; a reversible adiabatic process is isentropic.

Reversible process — ideal process producing no net entropy.

Carnot efficiency — maximum theoretical efficiency between two reservoir temperatures.

P–V diagram — graph of pressure against volume used to represent thermodynamic processes.

T–S diagram — graph of temperature against entropy.

Carnot's theorem — principle stating that no engine can exceed the efficiency of a reversible engine operating between the same temperatures.

Q_h — thermal energy absorbed from the hot reservoir.

Q_c — thermal energy rejected to the cold reservoir.

Net work (W) — useful work produced over one complete engine cycle.

Key Takeaways

• The Carnot cycle represents an ideal reversible heat-engine cycle.
• It consists of four stages.
• Stage 1 is isothermal expansion at T_h.
• Stage 2 is adiabatic expansion from T_h to T_c.
• Stage 3 is isothermal compression at T_c.
• Stage 4 is adiabatic compression from T_c to T_h.
• Heat enters only during the high-temperature isothermal expansion.
• Heat leaves during the low-temperature isothermal compression.
• No heat is transferred during either adiabatic stage.
• The working substance returns to its initial state after each complete cycle.
• The net work produced equals Q_h − Q_c.
• On a P–V diagram, the enclosed area represents the net work per cycle.
• The Carnot engine is completely reversible.
• It therefore produces no net entropy.
• Carnot efficiency is η_Carnot = 1 − T_c/T_h.
• Temperatures in the Carnot equation must be measured in kelvin.
• The Carnot efficiency depends only on the two reservoir temperatures.
• Increasing T_h increases the theoretical maximum efficiency.
• Decreasing T_c increases the theoretical maximum efficiency.
• A Carnot engine operating between finite temperatures cannot be 100% efficient.
• No real engine operating between the same temperatures can exceed Carnot efficiency.
• Real engines have lower efficiency because they contain irreversible processes.
• Friction, turbulence and finite temperature differences are important sources of irreversibility.
• Real engines must operate at finite rates to produce useful power.
• A perfectly reversible Carnot engine would operate extremely slowly and is therefore an idealization.
• Steam turbines, gasoline engines, diesel engines and gas turbines use practical cycles rather than the Carnot cycle.
• The Carnot cycle provides a benchmark against which real engines can be evaluated.
• Carnot's theorem is a fundamental consequence of the Second Law of Thermodynamics.

Check Your Understanding

1. What is the Carnot cycle?
2. Why is the Carnot cycle important in thermodynamics?
3. Name the four stages of the Carnot cycle in order.
4. What happens during isothermal expansion?
5. Why does the temperature remain constant during isothermal expansion?
6. What happens to the temperature during adiabatic expansion?
7. Why does the gas cool during adiabatic expansion?
8. What happens during isothermal compression?
9. What happens during adiabatic compression?
10. During which two stages is Q = 0?
11. What does the area enclosed by a Carnot cycle on a P–V diagram represent?
12. Write the equation for Carnot efficiency.
13. Why must temperatures in the Carnot efficiency equation be measured in kelvin?
14. A Carnot engine operates between 800 K and 400 K. Calculate its maximum efficiency.
15. A Carnot engine operates between 900 K and 300 K and absorbs 6000 J. Calculate its work output.
16. For the engine in Question 15, calculate Q_c.
17. Explain why a Carnot engine cannot reach 100% efficiency.
18. Why are real engines less efficient than Carnot engines operating between the same temperatures?
19. Explain why a perfectly reversible Carnot engine would be impractical as a real engine.
20. A real engine operating between 700 K and 350 K claims an efficiency of 55%. Use the Carnot limit to evaluate whether this claim is physically possible.

3. Refrigerators and Heat Pumps

Refrigerators and Heat Pumps

Refrigerators and heat pumps are devices that use external work to transfer thermal energy from a colder region to a warmer region.

This is opposite to the natural direction of spontaneous heat flow:

hot → cold

To force energy to move:

cold → hot

the device must receive energy as work, usually from electricity.

A refrigerator and a heat pump operate using essentially the same thermodynamic principle. The main difference is which energy transfer we consider useful.

 

 

 

7

---

Natural Heat Flow

If a hot object and a cold object are placed in thermal contact, energy naturally transfers:

hot → cold

This process is spontaneous and consistent with the:

Second Law of Thermodynamics

A refrigerator must do the opposite.

It removes thermal energy from a colder space and transfers it into warmer surroundings.

Therefore:

external work is required

---

Basic Refrigerator Energy Flow

A refrigerator operates between:

• a cold reservoir
• a hot reservoir

The refrigerator removes thermal energy:

Q_c

from the cold region.

Work:

W

is supplied to the refrigerator.

The refrigerator then releases:

Q_h

to the warmer surroundings.

Energy conservation gives:

Q_h = Q_c + W

---

Energy Is Conserved

Suppose a refrigerator removes:

500 J

from its interior and receives:

200 J

of electrical work.

Then:

Q_h = Q_c + W

Q_h = 500 + 200

Q_h = 700 J

Therefore the refrigerator releases:

700 J

into the room.

This is more energy than it removed from inside because the electrical work also ultimately contributes energy.

---

Why the Back of a Refrigerator Feels Warm

A refrigerator does not simply "destroy" thermal energy inside the cabinet.

It transfers that energy into the room.

The coils at the back or underneath the refrigerator release:

Q_h

which includes:

energy removed from food + electrical work supplied

Therefore the condenser coils can feel warm or hot.

 

 

---

Why Work Is Required

The spontaneous direction of heat transfer is:

hot → cold

A refrigerator forces energy to move:

cold → hot

This cannot happen continuously by itself.

External work must therefore be supplied.

Usually:

electrical energy → compressor work

The compressor is one of the most important components of a refrigerator or heat pump.

---

The Refrigeration Cycle

A typical vapor-compression refrigerator contains four major components:

• evaporator
• compressor
• condenser
• expansion valve

A refrigerant circulates continuously through these components.

 

 

 

9

The basic cycle is:

evaporator → compressor → condenser → expansion valve → evaporator

---

The Refrigerant

A refrigerant is a fluid that circulates through the system.

It is chosen because it can:

• evaporate at relatively low temperatures
• condense at suitable higher temperatures
• transport substantial thermal energy
• operate through repeated pressure changes

The refrigerant repeatedly changes:

pressure

temperature

and often:

state of matter

as it moves through the cycle.

---

Stage 1: Evaporator

The refrigerant enters the evaporator at relatively:

low pressure and low temperature

It absorbs thermal energy from the refrigerated space.

Therefore:

cold space → refrigerant

As it absorbs energy, the refrigerant evaporates.

 

 

 

8

The energy absorbed is:

Q_c

This is the useful effect of a refrigerator.

---

Why Evaporation Is Useful

Evaporation requires energy.

During the phase change:

liquid → gas

the refrigerant absorbs latent heat.

This allows the refrigerant to remove substantial amounts of thermal energy without requiring an enormous temperature increase.

The evaporator therefore cools the refrigerator interior.

---

Stage 2: Compressor

The refrigerant gas then enters the:

compressor

The compressor uses electrical energy to perform work on the gas.

Compression causes:

• pressure to increase
• temperature to increase

The refrigerant leaves the compressor as a:

hot, high-pressure gas

---

Why Compression Raises Temperature

When work is done on a gas, its internal energy can increase.

Therefore its temperature rises.

The compressor raises the refrigerant temperature above the temperature of the surrounding room.

This is essential because thermal energy must eventually flow:

refrigerant → room

For that to happen spontaneously, the refrigerant must be hotter than its surroundings.

---

Stage 3: Condenser

The hot, high-pressure refrigerant enters the:

condenser

The condenser is usually located where it can transfer energy to the surrounding air.

Because:

T_refrigerant > T_room

thermal energy flows:

refrigerant → room

The refrigerant cools and condenses:

gas → liquid

The energy released is:

Q_h

---

Condensation and Latent Heat

During condensation, the refrigerant releases latent heat.

