X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from ms-smtp-02.southeast.rr.com ([24.25.9.101] verified) by logan.com (CommuniGate Pro SMTP 5.1.10) with ESMTP id 2202266 for flyrotary@lancaironline.net; Fri, 27 Jul 2007 09:04:55 -0400 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.101; 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-02.southeast.rr.com (8.13.6/8.13.6) with SMTP id l6RD3p08014815 for ; Fri, 27 Jul 2007 09:03:52 -0400 (EDT) Message-ID: <002e01c7d04e$9a26f7a0$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Oil cooler inlet - what next? Date: Fri, 27 Jul 2007 09:04:00 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_002B_01C7D02D.12BF9680" 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_002B_01C7D02D.12BF9680 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Yep, I agree, Al, plenty of oil cooler core. Using a unit converter (rather than my own calculation) I show 160 HP = =3D 6791 BTU/Min also. That figure is the useful power produced. = However, we appear to differ in the area of waste heat we want to get = rid of through heat exchangers.=20 Here is my approach. A lb of gasoline has between 19000 -21000 BTU of = heat energy. At 160 HP the total heat energy produced by burning = gasoline (19,000 BTU/Lbm) is 28398 BTU/Min that is divided between = useful mechanical energy and waste heat of two forms - Exhaust and = coolant/oil. (Note: I use the lesser 19000 BTU because 100LL has less = energy content than a lower octane gasoline) That is the total heat energy of burning that amount of fuel/min. But, = naturally, we don't get all of that converted to power. Estimates and = figures vary, but here is what I used to conservatively proportion out = the heat energy (percentages may vary a few % depending on assumptions). 24% converted to useful mechanical energy 50 % out the tail pipe in from of exhaust heat 26% Waste heat that must be expelled by heat exchangers. So that gives me 0.26 *28398 (total) BTU/Min =3D 7383 BTU/Min of Waste = heat disposed of by radiators and oil cooler. Of that - for the rotary, = its traditional to assign 1/3 of that heat disposal to the oil cooler = =3D .1/3 * 7383 =3D 2461 BTU/min. So I was off about 200 BTU/min with my back of the envelope calculation = of 2668 for the oil heat. Also, if your assumptions about % conversion are different that would = make a difference but only of a few percent. We could have different = results because of differing on the % conversion of heat energy to which = category. But, regardless of the calculations, I fully agree, you have more than = sufficient oil cooler core. Ed ----- Original Message -----=20 From: Al Gietzen=20 To: Rotary motors in aircraft=20 Sent: Friday, July 27, 2007 1:48 AM Subject: [FlyRotary] Re: Oil cooler inlet - what next? I do have a question - what is the core volume of your oil cooler?=20 Cooler core is 375 cu. in.; should be quite capable of handling the = job. A stock 2 rotor color has a core area of approx 19x4/12x2 =3D 171 = Cubic Inches. That will cool a two rotor at 160 mph producing 160 Hp = with an oil BTU heat factor of 2688 BTU/min. A 3 rotor producing 1/3 = more HP (213 HP) rpm would create a corresponding rise in BTU/min to = 3360. Something wrong with the numbers there I think, ED; or else I'm way = off. My calcs say 160hp out results in about 1700 Btu/min to the oil. = 160 hp is 6770 Btu/min (2538 Btu/hr/hp), and is about 28% of the fuel = energy of 24,175 Btu/min. About 7% of the total goes to the oil. The = percentages will vary a bit depending on operating point - I picked = these off the graph at 6000 rpm. 215 hp out should result in somewhere = close to 2300 Btu/min to the oil. The reason I ask, is that if forcing more air through the core does = not decrease the oil temp - could there exist the possibility that the = core is saturated and can is simply not capable of transferring more oil = heat to the air? If that were the case, then, as you know, no amount = of additional air through the core would make any significant = difference. I don't know what you mean by "saturated". More air through removes = more heat - the limit is driving force (pressure) available to push it = through - until you get to the point that tube and fin surface temps are = the same as the average air temp. I think it's a long way from that. = But apparently it's close to the limit of the available pressure, and = the pressure drop of the core is a bit higher than expected. You'd think that if the scoop was even reasonable effective, it should = recover about 6" out of the 9.5" pressure available. I'm