In reply to Jed Rothwell's message of Fri, 17 Jun 2016 18:00:10 -0400: Hi, [snip] ><mix...@bigpond.com> wrote: > >Rubbish. The ignition of the fuel in a normal car engine cylinder results >> in a >> power production on the order of half a megawatt for a couple of >> milliseconds, >> and that's just chemical energy (with maybe few hydrinos thrown in ;). >> > >Really? Hmmm . . . How do you figure that? > > >At 60 mph a typical old car consumes ~20 mpg. 1 gallon is 3.8 L, so that's >0.19 L per mile (or per minute). 1 L of gasoline produces 34.2 MJ. Divide >by 20 to get 1.71 MJ per minute.
Using your figures of 34.2 MJ/L and 0.19 L/minute I get 6.5 MJ/min, however that's for 6 cylinders, so for 1 cylinder that would be about 1 MJ/min = 18 kJ/sec. >A single piston stroke produces only a >small fraction of that. I don't know how long it takes the gasoline to >burn, but I doubt such a small amount would reach the half-megawatt level. > >Here is a paper on the burn rate of gasoline, which you have to pay for: > >http://papers.sae.org/2008-01-0469/ > >At 60 mph, I think engines run at about 2500 rpm. A single piston stroke >with a 6-cylinder engine running at 2500 rpm would consume . . . ummm . . >.1.71 MJ / 15,000 = 114 joules. Right? That takes only 0.0002 s to burn? > >- Jed A 4 stroke engine has one power stroke/cylinder every 2 cycles, so at 2500 rpm, that's 1250 power strokes/cylinder/min = 21 power strokes/cylinder/sec. So each power stroke/cylinder produces 18 kJ/21 or about 860 J. At 2500 rpm, there is a revolution (360 deg) every 24 ms. If combustion happens within 10 deg. of TDC, then combustion time is 24 ms*(10/360) = 0.67 of a ms. 860 J / 0.67 ms is about 1.3 MW. The assumption I have made here is that combustion happens with 10 deg. of TDC, which I am unsure about. It may not actually be complete in that time. The figures I used initially were a little different to yours, hence my original calculation was also somewhat different. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