Therefore the condenser can transfer substantial thermal energy to the surroundings.

This explains why the outside coils of a refrigeration system become warm.

---

Stage 4: Expansion Valve

The high-pressure liquid refrigerant then passes through an:

expansion valve

or similar restriction.

Its pressure drops dramatically.

This lowers its temperature.

The refrigerant becomes a cold, low-pressure mixture ready to return to the evaporator.

The cycle then repeats.

---

The Complete Refrigeration Cycle

The four main stages are:

Evaporator

absorbs Q_c from cold space

Compressor

work W is supplied

Condenser

releases Q_h to warm surroundings

Expansion valve

reduces pressure and temperature

Then:

cycle repeats

---

Refrigerator Energy Diagram

The overall energy flow is:

cold reservoir

↓

Q_c

↓

refrigerator

← W

↓

Q_h

↓

hot reservoir

Energy conservation requires:

Q_h = Q_c + W

---

Refrigerators and the Second Law

At first, refrigerators might seem to violate the Second Law because they move energy:

cold → hot

But the Second Law does not say this is impossible.

It says such a transfer cannot occur as the sole result without other changes.

A refrigerator requires:

work input

and the complete process increases total entropy.

Therefore refrigerators are completely consistent with the Second Law.

---

Refrigerator vs Heat Engine

A heat engine and refrigerator operate in opposite thermodynamic directions.

Heat engine

hot → cold

while producing:

work output

Refrigerator

Uses:

work input

to transfer energy:

cold → hot

---

Comparison

Heat Engine	Refrigerator
Produces work	Requires work
Q_h enters from hot reservoir	Q_c enters from cold reservoir
Q_c leaves to cold reservoir	Q_h leaves to hot reservoir
Heat flow helps produce work	Work drives heat transfer
η measures performance	COP measures performance

---

Heat Pumps

A heat pump operates on essentially the same physical cycle as a refrigerator.

The major difference is:

which energy transfer is useful

For a refrigerator, the desired result is:

removing Q_c from the cold region

For a heat pump, the desired result is:

delivering Q_h to the warm region

 

 

 

8

---

Refrigerator

The goal is to keep something:

cold

Examples include:

• refrigerator
• freezer
• cold-storage room
• food display cooler

The useful energy transfer is:

Q_c

---

Heat Pump

The goal is to keep something:

warm

For example, a heat pump may remove thermal energy from cold outdoor air and transfer it into a building.

The useful energy transfer is:

Q_h

---

Same Machine, Different Purpose

Imagine a device between:

cold outdoors

and:

warm house

It extracts Q_c from outdoors.

It receives electrical work W.

It delivers:

Q_h = Q_c + W

into the house.

When the desired effect is heating the house, we call the device a:

heat pump

---

Air Conditioners

An air conditioner is thermodynamically very similar to a refrigerator.

Its useful purpose is:

removing thermal energy from a building

Therefore it operates like a large refrigerator.

Indoor air acts as the region being cooled.

Outdoor air receives the rejected thermal energy.

---

Reversible Heat Pumps

Many modern systems can reverse their operating direction.

During summer:

inside → outside

The system cools the building.

During winter:

outside → inside

The system heats the building.

This is commonly called a:

reversible heat pump

---

How Can a Heat Pump Get Energy from Cold Air?

A common misconception is that cold air contains no thermal energy.

Even cold outdoor air has internal energy as long as its temperature is above absolute zero.

A heat pump can use a refrigerant that is even colder than the outdoor air.

Therefore thermal energy can spontaneously transfer:

outdoor air → colder refrigerant

The compressor then raises the refrigerant's temperature so energy can be released indoors.

---

Example: Heating a House

Suppose:

outside air = 5°C

The refrigerant in the outdoor evaporator might be colder than 5°C.

Therefore:

outside air → refrigerant

Thermal energy enters the refrigerant.

The compressor then raises its temperature.

If the refrigerant reaches a temperature above the indoor temperature:

refrigerant → indoor air

Energy can then be delivered into the house.

---

Coefficient of Performance

Refrigerators and heat pumps are not normally described using heat-engine efficiency.

Instead, we use the:

coefficient of performance

abbreviated:

COP

COP compares the desired thermal-energy transfer with the work required.

---

Refrigerator COP

For a refrigerator, the desired effect is removing energy from the cold region.

Therefore:

COP_R = Q_c/W

where:

• COP_R = refrigerator coefficient of performance
• Q_c = thermal energy removed from cold region
• W = work input

---

Worked Example: Refrigerator COP

A refrigerator removes:

Q_c = 900 J

while requiring:

W = 300 J

Then:

COP_R = 900/300

COP_R = 3.0

This means the refrigerator removes:

3 J of thermal energy

from its cold interior for every:

1 J of work supplied

---

COP Can Be Greater Than 1

This is very important.

Suppose:

COP = 4

This does not mean:

400% efficiency

COP is not the same as efficiency.

The device is not converting 1 J into 4 J.

Instead, 1 J of work allows it to:

transfer additional energy from another location

Energy remains conserved.

---

Heat Pump COP

For a heat pump, the desired effect is delivering thermal energy to the warm region.

Therefore:

COP_HP = Q_h/W

Since:

Q_h = Q_c + W

a heat pump can deliver more thermal energy than the electrical work it consumes.

---

Worked Example: Heat Pump COP

A heat pump receives:

W = 1000 J

and extracts:

Q_c = 3000 J

from outdoors.

Then:

Q_h = Q_c + W

Q_h = 3000 + 1000

Q_h = 4000 J

Therefore:

COP_HP = 4000/1000

COP_HP = 4.0

For every 1 J of electrical work supplied, the system delivers 4 J of thermal energy indoors.

---

Relationship Between the COPs

Because:

Q_h = Q_c + W

divide everything by W:

Q_h/W = Q_c/W + 1

Therefore:

COP_HP = COP_R + 1

for the same idealized device operating between the same two reservoirs.

---

Worked Example

Suppose a refrigeration system has:

COP_R = 3.5

Then:

COP_HP = COP_R + 1

COP_HP = 3.5 + 1

COP_HP = 4.5

---

COP and Temperature Difference

A refrigerator or heat pump generally performs better when the temperature difference between hot and cold regions is:

smaller

Why?

Because moving thermal energy across a smaller temperature difference requires less work.

Therefore:

smaller temperature difference → higher possible COP

---

Ideal Refrigerator COP

For an ideal reversible refrigerator operating between T_c and T_h:

COP_R,max = T_c/(T_h − T_c)

Temperatures must be in:

kelvin

This gives the maximum theoretical refrigerator COP.

---

Ideal Heat Pump COP

For an ideal reversible heat pump:

COP_HP,max = T_h/(T_h − T_c)

Again:

temperatures must be in kelvin

These are theoretical maximum values based on reversible operation.

---

Worked Example: Ideal Refrigerator

Suppose:

T_c = 270 K

T_h = 300 K

Then:

COP_R,max = 270/(300 − 270)

COP_R,max = 270/30

COP_R,max = 9.0

This is the ideal maximum.

A real refrigerator would have a lower COP.

---

Worked Example: Ideal Heat Pump

Using the same temperatures:

COP_HP,max = 300/(300 − 270)

COP_HP,max = 300/30

COP_HP,max = 10.0

Notice:

COP_HP = COP_R + 1

10 = 9 + 1

---

Larger Temperature Difference

Suppose:

T_c = 250 K

T_h = 300 K

Then:

COP_R,max = 250/(300 − 250)

COP_R,max = 250/50

COP_R,max = 5.0

The larger temperature difference reduced the maximum COP.

---

Heat Pumps in Cold Weather

As outdoor temperature falls, a heat pump must move thermal energy across a larger temperature difference.

Therefore its COP generally:

decreases

This is one reason heat pumps can require more electrical energy during very cold weather.

System design and heat-pump type also strongly affect real performance.

---

Why Heat Pumps Can Be Efficient for Heating

Consider ordinary electric resistance heating.

