flyrotary@lancaironline.net Mailing List Archive

Chris,

I went over my notes again after reading other responses.

Every hp = 42.41 BTU

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.

George ( down under)

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