still = wondering if there could be enough air leakage around the cooler to lose = a significant amount of that pressure. =20 Kelly wrote: Uneducated guess but I will vote for a boundry layer problem......How = about extending the baffle below the bottom of the wing an inch or two and = retest.......It will be dirty and draggy but if that helps delta T it can be cleaned up with a proper = installation......IMHO That's an educated guess, or better yet, a fact - that it is A = problem. But I'm still thinking it is not THE problem since there is = 9.5" H2O dynamic pressure 12 - 5/8" off the surface, which is in the BL, = and should represent something like the average velocity into the 1 =BC" = opening of the scoop. Not sure what you mean extending the baffle below = the bottom of the wing an inch or two? Do you mean opening the scoop up = wider? I think making a larger radius on the entry edge of the scoop = could help, and would give an indication whether opening the scoop more = would help. Thanks, guys. Let's see; was it Steve who suggested the solution is moving to a = cooler climate?? Or I guess I could just fly between Nov. and MarchJ. Al --- Original Message -----=20 From: Al Gietzen=20 To: Rotary motors in aircraft=20 Sent: Thursday, July 26, 2007 7:06 PM Subject: [FlyRotary] Re: Oil cooler inlet - what next? Installed sheet metal 'baffle' to form new upper wall of the = diffuser as shown in the photo. The idea was to assist in maintaining = attached flow, and to block leakage through the gap at the top. The = baffle was done in 3 pieces in order to insert past the divider/supports = in the scoop; each piece is about 7 =BD" wide. There are gaps between = pieces of about 3/8 - =BD" inch. The inlet pressure probe was placed at = point "D". Test flight showed no noticeable difference in delta T on the oil. = The pressure measured at "D" was 3" H2O. Pressure behind the exit = fairing was again -3/4". The pressures at C and D (measured at = different times) are at about 6" from the inboard end of the 22" long = cooler. There may be some variation axially. Because of the gaps in = the baffle, and fitting around the end tanks, there is still some air = bypassing the cooler; but I don't know how significant. Given the 9+" = H2O dynamic pressure out in front of the scoop still indicates not good = pressure recovery. Nonetheless; it is certainly disappointing that there was no change = (within the accuracy of the temp measurements) in the effective cooling. = This suggests that the wall shape and the air leakage are not the = problem. Calculating back from the temp changes in oil and air suggest there = is only about 1000 cfm going through the cooler core. The extrapolation = of my measured data on air flow vs pressure drop across the core = suggests that at 3" H2O there should be about 2000 cfm through the core. = Because of the centrifugal blower I was using for flow tests I was only = able to get data up to about 0.6" H2O and 700 cfm. I fit the data to = Y=3Dax+bx2 using regression analysis, which gave a very good fit up to = that point; but extrapolating out to 3" may be stretching it. If I = assume the pressure drop goes as the cube of the flow velocity, the = extrapolation is considerable different - about 1330 cfm at 3" H2O. Al -----Original Message----- From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] = On Behalf Of Al Gietzen Sent: Friday, July 20, 2007 10:42 AM To: Rotary motors in aircraft Subject: [FlyRotary] Re: Oil cooler inlet Well; I may end up with VGs and change in upper duct wall shape. My = intention yesterday was to install VGs as a first step, test fly, = measure pressure and temps; then proceed with installing sheet metal = upper duct wall change. In deciding where to put the VGs, I looked at things with the gear = up (Photo 1). The gear door has a bump, and there is some gap around = the door. Don't know what all this does to BL. Ended up putting VG = toward the left side about 26" in font of scoop, and toward the right = side right on the gear door bump. I then spent a bunch of time trying to get the pressure measuring = tube situated. The only access is through the scoop opening, and I = can't get my hands in there; so it is very tough. Plus the tube going = in there, or along the surface in front will affect the flow behavior, = so what affect are we going to measure. Having multiple measurements = would be great; but very difficult to achieve. While doing that, I spent some time looking in there with a small = mirror. What I noted was that initial gaps above and below the cooler = (required