Table of Contents
Thermal Efficiency and Heat Engines
1. Introduction
A heat engine is a device that converts thermal energy into mechanical work by operating through a thermodynamic cycle.
Heat engines are central to modern energy systems. They are used in power stations, automobiles, aircraft, ships and industrial machinery.
However, no heat engine can convert all the heat it absorbs into useful work. Some energy must be rejected to a lower-temperature environment.
Thermal efficiency measures how effectively a heat engine converts supplied heat into useful work.
Understanding thermal efficiency and heat engines is important in thermodynamics, mechanical engineering, energy production and environmental studies.
2. Heat Engine
A heat engine is a device that operates cyclically, absorbing heat from a high-temperature source, converting part of that energy into work and rejecting the remaining heat to a lower-temperature sink.
A basic heat engine operates between two thermal reservoirs.
2.1 Main Components
A conventional heat-engine model includes:
Hot reservoir
The source from which the engine absorbs heat.
Working substance
The substance that undergoes thermodynamic changes, such as gas or steam.
Engine mechanism
The arrangement that converts part of the thermal energy into mechanical work.
Cold reservoir
The environment or sink to which the engine rejects unused heat.
3. Working Principle of a Heat Engine
A heat engine operates through a repeated thermodynamic cycle.
The general sequence is:
- The engine absorbs heat from a high-temperature source.
- The working substance undergoes thermodynamic changes.
- Part of the supplied energy is converted into mechanical work.
- The remaining heat is rejected to a lower-temperature sink.
- The working substance returns to its initial thermodynamic state.
- The cycle repeats.
Because the working substance returns to its initial state after each complete cycle, its net change in internal energy over one complete cycle is zero.
[
\Delta U_{\text{cycle}}=0
]
Using the First Law:
[
Q_{\text{net}}=W_{\text{net}}
]
For a heat engine absorbing heat (Q_H) and rejecting heat (Q_C):
[
W=Q_H-Q_C
]
Where:
- (Q_H) = heat absorbed from the hot reservoir
- (Q_C) = heat rejected to the cold reservoir
- (W) = net work output
Here, (Q_H), (Q_C) and (W) represent positive magnitudes.
4. Thermal Efficiency
Thermal efficiency measures the proportion of heat supplied to an engine that is converted into useful work.
It is represented by:
[
\eta=\frac{\text{Useful work output}}{\text{Heat supplied}}
]
For a heat engine:
[
\eta=\frac{W}{Q_H}
]
Since:
[
W=Q_H-Q_C
]
Therefore:
[
\eta=\frac{Q_H-Q_C}{Q_H}
]
Or:
[
\boxed{\eta=1-\frac{Q_C}{Q_H}}
]
To express efficiency as a percentage:
[
\eta(%)=\frac{W}{Q_H}\times100
]
4.1 Interpretation
If a heat engine receives 1,000 J of heat and produces 300 J of useful work, its efficiency is:
[
\eta=\frac{300}{1000}=0.30
]
[
\eta=30%
]
The remaining 700 J is rejected to the surroundings or otherwise leaves the engine as waste energy.
4.2 Why Is Thermal Efficiency Always Below 100%?
The Second Law of Thermodynamics states that a cyclic heat engine cannot convert all heat absorbed from a single reservoir completely into work.
A heat engine must reject some heat to a colder reservoir.
Therefore:
[
Q_C>0
]
And:
[
\eta<1
]
This limitation applies even to an ideal heat engine operating between two reservoirs at finite temperatures.
5. Heat Engines and Energy Conservation
A heat engine follows the First Law of Thermodynamics.
For one complete cycle:
[
\Delta U=0
]
Therefore:
[
Q_H-Q_C=W
]
This shows that the work output equals the difference between the heat absorbed and the heat rejected.
Energy is not destroyed during engine operation. Some of the supplied energy becomes useful work, while the remainder is transferred to the surroundings.
6. Carnot Engine
A Carnot engine is an idealised reversible heat engine operating between two thermal reservoirs.
It represents the maximum possible thermal efficiency for any engine operating between the same hot and cold reservoir temperatures.
The Carnot cycle consists of four reversible processes.
