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# Cyclic processes

#### Cyclic process

A cyclic process is a series of transformations that take the gas back to its original state. These form a closed loop on a $$pV$$ diagram.

A simple cyclic process involves:

1. Isobaric expansion - work is done on the surroundings
2. Isovolumetric cooling
3. Isobaric compression - work is done on the gas
4. Isovolumetric heating

Since $$W=p\Delta V$$, the area between a line on the $$pV$$ diagram and the $$V$$ axis gives work done. The net work done during a cyclic process is given by the enclosed area.

We can show the net work done using a Sankey diagram.

#### Thermal efficiency

The dimensionless quantity, thermal efficiency, of any mechanical process is defined as the ratio of the useful work done to the total energy input:

$$\eta={\textrm{useful work done}\over \textrm{energy input}}$$

For a heat engine:

• Energy input is $$Q_\textrm{H}$$
• Work done is $$W_\textrm{out}=Q_\textrm{H}-Q_\textrm{C}$$

$$\eta={Q_\textrm{H}-Q_\textrm{C}\over Q_\textrm{H}}=1-{Q_\textrm{C}\over Q_\textrm{H}}$$

#### Carnot cycle

The most efficient cyclic process is known as the Carnot cycle. This involves:

1. Isothermal expansion - work is done on the surroundings
2. Adiabatic expansion - work is done on the surroundings
3. Isothermal compression - work is done on the gas
4. Adiabatic compression - work is done on the gas

Note that you may be required to calculate the work done during any part of the process using the 'counting squares' method. This is a little arduous!

• Calculate the size represented by one square using the bottom left square on the axes
• Count the squares enclosed in the area required and multiply

To calculate the thermal efficiency we must consider entropy.

The total change in entropy, $$\Delta S=\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4$$

Since stages 2 and 4 are adiabatic, $$Q=0$$ and so $$\Delta S _2 = \Delta S_4=0$$

Stages 1 and 3 take place at constant temperature, so $$\Delta S_1 = {Q_\textrm{H} \over T_\textrm{H}}$$ and $$\Delta S_3 = {Q_\textrm{C} \over T_\textrm{C}}$$

Since the Carnot cycle is a reversible process, $$\Delta S = {Q_\textrm{H} \over T_\textrm{H}} - {Q_\textrm{C} \over T_\textrm{C}} = 0$$

$${Q_\textrm{H} \over T_\textrm{H}} = {Q_\textrm{C} \over T_\textrm{C}} \Rightarrow {Q_\textrm{C} \over Q_\textrm{H}} = {T_\textrm{C} \over T_\textrm{H}}$$

Substituting into the general equation for thermal efficiency of a heat engine:

$$\eta_\textrm{Carnot}=1-{T_\textrm{cold}\over T_\textrm{hot}}$$

The equation for thermal efficiency for a Carnot cycle makes clear that we can approach but never attain 100% efficiency. To do so we would need to have a high temperature thermal energy store approaching infinity or a low temperature thermal energy sink approaching absolute zero:

• If $$T_\textrm{hot} \rightarrow \infty$$$$\eta \rightarrow1$$
• If $$T_\textrm{cold}\rightarrow 0 \textrm{K}$$$$\eta \rightarrow1$$

At the operating temperatures provided, the Carnot cycle is the most efficient process possible.