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The Otto cycle & compression ratio

Strip a petrol engine of every practical complication — valves, flame speed, heat loss — and what remains is the air-standard Otto cycle: two adiabatic strokes and two constant-volume heat exchanges. It is simple enough to solve on one page, yet it delivers the single most important result in engine design: efficiency depends on compression ratio, and on almost nothing else.

The idealisation

“Air-standard” means we pretend the working fluid is plain air behaving as an ideal gas with constant γ=1.4\gamma = 1.4, that combustion is replaced by heat qinq_{in} added from outside, and that the exhaust/intake pair is replaced by heat qoutq_{out} rejected at constant volume. Four processes, four numbered states:

  • 1 → 2: isentropic (adiabatic, frictionless) compression from V1V_1 to V2V_2.
  • 2 → 3: heat added at constant volume — the idealised spark-ignition burn, so fast the piston hasn’t moved.
  • 3 → 4: isentropic expansion back to V1V_1 — the power stroke.
  • 4 → 1: heat rejected at constant volume — blowdown and gas exchange, collapsed to an instant.
12341→2 isentropic compression2→3 heat in (const V)3→4 isentropic expansion4→1 heat out (const V)p/p1V/VTDC (1 → r)
Fig. 1. The ideal Otto cycle for r = 10. Work in is the area under 1→2; work out is the larger area under 3→4; the enclosed loop is the net.

Deriving the efficiency

Both heat exchanges happen at constant volume, where for an ideal gas q=cv ΔTq = c_v\,\Delta T. So with the compression ratio r=V1/V2r = V_1 / V_2:

qin=cv(T3−T2),qout=cv(T4−T1)q_{in} = c_v (T_3 - T_2), \qquad q_{out} = c_v (T_4 - T_1)(1)

Efficiency is work out over heat in, and the first law makes net work w=qin−qoutw = q_{in} - q_{out}:

η=wqin=1−qoutqin=1−T4−T1T3−T2\eta = \frac{w}{q_{in}} = 1 - \frac{q_{out}}{q_{in}} = 1 - \frac{T_4 - T_1}{T_3 - T_2}(2)

Now use the isentropic relation TVγ−1=constT V^{\gamma-1} = \text{const} on both adiabats. They span the same volume ratio rr, so:

T2T1=rγ−1=T3T4\frac{T_2}{T_1} = r^{\gamma-1} = \frac{T_3}{T_4}(3)

That means T3=T4 rγ−1T_3 = T_4\,r^{\gamma-1} and T2=T1 rγ−1T_2 = T_1\,r^{\gamma-1}; substitute into (2) and the temperature differences cancel beautifully:

ηOtto=1−T4−T1(T4−T1) rγ−1=1−1rγ−1\eta_{\text{Otto}} = 1 - \frac{T_4 - T_1}{(T_4 - T_1)\,r^{\gamma-1}} = 1 - \frac{1}{r^{\gamma-1}}(4)
rr
compression ratio V₁/V₂ = V_BDC/V_TDC [–]
γ\gamma
ratio of specific heats cp/cv (1.4 for cold air) [–]
cvc_v
specific heat at constant volume [J/(kg·K)]
20%40%60%petroldiesel4812162024compression ratio rη (air-standard)
Fig. 2. Equation (4) plotted. The curve is steep at low r and flattens past ~14 — the shaded bands mark where petrol (knock-limited) and diesel engines actually live.

Try it — the demo below evaluates (4) live:

η=1−10.5 1−γ\eta = 1 - 10.5^{\,1-\gamma}η=61.0 %\eta = 61.0\,\%

Modern petrol range. Beyond ~12:1 pump petrol knocks — the mixture detonates before the spark-timed flame arrives.

Why not crank r to 20?

Because of knock. Compressing the charge heats it (that is equation (3) working against you): squeeze petrol–air mixture past roughly 11–12:1 and the unburned gas ahead of the flame front — the end gas — gets hot enough to detonate spontaneously. The resulting pressure spikes hammer the piston like a bell and can destroy an engine in minutes. Octane rating measures a fuel’s resistance to exactly this. Direct injection, charge cooling and knock sensors have pushed modern engines to ~12–14:1, but the ceiling is chemical, not mechanical.