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 2181808 for flyrotary@lancaironline.net; Tue, 17 Jul 2007 00:01:06 -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 l6H1SnF7002286 for ; Mon, 16 Jul 2007 21:28:50 -0400 (EDT) Message-ID: <000c01c7c811$e5401590$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Misconception? [FlyRotary] Re: FW: Oil cooler air flow Date: Mon, 16 Jul 2007 21:29:17 -0400 MIME-Version: 1.0 Content-Type: multipart/mixed; boundary="----=_NextPart_000_0008_01C7C7F0.5D792F50" 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_0008_01C7C7F0.5D792F50 Content-Type: multipart/alternative; boundary="----=_NextPart_001_0009_01C7C7F0.5D792F50" ------=_NextPart_001_0009_01C7C7F0.5D792F50 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Well, I wouldn't say a misconception, Al, converting kinetic energy to = static pressure is exactly what happens. Besides, if I am screwed up in = my logic then by presenting it fully, someone can catch and correct it = and that will enhance my understanding Part of what causes some confusion (I believe) is whether you are using = an absolute or differential/gauge pressure instrument to measure = pressure. IF using an absolute pressure gage then the static pressure = measured in the tube (or duct) would be indeed be ambient + dynamic. = But, using a differential pressure gauge( a manometer for example ) only = measures the localized static pressure inside your "pitot" measuring = tube which is the dynamic pressure component above ambient. In other words, air velocity is only potentially static pressure. It = does not increase the localized ambient pressure until the kinetic = energy is converted to pressure. This can be done by the air being = slowed or impacting a non-moving object. Now the 84% is straight out of K&W for the streamline duct. Attached = is the graph showing (upper left hand corner of lower graph) an equation = (PB1 - Pi)/(1/2pVi^2) =3D 0.84. The denominator of the equation is = our old friend dynamic pressure 1/2pVi^2. So re-arranging the equation = slightly we have PB1-Pi =3D 0.84 * 1/2pVi^2. Vi is the velocity of the = air entering the inlet of the duct. p is the air density. PB1 is the duct localized static pressure (above ambient) right before = the core and Pi is the ambient pressure at the entrance of the duct. So = the difference shown is the increase in static pressure (NOT ambient + = Dynamic, but Dynamic increase over ambient, with ambient being the = reference or zero point) from the duct entrance to the core - or your = pressure increase due to converting the moving air's kinetic energy to a = local static pressure increase. IF PB1 =3D Pi then that says there is = either no air flow OR there is no conversion of kinetic energy to static = pressure increase. If PB1>Pi then some dynamic potential is being = converted to static, so that leads back to my taking of your 9.5" H20 at the entrance and since that is the amount of dynamic = pressure available at the entrance Pi and you measure 3.25" at the core = (PB1). I would have expected with the perfect streamline duct that you = would have measure PB1 =3D 0.84 * 9.5 =3D 7.98 " H20 vice 3.25". Now, = the tube was not pointed directly into the flow from the best I can = tell, but that would imply that the 9.5" H20 was less than that = available in the free flow (12.0") which could mean you actually have = more dynamic potential at the entrance than the 9.5" reflects. But, if = true then that only reinforces my speculation that something evil is = going on in your duct. You are recovering 9.5" localized static = pressure in your measuring tube at the entrance but by the time you = measure it next to the core if has decreased considerably. But, I'll stop here, before I confuse myself. K&W makes good go-to-bed = reading {:>) Good luck on your modification. I will be very interested in seeing = what a van does for you. Best Regards Ed =20 ----- Original Message -----=20 From: Al Gietzen=20 To: Rotary motors in aircraft=20 Sent: Monday, July 16, 2007 9:16 PM Subject: [FlyRotary] Re: FW: Oil cooler air flow 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.=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_001_0009_01C7C7F0.5D792F50 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Well, I wouldn't say a misconception, Al, = converting=20 kinetic energy to static pressure is exactly what happens.  = Besides, if I=20 am screwed up in my logic then by presenting it fully, someone can catch = and=20 correct it and that will enhance my understanding
 
