This notes sheet introduces the second law of thermodynamics, used for a number of important calculations: finding temperature differences, reversibility, and efficiency of reversible vs irreversible heat engines.

Reversible & Irreversible Processes

Let us take the classic example of a quasi-equilibrium expansion:

The second law of thermodynamics, reversible vs irreversible processes, Claudius' Statement, heat engines, efficiency, Carnot Cycle

If the mass on the piston is made up of a vast number of minuscule weights, like sand, removing a single grain pushes the piston up ever so slightly. The change is so small, that the return to equilibrium is almost instant.

For each grain of sand removed, we have an additional point on the P-V diagram – as pressure decreases, volume increases.

This process is reversible

It is reversible, because if we return the grains of sand one at a time, the piston will move down by the same amount. We neglect friction or leakage.

The second law of thermodynamics, reversible vs irreversible processes, Claudius' Statement, heat engines, efficiency, Carnot Cycle

If, however, the weight consist of one large mass, and this is removed, the process is not in quasi-equilibrium.

The process is irreversible

The second law of thermodynamics, reversible vs irreversible processes, Claudius' Statement, heat engines, efficiency, Carnot Cycle

In this example, the expansion itself is quasi-equilibrium, as we have returned to the grains of sand. However, when the piston hits the top or bottom stoppers, it makes a sound, releasing energy. This energy cannot be gotten back, so

The process is irreversible

What makes a process irreversible?

  • Friction
  • Unresisted expansion
  • Heat transfer to surroundings
  • Combustion
  • Mixing of different fluids

Clearly, then, the majority of real-life processes are irreversible.

The Second Law

Take the example of a spindle rotating in a container:

The second law of thermodynamics, reversible vs irreversible processes, Claudius' Statement, heat engines, efficiency, Carnot Cycle

When the spindle rotates:

  • Gas in container heats up
  • Heat from gas is conducted through container and heats up the surroundings

If we reverse this and the surroundings are heated:

The second law of thermodynamics, reversible vs irreversible processes, Claudius' Statement, heat engines, efficiency, Carnot Cycle
  • Gas inside container heats up
  • Spindle does not start rotating

We know from experience that this is the case, but why?

This is where the second law comes in:

Heat cannot be transferred from a cold body to a hotter body without a work input.

This is the all-important idea of direction: some things are allowed backwards as well as forwards, but many things are only allowed forwards.

Heat Engines & Efficiency

A heat engine is a device that connects a hot and cold reservoir with a work input or output.

The second law of thermodynamics, reversible vs irreversible processes, Claudius' Statement, heat engines, efficiency, Carnot Cycle

From the first law:

\Sigma Q - \Sigma W = 0

\Sigma Q = \Sigma W

Q_H - Q_C = W_{net}

The second law is often used to describe efficiency:

\eta = \frac{output}{input} = \frac{W_{net}}{Q_H}

Using the equation from the first law above, the efficiency can also be given as:

\eta = 1 - \frac{Q_C}{Q_H}

Reversed Engine

For a reversed engine, like a refrigerator, work is the input not the output. The efficiency it known as the ‘Coefficient of Performance’ (COP).

In a fridge, the output is the cold reservoir, so COP is given as:

COP = \frac{Q_C}{W_{in}} = \frac{Q_C}{Q_H - Q_C}

Clausius’ Statement

The Clausius statement of the second law tells us that it is only possible to reverse a heat engine if there is a net work input from its surroundings:

The second law of thermodynamics, reversible vs irreversible processes, Claudius' Statement, heat engines, efficiency, Carnot Cycle

A reversible engine is always more efficient than an irreversible engine.

We can prove this by contradiction:

  • Take two engines, one reversible and one irreversible, and assume that the former has an efficiency of 10% and the latter 20%:
The second law of thermodynamics, reversible vs irreversible processes, Claudius' Statement, heat engines, efficiency, Carnot Cycle
  • Now reverse the reversible engine:
The second law of thermodynamics, reversible vs irreversible processes, Claudius' Statement, heat engines, efficiency, Carnot Cycle
  • Modelling both engines in a single control volume:
The second law of thermodynamics, reversible vs irreversible processes, Claudius' Statement, heat engines, efficiency, Carnot Cycle
  • According to the Clausius statement, this is impossible.

Similarly, we can use Clausius’ statement to show the 100% efficient engine is a physical impossibility:

  • Take a 100% efficient engine (E100 engine) and connect its output to a reversed engine:
The second law of thermodynamics, reversible vs irreversible processes, Claudius' Statement, heat engines, efficiency, Carnot Cycle
  • Model the two engines as a control volume:
The second law of thermodynamics, reversible vs irreversible processes, Claudius' Statement, heat engines, efficiency, Carnot Cycle
  • This, one again, violates Clausius’ statement

You cannot have an E100 engine.

This is summarised by the Kelvin-Planck statement:

No heat engine can deliver a work output equivalent to the heat input from a single reservoir.

