X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from ms-smtp-01.southeast.rr.com ([24.25.9.100] verified) by logan.com (CommuniGate Pro SMTP 5.1.10) with ESMTP id 2181486 for flyrotary@lancaironline.net; Mon, 16 Jul 2007 19:38:19 -0400 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.100; envelope-from=eanderson@carolina.rr.com Received: from edward2 (cpe-024-074-103-061.carolina.res.rr.com [24.74.103.61]) by ms-smtp-01.southeast.rr.com (8.13.6/8.13.6) with SMTP id l6GNbSvA003575 for ; Mon, 16 Jul 2007 19:37:29 -0400 (EDT) Message-ID: <001601c7c802$56de5690$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: FW: Oil cooler air flow Date: Mon, 16 Jul 2007 19:37:57 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0013_01C7C7E0.CF806B50" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.3138 X-MIMEOLE: Produced By Microsoft MimeOLE V6.00.2900.3138 X-Virus-Scanned: Symantec AntiVirus Scan Engine This is a multi-part message in MIME format. ------=_NextPart_000_0013_01C7C7E0.CF806B50 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable OK, Al, let me restate in a more comprehensive manner and see if that = helps. We know that "dynamic pressure" is actually measured by the increase it = causes in localized static pressure. So the term "dynamic pressure" is = actually just referring to the energy potential (Kinetic) of the moving = air to cause a localized increase in static pressure - if that air = movement were brought to a stop. In other words, if we had a flow of air with a specific velocity and = specific density, that air would have a ambient static pressure (say at = sea level of 29.92" HG). The moving air would also have a static = pressure potential (Dynamic pressure) based on its velocity and density. = So that if a tube were used to measure this "Dynamic Pressure" it must = first bring that part being measured to a stop the action of which = converts the dynamic pressure potential of the moving air to a localized = static pressure increase in the tube.=20 So the total static pressure at the measuring point would be the static = pressure of the ambient air (29.92"HG) plus whatever increase was caused = by stopping the moving air or converting its dynamic potential to static = pressure. So Pt =3D Pa + Pd with Dynamic Pressure component, Pd =3D = p*1/2V^2. So in case of a duct there is, of course, only ambient static pressure = in the duct if there is no air flow through the duct. Once there is = airflow then you also have potential pressure in the form of the kinetic = energy of the moving air. So that Pt =3D Pa + p1/2V^2. p being air = density, V being the velocity. The streamline duct (theoretically) can convert 84% of the moving air = potential dynamic pressure to static pressure increase. So that at the = widest part of the duct just before the core you would have a total = static pressure Pt =3D Pa + 0.84*p1/2V^2. =20 But, using differential pressure gauges with tubes pointed into the = moving air, we are not measuring total pressure, but the pressure = increase due solely to the moving air. In other words, if you were = measuring 5" H20 and then the air stopped moving , the gauge would read = zero. So with the manometer you are measuring the pressure above ambient = pressure or that resulting solely from the dynamic pressure potential of = the moving air being converted from kinetic energy to static pressure. = Yes, the ambient pressure is present but you are not measuring it. With = no moving air the water levels in you manometer would all be exactly at = the same level.. The fact is that you are measuring static pressure at both locations - = the 9.5" before the duct was a static pressure increase in your = measuring tube - cause by stopping the moving air so its refer to as = dynamic pressure. The fact is that you were also measuring static = pressure 3.25" at the location in the duct - but both resulted from the = transformation of the air's kinetic energy into a local static pressure = increase. Therefore, the fact that you were measuring considerably more = pressure before the duct than inside it indicates that the air stream's = velocity is not being efficiently transformed into static pressure in = the duct. This implies that perhaps there is less air velocity entering the duct = than your measurement a couple inches in front suggests OR there is = sufficient eddies and adverse flow situation inside the duct that = precludes the efficient transformation into a static pressure increase. = I do not have an aerodynamic or gas dynamics background, so I could = certainly be wrong. But, that is my understanding based on the somewhat = extensive reading I have done. Ed ----- Original Message -----=20 From: Al Gietzen=20 To: Rotary motors in aircraft=20 Sent: Monday, July 16, 2007 5:48 PM Subject: [FlyRotary] Re: FW: Oil cooler air flow if the free air velocity (160) converts to 12"H20 and you had a = streamline duct inlet actual had that coming in then theoretically you = could get approx 12 * .84 =3D 10.8" inside the duct. Since you measured = 3.25" static in front of the core, that would indicate a significant = lack of pressure recovery inside your duct (what ever the reason). = There are several reasons this might be happening. I think the confusion here is whether we're talking "dynamic" pressure = or "static" pressure. Are you saying that the maximum static pressure = in the duct is 0.84 of the dynamic pressure at the entrance to the duct? = If that is true, I have been under a misconception. I measured 9.5" = dynamic pressure out in front of the scoop; and 3.25" static pressure = near the face of the core - just below the midpoint. 1. The air flow and velocity is considerably reduced from what you = are expecting (too small opening/exit - which I don't believe to be the = case) 2. The boundary layer is a significant part of your duct total air = flow and as a consequence its lesser velocity has less dynamic pressure = potential. 3. A significant part of your duct flow is chaotic with eddies which = does not provide recoverable pressure - or it is much reduced. (The = boundary layer could be contributing to this) 4. Some combination of the above. Right, now I would suspect that the boundary layer could be the = culprit in that it can contribute to 2 and 3 above. But, as you know, = this is speculation on my part I'm sure you're right; a combination of 2 and 3. Yesterday I = measured the static pressure near the upper surface of the duct; an inch = or so in front of the core - less than 0.25" H2O. That confirmed to me = that the "flow is chaotic with eddies", as you say. I think the = addition of a vane is worth a try. Al ------=_NextPart_000_0013_01C7C7E0.CF806B50 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
OK, Al, let me restate in a more comprehensive = manner and=20 see if that helps.
 