to slide the unit in and out) had changed a bit. The cooler = is supported on pads of 'Cool-Mat' insulation. Those have compressed = just a little, so now there is very little gap at the bottom, and 1/8"+ = along the top. That is a fairly substantial leak, and the loss of = pressure at the top likely exacerbates the flow separation. I decided = it wasn't worth going to test the VGs as long as that leakage gap was = there. Taking the wing off (mostly getting it back on because of next to = impossible access to nuts), and removing the cooler looks a bit much = right now. I realized then; that by putting in a sheet metal 'false' = upper duct wall, I could extend it up into the gap at the top (photo 2), = thereby changing the shape, and (mostly) closing the gap at the top. The false wall has to be in three parts for the three openings, and = there will be gaps between because of the supporting dividers; but it = could make a substantial difference. I made the piece for the center, = and considered testing just that; but the upper gap concerns me enough = that I think I'll try to get all three fit in. Then go take a flight test. Unfortunately this combines three = changes, VGs, closing gap, and changing duct wall. I had hoped to test = these one at a time. If there is a substantial change; it will be easy = to remove the VGs to see what that effect was. Of course I'll let you know when I get some results. Oh, the price of innovationJ. Al -----Original Message----- From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] = On Behalf Of Thomas Jakits Sent: Thursday, July 19, 2007 10:31 AM To: Rotary motors in aircraft Subject: [FlyRotary] Re: Oil cooler inlet Okay, Monty thinks the emphasis is on the BL. I believe (don't know), the main-problem is the upper ductwall = shape. Even if you have perfect BL flow, the upper wall shape is still = not good and will stall the flow.=20 At the end of the game you want good flow at all speeds and be able = to close any ducts to limit excess cooling (when you hopefully get = there). Obviously BL will play a role in your installation as the intake is = rather narrow. However BL or not - BL does not mean there is no flow, just slower = and more turbulent, but still generally going towards the cooler. Aerodynamics in the duct should be much the same for laminar, = turbulent, any flow, as long as there is flow. When things stall is when flow pretty much ceases (in the stalled = area ....), no matter how well things where at the entrance. The stall in this case is rather "easy" to get, as the speed seems = rather low already. Still may be good enough if you can do away with the = stall. So I suggest to work on the duct wall first and optimize it. As suggested, with some kind of sheet, alu, fiberglass, etc. You can = curve it more and more until you peak. Maybe pinched ducts (copyright Ed!!) are not working here, but it = may as well - if they work a Ed's theory explains (energizes the = flow...) If this works, modify according to the best shape found. Then try to improve with VGs or sanding or turbolator tape. Then go for the exit - after all it is a differential pressure = game.... TJ =20 On 7/18/07, M Roberts wrote:=20 Al, I think you need to do something to energize the boundary layer. If = you can't divert it you need to put some energy into it. It is probably = getting slow and separating from the face of the duct. That is what your = data seems to indicate to me.=20 I like the shape that Thomas proposes better than what you have now, = however, I still think you will need some VG's in front of the inlet.=20 I know it may seem counter intuitive, but turbulence may actually = help in this case. You will not get very efficient internal diffusion, = but it will be a lot better than what you have now. I don't think that = putting a turning vane will help too much without doing something to = energize the boundary layer first. You'll just have a slow thick low = energy layer, and a high energy layer separated by a turning vane.=20 It is really easy to duct tape some aluminum VG's in front of the = inlet and see what it does.=20 You may need a combination of Thomas' contour, VG's and a turning = vane. Go with the easy fix and work your way up in complexity. Monty -------------------------------------------------------------------------= --- -- Homepage: http://www.flyrotary.com/ Archive and UnSub: = http://mail.lancaironline.net:81/lists/flyrotary/List.html ------=_NextPart_000_002B_01C7D02D.12BF9680 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Yep, I agree, Al, plenty of oil cooler = core.
 
