There was a german study about the "Influence of propulsion efficiency on contrail formation". In this study, an Airbus A340 and a Boeing 707 flew right next to each other.

The two contrail forming aircraft were

(i) a Boeing B707 equipped with four jet engines of type JT3D-3B with bypass-ratio of 1.4 and

(ii) an Airbus A340-300 with four jet engines of type CFM56-5C4 with bypass-ratio of 6.8.

The A340, with the high bypass-ratio, formed contrails behind it, whilst the Boeing with the low bypass-ratio did not form any contrails behind.

Formation flight A340 and B707


This has to do with the different exhaust gas temperatures of engines with different bypass-ratios.

  • How exactly is the formation of the contrails affected by the EGT (and therefore bypass-ratio), so whats the difference between low EGT and high EGT in terms of contrail formation? What is the physics behind it? Just why does the Airbus form a contrail, whilst the Boeing doesn't?

Note, 'How does contrail formation differ from turbofan to turbojet?' does not answer the question regarding the EGT.

  • 2
    $\begingroup$ Seems to be a duplicate of How does contrail formation differ from turbofan to turbojet? with an answer showing this same picture. $\endgroup$
    – mins
    Commented Jan 20, 2018 at 18:39
  • 2
    $\begingroup$ I think you have it reversed, high-bypass ratio engines have lower exhaust temps than low bypass-ratio engines because more air is bypassing the actual combustion in high-bypass ratio engines. tldr; Temperature and pressure are directly related; if the air is hot the amount of water it can hold decreases. The water molecules from the hot low-bypass exhaust move faster to lower pressure (ambient) air surrounding it causing a lower humidity meaning less or no contrail while the water in the colder high-bypass hang around longer $\endgroup$
    – Noah Wood
    Commented Jan 21, 2018 at 5:00
  • $\begingroup$ @NoahWood This already sounds like a good start for an answer. $\endgroup$ Commented Jan 21, 2018 at 9:47
  • 1
    $\begingroup$ Note that the linked answer does not talk about EGT, because the linked study does not directly either. Because what matters is the remaining energy, i.e. how far the water is from condensing. Higher temperature generally means more energy, but it's not the only factor. $\endgroup$
    – Jan Hudec
    Commented Jan 21, 2018 at 13:55
  • $\begingroup$ @mins No I'm pretty sure I said what I meant. Clouds are suspended water droplets. Hot air will evaporate the previously condensed water droplets and raise the pressure causing the water vapor to move towards a lower pressure where it will re-condense into water droplets and ice crystals. Perhaps 'amount of water it can hold' was a bad phrase though I can see why that might be confusing. Assuming an equalized pressure at all times in two separate containers of equal humidity, the heated container of air will flow out into the general atmosphere while the chilled container will condense. $\endgroup$
    – Noah Wood
    Commented Jan 22, 2018 at 3:10

1 Answer 1


There are some building-blocks to the answer:

Part 1: Assumption/Simplification
Let's assume for a moment that the only difference between the two planes is the exhaust gas temperature (EGT). A less efficient engine will utilise less of the available (chemical) energy of the fuel. This means a less efficient engine will have a hotter and faster exhaust jet (wasting caloric and kinetic energy).

Part 2:
The water-content of the exhaust gas contains water from two sources: first the water (very low fraction) from the ambient air humidity and second from the combustion of the fuel (carbon-hydrates).

Part 3: Assumption
When humid air (from an engine exhaust) mixes with the atmosphere it is a reasonable assumption that the state of the atmosphere will not change measurably given enough time for the exhaust to mix out.

Part 4:
The mass of water vapour which can be contained in a given mass of air depends on pressure and temperature of the air. This dependency is often visualized in either the Mollier-Diagram or Psychometric-Chart. The following simplification of a Mollier-Diagram shows two areas or states humid air can have. For a given temperature and low water content the water is in its gaseous phase. If the water content is increased (over the saturation point) than part of the water will be condensed (fog, small water droplets).

enter image description here

Part 5: Answer-Part I
When hot humid air exits the engine it will cool down an partially/continuously mix with the atmosphere. This means temperature and water content/ratio (per a given volume) will decrease. The following sketch illustrates this for two starting points. At the end of this process the jet reaches the state of the atmosphere (see part 3). In the case of a lower EGT the cooling of the exhaust can cause the humid air to cross the saturation-line which will result in a contrail.

enter image description here

Part 6: Answer-PartII
A lower by-pass-ratio-engine is likely to have a higher velocity difference between the ambience/atmosphere and the jet from the engine. This will increase the mixing of the exhaust gases which suppresses the condensation as well.

The model used in the paper (Schumann 2000) also includes basic thermodynamic considerations for the jet-engine which do increase the predictive accuracy of the model. But the thermodynamic effect should be captured by the parts 1-6 given here.

  • $\begingroup$ I would guess that the Fig.3 in the paper is a psychometric plot, which is a flipped Mollier-Diagram (used in part 4, 5). $\endgroup$
    – rul30
    Commented Jan 24, 2018 at 21:29

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