We know that "dynamic pressure" is actually = measured by=20 the increase it causes in localized static pressure. So=20 the term  "dynamic pressure" is actually just referring to the = energy=20 potential (Kinetic) of the moving air to cause a localized increase in = static=20 pressure - if  that air movement were brought to a = stop.
 
 In other words, if we had a flow of air = with a=20 specific velocity and specific density, that air would have a ambient = static=20 pressure (say at sea level of 29.92" HG).  The moving air would=20 also have a static pressure potential (Dynamic pressure) based on = its=20 velocity and density.  So that if a tube were used to measure this = "Dynamic=20 Pressure" it must first bring that part being measured to a = stop the action=20 of which converts the dynamic pressure potential of the moving air = to a=20 localized static pressure increase in the tube. 
 
 So the total static pressure at the = measuring point=20 would be the static pressure of the ambient air (29.92"HG) plus whatever = increase was caused by stopping the moving air or converting its dynamic = potential to static pressure.  So Pt =3D Pa  + Pd with Dynamic = Pressure=20 component, Pd =3D p*1/2V^2.
 
So in case of a duct there is, of course, only = ambient=20 static pressure in the duct if there is no air flow through the = duct.  Once=20 there is airflow then you also have potential pressure in the form of = the=20 kinetic energy of the moving air.  So that Pt =3D Pa + = p1/2V^2.  p being=20 air density, V being the velocity.
 
 The streamline duct (theoretically) can = convert 84%=20 of the moving air potential dynamic pressure to static pressure = increase. =20 So that at the widest part of the duct just before the core you would = have a=20 total static pressure  Pt =3D Pa + 0.84*p1/2V^2.  =
 
But, using differential pressure gauges with = tubes pointed=20 into the moving air, we are not measuring total pressure, but the = pressure=20 increase due solely to the moving air.  In other words, if you were = measuring 5" H20 and then the air stopped moving , the gauge would read=20 zero.
 
 So with the manometer you are measuring = the pressure=20 above ambient pressure or that resulting solely from the dynamic = pressure=20 potential of the moving air being converted from kinetic energy to = static=20 pressure.  Yes, the ambient pressure is present but you are not = measuring=20 it.  With no moving air the water levels in you manometer would all = be=20 exactly at the same level..
 
The fact is that you are measuring static = pressure at both=20 locations - the 9.5" before the duct was a static pressure increase in = your=20 measuring tube - cause by stopping the moving air so its refer to = as=20 dynamic pressure.  The fact is that you were also measuring static = pressure=20 3.25" at the location in the duct - but both resulted from the = transformation of=20 the air's kinetic energy into a local static pressure increase.  = Therefore,=20 the fact that you were measuring considerably more pressure before the = duct than=20 inside it indicates that the air stream's velocity is not being = efficiently=20 transformed into static pressure in the duct.
 
This implies that perhaps there is less air = velocity=20 entering the duct than your measurement a couple inches in front = suggests OR=20 there is sufficient eddies and adverse flow situation inside the = duct that=20 precludes the efficient transformation into a static pressure = increase. =20
 
I do not have an aerodynamic or gas dynamics = background,=20 so I could certainly be wrong.  But, that is my understanding based = on the=20 somewhat extensive reading I have done.
 
Ed
 
 
----- Original Message -----
From:=20 Al = Gietzen=20
Sent: Monday, July 16, 2007 = 5:48 PM
Subject: [FlyRotary] Re: FW: = Oil cooler=20 air flow

if the free air velocity = (160)=20 converts to 12"H20 and you had a streamline duct inlet actual had that = coming=20 in then theoretically you could get approx 12 * .84 =3D 10.8" inside = the=20 duct.  Since you measured 3.25" static in front of the core, that = would=20 indicate a significant lack of pressure recovery inside your duct = (what ever=20 the reason).  There are several reasons this   = might be=20 happening.

I think = the=20 confusion here is whether we=92re talking =93dynamic=94 pressure or = =93static=94=20 pressure.  Are you saying that the maximum static pressure in the = duct is=20 0.84 of the dynamic pressure at the entrance to the duct? If that is = true, I=20 have been under a misconception.  I measured 9.5=94 dynamic = pressure out in=20 front of the scoop; and 3.25=94 static pressure near the face of the = core =96 just=20 below the midpoint.

 

1.  The air flow = and velocity=20 is considerably reduced from what you are expecting (too small = opening/exit -=20 which I don't believe to be the case)

 

2.  The boundary = layer is a=20 significant part of your duct total  air flow  and as a = consequence=20 its lesser velocity has less dynamic pressure=20 potential.

 

3.  A significant = part of=20 your duct flow is chaotic with eddies which does not provide = recoverable=20 pressure - or it is much reduced.  (The boundary layer could be=20 contributing to this)

 

4.  Some = combination of the=20 above.

 

Right, now I=20 would suspect that the boundary layer could be the culprit in that = it can=20 contribute to 2 and 3 above.  But, as you know, this is = speculation on=20 my part

 

I=92m = sure you=92re=20 right; a combination of 2 and 3. Yesterday I measured the static = pressure=20 near the upper surface of the duct; an inch or so in front of the = core =96=20 less than 0.25=94 H2O.  That confirmed to me that the=20 =93flow is=20 chaotic with eddies=94, as you=20 say.  I think the = addition of a=20 vane is worth a try.

 

Al

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