Al Gietzen wrote:
Not that it is important to Bernie (his installation is done); but now that I'm back from a short
vacation I wanted to make some further comment.
It's all a matter of flow area. The air exiting the back of the rad is moving
relatively slowly, and I wouldn't expect it has much problem moving around objects that
might be there.
I thought the same thing going into my rad test. I was rather
surprised to find that the air speed coming out of the back of this
rad was 40% of the forward velocity despite having to turn 90 degrees.
I would not call that relatively slowly.
What I meant by relatively slowly is that with an inlet area 4 - 5 times smaller than the radiator
(as it should be) the average exit velocity would 4 - 5 times less. With decent pressure recovery
the effects of things behind the radiator in Bernie's case should be minimal.
20% to 25% area as you suggest probably won't work. Insufficient
air to carry away the heat during climb. We have lots of
real world experience to suggest that. I don't know of
any installation that is using the 20 to 25% ratio. My numbers
are a minimum of 30%. Tracy is up to about 50% or more.
Not that it is relevant but the P40 had 100% with real thick rads.
The actual exit velocity is also a function of the rad thickness.
A modern thin rad of about 2 inches flows more air than say a rad
of 3 inches thick. We are still gathering data on these
issues. PL
As far as radiant heat is concerned:
Get a 100 watt infrared heat lamp and shine it on a surface ten inches away.
Gets very hot right? It will burn your hand lickity split.
That's true; but not relavant. Radiant heat transfer goes as the fourth power of temp, and the
filament in the bulb is more than 2 times hotter than an exhaust pipe that has 150F air blowing on
it. So the lamp has about 15 - 20 times higher heat flux.
Well we will just have to put a camera in there to see what color
the exhaust pipes are won't we? :-) Actually come to think of it we already
have a data acquisition system capable of measuring the exhaust
pipe temps. PL
Now think of a wildly optimistic 33% efficient, 200 net HP rotary engine.
600 HP total heat in the fuel burned. 100%
400 HP total waste heat. 66%
132 HP out the water cooling system. 22%
68 HP out the oil cooling system. 11%
200 HP out the exhaust. 33%
One HP is 746 watts. 200 HP x 746 = 149,200 watts. Lets say only one
tenth of that exhaust heat is radiant energy from the exhaust pipes.
149,200 x .1 = 14,920 watts. Now think of an 14,920 watt infrared
heat lamp ten inches away from a surface. How hot do you think
that would be?
I know you'll agree that assuming a "tenth" radiated from the exhaust was a wild guess; and then
only a small fraction of the actual amount is intercepted by the rad. I just ran some quick
numbers as follows: The radiant heat flux is the emissivity of the material times the Stefan
Boltsman constant (5.67 x 10^-8) times the absolute temp to the fourth power.
The emissivity of SS is probably somewhere between .1 and .2; let's say .15. The temp of the
pipes is maybe 1500F (1075K). This gives a heat flux of 1.1 watts/cm^2. If the pipes are 2" dia,
and 10" is adjacent to rad, two pipes is about 800 sq cm. The form factor (the fraction
intercepted) is .5 for an infinite plane; so maybe about .3 for the rad in this case.
So unless I did something wrong; 1.1 x .3 x 800 = roughly 260 watts. With about 98,000 watts of
cooling being done by the rad, I don't think 260 watts will be noticed.
BTW, there is a neat website that covers heat transfer by conduction, convection and radiation,
and gives applets for plugging in numbers and getting results.
http://www.engr.colostate.edu/~allan/heat_trans/page4/page4f.html
Al Gietzen
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