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Ed,
If memory serves me correctly, the 124 mph figure was about what I got
when plugging my numbers into your spreadsheet. With the airflow
across the radiator being so important, I can see where cowl flaps could
be a big help in solving the cooling problem when attempting to optimize
for cruise and takeoff conditions. I have a cowl flap that I hope
will help to accomplish that. I guess I'll have to fly it to find
out for sure.
Mark
At 01:54 PM 8/13/2004 -0400, you wrote:
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 -----
- From: Mark Steitle
- To: Rotary motors in
aircraft
- 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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