Al, looking at photos
of Tracy’s
inlets, I would hazard a guess that the inlet may be more than 5” in
diameter. If Tracy misspoke and meant 5”
radius
then that would give
10” diameter or approx 78 sq inch = 0.547 sq ft which then
taking
4000 CFM /0.547 =
7322 ft/min = 83 MPH somewhat more realistic
But, then Tracy may mean 5”
diameter in which case I am going to fly down, pour him full of beer and
badger the secret out of him {:>)
Ed
From:
Rotary motors in aircraft
[mailto:flyrotary@lancaironline.net] On
Behalf Of Al Gietzen
Sent: Wednesday, December 23, 2009 11:21
AM
To: Rotary motors in aircraft
Subject: [FlyRotary] Air Flow
Question
Tracy
wrote:
My 5" round inlet for the radiator looks
ridiculously small compared to yours but so far it is cooling the 20B
OK.
Now that sort of
boggles my mind as it seems to violate the laws of physics. Let’s just take a
modest climb power of, say; 225 hp. At that power, the energy going into
the coolant is about 6000 Btu/min. In order to remove that amount of
heat, at a typical air temp increase of 75 degrees; takes about 4000 cfm air
flow. A 5” dia inlet is 0.14 sq feet, meaning an average inlet velocity
about 29,000 ft/min, or 330 mph. Even at 100 air temp increase (unlikely
on a 90F day) it’s 250 mph. And I’m guessing your climb speed is half
that. Similar math suggests you’d be limited to a steady state (cruise)
power of about 50%.
Of course, being a
pusher driver, I think of inlet air speeds in terms of the speed of the
airplane. So does the fact that the inlet is behind prop give a much
higher effective inlet velocity? I’ve been thinking that the turbulence
in the prop wash would negate a good portion of the extra mean air velocity
because of reduced inlet effectiveness.
It will be
interesting to know how it works out on a hot day.
Great that you have
your 20B in the air and working well.
Merry Christmas and
Happy New Year to all.
Al
G
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