Intake
θ 0°p 1 barT 300 K
IntakeCompressionPowerExhaust
Valve timing (cyl. 1)
T–s diagram (ideal)
Crankshaft torque
1. The four strokes
One thermodynamic cycle takes two crank revolutions (720°). With crank angle $\theta$ measured from top dead centre (TDC, piston at its highest point) at the start of intake; bottom dead centre (BDC) is its lowest point:
| Stroke | Crank angle | Valves | What happens |
|---|---|---|---|
| Intake | 0–180° | Inlet open | Piston descends, fresh air–fuel charge drawn in through the throttle. |
| Compression | 180–360° | Both closed | Charge compressed by the ratio $r$; spark fires shortly before TDC. |
| Power (expansion) | 360–540° | Both closed | Burning gas pushes the piston down: the only stroke that produces work. |
| Exhaust | 540–720° | Exhaust open | Piston rises and pushes the burnt gas out. |
In a real engine the valves open early and close late to use the gas momentum: here the inlet valve opens (IVO) 10° before TDC and closes (IVC) 40° after BDC; the exhaust valve opens (EVO) 40° before BDC and closes (EVC) 10° after TDC. The period when both valves are open around TDC is the valve overlap.
2. Slider-crank kinematics
The slider-crank turns the piston's up-and-down motion into rotation. With bore $B$ (cylinder diameter), stroke $S$ (piston travel), crank radius $r_c = S/2$, connecting rod length $l$, rod ratio $\lambda = l/r_c$. The piston's distance below TDC is exact geometry, not a sine wave:
$$x(\theta) = r_c(1-\cos\theta) + l - \sqrt{l^2 - r_c^2\sin^2\theta}$$
$$V(\theta) = V_c + \frac{\pi B^2}{4}x(\theta),\qquad V_d = \frac{\pi B^2}{4}S,\qquad r = \frac{V_c+V_d}{V_c}\Rightarrow V_c = \frac{V_d}{r-1}$$
Because the rod is inclined, the piston covers more than half the stroke in the first 90° after TDC: it moves faster near TDC than near BDC. The velocity is $\dot x = \omega\,r_c\sin\theta\left(1+\frac{r_c\cos\theta}{\sqrt{l^2-r_c^2\sin^2\theta}}\right)$ and the mean piston speed is $\bar U_p = 2SN/60$.
3. The ideal Otto cycle
Why it matters: this simple model explains why squeezing the charge harder (higher compression ratio) gets more work out of the same fuel. The air-standard model replaces combustion by heat addition at constant volume and uses an ideal gas with constant $\gamma = c_p/c_v$:
- 1→2 isentropic compression (fast, so no heat in or out, and no friction): $p_2 = p_1 r^{\gamma}$, $T_2 = T_1 r^{\gamma-1}$.
- 2→3 constant-volume heat addition: $Q_{in} = m c_v (T_3 - T_2)$.
- 3→4 isentropic expansion: $T_4 = T_3 / r^{\gamma-1}$.
- 4→1 constant-volume heat rejection: $Q_{out} = m c_v (T_4 - T_1)$.
$$\eta_{Otto} = 1 - \frac{Q_{out}}{Q_{in}} = 1-\frac{1}{r^{\gamma-1}},\qquad W = \oint p\,dV = \eta_{Otto}\,Q_{in}$$
Efficiency depends only on the compression ratio and $\gamma$. Petrol engines are limited to $r\approx 9$–$13$ by knock (auto-ignition of the end gas). The simulator uses $T_1 = 300$ K, $R = 287$ J/(kg·K) and adds $q_{in} = 1800$ kJ per kg of charge, a common textbook air-standard value. (A stoichiometric petrol–air mixture holds about 2.8 MJ of chemical energy per kg; the lower figure allows for residual exhaust gas, incomplete combustion and dissociation.)
