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By Kelvin's statement of second law of thermodynamics, we can't have an engine operating between two heat reservoir having efficiency 1.
I think that if we have infinite engines between two heat reservoirs then, the efficiency of overall engine can tend to 1.
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So we can see that $Q_{i+1}=Q_i-W_i \tag{1}$
Thus $Q_1>Q_2>Q_3>......>Q_n>...$
We can always adjust the temperature $T_c$ such that $Q_i\underline{>}0$ (every heat engine dumps heat to next one instead of taking). As $Q_i$ is a monotone decreasing sequence and has a lower boundso by monotone convergence theorem it tends to its infimum.
Suppose all the engines have same efficiency, $\eta$, where $\eta=\frac{W_i}{Q_i}\;\forall i$.
So, (1) becomes, $Q_{i+1}=Q_i-\eta Q_i$
$\implies \lim_\limits{n\to\infty}Q_{i+1}=\lim_\limits{n\to\infty}Q_i-\eta\lim_\limits{n\to\infty}Q_i$
$\implies \eta\lim_\limits{n\to\infty}Q_i=0$
$\implies \lim_\limits{n\to\infty}Q_i=0$

So, theoretically, we can take infinite heat engines between two heat reservoirs such that each heat engine dumps heat to the next one. Then we get an equivalent heat engine which absorbs $Q_1$ amount of heat and does work $\sum_i W_i$ which tends to $Q_1$, thus violating the Kelvin's statement of second law of thermodynamics. Is my reasoning correct or I have made some mistake? Does it theoretically violates the Kelvin's statement?

$\mathbf{Edit}$ - I have found that the question which I ask is not so precise. So I am rephrasing my question.

We know that a process is said to be thermodynamic if it follows first and second law of thermodynamics (I haven't studied 3rd law). If I have two heat reservoirs at temperature Th and Tc with Th>Tc (say 1000K and 500K respectively) having infinite heat engines with each having same efficiency (so they are not carnot engines as efficiency is independent of temperature of heat reservoirs), operating between them such that each heat engine dumps heat to the next. So the subsequent difference of temperature at which heat exchange takes place between them tend to zero. So, heat dumped by the engines also tend to zero and net work tend to Q1 as shown in my question.
So in the overall heat engine, heat taken by engine completely gets converted into work done by the engine. So we can see that the overall heat engine does not follow Kelvin's statement thus not undergoing thermodynamic process. But each engine in infinite set of engines follows first law and Kelvin's statement. So can we make a conclusion that if we consider a infinite sequence of thermodynamic processes then the net process which consists of all those thermodynamic processes may not be thermodynamic?
This is my exact question. I am extremely sorry for asking less precise question.

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  • $\begingroup$ Hm, doesn't the temperature difference between two engines also go to zero in this limit, which would bring the heat flow/engine asap to a stop ? $\endgroup$
    – Hans Wurst
    Mar 24, 2021 at 15:40
  • $\begingroup$ @Hans, may you please read my comment to Bob D's answer and clarify the doubt. $\endgroup$
    – Iti
    Mar 24, 2021 at 16:11

2 Answers 2

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By Kelvin's statement of second law of thermodynamics, we can't have an engine operating between two heat reservoir having efficiency 1.

Correct, and the reason is the Kelvin-Planck statement says no heat engine can operate in a cycle while transferring heat with a single heat reservoir, meaning some heat must be rejected to another reservoir resulting in less heat available to do work. For the efficiency to be 1, all of the heat taken from the heat reservoir would be converted to work in a cycle.

The corollary to the Kelvin-Planck statement is no heat engine can have a higher efficiency than a Carnot Cycle operating between the same reservoirs. The Carnot efficiency is

$$\eta=1-\frac{T_L}{T_H}$$

So in theory the efficiency of a Carnot heat engine it can approach 1 for if either $T_{L}\rightarrow 0$ or if $T_H$ $\rightarrow \infty$.

So, theoretically, we can take infinite heat engines between two heat reservoirs such that each heat engine dumps heat to the next one.

Sure, but I fail to see how the overall combined efficiency of your infinite number of heat engines is any different than one Carnot heat engine operating between the same two reservoirs which, as already pointed out, can theoretically approach 1.

𝐄𝐝𝐢𝐭 - I have found that the question which I ask is not so precise. So I am rephrasing my question.

Your edit has not helped me understand your point. Be that as it may, would you not agree that the more heat engines you operate between $T_h$ and $T_c$ the smaller the temperature difference between the input and output of each engine and the less the difference between the heat input and heat output for each engine.

The work done by any of your heat engines is

$$W_{i}=Q_{i}-Q_{i+1}$$

And the efficiency of each heat engine is

$$\eta_{i}=\frac{W_{i}}{Q_i}=1-\frac{Q_{i+1}}{Q_i}$$

As you place more engines between the two reservoirs the less the difference in input and output temperatures and heat transfers for each engine, so that

$$W_{i}=Q_{i}-Q_{i+1}\rightarrow 0$$

and

$$\eta=1-\frac{Q_{i+1}}{Q_i}\rightarrow 0$$

So while the addition of each heat engine adds work to the total, the work done by each additional engine approaches zero. The net effect is the work done is equal to that of a single engine operating between the two reservoirs, and the efficiency of that engine is

$$\eta=1-\frac{Q_L}{Q_{H}}$$

There is no violation of the second law.

