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

 

 

Tracy wrote:

 

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

 

 

 

__________ Information from ESET NOD32 Antivirus, version of virus signature database 3267 (20080714) __________

The message was checked by ESET NOD32 Antivirus.

http://www.eset.com