Fuel-air cycles and actual cycle losses

Why it matters

The air-standard Otto cycle promises about 60 % efficiency at a compression ratio of 10, yet a real petrol engine converts only 30–35 % of its fuel energy into indicated work. The fuel-air cycle and the list of actual-cycle losses explain where the difference goes, and that tells the engine designer which loss is worth attacking: heat transfer, combustion timing, throttling or the exhaust process.

Key ideas

Three levels of idealisation.

1. Air-standard cycle: air only, constant specific heats (γ = 1.4), instantaneous heat addition, no losses.
2. Fuel-air cycle: still no heat transfer, no friction and instantaneous combustion, but the working fluid is the real mixture of air, fuel vapour and residual gas before combustion, and the real combustion products after it.
3. Actual cycle: what the indicator diagram of a running engine shows, including all the losses below.

What the fuel-air cycle adds.

• Variable specific heats. cp and cv of gases rise with temperature, so γ falls (to roughly 1.25–1.3 for hot products). Less temperature rise for the same heat input and a lower effective γ both lower efficiency and peak temperature.
• Dissociation. Above about 1500 K, CO2 and H2O partly split into CO, H2 and O2, absorbing heat. Some of it is recovered during expansion, but too late to do full work. Dissociation lowers peak temperature by a few hundred kelvin and is strongest near the stoichiometric mixture.
• Change in number of moles. For liquid hydrocarbons the products contain slightly more moles than the reactants, which raises pressure a little.
• Real fuel properties and residual gas. Residual exhaust left in the clearance volume dilutes the fresh charge.

Effect of mixture strength (equivalence ratio φ). In the fuel-air cycle, efficiency increases as the mixture gets leaner, because the lower peak temperature means less variable-specific-heat and dissociation penalty. Peak temperature and pressure, and maximum power, occur with a slightly rich mixture (φ ≈ 1.1–1.2), not at stoichiometric, because dissociation is lower there. Efficiency still rises with compression ratio, and the ratio of fuel-air to air-standard efficiency is roughly constant over the usual range of r.

Actual-cycle losses (indicator diagram).

• Time loss (finite combustion). Combustion takes 40–60° of crank angle, so it is not at constant volume; the spark must be advanced. Too little advance loses work after TDC; too much makes the piston compress burning gas.
• Heat loss. Heat flows to cylinder walls, head and piston during combustion and expansion; lowers pressure and work. It is a large loss at low speed (more time per cycle).
• Exhaust blowdown loss. The exhaust valve opens about 40–60° before BDC so the gas can escape; the pressure drop before BDC loses some expansion work, but it reduces the work needed to push gas out.
• Pumping loss. Work spent drawing in the charge (below atmospheric, especially with the throttle partly closed) and expelling exhaust. It is the negative "pumping loop" of the p–V diagram and is large at part load in SI engines; diesels have no throttle and so a small pumping loss.
• Incomplete combustion and crevice/leakage losses. Some fuel leaves unburnt (HC and CO) and some gas leaks past the rings.
• Rubbing friction is mechanical rather than a cycle loss: it separates indicated from brake power and is treated in performance testing.

Typical magnitudes (indicative; values vary by engine and are given in your textbook): fuel-air cycle efficiency is roughly 75–85 % of the air-standard value, and the actual indicated efficiency is roughly 80–85 % of the fuel-air value.

Formulas

η_air = 1 − 1/r^(γ−1) (Otto) Air-standard efficiency; r compression ratio (–), γ = cp/cv (–).

cp = a + k·T, cv = b + k·T, γ = cp/cv Linear variable specific heats (kJ/kg·K), T in K; a, b, k empirical constants from your data book; cp − cv = R remains constant.

η_rel = η_actual / η_air Relative (efficiency) ratio, dimensionless; η_actual usually the indicated thermal efficiency.

η_ith = W_i / (m_f · CV) Indicated thermal efficiency; W_i indicated work (J), m_f fuel mass (kg), CV calorific value (J/kg).

W_pump = (p_exh − p_int) · V_s Approximate pumping work per cycle (J); p in Pa, V_s swept volume (m³). Valid when intake and exhaust pressures are roughly constant during their strokes.

imep_net = imep_gross − pmep Mean effective pressures (Pa or bar); pmep = pumping mean effective pressure ≈ p_exh − p_int.

