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In an imagined scenario, $n$ moles of steam at a temperature $T_h$ and pressure $p$ enter an engine. The steam leaving the engine is at a temperature $T_c$ and pressure $p$.

In this scenario, the heat lost by the steam is equal to $Cp(T_c-T_h)n$ , which is also equal to the net work done by the engine.

I am confused by this equality as it brings up the question of which aspect of heat transfer is the limiting factor of the efficiency of this engine: Is the limiting factor the amount of heat that can be extracted from the n moles of steam(which is equal to $Q_h+Q_c$), or is the limiting factor the amount of work the extracted heat can be converted into? The equation for the work output of a heat engine is given by $Q_h+Q_c$, but how is it known for sure that this quantity of heat is completely transformed into work?

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    $\begingroup$ "In this scenario, the heat "lost" by the steam is...equal to the net work done by the engine." This is incorrect; the Second Law precludes turning heat entirely into work, as this would entail destroying entropy. You must multiply the heat by the Carnot factor $1-T_\mathrm{cold}/T_\mathrm{hot}$ to obtain the maximum work obtainable. The remaining energy must be rejected to the cold reservoir via heating to balance the entropy books. $\endgroup$ May 6 at 22:57
  • $\begingroup$ @Chemomechanics, but the heat that can be extracted from the steam that enters the engine isn't entirely used as the output temperature of the steam isn't equal to 0. Doesn't this satisfy the requirement that all heat input isn't converted into work? $\endgroup$
    – Piksiki
    May 6 at 23:24
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    $\begingroup$ No; the requirement stands even if the steam is cooled only slightly. You still can't convert this thermal energy entirely into work, as heating carries entropy but work doesn't. $\endgroup$ May 6 at 23:30
  • $\begingroup$ Are you describing a Carnot heat engine? Also note that the amount of work done by a Carnot heat engine is equal to $Q_h - Q_c$, not $Q_h + Q_c$. $\endgroup$ May 7 at 2:39
  • $\begingroup$ Is your heat engine a steam turbine? And if so, is it reversible? $\endgroup$
    – Bob D
    May 7 at 12:34

2 Answers 2

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What exactly is the limiting factor of the efficiency of a heat engine?

The short answer to the title of your post, the limiting factors are the maximum and minimum temperatures between which the heat engine operates, whether or not those temperatures are fixed, and whether or not the heat engine cycle is reversible.

For a reversible cycle between two fixed thermal reservoirs the maximum efficiency is the Carnot efficiency $\eta=1-T_{c}/T_h$.

Is the limiting factor the amount of heat that can be extracted from the n moles of steam (which is equal to $Q_h+Q_c$), or is the limiting factor the amount of work the extracted heat can be converted into?

For conservation of energy, for any heat engine the work done by the engine is always the difference between the heat added, $Q_h$, and the heat rejected $Q_c$. Or, where it is understood that $Q_h$ is positive and $Q_c$ is negative,

$$W=Q_{h}+Q_c$$

The limiting factor is to how much heat, $Q_c$, is rejected. For a reversible engine, $Q_c$ is a minimum. For an irreversible engine, $Q_c$ is greater than that for a reversible engine.

The equation for the work output of a heat engine is given by $Q_h+Q_c$, but how is it known for sure that this quantity of heat is completely transformed into work?

We know for sure because energy must be conserved. The important point is that not all the heat added $Q_h$ is converted to work in the cycle. That would violate the Kelvin-Planck statement of the second law:

No heat engine can operate in a cycle while transferring heat with a single heat reservoir.

In other words, some of the heat taken from the hot reservoir must be rejected to a cold reservoir. With regard to the maximum possible efficiency of the engine, the following Corollary to the Kelvin-Plank statement applies:

No heat engine can have a higher efficiency than a Carnot Cycle operating between the same reservoirs

The above said, I have a couple of comments with respect to your steam engine example.

  1. It is important to note that $C_p$ for steam is a function of temperature, i.e., it is not a constant. The greater the range in temperatures the greater the variation.

  2. If the steam entering and exiting the engine are at different temperatures but the same pressure, then there are two possibilities: (1) The entering and exiting steam are both superheated or (2) the entering steam is superheated while the exiting steam is saturated vapor.

  3. For a steam engine (e.g., steam turbine), rather than trying to determine the work done using your equation, the work done by the turbine equals the difference between the inlet and exit specific enthalpies of the steam, or

$$w=h_{i}-h_e$$

You can obtain the specific enthalpies from the superheated water tables.

Hope this helps.

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  • $\begingroup$ So is the heat lost by the steam indeed equal to the net work done by the system? $\endgroup$
    – Piksiki
    May 7 at 23:28
  • $\begingroup$ The heat added to the heat engine minus the heat rejected by the engine equals the work done by the engine $\endgroup$
    – Bob D
    May 7 at 23:35
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Let us consider gas, cylinder, and a piston.

  1. The only energy that the gas has is thermal energy.
  2. The only work the gas does is pushing the piston

From 1 and 2 it follows that: lost thermal energy = energy used to push the piston

So the efficiency of the energy conversion is 1. Carnot-efficiency of the machine is less than 1, if any thermal energy is left unconverted, which is the case if the machine stops in finite time. That is, if the only moving part of the machine stops. I mean the piston.

Let us now consider gas, infinitely long cylinder, and a piston, in the vacuum of space.

After the gas has pushed the piston an infinite distance, then the situation is:

lost thermal energy = all thermal energy = energy used to push the piston

efficiency in this case is:

efficiency = energy used to push the piston / all thermal energy = 1

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