flyrotary@lancaironline.net Mailing List Archive

I’m sorry; in my haste I made an error; inlet area is 50 sq. in. (see correction in red below)

-----Original Message-----
From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Al Gietzen
Sent: Wednesday, September 01, 2010 8:17 AM
To: Rotary motors in aircraft
Subject: [FlyRotary] Re: Radiator Math

Here’s a brief summary of cooling air intake size; a.k.a., air flow requirements.

The inlet area requirement is not determined by first choosing the radiator area. It is computed from the engine power, and speed of the aircraft. The engine power determines the amount of heat to be rejected, and the speed determines how much air can flow into the scoop.

Let’s assume your engine output in full power climb (which is the highest demand on the cooling system) is 180 hp. (Skipping a couple of steps) The amount of heat transferred to the coolant, and hence to the air, is about 4800 BTU/min. Generally we’d like to design for an air temp rise through the rad of 50 – 80 F. Knowing the specific heat of air we can compute the air flow. Assuming a 70F DT; it is about 3500 cfm.

Now let’s assume your climb speed is 90 kts. Knowing the volume of air, and the speed it is coming into the scoop, we can compute the inlet area of about 50 sq. in. – assuming a very effective scoop. That’s a good starting point; but you can scale that to some different assumption; e.g., you might consider that your actual hp for climb-out is something less due to altitude/temp .etc. The inlet area will scale directly with power or aircraft speed.

Now you can consider how much you need to expand (thereby slowing) the air to ensure that you have sufficient static pressure to overcome the pressure drop of the core. For a typical rad of maybe 2 ½” thickness and 16 fins/in a ratio of roughly 1:4 has been shown to be pretty good. For a thicker rad you may need more pressure recovery (more expansion) to overcome a bigger pressure drop; a thinner rad can take a lower ratio. So here you see a dilemma – larger ratio, bigger rad face area – hum-m-m, leads to thinner rad for same volume. All this assumes that the pressure at the rad exit is ambient or lower.

I skipped all the formulas cuz this is all the time I have for this right now. But maybe gives different insight.

Al G

-----Original Message-----
From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Chris and Terria
Sent: Tuesday, August 31, 2010 6:40 PM
To: Rotary motors in aircraft
Subject: [FlyRotary] Radiator Math

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