6.1 Four Processes of the Carnot Cycle
1. Isothermal Expansion
The working substance expands at the hot-reservoir temperature and absorbs heat.
2. Adiabatic Expansion
The working substance continues to expand without heat exchange, and its temperature decreases.
3. Isothermal Compression
The working substance is compressed at the cold-reservoir temperature and rejects heat.
4. Adiabatic Compression
The working substance is compressed without heat exchange, and its temperature increases until it returns to the initial state.
7. Carnot Efficiency
The efficiency of an ideal Carnot engine operating between two reservoirs is:
[
\eta_C=1-\frac{T_C}{T_H}
]
Where:
- (\eta_C) = Carnot efficiency
- (T_H) = absolute temperature of the hot reservoir
- (T_C) = absolute temperature of the cold reservoir
Both temperatures must be expressed in kelvin.
7.1 Important Features
- Carnot efficiency depends only on the temperatures of the two reservoirs.
- It does not depend on the working substance.
- No engine operating between the same two reservoirs can be more efficient than a reversible Carnot engine.
- A real engine has lower efficiency because of irreversible processes and practical losses.
7.2 Factors Affecting Carnot Efficiency
Carnot efficiency increases when:
- The hot-reservoir temperature increases.
- The cold-reservoir temperature decreases.
It approaches 100% only in the theoretical limit where the cold-reservoir temperature approaches absolute zero, which cannot be achieved in practice.
8. Real Heat Engines
Real engines operate with irreversible processes and practical limitations.
Sources of energy losses include:
- Friction.
- Heat leakage.
- Exhaust energy.
- Incomplete combustion.
- Fluid resistance.
- Mechanical inefficiencies.
- Heat transfer across finite temperature differences.
Consequently, real engines have lower thermal efficiency than the Carnot limit for the same reservoir temperatures.
9. Types of Heat Engines
9.1 Internal Combustion Engine
An internal combustion engine burns fuel inside the engine’s working cylinders or combustion chamber.
Examples include:
- Petrol engines.
- Diesel engines.
- Gas engines.
The high-temperature gases produced by combustion expand and perform mechanical work.
9.2 External Combustion Engine
An external combustion engine receives heat from a source outside the working cylinder or working-fluid chamber.
Examples include:
- Steam engines.
- Steam turbine power systems.
- Stirling engines.
In a steam power station, fuel or another heat source heats water to produce steam, which drives a turbine.
10. Petrol Engine and Otto Cycle
The ideal Otto cycle is used to model spark-ignition petrol engines.
It consists of four idealised processes:
- Isentropic compression.
- Constant-volume heat addition.
- Isentropic expansion.
- Constant-volume heat rejection.
The Otto cycle provides a theoretical framework for understanding the operation of petrol engines.
10.1 Otto Cycle Efficiency
For an ideal air-standard Otto cycle:
[
\eta_{\text{Otto}}=
1-\frac{1}{r^{\gamma-1}}
]
Where:
- (r) = compression ratio
- (\gamma=C_p/C_v) = ratio of specific heat capacities
The ideal efficiency increases with compression ratio, assuming the same value of (\gamma).
Actual engine efficiency is lower because of real combustion, heat transfer, friction and other losses.
11. Diesel Engine and Diesel Cycle
The ideal Diesel cycle models compression-ignition engines.
Its four idealised processes are:
- Isentropic compression.
- Constant-pressure heat addition.
- Isentropic expansion.
- Constant-volume heat rejection.
Diesel engines compress air to a high temperature and inject fuel into the hot compressed air, causing ignition.
The ideal Diesel cycle differs from the Otto cycle mainly in the model used for heat addition.
12. Steam Turbine Power Plants
Steam power plants convert thermal energy into mechanical work and then into electrical energy.
The basic process is:
- A boiler heats water to produce high-pressure steam.
- Steam expands through a turbine.
- The turbine drives an electrical generator.
- Exhaust steam enters a condenser.
- The condenser converts steam back into water.
- A pump returns the water to the boiler.
This arrangement is commonly analysed using the Rankine cycle.
