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