0
$\begingroup$

This is the proof I've found for the Carnot Factor:

$$S = \frac{Q}{T} \qquad \qquad (1)$$

$$S_{in} = S_{out}\qquad \qquad (2)$$

Because entropy stays the same when using an ideal engine: $$ (1) \& (2) \Rightarrow \frac{Q_{in}}{T_{in}} = \frac{Q_{out}}{T_{out}} \qquad \qquad (3)$$ $$ (1) \& (2) \Rightarrow Q_{out}= Q_{in}\frac{T_{out}}{T_{in}} \qquad \qquad (3)$$

From the energy conservation law

$$ W = Q_{in} - Q_{out}\qquad \qquad (4) $$

Then from (5) (3) & (4), the Carnot factor $\eta $ is calculated as:

$$ W = Q_{in} - Q_{in}\frac{T_{out}}{T_{in}}$$ $$ W = Q_{in}\left(1 - \frac{T_{out}}{T_{in}}\right) $$ $$ \eta = \frac{W}{Q _{in}} = \left(1 - \frac{T_{out}}{T_{in}}\right) $$

I don't see, why $T_{out}$ must be smaller than $T_{in}$. There are no restrictions that says $T_{in} > T_{out}$. It's just necessary, that $S_{in} = S_{out}$. Maybe my book simplifies certain aspects?

$\endgroup$
3
  • $\begingroup$ Which book is that? $\endgroup$
    – Solar Mike
    Mar 21 at 6:18
  • 1
    $\begingroup$ Search for Carnot Efficiency, Carnot Theorem or Carnot Limit for more info. But T_in exceeds T_out because heat travels from hot to cold unless you apply energy to move heat the other way. $\endgroup$
    – Solar Mike
    Mar 21 at 6:24
  • $\begingroup$ @SolarMike German book: Energie - Wie verschwendet man etwas, das nicht weniger werden kann? $\endgroup$
    – DarkTrick
    Mar 28 at 5:01
2
$\begingroup$

There are a lot of misconceptions in your question. Let's first start by the Carnot cycle.

Given a two reservoirs at temperatures $T_H$ (hot reservoir) and $T_L$ (cold reservoir) where $T_H > T_L$, what is the upper limit of the mechanical work the can be generated if we added a heat engine between the two reservoirs? or in other words, what is the maximum efficiency of that engine?

Naturally, the cycle that consists of process that require the least amount of work, will deliver the the maximum work, such process are the reversible process, so Carnot proposed a theoretical cycle working on ideal gas that consists merely of reversible process: reversible isothermal expansion (A to B), , reversible adiabatic expansion (B to C), reversible isothermal compression (C to D) and reversible adiabatic compression (D to A). And on the $T-S$ diagram, the cycle will look like this.

Carnot T-S Diagram

Let's calculate the heat added and rejected by the system: $$ Q_H = W_{AB} = NR\ T_H \ln(\frac{V_B}{V_A})$$ $$ Q_L = W_{CD} = NR\ T_L \ln(\frac{V_D}{V_C}) = - NR\ T_L \ln(\frac{V_C}{V_D})$$

By substituting in efficiency, noting that volume of cylinder at A = volume at D and same for C & B: $$ \eta = 1 - \frac{Q_L}{Q_H} = 1 - \frac{T_L}{T_H}$$

Now back to your question:

$$S = \frac{Q}{T} \qquad \qquad (1)$$

$$S_{in} = S_{out}\qquad \qquad (2)$$

Actually, $ds \ge \frac{\delta Q}{T}$, and for the Carnot cycle $\Delta S_H = \Delta S_C$. You need to review Entropy and second law of thermodynamics.

I don't see, why $T_{out}$ must be smaller than $T_{in}$. There are no restrictions that says $T_{in} > T_{out}$.

Those are temperature reservoirs, by definition, one is hot and the other is cold.

$\endgroup$
1
  • $\begingroup$ My question was aiming for the part you mathematically skipped with By substituting in efficiency,: Why would Q_L go on top in the first place? As far as I understand (now), that is just a convention. I.e. talking about "Carnot Factor" says "Q_L" goes on top, but talking about the math behind it, Q_L can go in the bottom. That way the result will e.g. also tell you about the exergy flow (negative => from cold to warm). $\endgroup$
    – DarkTrick
    Mar 28 at 7:28
0
$\begingroup$

It is as fundamental as the proof you've given.

The reason why $T_{in}$ has to be smaller than a $T_{out}$ is due to the second Law of thermodynamics.

There exists for every thermodynamic system in equilibrium an extensive scalar property called the entropy, $ S$ , such that in an infinitesimal reversible change of state of the system, $ dS = dQ/T$ , where $ T$ is the absolute temperature and $ dQ$ is the amount of heat received by the system. The entropy of a thermally insulated system cannot decrease and is constant if and only if all processes are reversible.

This is essentially a statement of the everyday observation that heat (energy) naturally flows from areas with higher temperature to areas with lower temperature.

Another way to look at it, is that areas with high thermal energy density dissipate energy more quickly than they receive to adjacent areas with lower thermal energy density.

In a more philosophical tone, As I've read somewhere (too long ago so I can give a proper reference), Nature abhors a vacuum (whether that is a pressure or energy vacuum) and so it rushes to fill it. In that natural process, everything will eventually become equal. But becoming equal (or a better term same), means that there is no reason for exchanging information/energy. There needs to be a type of gradient to instigate the exchange.

The word machine (μηχανή), comes from a word that means to think way of doing something (μηχανευομαι). So machines were tools invented and used to perform contrary to the natural way for the benefit of the man.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.