Mailing List flyrotary@lancaironline.net Message #52026
From: Tracy <tracy@rotaryaviation.com>
Sender: <rwstracy@gmail.com>
Subject: Re: [FlyRotary] Re: Radiator Math
Date: Fri, 3 Sep 2010 22:08:23 -0400
To: Rotary motors in aircraft <flyrotary@lancaironline.net>
One detail that might be a factor on my cooling results (and possibly others) is the effect of  prop blast on inlet air volumes.  I do see a considerable increase in front side of rad & oil cooler air pressure when going from idle to full throttle (while holding airspeed steady).
My inlets are only about 3/4" behind the prop and out at the far edge of the cowl instead of right next to the spinner.

This factor may affect the required square inches of inlet.   Without prop 'boost' to inlet air the size might have to be larger.   On canards especially.

Tracy

On Fri, Sep 3, 2010 at 7:17 PM, George Lendich <lendich@aanet.com.au> wrote:
Chris,
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.
George ( down under)
 
Chris,
Sorry
180hp = 7633.8 BTU
George (down under)

Chris,
I went overmy 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


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