Using a unit converter (rather than my own = calculation) I=20 show 160 HP =3D 6791 BTU/Min also.  That figure is the useful power = produced.  However, we appear to differ in the area of waste=20 heat  we want to get rid of through heat=20 exchangers. 
 
 
Here is my approach.  A lb of gasoline has = between=20 19000 -21000 BTU of heat energy.  At 160 HP the total heat energy = produced=20 by burning gasoline (19,000 BTU/Lbm)  is 28398 BTU/Min that is = divided=20 between useful mechanical energy and waste heat of two forms - Exhaust = and=20 coolant/oil. (Note: I use the lesser 19000 BTU because 100LL has less = energy=20 content than a lower octane gasoline)
 
That is the total heat energy of burning that = amount of=20 fuel/min.  But, naturally, we don't get all of that converted to=20 power.  Estimates and figures vary, but here is what I used to=20 conservatively proportion out the heat energy (percentages may vary a = few %=20 depending on assumptions).
 
24% converted to useful mechanical = energy
50 % out the tail pipe = in from of=20 exhaust heat
26% Waste heat that must be expelled by heat=20 exchangers.
 
So that gives me 0.26 *28398 (total) BTU/Min =3D = 7383=20 BTU/Min of Waste heat disposed of by radiators and oil cooler.  Of = that=20 - for the rotary, its traditional to assign 1/3 of that heat = disposal to=20 the oil cooler =3D .1/3 * 7383 =3D 2461 BTU/min.
 
So I was off about 200 BTU/min with my back of = the=20 envelope calculation of 2668 for the oil heat.
Also, if your assumptions about % conversion are = different=20 that would make a difference but only of a few percent. We could = have=20 different results because of differing on the % conversion of heat = energy to=20 which category.
 
But, regardless of the calculations, I fully = agree, you=20 have more than sufficient oil cooler core.
 
 
Ed
 
----- Original Message -----
From:=20 Al = Gietzen=20
Sent: Friday, July 27, 2007 = 1:48 AM
Subject: [FlyRotary] Re: Oil = cooler inlet=20 - what next?

 

 

I do have a question - = what is the=20 core volume of your oil cooler? 

 

Cooler = core is 375=20 cu. in.; should be quite capable of handling the = job.

 

 A stock 2 rotor = color has a=20 core area of approx 19x4/12x2 =3D 171 Cubic Inches.  That will = cool a two=20 rotor at 160 mph producing 160 Hp with an oil BTU heat factor of 2688=20 BTU/min.  A 3 rotor producing 1/3 more HP (213 HP) rpm would = create=20 a corresponding rise in BTU/min to 3360.

 

Something = wrong=20 with the numbers there I think, ED; or else I=92m way off.  My = calcs say=20 160hp out results in about 1700 Btu/min to the oil.  160 hp is = 6770=20 Btu/min (2538 Btu/hr/hp), and is about 28% of the fuel energy of = 24,175=20 Btu/min.  About 7% of the total goes to the oil.  The = percentages=20 will vary a bit depending on operating point =96 I picked these off = the graph at=20 6000 rpm.  215 hp out should result in somewhere close to 2300 = Btu/min to=20 the oil.

 

The reason I ask, is = that if=20 forcing more air through the core does not decrease the oil temp - = could=20 there  exist the possibility that the core is saturated and=20 can is simply not capable of transferring more oil heat to the = air? =20 If that were the case, then, as you know,  no amount of = additional air=20 through the core would make any significant = difference.

 

I don=92t = know what=20 you mean by =93saturated=94.  More air through removes more heat = =96 the limit=20 is driving force (pressure) available to push it through =96 until you = get to=20 the point that tube and fin surface temps are the same as the = average=20 air temp.  I think it=92s a long way from that.  But = apparently it=92s=20 close to the limit of the available pressure, and the pressure drop of = the=20 core is a bit higher than expected.

 

You=92d = think that if=20 the scoop was even reasonable effective, it should recover about 6=94 = out of the=20 9.5=94 pressure available.  I=92m still wondering if there could = be enough=20 air leakage around the cooler to lose a significant amount of that = pressure.=20  

 

Kelly=20 wrote:

Uneducated guess but I will vote for a = boundry layer=20 problem......How about extending

the baffle below the bottom of the wing an = inch or two=20 and retest.......It will be dirty and

draggy but if that helps delta T = it can be=20 cleaned up with a proper installation......IMHO

 

That=92s = an educated=20 guess, or better yet, a fact =96 that it is A problem.  But I=92m = still=20 thinking it is not THE problem since there is 9.5=94 H2O dynamic = pressure 12 =96=20 5/8=94 off the surface, which is in the BL, and should represent = something like=20 the average velocity into the 1 =BC=94 opening of the scoop.  Not = sure what=20 you mean extending the baffle below the bottom of the = wing an=20 inch or two?  Do you = mean opening=20 the scoop up wider?  I think making a larger radius on the entry = edge of=20 the scoop could help, and would give an indication whether opening the = scoop=20 more would help.