P = imep · V_s · (N/2) · z (four-stroke) Power in W; imep in Pa; V_s per cylinder in m³; N in rev/s; z number of cylinders (N/1 for two-stroke).

Worked examples

Example 1 (standard): where the efficiency goes. An SI engine has r = 9. The fuel-air cycle efficiency for its mixture is 0.80 of the air-standard value, and the measured indicated thermal efficiency is 33 %. Find the air-standard and fuel-air efficiencies and the relative efficiencies.

1. η_air = 1 − 1/r^(γ−1) = 1 − 1/9^0.4 = 1 − 1/2.4082 = 1 − 0.4152 = 58.5 %.
2. Fuel-air efficiency = 0.80 × 58.5 = 46.8 %. The 11.7 percentage points are lost to variable specific heats and dissociation.
3. Relative to air-standard: η_rel = η_actual/η_air = 33/58.5 = 0.564.
4. Relative to fuel-air: 33/46.8 = 0.705. The remaining 13.8 points are actual-cycle losses: time, heat transfer, blowdown, pumping and incomplete combustion.

For comparison, lowering γ from 1.4 to 1.3 alone at r = 10 drops the ideal efficiency from 1 − 10^(−0.4) = 60.2 % to 1 − 10^(−0.3) = 49.9 %, showing how strongly the working-fluid properties matter.

Example 2 (GATE level): pumping loss at part throttle. A four-cylinder, four-stroke SI engine has a swept volume of 0.5 L per cylinder and runs at 3000 rpm at part load. The gross indicated mean effective pressure (compression and expansion strokes only) is 6.0 bar. Intake manifold pressure is 0.50 bar and exhaust back pressure 1.05 bar. Find the pumping power, net indicated power and the fraction of gross work lost to pumping.

1. pmep ≈ p_exh − p_int = 1.05 − 0.50 = 0.55 bar = 55 kPa.
2. Power cycles per second per cylinder = N/2 = (3000/60)/2 = 25 s⁻¹.
3. Gross indicated power: P = imep·V_s·(N/2)·z = 600 000 Pa × 0.5 × 10⁻³ m³ × 25 × 4 = 30 000 W = 30.0 kW.
4. Pumping power = 55 000 × 0.5 × 10⁻³ × 25 × 4 = 2750 W = 2.75 kW.
5. Net indicated power = 30.0 − 2.75 = 27.25 kW (equivalently imep_net = 5.45 bar).
6. Fraction lost to pumping = 0.55/6.0 = 9.2 % of gross indicated work. At wide-open throttle p_int would be close to 1 bar and this loss would nearly vanish, which is why SI engines are least efficient at part load.

Common mistakes

• Treating the fuel-air cycle as "the actual cycle". It still ignores heat transfer, friction and combustion time; it only uses real gas properties.
• Saying maximum efficiency and maximum power occur at the same mixture. Efficiency is highest lean; power and peak temperature are highest slightly rich.
• Counting rubbing friction as a cycle (indicated) loss. It affects brake power, not the indicator diagram.
• Thinking blowdown loss is pure waste. Opening the exhaust valve early costs some expansion work but saves more exhaust-stroke pumping work at high speed.
• Using rev/min directly in the power formula, or forgetting the factor 1/2 for a four-stroke engine.
• Mixing bar and Pa in pumping work; 1 bar = 10⁵ Pa.

For GATE ME

Expect conceptual questions on why actual efficiency is below air-standard, the effect of variable specific heats and dissociation, and which mixture gives maximum power versus maximum efficiency. Numericals combine air-standard efficiency with relative efficiency, compute indicated power from mep, or separate gross and pumping work. Practise reading a p–V diagram: the positive power loop, the negative pumping loop and the rounding at TDC and BDC caused by time and blowdown losses.

Quick check

1. Name two effects the fuel-air cycle includes that the air-standard cycle ignores.
2. In a fuel-air cycle, does efficiency rise or fall as the mixture is made leaner?
3. Why is the pumping loss of a diesel engine small at part load?
4. An engine with r = 8 (η_air = 56.5 %) has an indicated efficiency of 28 %. What is its relative efficiency?
5. Which loss is caused by the exhaust valve opening before BDC?

Answers: 1. Variable specific heats and dissociation (also change in moles and residual gas). 2. It rises. 3. It has no throttle, so intake pressure stays near atmospheric. 4. 28/56.5 = 0.50. 5. Exhaust blowdown loss.