4. Work, mean effective pressure, power and torque
$$\text{MEP} = \frac{W_{cycle}}{V_d},\qquad P_i = \frac{W_{cycle}\,n_{cyl}\,N}{2\times60},\qquad \bar T = \frac{W_{cycle}\,n_{cyl}}{4\pi}$$
The factor 2 (or $4\pi$ rad) appears because a four-stroke engine fires each cylinder once every two revolutions. MEP is the constant pressure that would do the same work in one stroke; it lets you compare engines of different size (good naturally aspirated petrol engines: 10–13 bar indicated).
The instantaneous torque from one cylinder (gas forces only, ignoring inertia and friction) is
$$T(\theta) = (p - p_{cc})\,A_p\,\frac{dx}{d\theta},\qquad \frac{dx}{d\theta}=r_c\sin\theta\left(1+\frac{r_c\cos\theta}{\sqrt{l^2-r_c^2\sin^2\theta}}\right)$$
5. The realistic cycle
The realistic curve is a single-zone model. Between inlet closing and exhaust opening the cylinder is a closed system, and the first law gives
$$\frac{dp}{d\theta} = -\gamma\frac{p}{V}\frac{dV}{d\theta} + \frac{\gamma-1}{V}\frac{dQ}{d\theta},\qquad x_b(\theta) = 1-\exp\!\left[-5\left(\frac{\theta-\theta_s}{\Delta\theta}\right)^{3}\right]$$
where $x_b$ is the Wiebe mass-fraction-burned curve, starting at spark timing $\theta_s$ and lasting $\Delta\theta = 50°$. About 20% of the fuel energy is lost to heat transfer and incomplete combustion. Burning takes time, so the peak pressure is lower and later than in the ideal cycle. Firing too early wastes work pushing against the rising piston; too late wastes expansion. The best compromise is MBT timing (minimum advance for best torque).
The lower loop is the pumping loop: the piston draws charge in at manifold pressure below atmospheric (because of the throttle) and pushes exhaust out above atmospheric. It is traced anticlockwise and its area is negative work. At part throttle it becomes large: this is one reason petrol engines are inefficient at light load, and why diesels, which control load by fuel quantity rather than a throttle, lose much less at part load.
6. The Diesel cycle
In a compression-ignition engine fuel is injected into hot compressed air and burns roughly at constant pressure. With cut-off ratio $\rho = V_3/V_2$:
$$\eta_{Diesel} = 1 - \frac{1}{r^{\gamma-1}}\,\frac{\rho^{\gamma}-1}{\gamma(\rho-1)}$$
The bracket is greater than 1, so at the same compression ratio the Otto cycle is more efficient. But diesels run at $r = 15$–$22$ without knock, so in practice they are more efficient. Turn on the Diesel comparison to see both effects.
Model limits. Constant $\gamma$, no dissociation, a single gas zone, a fixed 20% heat loss, simplified gas exchange (prescribed pressures instead of flow through valves) and no friction. Torque is indicated torque from gas pressure; brake torque is lower by friction and pumping losses, and real engines also have inertia torques from the reciprocating masses.
Challenges
Checked live as you change the engine.
Solved 0 of 0.
Guided experiments
- Efficiency vs compression. Record ηOtto at r = 6, 8, 10, 12, 14. Plot it. Why does the curve flatten?
- γ matters. At r = 10, change γ from 1.40 (air) to 1.30 (closer to real burning mixture). How much efficiency is "lost"?
- Part load. Select Realistic. Close the throttle from 100% to 20% and watch the pumping loop grow (use log axes to see it clearly). Note how indicated efficiency drops.
- Spark timing. Sweep spark advance from 0° to 45° at full throttle and note net work. Where is the maximum? What happens to peak pressure?
- Rod ratio. At 90° crank angle compare the piston position for l/r = 2.6 and 5. Which is closer to half stroke, and why?
- Smoothness. Compare the crankshaft torque chart for 1, 2 and 4 cylinders at the same total displacement. Why do larger engines usually have more cylinders?