Approximately:

1 J electrical energy → 1 J thermal energy

at the point of use.

A heat pump does something different.

It uses electrical work to move additional thermal energy from outdoors.

For example:

1 J electrical work

•  

3 J transferred from outdoors

=

4 J delivered indoors

Therefore:

COP_HP = 4

This can make heat pumps highly effective for space heating.

---

Household Refrigerator

A household refrigerator uses the vapor-compression cycle to maintain a low interior temperature.

The main components include:

• compressor
• condenser
• expansion device
• evaporator
• refrigerant
• thermostat

The thermostat controls operation so the compressor does not need to run continuously.

---

Freezers

Freezers operate on the same basic principle as refrigerators.

However, they maintain:

lower temperatures

This usually requires the refrigeration system to operate across a larger temperature difference.

That can reduce COP and increase energy consumption.

---

Air Conditioners

Air conditioners remove thermal energy and often moisture from indoor air.

The indoor evaporator:

absorbs thermal energy

The outdoor condenser:

releases thermal energy

 

 

 

8

The compressor requires electrical work.

Thus:

air conditioner = refrigeration system for a building

---

Dehumidification

When warm humid air passes over a cold evaporator coil, water vapor can:

condense

The phase change:

gas → liquid

removes water from the air.

Therefore air conditioners often:

• cool the air
• reduce humidity

This can improve comfort even without an enormous temperature decrease.

---

Domestic Heat Pumps

Heat pumps can provide:

• space heating
• space cooling
• water heating

Common types include:

• air-source heat pumps
• ground-source heat pumps
• water-source heat pumps

---

Air-Source Heat Pumps

An air-source heat pump exchanges thermal energy with outdoor air.

During heating:

outdoor air → refrigerant → indoor air

During cooling:

indoor air → refrigerant → outdoor air

These systems are widely used because installation is relatively straightforward compared with ground-source systems.

---

Ground-Source Heat Pumps

A ground-source heat pump exchanges thermal energy with the ground.

 

 

 

8

Underground temperatures often vary less throughout the year than outdoor air temperatures.

This can allow a smaller operating temperature difference and potentially improve performance.

However, installation can be more complex and expensive.

---

Heat-Pump Water Heaters

A heat-pump water heater uses a refrigeration cycle to transfer thermal energy:

surrounding air → water

Rather than generating all the heating directly from electrical resistance, it moves existing thermal energy into the water.

This can reduce electrical energy consumption.

---

Refrigerated Transport

Refrigeration systems are also used in:

• refrigerated trucks
• shipping containers
• food warehouses
• supermarket displays
• vaccine storage
• medical storage

All use the same fundamental idea:

use work to transfer thermal energy from cold to warm

---

Refrigerator Door Open in a Room

Suppose you leave a refrigerator door open in a sealed room.

Will the refrigerator cool the room?

No.

The refrigerator removes Q_c from nearby air but releases:

Q_h = Q_c + W

back into the same room.

Since:

Q_h > Q_c

the net effect is:

room warms slightly

because the electrical work ultimately becomes thermal energy.

---

Worked Example: Open Refrigerator

Suppose the refrigerator removes:

600 J

from nearby air.

The compressor uses:

200 J

of electrical work.

It releases:

Q_h = 600 + 200

Q_h = 800 J

to the room.

Net effect on room:

800 − 600

= 200 J added

Therefore the room becomes warmer overall.

---

Why the Refrigerator Interior Gets Cold

The refrigerator does not "produce cold."

Instead, it:

removes thermal energy

from its interior.

The interior temperature falls because its internal energy decreases.

It is therefore more accurate to say:

the refrigerator transfers heat out

rather than:

the refrigerator creates cold

---

Insulation

Refrigerators and buildings use insulation to reduce unwanted heat transfer.

For a refrigerator:

warm room → cold interior

is the natural direction of heat flow.

Insulation slows this transfer.

Better insulation means the compressor needs to operate less often.

Therefore:

better insulation → lower energy consumption

---

Door Seals

Damaged refrigerator door seals allow warm room air to enter.

This increases:

• heat transfer into refrigerator
• compressor operating time
• electrical energy consumption

Maintaining good seals therefore improves practical efficiency.

---

Refrigerator Placement

A refrigerator must release Q_h through its condenser.

If airflow around the condenser is poor, heat transfer becomes more difficult.

The compressor may then need to work harder.

Therefore adequate ventilation around refrigeration equipment can improve performance.

---

Real Refrigerators and Irreversibility

Real refrigeration systems contain:

• compressor friction
• electrical resistance
• pressure losses
• heat transfer across finite temperature differences
• refrigerant flow resistance

These processes produce entropy.

Therefore:

COP_real < COP_ideal

---

Refrigerator vs Heat Pump

Feature	Refrigerator	Heat Pump
Main purpose	Cooling	Heating
Useful transfer	Q_c	Q_h
Work required	W	W
Cold-side energy	Removed	Collected
Hot-side energy	Rejected	Useful
COP	Q_c/W	Q_h/W
Example	Kitchen refrigerator	Home heating system

The underlying physical device can be almost identical.

---

Refrigerator vs Air Conditioner

A refrigerator and air conditioner are also fundamentally similar.

Refrigerator

Cools:

small insulated compartment

Air conditioner

Cools:

room or building

Both transfer thermal energy from a colder region to warmer surroundings using external work.

---

Heat Pump vs Electric Heater

Resistance heater

Electrical energy is directly converted into thermal energy.

Approximately:

1 J electrical → 1 J heat

Heat pump

Electrical energy operates the compressor while additional energy is transferred from the environment.

For example:

1 J electrical + 3 J environment → 4 J indoor heating

Therefore the heat pump may have:

COP > 1

---

COP Is Not Efficiency

This distinction is important.

Efficiency is generally:

useful energy output / energy input

and cannot exceed 1 for an ordinary energy-conversion device.

COP measures:

desired thermal-energy transfer / work input

Because much of the transferred energy comes from another reservoir rather than being created from the work input:

COP can exceed 1

There is no violation of energy conservation.

---

Worked Example: Complete Refrigerator Problem

A refrigerator removes:

Q_c = 2400 J

from its interior and requires:

W = 800 J

Find Q_h

Q_h = Q_c + W

Q_h = 2400 + 800

Q_h = 3200 J

Find COP_R

COP_R = Q_c/W

COP_R = 2400/800

COP_R = 3.0

Therefore:

• energy removed = 2400 J
• work input = 800 J
• energy released = 3200 J
• refrigerator COP = 3.0

---

Worked Example: Heat Pump

A heat pump has:

COP_HP = 4.5

and uses:

W = 2000 J

Find Q_h.

COP_HP = Q_h/W

Therefore:

Q_h = COP_HP × W

Q_h = 4.5 × 2000

Q_h = 9000 J

Now find Q_c:

Q_c = Q_h − W

Q_c = 9000 − 2000

Q_c = 7000 J

The heat pump transfers 7000 J from outside and uses 2000 J of work to deliver 9000 J indoors.

---

Worked Example: Finding Work

A refrigerator removes:

3600 J

and has:

COP_R = 4.0

Use:

COP_R = Q_c/W

Rearrange:

W = Q_c/COP_R

W = 3600/4

W = 900 J

---

Worked Example: Comparing Refrigerators

Refrigerator A:

Q_c = 1200 J

W = 400 J

COP_A = 1200/400

COP_A = 3

Refrigerator B:

Q_c = 1600 J

W = 400 J

COP_B = 1600/400

COP_B = 4

Therefore:

Refrigerator B has the higher COP

because it removes more thermal energy for the same work input.

---

Common Mistakes

Mistake: Saying refrigerators create cold.
They remove thermal energy from a colder region.

Mistake: Saying heat naturally moves from cold to hot.
External work is required for sustained cold-to-hot transfer.

Mistake: Forgetting the compressor requires energy.
The compressor supplies the work needed to drive the cycle.

Mistake: Saying Q_h = Q_c.
Actually, Q_h = Q_c + W.

