For a steady-running engine, the energy balance can be expressed approximately as:

Fuel energy input = Useful work output + Energy losses

For rates of energy transfer:

P_fuel = P_brake + P_losses

where all terms are measured in watts.

Thermal Efficiency

Thermal efficiency measures the fraction of supplied thermal or fuel energy converted into useful work.

For a real engine, brake thermal efficiency is commonly defined as:

η_b = Useful shaft power / Fuel energy input rate

Therefore:

η_b = P_b / (ṁ_f × LHV)

where:

• η_b = brake thermal efficiency.

• P_b = brake power output, in watts.

• ṁ_f = fuel mass flow rate, in kg/s.

• LHV = lower heating value of the fuel, in J/kg.

The lower heating value excludes the energy that would be recovered by condensing the water vapour in the exhaust.

Worked Example 2: Calculating Real Engine Efficiency

An engine receives fuel energy at a rate of 120 kW.

It produces 36 kW of useful shaft power.

Calculate its brake thermal efficiency.

Solution

η_b = P_b / P_fuel

η_b = 36 / 120

η_b = 0.30

Percentage efficiency:

η_b = 30%

Answer: 30%

The remaining 70% of the input energy is not converted into useful shaft power.

It leaves through exhaust, heat transfer, frictional effects, and other losses.

Work Done by Expanding Gases

When gases expand against a piston, they can perform mechanical work.

For a constant external pressure:

W = PΔV

where:

• W = work done, in joules.

• P = pressure, in pascals.

• ΔV = change in volume, in cubic metres.

For variable pressure:

W = ∫ P dV

The work corresponds to the area under the pressure–volume curve for a quasistatic process.

For a complete cycle, the net work is represented by the area enclosed by the cycle on a P–V diagram.

Worked Example 3: Work During Expansion

A gas expands from 0.002 m³ to 0.006 m³ at a constant pressure of 300 kPa.

Calculate the work done.

Solution

W = PΔV

Convert pressure:

300 kPa = 300,000 Pa

Change in volume:

ΔV = 0.006 − 0.002

ΔV = 0.004 m³

Therefore:

W = 300,000 × 0.004

Answer: W = 1,200 J

The gas performs 1,200 J of work during expansion.

This calculation assumes constant pressure throughout the process.

Engine Power

Power is the rate at which work is done.

P = W/t

where:

• P = power, in watts.

• W = work, in joules.

• t = time, in seconds.

For rotating machinery, power can also be calculated using:

P = τω

where:

• τ = torque, in N·m.

• ω = angular velocity, in rad/s.

If engine speed is measured in revolutions per minute (rpm):

ω = 2πN/60

Therefore:

P = 2πτN/60

Worked Example 4: Engine Power from Torque

An engine produces a torque of 200 N·m at 3,000 rpm.

Calculate its mechanical power output.

Solution

P = 2πτN/60

Substitute:

P = 2π(200)(3,000)/60

P = 20,000π

P ≈ 62,832 W

Convert to kilowatts:

P ≈ 62.8 kW

Answer: Approximately 62.8 kW

This is the power corresponding to the stated torque and rotational speed.

Torque and Engine Performance

Torque is a measure of the turning effect of a force.

τ = Fr

for a force acting perpendicular to a lever arm of length r.

In an engine, torque is produced by forces transmitted from the pistons through the connecting rods to the crankshaft.

Torque Versus Power

Torque and power describe different aspects of engine performance.

Torque describes rotational force.

Power describes the rate of energy transfer or work.

An engine may produce high torque at relatively low rpm but reach maximum power at a higher rpm.

The curves illustrate that maximum torque and maximum power may occur at different engine speeds.

Because the two quantities have different units, their numerical heights should not be directly compared. Each curve should be interpreted according to its own units.

Fuel Consumption and Engine Efficiency

Fuel consumption is another important measure of engine performance.

An engine that uses less fuel to produce the same useful work is generally more fuel-efficient.

Brake Specific Fuel Consumption

Brake specific fuel consumption (BSFC) measures the mass of fuel consumed per unit of useful shaft energy produced.

BSFC = Fuel mass flow rate / Brake power

A common unit is:

g/kWh

Lower BSFC generally indicates better fuel efficiency for a given fuel.

Worked Example 5: Calculating BSFC

An engine consumes 12 kg of fuel per hour while producing 50 kW of shaft power.

Calculate BSFC.

Solution

Convert fuel consumption:

12 kg/h = 12,000 g/h

BSFC = 12,000 / 50

Answer: 240 g/kWh

This means the engine consumes 240 grams of fuel for every kilowatt-hour of shaft energy produced at this operating condition.

Worked Example 6: Calculating Efficiency from BSFC

An engine has a BSFC of 250 g/kWh.

