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?
Best Answer
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.
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