X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from cdptpa-omtalb.mail.rr.com ([75.180.132.120] verified) by logan.com (CommuniGate Pro SMTP 5.3c4) with ESMTP id 4034187 for flyrotary@lancaironline.net; Mon, 21 Dec 2009 17:17:47 -0500 Received-SPF: pass receiver=logan.com; client-ip=75.180.132.120; envelope-from=eanderson@carolina.rr.com Return-Path: X-Authority-Analysis: v=1.0 c=1 a=ayC55rCoAAAA:8 a=arxwEM4EAAAA:8 a=QdXCYpuVAAAA:8 a=7g1VtSJxAAAA:8 a=ekHE3smAAAAA:20 a=UretUmmEAAAA:8 a=Ia-xEzejAAAA:8 a=nUuTZ29dAAAA:8 a=eTsptyESEjRY4RYb6IwA:9 a=uWVA4YRI9jwlIATvXz8A:7 a=xN55kFkWn8P_OKa1jwm1fhdhdr0A:4 a=1vhyWl4Y8LcA:10 a=EzXvWhQp4_cA:10 a=H6fgXQeuygc4TgYS:21 a=F8j2fCvbl5NH-tLa:21 a=Y2VNeNrzAAAA:8 a=yMhMjlubAAAA:8 a=TW66zc2HAAAA:8 a=SSmOFEACAAAA:8 a=HQ31llbKAAAA:8 a=LI2dLk-wI8WLKrUxcgcA:9 a=TyLQNe07IjnkfYWX-ZwA:7 a=d-wbSZ5HOnc8CxPu3_kasIHD2GoA:4 a=rd0Vg5jzDUiXwG_F:21 a=4v0ttkRkTElL44ff:21 X-Cloudmark-Score: 0 X-Originating-IP: 75.191.186.236 Received: from [75.191.186.236] ([75.191.186.236:2087] helo=computername) by cdptpa-oedge04.mail.rr.com (envelope-from ) (ecelerity 2.2.2.39 r()) with ESMTP id E0/D9-01550-7E3FF2B4; Mon, 21 Dec 2009 22:17:12 +0000 From: "Ed Anderson" Message-ID: To: "'Rotary motors in aircraft'" Subject: RE: [FlyRotary] Re: Air Flow Question Date: Mon, 21 Dec 2009 17:17:17 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_000C_01CA8261.72CD0D80" X-Mailer: Microsoft Office Outlook, Build 11.0.5510 Thread-Index: AcqChh3C+vtrI/uxT8WNVVydel6c6gAAdFMg In-Reply-To: X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5579 This is a multi-part message in MIME format. ------=_NextPart_000_000C_01CA8261.72CD0D80 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: 7bit Thomas, you are correct in that you would like to decrease the air velocity through your core to minimum possible consistent with adequate cooling because that (for one thing) reduces your cooling drag. As Tracy indicated getting cooling right tends to be a combination of understanding, testing and some luck. Trying to figure out cooling can drive you bonkers, what seems intuitive can turn out to be just the opposite of what you may think. Mr. Bernoulli's equation can offer some understanding of what happens in a duct with airflow (but not necessarily the complete picture). P1A1V1 = P2A2V2 where P stands for density which is essentially constant at our airspeeds so can be left out simplifying to A1V1 =A2V2. Or the product of area and velocity is constant through out a duct. So if Area increases, velocity decreases, if area decreases velocity increases. There are other non-intuitive factors, For example in a duct When you slow down airflow through an area expansion 1. Pressure in a duct increases 2. Temperature increases (very small amount at our speeds) 3. Slower air will increase in temperature (due to longer exposure to the hot metal) 4. Slower air will decrease your air mass flow which you need to carry the heat away Boundary layer separation in a duct generally leads to poorer cooling, rapid expansion leads to boundary layer separation, and the secret is to prevent the boundary layer separation until the last possible moment. The closer it happens to the core the less area of the core is adversely affected. It's all about compromising conflicting factors in a cooling system. Slower air picks up more heat per unit mass, but faster air carries away more heat per unit time. More air mass flow cools better, higher velocity gives more air mass flow, however, higher velocity means more cooling drag, etc. Stick with this group, we've all been there and have conquered the cooling beast - well, at least tamed it a bit. Ed Ed Anderson Rv-6A N494BW Rotary Powered Matthews, NC eanderson@carolina.rr.com http://www.andersonee.com http://www.dmack.net/mazda/index.html http://www.flyrotary.com/ http://members.cox.net/rogersda/rotary/configs.htm#N494BW http://www.rotaryaviation.com/Rotorhead%20Truth.htm _____ From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Thomas Mann Sent: Monday, December 21, 2009 4:39 PM To: Rotary motors in aircraft Subject: [FlyRotary] Re: Air Flow Question The numbers I'm looking at are for relation reference only so 180kt inlet speed does not indicate the airspeed of the aircraft (for sake of argument.) I have read that the airspeed needs to be reduced, hence the expansion from a 16 x 3 opening to a core frontal area of 200 sq in.) As a result of the expansion and drop in pressure, I realize that the air that reaches the core will be cooler than the air that enters the inlet. From the radiator to the exhaust I'm looking at reducing to an opening size of approximately 75 sq inches. Based on what I have read this should slow the air down sufficiently to get it closer to a speed where it can absorb the most heat from the core. Some of the speeds I've been quoted have been in the neighborhood of 35 kts. I don't know how accurate those numbers are. The air is expanding to a greater volume after it passes through the radiator and is compressed due to the ducting which is supposed to result in enough thrust gain to cancel out 90+% of the drag created by the scoop. I know that a 180 > 45 is close. I don't know if I could get away with a smaller inlet or if a 4 to one ration is as far as I can push it. T Mann __________ Information from ESET NOD32 Antivirus, version of virus signature database 3267 (20080714) __________ The message was checked by ESET NOD32 Antivirus. http://www.eset.com ------=_NextPart_000_000C_01CA8261.72CD0D80 Content-Type: text/html; charset="us-ascii" Content-Transfer-Encoding: quoted-printable