The fuel has a lower heating value of 43 MJ/kg.

Calculate its brake thermal efficiency.

Solution

For 1 kWh of useful output:

Useful energy = 3.6 MJ

Fuel consumed:

250 g = 0.250 kg

Fuel energy supplied:

E_in = 0.250 × 43

E_in = 10.75 MJ

Efficiency:

η = 3.6 / 10.75

η ≈ 0.3349

Answer: Approximately 33.5%

This shows how fuel consumption data can be used to evaluate engine performance.

Why Efficiency Changes with Engine Speed and Load

An engine does not operate at one constant efficiency.

Efficiency changes with:

• Engine speed.

• Load.

• Air-fuel ratio.

• Compression ratio.

• Combustion timing.

• Engine temperature.

• Friction.

• Pumping losses.

• Turbocharger operation.

• Auxiliary power requirements.

At very low loads, fixed friction and pumping losses can represent a large proportion of the useful output.

At high loads, some engines may require changes to combustion timing or fuel mixture that reduce efficiency.

An engine's most efficient operating point is often somewhere between very low and maximum load.

Brake Thermal Efficiency Maps

Engineers use efficiency maps to compare performance across different combinations of speed and torque.

Efficiency maps help engineers choose operating conditions and design transmissions and hybrid systems.

Mechanical Efficiency

Not all power developed by the gases inside the cylinders reaches the output shaft.

Some is lost to friction and the operation of engine components.

Two useful quantities are:

Indicated power (Pᵢ): Power developed by the gases acting on the pistons.

Brake power (Pᵦ): Useful power measured at the engine output shaft.

Mechanical efficiency is:

ηₘ = Pᵦ/Pᵢ

Worked Example 7: Mechanical Efficiency

An engine develops 80 kW of indicated power.

Its measured brake power is 68 kW.

Calculate its mechanical efficiency.

Solution

ηₘ = Pᵦ/Pᵢ

ηₘ = 68/80

ηₘ = 0.85

Answer: 85%

The difference between indicated and brake power is:

80 − 68 = 12 kW

This is the friction and mechanical-loss power under the stated measurement convention.

Comparing Ideal and Real Engines

Ideal models remain valuable because they establish reference behaviour and help engineers understand how changes in compression ratio, temperature, and other parameters affect performance.

The Second Law of Thermodynamics and Engine Efficiency

The first law of thermodynamics states that energy is conserved.

However, energy conservation alone does not explain why engines cannot convert all fuel energy into useful work.

The second law of thermodynamics introduces limits on energy conversion.

A cyclic heat engine operating between a hot reservoir and a cold reservoir cannot convert all the heat it receives into work.

For an ideal reversible engine operating between two constant-temperature reservoirs, the maximum thermal efficiency is the Carnot efficiency:

η_Carnot = 1 − T_c/T_h

where:

• T_c = cold reservoir temperature in kelvin.

• T_h = hot reservoir temperature in kelvin.

This expression is a theoretical limit for the specified reservoirs. A real internal combustion engine does not operate between two constant-temperature reservoirs, so substituting a single combustion temperature into this formula does not directly predict its actual efficiency.

Worked Example 8: Carnot Efficiency

A theoretical heat engine operates between:

T_h = 900 K

T_c = 300 K

Calculate the maximum efficiency.

Solution

η = 1 − T_c/T_h

η = 1 − 300/900

η = 1 − 1/3

η = 2/3

Answer: Approximately 66.7%

A real engine operating under comparable thermal limits would achieve a lower efficiency because of irreversibilities and other losses.

Environmental Performance

Engine performance is not measured only by power and efficiency.

Environmental effects are also important.

Carbon Dioxide

Burning hydrocarbon fuels produces carbon dioxide.

CO₂ is a greenhouse gas that contributes to climate change.

Reducing fuel consumption generally reduces CO₂ emissions for a given fuel and operating task.

Nitrogen Oxides

High combustion temperatures can promote the formation of nitrogen oxides (NOₓ).

These pollutants contribute to air-quality problems.

Particulate Matter

Diesel engines, particularly without effective emissions controls, can produce particulate matter.

Modern diesel particulate filters substantially reduce particulate emissions.

Carbon Monoxide and Unburned Hydrocarbons

Incomplete combustion can produce carbon monoxide and unburned hydrocarbons.

Catalytic converters and other emissions-control systems help reduce these pollutants.

Efficiency Versus Emissions

Improving thermal efficiency does not automatically minimise every pollutant.

For example, combustion conditions that improve efficiency may also increase NOₓ formation.

Engine design therefore involves balancing:

• Efficiency.

• Power.

• Fuel consumption.

• Reliability.