Hope this helps.

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  • $\begingroup$ Thanks for the reply. Here $\eta$ can be anything it is just a number between 0 and 1, which may not be equal to that of carnot engine. $\endgroup$
    – Iti
    Mar 24, 2021 at 15:59
  • $\begingroup$ Also, I have studied that we can't reach 0 K thermodynamically because then that process is not thermodynamic as it violates the Kelvin's statement (as carnot engine then can have efficiency 1). But the framework which I proposed of infinite engines (which may not be carnot engine) , I think it is thermodynamic because each engine follows Kelvin's statement and first law of thermodynamics but overall engine violates the Kelvin's statement. Please clarify the doubt I am very confused. $\endgroup$
    – Iti
    Mar 24, 2021 at 16:10
  • $\begingroup$ @Iti I'm sorry, but I can't follow your logic. Good luck. $\endgroup$
    – Bob D
    Mar 24, 2021 at 16:19
  • $\begingroup$ I have tried to explain my question more precisely. We say that a process is thermodynamic if it follows first and second law of thermodynamics (I haven't studied 3rd law). If I have a heat reservoir at Th and Tc (say 1000K and 500K respectively) having infinite heat engines(which may not be carnot), operating between them such each heat engine dumps heat to the next. So the subsequent difference of temperature at which heat exchange takes place between them tend to zero. So, heat dumped by the engines also tend to zero and net work tend to Q1 as shown in my question. $\endgroup$
    – Iti
    Mar 24, 2021 at 16:30
  • $\begingroup$ So in the overall heat engine, heat taken by engine completely gets converted into work done by the engine. So the question is does this overall heat engine not follow a thermodynamic process (as Kelvin's statement is violated)? But each engine follows the thermodynamic process (follows first law and Kelvin's statement)? Sorry for asking the less precise question. $\endgroup$
    – Iti
    Mar 24, 2021 at 16:34
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You can't have all the engines have the same fixed efficiency in the limit of an infinite number of heat engines. No matter what the final temperature is, the temperature difference $\Delta T_i = T_{i} - T_{i+1}$ at each stage must approach zero as the number of engines goes to infinity (since $\sum_{i=0}^n \Delta T_i = T_h - T_c$.) This means that the efficiency $\eta_i$ of each stage must also go to zero as $n \to \infty$, since the efficiency at each stage $i$ will be $$ \eta_i < 1 - \frac{T_{i+1}}{T_i} = \frac{\Delta T_i}{T_i} \to 0 \text{ as $n \to \infty$.} $$ This means that your logic should be \begin{align*} Q_{i+1} &= Q_i - \eta_i Q_i \\ \lim_{n \to \infty}Q_{i+1} &= \lim_{n \to \infty}Q_i - \lim_{n \to \infty}\eta_i Q_i \\ \lim_{n \to \infty}\eta_i Q_i &= 0 \end{align*} But since $\lim_{n \to \infty} \eta_i = 0$, this does not imply that $\lim_{n \to \infty} Q_i = 0$. In fact, it is entirely plausible mathematically that $\lim_{n \to \infty} Q_i > 0$; and indeed, it is required by the laws of thermodynamics.

Concerning the invocation of the monotone convergence theorem: I think you have confused a lower bound on a sequence of numbers with the infimum of that sequence. As an example, consider the following monotonically decreasing sequence of positive numbers: $$2, \frac{3}{2}, \frac{5}{4}, \frac{9}{8}, \frac{17}{16}, \frac{33}{32}, ..., 1 + \frac{1}{2^n}, ...$$ This is a monotonically decreasing positive sequence whose infimum is not zero. The number 0 is less than all numbers in the sequence, but it is not the largest such number with that property, and so it is not the infimum of the sequence. (It shouldn't be too hard to see that 1 is the infimum instead.)

Similarly, in your construction, it is true that all of the $Q_i$ numbers satisfy $Q_i > 0$; but that doesn't mean that their infimum is zero. In fact, your calculation is a pretty good demonstration that their infimum must be positive, not zero.

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  • $\begingroup$ thank you for pointing out the mistake, but still $Q_i$ tend to zero as $Q_i$ is monotonically decreasing sequence and is greater than or equal to 0(since each heat engine dump heat to the next one instead of taking), so by monotone convergence theorem $Q_i$ tend to 0, and still the overall heat engine violates the Kelvin's statement. $\endgroup$
    – Iti
    Mar 24, 2021 at 18:07
  • $\begingroup$ @lti: 0 is a lower bound on the sequence, but it's not necessarily the infimum. See my edit. $\endgroup$ Mar 24, 2021 at 19:21
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    $\begingroup$ thanks a lot now I have understand. Basically my mistake is that I am considering $\eta$ to be constant and thus get the absurd result. As $\eta$ gets decreased the heat taken by engine mostly get dumped into the next engine with very little work done. And as a result, when $i$ gets higher, heat taken by engine approximately gets equal to the heat dumped by that and thus $Q_i$ can't tend to 0. $\endgroup$
    – Iti
    Mar 25, 2021 at 6:02

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