Part of what causes some confusion (I = believe) is=20  whether you are using an absolute or differential/gauge pressure=20 instrument to measure pressure.  IF using an absolute pressure =  gage=20 then the static pressure measured in the tube (or duct)  would be = indeed be=20  ambient + dynamic.  But, using a differential pressure gauge( = a=20 manometer for example ) only measures the localized static pressure = inside your=20 "pitot" measuring tube which is the dynamic  pressure component = above=20 ambient.
 
In other words, air velocity is only potentially = static=20 pressure.  It does not increase the localized ambient pressure = until the=20 kinetic energy is converted to pressure.  This can be done by the = air being=20 slowed or impacting a non-moving object.
 
  Now the 84% is straight out of K&W = for the=20 streamline duct.   Attached is the graph showing (upper left = hand=20 corner of lower graph) an equation  (PB1 - Pi)/(1/2pVi^2) = =3D=20 0.84.   The denominator of the equation is our old = friend=20 dynamic pressure 1/2pVi^2.  So re-arranging the = equation=20 slightly we have PB1-Pi =3D 0.84 * 1/2pVi^2. =20 Vi is the velocity of the air entering the inlet of the = duct.=20 p is the air density.
 
PB1 is the duct localized = static pressure=20 (above ambient)  right before the core and Pi is = the=20 ambient pressure at the entrance of the duct.  So the = difference shown=20 is  the increase in static pressure (NOT ambient + Dynamic, but = Dynamic=20 increase over ambient, with ambient being the reference or zero point) = from the=20 duct entrance to the core - or your pressure increase due to converting = the=20 moving air's kinetic energy to a local static pressure increase. =20 IF PB1 =3D Pi then that says there is either no air = flow OR there=20 is no conversion of kinetic energy to static pressure increase.  If = PB1>Pi then some dynamic potential is being = converted to=20 static, so that leads back to my  taking of your
 
9.5" H20 at the entrance and since that is the = amount of=20 dynamic pressure available at the entrance Pi and you = measure=20 3.25" at the core (PB1).  I would have expected = with the=20 perfect streamline duct that you would have measure PB1 =3D 0.84 = * 9.5 =3D=20 7.98 " H20 vice 3.25".  Now, the tube was not pointed = directly=20 into the flow from the best I can tell, but that would imply that the = 9.5" H20=20 was less than that available in the free flow (12.0") which could mean = you=20 actually have more dynamic potential at the entrance than the 9.5"=20 reflects.  But, if true then that only reinforces my speculation = that=20 something evil is going on in your duct.  You are recovering 9.5" = localized=20 static pressure in your measuring tube at the entrance  but by the = time you=20 measure it next to the core if has decreased considerably.
 
But, I'll stop here, before I confuse = myself. =20 K&W makes good go-to-bed reading {:>)
 
Good luck on your modification.  I will be = very=20 interested in seeing what a van does for you.
 
Best Regards
 
Ed
 
 
 
 
----- Original Message -----
From:=20 Al = Gietzen=20
Sent: Monday, July 16, 2007 = 9:16 PM
Subject: [FlyRotary] Re: FW: = Oil cooler=20 air flow

Ed;

 

I = appreciate your=20 thorough presentation.  I guess you could have been brief; and = said =93Yep;=20 you have a misconception=94 J In any = case I was=20 not aware of the 0.84 maximum.

 

Even = though I guess=20 I knew at some level it wasn=92t correct; somewhere along the way I = had gotten=20 it into my head that in converting the =91dynamic=92 to =91static=92 = the static=20 pressure could be greater =96 something about conservation of energy; = or who=20 knows what; but clearly that was a =91misconception=92 (having one of = those is=20 much better than being completely screwed up)J.

