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 time-derivative 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). > 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. > m_dot_air could also be > estimated as a function of RPM, engine displacement, and manifold air > density. Something like: > > m_dot_air = > (1/2)(RPM/60)(rho_manifold)(displacement)(volumetric_efficiency) Sounds very reasonable to me. > [2] gives us a formula for v_e: > v_e = sqrt( (T*R/M) (2k/(k-1)) [1 - (P_e/P)^((k-1)/K)]) > > See the wikipedia article for a full explanation of the formula. 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. > k~= 1.4 > > for exhaust temperatures up to 1000 degrees C. > so the second term becomes: > > (2k/(k-1)) ~= 7.0 > > The third term > > [1 - (P_e/P)^((k-1)/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 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 ------------------------------------------------------------------------------ This SF.net Dev2Dev email is sponsored by: Show off your parallel programming skills. Enter the Intel(R) Threading Challenge 2010. http://p.sf.net/sfu/intel-thread-sfd _______________________________________________ Flightgear-devel mailing list Flightgear-devel@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/flightgear-devel
