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The example is just fine but it has little to do with an aircraft
engine cooling system. I was waiting to pounce on you for suggesting
that the lowest drag would result from a rad with the lowest
restriction but you didn't go there.
Consider this example: The drag brakes installed on some aircraft are
essentially small flat plates extended into the airstream. Those
plates have large HOLES in them. They make more drag that way. Is
that intuitive?
What I'm getting at is that ANY practical radiator is going to chew up
virtually all of the energy in the airstream that passes through it.
It is only a pipe dream that a free(er) flowing radiator is going to
give less drag than one with a higher pressure drop. For other
reasons, the opposite is usually true.
Tracy (brain dead at 02:21 )
On Nov 13, 2007 9:17 PM, Ron Springer <ron2369@sbcglobal.net> wrote:
> How's this example?
>
> Let's say you have an inlet opening of a given area
> and downstream of that is a constant area duct at the
> same area. At the exit of the duct is a door flap that
> can be set to anything from fully open to fully
> closed.
>
> When the door is fully open you will get the max flow
> through the duct. It might be 99%+ of the freestream
> flow at that same cross-sectional area.
>
> When the door is fully closed you get zero flow
> through the duct. All flow streamlines steer right
> around the inlet opening. This is a pitot tube
> basically.
>
> When the door is set to an intermediate position, you
> can get any flow you want through the duct (between
> zero and max) and the inlet area is the same.
>
> Changing door positions is like swapping in and out
> different radiators and ductwork that have different
> flow resistances. When going to a higher resistance
> radiator (for instance, one with smaller frontal area,
> a thicker core, and the same fin spacing) it is like
> closing the door a bit. The air flow will decrease.
>
> So, my point is that you can get any flow from zero to
> max flow through an inlet by changing the downstream
> flow losses.
>
> Ron
>
> --
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