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The Otto cycle's thermodynamic efficiency is often described using the compression ratio:

$$\eta_{th} = 1 - \left( \frac{1}{CR} \right ) ^ {k-1},$$

(where CR is the compression ratio, and k the ratios of specific heat).

Now I tried describing this using a pressure ratio (So instead of $\frac{V_1}{V_2}$ using $\frac{p_1}{p_2}$. Taking into account the compression in a cycle is (in the ideal case) an isentropic process, one could use $$p_1V_1^k = p_2V_2^k$$ $$\frac{P_2}{P_1} = \left ( \frac{V_1}{V_2} \right ) ^ {k} = \Pi$$

Combining these I would get: $$\eta_{th} = 1 - \left( \frac{1}{\Pi ^ {\frac{k-1}{k}}} \right ) $$

However this is exactly the same as the Brayton efficiency.. While the cycles are different (Otto uses isochoric combustion, while Brayton uses an isobaric expansion during combustion). I wanted to compare those two, what did go wrong?

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  • $\begingroup$ \frac does work for me. Can you give a specific example that doesn't work for you? $\endgroup$ – Davide Cervone Feb 24 '14 at 12:34
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You did everything correctly. For the same compression ratio, the Brayton cycle efficiency is equal to Otto cycle efficiency. You're right that the cycles are different, however both start with isentropic compression, and efficiency of both in ideal case can be expressed as $\eta_{th}=1-\frac{T_1}{T_2}$, so only compression process matters. That's the case with theory and ideal processes: they're often simple but they don't really exsist.

Reference

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  • $\begingroup$ So that would now mean there is no inherent difference -> and that the efficiency is equal to the carnot cycle's efficiency? $\endgroup$ – paul23 Mar 1 '14 at 0:21
  • $\begingroup$ @paul23 I don't know what You mean by "inherent difference", but it won't be equal to Carnot cycle's efficiency. The formulas may look similar, But in Carnot cycle compression is isothermal, so there won't be any change in temperature during the compression. In Carnot cycle efficiency formula You have to use cold and hot reservoir temperatures. $\endgroup$ – Wojciech Mar 1 '14 at 11:00

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