Mistake: Confusing refrigerator and heat-pump COP.
For a refrigerator, COP_R = Q_c/W. For a heat pump, COP_HP = Q_h/W.

Mistake: Saying COP > 1 violates energy conservation.
COP measures transferred energy relative to work input, not energy created.

Mistake: Treating COP as percentage efficiency.
A COP of 4 is normally written as 4, not 400% efficiency.

Mistake: Saying a refrigerator with its door open cools a closed room.
It adds net thermal energy to the room because of compressor work.

Mistake: Saying cold outdoor air contains no thermal energy.
Matter above absolute zero contains internal energy.

Mistake: Saying heat pumps and refrigerators use completely different principles.
They use essentially the same thermodynamic cycle but have different desired outputs.

---

Key Terms

Refrigerator — device using work to remove thermal energy from a colder region and transfer it to warmer surroundings.

Heat pump — device using work to transfer thermal energy into a warmer region for heating.

Refrigerant — working fluid circulating through a refrigeration or heat-pump system.

Evaporator — component where refrigerant absorbs thermal energy and evaporates.

Compressor — component that performs work on refrigerant, increasing its pressure and temperature.

Condenser — component where refrigerant releases thermal energy and condenses.

Expansion valve — device that lowers refrigerant pressure before it enters the evaporator.

Cold reservoir — low-temperature region from which Q_c is removed.

Hot reservoir — high-temperature region receiving Q_h.

Coefficient of performance (COP) — ratio of desired thermal-energy transfer to required work input.

COP_R — refrigerator coefficient of performance, Q_c/W.

COP_HP — heat-pump coefficient of performance, Q_h/W.

Vapor-compression cycle — common refrigeration cycle using evaporation, compression, condensation and expansion.

Air-source heat pump — heat pump exchanging energy with outdoor air.

Ground-source heat pump — heat pump exchanging energy with the ground.

Latent heat — energy transferred during a phase change without a temperature change at constant pressure for a pure substance.

---

Key Takeaways

• Refrigerators and heat pumps transfer thermal energy from colder regions to warmer regions.
• Cold-to-hot heat transfer requires external work.
• Most household systems obtain this work from electrical energy.
• A refrigerator removes Q_c from its cold interior.
• Work W is supplied to the compressor.
• The condenser releases Q_h to the warmer surroundings.
• Energy conservation gives Q_h = Q_c + W.
• The evaporator absorbs thermal energy.
• The compressor raises refrigerant pressure and temperature.
• The condenser releases thermal energy.
• The expansion valve lowers refrigerant pressure and temperature.
• Refrigerators use phase changes of a refrigerant to transport thermal energy effectively.
• A refrigerator's useful effect is removing Q_c.
• A heat pump's useful effect is delivering Q_h.
• Refrigerators and heat pumps therefore use essentially the same physics.
• Air conditioners operate as refrigeration systems for buildings.
• Reversible heat pumps can provide both heating and cooling.
• Refrigerator performance is measured using COP_R = Q_c/W.
• Heat-pump performance is measured using COP_HP = Q_h/W.
• For the same device, COP_HP = COP_R + 1.
• COP can be greater than 1 without violating energy conservation.
• COP is not the same as efficiency.
• Smaller temperature differences generally allow higher COP values.
• Ideal reversible COP values depend on the hot and cold reservoir temperatures.
• Real devices have lower COP because of irreversibility.
• Heat pumps can deliver several joules of thermal energy for each joule of electrical work consumed.
• Household applications include refrigerators, freezers, air conditioners, space-heating heat pumps and heat-pump water heaters.
• Leaving a refrigerator door open cannot cool a closed room; it causes net heating.
• Insulation reduces unwanted thermal-energy transfer and lowers refrigeration energy use.
• Refrigerators and heat pumps obey both the First and Second Laws of Thermodynamics.

Check Your Understanding

1. What is the main purpose of a refrigerator?
2. Why is work required to operate a refrigerator?
3. What do Q_c, Q_h and W represent?
4. Write the energy equation for a refrigerator.
5. Name the four main components of a vapor-compression refrigeration system.
6. What happens to the refrigerant in the evaporator?
7. What does the compressor do to the refrigerant?
8. Why can the refrigerant release energy to the room after compression?
9. What happens in the condenser?
10. What is the purpose of the expansion valve?
11. Explain the main difference between a refrigerator and a heat pump.
12. Write the equation for refrigerator COP.
13. Write the equation for heat-pump COP.
14. A refrigerator removes 1500 J while requiring 500 J of work. Calculate its COP.
15. For Question 14, calculate the energy released to the room.
16. A heat pump has COP_HP = 4 and uses 1500 J of electrical work. Calculate the energy delivered to the building.
17. Explain why a COP greater than 1 does not violate energy conservation.
18. Explain why a heat pump generally has a lower COP when the outdoor temperature becomes much colder.
19. Predict what happens to the temperature of a closed room if a refrigerator is operated with its door open. Explain your answer.
20. Compare a refrigerator, heat pump and heat engine in terms of the directions of heat transfer and work.

 

 

 

4. Real Engines

Real Engines

Engines are among the most important applications of thermodynamics. They power cars, trucks, ships, aircraft, generators, agricultural machinery, and industrial equipment.

An engine converts energy from one form into another. In a typical internal combustion engine, the chemical energy stored in fuel is released through combustion. Some of this energy is converted into useful mechanical work, while the remainder is transferred to the surroundings.

Although ideal thermodynamic models help us understand the maximum possible performance of heat engines, real engines are never perfectly efficient. Friction, heat transfer, exhaust gases, incomplete combustion, and mechanical limitations all reduce the amount of useful work produced.

Understanding real engines requires knowledge of energy conservation, thermodynamic cycles, pressure, temperature, work, power, and efficiency.

In this lesson, we will investigate how internal combustion engines operate, compare gasoline and diesel engines, explain why real engines are less efficient than ideal engines, identify major energy losses, and evaluate engine performance using mathematical calculations.

What Is an Internal Combustion Engine?

An internal combustion engine (ICE) is a heat engine in which fuel burns inside the engine, producing high-temperature, high-pressure gases that perform mechanical work.

In most automotive engines, the expanding gases push a piston inside a cylinder. The piston is connected to a crankshaft, which converts the piston's back-and-forth motion into rotational motion.

This rotation can drive a vehicle's wheels or operate machinery.

Main Components

Energy Transformations

The principal energy transformations are:

Chemical energy in fuel → Thermal energy from combustion → Mechanical work

However, not all the fuel's chemical energy becomes useful work.

Some energy is transferred to:

• Exhaust gases.

• Engine coolant.

• Surrounding air.

• Frictional heating.

• Sound and vibration.

• Auxiliary equipment.

The law of conservation of energy requires that all energy entering the engine must be accounted for.

The Four-Stroke Engine Cycle

Most conventional car engines operate using a four-stroke cycle.

A stroke is one movement of the piston from one end of the cylinder to the other.

The four strokes are intake, compression, power, and exhaust.

A complete four-stroke cycle requires two crankshaft revolutions, or 720° of rotation.

Each cylinder produces one power stroke during these two revolutions.

In a multi-cylinder engine, the power strokes are staggered to provide smoother operation.

Gasoline Engines

A gasoline engine generally operates using spark ignition.

Air and fuel are mixed, compressed, and ignited by a spark plug.

The resulting combustion increases the temperature and pressure of the gases inside the cylinder.

These gases expand and push the piston downward.

The Ideal Otto Cycle

The ideal Otto cycle is a simplified thermodynamic model commonly associated with spark-ignition engines.

It consists of four idealised processes:

1. Adiabatic, reversible (isentropic) compression.

2. Constant-volume heat addition.

3. Isentropic expansion.

4. Constant-volume heat rejection.

The ideal Otto cycle is a closed-system model. Actual engines exchange gases with their surroundings and do not follow these processes exactly.

