Air-standard efficiency of the Otto cycle

η_Otto = 1 − 1/r^(γ−1).

Worked answer.

The Otto cycle has isentropic compression, constant-volume heat addition, isentropic expansion and constant-volume heat rejection. Its efficiency depends only on the compression ratio r and the ratio of specific heats γ:
η = 1 − 1 / r^(γ − 1).

Step 1: r = 8.
r^(γ−1) = 8^0.4. Since 8 = 2³, 8^0.4 = 2^1.2 ≈ 2.297.
η = 1 − 1/2.297 ≈ 1 − 0.4353 = 0.5647, or about 56.5%.

Step 2: r = 10.
10^0.4 ≈ 2.512.
η = 1 − 1/2.512 ≈ 1 − 0.3981 = 0.6019, or about 60.2%.

Step 3: Improvement. 60.2 − 56.5 ≈ 3.7 percentage points.

Check. The efficiency must increase with r and must be below the Carnot efficiency between the cycle's peak and minimum temperatures. Both conditions hold. Also note that the gain from r = 8 to 10 is smaller than from r = 6 to 8 (51.2% to 56.5%), showing the diminishing return as r rises.

What limits the compression ratio in petrol engines? Knock (detonation). At high compression, the unburnt end gas ahead of the flame gets hot enough to self-ignite. This produces pressure waves, a pinging sound, high local heat transfer and possible piston damage. Fuel octane rating, combustion chamber design, cooling and spark timing set the knock limit. Typical modern petrol engines use roughly 9 to 13, with direct injection and knock sensors pushing toward the upper end.

Real engines vs air-standard. Actual brake thermal efficiencies are much lower (often 25–35%) because of variable specific heats, heat loss, incomplete combustion, pumping and friction losses.

Common mistake. Using γ instead of γ − 1 as the exponent, which gives 1 − 1/8^1.4 ≈ 94.6%, an obviously unrealistic number.

Likely follow-up. "Why can diesel engines use compression ratios of 15 to 22?" They compress only air, and fuel is injected at the end of compression, so there is no end gas to knock.