What is the difference between otto and carnot cycle

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Research Article January 01 Shaw John E. This Site. Google Scholar. Transactions of the Missouri Academy of Science 42 : 1—6. Get Permissions. Cite Icon Cite. This content is only available as a PDF. Send Email Recipient s will receive an email with a link to 'Comparing Carnot, Stirling, Otto, Brayton and Diesel Cycles' and will not need an account to access the content. Recipient Then the Otto cycle is more efficient, but as far as I know the most efficient is Carnot. Otto cycle consists of two isochoric and two isentropic processes.

Thus, although the numerator of the Otto efficiency approaches to zero, but the fraction doesn't approach to zero, because the denominator approaches to zero too and we will have an indeterminate form. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. As you drink more of the beer, the pitch goes down.

As air rushes in and out of the bottle, its velocity is highest at the bottleneck, and since kinetic energy is proportional to the square of the velocity, essentially all of the kinetic energy is that of the air that's in the neck. Substituting into the earlier result, we find that the resonant frequency is.

This is known as a Helmholtz resonator. As shown in figure d , a violin or an acoustic guitar has a Helmholtz resonance, since air can move in and out through the f-holes. Problem 10 is a more quantitative exploration of this. There are a number of different resonance peaks, some strong and some weak; the ones near and Hz are vibrations of the wood, but the one near Hz is a resonance of the air moving in and out through those holes shaped like the letter F.

The white lines show the frequencies of the four strings. We have already seen, based on the microscopic nature of entropy, that any Carnot engine has the same efficiency, and the argument only employed the assumption that the engine met the definition of a Carnot cycle: two insulated strokes, and two constant-temperature strokes. Since we didn't have to make any assumptions about the nature of the working gas being used, the result is evidently true for diatomic or polyatomic molecules, or for a gas that is not ideal.

How can it always turn out so simple, given the kind of mathematical complications that were swept under the rug in example 22?