X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from mail-wy0-f180.google.com ([74.125.82.180] verified) by logan.com (CommuniGate Pro SMTP 5.3.9) with ESMTP id 4455290 for flyrotary@lancaironline.net; Fri, 03 Sep 2010 22:09:01 -0400 Received-SPF: pass receiver=logan.com; client-ip=74.125.82.180; envelope-from=rwstracy@gmail.com Received: by wyb40 with SMTP id 40so2413789wyb.25 for ; Fri, 03 Sep 2010 19:08:24 -0700 (PDT) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=gamma; h=domainkey-signature:mime-version:received:sender:received :in-reply-to:references:date:x-google-sender-auth:message-id:subject :from:to:content-type; bh=hg9l84GV09IndaHAkYFKHqEHhh87mWr6h/EBkaqNYWQ=; b=wS7FUdLvxXQLf54TqJ7jiWuammruIhGD9wcTtV95k2yFHPqKghx/GTNnLkO9/oeeDI W4qEYx6WWUHT08ea4y7D4D57s6zJzMF+wKdMxt1GNgoTkjw7KaxpsgVx3A58Hf9BUDqG /VVimsy7DlOetioi4JgB56gZJPCA9D5fU7JZs= DomainKey-Signature: a=rsa-sha1; c=nofws; d=gmail.com; s=gamma; h=mime-version:sender:in-reply-to:references:date :x-google-sender-auth:message-id:subject:from:to:content-type; b=OpKtFrg+wZg9ER3zUcODasf/U61DzuOsw7c2YoIupZQMbCY7y4+LqGP9/SOPqf9A32 ovtc3fcafNbA2AISrcGxlLa2nKpdf7RffInojttWS+qxSYzVT5egnoxeHDHW+SSQS9SC 9Eq1OGrL+A/KifSgZXUSQCK2zutWi7ZQ1RTR0= MIME-Version: 1.0 Received: by 10.216.90.209 with SMTP id e59mr36854wef.9.1283566103977; Fri, 03 Sep 2010 19:08:23 -0700 (PDT) Sender: rwstracy@gmail.com Received: by 10.216.47.73 with HTTP; Fri, 3 Sep 2010 19:08:23 -0700 (PDT) In-Reply-To: References: Date: Fri, 3 Sep 2010 22:08:23 -0400 X-Google-Sender-Auth: teUgygxkRCULHnpVE6crlwM6m3w Message-ID: Subject: Re: [FlyRotary] Re: Radiator Math From: Tracy To: Rotary motors in aircraft Content-Type: multipart/alternative; boundary=0016e6d7e363ac5fa5048f65858b --0016e6d7e363ac5fa5048f65858b Content-Type: text/plain; charset=windows-1252 Content-Transfer-Encoding: quoted-printable 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 th= e 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 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 coolin= g > might look like this. > > 3 rotor=3D 270hp x 42.41=3D 11450.7 Btu x 66%(Mistral) =3D 7557.46Btu wat= er > 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" migh= t > cool for 7213.5Btu. > Tracy's Rad 18x12x2.625 =3D 567cu" ( 216 sq") > 25% ( 25% to 40% Rule) of 216 sq" is 54 sq" > Tracy's Inlet 6"dia =3D 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" th= ick > rad, it's likely that your not comparing apples with apples in your set-u= p. > > However I do find it very interesting when Tracy is doing so well with wh= at > might be considered the minimum requirements if well engineered cooling > system. > I have included them in my notes. > George ( down under) > > > Chris, > Sorry > 180hp =3D 7633.8 BTU > George (down under) > > > Chris, > I went overmy notes again after reading other responses. > Every hp =3D 42.41 BTU > 180 BTU =3D 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 =3D 540 cu" > Suggested inlet opening is 25 -40% of Rad surface area. > If Rad is 2.5" thick =3D 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 jus= t > 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 =3D 143.64 sq in face > > 2.1 cubic in per HP > > 2.1*210 =3D 441 cubic in > > 2.48 cubic in per HP > > 2.48*210=3D521 cubic in > > Inlet should be 15% of face > > 36 sq in inlet =3D 240 sq in face > > Since my oil cooler is on the side, and works fine, I get to use the enti= re > 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 =3D 720 cubic in o= f > radiator. This should do the trick. I could even make it a little small= er > to ensure easy clearance. My inlet may be a little smaller than the > radiator can handle, but I don=92t 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(9= 3C) > coolant and a hot day, 90F(30C) I get: > > (13.7/144) inlet *63 =3D 210 so inlet =3D 35 sq in > > So up to now I=92m 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% =3D 43.2 > > 43.2 / 16 =3D 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 > > --0016e6d7e363ac5fa5048f65858b Content-Type: text/html; charset=windows-1252 Content-Transfer-Encoding: quoted-printable One detail that might be a factor on my cooling results (and possibly other= s) is the effect of=A0 prop blast on inlet air volumes.=A0 I do see a consi= derable increase in front side of rad & oil cooler air pressure when go= ing 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 af= fect the required square inches of inlet. =A0 Without prop 'boost' = to inlet air the size might have to be larger.=A0=A0 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=20 cooling configuration and performance outcomes, it=A0might reveal that his= =20 calculation for cooling might look like this.
=A0
3 rotor=3D 270hp x 42.41=3D 11450.7 Bt= u x 66%(Mistral)=20 =3D 7557.46Btu water cooling. Al's figures might suggest 63% or 7213.5= =20 Btu=A0
=A0
If Al's 50 sq" INLET cools fo= r 4,800Btu, then (=20 extrapolated) 75 sq" might cool for=A0 7213.5Btu.
Tracy's Rad 18x12x2.625 =3D 567cu&= quot; ( 216=20 sq")
25% ( 25% to 40% Rule) of 216 sq"= is 54=20 sq"
Tracy's Inlet 6"dia =3D 28.27= sq" or 13% of Rad face=20 and only 38% of my extrapolation of Al's calculations.
Tracy may have estimated 2.1 cu" = per HP (as in your=20 original calculations).
=A0
Remembering that Tracy has been at thi= s for a long=20 time and come to his conclusions through trial and error and you have opted= for=20 a 2 pass 5" thick rad, it's likely that your not comparing apples = with apples in=20 your set-up.
=A0
However I do find it very interesting = when Tracy is=20 doing so well with what might be considered=A0the minimum requirements if= =20 well engineered cooling system.
I have included them in my notes.
George ( down under)
=A0
Chris,
Sorry
180hp =3D 7633.8 BTU
George (down under)

