30
$\begingroup$

I was recently reading about the MiG-29 and its many upgrades, one of which (MiG-29M) was to smokeless engines.

What makes a smokeless jet engine? Is it just the fuel, or something about the way the engine works itself?

I've observed smokey engines were more common in the past and I'm wondering why that is too. Were they just easier to build for some reason?

$\endgroup$
4
  • 14
    $\begingroup$ Just as a guess: more efficient combustion. Smoke is basically not-fully-burned bits, so the more of the fuel you burn, the less smoke you have. (How you get that, I don't know, which is why I'm not posting an answer :) ). $\endgroup$
    – yshavit
    May 16, 2016 at 3:08
  • 7
    $\begingroup$ Compressor stall will make a jet engine smokeless, but this condition is to be avoided during flight. $\endgroup$ May 16, 2016 at 8:29
  • 1
    $\begingroup$ @HowardMiller I'd think flame-out would be more likely to produce a smokeless engine than compressor stall (though the latter could lead to the former, but tends to produce significant amounts of smoke in the process.) $\endgroup$
    – reirab
    May 16, 2016 at 16:19
  • 1
    $\begingroup$ @reirab, I have to admit I haven't seen many flameouts, but yeah, compressor stalls tend to be pretty dramatic. When I was at Cam Ranh Bay, the usual aircraft launch plan tended to be to get going as fast as you can on the runway, rotate and climb away as fast as possible. For supersonic fighters, this meant launching on afterburners, and I never saw much smoke. C-130s, which have turbine engines, pretty much the same. C-124s inject a water alcohol mix into the engines on takeoff and the tended to have more of a smoke tail. $\endgroup$ May 16, 2016 at 16:59

2 Answers 2

26
$\begingroup$

Generally, smoke comes from incomplete burned fuel.

Normally, a jet engine burns a very lean mixture (due to turbine temperature constraints). Nevertheless, getting all the fuel to burn is a considerable challenge, primarily because the air/mixture flows faster in the engine than the flame front. So in order to provide stable burn, many tricks come into play: the combustion chamber expands after the compressor; there are special vortex generators that force the mixture to backflow and circulate; the air may be fed in stages along the chamber, etc.

This complicated gas dynamics is difficult to optimise, especially in the 70s when MiG-29's engines (RD-33) were designed. What usually happens is that there are pockets of over-rich mixture in the combustion chamber, and the extra fuel decomposes due to high temperature before it has an opportunity to burn.

From what I know, RD-33MK got a redesigned combustion chamber (amongst other things), but what exactly the changes are is hard to say. I can speculate that the source of the original problem was that RD-33 had an early example of the annular combustion chamber, which was not well studied then. But the problem was bad enough to make MiG-29 more easily detectable in visual fight.

$\endgroup$
3
  • 4
    $\begingroup$ Isn't lean burn hotter than rich one? $\endgroup$
    – Agent_L
    May 16, 2016 at 9:26
  • 4
    $\begingroup$ @Agent_L Not if you have enough excess air that it dilutes the exhaust gasses 5-10:1. I believe it's a stoichiometric (ideal) mixture that burns hottest, because there is no excess fuel or air to heat. $\endgroup$ May 16, 2016 at 20:14
  • 4
    $\begingroup$ @Agent_L, carburated, spark-ignited piston engines, like still used in many GA planes, are always operated rich-of-peak, because they tend to run rough lean of peak due to unevenness in the fuel distribution. So there leaner means hotter, because it is closer to stoichiometric. But turbines (and compression-ignition engines) run lean with a lot of excess air and leaner means colder there. $\endgroup$
    – Jan Hudec
    May 17, 2016 at 6:37
18
$\begingroup$

Smoke results from soot in the combustion gasses, so the carbon part of the hydrocarbon fuel was incompletely burned.

The literature (sorry, paywalled, but you can read the first page) reports that in the first decades of jet engine design much combustor development was trial and error. Specifically, to arrive at a smokeless combustor, the JT9D team had to perform 465 modifications and 140 full engine tests. Mainly, these tricks help:

  • Slow flow speeds by careful diffusor design at the compressor exit
  • Swirling of the flow at the combustor entry to improve fuel-air mixing
  • Measured inflow of air into the combustor, such that the fuel injection area has an essentially stoichiometric mixture for best flame stability. Then more air is added in a secondary and a third dilution region. This dilution helps to lower the gas temperature so the gas stream is palatable to the turbine.
  • Longer combustors, so that 99% combustion can be achieved at the end of the secondary region.

Selecting the right fuel helps, too. If you would burn methanol or ethanol, a smokeless combustor is easier to achieve, but the energy density of the fuel suffers. Lighter fuels with shorter hydrocarbon chains help, but are more expensive to produce and reduce energy density, so kerosene was preferred even if its combustion produces soot.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .