Using Entropy

The question

Mix hot and cold water and you get warm water — never the reverse. Open a valve and gas rushes out of a tank — never back in. Something is being counted that only ever goes up. What is entropy, and why does it only ever grow?

From an inequality to a property

Lesson 5 left the second law as a set of prohibitions. Clausius turned it into something you can compute. For any cycle, he showed, the heat received at each point divided by the boundary temperature there sums to something that can never be positive:

$\oint \left(\frac{\delta \cheat{Q}}{T}\right)_b \le 0,$

with equality only for a reversible cycle. And a quantity whose integral around every reversible cycle is zero must be a property — its change depends only on the end states, like the change of height between two points on a hill. That property is , $\cent{S}$ :

$d\cent{S} = \left(\frac{\delta \cheat{Q}}{T}\right)_\text{int rev}.$

For a reversible process, heat in raises entropy, heat out lowers it, and the heat is the area under the path on a T–s diagram: $\cheat{Q} = \int T\,d\cent{S}$ . That is why engineers draw cycles on T–s diagrams, as the journey on the course home page did: the area under the heating line is the heat in, the area under the cooling line is the heat out, and the area inside the loop is the net work.

Looking entropy up

Because entropy is a property, it is tabulated alongside $u$ and $h$ in the steam tables — you dragged it around in Lesson 3. For other substances, two relations do the work. Combining the first law with $\delta Q = T\,dS$ for a reversible process gives the T ds equations, which hold for any process because they relate properties only:

$T\,ds = du + p\,dv, \qquad T\,ds = dh - v\,dp.$

For an ideal gas, these integrate to

$s_2 - s_1 = s^\circ(T_2) - s^\circ(T_1) - R\ln\frac{p_2}{p_1},$

where $s^\circ(T)$ is tabulated for each gas. For a liquid, simply $s_2 - s_1 = c\ln(T_2/T_1)$ .

Entropy generation: the accountant of irreversibility

Here is the heart of it. For a closed system, the entropy balance is

$\Delta \cent{S} = \int \left(\frac{\delta \cheat{Q}}{T}\right)_b + \cent{\sigma}, \qquad \cent{\sigma} \ge 0.$

Entropy changes because entropy is transferred with heat — and because it is generated, $\cent{\sigma}$ , by every irreversibility inside the system. Entropy generation is zero for reversible processes and positive for all real ones. It is never negative. Watch it appear:

ExperimentWhere entropy is born

Hot block500 K

Cold block300 K

Entropy generated0.0 J/Kalways positive

Work extracted0.0 kJmaximum 9.8 kJ

Touching the blocks generates entropy and throws away the chance to make up to 9.8 kJ of work — what a reversible engine between them would deliver.

Challenges0/2

Let two blocks touch and watch entropy appear from nowhere.
Bring the blocks to a common temperature while generating (almost) no entropy.

Two 1-kg copper blocks at different temperatures, isolated from everything else. Let them touch, and heat flows until they share a temperature. Or connect them through a reversible heat engine, which turns some of that flow into work. The bars show each block’s entropy change, ΔS = mc ln(T/T₀), and their sum.

When the blocks simply touch, the hot one loses entropy (heat leaves it at a high temperature) and the cold one gains more (the same heat arrives at a low temperature, and $Q/T$ is bigger). The total goes up: entropy has been generated, by heat transfer across a finite temperature difference. Run a reversible engine between the blocks instead, and the total stays at zero — but now the final temperature is lower, because some energy left as work.

What is entropy, really?

The classical definition is precise but abstract. Ludwig Boltzmann gave it a meaning. A macroscopic state — say, "gas filling the whole box" — can be realised by an astronomical number $W$ of microscopic arrangements of molecules. Boltzmann's formula, carved on his tombstone, is

$\cent{S} = k\ln W.$

ExperimentEntropy is counting

Molecules6

In the left half6 of 6

ln W (arrangements)0.00maximum 4.16 = N ln 2

Chance all are on the left1 in 64

For a mole of gas, N ≈ 6×10²³, and the chance that all the molecules gather back in one half is 1 in 10^(1.8×10²³). The entropy increase is ΔS = N k ln 2 = R ln 2 per mole — exactly what the classical formula ΔS = R ln(V₂/V₁) gives.

Challenges0/2

With the partition gone, catch every molecule back in the left half at once.
Try the same with 30 or more molecules. How long would you wait?

