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X-Color: red Return-Path: Sender: To: lml@lancaironline.net Date: Sun, 29 Apr 2007 12:19:55 -0400 Message-ID: X-Original-Return-Path: Received: from elasmtp-mealy.atl.sa.earthlink.net ([209.86.89.69] verified) by logan.com (CommuniGate Pro SMTP 5.1.8) with ESMTP id 2016062 for lml@lancaironline.net; Sun, 29 Apr 2007 09:11:04 -0400 Received-SPF: none receiver=logan.com; client-ip=209.86.89.69; envelope-from=artbertolina@earthlink.net DomainKey-Signature: a=rsa-sha1; q=dns; c=nofws; s=dk20050327; d=earthlink.net; b=bMjTJftBDI/aJhLQe+3aIjOlgwCsf7lBcPmHa7IvFfGSIIQtRioSKAqkucKhymbl; h=Received:Message-ID:From:To:Subject:Date:MIME-Version:Content-Type:X-Priority:X-MSMail-Priority:X-Mailer:X-MimeOLE:X-ELNK-Trace:X-Originating-IP; Received: from [24.121.161.23] (helo=LAPTOP2) by elasmtp-mealy.atl.sa.earthlink.net with asmtp (Exim 4.34) id 1Hi9AI-0002RJ-OI for lml@lancaironline.net; Sun, 29 Apr 2007 09:10:12 -0400 X-Original-Message-ID: <000901c78a5f$b6e6be50$4e00000a@LAPTOP2> From: "Art Bertolina" X-Original-To: "Lancair list Lancair list" Subject: Fw: Cooling Drag X-Original-Date: Sun, 29 Apr 2007 06:10:08 -0700 MIME-Version: 1.0 Content-Type: multipart/mixed; boundary="----=_NextPart_000_0005_01C78A25.09F8ECC0" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.3028 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3028 X-ELNK-Trace: 1c57f4aab304e4a6fc8cc707cfd6285a4d2b10475b571120c21e489b13de2806a6b28ae89ec156b09072ae4777b98cc8350badd9bab72f9c350badd9bab72f9c X-Originating-IP: 24.121.161.23 This is a multi-part message in MIME format. ------=_NextPart_000_0005_01C78A25.09F8ECC0 Content-Type: multipart/alternative; boundary="----=_NextPart_001_0006_01C78A25.09F8ECC0" ------=_NextPart_001_0006_01C78A25.09F8ECC0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Hi Marv hope this works ----- Original Message -----=20 From: Fred Moreno=20 To: 'Art Bertolina'=20 Sent: Saturday, April 28, 2007 7:10 PM Subject: RE: Cooling Drag Hi Art:=20 I composed the letter below and the attachment to respond to you = questions and those from others. My emails to LML keep bouncing. Marv = and I are grappling with the problem, but to help with the solution, = please forward to LML from your email, and let's see if that works. = Let me know if I answered your questions.=20 Cheers,=20 Fred ____________________________________________________ I have received questions both on the forum and off concerning my = earlier comments about cooling drag. They sent me scurrying back to my = books and calculator. =20 I spent some time generating a self-consistent set of numbers for = cooling drag, pressures, temperatures, etc. with a uniform set of = assumptions. Some guesstimates are required, particularly concerning = losses from friction in various locations, so I have listed everything = assumed and calculated, and invite investigation and questions. The = results are summarized in the ugly sketch attached. It shows the flow = conditions at various points along the cooling air pathway assuming you = have the ultimate set up, no leaks, cowl flaps perfectly adjusted to get = the right exit area, etc. =20 Air enters from the left, is partially compressed in front of the inlet = by the slowing of the air flow, goes through the inlet, then through a = short diffuser where it is slowed more, and then dumps into a large = plenum over the engine where the flow is a fairly slow moving mess. = Then the air flows through cylinders and oil cooler (which can take a = lot of flow) into the space below the engine (where the flow is again a = mess, but a slow moving mess), and then accelerates through the outlets, = the area of which is controlled by cowl flaps which are optimally = adjusted. I have assumed 9000 feet, 200 knots IAS, 230 knots TAS as an = interesting speed for us aspirated engine guys to investigate. I have = used pressures in inches of water since the engine pressure drop charts = use these units of measure. The inlet velocity ratio