# What is entropy?

This post is concerned with a common question I get from my supervisees — “what is entropy?” Before going any further, I should say I am talking about classical thermodynamics, not statistical thermodynamics or information theory.

If I ask the student, “what do you know already?”, they can usually come up with something qualitative, such as, entropy is a measure of,

- disorder;
- irreversibility;
- quality of energy; or
- energy unavailable for work.

More rarely, a student might mention the Second Law of Thermodynamics: “heat cannot of itself pass from one body to a hotter body.” The First Law of Thermodynamics only states that energy is conserved, and does not prevent heat flowing from cold to hot. Our experience is that naturally (without any work input), heat flows the other way, from hot to cold, and we need the Second Law to model this. This is related to the arrow of time, in that some processes are inherently ‘one-way’ and cannot be rewound.

All these ideas are correct, but such an explanation will not satisfy a good student. They are happy enough doing calculations given a formula for the Second Law, but the link between the qualitative description and the definition of entropy, \(S=\int \mathrm{d}{Q}/T\) for a reversible process, is still opaque to them.

In this post, I will explain how the definition of entropy arises, and why the quantity is a useful concept in Engineering.

## Where does the definition come from?

Imagine we have a heat engine operating between a hot reservoir at temperature \(T_\mathrm{h}\) and a cold reservoir at \(T_\mathrm{c}\).

If we assume the working fluid is an ideal gas, and construct a reversible cycle after Carnot, we find an efficiency,

$$\eta_\mathrm{Carnot} = \frac{{W}_x}{{Q}_\mathrm{h}} = 1 - \frac{{Q}_\mathrm{c}}{{Q}_\mathrm{h}} = 1 - \frac{T_\mathrm{c}}{T_\mathrm{h}} , \tag{I}$$

where \({W}_x\) is the shaft work, \({Q}_\mathrm{h}\) is the heat flow
*from* the hot reservoir, and \({Q}_\mathrm{c}\) is the heat flow *to* the
cold reservoir.
This expression is generally true for all reversible heat engines with any
working fluid. If it were not, we could use an engine with higher efficiency
to drive a less efficient one between the same reservoirs, resulting in a net
transfer of heat from cold to hot reservoirs without any work input. This would
be forbidden by our qualitative statements of the Second Law.

Students are familiar with Eqn. (I), and happy to take it as a starting point for defining entropy. Now we consider a real heat engine, which may be irreversible, operating between \(T_\mathrm{h}\) and \(T_\mathrm{c}\). The efficiency of a real heat engine is less than or equal to the Carnot efficiency. From Eqn. (I), $$\eta \le \eta_\mathrm{Carnot} \quad \Rightarrow \quad 1 - \frac{{Q}_\mathrm{c}}{{Q}_\mathrm{h}} \le 1 - \frac{T_\mathrm{c}}{T_\mathrm{h}} \quad \Rightarrow \quad \frac{{Q}_\mathrm{h}}{T_\mathrm{h}} - \frac{{Q}_\mathrm{c}}{{T}_\mathrm{c}} \le 0 . $$ This a form of the Clausius inequality. Our example has only two heat flows, but the Clausius inequality generalises according to, $$\oint \frac{\mathrm{d}{Q}}{T} \le 0 ,\tag{II}$$ for a system in a cyclic process receiving many infinitesimal heat transfers \(\mathrm{d}{Q}\) at temperatures \(T\). Equation (II) is a very compact statement of the Second Law, applicable to reversible or irreversible cyclic processes.

Restricting ourselves to reversible processes again, the inequality in Eqn. (II) becomes an equals, $$\oint \frac{\mathrm{d}{Q}_\mathrm{rev} }{T} = 0 .$$ Think of taking two arbitrary points in the cycle, \(a\) and \(b\), and splitting this integral into sections. Both sections must sum to zero, or, correcting for the direction of integration, going clockwise or anti-clockwise yields the same result. This implies that the value of a line integral between two states $$\int_a^b\frac{\mathrm{d}{Q}_\mathrm{rev}}{T} ,$$ is path-independent. The integral depends only on the start and end states, and is eligible to be a thermodynamic property.

