Ignition and Combustion

In a CI engine, air is compressed to about 30 to 40 bar and 500 to 600 degrees C. Diesel is injected near TDC and burns by auto-ignition. Combustion occurs in four stages.

 P
 |            C
 |          /  \  D
 |        /B    \___
 |  A----/            .
 |  inj starts    TDC
 +------------------------ crank angle
   A-B delay  B-C rapid  C-D controlled  D- after

1. Ignition delay (A-B): from start of injection to first flame. Physical delay (atomisation, vaporisation, mixing) plus chemical delay (pre-flame reactions). No pressure rise above the motoring curve.
2. Rapid (uncontrolled) combustion (B-C): the fuel injected during the delay burns almost at once, giving a steep pressure rise. This is the source of the diesel noise.
3. Controlled (mixing-controlled) combustion (C-D): fuel injected now burns as it enters; the rate depends on injection and mixing. Peak temperature is reached.
4. After-burning: remaining fuel burns late in the expansion stroke; undesirable, since it raises exhaust temperature and smoke.

Diesel knock and ignition delay

If the delay is long, a large amount of fuel accumulates and burns suddenly. This causes a very high rate of pressure rise, shock waves and a metallic knocking sound called diesel knock.

Remedies: use a high cetane fuel, higher compression ratio, higher intake temperature/pressure, pilot injection, a small injection quantity at start (pintle nozzle) and warm engine.

(a) Normal combustion and knock

In an SI engine a spark ignites the compressed mixture. A thin flame front spreads outward from the plug at 20 to 30 m/s, burning the charge progressively. Pressure rises smoothly and peaks at about 10 to 15 degrees after TDC.

Knocking occurs when the unburnt end gas ahead of the flame front is compressed and heated by the burnt gas so that it auto-ignites before the flame front reaches it. The end gas burns almost instantly, creating pressure waves that hit the cylinder walls with a sharp metallic "pinging" sound. It causes loss of power, overheating, and may damage pistons and bearings.

(b) Differences and remedies

Factors reducing knock

• Use higher octane fuel.
• Lower compression ratio and lower intake temperature.
• Retard ignition timing.
• Higher turbulence and compact combustion chamber, spark plug at centre (short flame travel).
• Richer or leaner mixture (not near the worst-case 10 to 12 percent rich).
• Better cooling and clean combustion chamber.

(a) Battery (coil) ignition system

 +-------[Ignition switch]--[Ballast R]--+
 |                                       |
 Battery 12V                    Primary winding (N1)
 |                                       |--+-- Contact breaker
 |                              Secondary (N2) |  + condenser
 |                                       |   Cam
 +------ ground --------------------------+
                  Secondary --> Distributor --> Spark plugs

Working

1. With the breaker points closed, current flows from the battery through the primary winding and builds a magnetic field in the iron core.
2. The cam on the distributor shaft opens the points. Primary current collapses suddenly.
3. The collapsing field induces about 250 to 300 V in the primary and, because of turns ratio, 20 to 30 kV in the secondary.
4. The high voltage goes through the distributor to the correct spark plug, which jumps a spark across its gap.
5. The condenser across the points absorbs the primary induced voltage, prevents arcing at the points and makes the current collapse faster.

In an electronic system the points are replaced by a transistor switch triggered by a Hall-effect or pickup coil, so no contact wear occurs.

(b) Numerical

Sparks per second: a 4-stroke engine fires each cylinder once per 2 revolutions.

$$\text{sparks/s} = \frac{6 \times 4000}{2 \times 60} = 200$$

Interval $= 1/200 = 5$ ms.

Time for 24 degrees advance: crank speed at 4000 rpm $= 4000 \times 360/60 = 24\,000$ deg/s.

$$t_{4000} = \frac{24}{24\,000} = 1.0\ \text{ms}, \qquad t_{3000} = \frac{24}{18\,000} = 1.33\ \text{ms}$$

At a fixed angle, the spark occurs earlier in time at lower speed, so a centrifugal advance mechanism increases the angle at higher speed.

Coil energy:

$$E = \tfrac{1}{2} L I^2 = \tfrac{1}{2}(8\times10^{-3})(4)^2 = 0.064\ \text{J} = 64\ \text{mJ}$$

Secondary voltage:

$$V_2 = 120 \times 280 = 33\,600\ \text{V} = 33.6\ \text{kV}$$

Answer: 200 sparks/s (one every 5 ms); advance time 1.0 ms at 4000 rpm and 1.33 ms at 3000 rpm; $E = 64$ mJ; $V_2 = 33.6$ kV.