12.1 Rankine Cycle
The ideal Rankine cycle consists of four main processes:
- Isentropic pumping of liquid water.
- Constant-pressure heat addition in the boiler.
- Isentropic expansion through the turbine.
- Constant-pressure heat rejection in the condenser.
The Rankine cycle is fundamental to conventional steam power generation.
13. Refrigerator and Heat Pump
A refrigerator is a device that transfers heat from a low-temperature region to a higher-temperature environment using external work.
A heat pump operates on the same general thermodynamic principle but is designed primarily to deliver useful heat to a warmer space.
Both are opposite in purpose to a heat engine.
13.1 Refrigerator
A refrigerator removes heat from a cold space and rejects it to warmer surroundings.
Its operation requires work supplied by an electric motor or another power source.
13.2 Heat Pump
A heat pump extracts energy from a relatively cool source and transfers it to a warmer space.
Examples include:
- Residential heating systems.
- Commercial building heating.
- Industrial heat recovery.
- Water heating systems.
14. Coefficient of Performance
The performance of refrigerators and heat pumps is measured using the coefficient of performance (COP), rather than thermal efficiency.
14.1 Refrigerator COP
[
COP_R=\frac{Q_C}{W}
]
Where:
- (Q_C) = heat removed from the cold space
- (W) = work supplied
Since:
[
W=Q_H-Q_C
]
The refrigerator COP can also be written as:
[
COP_R=\frac{Q_C}{Q_H-Q_C}
]
14.2 Heat Pump COP
[
COP_{HP}=\frac{Q_H}{W}
]
Or:
[
COP_{HP}=\frac{Q_H}{Q_H-Q_C}
]
For the same operating cycle:
[
COP_{HP}=COP_R+1
]
14.3 Difference Between Efficiency and COP
Thermal efficiency measures the fraction of supplied heat converted into work by a heat engine.
COP measures the useful heating or cooling effect obtained per unit of work supplied.
COP can be greater than 1 because a refrigerator or heat pump moves thermal energy rather than converting all supplied work directly into heat at the desired location.
15. Carnot Refrigerator and Heat Pump
A reversible Carnot refrigerator operating between cold and hot reservoirs has:
[
COP_{R,C}=\frac{T_C}{T_H-T_C}
]
A reversible Carnot heat pump has:
[
COP_{HP,C}=\frac{T_H}{T_H-T_C}
]
Where:
- (T_C) = absolute temperature of the cold reservoir
- (T_H) = absolute temperature of the hot reservoir
These equations give the maximum theoretical COP between the two specified temperatures.
16. Comparison of Heat Engines, Refrigerators and Heat Pumps
| Device | Main purpose | Energy input | Useful output |
|---|---|---|---|
| Heat engine | Convert heat into work | Heat from hot reservoir | Mechanical work |
| Refrigerator | Remove heat from a cold space | External work | Cooling |
| Heat pump | Deliver heat to a warm space | External work | Heating |
17. Importance of Thermal Efficiency
Improving thermal efficiency helps:
- Reduce fuel consumption.
- Lower operating costs.
- Increase electricity generation from available energy.
- Reduce waste heat.
- Improve industrial productivity.
- Reduce greenhouse gas emissions when less fossil fuel is burned for the same useful output.
Efficiency improvements must also consider safety, reliability, capital cost and environmental impacts.
18. Worked Examples
Example 1: Thermal Efficiency of a Heat Engine
A heat engine absorbs 1,000 J of heat and rejects 600 J to its surroundings. Calculate its work output and thermal efficiency.
Step 1: Calculate work output
[
W=Q_H-Q_C
]
[
W=1000-600
]
[
W=400\ \text{J}
]
Step 2: Calculate efficiency
[
\eta=\frac{W}{Q_H}
]
[
\eta=\frac{400}{1000}
]
[
\boxed{\eta=0.40=40%}
]
Example 2: Carnot Engine Efficiency
A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. Calculate its maximum theoretical efficiency.
Using:
[
\eta_C=1-\frac{T_C}{T_H}
]
[
\eta_C=1-\frac{300}{600}
]
[
\eta_C=1-0.5
]
[
\boxed{\eta_C=0.50=50%}
]
Example 3: Refrigerator COP
A refrigerator removes 900 J of heat from its interior while consuming 300 J of work. Calculate its COP.
