On Thursday 02 September 2010 01:39:52 thorsten.i.r...@jyu.fi 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 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).

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)^((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 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 fall-off 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


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