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From: <thorsten.i.renk@jy...>  20100903 08:49:57

> Don't we need to keep R, the universal gas constant, to make the equation > work? > > v ~ sqrt(TR/M) ? Not when you write it like this, because it's just a constant, so you can pull it out and the result is still proportional to the rest if you drop it, like v ~ sqrt(R) * sqrt(T/M) ~ sqrt(T/M) (you have to put it back in when you want v_E = xxx instead of v_E ~ xxx in the end  but you can gather all numerical constants into a single number in the end, and I guess that's precisely what you'd want to do anyway, see below). > It was my thinking that an increase in RPM would cause a pressure > increase > because: >  The pressure from the exhaust valve falls off over time and having the > valves open more often would increase the average pressure. (untested > hypothosis) >  The piston ejecting the burnt air from the cylinder moves faster as RPM > increases, causing the air to be ejected faster and leading to an > increase > in total pressure. (another untested hypothosis) Air pressure adjusts to changing conditions with the speed of sound (any deviation from equilibrium propagates with the speed of sound). Unless the piston moves supersonically, its movement cannot cause the air to be ejected faster than it would naturally do given its thermodynamics. If it moves supersonically, things become complicated, because then you're into shockwave propagation and then the precise geometry of the engine matters  thermodynamics is then insufficient to provide an answer, you'd need to run fluid dynamics. > Would p_E be the ambient air pressure? And p would not be the intake > manifold > pressure, which we have, it is the exhaust manifold pressure which we do > not > have. [1] I also assumed that a falloff of ambient pressure would be > accompanied by a falloff in mass flow. *shrugs* Your proposal was to replace the term by a constant. Mine was to use manifold pressure and ambient pressure as proxies for the real conditions (which we do not have). Of course that's not exact  but in all likelihood captures qualitatively what is going on and is better than a constant. In the end, there are hundreds of things you don't know  friction in the exhaust tube for example... its geometrical shape (exhaust velocity isn't actually a constant  there's some spatial profile to the velocity field)... turbulence when the exhaust airstream enters ambient air... and so on. So, you want to write down an equation which gives you the right qualitative behaviour, and have an adjustable constant in front of it which gives you the number right which is supposed to take care of all the things you don't know or have neglected with the hope that enginespecific constant and qualitative equations are a good enough model for the physics that goes into the problem. If the air mass flow drops with ambient pressure or not is a question of having a turbocharger for the engine or not I guess. Plus, there is of course the dynamic ram pressure due to the aircraft motion which increases air density in the engine. I would guess as long as manifold pressure does not change, the air mass flow does not change, because that measures at the intake. Cheers, * Thorsten 
From: Ron Jensen <wino@je...>  20100903 05:46:40

On Thursday 02 September 2010 01:39:52 thorsten.i.renk@... wrote: > A few comments to your math: > > I haven't done anything (yet) with Hal's exhaust thrust function. I was > > waiting for someone else to work out the math... > > > > The exhaust thrust function should return a force. Newton tells us: > > > > F=ma > > Indeed, but not very helpful here... force as a timederivative of the > momentum is what you need, i.e. > > F = dp/dt = dm/dt v_E + m * d v_E/dt = dm/dt * v_E > > using > > p = m * v_E > > and > > v_E = const. > > (as you also use yourself below). I was just introducing the concept of dimension analysis with that formula. > > Hal had said something like > > > > F=m*v^2 > > That would be the kinetic energy  as you point out correctly, it has the > wrong dimensions. > > > m_dot = m_dot_air + m_dot_fuel for our model. > > I am prepared to bet that for an aircraft engine (unlike a rocket) > dm_air/dt >> dm_fuel/dt is always true, i.e. you can forget about the fuel > content. Yes, m_dot_air : m_dot_fuel is 12:1 or higher > Yes, this says that the thermal velocity is proportional to the root of > the temperature. That is so because temperature (in suitable units) is a > measure of the mean kinetic energy of gas molecules, so if you write > > T ~ E_kin = M v^2 > > you can solve it for v and get > > v ~ sqrt(T/M) > > which is what the formula says. The rest gathers the physics of expansion > and pressure gradient. Note that units matter here! The temperature must > be given in Kelvin [K] in order to agree with the interpretation as unit > of energy measure. I doubt that the egt property is in K  so you need to > convert to proper units before inserting into the formula. True. JSBSim tends to be in US units, I believe its in Fahrenheit so T_Rankine = (T_Fahrenheit+459.67) T_Kelvin = (5/9)(T_Rankine) Don't we need to keep R, the universal gas constant, to make the equation work? v ~ sqrt(TR/M) ? > > The third term > > > > [1  (P_e/P)^((k1)/K)] > > > > Can be thought of as an efficiency factor, It should always be greater > > than or > > equal to zero and less than or equal to 1. Its value probably increases > > with > > pressure altitude as I believe that will lower P_e (pressure at the exit > > end > > of the exhaust manifold); P (pressure from the cylinders when the exhaust > > valves open) is probably related to manifold pressure and RPM. > > I guess that's a good starting point  the lower pressure would be outside > pressure, the higher pressure manifold pressure. I don't see how RPM come > in though. It was my thinking that an increase in RPM would cause a pressure increase because:  The pressure from the exhaust valve falls off over time and having the valves open more often would increase the average pressure. (untested hypothosis)  The piston ejecting the burnt air from the cylinder moves faster as RPM increases, causing the air to be ejected faster and leading to an increase in total pressure. (another untested hypothosis) > > It may be > > appropriate to think of this as a constant, too. > > Why  you have outside pressure and manifold pressure, so you can compute > it dynamically. It's not going to be a big effect though... p_E/p is > usually a small number. > > Cheers, > > * Thorsten Would p_E be the ambient air pressure? And p would not be the intake manifold pressure, which we have, it is the exhaust manifold pressure which we do not have. [1] I also assumed that a falloff of ambient pressure would be accompanied by a falloff in mass flow [1] http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/fig5OttoReal_web.jpg from http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node26.html 