is assumed to be = 0.4 (velocity through cowl inlets divided by TAS) which seems to be = roughly what the latest and fastest designs are using in the fastest = production aircraft and speed mods offered by LoPresti. Comments:=20 1) Flow is controlled at the outlet with cowl flaps, so we can get a = lot of compression in front of the inlets where the compression process = is frictionless. With an inlet velocity of 0.4, we get about 86% of the = total dynamic pressure in front of the inlet. =20 2) We are forced to accept a short diffuser behind the inlets which = further slows the flow, but friction losses eat up an assumed 2 inches = of water pressure.=20 3) The velocity in the plenum is anybody's guess as the flow is a = turbulent mess, but the velocity is low, so losses are low. But there = are losses in the plenum, particularly with the 550 Continental engines = because of the intake manifold and other clap trap above the cylinders.=20 4) Note that the compression raises the temperature about 12F so the = assumed temperature above the fins is about 40F. We can use this when = we go to the engine cooling charts to make an estimate of the air flow = required.=20 5) Engine cooling charts are published assuming full power and 475F = CHT (red line) and cruise with 435F CHT. The charts suggest that the = engine pressure drop will be about 5 inches of water at 435f CHT. But = we like 350F or thereabouts for our CHT and that requires more airflow = for more cooling. A real rough guess is about 20-25% more air flow, and = this requires about 8 inches of pressure drop across the engine to force = the increased air flow through the fins. =20 6) The flow below the engine is chaotic, but slow so losses are low. = It is also hot. With 350F CHT, it is probably 100-120F higher than = ambient, or perhaps 150F. (There is a greater temperature rise across = cylinder heads, and much less across the oil cooler, and it all mixes = together.) Because there is a lot of clap trap below the engine, there = are more losses which I guesstimate at around 1.0 inch of water.=20 7) As you approach the exit ("discharge nozzle") the flow is = accelerated, hopefully smoothly, efficiently, and directly aft to regain = as much momentum as possible. However, our exit nozzle is filled with = exhaust pipes, pipe brackets, and other clap trap in the higher velocity = area as well as friction losses along the walls. My guess is we lose = another 2 inches of water pressure. This leaves about 12 inches of = static pressure converted into dynamic pressure in the exiting flow. = With the warm, lower density air I get about 170 knots discharge = velocity directly aft for our perfect cooling system. The loss in = velocity for the cooling flow is then 230 KTAS minus 170 KTAS outlet, or = 60 knots. =20 Cooling drag is equal to the momentum loss of the cooling air. It is = cooling air mass flow (pounds per second) times the velocity lost during = passage through the cowl and engine. If we had lower pressure losses, = we would have more pressure at the exit to accelerate the flow aft = faster. =20 If we could get the exit velocity =3D free stream velocity (the true air = speed), cooling drag would be zero. =20 If exit flow was faster than the free stream velocity we would get = thrust. =20 If we are moving really fast with lots of ram pressure and low losses = and we raised the temperature of the air stream up very high to lower = the air density a lot and allow it to be accelerated aft really fast, we = would have a ram jet.=20 But for our air cooled engine with lots of friction inside the cowl and = engine and only modest temperature rises, we can not get thrust, even = for our prefect case above.