• Manufacturing cost.

• Emissions.

• Noise and vibration.

Technologies That Improve Engine Efficiency

Common Mistakes

Mistake 1: Assuming engines convert all fuel energy into useful work

Real engines lose energy through exhaust gases, cooling, friction, and other processes.

Mistake 2: Confusing gasoline and diesel ignition methods

Typical gasoline engines use spark ignition.

Diesel engines use compression ignition.

Mistake 3: Assuming diesel engines are always more efficient

Diesel engines often achieve higher efficiency, but actual performance depends on engine design and operating conditions.

Mistake 4: Confusing torque with power

Torque measures rotational turning effect.

Power measures the rate of doing work.

Mistake 5: Assuming maximum power occurs at maximum efficiency

An engine may produce maximum power at a different speed and load from its most efficient operating point.

Mistake 6: Using Celsius temperatures in the Carnot efficiency formula

Absolute temperatures in kelvin must be used.

Mistake 7: Confusing ideal-cycle efficiency with brake thermal efficiency

Ideal-cycle efficiency describes a simplified thermodynamic model.

Brake thermal efficiency measures useful shaft output relative to fuel energy input.

Mistake 8: Assuming a higher compression ratio always produces better real-world performance

Higher compression ratios can improve theoretical efficiency but may introduce combustion, mechanical, and thermal limitations.

Mistake 9: Ignoring units

Power, energy, torque, and fuel consumption are different quantities.

Always check unit consistency before calculating.

Mistake 10: Assuming low fuel consumption automatically means low emissions of every pollutant

CO₂ emissions and local air pollutants depend on different factors, including combustion conditions and emissions-control technology.

Key Terms

• Internal combustion engine: An engine in which fuel burns inside the engine to produce mechanical work.

• Heat engine: A device that converts part of a thermal energy input into work while rejecting energy to its surroundings.

• Piston: A moving component that transfers force from expanding gases.

• Cylinder: The chamber in which the piston moves.

• Crankshaft: A rotating shaft that converts reciprocating piston motion into rotation.

• Four-stroke cycle: A sequence of intake, compression, power, and exhaust strokes.

• Spark ignition: Ignition initiated by a spark plug.

• Compression ignition: Ignition caused by fuel entering sufficiently hot compressed air.

• Otto cycle: An idealised thermodynamic cycle associated with spark-ignition engines.

• Diesel cycle: An idealised thermodynamic cycle associated with compression-ignition engines.

• Compression ratio: The ratio of maximum to minimum cylinder volume.

• Cutoff ratio: The volume ratio during constant-pressure heat addition in the ideal Diesel cycle.

• Thermal efficiency: The ratio of useful work output to energy supplied.

• Brake thermal efficiency: Useful shaft power divided by the rate of fuel chemical energy input.

• Indicated power: Power developed by gases acting on the pistons.

• Brake power: Useful mechanical power measured at the output shaft.

• Mechanical efficiency: Brake power divided by indicated power.

• Torque: The turning effect of a force.

• Power: The rate of energy transfer or work.

• Friction loss: Mechanical energy converted into thermal energy through friction.

• Pumping loss: Work required to move gases into and out of an engine.

• Exhaust energy loss: Energy carried away by exhaust gases.

• Brake specific fuel consumption: Fuel mass consumed per unit of useful shaft energy output.

• Lower heating value: Fuel energy content excluding the latent heat recoverable by condensing combustion-produced water.

• Carnot efficiency: The maximum efficiency of a reversible heat engine operating between two constant-temperature reservoirs.

• Turbocharger: A device that uses exhaust gas energy to drive an intake-air compressor.

• Nitrogen oxides: Pollutants formed during combustion, particularly at high temperatures.

• Particulate matter: Small solid or liquid particles emitted in exhaust gases.

• Irreversibility: A real-process effect that reduces the ability to convert energy into useful work.

Key Takeaways

1. Internal combustion engines convert fuel chemical energy into mechanical work.

2. Four-stroke engines operate through intake, compression, power, and exhaust strokes.

3. Gasoline engines generally use spark ignition, while diesel engines use compression ignition.

4. The Otto and Diesel cycles provide idealised thermodynamic models.

5. Real engines are less efficient because of friction, heat transfer, exhaust energy, pumping losses, and non-ideal combustion.

6. Brake thermal efficiency measures useful shaft output relative to fuel energy input.

7. Torque and rotational speed determine mechanical power.

8. BSFC measures fuel consumption per unit of useful shaft energy.

9. Engine efficiency varies with operating speed, load, and design.

10. Improving engine performance involves balancing power, efficiency, emissions, cost, and reliability.

Check Your Understanding

Question 1: Engine Operation

Describe the four strokes of a typical four-stroke internal combustion engine.

