Questions: Otto Cycle and Spark-Ignition Reciprocating Engines

Question 1

An automotive engineer proposes increasing a gasoline engine's fuel injection by 30% at the same compression ratio, expecting to improve thermal efficiency. Based on the ideal Otto cycle, what is the effect on efficiency?

• A. Efficiency increases because more heat input raises the peak cycle temperature
• B. Efficiency decreases because the higher heat input creates proportionally more waste heat
• C. Efficiency is unchanged because it depends only on compression ratio and γ, not on heat input
• D. Efficiency increases up to a stoichiometric limit, then decreases

Answer: (C)

Otto cycle efficiency η = 1 − 1/r_c^(γ−1) contains no term for heat input Q_in. Increasing fuel injection changes the total work output and power, but not the fraction of heat converted to useful work. More fuel means more total work AND more waste heat in the same ratio — efficiency stays constant. This is the key insight: efficiency is a property of the cycle geometry (compression ratio) and working fluid (γ), not the fuel quantity. To improve efficiency you must change r_c or γ, not the amount of fuel.

Question 2

Two ideal Otto cycles have compression ratios of r_c = 6 and r_c = 12 respectively, with the same working fluid (γ = 1.4). Compared to the r_c = 6 engine, the r_c = 12 engine:

• A. Has exactly twice the thermal efficiency, because efficiency scales linearly with compression ratio
• B. Has higher thermal efficiency, but less than twice as high, because efficiency scales as 1 − 1/r_c^(γ−1)
• C. Has the same efficiency if both operate on the same fuel
• D. Has lower efficiency because higher compression raises cylinder temperatures and heat losses

Answer: (B)

η = 1 − 1/r_c^(γ−1) is nonlinear in r_c. For γ = 1.4: η(r_c=6) = 1 − 1/6^0.4 ≈ 0.512; η(r_c=12) = 1 − 1/12^0.4 ≈ 0.630. Doubling the compression ratio improved efficiency from ~51% to ~63% — a meaningful gain, but not a doubling. Each increment in r_c yields diminishing returns, which is why the gain from r_c = 14 to r_c = 16 is smaller than the gain from r_c = 6 to r_c = 8. Option C is wrong because efficiency does not depend on fuel type under the air-standard assumption.

Question 3

According to the ideal Otto cycle model, a spark-ignition engine running on hydrogen and an engine with identical compression ratio running on gasoline would have the same thermal efficiency (assuming the same γ).

• T. True
• F. False

Answer: True

Ideal Otto cycle efficiency η = 1 − 1/r_c^(γ−1) depends only on compression ratio and γ, not on the fuel's chemical identity or heating value. Under the air-standard assumption, the working fluid is treated as an ideal gas regardless of the actual fuel. Two engines with identical r_c and γ have identical ideal efficiency. In practice, hydrogen's higher flame speed and different combustion properties affect real engine performance — but these are deviations from the ideal cycle, not predictions of the Otto cycle formula.

Question 4

The constant-volume heat addition process in the Otto cycle accurately describes real combustion in gasoline engines — the fuel burns so quickly that the piston barely moves during the process.

• T. True
• F. False

Answer: False

Constant-volume heat addition is an idealization. In reality, combustion takes a finite time — typically 15–40 degrees of crank rotation — during which the piston is moving, so volume changes throughout combustion. The Otto cycle assumes instantaneous heat release at fixed volume to produce a tractable analytical model. Real engines also lose heat through cylinder walls, experience friction, and use fuel-air mixtures rather than pure air. The Otto cycle captures the correct qualitative trends (efficiency rises with r_c) but systematically overpredicts actual efficiencies.

Question 5

Why does ideal Otto cycle thermal efficiency depend only on compression ratio and not on the amount of heat input (how much fuel is burned)? Explain using the cycle's structure.

Model answer: In the Otto cycle, both heat addition (2→3) and heat rejection (4→1) occur at constant volume. The isentropic compression and expansion processes connect the same two volumes (V_max and V_min), so the temperature ratios T₂/T₁ and T₄/T₃ both equal r_c^(γ−1) regardless of how much heat was added. When efficiency is computed as η = 1 − Q_out/Q_in = 1 − (T₄−T₁)/(T₃−T₂), both temperature differences scale proportionally with the heat added — Q_in cancels out, leaving only the compression ratio. Adding more heat raises all four temperatures proportionally; the fraction rejected stays the same.

This result is analogous to Carnot efficiency depending only on temperature limits. The Otto cycle's efficiency is set by how much the isentropic compression amplifies temperature (determined by r_c), not by absolute temperatures or fuel quantities. A practical implication: you cannot make a gasoline engine more efficient by using a richer mixture. The only paths to higher efficiency are increasing r_c (limited by knock), increasing γ (limited by mixture properties), or redesigning the cycle itself — for example, the Atkinson cycle uses a longer expansion stroke to extract more work from the same heat.