Molecules in a box, held in the left half by a partition. Remove it, and they spread out. Nothing pushes them right; there are simply vastly more arrangements with the molecules spread out than bunched up. Boltzmann’s entropy is S = k ln W, where W counts those arrangements.

The molecules don't "want" to spread out. There are just overwhelmingly more ways to be spread out than bunched up, and random motion samples all of them. For a handful of molecules, you will see them all wander back to one side now and then. For a mole, it would not happen in a time vastly longer than the age of the universe. The second law is a law of statistics — but statistics with $10^{23}$ players is as certain as anything in physics.

Isentropic efficiency

Real turbines, compressors, nozzles and pumps are nearly adiabatic, so the entropy of the fluid passing through them can only rise — by the entropy generated inside. The ideal they are measured against is the isentropic process: adiabatic and reversible, with $s_2 = s_1$ . The compares the real device with that ideal:

$\eta_t = \frac{h_1 - h_2}{h_1 - h_{2s}} \ \ \text{(turbine)}, \qquad \eta_c = \frac{h_{2s} - h_1}{h_2 - h_1} \ \ \text{(compressor or pump)}.$

ExploreThe price of friction

Isentropic efficiency85.0%

Expansion ratio1000 · exit 0.100 bar

Work out1083 kJ/kgideal 1274 kJ/kg

Entropy generated0.599 kJ/(kg·K)

Lost work T₀·s_gen179 kJ/kgat T₀ = 25 °C

Exit quality87.1%

Challenges0/3

Make the process generate no entropy at all.
Make a compressor deliver air hotter than 800 K.
Find a turbine efficiency at which the exhaust is dry (superheated), even at 1000:1 expansion.

Adiabatic expansion through a steam turbine (100 bar, 500 °C in) or compression in an air compressor (1 bar, 300 K in), on the T–s diagram. The dashed line is the ideal, isentropic process; the solid one is real. The gap in entropy is entropy generated by irreversibilities inside the machine.

Large modern steam and gas turbines reach 88–92%, compressors 85–90%, small pumps much less. Notice how a real turbine leaves its steam drier and a real compressor leaves its air hotter than the ideal ones would: the "lost" work has become internal energy of the fluid, and — via the second law — part of it can never be recovered.

Estimate first

A steam turbine generates 0.2 kJ/(kg·K) of entropy, with the environment at 25 °C. How much work, in kJ per kg of steam, has been lost for ever?

1101001,00010,000

Drag along the line to place your guess — each tick is ten times the last.

Check your intuition

Steam expands adiabatically through a turbine, and its entropy decreases from inlet to exit. What can you conclude?

Going deeper
Going deeper: entropy balance for control volumes

For open systems, entropy also travels with mass. The steady-state rate balance is

$0 = \sum_j \frac{\dot Q_j}{T_j} + \sum_\text{in} \dot m s - \sum_\text{out} \dot m s + \dot\sigma_\text{cv}.$

For a one-inlet, one-exit adiabatic device, $\dot\sigma = \dot m(s_2 - s_1)$ : the entropy generated is just the rise in the fluid's specific entropy. Pair it with the energy balance and you can locate irreversibilities device by device — the method Lesson 7 turns into a full exergy analysis.

For a polytropic process of an ideal gas, the same tools show that $n = \gamma$ with no friction is isentropic, and the work of an internally reversible steady-flow device is $w = -\int v\,dp$ : big for gases with large $v$ , tiny for liquids — the pump-versus-compressor lesson again, now from the second law.

Recap

What to carry forward

Entropy is a property: $dS = (\delta Q/T)_\text{int rev}$ . On a T–s diagram, the area under a reversible path is the heat.
Look it up in tables, or compute it from the T ds equations: $T\,ds = dh - v\,dp$ .
Entropy balance: $\Delta S = \int \delta Q/T + \sigma$ with $\sigma \ge 0$ . Irreversibilities generate entropy; isolated systems can only gain it.
Boltzmann: $S = k \ln W$ — entropy counts microscopic arrangements; the second law is overwhelming statistics.
Isentropic efficiency compares real turbines, compressors and pumps with reversible adiabatic ones. Lost work $= T_0\,\sigma$ .

Next: if every irreversibility destroys some work potential, can we measure how much, and where? Which losses really matter?