 

Thanks,=20 guys.

 

Let=92s = see; was it=20 Steve who suggested the solution is moving to a cooler climate??  = Or I=20 guess I could just fly between Nov. and MarchJ.

 

Al

 

 

 

 

 

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

From: Al = Gietzen=20

To: Rotary motors in = aircraft=20

Sent:=20 Thursday, July 26, 2007 7:06 PM

Subject:=20 [FlyRotary] Re: Oil cooler inlet - what = next?

 

 Installed sheet=20 metal =91baffle=92 to form new upper wall of the diffuser as shown = in the=20 photo.  The idea was to assist in maintaining attached flow, = and to=20 block leakage through the gap at the top.  The baffle was done = in 3=20 pieces in order to insert past the divider/supports in the scoop; = each piece=20 is about 7 =BD=94 wide.  There are gaps between pieces of about = 3/8 =96 =BD=94=20 inch.  The inlet pressure probe was placed at point=20 =93D=94.

 

Test=20 flight showed no noticeable difference in delta T on the oil.  = The=20 pressure measured at =93D=94 was 3=94 H2O.  Pressure behind the = exit fairing=20 was again -3/4=94.  The pressures at C and D (measured at = different=20 times) are at about 6=94 from the inboard end of the 22=94 long = cooler. =20 There may be some variation axially.  Because of the gaps in = the=20 baffle, and fitting around the end tanks, there is still some air = bypassing=20 the cooler; but I don=92t know how significant.  Given the = 9+=94 H2O=20 dynamic pressure out in front of the scoop still indicates not good = pressure=20 recovery.

 

Nonetheless; it=20 is certainly disappointing that there was no change (within the = accuracy of=20 the temp measurements) in the effective cooling.  This suggests = that=20 the wall shape and the air leakage are not the = problem.

 

Calculating back=20 from the temp changes in oil and air suggest there is only about = 1000 cfm=20 going through the cooler core. The extrapolation of my measured data = on air=20 flow vs pressure drop across the core suggests that at 3=94 H2O = there should=20 be about 2000 cfm through the core.  Because of the centrifugal = blower=20 I was using for flow tests I was only able to get data up to about = 0.6=94 H2O=20 and 700 cfm.  I fit the data to Y=3Dax+bx2 using = regression=20 analysis, which gave a very good fit up to that point; but = extrapolating out=20 to 3=94 may be stretching it.  If I assume the pressure drop = goes as the=20 cube of the flow velocity, the extrapolation is considerable = different =96=20 about 1330 cfm at 3=94 H2O.

 

Al

 

 

-----Original=20 Message-----
From: Rotary=20 motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Al = Gietzen
Sent: Friday, July 20, 2007 = 10:42=20 AM
To: Rotary = motors in=20 aircraft
Subject: = [FlyRotary] Re: Oil cooler inlet

 

Well; I = may end=20 up with VGs and change in upper duct wall shape. My intention = yesterday was=20 to install VGs as a first step, test fly, measure pressure and = temps; then=20 proceed with installing sheet metal upper duct wall=20 change.

 

In=20 deciding where to put the VGs, I looked at things with the gear up = (Photo=20 1).  The gear door has a bump, and there is some gap around the = door.  Don=92t know what all this does to BL.  Ended up = putting VG=20 toward the left side about 26=94 in font of scoop, and toward the = right side=20 right on the gear door bump.

 

I=20 then spent a bunch of time trying to get the pressure measuring tube = situated.  The only access is through the scoop opening, and I = can=92t=20 get my hands in there; so it is very tough.  Plus the tube = going in=20 there, or along the surface in front will affect the flow behavior, = so what=20 affect are we going to measure.  Having multiple measurements = would be=20 great; but very difficult to achieve.

 

While = doing that,=20 I spent some time looking in there with a small mirror.  What I = noted=20 was that initial gaps above and below the cooler (required to slide = the unit=20 in and out) had changed a bit.  The cooler is supported on pads = of=20 =91Cool-Mat=92 insulation.  Those have compressed just a = little, so now=20 there is very little gap at the bottom, and 1/8=94+ along the = top.  That=20 is a fairly substantial leak, and the loss of pressure at the top = likely=20 exacerbates the flow separation.  I decided it wasn=92t worth = going to=20 test the VGs as long as that leakage gap was = there.