=20 Now for a cross check. Let's assume our IO-550 is pumping out about 225 = HP (about 73%). If you go to a Lycoming power chart and assume it is = the same for the Continental (pretty good assumption), for 40F above the = engine and 435F CHT the chart shows you need 3 lbs/sec. of cooling air = and 5 inches of water pressure drop across the engine to get this flow. = But we want 350F CHT to let our jugs live longer, so the required air = flow is (rough guesstimate) 3.7 lbs/sec. and then the required pressure = drop to get this flow is about 8 inches of water, the figure I used = above. And if you calculate the cowl inlet area required to admit this = flow rate at 40% of the free stream velocity (TAS), you get two 6 inch = diameter inlets.=20 Moving right along, if we assume that the prop efficiency at 230 KTAS = and 225 HP is 85%, then the thrust =3D drag =3D 272 pounds. Now we = compute the cooling drag which is the momentum loss which is the cooling = air flow rate times the velocity loss of 60 knots times some constants = to make the units work and we get 12 pounds for the cooling drag or = about 4% of total drag for our perfect zero leakage cooling system. =20 In my earlier message I noted that your run-of-the-mill spam can had = cooling drag of about 6-7%. Our hot rods have about half the drag as a = typical Bonanza or C210, so cooling drag becomes maybe 14%. If we work = really hard maybe we can get it down to 4%. You can now see what it = takes.=20 Your turbo guys at 25,000 feet have a much more serious problem because = there is little to work with. Assuming 280 knots TAS, the IAS =3D 190 = knots and you thus have about 10% less ram pressure available to start = with, maybe 23 inches of water. The air is colder, but it is also a lot = less dense so if I really extrapolate off my power chart, and assume 262 = HP (75% for a TSIO 550) and 350 CHT, the pressure drop required across = the engine is 12-14 inches of water, call it 5 inches more than at low = altitude. So the high altitude means that you lose 3+5=3D 8 inches of = water pressure compared to the low altitude case above as the flow = approaches the exit. Instead of 12 inches of water to accelerate the = flow aft, you have about 4 inches, so the exit velocity is much lower = requiring a larger exit area, and the momentum loss much higher which = means cooling drag is much higher. And that is the ideal case. Most = LIV cooling systems are far from ideal, so my guess is that for the = stockers (no mods) cooling drag is somewhere around 20-25% of total = drag.=20 You can see for the high altitude case cooling drag will always be a = bigger piece of the drag pie even if we assume a perfect, leak free = cooling system. However, if you run this configuration (big exit area) = at lower altitude, you will have a lot of excess cooling drag you could = shed - if you could close cowl flaps, shrink the exit area and thereby = accelerate the flow aft to a higher velocity (lower momentum loss).=20 Closing comments: the above discussion and figures assume a perfect = leak-free system with optimal inlet and exit areas. Optimal exit area = for cruise flight is too small for slow, high power climb on a hot day = (you will cook the heads), so it requires adjustable exit area meaning = cowl flaps. And low leakage means a plenum, careful attention to = plugging leaks, and also plugging leaks from inside the cowl to the = outside. =20 Bottom line: the best you can get over stock is probably 10-12 knots. = If you just minimize leakage with a plenum and closing off leaks, you = might get 5 knots which has been reported by Chris Z. If you don't want = cowl flaps, you can close down the inlets (velocity ratio much higher = than 0.4) to minimize flow rate and get another 2-3 knots, but you will = have to climb at high IAS to stay cool. =20 If you want all you can get (10-12 knots) then you will have to do it = all, and do it to perfection. =20 Cautions: I am getting old and forgetful and probably forgot something. = Your mileage may vary.=20 Questions and comments welcome.=20 Fred Moreno ------=_NextPart_001_0006_01C78A25.09F8ECC0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Hi Marv
hope this works
----- Original Message -----=20
From: Fred=20 Moreno
Sent: Saturday, April 28, 2007 7:10 PM
Subject: RE: Cooling Drag

