RobertE Posted July 1, 2013 Report Posted July 1, 2013 All things being equal, the same manifold pressure and rpm out to imply the same HP, right? But my manual for an M20J doesn't show that. At first I figured I understood why, which is that different altitudes in the manual are premised on different temperatures. But then I noticed the differences go the wrong way! So can someone tell me why, at best economy, 21.1 inches and 2600 rpmproduces 65% of power at 10,000 feet whereas at 2000 feet it takes 21.6 inches? Yet the assumed temp at the higher altitude is lower (which ought to mean more power at a given MP level). So, anyone got an idea?? Quote
Super Dave Posted July 1, 2013 Report Posted July 1, 2013 You're on the right track, but the temps are going the right way. Warmer/less dense air (for a given MP) at 2000', so it takes a little more MP to get the same HP. Quote
RobertE Posted July 1, 2013 Author Report Posted July 1, 2013 Ahh, yes. I see I got my thinking a little crooked. The differences do go the right way. So is the answer purely the different assumed temperatures? And if the answer is buried in "PV=nRT" it would nice to know what variables that equation is premised upon. Quote
carusoam Posted July 1, 2013 Report Posted July 1, 2013 Ideally speaking, it's the Ideal gas law from the professor's deep rooted sense of humor department. It essentially gives the relationship of the amount of oxygen per volume of air entering a cylinder as the temperature and pressure decrease with the increase of altitude.... Pressure, volume, temperature, and the constant aka Avagadro's number. If you like Bernoulli, then you will love his friends Boyle and Lord Kelvin. EB is no doubt right, but is he funny? I think so. Best regards, -a- Quote
aviatoreb Posted July 1, 2013 Report Posted July 1, 2013 Yes, I am being a jokester by writing an equation and no words. Forgive me - I am a bit sillier than most math professors you might meet. carusoam, if I am funny it is simply a consequence of procrastination as I am supposed to be doing other things besides mooneyspace right now! I don't know much about the gas environment of an engine, and there are chemical reactions taking place as well that ruin the assumptions of the ideal gas law. But most definitely, for the onset of the combustion event, the equation tells you the right monotonicity relationship (n decreases with respect to increasing temp, but increases with respect to increasing pressure) of the oxygen at the onset of the event. But the law assumes well mixed I believe and I do not know if the air is homogeneous and reset between events. Yes yes and Boyle and Kelvin - by the way - PV=nRT is really not owned by math guys - it is really considered the realm of the chem profs or maybe physics profs. If I know any ideal gas law, it is simply from my own undergrad education. Quote
bnicolette Posted July 1, 2013 Report Posted July 1, 2013 PV=nRT Dammit Erik, there's a few minutes of my life I ain't gettin back. Quote
aviatoreb Posted July 1, 2013 Report Posted July 1, 2013 Silly math teachers! Screen Shot 2013-07-01 at 3.06.55 PM.png My people! You GET me! Quote
aviatoreb Posted July 1, 2013 Report Posted July 1, 2013 Dammit Erik, there's a few minutes of my life I ain't gettin back. I will send you a refund: when you on at the end, just remind me and I will put in a good word for you so that you will be provided with another 2:32. Are we even then? Or just odd? Quote
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