The Carnot Cycle consists of 2 pairs of curves. The first pair are the Adiabatic or Work Curves.
The Work Curves (aka adiabatic, insulated) rise and fall exactly equal amounts of temperature The definition of the Carnot Cycle.
This means that the falling (expanding) curve does work equal to K * Delta T, where
Delta T = Temperature (T) Hot - Temperature (T) Cold. However, the rising/compressing curve does exactly the same magnitude of work, but negative, it equals -K *Delta T.
So the net work done by both Work Curves is zero, for all possible Carnot Cycles.
And ironically, this means the Carnot Cycle is the ONLY cycle for which the Work Ratio has no part in computing the amount of work it does.
No Cold Body?
Although it seems to require a Cold Body, it does not. How do you cool it off?
How does the Earth cool off every night? Earth is wrapped in vacuum, like a giant Thermos bottle. How can it cool off? It radiates heat. And during the day, Heat is added from the Sun, at Night, some of that heat leaves as radiation. A Huge Carnot Cycle. Air heats and expands during the day, and contracts at night, creating wind and weather.
How could you make a Carnot Cycle Machine in a vacuum? Put it in orbit around the sun, and make it rotate once every 24 hours. Needs a heat source, but not a cold source.
The work done in a Carnot cycle all happens during the isothermal portions of the curve.
Since the pressure (assuming no state changes) is proportional to the temperature, the ratio of the Work during expansion vs during compression can precisely be calculated by temperature ratios. One of few examples the temperature ratio is not approximate.
Last, there are an infinite number of cycles for which the Temperature ratio solution, identical to the Carnot Cycle solution applies. A simple example is to replace the adiabaitic curves with vertical lines, meaning heat is added or subtracted with no volume change. As long as the area under the temperature cahange curves are equal, the temperature ratio of the isothermal corves is the solution. This is an infinite class of possible engine cycles. Nothing Unique about the carnot cycle.
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The second pair of curves is the isothermal or uninsulated curves,
of the form
T:Hot/Volume and
T:Cold/Volume.
To calculate work, integrate ∫AB T:Hot/Volume - ∫A1B1 T:Cold/Volume to get work area where A is the smaller volume and B is the larger volume at ends of isothermal curve..
The solution to the net work, or area inside all the curves simplifies to
Pressure:Cold x (Ln(SmallVolume+Delta)*Expansion - Ln(SmallVolume+Delta))
Pressure:Hot x (Ln(SmallVolume)*Expansion - Ln(SmallVolume)) -
and further simplifies to
Pressure:Cold x Ln(Expansion)
Pressure:Hot x Ln(Expansion)
And finally the Ln()'s cancel out leaving:
W=K*(T:Hot-T:Cold)/T:Hot
Which looks like the Work Ratio, but isn't. The Work Ratio can be used to calculate the total amount of work done, by multiplying times T:Hot.
The solution can be couched in terms of a temperature ratio, or a pressure ratio, or the ratio of two quantities of work. It is the ratio of work done during isothermal compression to work done in isothermal expansion, because everything else cancelled out. Temperature is proportional to pressure (for gases), which is often proportional to total force or total energy. So it is unsurprising that temperature ratios can show up all over the place, pretty much any two measured quantities about a vapor undergoing change can in some fashion be expressed as proportional to temperature.
Here: As the Volume Change goes to zero, the work falls to zero, no matter the extremes of temperature.
As the Volume Change approaches infinity, the formula simplifies to
(T:Hot - T:Cold)
x Ln(Infinity) or just
(T:Hot - T:Cold) * Infinity, or just
Infinity
for any temperature difference, no matter how small.
In other words, the bounds on the amount of work that can be done by
any possible Carnot Cycle, is 0 to Infinity,
or completely independent of T:Hot and T:cold.
Consequently, the "Carnot"-Clapeyron-Kelvin-Work Ratio, does not solve the Carnot Cycle, in the sense it can be used to calculate total work from temperature. The total work must be calculated by measurement or calculation of the actual area's under the curve.
Lesson, every temperature ratio, is NOT a Work Ratio, even if they look the same. A Work Ratio can be used to calculate the amount of work done, from the vapor quantity and type, and maximum heat. This ratio cannot. Work Ratio is the division of a fixed quantity of energy between work and heat, and total energy is constant, and computable from the vapor properties and the ratio, even if the expansion factor changes. This is the ratio of energy put in the system to energy that has left the system. The total quantity of energy or total work is un computable from the ratio and vapor properties. It is also variable if the expansion factor changes. A subtle difference.
All of the above is a character of this design. Its only limit of expansion is the pressure Envelope. One could put compressed air in it, say 1000 times Atmospheric density, and it could expand far beyond Ambient Temperature on the cold side. Ambient Temperature plays no part in limiting efficiency. The "temperature" in T:Cold is determined solely by density and expansion, and expansion limit by the pressure envelope. For extremely high densities, a 1 degree rise above ambient of 300ish Kelvin, could expand to just above absolute zero. (Which would also liquefy or solidify the gases, another topic there.)
Granted, because of the design, some means of removing heat at the bottom is still needed, but read left panel for a "Temperature-less" alternative to a Cold Body.
Myths
Is it the best possible cycle? No. Its easy to show examples of "better" cycles. Here is a better cycle using steam.
Is it a "limit" of efficiency? No, this was disproved by Lord Kelvin in 1848, by the simple means of using another Carnot Cycle whose heat source is the isothermal compression curve. More here.
Does it prove heat is only partially converted to work? Again no. The work it does is the area under hotter isothermal expansion curve, which consumes an exactly equal amount of heat energy (because it is 100% efficient), minus the area under the cooler isothermal compression curve, which produces an exactly equal amount of heat energy (again, because its 100% efficient). Doesn't that "prove" an effective limit? No, the above two items show it is neither the "best" cycle nor is it necessary to discard the heat generated during isothermal compression.
When Sadi Carnot and Lord Kelvin both refer to reversibility as proof that all engines are EQUAL in being 100% efficient at converting heat to work, they refer to the transfer of heat to and from work itself. Engines do differ on how they add and subtract heat, but not on how much work they get for a given amount of heat. For all values of Gamma, the heat energy is equal to the work energy, forwards or backwards, for all engines. Engine "efficiency" is less than 100% only if heat is discarded by the engine.
A final word, at the same point the air Carnot Cycle is described, a steam cycle is described. In an air cycle, work is done to compress the air back into the original state, consuming energy. In a steam cycle, steam is cooled to condense back to liquid to return to the original state. In other words, energy is produced, instead of consumed, throughout the cycle. It takes no work to compress steam to liquid. In fact work could be produced during this step by creating a vacuum, and reclaiming the work done lifting the atmosphere as steam was produced. There is no fundamental requirement to do work to reset an engine cycle, hence no such "limit".
But my Physics book says... Yep, that's why this chapter is here.
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