There is another point to clarify
regarding the inlet scoop area, which I probably should have made earlier.
Since I fly a pusher; my thinking is in terms free-stream air velocity into the
scoop, and that is the ref case for the 50 sq.in. number. The case for an
intake strategically located behind the prop; that velocity is higher; and
considerably higher during takeoff roll.
In the case of Tracy’s
setup, that is a significant factor. Still; I’d suggest he has an
issue making a full power, extended climb on a 90F day without overheating the
coolant. I think Tracy will agree (right, Tracy?)
But he doesn’t have to. As
he says, after he has climbed to maybe pattern altitude (or whatever) he can
start pulling back the power. During that climb he also can take
advantage of the heat capacity of the all the steel, aluminum, and coolant. It
takes a while to heat that up. And he has mentioned considering installing a
water spray augment system.
There is good reason to have the minimum
air intake that allows you to complete your climb – lower drag for the
entire flight. At cruise, running at, say, 55% power, the 28 sq in is
entirely adequate.
Al
-----Original Message-----
From: Rotary motors in aircraft
[mailto:flyrotary@lancaironline.net] On
Behalf Of George Lendich
Sent: Friday, September 03, 2010
3:18 PM
To: Rotary motors in aircraft
Subject: [FlyRotary] Re: Radiator
Math
From Tracy's response to Steve's
request on Tracy's cooling configuration and performance outcomes,
it might reveal that his calculation for cooling might look like this.
3 rotor= 270hp x 42.41= 11450.7 Btu
x 66%(Mistral) = 7557.46Btu water cooling. Al's figures might suggest 63% or
7213.5 Btu
If Al's 50 sq" INLET cools for
4,800Btu, then ( extrapolated) 75 sq" might cool for 7213.5Btu.
Tracy's Rad 18x12x2.625 =
567cu" ( 216 sq")
25% ( 25% to 40% Rule) of 216
sq" is 54 sq"
Tracy's Inlet 6"dia = 28.27
sq" or 13% of Rad face and only 38% of my extrapolation of Al's
calculations.
Tracy may have estimated 2.1
cu" per HP (as in your original calculations).
Remembering that Tracy has been at
this for a long time and come to his conclusions through trial and error and
you have opted for a 2 pass 5" thick rad, it's likely that your not
comparing apples with apples in your set-up.
However I do find it very
interesting when Tracy is doing so well with what might be considered the
minimum requirements if well engineered cooling system.
I have included them in my notes.
I went overmy notes again after
reading other responses.
180 BTU = 7633.8 (approx)
Mistral suggests 2/3 (66%) heat
rejection for water or 5,089.2 BTU
Al suggests approx 4,800 or 63% ( I
think Al is more accurate from memory of previous maths examples), however not
much in it.
Suggested ( my notes) Rad size is 3
cu" per hp = 540 cu"
Suggested inlet opening is 25 -40%
of Rad surface area.
If Rad is 2.5" thick = 216
sq" (18" x 12").
Then Inlet is 54 sq" - 86.2
sq"
Al suggests 50 sq" or 23%
approx. Bill suggest 20% min or 43 sq"
I've used some rule of thumb figures
to get to this ( bottom line) but I believe Al's maths may be more accurate.
But from where I'm sitting they are
pretty much saying the same thing.
Let me start with my old radiator. The core is 8
x 15 x 5. It is a double pass radiator. My inlet is 36 sq in.
I think the radiator is just too thick for good enough air flow at the speeds
we are climbing and cruising, not to mention ground ops.
So I geeked out, and did the math that I have found to
figure out the correct size. Those who have done this before, please
check my math.
I found the following requirements/suggestions during
my research:
1.2 sq in of face per cubic in of displacement
1.2*3*39.9 = 143.64 sq in face
2.1 cubic in per HP
2.1*210 = 441 cubic in
2.48 cubic in per HP
2.48*210=521 cubic in
Inlet should be 15% of face
36 sq in inlet = 240 sq in face
Since my oil cooler is on the side, and works fine, I
get to use the entire bottom of the engine for the radiator. This gives
me a max space of 16 x 18 for 288 sq in face. Using a 2.5 in thick core
288*2.5 = 720 cubic in of radiator. This should do the trick. I
could even make it a little smaller to ensure easy clearance. My inlet
may be a little smaller than the radiator can handle, but I don’t see how
it can hurt to have a slightly larger radiator than the inlet can handle.
Along the way I found a reference that said the heat
from 13.7 HP is shed for every 1*C temp differential for every sq ft of
intake. Assume 200F(93C) coolant and a hot day, 90F(30C) I get:
(13.7/144) inlet *63 = 210 so inlet = 35 sq in
So up to now I’m feeling pretty good about the
math, but please let me know if I messed it up.
I will have to use the wedge shaped duct to move the
air through the radiator. So to figure out the height of the duct from
the radiator, I again used 15% of the facial area, then divided by the width.
16 x 18 x 15% = 43.2
43.2 / 16 = 2.7
So Do I really make the front of the wedge only 2.7
inches tall? This seems pretty small. It would result in a really
long, thin triangle. I have a max of 6.5 inches available, so can easily
make it bigger.
Thanks for the help.
Chris