Pressure–Volume Diagram

Otto Cycle Efficiency

For an ideal Otto cycle with a constant specific heat ratio:

η = 1 − 1/r^(γ − 1)

where:

• η = thermal efficiency.

• r = compression ratio.

• γ = ratio of specific heats, Cₚ/Cᵥ.

For air near room temperature, γ is approximately 1.4. In real engines, the effective value changes with temperature and gas composition.

The compression ratio is:

r = Vₘₐₓ / Vₘᵢₙ

A higher compression ratio generally increases ideal Otto-cycle efficiency.

However, gasoline engines are limited by abnormal combustion, particularly knock, as well as temperature, pressure, and material constraints.

Worked Example 1: Otto Cycle Efficiency

An ideal Otto-cycle engine has a compression ratio of 9.

Assume γ = 1.4.

Calculate the theoretical thermal efficiency.

Solution

η = 1 − 1/r^(γ − 1)

Substitute:

η = 1 − 1/9^0.4

9^0.4 ≈ 2.408

η = 1 − 1/2.408

η ≈ 0.5847

Answer: Approximately 58.5%

This is an ideal-cycle efficiency, not the actual brake thermal efficiency of a gasoline engine.

Diesel Engines

A diesel engine generally uses compression ignition.

Unlike a typical gasoline engine, a diesel engine does not rely on a spark plug to ignite the main fuel charge.

Instead:

1. Air enters the cylinder.

2. The air is strongly compressed.

3. Compression raises its temperature.

4. Diesel fuel is injected into the hot compressed air.

5. The fuel ignites and combustion occurs.

Diesel engines commonly operate at higher compression ratios than gasoline engines.

The Ideal Diesel Cycle

The ideal Diesel cycle consists of:

1. Isentropic compression.

2. Constant-pressure heat addition.

3. Isentropic expansion.

4. Constant-volume heat rejection.

Actual diesel combustion does not occur entirely at constant pressure. The ideal Diesel cycle is a mathematical approximation.

Diesel Cycle Efficiency

For an ideal Diesel cycle with constant specific heats:

η = 1 − [1/r^(γ − 1)] × [(ρ^γ − 1)/(γ(ρ − 1))]

where:

• r = compression ratio.

• ρ = cutoff ratio.

• γ = specific heat ratio.

The cutoff ratio is the volume at the end of ideal heat addition divided by the volume at its beginning.

For the same compression ratio and γ, the ideal Otto cycle has a higher efficiency than the ideal Diesel cycle when ρ > 1.

However, real diesel engines often achieve higher fuel efficiency than comparable conventional gasoline engines because they can operate with higher compression ratios and generally have lower throttling losses.

Comparing Gasoline and Diesel Engines

These are broad tendencies, not universal rules. Modern engine technologies, including turbocharging, direct injection, hybridisation, variable valve timing, and emissions aftertreatment, can significantly change performance.

Why Are Real Engines Less Efficient?

Ideal thermodynamic cycles assume simplified, reversible processes.

Real engines operate under conditions that introduce irreversible energy transfers and mechanical losses.

The main reasons for reduced efficiency include:

Friction

Moving components experience friction.

Examples include:

• Piston rings sliding against cylinder walls.

• Crankshaft bearings.

• Valve mechanisms.

• Oil pumps and other moving parts.

Some mechanical energy is converted into thermal energy through friction.

Heat Transfer to the Surroundings

Combustion gases reach high temperatures.

Energy is transferred to the cylinder walls, cylinder head, piston, coolant, and surrounding air.

This energy is generally unavailable for useful shaft work.

Exhaust Energy

Exhaust gases leave the engine at temperatures and pressures above ambient conditions.

They carry thermal energy and, in some cases, pressure energy that has not been converted into useful work.

Turbochargers can recover part of the exhaust gas energy to drive a compressor, although this does not eliminate exhaust losses.

Incomplete Combustion

Not all fuel necessarily burns completely.

Incomplete combustion can produce carbon monoxide, unburned hydrocarbons, and soot.

Some of the fuel's chemical energy remains in these products.

Pumping Losses

Engines must move air into the cylinders and exhaust gases out.

This requires work.

In throttled gasoline engines, restrictions to intake airflow can cause significant pumping losses, especially at low loads.

Auxiliary Equipment

Some useful engine output is consumed by systems such as:

• Oil pumps.

• Coolant pumps.

• Alternators.

• Cooling fans.

• Fuel delivery equipment.

These reduce the power available at the output shaft or to the driven load.

Non-Ideal Combustion and Expansion

Real combustion takes a finite amount of time.

The heat release does not occur exactly according to ideal constant-volume or constant-pressure processes.

Other effects include:

• Heat transfer during compression and expansion.

• Gas leakage past piston rings.

• Changing gas properties.

• Irreversible flow.

• Pressure losses.

• Combustion occurring away from the most favourable crank angle.

These effects further reduce performance.

Energy Flow Through a Real Engine

The chemical energy supplied by fuel is distributed among useful work and several forms of energy loss.

5. Efficiency and Sustainability

Real Engines

A real engine converts energy from a fuel or another energy source into useful mechanical work under practical operating conditions.

A common example is the internal combustion engine, where fuel burns inside the engine itself. The hot gases produced by combustion expand and exert forces on pistons or turbine components.

Two important types of internal combustion engine are:

• gasoline engines
• diesel engines

Unlike an ideal Carnot engine, real engines contain friction, rapid heat transfer, turbulence, combustion, exhaust losses and other irreversible processes. Their efficiency is therefore always below the maximum theoretical Carnot efficiency.

 

 

 

5

Internal Combustion Engines

In an internal combustion engine, fuel burns inside a combustion chamber.

The general energy pathway is:

chemical energy in fuel

↓

thermal energy from combustion

↓

expanding hot gases

↓

mechanical work

The expanding gases exert pressure on a piston.

The piston moves and turns a:

crankshaft

which produces rotational mechanical motion.

Main Engine Components

A typical piston engine contains:

• cylinder
• piston
• connecting rod
• crankshaft
• intake valve
• exhaust valve
• combustion chamber
• fuel system
• cooling system
• lubrication system

Gasoline engines also normally use:

spark plugs

Diesel engines rely primarily on:

compression ignition

rather than spark plugs.

The Piston and Cylinder

The piston moves up and down inside the cylinder.

Gas pressure acting on the piston produces a force.

This linear piston motion is transferred through the connecting rod to the crankshaft.

The crankshaft converts:

reciprocating motion → rotational motion

This rotation can ultimately drive the vehicle's wheels.

The Four-Stroke Cycle

Most modern gasoline and diesel piston engines operate using a four-stroke process:

intake → compression → power → exhaust

 

Each stroke corresponds to movement of the piston through the cylinder.

A complete four-stroke cycle requires:

two crankshaft revolutions

Intake Stroke

During the intake stroke:

• piston moves downward
• intake valve opens
• material enters the cylinder

In a conventional gasoline engine, this is typically an air-fuel mixture or air with fuel added by injection.

In many diesel engines:

air enters first

and fuel is injected later.

Compression Stroke

The valves close.

The piston moves upward and compresses the contents of the cylinder.

Compression causes:

• volume to decrease
• pressure to increase
• temperature to increase

The compression stage prepares the mixture or air for combustion.

Power Stroke

During the power stroke, combustion rapidly increases the temperature and pressure of the gases.

The high-pressure gases push the piston downward.

This produces:

mechanical work

The power stroke is the main stage in which chemical energy from the fuel ultimately produces useful mechanical output.

Exhaust Stroke

The exhaust valve opens.

The piston moves upward.

Combustion products are forced out of the cylinder.

The cycle can then begin again.

The exhaust gases still contain significant:

thermal energy

which represents one important pathway by which energy leaves the engine without becoming useful shaft work.

Gasoline Engines

A gasoline engine generally uses spark ignition.

During operation:

• air and fuel are introduced
• the mixture is compressed
• a spark plug creates a spark
• combustion begins
• expanding gases push the piston

 

 

 

7

The timing of ignition is carefully controlled so combustion produces useful pressure at the appropriate point in the cycle.

