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It is pretty straightforward to compute the heat and power generated by the combustion process, given the displacement, RPM and fuel mass flow rate. Similarly, the power exhausted can be computed with knowledge of EGT and conservation of mass. The difference is dissipated as heat, from air cooling of cylinders and oil cooler. I'm looking for a rough ratio between oil cooler and cylinder heat dissipation.

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    $\begingroup$ Welcome to Av.SE. Interesting question! $\endgroup$ – Ralph J Oct 2 '18 at 19:33
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According to Fundamentals of Powerplants for Aircraft, quoted in Mike Busch on Engines (chapter 2), "Other thermal losses" account for 12.2% of fuel energy:

  • Conduction to air: 7.2%
  • Conduction to oil: 1.6%
  • Radiation and misc: 3.4%

The other two areas of loss are exhaust (51.6%) and mechanical (36.2%).

I don't know how "cylinder heat dissipation" would fit into this classification (some combination of air conduction and friction losses?), but looking only at conduction losses gives a ratio of air to oil of 4.5:1.

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  • $\begingroup$ Agreed. I would like to point out that 'friction losses' do not completely go into an increase of temperature of the environment. Some energy is invested in 'destruction work' of the inevitable wear of the engine... $\endgroup$ – xxavier Oct 3 '18 at 10:23
  • $\begingroup$ From 'Internal Combustion Engine Fundamentals', by John Heywood: imgur.com/Mh5f9Cn $\endgroup$ – xxavier Oct 3 '18 at 11:39
  • $\begingroup$ Interesting to define mechanical energy as a "loss" ;) $\endgroup$ – Sanchises Oct 3 '18 at 19:03
  • $\begingroup$ @Sanchises - Similarly, 75% of a turbojet's energy is wasted as heat :) $\endgroup$ – ymb1 Oct 3 '18 at 20:09
  • $\begingroup$ I was trying to locate that data for my post but was unable to. Those numbers are pretty close to what I had always understood in ICEs. Thanks for posting. $\endgroup$ – Carlo Felicione Oct 3 '18 at 22:29
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Typically in an air cooled, reciprocating engine, for the total amount of chemical energy released during combustion, about 40-45% exits the tailpipe as heat in the exhaust gases, 25-30% of that energy is transferred to the propeller through the crankshaft as mechanical energy, 15-20% or so exits the engine from cool air flowing through the cowling over the engine block, and the final 5-10% is removed through the oil cooler.

SOURCE: THE AIRCRAFT ENGINE AND ITS OPERATION - PWA 109702, pp 19 - Pratt & Whitney Aircraft Corporation, 1949.

I knew I had this in my library somewhere.

If we take a look at an airplane eg an SR-22 on takeoff, the IO-540 engine consumes about 25 gal/hr of 100LL and produces 310 BHP, or is outputting 8.3E8 Joules of mechanical energy per hour. With a chemical energy density of 40MJ/kg for 100LL, this means that about 2.6E9 Joules of chemical energy are supplied to the engine every hour, thus giving us around a 31% efficiency in this example, and that’s about what we would expect.

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    $\begingroup$ Do you have a source for this? Seems pretty definitive but could use some supporting evidence. $\endgroup$ – Ron Beyer Oct 3 '18 at 1:44
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    $\begingroup$ donwvoted as the answer seems to throw random figures without any calculation nor references. $\endgroup$ – Manu H Oct 3 '18 at 15:05
  • $\begingroup$ Those figures are very typical for internal combustion engines. They will typically only achieve about a 25-35% thermal efficiencies (you just can’t get around the Second Law of Thermodynamics). More exotic combined cycle engines approach higher thermal efficiencies - in the neighborhood of 50% or so and that’s about as good as you’re going to get. Yes the rest of that chemical energy just leaves the engine in heat. $\endgroup$ – Carlo Felicione Oct 3 '18 at 17:31
  • $\begingroup$ If we take a look at an airplane eg an SR-22 on takeoff, the IO-540 engine consumes about 25 gal/hr of 100LL and produces 310 BHP, or is outputting 8.3E8 Joules of mechanical energy per hour. With a chemical energy density of 40MJ/kg for 100LL, this means that about 2.6E9 Joules of chemical energy are supplied to the engine every hour, thus giving us around a 31% efficiency in this example, and that’s about what we would expect. $\endgroup$ – Carlo Felicione Oct 3 '18 at 17:48

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