The Otto cycle | Engimations

By the end of this topic you will be able to

Describe the four processes of the ideal Otto cycle on a P–v diagram.
Calculate the temperatures and pressures at the four states.
Calculate thermal efficiency and net work, and explain the effect of compression ratio.

The engine idealised

The Otto cycle models a spark-ignition (petrol) engine. In the air-standard analysis we treat the working fluid as air behaving as an ideal gas, with heat supplied from outside instead of combustion. The cycle has four processes:

1→2 Isentropic compression. The piston compresses the charge; temperature and pressure rise.
2→3 Heat addition at constant volume. Combustion adds heat $q_{\text{in}}$ almost instantly.
3→4 Isentropic expansion. The power stroke; the gas does work on the piston.
4→1 Heat rejection at constant volume. The exhaust carries away heat $q_{\text{out}}$ .

The enclosed area on a P–v diagram is the net work per kilogram.

State calculations

With the compression ratio $r = v_1/v_2$ (typically 8 to 11 for petrol engines):

T_2 = T_1 r^{\gamma - 1}, \qquad T_3 = T_2 + \frac{q_{\text{in}}}{c_v}, \qquad T_4 = T_3 r^{1-\gamma}

Pressures follow from the ideal gas law or from $Pv^\gamma =$ const along the isentropes.

Efficiency

Because the two heat-transfer processes are at constant volume, the thermal efficiency simplifies to

\eta = \frac{w_{\text{net}}}{q_{\text{in}}} = 1 - \frac{1}{r^{\gamma - 1}}

It depends only on $r$ and $\gamma$ . Raising the compression ratio improves efficiency, but real petrol engines are limited by knock (premature ignition of the charge), and higher peak pressures need a stronger, heavier engine.

Interactive modelRaise the compression ratio and heat input. Watch the cycle area and efficiency.

This interactive model needs JavaScript. The theory, worked example and problems on this page work without it.

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Worked example

An Otto cycle has $r = 8$ and draws in air at 100 kPa and 300 K. Heat added is 1200 kJ/kg. Take $\gamma = 1.4$ and $c_v = 0.7175$ kJ/(kg·K).

State 2. $T_2 = 300\times8^{0.4} = 689$ K and $P_2 = 100\times8^{1.4} = 1838$ kPa.

State 3. $T_3 = 689 + 1200/0.7175 = 2362$ K, and $P_3 = P_2 T_3/T_2 = 1838\times2362/689 = 6298$ kPa.

State 4. $T_4 = 2362\times8^{-0.4} = 1028$ K.

Efficiency. $\eta = 1 - 8^{-0.4} = 0.565$ , or 56.5%. Net work $= 0.565\times1200 = 678$ kJ/kg and the rejected heat $= 522$ kJ/kg.

The model is set to this case. Real engines achieve about 25–35% because of friction, heat loss, incomplete combustion and gases that are not ideal.

Practice problems

Calculate the ideal Otto efficiency for $r = 10$ and $\gamma = 1.4$ .

Show answer

$\eta = 1 - 10^{-0.4} = 1 - 0.398 = 0.602$ (60.2%).
If the compression ratio rises from 8 to 12, find the new ideal efficiency.

Show answer

$\eta = 1 - 12^{-0.4} = 1 - 0.370 = 0.630$ (63.0%). Higher, but with diminishing returns.
A cycle receives 900 kJ/kg and has $\eta = 0.55$ . Find net work and heat rejected.

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$w = 0.55\times900 = 495$ kJ/kg; $q_{\text{out}} = 900 - 495 = 405$ kJ/kg.
Why is the Otto cycle's efficiency lower when $\gamma$ is lower? (Try 1.3 in the model.)

Show answer

$1 - r^{1-\gamma}$ falls as $\gamma$ falls: with $\gamma = 1.3$ and $r = 8$ , $\eta = 1 - 8^{-0.3} = 46.4\%$ . Hot combustion gases have a lower $\gamma$ than cold air, which is one reason the real cycle is less efficient than the air-standard one.

Key formulae

Compression ratio: $r = \frac{v_1}{v_2}$
Temperatures after adiabatic processes: $T_2 = T_1 r^{\gamma - 1}, \quad T_4 = T_3 r^{1 - \gamma}$
Heat added at constant volume: $q_{\text{in}} = c_v (T_3 - T_2)$
Thermal efficiency: $\eta = 1 - \frac{1}{r^{\gamma - 1}}$
Net work: $w_{\text{net}} = \eta\, q_{\text{in}}$

Lecturer notes: suggested plan, activity and assessment

Suggested plan (45 min). 10 min: show a cutaway of a four-stroke engine; 10 min: map the strokes onto the P–v diagram; 15 min: worked example; 10 min: model, compression ratio sweep.

Activity. Plot $\eta$ against $r$ for $r = 4$ to $14$ from the model's readout, then discuss why practical engines stop around 10 to 12.

Discussion. How does the Diesel cycle differ? (Compression ignition, heat added at constant pressure, no knock limit.)

Assessment idea. Compare an Otto cycle with a Carnot cycle between the same temperature limits.