Spark Ignition

The spark plug produces an electrical discharge across a small gap.

This initiates combustion of the compressed air-fuel mixture.

The combustion front spreads through the chamber.

Pressure rises rapidly and drives the piston during the power stroke.

Diesel Engines

A diesel engine normally uses:

compression ignition

Rather than using a spark plug to ignite a premixed charge, the engine strongly compresses air.

This raises the air temperature.

Fuel is then injected into the hot compressed air.

The fuel ignites because the air temperature is sufficiently high.

 

 

 

8

Diesel Compression

Diesel engines typically operate with a higher compression ratio than conventional gasoline engines.

Compression ratio compares the cylinder's maximum volume with its minimum volume:

compression ratio = V_max/V_min

A higher compression ratio generally allows greater thermodynamic efficiency, although real engine design involves many additional factors.

Gasoline vs Diesel

These are broad comparisons; modern engine designs vary considerably.

Why Diesel Engines Can Be More Efficient

Several factors can contribute to diesel-engine efficiency.

Diesel engines commonly use:

• higher compression ratios
• leaner operation
• reduced throttling losses
• combustion strategies suited to high-pressure operation

Higher compression ratios can increase the amount of useful work obtained from the expanding gases.

However:

diesel engines are still real engines

and therefore remain far below perfect efficiency.

The Otto Cycle

The idealized thermodynamic model commonly associated with a conventional gasoline engine is the:

Otto cycle

 

 

 

8

The real gasoline engine does not exactly follow the ideal Otto cycle.

The model simplifies processes such as:

• compression
• combustion
• expansion
• exhaust

It allows physicists and engineers to study the thermodynamic principles without modeling every detail of a real engine.

The Diesel Cycle

The idealized model commonly associated with a diesel engine is the:

Diesel cycle

Like the Otto cycle, it is a simplified thermodynamic model.

Real diesel engines differ from the ideal model because combustion and gas flow occur over finite times and involve many irreversible processes.

Real Engines vs Ideal Engines

An ideal engine may assume:

• no friction
• reversible processes
• perfect insulation where needed
• controlled heat transfer
• ideal gas behavior
• no mechanical losses

A real engine contains:

• friction
• turbulence
• rapid combustion
• heat transfer
• exhaust flow
• pumping losses
• incomplete idealization of compression and expansion

Therefore:

η_real < η_ideal

Efficiency of a Real Engine

Engine efficiency can be expressed as:

η = useful energy output / total energy input

For an internal combustion engine:

η = W_useful/E_fuel

where:

• W_useful = useful mechanical work
• E_fuel = chemical energy supplied by the fuel

Percentage efficiency:

η = (W_useful/E_fuel) × 100%

Worked Example: Engine Efficiency

An engine receives:

E_fuel = 5000 J

and produces:

W_useful = 1750 J

Then:

η = 1750/5000

η = 0.35

Therefore:

η = 35%

This means:

35% of the input energy becomes useful mechanical output

The remaining 65% has not disappeared.

It has been transferred through other pathways.

Where Does the Other Energy Go?

Energy leaving an engine may be transferred through:

• hot exhaust gases
• engine cooling system
• friction
• thermal radiation
• sound
• pumping gases
• mechanical accessories

 

 

 

7

Energy is conserved.

The issue is that much of it becomes:

dispersed thermal energy

rather than useful mechanical work.

Exhaust Losses

Exhaust gases leave the engine at relatively high temperatures.

Therefore they carry away:

internal energy

This energy is no longer available to produce work in the piston engine unless additional technology is used to recover some of it.

Hot exhaust is often one of the major energy-loss pathways in combustion engines.

Cooling Losses

Combustion produces extremely high temperatures.

Engine components cannot safely remain at those temperatures.

A cooling system therefore removes energy from:

• cylinder walls
• cylinder head
• engine block

This protects the engine from:

• overheating
• deformation
• lubricant breakdown
• material failure

However, energy transferred to the cooling system is generally unavailable as useful shaft work.

Engine Cooling System

Many vehicle engines use liquid cooling.

A coolant circulates through passages around the engine.

The coolant absorbs thermal energy and transports it to a:

radiator

The radiator transfers energy to the surrounding air.

 

 

 

7

Friction Losses

Engine components move against one another.

Examples include:

• pistons against cylinder walls
• crankshaft bearings
• camshaft components
• valve mechanisms

Friction converts mechanical energy into:

thermal energy

This reduces useful output and produces entropy.

Lubrication

Engine oil reduces friction by forming a layer between moving surfaces.

Lubrication helps:

• reduce friction
• reduce wear
• remove some heat
• increase component life

Reducing friction can improve engine efficiency.

However, friction can never be completely eliminated in a practical engine.

Pumping Losses

An engine must move gases:

into and out of cylinders

This requires work.

Energy is required to:

• draw in air
• compress gases
• push exhaust gases out
• overcome restrictions in intake and exhaust systems

These are called:

pumping losses

Gasoline Engine Throttling Losses

Many gasoline engines control power partly using a:

throttle

At partial load, the throttle restricts incoming air.

The engine must work to draw air past this restriction.

This creates:

pumping losses

Diesel engines often operate with less intake throttling, which can contribute to better efficiency under some operating conditions.

Combustion Losses

Real combustion is not an ideal reversible process.

Combustion occurs:

• rapidly
• at finite temperature differences
• with turbulent gas motion
• through complex chemical reactions

These processes produce:

entropy

Therefore some of the fuel's potential to perform useful work is lost through irreversibility.

Incomplete Combustion

Ideally, hydrocarbon fuel burns completely to form primarily:

CO₂ + H₂O

In real engines, combustion may not always be complete.

Incomplete combustion can produce substances such as:

• carbon monoxide
• unburned hydrocarbons
• soot

This means some chemical energy may leave without being fully converted through combustion.

Sound and Vibration

Engines produce:

• sound
• vibration

These energy transfers usually represent only part of the total energy balance, but they are still energy pathways that do not contribute to useful vehicle motion.

Eventually much of this energy becomes:

thermal energy in the surroundings

Accessories

Some engine output is used to operate components such as:

• oil pump
• coolant pump
• alternator
• air-conditioning compressor
• power-assisted systems

This means not all mechanical work produced at the crankshaft reaches the wheels.

Drivetrain Losses

In a vehicle, mechanical energy must pass through components such as:

• transmission
• differential
• bearings
• drive shafts

These components also have friction and other losses.

Therefore:

engine efficiency

and:

overall vehicle efficiency

are not necessarily the same thing.

Why Real Engines Are Less Efficient Than Carnot Engines

A Carnot engine is:

perfectly reversible

A real engine is:

irreversible

Real engines produce entropy through:

• combustion
• friction
• heat transfer across finite temperature differences
• turbulent flow
• mixing
• exhaust expansion
• pressure losses

Therefore real engines cannot reach Carnot efficiency.

The Carnot Limit

For any heat engine operating between hot and cold reservoir temperatures:

η_Carnot = 1 − T_c/T_h

This gives the theoretical maximum efficiency for a reversible engine operating between those temperatures.

Temperatures must be in:

kelvin

Worked Example: Carnot Limit

Suppose an engine operates approximately between:

T_h = 1000 K

and:

T_c = 300 K

Then:

η_Carnot = 1 − 300/1000

η_Carnot = 0.70

η_Carnot = 70%

If the real engine operates at:

35%

then:

35% < 70%

which is consistent with the Second Law.

Carnot Efficiency Is Not the Expected Real Efficiency

If the Carnot limit is 70%, this does not mean engineers should expect the real engine automatically to reach 70%.

Carnot efficiency assumes:

completely reversible operation

A real combustion engine cannot meet those conditions.

Therefore Carnot efficiency is best viewed as:

an upper theoretical benchmark

Temperature and Efficiency

In general, increasing the maximum operating temperature can increase the theoretical opportunity for higher efficiency.

However, high temperatures create engineering problems.