We are now ready to define entropy, \(S\), $${S}_b - {S}_a =\int_a^b \frac{\mathrm{d}{Q}_\mathrm{rev} }{T} ,\tag{III}$$ To be consistent with our qualitative notions of what entropy is, it should increase during irreversible processes. To check this, we construct a different cycle going from \(a\) to \(b\) and back again: the outward path is irreversible, and the return journey is reversible. Applying Eqn. (II), $$ \oint \frac{\mathrm{d}{Q}}{T} = \underbrace{\int_a^b \frac{\mathrm{d}{Q}}{T}}_\text{outward} + \underbrace{\int_b^a \frac{\mathrm{d}{Q}_\mathrm{rev}}{T}}_\text{return} \le 0 . $$ Eliminating the integral over the reversible branch with the definition of entropy, Eqn. (III), $$ {S}_b-{S}_a \ge \int_a^b \frac{\mathrm{d}{Q}}{T} $$ If irreversibility in the process from \(a\) to \(b\) increases, this expression is is further from equality and the change in entropy as defined will also increase. To make this definite, $$ {S}_b-{S}_a = \int_a^b \frac{\mathrm{d}{Q}}{T} + \Delta{{S}}_\mathrm{irrev} ,\tag{IV} $$ where \(\Delta{{S}}_\mathrm{irrev}\ge 0\) is a non-negative quantity that measures irreversibility, as we desire. Equation (IV) is the usual form of the Second Law as given in a Data Book. The different terms on the right-hand side show that there are two ways of altering entropy: heat transfer, which can in principle be reversible; and other irreversible generation processes, which by definition are not reversible.

## Why is entropy a useful quantity?

This section uses a simple example to illustrate why the concept of entropy is a useful tool in an Engineer’s toolbox. We consider the flow of a perfect gas, supplied at a temperature \(T_1\) and pressure \(p_1\), expanded through a turbine to a fixed exit pressure \(p_2\). The task for the Engineer designing the turbine, is to extract the maximum work possible.

Start by writing the First and Second Laws for a control volume surrounding the turbine, on a per unit mass basis, $$ h_2 - h_1 = \dot{q} - \dot{w}_x , $$$$ s_2 - s_1 = \int \frac{\mathrm{d}\dot{q}}{T} + \Delta \dot{s}_\mathrm{irrev} , $$ and assuming the turbine is adiabatic, \(\dot{Q}=\mathrm{d}\dot{Q}=0\), and, $$ \dot{w}_x = h_1 - h_2 , $$$$ \Delta \dot{s}_\mathrm{irrev} = s_2 - s_1 . $$ Now using standard expressions for enthalpy and entropy change of a perfect gas available in a Data Book, $$ \dot{w}_x = c_p\left(T_1 - T_2\right) , \tag{V.i} $$$$ \Delta \dot{s}_\mathrm{irrev} = c_p \log \frac{T_2}{T_1} - R \log \frac{p_2}{p_1} ,\tag{V.ii}$$ where \(c_p\) is specific heat capacity at constant pressure, \(R\) the specific gas constant. Eliminating the unknown \(T_2\) from Eqns. (V), $$ \dot{w}_x = c_pT_1\left[1-\exp\left(\frac{\Delta \dot{s}_\mathrm{irrev}}{c_p}\right)\left(\frac{p_2}{p_1}\right)^\tfrac{\gamma-1}{\gamma} \right] , $$ where we have used \(R/c_p = (\gamma -1)/\gamma\), the ratio of specific heats \(\gamma\) being a property of the gas. If the turbine designer is doing a good job, then \(\Delta \dot{s}_\mathrm{irrev} \ll c_p\) and we can approximate the exponential with a Taylor series, \(\exp (z) \approx 1 + z \), finally giving, $$ \dot{w}_x \approx \underbrace{c_pT_1\left[1-\left(\frac{p_2}{p_1}\right)^\tfrac{\gamma-1}{\gamma} \right]}_\text{work if reversible} - \underbrace{\Delta \dot{s}_\mathrm{irrev} T_1\left(\frac{p_2}{p_1}\right)^\tfrac{\gamma-1}{\gamma}\vphantom{\left[\left(\frac{p_2}{p_1}\right)^\tfrac{\gamma-1}{\gamma}\right]}}_\text{lost work due to irreversibility} . \tag{VI} $$

The first term in Eqn. (VI) is a function of the inlet state, exit pressure, and fluid properties, which the designer has no control over. Only the second term, a decrement in work proportional to irreversible entropy generation, is under the control of the designer. Therefore, they should seek to minimise \(\Delta \dot{s}_\mathrm{irrev}\) to obtain maximum useful work.

## Summary

In the first section of this post, I explained how the definition of entropy arises. Starting from the Carnot efficiency as an upper bound to a real heat engine, a generalisation can be made to the Clausius inequality, which implies that \(\Delta S = \int \mathrm{d}Q/T\) is path-independent for a reversible process. For a general process, the entropy change is a sum of heat transfer and irreversible generation terms.

The second section of this post contained an example of the implications of entropy in an Engineering situation. To maximise the work output of a turbine, running between a given inlet state and fixed exit pressure, a designer should minimise entropy generation due to irreversibility.