Actual vs Ideal Efficiencies

Since no heat engine can be 100% efficient, it is often unhelpful to talk about an engine’s actual efficiency. Instead, when talking about its performance, we compare its efficiency to that of its ideal (reversible) counterpart:

The second law of thermodynamics, reversible vs irreversible processes, Claudius' Statement, heat engines, efficiency, Carnot Cycle

The efficiency of a reversible engine is a function of the temperatures between which the engine operates. Therefore, we call it thermal efficiency:

\eta_{th} = 1 - \frac{T_C}{T_H}

This is due to the absolute temperature scale, as defined by Kelvin: the scale that sets the triple point of water (0°C) as 273.16 K. This scale is identical to the ideal gas temperature scale, from which the relationship Pv=RT is derived. This can be reworked to show that for a reversible engine, heat is a function of temperature.

Therefore:

\frac{Q_C}{Q_H} = \frac{f(T_C)}{f(T_H)}

Since the increment of the scale is 1K, this simply becomes:

\frac{Q_C}{Q_H} = \frac{T_C}{T_H}

Remember, this is only the case for ideal, reversible engines (that do not exist irl).

There is a huge difference in the actual efficiency of a heat engine and its efficiency compared to the maximum achievable efficiency from its reversible counterpart. Take, for example, a heat engine operating between 500K and 400k, producing 10kJ or work for 100kJ of heat input:

\eta_{actual} = 1 - \frac{Q_C}{Q_H} = 1-\frac{90}{100} = 10\%

\eta_{reversible} = 1 - \frac{T_C}{T_H} = 1 - \frac{400}{500} = 20\%

The actual efficiency appears to be just 10%, but really it is far higher, as we need to compare it with the maximum achievable (reversible) efficiency:

\eta_{relative} = \frac{\eta_{actual}}{\eta_{reversible}} = \frac{0.1}{0.2} = 50 \%

Reversed Engine

Back to the fridge example, the ideal Coefficient of Performance (COP) for a reversible reversed engine is given as a function of the temperatures, not heats:

COP_{reversible} = \frac{T_C}{T_H - T_C}

Improving Efficiency

As you can probably tell from the numbers in the example above, efficiencies of heat engines can often be quite low. Therefore, there is a constant push to increase this to boost the performance of a system, to make it more economical, or to make it more environmentally friendly.

There are a few key ways of boosting a heat engine’s efficiency:

  • Find a lower temperature cold reservoir
  • This is difficult, however. Generally, rivers, lakes or the ocean are used for cold reservoirs, but these are rarely lower than around 10 °C
  • Increase the temperature of the hot reservoir
  • Increasing the temperature difference between the two reservoirs increases the maximum possible efficiency
  • Minimise losses
  • Insulation, less friction etc. brings the actual engine closer to its reversible counterpart
  • Find a use for the losses
  • Can you use the output heat for a practical purpose, such as heating?
  • Generally, this is difficult because the cold reservoir is around 20 °C, which is not helpful for anyone.

Carnot Cycle

The Carnot cycle is an unobtainable, ideal cycle of compression and expansion. An ideal petrol piston engine can be modelled as a Carnot engine.

Carnot cycle, reversible cycle, second law of thermodynamics

It consists of four stages:

  1. Isothermal Compression
Carnot cycle, reversible cycle, second law of thermodynamics
  • There is a heat input
  • Since the process is isothermal, there is no change in internal energy: \Delta Q = \Delta W (from the first law)
  • The piston moves down
  • The temperature does not change, so T_1 = T_2, the cold reservoir temperature

2. Adiabatic Compression

Carnot cycle, reversible cycle, second law of thermodynamics
  • There is a work input
  • Adiabatic, so first law becomes -\Delta W = \Delta U
  • The piston still moves down
  • T_3 is the hot reservoir temperature

3. Isothermal Expansion

Carnot cycle, reversible cycle, second law of thermodynamics
  • There is a heat input
  • Isothermal, so \Delta Q = \Delta W
  • The piston moves up
  • T_3 = T_4, the hot reservoir temperature

4. Adiabatic Expansion

Carnot cycle, reversible cycle, second law of thermodynamics
  • There is a work output
  • Adiabatic, so -\Delta W = \Delta U
  • The piston continues to move up
  • T_1 = T_2, the cold reservoir temperature

All four processes are fully reversible (the Carnot cycle is a theoretical cycle only).

  • Processes are either reversible or irreversible
    • Real-life processes cannot be reversible
  • The second law states that heat cannot be transferred from a cold body to a hotter body without a work input (Clausius’ statement).
  • According to the Kelvin-Planck statement, no heat engine can deliver a work output equivalent to the heat input from a single reservoir – you cannot have an E100 engine.
  • The actual efficiency is always than the thermal efficiency:
    • \eta_{actual} = \frac{W_{net}}{Q_H} = 1 - \frac{Q_C}{Q_H}
    • \eta_{th} = 1 - \frac{T_C}{T_H}
  • Efficiency can be improved by finding a lower T_C, higher T_H, minimising losses or finding a use for the lost heat.
  • The Carnot cycle is a theoretical, fully reversible compression and expansion cycle in four stages:
    1. Isothermal heat rejection (compression)
    2. Adiabatic compression (work input)
    3. Isothermal heat addition (expansion)
    4. Adiabatic expansion (work output)