Materials may:

• weaken
• oxidize
• deform
• melt
• lose protective coatings

Lubricants may also degrade.

Therefore practical engine temperature is limited by:

materials and design

Compression Ratio and Efficiency

Increasing compression ratio can improve the theoretical efficiency of certain engine cycles.

A greater compression ratio can allow more useful work to be extracted during expansion.

However, compression ratio cannot simply be increased without limit.

Potential problems include:

• abnormal combustion
• high mechanical stresses
• excessive temperatures
• emissions
• material limitations

Engine design requires compromises.

Knock in Gasoline Engines

In gasoline engines, excessive pressure and temperature can cause portions of the air-fuel mixture to ignite in an uncontrolled manner.

This is associated with:

engine knock

Knock can:

• reduce performance
• increase mechanical stress
• damage components

Fuel properties, compression ratio, ignition timing and engine controls are therefore important.

Octane Rating

Gasoline's octane rating describes its resistance to unwanted autoignition under specified test conditions.

Higher-octane fuel can support engine designs or operating conditions with greater resistance to knock.

It does not simply mean:

the fuel contains more energy

Octane rating primarily relates to:

knock resistance

Diesel Fuel and Cetane

Diesel fuels can be characterized using a:

cetane number

This relates to ignition behavior in compression-ignition engines.

A diesel engine requires fuel to ignite reliably after injection into hot compressed air.

Thus gasoline and diesel fuels are optimized for different combustion processes.

Turbochargers

A turbocharger uses energy from exhaust gases to compress incoming air.

 

 

 

7

Hot exhaust drives a turbine.

The turbine drives a compressor.

The compressor forces more air into the engine.

This can allow:

• more fuel to be burned
• greater power from a smaller engine
• improved performance
• recovery of some otherwise unused exhaust energy

Turbocharging and Efficiency

A turbocharger does not recover all exhaust energy.

However, it can use some energy that would otherwise leave directly with the exhaust.

Modern engine designs may combine:

• turbocharging
• direct fuel injection
• variable valve timing
• advanced combustion control

to improve performance and efficiency.

Waste Heat Recovery

Some systems attempt to recover additional energy from:

hot exhaust

For example, exhaust energy can potentially be used to:

• drive turbochargers
• generate electrical energy
• heat other fluids
• support combined thermodynamic cycles

Waste-heat recovery can increase overall system efficiency.

Engine Performance

Efficiency is only one measure of engine performance.

Engineers may also consider:

• power
• torque
• fuel consumption
• emissions
• reliability
• mass
• durability
• cost
• noise
• operating range

A high-performance engine must often balance several competing requirements.

Power

Power is the rate at which work is done.

P = W/t

where:

• P = power
• W = work
• t = time

SI unit:

watt (W)

An engine that performs more work each second produces greater power.

Efficiency vs Power

A more powerful engine is not necessarily:

more efficient

For example:

Engine A may produce:

200 kW

at 30% efficiency.

Engine B may produce:

100 kW

at 40% efficiency.

Engine A produces more power.

Engine B converts a larger fraction of its input energy into useful output.

Therefore:

power ≠ efficiency

Torque

Torque describes the turning effect of a force.

For a force acting perpendicular to a lever arm:

τ = Fr

where:

• τ = torque
• F = force
• r = perpendicular distance from the axis

SI unit:

N·m

Engines produce torque at the crankshaft.

Torque and Power

For rotational motion:

P = τω

where:

• P = power
• τ = torque
• ω = angular velocity

This means engine power depends on both:

torque

and:

rotational speed

An engine can produce high torque at relatively low rotational speed or lower torque at high rotational speed and still produce substantial power.

Fuel Consumption

Fuel consumption provides another way of evaluating an engine.

A more efficient engine generally requires less fuel to produce a given amount of useful work.

However, actual vehicle fuel consumption also depends on:

• vehicle mass
• speed
• aerodynamic drag
• tire resistance
• driving behavior
• road gradient
• transmission
• accessories

Therefore engine efficiency and vehicle fuel economy are related but not identical.

Evaluating Two Engines

Suppose Engine A receives:

100 MJ

of fuel energy and produces:

35 MJ

of useful work.

η_A = 35/100

η_A = 35%

Engine B receives:

80 MJ

and produces:

32 MJ

of useful work.

η_B = 32/80

η_B = 40%

Engine B is more efficient even though Engine A produces more total useful energy in this example.

Energy-Flow Evaluation

Suppose a gasoline engine receives:

100 units

of fuel energy.

Imagine:

• 35 units → useful mechanical work
• 30 units → exhaust
• 25 units → cooling
• 7 units → friction and accessories
• 3 units → other transfers

Total:

35 + 30 + 25 + 7 + 3 = 100

Efficiency:

η = 35/100

η = 35%

This type of energy-flow analysis helps engineers identify where improvements may be possible.

Sankey Diagrams

A Sankey diagram can represent engine energy transfers.

The width of each arrow represents the amount of energy transferred along that pathway.

 

 

 

8

A Sankey diagram can show:

fuel energy → useful work + exhaust + cooling + friction

It provides a useful visual tool for evaluating engine performance.

Evaluating Improvements

Suppose engineers reduce friction losses from:

10% → 6%

If other energy transfers remain unchanged, more input energy may become available for useful work.

But improving one component can sometimes affect others.

For example:

• higher temperatures may improve thermodynamic efficiency
• but increase cooling requirements
• lighter components may reduce mass
• but reduce durability
• emissions controls may affect exhaust flow

Engineering involves:

trade-offs

Emissions and Performance

Engine performance cannot be judged solely by efficiency.

Combustion engines can produce pollutants including:

• carbon monoxide
• nitrogen oxides
• unburned hydrocarbons
• particulate matter

They also commonly produce:

CO₂

when carbon-containing fuels burn.

Therefore engine evaluation increasingly considers both:

energy performance and environmental impact

Catalytic Converters

Gasoline vehicles commonly use catalytic converters to reduce harmful exhaust pollutants.

They promote chemical reactions that reduce emissions such as:

• carbon monoxide
• unburned hydrocarbons
• nitrogen oxides

Catalytic converters improve emissions performance but do not eliminate all environmental impacts of fuel combustion.

Diesel Emission Controls

Modern diesel vehicles may use technologies such as:

• diesel particulate filters
• oxidation catalysts
• selective catalytic reduction

These systems help reduce pollutants.

However, they add:

• complexity
• cost
• maintenance requirements

Again, real engineering involves balancing multiple objectives.

Gasoline vs Diesel Performance

Gasoline engines often offer:

• relatively light construction
• high rotational speeds
• smooth operation
• strong power-to-mass performance

Diesel engines often offer:

• higher thermal efficiency
• strong low-speed torque
• durability
• efficient operation under heavy loads

Neither engine type is automatically "better."

The appropriate choice depends on the:

application

Passenger Cars

Passenger cars may prioritize:

• low mass
• smooth operation
• acceleration
• low noise
• low emissions
• efficiency

These factors historically contributed to widespread gasoline-engine use in many passenger vehicles.

Trucks and Heavy Equipment

Heavy vehicles often require:

• high torque
• long operating life
• efficiency under sustained load
• durability

These characteristics have made diesel engines common in:

• trucks
• buses
• construction equipment
• agricultural machinery

Ships and Generators

Large diesel engines are also widely used in:

• ships
• stationary generators
• industrial machinery

Large engines can achieve high efficiencies compared with many smaller internal combustion engines, although they remain subject to Second Law limitations.

Hybrid Vehicles

Hybrid vehicles combine an internal combustion engine with:

• electric motor
• battery
• electronic control systems

The electric system can help operate the combustion engine closer to efficient conditions.

It can also recover some vehicle kinetic energy through:

regenerative braking

Regenerative Braking

Traditional friction braking converts kinetic energy mainly into:

thermal energy

Regenerative braking instead converts some kinetic energy into:

electrical energy

which can be stored in a battery.

This improves overall vehicle energy efficiency.

However:

not all kinetic energy can be recovered

because real systems contain irreversible losses.