Answer

1. Intake: Air or an air-fuel mixture enters the cylinder.

2. Compression: The piston compresses the gas.

3. Power: Combustion raises pressure and expanding gases perform work.

4. Exhaust: Combustion products leave the cylinder.

A complete cycle requires two crankshaft revolutions.

Question 2: Gasoline Versus Diesel

Explain two major differences between conventional gasoline and diesel engines.

Answer

First, gasoline engines typically use spark plugs to initiate combustion, while diesel engines use compression ignition.

Second, diesel engines generally operate at higher compression ratios.

These differences influence efficiency, combustion behaviour, engine design, and emissions.

Question 3: Thermal Efficiency

An engine receives 200 kW of fuel energy and produces 70 kW of useful power.

Calculate its efficiency.

Solution

η = P_out/P_in

η = 70/200

η = 0.35

Answer: 35%

Question 4: Energy Losses

An engine receives 500 kJ of fuel energy during a specified operating interval.

It produces 150 kJ of useful work.

Calculate the total energy not converted into useful work.

Solution

Energy losses = Input − Useful output

= 500 − 150

Answer: 350 kJ

Question 5: Engine Power

An engine performs 90,000 J of work in 3 seconds.

Calculate its average power.

Solution

P = W/t

P = 90,000/3

P = 30,000 W

Answer: 30 kW

Question 6: Torque and Rotational Speed

An engine produces 150 N·m of torque at 4,000 rpm.

Calculate its power output.

Solution

P = 2πτN/60

P = 2π(150)(4,000)/60

P = 20,000π

P ≈ 62,832 W

Answer: Approximately 62.8 kW

Question 7: Brake Specific Fuel Consumption

An engine consumes 15 kg of fuel per hour while producing 60 kW.

Calculate BSFC.

Solution

Convert:

15 kg/h = 15,000 g/h

BSFC = 15,000/60

Answer: 250 g/kWh

Question 8: Mechanical Efficiency

An engine develops 100 kW of indicated power and delivers 88 kW of brake power.

Calculate mechanical efficiency.

Solution

ηₘ = Pᵦ/Pᵢ

ηₘ = 88/100

ηₘ = 0.88

Answer: 88%

Question 9: Ideal Otto Cycle

An ideal Otto-cycle engine has a compression ratio of 10.

Assume γ = 1.4.

Calculate its theoretical efficiency.

Solution

η = 1 − 1/r^(γ − 1)

η = 1 − 1/10^0.4

10^0.4 ≈ 2.512

η = 1 − 1/2.512

η ≈ 0.6019

Answer: Approximately 60.2%

This is a theoretical ideal-cycle efficiency, not the measured efficiency of a real engine.

Question 10: Comprehensive Engine Performance

An engine operates with the following characteristics:

• Fuel consumption: 18 kg/h

• Fuel lower heating value: 42 MJ/kg

• Brake power: 75 kW

• Indicated power: 90 kW

Calculate:

a) Fuel energy input rate.

b) Brake thermal efficiency.

c) Mechanical efficiency.

d) Brake specific fuel consumption.

e) Total energy loss rate relative to useful shaft output.

Solution

a) Fuel energy input rate

Fuel mass flow:

ṁ = 18/3,600

ṁ = 0.005 kg/s

Fuel power:

P_fuel = ṁ × LHV

P_fuel = 0.005 × 42,000,000

P_fuel = 210,000 W

Answer: 210 kW

b) Brake thermal efficiency

η_b = 75/210

η_b ≈ 0.3571

Answer: 35.7%

c) Mechanical efficiency

ηₘ = 75/90

ηₘ ≈ 0.8333

Answer: 83.3%

d) BSFC

Fuel consumption:

18 kg/h = 18,000 g/h

BSFC = 18,000/75

Answer: 240 g/kWh

e) Total energy loss rate

P_losses = P_fuel − P_brake

P_losses = 210 − 75

Answer: 135 kW

This includes exhaust energy, cooling losses, frictional effects, and other losses.

Final Reflection

Real engines demonstrate the practical importance of thermodynamics.

Although fuel combustion releases large amounts of energy, only part of that energy can be converted into useful mechanical work. Ideal thermodynamic cycles help us understand theoretical performance, while real engine measurements reveal the effects of friction, heat transfer, exhaust losses, and irreversible processes.

Gasoline and diesel engines use different ignition methods and operating characteristics, but both must obey the laws of thermodynamics.

Evaluating engine performance requires more than calculating efficiency. Engineers must also consider torque, power, fuel consumption, operating conditions, reliability, and environmental impact.

By understanding these relationships, we can explain how engines operate, why energy losses occur, and how engineering improvements can make energy conversion more effective.