Hi
George,
Like many results of
using equations what you get out depends on what you put in
{:>).
1st there
is 42.41 BTU/min per HP. So 42.41 *270 = 11450.7 BTU/Min - as you
calculated BUT (big but)
that is the heat energy needed to produce 270 HP that is NOT the waste heat energy you need to get rid
of. The amount you calculated is basically the heat energy
to produce 270 HP of mechanical work (turning the prop). What you want
to figure out is how much waste heat you need to get rid of (that
is - the heat energy not
used to rotate the prop). That waste heat is the heat that has not/can not be used to produce
mechanical work. Waste heat is generally gotten rid of two ways out
the exhaust stack and through convection (coolant and oil) system in our
engines. So you are primarily interested in how much heat you must get
rid of by convection (coolant and oil) systems.
Here is my method and
how I got the 8288 BTU/min . Again, your calculation does gives
you the amount of heat energy in BTU that produces 270 HP - that heat
has been used doing work by your engine, however, that is NOT the heat you are getting rid of
through your coolers . For cooling purposes, you want to find the
BTUs of heat energy you need to get rid of that is NOT producing power -
or in other words - the waste heat (I think I repeated myself).
Again,
the HP of work your engine is producing is NOT the heat energy that you have to get
rid of through your two cooling systems (radiators and oil coolers) (oops
repeated myself again but the distinction is crucial!)
Here is how I arrived
at my figures for the waste
heat. There is more than one way to do this for sure and the
results depends a lot on your heat allocation (more on that in a
bit).
Using a air/flow and
air density and a formula for a 3 rotor, I calculated that at a 12.65:1
air/fuel ratio (best power so not cruise {:>)) that you would burn around
25.2 GPH at 270 HP.
Using the old and
simply (but good) approx power formula, I calculate HP = 25.2 * 6 /0.55
= 274.9 HP which is pretty close to the results (270 HP) of my more
complex calculation
which is based on
engine air/flow and fuel mass consumed at that ratio. So 25.2 GPH is a pretty
valid figure for our fuel consumption at that
power.
However, I
calculate engine HP from a different approach (got to make it more complex,
you know {:>)). My approach is based on the total
number of BTU in the amount of fuel
ingested by the
engine to meet the power requirement (270 HP) with following heat energy
allocation: If your percentage of energy allocation is different, then
you will get a different results:
24% = HP (useful
mechanical work) some folks may think this is too conservative, but I tend
to be that way when calculating HP.
50%= out the
Exhaust
26% = Cooling Waste
(coolant + oil)
We know coolant = 2/3
of 26% Cooling Waste and the oil = 1/3.
There is approx
19000-20000 BTU/lbm in gasoline (I use 19000 BTU/pound to be a bit
conservative besides 100LL has less energy than 87 Mogas
{:>)
So with 25.2 GPH * 6
lbm/gallon = 151 lbm/hour/60 = 2.52 lbm/min of gasoline is being
burned to produce the energy so out engine produces 270
HP.
So 2.52 *19000
BTU/Lbm = 47880 BTU/min total
energy of the fuel consumed per minute at the 25.2 GPH rate.
This total includes the work energy (HP) and the waste
energy.
So to find out the
heat used to produce work 24% *47880 BTU/min = 11491.2 BTU/min Now
that is not much different from what you got 11450.7 BTU/Min. But, as I
stated this is
NOT the waste heat you
need to reject.
So Allocation of heat
energy we need to get rid of through the cooling system, we
have
26% = cooling waste =
0.26 * 47880 = 12448.8 BTU/Min of which 2/3 is rejected through
radiator , therefore
0.6666 * 12448.8 =
8298 BTU/Min of waste heat
through the radiators. (does not include waste heat through the oil
which is 12448.8 8298 = 4150.8 BTU/min).
So altogether the
radiator and oil cooler has to get rid of around 12500 BTU/min at that power
level so you can see that if you do not have adequate cooling at 270 HP
power production you are going to fry your engine fairly quickly.
This is how it
appears to me, George. Hope it helps.
Ed
From:
Rotary motors in aircraft
[mailto:flyrotary@lancaironline.net] On
Behalf Of George Lendich
Sent: Tuesday, January 05, 2010 6:01
PM
To: Rotary motors in aircraft
Subject: [FlyRotary] Re: Air Flow
Question
I have been terribly busy, but
wanted to follow this thread and finally got around to it, but can't
understand where you got 8288 BTU/min.
If 1hp = 2545BTU/hr /60
=42.41BTU/min x 270hp=1145.7 btu/min of which we are needing only 2/3 of water
cooling, as 1/3 is done by the oil . Therefore 2/3 of 1145.7=7633.8
btu/min. Did I go wrong somewhere ?
Also I didn't know we were looking
for 18knots as optimum air speed flow through the rad. I did a quick copy of
your calculations and got 28 knots for 125 hp, so I will have to redefine
my cooling set-up.
Ed, I like the maths approach. Is
Tracy
following this line of thought i.e. calculating for cooling, as an
Engineer I assumed he would be doing something following this line of
approach?
Thomas, it doesnt
work quite that way. Old Bernoullis 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 (cant 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 lets 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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