Vivian Meazza wrote
>
> Andy Ross
>
> > [Starting a new thread. The reply nesting level in my
> > mozilla window was getting freaky.]
> >
> > Vivian Meazza wrote:
> > > The engine I'm trying to specify developed 1140 HP at engine
> > > revolutions of 2850 rpm at a boost pressure of 9 psi. It
> was fitted
> > > with 1:0.477 reduction gearing, which I think means that
> > the propeller
> > > turned at 1360 rpm.
> >
> > Hrm, 1360 RPM is very slow for a cruise value, just over idle
> > speed for a smaller plane. Likewise, 2850 RPM really isn't
> > that fast for a piston engine. It's at the top end of
> > ungeared engines like a Lycoming O-360 or whatnot, but not
> > really very fast for four stroke engines as a whole (my
> > Saturn redlines at 6000, for example).
> >
> > Is it possible that the 2850 number is a *propeller* RPM at
> > max power? Then you'd get a max power engine speed of 5975,
> > which seems plausible to me and avoids the problems with
> > solving for a propeller which "cruises" at a pitch where
> > normal props would be windmilling.
> >
> > Does anyone have good info on whether the cockpit engine
> > speed gauge in a Spitfire (which is presumably what most
> > sources will quote for
> > "RPM") reads engine or propeller speed?
> >
> > Andy
> >
>
....
>
> I believe the low propeller rpm was to do with tip speeds
> approaching or exceeding Mach 1 at high aircraft speeds. I
> will research that next.
>
> Regards
>
> Vivian
>
Here are some calculations on propeller rpm.
The propeller the tip speed should be as high as possible with the only
limitation being that the tip should not get into the region of aerodynamic
compressibility. Typically a figure of Mach 0.85 is used as the magic number
that should not be exceeded. (This makes some allowance for the speed
increase as the air passes over the aerofoil curved surface and the increase
in air velocity caused by the propeller operation.)
If we take 8000 ft as the operating altitude then Mach 1 = 1085 ft/sec
(approx)
Assuming that the forward velocity of the aircraft is 300 mph = 440 ft/sec
Then the maximum rotational velocity may be calculated by Pythagoras:
Max Rotational Velocity = ((M *1085)^2 - (V)^2)^0.5
-----(1)
where M is the designed Mach Number (0.85) and V is the
aircraft forward velocity
= ((0.85*1085)^2 -(440)^2)^0.5 =
810.52 ft/sec
RPM at Max rotational velocity is given by:
RPM = Max rotational velocity*60/(PI * D)
-----(2)
Where D is the propeller diameter (ft)
= 810.52*60/(PI * 10.75) = 1420 rpm
Thus we can see that 1360 rpm is more appropriate for this application than
2850
We can also calculate the Max Rotational Velocity @ 2850
Max rotational velocity (PI * D) = (RPM/60) * (PI * D)
= (2850/60) * (PI * 10.75)
= 1604 ft/sec
We can also calculate the Mach Number of the tip by rearranging and
substituting in (1)
M = ((1604^2+440^2)^0.5)/1085
where M is the Mach Number of the tip
= 1.5329
We can see that 2850 is unlikely to be the rpm of a 10.75 diameter propeller
Well, I hope I've got the math right! Please pick holes in it.
Regards
Vivian
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