Real Engines and Entropy

Real engines produce entropy.

Important sources include:

• combustion
• friction
• mixing
• heat transfer across temperature differences
• exhaust flow
• turbulence

Therefore:

ΔS_total > 0

for real engine operation.

This entropy production is one fundamental reason why not all fuel energy can become useful work.

The First and Second Laws

The First Law tells us:

energy is conserved

If fuel supplies 100 MJ, all 100 MJ must eventually be accounted for.

The Second Law tells us:

not all of that energy can be converted into useful work

Some becomes dispersed in forms that are less useful for performing mechanical work.

Real Engines vs Carnot Engines

Evaluating an Engine

When evaluating a real engine, ask:

What is the energy input?

How much useful work is produced?

What is the efficiency?

Where does the remaining energy go?

How much power does it produce?

What torque does it produce?

How much fuel does it consume?

What pollutants are produced?

How reliable is it?

How expensive is it to operate?

A complete evaluation considers much more than a single number.

Worked Example: Engine Energy Balance

An engine receives:

E_in = 20 MJ

It produces:

W = 7 MJ

Efficiency

η = W/E_in

η = 7/20

η = 0.35 = 35%

Energy transferred elsewhere

E_other = 20 − 7

E_other = 13 MJ

Therefore:

13 MJ

is transferred through exhaust, cooling, friction and other pathways.

Worked Example: Power

An engine produces:

600 kJ

of useful work in:

4 s

Use:

P = W/t

P = 600,000/4

P = 150,000 W

Therefore:

P = 150 kW

Worked Example: Fuel Comparison

Engine A:

• fuel input = 50 MJ
• useful output = 15 MJ

η_A = 15/50

η_A = 30%

Engine B:

• fuel input = 50 MJ
• useful output = 20 MJ

η_B = 20/50

η_B = 40%

Engine B produces:

5 MJ more useful work

from the same fuel-energy input.

Therefore Engine B is more efficient.

Worked Example: Real vs Carnot

A real engine operates between approximate temperatures:

T_h = 1200 K

T_c = 300 K

Carnot limit:

η_Carnot = 1 − 300/1200

η_Carnot = 0.75

η_Carnot = 75%

Suppose the real engine efficiency is:

40%

This is physically reasonable because:

40% < 75%

The difference reflects the fact that the real engine contains substantial irreversibility.

Common Mistakes

Mistake: Saying gasoline and diesel engines ignite fuel in exactly the same way.
Gasoline engines normally use spark ignition; diesel engines use compression ignition.

Mistake: Saying diesel engines have spark plugs for normal ignition.
Normal diesel combustion relies on fuel injection into hot compressed air.

Mistake: Saying unused energy disappears.
Energy is conserved and transferred through exhaust, cooling, friction and other pathways.

Mistake: Saying all heat leaving an engine is simply caused by bad engineering.
Some losses can be reduced, but the Second Law imposes fundamental limits on heat-engine efficiency.

Mistake: Saying the most powerful engine is the most efficient.
Power and efficiency measure different things.

Mistake: Saying higher compression is always better without limits.
Material stresses, combustion behavior, emissions and other factors impose practical constraints.

Mistake: Saying a turbocharger creates free energy.
It recovers some energy from exhaust flow to drive a compressor.

Mistake: Assuming the Carnot efficiency is the expected efficiency of a real engine.
It is the maximum theoretical limit between two temperatures.

Mistake: Treating engine efficiency and overall vehicle efficiency as identical.
The drivetrain and vehicle itself introduce additional energy transfers and losses.

Key Terms

Internal combustion engine — engine in which fuel burns inside the engine's combustion chamber.

Gasoline engine — internal combustion engine normally using spark ignition.

Diesel engine — internal combustion engine normally using compression ignition.

Spark ignition — initiation of combustion using an electrical spark.

Compression ignition — ignition caused when fuel is injected into sufficiently hot compressed air.

Piston — moving component that receives force from expanding gases.

Cylinder — chamber in which the piston moves.

Crankshaft — rotating shaft that converts piston motion into rotational motion.

Compression ratio — ratio V_max/V_min describing the change in cylinder volume during compression.

Otto cycle — idealized thermodynamic cycle commonly used to model spark-ignition engines.

Diesel cycle — idealized thermodynamic cycle commonly used to model compression-ignition engines.

Efficiency — fraction of input energy converted into useful output.

Power — rate at which work is performed.

Torque — turning effect of a force.

Pumping loss — energy required to move gases through an engine.

Friction loss — conversion of mechanical energy into thermal energy through friction.

Waste heat — thermal energy transferred away rather than becoming the desired useful output.

Turbocharger — device using exhaust flow to drive a compressor that increases intake-air pressure.

Sankey diagram — diagram showing energy-transfer pathways using arrow widths proportional to energy amounts.

Regenerative braking — recovery of some vehicle kinetic energy into stored electrical energy.

Key Takeaways

• Internal combustion engines burn fuel inside the engine.
• Chemical energy from fuel ultimately produces mechanical work.
• Most piston engines use intake, compression, power and exhaust strokes.
• A four-stroke engine completes one operating cycle in two crankshaft revolutions.
• Gasoline engines normally use spark ignition.
• Diesel engines normally use compression ignition.
• Diesel engines generally operate at higher compression ratios than conventional gasoline engines.
• Higher compression ratios can contribute to greater thermodynamic efficiency.
• Diesel engines are often more efficient than comparable gasoline engines, although actual performance depends on design and operating conditions.
• Real engines are always less efficient than ideal reversible engines.
• Real engines contain irreversible processes that produce entropy.
• Major energy-transfer pathways include exhaust heat, cooling, friction and pumping losses.
• Energy lost from useful output is not destroyed.
• The First Law requires all input energy to be accounted for.
• The Second Law limits how much input energy can become useful work.
• Engine efficiency is η = useful output/input energy.
• Carnot efficiency provides an upper theoretical benchmark.
• Real engines cannot achieve Carnot efficiency.
• Higher operating temperatures can increase theoretical efficiency but create material and cooling challenges.
• Friction converts useful mechanical energy into dispersed thermal energy.
• Lubrication reduces friction and wear.
• Turbochargers recover some energy from exhaust gases.
• Waste-heat recovery can improve overall system efficiency.
• Power measures the rate of doing work and is not the same as efficiency.
• Torque measures rotational turning effect.
• Engine evaluation should consider efficiency, power, torque, fuel consumption, emissions, reliability and cost.
• Gasoline and diesel engines have different strengths for different applications.
• Hybrid systems can improve overall vehicle efficiency by combining combustion and electric technologies.
• Regenerative braking recovers some energy that would otherwise become thermal energy through friction.
• Real engine design involves trade-offs between efficiency, performance, emissions, durability and cost.

Check Your Understanding

1. Define an internal combustion engine.
2. Name the four strokes of a four-stroke engine in order.
3. What happens during the power stroke?
4. How is piston motion converted into rotational motion?
5. Explain how fuel is normally ignited in a gasoline engine.
6. Explain how fuel is normally ignited in a diesel engine.
7. Give two differences between gasoline and diesel engines.
8. Why can higher compression ratios contribute to improved engine efficiency?
9. Identify four major energy-loss pathways in a real engine.
10. Explain why hot exhaust represents an energy loss from useful engine output.
11. Why does an engine require a cooling system?
12. Explain how friction reduces engine efficiency.
13. An engine receives 5000 J of fuel energy and produces 1800 J of useful work. Calculate its efficiency.
14. Where does the remaining energy in Question 13 go?
15. Explain why a real engine cannot reach Carnot efficiency.
16. Distinguish between engine power and engine efficiency.
17. Explain how a turbocharger can improve engine performance.
18. How can regenerative braking improve the overall efficiency of a vehicle?
19. Compare a real internal combustion engine with an ideal Carnot engine.
20. Two engines have different efficiencies, power outputs, fuel consumption and emissions. Explain what information you would consider when deciding which engine has the better overall performance.