 

Taking = the wing=20 off (mostly getting it back on because of next to impossible access = to=20 nuts), and removing the cooler looks a bit much right now.  I = realized=20 then; that by putting in a sheet metal =91false=92 upper duct wall, = I could=20 extend it up into the gap at the top (photo 2), thereby changing the = shape,=20 and (mostly) closing the gap at the top.

 

The=20 false wall has to be in three parts for the three openings, and = there will=20 be gaps between because of the supporting dividers; but it could = make a=20 substantial difference.  I made the piece for the center, and=20 considered testing just that; but the upper gap concerns me enough = that I=20 think I=92ll try to get all three fit in.

 

Then=20 go take a flight test. Unfortunately this combines three changes, = VGs,=20 closing gap, and changing duct wall.  I had hoped to test these = one at=20 a time.  If there is a substantial change; it will be easy to = remove=20 the VGs to see what that effect was.

 

Of=20 course I=92ll let you know when I get some = results.

 

Oh,=20 the price of innovationJ.

 

Al

 

 

-----Original=20 Message-----
From: Rotary=20 motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Thomas = Jakits
Sent: Thursday, July 19, 2007 = 10:31=20 AM
To: Rotary = motors in=20 aircraft
Subject: = [FlyRotary] Re: Oil cooler inlet

 

Okay,

 

Monty thinks the emphasis = is on the=20 BL.

I believe = (don't know),=20 the main-problem is the upper ductwall shape. Even if you have = perfect BL=20 flow, the upper wall shape is still not good and will stall the = flow.=20

At the end of the game you = want good=20 flow at all speeds and be able to close any ducts to limit excess = cooling=20 (when you hopefully get there).

Obviously BL will play a = role in your=20 installation as the intake is rather narrow.

However BL or not - BL does = not mean=20 there is no flow, just slower and more turbulent, but still = generally going=20 towards the cooler.

Aerodynamics in the duct = should be much=20 the same for laminar, turbulent, any flow, as long as there is=20 flow.

When things stall is when = flow pretty=20 much ceases (in the stalled area ....), no matter how well things = where at=20 the entrance.

The stall in this case is = rather "easy"=20 to get, as the speed seems rather low already. Still may be good = enough if=20 you can do away with the stall.

So I suggest to work on the = duct wall=20 first and optimize it.

 

As suggested, with some = kind of sheet,=20 alu, fiberglass, etc. You can curve it more and more until you=20 peak.

Maybe pinched ducts = (copyright Ed!!)=20 are not working here, but it may as well - if they work a Ed's = theory=20 explains (energizes the flow...)

 

If this works, modify = according to the=20 best shape found.

Then try to improve with = VGs or sanding=20 or turbolator tape.

Then go for the exit - = after all it is=20 a differential pressure game....

 

TJ

 

On = 7/18/07,=20 M Roberts <montyr2157@alltel.net>=20 wrote:

Al,

 

I think you need to do = something=20 to energize the boundary layer. If you can't divert it you need to = put some=20 energy into it. It is probably getting slow and separating from the = face of=20 the duct. That is what your data seems to indicate to me.=20

 

I like the shape that = Thomas=20 proposes better than what you have now, however, I still think you = will need=20 some VG's in front of the inlet.

 

I know it may seem = counter=20 intuitive, but turbulence may actually help in this case. You will = not get=20 very efficient internal diffusion, but it will be a lot better than = what you=20 have now. I don't think that putting a turning vane will help too = much=20 without doing something to energize the boundary layer first. You'll = just=20 have a slow thick low energy layer, and a high energy layer = separated by a=20 turning vane.

 

It is really easy to = duct tape=20 some aluminum VG's in front of the inlet and see what it does.=20

 

You may need a = combination of=20 Thomas' contour, VG's and a turning vane. Go with the easy fix and = work your=20 way up in complexity.

 

Monty

 

--
Homepage: =20 http://www.flyrotary.com/
Archive and UnSub:  =20 = http://mail.lancaironline.net:81/lists/flyrotary/List.html<= /P>

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