Hi Art:=20

 

I composed = the letter=20 below and the attachment to respond to you questions and those from=20 others.  My emails to LML keep bouncing.  Marv and I are = grappling=20 with the problem, but to help with the = solution,=20 please forward to LML from your email, and let=92s see if that works.=20   Let me know if I answered your questions.

 

Cheers,=20

 

Fred

____________________________________________________=

 

I have received questions = both on=20 the forum and off concerning my earlier comments about cooling = drag.  They=20 sent me scurrying back to my books and calculator.  =

 

I spent some time = generating a=20 self-consistent set of numbers for cooling drag, pressures, = temperatures, etc.=20 with a uniform set of assumptions.  Some guesstimates are required, = particularly concerning losses from friction in various locations, so I = have=20 listed everything assumed and calculated, and invite investigation and=20 questions.  The results are summarized in the ugly sketch = attached. =20 It shows the flow conditions at various points along the cooling air = pathway=20 assuming you have the ultimate set up, no leaks, cowl flaps perfectly = adjusted=20 to get the right exit area, etc.   

 

Air enters from the left, = is=20 partially compressed in front of the inlet by the slowing of the air = flow, goes=20 through the inlet, then through a short diffuser where it is slowed = more, and=20 then dumps into a large plenum over the engine where the flow is a = fairly slow=20 moving mess.  Then the air flows through cylinders and oil cooler = (which=20 can take a lot of flow) into the space below the engine (where the flow = is again=20 a mess, but a slow moving mess), and then accelerates through the = outlets, the=20 area of which is controlled by cowl flaps which are optimally = adjusted. =20  I have assumed 9000 feet, 200 knots IAS, 230 knots TAS as an = interesting=20 speed for us aspirated engine guys to investigate.  I have used = pressures=20 in inches of water since the engine pressure drop charts use these units = of=20 measure.  The inlet velocity ratio is assumed to be 0.4 (velocity = through=20 cowl inlets divided by TAS) which seems to be roughly what the latest = and=20 fastest designs are using in the fastest production aircraft and speed = mods=20 offered by LoPresti.

 

Comments: =

1)     = Flow is = controlled at the=20 outlet with cowl flaps, so we can get a lot of compression in front of = the=20 inlets where the compression process is frictionless.  With an = inlet=20 velocity of 0.4, we get about 86% of the total dynamic pressure in front = of the=20 inlet. 

2)     = We are = forced to accept a=20 short diffuser behind the inlets which further slows the flow, but = friction=20 losses eat up an assumed 2 inches of water pressure.

3)     = The = velocity in the=20 plenum is anybody=92s guess as the flow is a turbulent = mess, but the=20 velocity is low, so losses are low.  But there are losses in the = plenum,=20 particularly with the 550 Continental engines because of the intake = manifold and=20 other clap trap above the cylinders.

4)     = Note that = the compression=20 raises the temperature about 12F so the assumed temperature above the = fins is=20 about 40F.  We can use this when we go to the engine cooling charts = to make=20 an estimate of the air flow required.

5)     = Engine = cooling charts are=20 published assuming full power and 475F CHT (red line) and cruise with = 435F=20 CHT.  The charts suggest that the engine pressure drop will be = about 5=20 inches of water at 435f CHT.  But we like 350F or thereabouts for = our CHT=20 and that requires more airflow for more cooling.  A real rough = guess is=20 about 20-25% more air flow, and this requires about 8 inches of pressure = drop=20 across the engine to force the increased air flow through the = fins. =20  

6)     = The flow = below the engine=20 is chaotic, but slow so losses are low.  It is also hot.  With = 350F=20 CHT, it is probably 100-120F higher than ambient, or perhaps 150F.  = (There=20 is a greater temperature rise across cylinder heads, and much less = across the=20 oil cooler, and it all mixes together.)  Because there is a lot of = clap=20 trap below the engine, there are more losses which I guesstimate at = around 1.0=20 inch of water.

7)     = As you = approach the exit=20 (=93discharge nozzle=94) the flow is accelerated, hopefully = smoothly,=20 efficiently, and directly aft to regain as much momentum as=20 possible.  However, our exit nozzle is filled with exhaust = pipes, pipe=20 brackets, and other clap trap in the higher velocity area as well as = friction=20 losses along the walls.  My guess is we lose another 2 inches of = water=20 pressure.  This leaves about 12 inches of static pressure converted = into=20 dynamic pressure in the exiting flow.  With the warm, lower density = air I=20 get about 170 knots discharge velocity directly aft for our perfect = cooling=20 system.   The loss in velocity for the cooling flow is then = 230 KTAS=20 minus 170 KTAS outlet, or 60 knots. 

 

Cooling drag is equal to = the=20 momentum loss of the cooling air.  It is cooling air mass = flow (pounds=20 per second) times the velocity lost during passage through the cowl and=20 engine.  If we had lower pressure losses, we would have more = pressure at=20 the exit to accelerate the flow aft faster. 

 

If we could get the exit = velocity =3D=20 free stream velocity (the true air speed), cooling drag would be = zero. =20

 

If exit flow was faster = than the=20 free stream velocity we would get thrust. 