Thomas, you are correct in that you = would like to decrease the air velocity through your core to minimum possible = consistent with adequate cooling because that (for one thing)  reduces your = cooling drag. As Tracy indicated getting cooling right tends to be a combination of = understanding, testing and some luck.

 

Trying to figure out cooling can = drive you bonkers, what seems intuitive can turn out to be just the opposite of = what you may think.  Mr. Bernoulli’s equation can offer some = understanding of what happens in a duct with airflow (but not necessarily the complete = picture).  P1A1V1 =3D P2A2V2 where P stands for density which is essentially = constant at our airspeeds so can be left out simplifying to A1V1 =3DA2V2.  Or the = product of area and velocity is constant through out a duct.  So if Area = increases, velocity decreases, if area decreases velocity = increases.

 

There are other non-intuitive = factors, For example in a duct

 

When you slow down airflow through = an area expansion

 

  1. Pressure in a duct increases
  2. Temperature increases (very small amount at our = speeds)
  3. Slower air will increase in temperature (due to longer exposure to the hot = metal)
  4. Slower air will decrease your air mass flow which you need to carry the = heat away

 

Boundary layer separation in a duct generally leads to poorer cooling, rapid expansion leads to boundary = layer separation, and the secret is to prevent the boundary layer separation until the = last possible moment.  The closer it happens to the core the less area = of the core is adversely affected.

 

It’s all about compromising conflicting factors in a cooling system. Slower air picks up more heat = per unit mass, but faster air carries away more heat per unit time.  More = air mass flow cools better, higher velocity gives more air mass flow, however, = higher velocity means more cooling drag, etc.

 

Stick with this group, we’ve = all been there and have conquered the cooling beast – well, at least = tamed it a bit.

 

Ed

From: = Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Thomas Mann
Sent: Monday, December = 21, 2009 4:39 PM
To: Rotary motors in aircraft
Subject: [FlyRotary] Re: = Air Flow Question

 

The numbers I’m looking at are for relation reference only so 180kt inlet = speed does not indicate the airspeed of the aircraft (for sake of = argument.)

 <= /o:p>

I have read = that the airspeed needs to be reduced, hence the expansion from a 16 x 3 opening = to a core frontal area of 200 sq in.) As a result of the expansion and drop = in pressure, I realize that the air that reaches the core will be cooler = than the air that enters the inlet.

 <= /o:p>

From the = radiator to the exhaust I’m looking at reducing to an opening size of = approximately 75 sq inches. Based on what I have read this should slow the air down sufficiently to get it closer to a speed where it can absorb the most = heat from the core. Some of the speeds I’ve been quoted have been in the neighborhood of 35 kts. I don’t know how accurate those numbers = are.

 <= /o:p>

The air is = expanding to a greater volume after it passes through the radiator and is = compressed due to the ducting which is supposed to result in enough thrust gain to = cancel out 90+% of the drag created by the scoop.

 <= /o:p>

I know that = a 180 > 45 is close. I don’t know if I could get away with a smaller = inlet or if a 4 to one ration is as far as I can push = it.

 <= /o:p>

T Mann =



__________ Information from ESET NOD32 Antivirus, version of virus = signature database 3267 (20080714) __________

The message was checked by ESET NOD32 Antivirus.

http://www.eset.com

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