Otto’s four-stroke cycle (1876)

Nicolaus Otto’s 1876 engine compressed its charge before igniting it, a principle Alphonse Beau de Rochas had described in 1862. Its four strokes (intake, compression, power, exhaust) are idealized as the air-standard Otto cycle: isentropic compression through a ratio \(r\), heat added at constant volume, isentropic expansion and heat rejected at constant volume. The efficiency \(1 - r^{1-\gamma}\) (otto_efficiency(), OttoCycle) depends only on the compression ratio. Otto’s first engine had \(r \approx 2.5\); knock limits petrol engines to about 10-12 today.

import matplotlib.pyplot as plt
import numpy as np

from engineeringkit.thermodynamics import OttoCycle, isentropic_pressure_volume, otto_efficiency, thermal_efficiency
from engineeringkit.thermodynamics.visualizers import plot_pv_diagram

The cycle on the p-v diagram

One kilogram of air at 100 kPa and 300 K, compression ratio 8, 800 kJ/kg added by combustion.

R, gamma, r = 287.0, 1.4, 8.0
cv = R / (gamma - 1)
p1, T1, q_in = 100e3, 300.0, 800e3
v1 = R * T1 / p1
v2 = v1 / r
p2 = isentropic_pressure_volume(p1, v1, v2, gamma)
T2 = p2 * v2 / R
T3 = T2 + q_in / cv
p3 = R * T3 / v2
p4 = isentropic_pressure_volume(p3, v2, v1, gamma)
T4 = p4 * v1 / R
q_out = cv * (T4 - T1)

compress = np.linspace(v1, v2, 60)
expand = np.linspace(v2, v1, 60)
v_path = np.concatenate([compress, expand, [v1]])
p_path = np.concatenate([isentropic_pressure_volume(p1, v1, compress, gamma), isentropic_pressure_volume(p3, v2, expand, gamma), [p1]])
ax = plot_pv_diagram(v_path, p_path, label=f"r = {r:g}")
ax.set(xlabel="specific volume [m$^3$/kg]", title="Air-standard Otto cycle")

cycle = OttoCycle(r)
print(cycle.summary())
print(f"peak temperature {T3:.0f} K, peak pressure {p3 / 1e6:.1f} MPa")
print(f"efficiency from the four states {thermal_efficiency(q_in, q_out):.4f}")
assert np.isclose(thermal_efficiency(q_in, q_out), cycle.efficiency)
assert np.isclose(T2 / T1, T3 / T4)

{'compression_ratio': 8.0, 'gamma': 1.4, 'efficiency': np.float64(0.5647247183519378)}
peak temperature 1804 K, peak pressure 4.8 MPa
efficiency from the four states 0.5647

The compression ratio is everything

The efficiency does not depend on how much fuel is burned, only on how far the charge is compressed. Higher octane fuels resist knock and allow higher compression ratios.

ratios = np.linspace(1, 16, 151)
fig, ax = plt.subplots(figsize=(6, 3.5))
ax.plot(ratios, otto_efficiency(ratios))
for label, rr in (("Otto 1876", 2.5), ("1950s car", 7.0), ("modern petrol", 11.0)):
    ax.plot(rr, otto_efficiency(rr), "o", label=f"{label}: r = {rr:g}, η = {otto_efficiency(rr):.2f}")
ax.set(xlabel="compression ratio", ylabel="ideal efficiency", ylim=(0, 0.7))
ax.legend(fontsize=8)
fig.tight_layout()

plt.show()