From an earlier example:
Quote:
Originally Posted by co_driver
Numbers to use:
L* = effective diameter of your car [1990 Firebird: (72.4" (width) * 49.8" (height) )^0.5 = ~60" = 1.53 m]
v = velocity [50 mph = 22.35 m/s]
'nu' = 1.6*10^-5 m^2/s
Thus: Re = 1.53 * 22.35 / 1.6*10^-5 = ~2,100,000
(*) Note:
A better reference would be the effective diameter rather than its length, as length is used for flat plates.
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Quote:
Originally Posted by apgrok1
Ok, I reworked the numbers and came up with 494,000 for 60mph. My discrepancy is that we have been using a diameter for characteristic length. That is ok for tubes, but for flow over a flat plate, the characteristic length is the car's length. Am I wrong?
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As I believe that we may consider practically all vehicles "bluff bodies", an effective length should be considered as the square root of the frontal area. Only in extreme cases could one start to consider a car as an aerofoil, in which case the length of the aerofoil would actually be an effective length for use in the Reynolds Number. Remember, the Re is meant for modelling full-size objects to a smaller scale in order to simulate its characteristics.
BTW: a sub-500k Re @ 60 mph is a very small frontal area (3.2 sqft)...