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 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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