Cyclic processes

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  3. 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.