Engine fuels

Requirements of an engine fuel

• High calorific value per unit mass and volume.
• Easy vaporisation (SI fuel) or good atomisation and self-ignition (CI fuel).
• Good anti-knock quality: high octane number for SI engines, high cetane number for CI engines.
• Complete combustion with little deposit and low pollutant formation.
• Chemically stable in storage, no gum formation.
• Non-corrosive (low sulphur) and non-toxic as far as possible.
• Low freezing point, suitable viscosity (diesel), safe flash point.
• Cheap and easily available.

Chemical structure of petroleum

Petroleum is a mixture of hydrocarbons (C and H), with small amounts of S, N and O compounds. The main families are:

Gasoline is mostly C$_5$-C$_{12}$ paraffins, naphthenes and aromatics. Diesel is C$_{12}$-C$_{20}$, rich in straight-chain paraffins, which is why it ignites easily. Compact, branched and aromatic molecules resist auto-ignition, so they suit SI engines. Long straight chains break down easily, so they suit CI engines.

Octane number

The octane number (ON) is the percentage by volume of iso-octane (2,2,4-trimethylpentane) in a blend with n-heptane that gives the same knocking behaviour as the fuel under test in a standard engine. Iso-octane (very knock-resistant) is given ON = 100 and n-heptane (very knock-prone) ON = 0. A fuel of ON 90 behaves like a mixture of 90% iso-octane and 10% n-heptane.

Rating method

The fuel is run in a standard variable-compression-ratio CFR engine. The compression ratio is raised until a standard knock intensity is shown by a knockmeter. A blend of reference fuels is then prepared that gives the same knock at the same compression ratio. The blend percentage is the ON.

MON is usually 8-10 units lower than RON. The anti-knock index is $(\text{RON} + \text{MON})/2$.

Improving anti-knock quality

• Blend high-octane components: aromatics, iso-paraffins, ethanol, MTBE.
• Catalytic reforming and alkylation in the refinery.
• Anti-knock additives: tetra-ethyl lead (TEL) with ethylene dibromide scavenger (now banned in most countries because of lead pollution); modern fuels use oxygenates and aromatics.

Highest useful compression ratio (HUCR)

It is the highest compression ratio at which a fuel can be used in a given engine, at the given speed and with the mixture and spark timing set for best power, without detectable knock. A higher HUCR gives higher efficiency.

Performance number (PN)

Used for fuels better than iso-octane (ON above 100). It is the ratio of the knock-limited power of the fuel to that of iso-octane, multiplied by 100. For example, a fuel with PN 130 allows 30% more knock-limited power than iso-octane. Approximately, $\text{ON} = 100 + (\text{PN} - 100)/3$.

Cetane number

The cetane number (CN) of a diesel fuel is the percentage by volume of cetane (n-hexadecane, C$_{16}$H$_{34}$) in a mixture with alpha-methylnaphthalene that gives the same ignition delay as the fuel under test. Cetane (ignites very easily) has CN = 100 and alpha-methylnaphthalene (ignites poorly) has CN = 0. In current practice heptamethylnonane (HMN, CN = 15) is often used in place of alpha-methylnaphthalene, so that $\text{CN} = \%\ \text{cetane} + 0.15 \times \%\ \text{HMN}$.

Rating of CI fuels

• The fuel is tested in a CFR diesel engine at 900 rpm with fixed injection timing (13 degrees before TDC).
• The compression ratio is changed until ignition starts exactly at TDC, i.e. the ignition delay is 13 degrees.
• Reference blends are run to find the blend with the same delay at the same compression ratio.
• Another quick estimate is the diesel index from aniline point and API gravity: $\text{DI} = \text{aniline point}(^\circ\text{F}) \times \text{API}/100$.

Normal diesel has CN 40-55. A high CN means a short ignition delay.

Octane number versus cetane number

Effect on diesel knock

A low CN gives a long ignition delay. A large amount of fuel collects before burning and then burns suddenly, giving a very rapid pressure rise and the harsh knock. A high-CN fuel has a short delay, smooth combustion, easier starting and less noise.

Stoichiometric reaction

$$\text{C}_8\text{H}_{18} + 12.5\,(\text{O}_2 + 3.76\,\text{N}_2) \rightarrow 8\,\text{CO}_2 + 9\,\text{H}_2\text{O} + 47\,\text{N}_2$$

Oxygen needed: 8 (for C) + 4.5 (for H) = 12.5 mol per mol of fuel.

(a) Air-fuel ratio

$$\begin{aligned} (A/F)_{stoich} &= \frac{12.5 \times 4.76 \times 28.97}{114.23} = \frac{1723.7}{114.23} = 15.09 \\ (A/F)_{actual} &= 1.20 \times 15.09 = 18.11 \end{aligned}$$

Reaction with 120% theoretical air

Oxygen supplied $= 1.2 \times 12.5 = 15$ mol, nitrogen $= 15 \times 3.76 = 56.4$ mol, excess O$_2$ $= 15 - 12.5 = 2.5$ mol.

$$\text{C}_8\text{H}_{18} + 15\,(\text{O}_2 + 3.76\,\text{N}_2) \rightarrow 8\,\text{CO}_2 + 9\,\text{H}_2\text{O} + 2.5\,\text{O}_2 + 56.4\,\text{N}_2$$

(b) Dry products

Water is removed, so the dry mole total is $8 + 2.5 + 56.4 = 66.9$ mol.

Total wet products are 75.9 mol, so the water vapour is $9/75.9 = 11.9\%$ of the wet gas.

Answer: $(A/F)_{stoich} = 15.09$, $(A/F)_{actual} = 18.11$ kg air per kg fuel; dry products 11.96% CO$_2$, 3.74% O$_2$, 84.30% N$_2$.

HCV and LCV

• Higher (gross) calorific value, HCV: heat released when 1 kg of fuel is burned completely and the products are cooled to the original temperature, so that the water vapour is condensed and its latent heat is recovered.
• Lower (net) calorific value, LCV: the same, but the water leaves as vapour, so its latent heat is not available. Engines exhaust hot gas, so LCV is used for efficiency calculations.

$$\text{LCV} = \text{HCV} - m_w h_{fg}$$

where $m_w = 9\,\text{H}$ kg of water is formed per kg of fuel (H is the mass fraction of hydrogen).

Dulong's formula

$$\text{HCV} = 33\,800\,\text{C} + 144\,200\left(\text{H} - \frac{\text{O}}{8}\right) + 9400\,\text{S}\ \ \text{kJ/kg}$$

Here the oxygen in the fuel is assumed to be already combined with part of the hydrogen (O/8).

Calculation

$$\begin{aligned} 33\,800 \times 0.84 &= 28\,392 \\ 144\,200 \times (0.125 - 0.015/8) &= 144\,200 \times 0.123125 = 17\,754.6 \\ 9400 \times 0.01 &= 94 \\ \text{HCV} &= 28\,392 + 17\,754.6 + 94 = 46\,240.6\ \text{kJ/kg} \end{aligned}$$ $$\begin{aligned} m_w &= 9 \times 0.125 = 1.125\ \text{kg water/kg fuel} \\ \text{LCV} &= 46\,240.6 - 1.125 \times 2442 = 43\,493.4\ \text{kJ/kg} \end{aligned}$$

Answer: HCV $\approx 46.2$ MJ/kg and LCV $\approx 43.5$ MJ/kg.