Chris,
I went overmy notes again after re= ading other=20 responses.
Every hp =3D 42.41 BTU
180 BTU =3D 7633.8 (approx)=
Mistral suggests 2/3 (66%) heat re= jection for=20 water or 5,089.2 BTU
Al suggests approx 4,800 or 63% ( = I think Al is=20 more accurate from memory of previous maths examples), however not much= in=20 it.
Suggested ( my notes) Rad size is = 3 cu" per hp=20 =3D 540 cu"
Suggested inlet opening is 25 -40%= of Rad=20 surface area.
If Rad is 2.5" thick =3D 216 = sq" (18" x=20 12").
Then Inlet is 54 sq" - 86.2 s= q"
Al suggests 50 sq" or 23% app= rox. Bill suggest=20 20% min or 43 sq"
I've used some rule of thumb f= igures to get to=20 this ( bottom line) but I believe Al's maths may be more=20 accurate.
But from where I'm sitting the= y are pretty much=20 saying the same thing.
George ( down under)
=A0

Let me start with my old radiator.=A0 The core= is 8=20 x 15 x 5.=A0 It is a double pass radiator.=A0 My inlet is 36 sq=20 in.=A0 I think the radiator is just too thick for good enough air flo= w=20 at the speeds we are climbing and cruising, not to mention ground=20 ops.

=A0

So I geeked out, and did the math that I have = found to=20 figure out the correct size.=A0 Those who have done this before, plea= se=20 check my math.

=A0

I found the following requirements/suggestions= during=20 my research:

=A0

1.2 sq in of face per cubic in of=20 displacement

=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =20 1.2*3*39.9 =3D 143.64 sq in face

2.1 cubic in per HP

=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =20 2.1*210 =3D 441 cubic in

2.48 cubic in per HP

=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =20 2.48*210=3D521 cubic in

Inlet should be 15% of face

=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =20 36 sq in inlet =3D 240 sq in face

Since my oil cooler is on the side, and works = fine, I=20 get to use the entire bottom of the engine for the radiator.=A0 This= =20 gives me a max space of 16 x 18 for 288 sq in face.=A0 Using a 2.5 in= =20 thick core 288*2.5 =3D 720 cubic in of radiator.=A0 This should do th= e=20 trick.=A0 I could even make it a little smaller to ensure easy=20 clearance.=A0 My inlet may be a little smaller than the radiator can= =20 handle, but I don=92t see how it can hurt to have a slightly larger r= adiator=20 than the inlet can handle.

=A0

Along the way I found a reference that said th= e heat=20 from 13.7 HP is shed for every 1*C temp differential for every sq ft = of=20 intake.=A0 Assume 200F(93C) coolant and a hot day, 90F(30C) I=20 get:

=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =20 (13.7/144) inlet *63 =3D 210=A0 so inlet =3D 35 sq in

So up to now I=92m feeling pretty good about t= he math,=20 but please let me know if I messed it up.

=A0

I will have to use the wedge shaped duct to mo= ve the=20 air through the radiator.=A0 So to figure out the height of the duct= =20 from the radiator, I again used 15% of the facial area, then divided = by=20 the width.

=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =20 16 x 18 x 15% =3D 43.2

=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =20 43.2 / 16 =3D 2.7

So Do I really make the front of the wedge onl= y 2.7=20 inches tall?=A0 This seems pretty small.=A0 It would result in a=20 really long, thin triangle.=A0 I have a max of 6.5 inches available, = so=20 can easily make it bigger.

=A0

Thanks for the help.

=A0

Chris


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