Thomas, it doesn’t
work quite that way. Old Bernoulli’s law is (simplified by removing
density) is A1V1 = A2V2 meaning the product
of the area and velocity anywhere in your duct is equal.
Something to do with conservation of mass (can’t created/destroy it). So it
based on area expansion rather than volume.
So to determine (more
or less) the air velocity you need to know the velocity of the air entering
your duct (and the inlet area) and the area down stream that you are
expanding. However, determining the velocity of air entering your duct
may not be as simple as it first seems due to a condition known as external
diffusion. This is the air streaming being slowed down in front of the
inlet due to a pressure gradient extending out of your duct opening (you
can sort of think of it as air molecules piling up before your core and
increasing the pressure back out your duct).
Where
to Start? Either find an
installation very similar to yours that is cooling adequate and copy that OR
you can do some figuring on the back of an
envelope.
You gotta start
somewhere and Mr. Horner indicated that you need to have the airflow through
your core either 10% of your cruise speed or 30% of your climb speed.
So if your cruise
speed is 180 knots then you would want the airflow through your core to be
ideally around 18knots.
Just as an example
(disregarding external diffusion) lets say your inlet opening A1 was 20
x2 = 40 sq inches = 40/144 = 0.277 sq ft then if you want 18.4 knots the
area in the duct at A2 then you solve for A2
A1V1 = A2V2 so
solving for A2 = A1V1/V2 = 0.277 * 180/18 = 2.777 sq ft or 2.777 * 144 = 400
square inches or an appox 10:1 difference between opening and expanded
area.
However, another
couple of wizards (Kuchumman and Weber) indicated that for good diffusion,
your ratio of inlet area to area before your core should be between 0.25 and
0.40 - going beyond that you start to go bad.
So lets say you need
400 sq inches to accommodate your radiator core, then according to K&W
your inlet would need to be between 0.25 and 0.40 X 400 = 100 – 160 sq
inches The larger inlet would also tend to diminish the external
diffusion effect but not slow the air velocity as much as the smaller
opening..
However we still have
A1V1 = A2V2 so with A1 at 100 sq inch (0.694 sq ft) and assuming inlet
air velocity is 180 knots and now having A2 fixed at 400 sq inch or 2.777 sq
ft we have V2 = A1V1/A2 = 0.694 * 180/2.777 = 45
kts.
So our air velocity
at the core is a bit higher than we would like (according to Horner) so while
it will cool, we may be encountering more cooling drag due to the higher
velocity air through the core than we would like.
But, this is just a
back of the envelope calculation. So many things can affect cooling, we
should be so lucky that it would be just one major thing.
Where
I would Start:
I personally think
the place to start is to 1st size your radiator core based on the
heat you want to get rid of in your worst case situation (probably take
off/climb). Since few of us have wind tunnels, starting with a rule of
thumb for core volume to HP would probably be a good place to
start.
Then looking at your
space constraints to determine your radiator size. I would not go much
thicker than 3” . NASCAR car radiators are typically around 3” in
thickness with some going up to 7” thick for the higher speed long tracks at
speeds comparable to ours. Also whatever the radiator builders have sort
of mandates what you use.
So I think you
mentioned a rule of thumb of 1.8 cubic inch of core/HP ( I personally feel
this may be a little on the low side). Assuming 270 HP max engine power
then that would indicate a core volume of approx 1.8 * 270 = 486 cubic
inches (lets round it up to an “even” 500 cubic inches). Assuming you
find a core 3” thick then its front area would need to be 500/3 = 166.6 sq
inches. So that could be 16 wide and 10” high or 27.7 inches wide
by 6” high or what ever combination – again likely constrained by what the
manufactures build.
But let’s say you
chose 16 x10 x3 radiator. The next thing you need to know is how much
airflow you must have through it to dissipate the heat (coolant only in this
example). For a three rotor producing 270 hp the coolant needs to get
rid of approx 8288 BTU/Minute.
Assuming we can add heat to the cooling air increasing its temperature by 80F
(might get 100),
Then the air mass required can be
found from Q (BTU) = M(mass)*Dt*Cp
rearranging the formula
M = Q/
Dt*Cp =
8288/(80*0.25) = 414.4 lbm/min of air. One Cubic foot of air at sea
level = 0.0765 lbm
So air flow in CFM = 414.4/0.0765
= 5416 CFM of cooling air. We need to pass that through the
frontal area of our core (166.6 sq inches /144 = 1.1569 sq ft).
5416 / 1.1569 = 4681 ft/min of air velocity or dividing by 60 = 78.01 ft/sec =
46 knots air velocity through your core (166.6 sq inch). Not quite the
18 kts Horner wanted but at least a start. So what does this tell
us. That if we want to get by with the ideal (slower) airflow
through the core (18 kts) then our core frontal area needs to be larger than
166 sq inches.
Back to A1V1 = A2V2 if we need 46
knots through 1.1569 sq ft of core frontal area and V1 (assuming no external
diffusion) = 180 kts then the inlet A1 = 46 * 1.569/180 = 0.4 sq ft inlet =
0.4 * 144 = 57.7 sq inch inlet opening. Now remember this is all looking
at cooling a cruise – where things are best. Conditions during
take off and climbout are going to be worst case. So you need to do all
of this for those airspeeds as well
Note there are a whole bunch of
assumptions made to simplify things which may not hold true in all
cases.
The first major one is the rule of
thumb 1.8 cubic inch /Hp. IF your core is fabricated similar to the core
from which this rule of thumb was drawn then you are probably OK. But,
if substantive different then this rule of thumb may not be valid and then
your basic assumption is flawed and need I add all following that
is now garbage.
The alternative to all of this
stuff – is to find an installation as close as possible to yours that is
cooling adequately and use that as you cooling system design
basis.
Sorry – got carried
away.
Ed
From:
Rotary motors in aircraft
[mailto:flyrotary@lancaironline.net] On
Behalf Of Thomas Mann
Sent: Monday, December 21, 2009 2:04
PM
To: Rotary motors in aircraft
Subject: [FlyRotary] Air Flow
Question
If I have a
volume of air entering my scoop at 180 kts and expand the volume of the
chamber by 400% can I expect the speed of the airflow to drop to 45 kts at
that point?
T
Mann
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