 

Al

 

-----Original=20 Message-----
From: = Rotary=20 motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Ed = Anderson
Sent: Monday, July 16, 2007 = 3:38=20 PM
To: Rotary = motors in=20 aircraft
Subject: = [FlyRotary]=20 Re: FW: Oil cooler air flow

 

OK, Al, let me restate = in a more=20 comprehensive manner and see if that helps.

 

We know that "dynamic = pressure" is=20 actually measured by the increase it causes in = localized static=20 pressure. So the term  "dynamic pressure" is actually = just=20 referring to the energy potential (Kinetic) of the moving air to cause = a=20 localized increase in static pressure - if  that air movement = were=20 brought to a stop.

 

 In other words, if = we had a=20 flow of air with a specific velocity and specific density, that air = would have=20 a ambient static pressure (say at sea level of 29.92" HG).  The = moving=20 air would also have a static pressure potential (Dynamic = pressure) based=20 on its velocity and density.  So that if a tube were used to = measure this=20 "Dynamic Pressure" it must first bring that part being measured to a=20 stop the action of which converts the dynamic pressure = potential of=20 the moving air to a localized static pressure increase in the=20 tube. 

 

 So the total = static pressure=20 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=20 converting its dynamic potential to static pressure.  So Pt =3D = Pa  +=20 Pd with Dynamic Pressure component, Pd =3D = p*1/2V^2.

 

So in case of a duct = there is, of=20 course, only ambient static pressure in the duct if there is no air = flow=20 through the duct.  Once there is airflow then you also have = potential=20 pressure in the form of the kinetic energy of the moving air.  So = that Pt=20 =3D Pa + p1/2V^2.  p being air density, V being the=20 velocity.

 

 The streamline = duct=20 (theoretically) can convert 84% of the moving air potential dynamic = pressure=20 to static pressure increase.  So that at the widest part of the = duct just=20 before the core you would have a total static pressure  Pt =3D Pa = +=20 0.84*p1/2V^2. 

 

But, using differential = pressure=20 gauges with tubes pointed into the moving air, we are not measuring = total=20 pressure, but the pressure increase due solely to the moving = air.  In=20 other words, if you were measuring 5" H20 and then the air stopped = moving ,=20 the gauge would read zero.

 

 So with the = manometer you=20 are measuring the pressure above ambient pressure or that resulting = solely=20 from the dynamic pressure potential of the moving air being converted = from=20 kinetic energy to static pressure.  Yes, the ambient pressure is = present=20 but you are not measuring it.  With no moving air the water = levels in you=20 manometer would all be exactly at the same = level..

 

The fact is that you are = measuring=20 static pressure at both locations - the 9.5" before the duct was a = static=20 pressure increase in your measuring tube - cause by stopping the = moving=20 air so its refer to as dynamic pressure.  The fact is that you = were also=20 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=20 static pressure increase.  Therefore, the fact that you were = measuring=20 considerably more pressure before the duct than inside it indicates = that the=20 air stream's velocity is not being efficiently transformed into = static=20 pressure in the duct.

 

This implies that = perhaps there is=20 less air velocity entering the duct than your measurement a couple = inches in=20 front suggests OR there is sufficient eddies and adverse flow=20 situation inside the duct that precludes the efficient = transformation=20 into a static pressure increase. 

 

I do not have an = aerodynamic or=20 gas dynamics background, so I could certainly be wrong.  But, = that is my=20 understanding based on the somewhat extensive reading I have=20 done.

 

Ed

 

 

----- Original Message = -----=20

From: Al = Gietzen=20

To: Rotary motors in = aircraft=20

Sent: Monday,=20 July 16, 2007 5:48 PM

Subject:=20 [FlyRotary] Re: FW: Oil cooler air flow

 

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

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

 

1.  The air flow = and=20 velocity is considerably reduced from what you are expecting (too = small=20 opening/exit - 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=20 consequence 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 = 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=20 on 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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