 

If we are moving = really fast=20 with lots of ram pressure and low losses and we raised the = temperature of=20 the air stream up very high to lower the air density a lot and = allow it=20 to be accelerated aft really fast, we would have a ram jet. =

 

But for our air cooled = engine with=20 lots of friction inside the cowl and engine and only modest temperature = rises,=20 we can not get thrust, even for our prefect case above. =

 

Now for a cross = check.  Let=92s=20 assume our IO-550 is pumping out about 225 HP (about 73%).  If you = go to a=20 Lycoming power chart and assume it is the same for the Continental = (pretty good=20 assumption), for 40F above the engine and 435F CHT the chart shows you = need 3=20 lbs/sec. of cooling air and 5 inches of water pressure drop across the = engine to=20 get this flow. 

 

But we want 350F CHT to = let our jugs=20 live longer, so the required air flow is (rough guesstimate) 3.7 = lbs/sec. and=20 then the required pressure drop to get this flow is about 8 inches of = water, the=20 figure I used above.   And if you calculate the cowl inlet = area=20 required to admit this flow rate at 40% of the free stream velocity = (TAS), you=20 get two 6 inch diameter inlets.

 

Moving right along, if we = assume=20 that the prop efficiency at 230 KTAS and 225 HP is 85%, then the thrust = =3D drag =3D=20 272 pounds.  Now we compute the cooling drag which is the momentum = loss=20 which is the cooling air flow rate times the velocity loss of 60 knots = times=20 some constants to make the units work and we get 12 pounds for the = cooling drag=20 or about 4% of total drag for our perfect zero leakage cooling = system. =20

 

In my earlier message I = noted that=20 your run-of-the-mill spam can had cooling drag of about 6-7%.  Our = hot rods=20 have about half the drag as a typical Bonanza or C210, so cooling drag = becomes=20 maybe 14%.  If we work really hard maybe we can get it down to = 4%. =20 You can now see what it takes.

 

Your turbo guys at 25,000 = feet have=20 a much more serious problem because there is little to work with.  = Assuming=20 280 knots TAS, the IAS =3D 190 knots and you thus have about 10% less = ram pressure=20 available to start with, maybe 23 inches of water.  The air is = colder, but=20 it is also a lot less dense so if I really extrapolate off my power = chart, and=20 assume 262 HP (75% for a TSIO 550) and 350 CHT, the pressure drop = required=20 across the engine is 12-14 inches of water, call it 5 inches more than = at low=20 altitude.  So the high altitude means that you lose 3+5=3D 8 inches = of water=20 pressure compared to the low altitude case above as the flow approaches = the=20 exit.  Instead of 12 inches of water to accelerate the flow aft, = you have=20 about 4 inches, so the exit velocity is much lower requiring a larger = exit area,=20  and the momentum loss much higher which means cooling drag is much = higher.  And that is the ideal case.  Most LIV cooling systems = are far=20 from ideal, so my guess is that for the stockers (no mods) cooling drag = is=20 somewhere around 20-25% of total drag.

 

You can see for the high = altitude=20 case cooling drag will always be a bigger piece of the drag pie even if = we=20 assume a perfect, leak free cooling system.  However, if you run = this=20 configuration (big exit area) at lower altitude, you will have a lot of = excess=20 cooling drag you could shed =96 if you could close cowl flaps, shrink = the exit=20 area and thereby accelerate the flow aft to a higher velocity (lower = momentum=20 loss).

 

Closing comments: the = above=20 discussion and figures assume a perfect leak-free system with optimal = inlet and=20 exit areas.  Optimal exit area for cruise flight is too small for = slow,=20 high power climb on a hot day (you will cook the heads), so it requires=20 adjustable exit area meaning cowl flaps.  And low leakage means a = plenum,=20 careful attention to plugging leaks, and also plugging leaks from inside = the=20 cowl to the outside. 

 

Bottom line: the best you = can get=20 over stock is probably 10-12 knots.  If you just minimize leakage = with a=20 plenum and closing off leaks, you might get 5 knots which has been = reported by=20 Chris Z.  If you don=92t want cowl flaps, you can close down the = inlets=20 (velocity ratio much higher than 0.4) to minimize flow rate and get = another 2-3=20 knots, but you will have to climb at high IAS to stay cool. =20

 

If you want all you can = get (10-12=20 knots) then you will have to do it all, and do it to perfection.=20   

 

Cautions: I am getting old = and=20 forgetful and probably forgot something.  Your mileage may vary.=20

 

Questions and comments = welcome.=20

 

Fred = Moreno

 

 

 

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