X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from fed1rmmtao101.cox.net ([68.230.241.45] verified) by logan.com (CommuniGate Pro SMTP 5.1.10) with ESMTP id 2181560 for flyrotary@lancaironline.net; Mon, 16 Jul 2007 20:15:30 -0400 Received-SPF: none receiver=logan.com; client-ip=68.230.241.45; envelope-from=alventures@cox.net Received: from fed1rmimpo02.cox.net ([70.169.32.72]) by fed1rmmtao101.cox.net (InterMail vM.7.08.02.01 201-2186-121-102-20070209) with ESMTP id <20070717001452.NZPZ1349.fed1rmmtao101.cox.net@fed1rmimpo02.cox.net> for ; Mon, 16 Jul 2007 20:14:52 -0400 Received: from BigAl ([72.192.132.90]) by fed1rmimpo02.cox.net with bizsmtp id QQEs1X0011xAn3c0000000; Mon, 16 Jul 2007 20:14:52 -0400 From: "Al Gietzen" To: "'Rotary motors in aircraft'" Subject: RE: [FlyRotary] Re: FW: Oil cooler air flow Date: Mon, 16 Jul 2007 17:16:22 -0800 Message-ID: <000001c7c810$16ae44a0$6400a8c0@BigAl> MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0001_01C7C7CD.088B04A0" X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook, Build 10.0.6626 Importance: Normal In-Reply-To: X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3138 This is a multi-part message in MIME format. ------=_NextPart_000_0001_01C7C7CD.088B04A0 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: quoted-printable Ed; =20 I appreciate your thorough presentation. I guess you could have been = brief; and said "Yep; you have a misconception" :-) In any case I was not aware = of the 0.84 maximum. =20 Even though I guess I knew at some level it wasn't correct; somewhere = along the way I had gotten it into my head that in converting the 'dynamic' to 'static' the static pressure could be greater - something about = conservation of energy; or who knows what; but clearly that was a 'misconception' = (having one of those is much better than being completely screwed up):-). =20 Al =20 -----Original Message----- From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Ed Anderson Sent: Monday, July 16, 2007 3:38 PM To: Rotary motors in aircraft Subject: [FlyRotary] Re: FW: Oil cooler air flow =20 OK, Al, let me restate in a more comprehensive manner and see if that = helps. =20 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. =20 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 =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. =20 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. =20 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 =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. =20 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.. =20 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. =20 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. =20 =20 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. =20 Ed =20 =20 ----- 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 =20 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. =20 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) =20 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. =20 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) =20 4. Some combination of the above. =20 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 =20 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. =20 Al ------=_NextPart_000_0001_01C7C7CD.088B04A0 Content-Type: text/html; charset="us-ascii" Content-Transfer-Encoding: quoted-printable

Ed;

 

I appreciate your thorough presentation.  I guess you could have been brief; and said = “Yep; you have a misconception” J In any case I was not aware of the 0.84 = maximum.

 

Even though I guess I knew at = some level it wasn’t correct; somewhere along the way I had gotten it into my = head that in converting the ‘dynamic’ to ‘static’ the = static pressure could be greater – something about conservation of = energy; or who knows what; but clearly that was a ‘misconception’ = (having one of those is much better than being completely screwed = up)J.=

 

Al

 

-----Original = Message-----
From: Rotary motors in = aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Ed Anderson
Sent: Monday, July 16, = 2007 3:38 PM
To: Rotary motors in = aircraft
Subject: [FlyRotary] Re: = FW: Oil cooler air flow

 

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. 

 

 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. 

 

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 = -----

From: Al = Gietzen

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

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