Using:
[
COP_R=\frac{Q_C}{W}
]
[
COP_R=\frac{900}{300}
]
[
\boxed{COP_R=3}
]
The refrigerator removes 3 J of heat for every 1 J of work supplied.
Example 4: Heat Pump COP
A heat pump delivers 1,200 J of heat to a room while consuming 300 J of work. Calculate its COP.
Using:
[
COP_{HP}=\frac{Q_H}{W}
]
[
COP_{HP}=\frac{1200}{300}
]
[
\boxed{COP_{HP}=4}
]
The heat pump delivers 4 J of heat for every 1 J of work supplied.
Example 5: Carnot Refrigerator COP
A refrigerator operates between a cold reservoir at 270 K and a hot reservoir at 300 K. Calculate its maximum theoretical COP.
Using:
[
COP_{R,C}=\frac{T_C}{T_H-T_C}
]
[
COP_{R,C}=\frac{270}{300-270}
]
[
COP_{R,C}=\frac{270}{30}
]
[
\boxed{COP_{R,C}=9}
]
19. Common Examination Mistakes
- Confusing heat engine efficiency with refrigerator COP.
- Using Celsius temperatures in Carnot efficiency calculations.
- Forgetting that a heat engine must reject some heat.
- Assuming a real engine can reach Carnot efficiency.
- Confusing a refrigerator with a heat engine.
- Assuming COP must always be less than 1.
- Forgetting that a complete engine cycle returns the working substance to its initial state.
- Confusing the Otto cycle with the Diesel cycle.
- Assuming that the working substance in every heat engine is steam.
- Ignoring the role of the cold reservoir in a heat engine.
36 Key Takeaways for Exams
- A heat engine converts part of supplied thermal energy into mechanical work.
- A heat engine operates through a thermodynamic cycle.
- A conventional heat engine absorbs heat from a hot reservoir.
- A heat engine rejects some heat to a colder reservoir.
- The working substance undergoes thermodynamic changes during engine operation.
- The First Law governs energy conservation in heat engines.
- The net change in internal energy over a complete cycle is zero.
- Heat-engine work output is (W=Q_H-Q_C).
- Thermal efficiency is the ratio of useful work output to heat supplied.
- Heat-engine efficiency is (\eta=W/Q_H).
- Heat-engine efficiency can be written as (\eta=1-Q_C/Q_H).
- No cyclic heat engine can convert all absorbed heat from a single reservoir into work.
- A heat engine must reject heat to a lower-temperature reservoir.
- The Carnot engine is an ideal reversible heat engine.
- Carnot efficiency is the maximum theoretical efficiency between two specified reservoir temperatures.
- Carnot efficiency is (\eta_C=1-T_C/T_H).
- Temperatures in Carnot calculations must be expressed in kelvin.
- Carnot efficiency depends on the temperatures of the hot and cold reservoirs.
- Real engines are less efficient than the Carnot limit.
- Friction, heat leakage and exhaust losses reduce real engine efficiency.
- Internal combustion engines burn fuel within the engine’s working chamber.
- Petrol engines are commonly modelled using the Otto cycle.
- Diesel engines are commonly modelled using the Diesel cycle.
- The ideal Otto cycle involves constant-volume heat addition.
- The ideal Diesel cycle involves constant-pressure heat addition.
- Steam power plants commonly operate using the Rankine cycle.
- A refrigerator transfers heat from a colder region to a warmer environment using external work.
- A heat pump delivers useful heat to a warmer space.
- Refrigerator performance is measured by its coefficient of performance.
- Refrigerator COP is (COP_R=Q_C/W).
- Heat-pump COP is (COP_{HP}=Q_H/W).
- For the same cycle, heat-pump COP equals refrigerator COP plus one.
- COP can be greater than one.
- The Carnot refrigerator COP is (T_C/(T_H-T_C)).
- The Carnot heat-pump COP is (T_H/(T_H-T_C)).
- Improving thermal efficiency can reduce fuel consumption, operating costs and waste heat.