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Wait!!! Mark, remember this was all for the
example of Tracy Crook burning 7.0 gallons/hour of fuel. Now, that is near
the low end of what a two rotor will see and as Tracy stated that was for a
Cruise condition - which is the regime of flight that normally is the least
demanding so far as cooling requirements. You really need to look at the
most severe regime which is generally take off and initial climb. There
you are putting out full power and the air speed is slowest (therefore you
cooling mass flow is lowest).
But, yes what ever the two rotor BTU cooling needs
are (all other things being equal) for a specific power output then
the 3 rotor's needs would be approx 1.5 times that.
Remember this cooling stuff is far from as
simple as you might have surmised from my calculations. These equations
will give you the coolant and airflow capable of removing your waste heat -
however, whether you particular system provides that air mass flow through your
radiators and heat transfer is a different problem. Your ducting and type
diffuser and cooler core are significant factors as to
whether
1. you get your minimal airflow required and
2. That your cores effectively transfer the heat to
that cooling air mass.
For instance you could hang your radiators out in
the slip stream and possible get the required minimum airflow through them (if
you fly fast enough) but the cooling drag would be extremely high - not
desirable. To get maximum cooling with minimum drag takes a bit of
more.
Its generally accepted that you should ideally
design your cooling system to either
1. Provide adequate cooling with an airflow
velocity around 10% of CRUISE velocity
2 Or 30% of your climb velocity
In some cases the two rules produce
approx the same air velocity figure for the cooler.
So if your climb velocity was say 120 MPH IAS then
your "Optimum" airflow through your radiator would be approx 120 *.3 = 36
MPH
If your cruise airspeed was 200 mph then your
design optimum (assuming you chose cruise as you design point) would be
approx 20 MPH.
So first you need to decide on which regime you
want to design for and the trade offs involved. That establishes your
airspeed and power parameters. I choose climb cruise which leaves me with
a slight cooling deficit until I hit approx 120 mph in climb. That means
my temperature has continue to climb until I hit my design airspeed. But,
I accept I will have increased temps for a short periods I spent very little
time in take off and initial climb (to pattern altitude or maybe 2500
AGL.
Then you need to estimate your power produced
during that phase of operation, in my cause I use 170 HP - I may not make that
but if its less then my cooling problem is less severe {:>). Then you
figure your waste BTU you need to get rid of. The spreadsheet I sent will
do that for you for either a 2 rotor or 3 rotor - but you have to change the
value from 2 to 3 on the spread sheet.
So you find the rpm you are operating at and the
power/BTU rejection you must achieve. Figure that the number of BTU you
must reject through oil and coolant to be the same amount of BTU as the HP
produced (that's what the spreadsheet does). It breaks out that into the
coolant and oil BTU rejection requirements.
Then once you have the BTUs (for that set of
operating parameters) you can start getting a handle on your cooling
challenge.
Since your total surface area is appox 165 sq
inches greater that 360 vs 195 (for 2 gm cores used on the two rotors), I would
say you have the basic radiator needed to cooling your 3 rotor provided you get
the minimum required airflow.
Lets see if I use 170 HP for my climb power figure
lets assume you will use 1.5 times that or 255 HP for your 3 rotor. That
would mean you need to reject approx 7834 BTU (not the 2400 BTU of the two rotor
at lean cruise conditions) with your radiator (and another 3917 BTU from your
oil).
So we need an airmass flow through your 360 sq inch
to carry away 7834 BTU/min.
So again turning to our heat transfer equation we
have Q = W DeltaT cp. Lets assume the deltaT (increased in air temp
caused by dumping the heat into it) is around 20 - 50F(it will be more when
you are operating at a higher temps of full power situations, so lets pick
30F). The air mass flow required W = Q/(DeltaT *0.25)
W = 7834 /(50*.25) = 626 lbms/min. 626
lbm/0.076 = 8236.8 CFM (already factored in the 0.25 cp). So
changing that to cubic ft/sec we divide by 60 , 8237/60 = 137 cubic ft/sec
. Taking your radiator core surface are of 360/144 = 2.5 sq ft and we
have
( 137 cubic ft/sec)/(2.5 sq ft) = 54.31
ft/sec air velocity required. = 37.4 mph air flow. Using the
30% climb rule this means you would need to climb (while producing 255 hp) at an
airspeed of around 37/0.3 = 124 mph to maintain cooling at that power
setting and your current radiator size. Now I fudged a little bit by assuming a
DeltaT of 50F.
I don't know what delta T your system will produce
having no data points on your radiator core. I purposely picked 50F as I
knew it would produce an airflow close to what you had calculated. (See
you can make the figures provide any answer you want provided you get to make
the assumptions {:>)).
However, lets say your core's delta T is
closer to the delta Ts reported by GM core users which is generally
20-30F.
Generally the GM core users have been reporting a
20 -31F rise in air temps at high power settings. Lets say your radiator
produced the same increase in temp of 30F and not the 50F I earlier
assumed.
Then W = 7834/(30*.25) = 1045 lbm/min of air
flow needed. 1045/0.076 = 137434 Cubic ft/Min divide by 60 gives 229 cubic
ft/Sec
taking your 2.5 sq ft V = 229/2.5 = 92
ft/sec of air velocity = 62.7 MPH through your radiator. So this is approx
double the required airflow over the case where the air temp increase was
assumed to be 50F. Still all in all you radiator should have no
problem getting rid of the heat - although your cooling drag may be a bit higher
than optimum.
So just trying to point out that while these
calculations can give you a starting point, just some small differences in
assumptions (as I did with the deltaT ) can make the difference between success
and not. But since your radiator is sufficiently large, the worst penalty you
are likely to paid is a bit more drag. Assuming that your ducting and
diffuser provides airflow reasonably close to that calculated.
Hope this helps
Ed
Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
----- Original Message -----
Sent: Friday, August 13, 2004 11:02
AM
Subject: [FlyRotary] Re: Answer to when
is 2 gallons enough?Re: DeltaT Coolant was : [FlyRotary] Re: coolant
temps
Ed,
Wow, that was a great explanation. I
think I even understood most of it. ;-) I'm assuming that
for a 3-rotor I would just multiply the 2400 BTU figure by 1.5 to arrive at a
BTU requirement of 3600 for the coolant? Same for the air requirement
(240 * 1.5 = 360 lbs, or 5524.75 * 1.5 = 8287
CFM). My single radiator core is approx. 18" x 20"
or 360 sq. in. or 2.5 cubic feet. Then 8287/2.5 = 3314.8 ft/min
(55.25 ft/sec) air velocity, which equals 37.67 mph.
Does this
look right? I can cool my engine at 37 mph airflow across the
core? How does one determine the airspeed across the
radiator?
Mark
(See what you
started???)
-
- Ed Anderson
- RV-6A N494BW Rotary Powered
- Matthews, NC
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