X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from cdptpa-omtalb.mail.rr.com ([75.180.132.122] verified) by logan.com (CommuniGate Pro SMTP 5.3c4) with ESMTP id 4034104 for flyrotary@lancaironline.net; Mon, 21 Dec 2009 15:32:03 -0500 Received-SPF: pass receiver=logan.com; client-ip=75.180.132.122; envelope-from=eanderson@carolina.rr.com Return-Path: X-Authority-Analysis: v=1.0 c=1 a=ayC55rCoAAAA:8 a=arxwEM4EAAAA:8 a=QdXCYpuVAAAA:8 a=7g1VtSJxAAAA:8 a=ekHE3smAAAAA:20 a=UretUmmEAAAA:8 a=Ia-xEzejAAAA:8 a=nUuTZ29dAAAA:8 a=Lu7ihWn95FLIZWJNrb0A:9 a=hbc5QQT2yyI4P0NyM68A:7 a=Bed98Ud5tjmQ4lO3sd4y8eUDXOEA:4 a=1vhyWl4Y8LcA:10 a=jI77epmzR7sA:10 a=EzXvWhQp4_cA:10 a=_X4pQo3mz8z8h58v:21 a=ZVBytJWXxezybGYp:21 a=Y2VNeNrzAAAA:8 a=yMhMjlubAAAA:8 a=TW66zc2HAAAA:8 a=SSmOFEACAAAA:8 a=HQ31llbKAAAA:8 a=N5A2d2nPPwtTCiUshhoA:9 a=m6roDp6BJGsf9R51xrkA:7 a=KFkkPsbbTkMFwb0DombGymnoOrAA:4 a=nL4BnSq_J2GLUdNz:21 a=4v28zJi-6gTpdy30:21 X-Cloudmark-Score: 0 X-Originating-IP: 75.191.186.236 Received: from [75.191.186.236] ([75.191.186.236:2069] helo=computername) by cdptpa-oedge04.mail.rr.com (envelope-from ) (ecelerity 2.2.2.39 r()) with ESMTP id 35/1E-01550-02BDF2B4; Mon, 21 Dec 2009 20:31:29 +0000 From: "Ed Anderson" Message-ID: <35.1E.01550.02BDF2B4@cdptpa-omtalb.mail.rr.com> To: "'Rotary motors in aircraft'" Subject: RE: [FlyRotary] Air Flow Question Date: Mon, 21 Dec 2009 15:31:34 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_000C_01CA8252.ADC309D0" X-Mailer: Microsoft Office Outlook, Build 11.0.5510 Thread-Index: AcqCcGxlmkhEkJtlSyWTWleXAncmdwAAjmRw In-Reply-To: X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5579 This is a multi-part message in MIME format. ------=_NextPart_000_000C_01CA8252.ADC309D0 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: 7bit Thomas, it doesn't work quite that way. Old Bernoulli's law is (simplified by removing density) is A1V1 = A2V2 meaning the product of the area and velocity anywhere in your duct is equal. Something to do with conservation of mass (can't created/destroy it). So it based on area expansion rather than volume. So to determine (more or less) the air velocity you need to know the velocity of the air entering your duct (and the inlet area) and the area down stream that you are expanding. However, determining the velocity of air entering your duct may not be as simple as it first seems due to a condition known as external diffusion. This is the air streaming being slowed down in front of the inlet due to a pressure gradient extending out of your duct opening (you can sort of think of it as air molecules piling up before your core and increasing the pressure back out your duct). Where to Start? Either find an installation very similar to yours that is cooling adequate and copy that OR you can do some figuring on the back of an envelope. You gotta start somewhere and Mr. Horner indicated that you need to have the airflow through your core either 10% of your cruise speed or 30% of your climb speed. So if your cruise speed is 180 knots then you would want the airflow through your core to be ideally around 18knots. Just as an example (disregarding external diffusion) lets say your inlet opening A1 was 20 x2 = 40 sq inches = 40/144 = 0.277 sq ft then if you want 18.4 knots the area in the duct at A2 then you solve for A2 A1V1 = A2V2 so solving for A2 = A1V1/V2 = 0.277 * 180/18 = 2.777 sq ft or 2.777 * 144 = 400 square inches or an appox 10:1 difference between opening and expanded area. However, another couple of wizards (Kuchumman and Weber) indicated that for good diffusion, your ratio of inlet area to area before your core should be between 0.25 and 0.40 - going beyond that you start to go bad. So lets say you need 400 sq inches to accommodate your radiator core, then according to K&W your inlet would need to be between 0.25 and 0.40 X 400 = 100 - 160 sq inches The larger inlet would also tend to diminish the external diffusion effect but not slow the air velocity as much as the smaller opening.. However we still have A1V1 = A2V2 so with A1 at 100 sq inch (0.694 sq ft) and assuming inlet air velocity is 180 knots and now having A2 fixed at 400 sq inch or 2.777 sq ft we have V2 = A1V1/A2 = 0.694 * 180/2.777 = 45 kts. So our air velocity at the core is a bit higher than we would like (according to Horner) so while it will cool, we may be encountering more cooling drag due to the higher velocity air through the core than we would like. But, this is just a back of the envelope calculation. So many things can affect cooling, we should be so lucky that it would be just one major thing. Where I would Start: I personally think the place to start is to 1st size your radiator core based on the heat you want to get rid of in your worst case situation (probably take off/climb). Since few of us have wind tunnels, starting with a rule of thumb for core volume to HP would probably be a good place to start. Then looking at your space constraints to determine your radiator size. I would not go much thicker than 3" . NASCAR car radiators are typically around 3" in thickness with some going up to 7" thick for the higher speed long tracks at speeds comparable to ours. Also whatever the radiator builders have sort of mandates what you use. So I think you mentioned a rule of thumb of 1.8 cubic inch of core/HP ( I personally feel this may be a little on the low side). Assuming 270 HP max engine power then that would indicate a core volume of approx 1.8 * 270 = 486 cubic inches (lets round it up to an "even" 500 cubic inches). Assuming you find a core 3" thick then its front area would need to be 500/3 = 166.6 sq inches. So that could be 16 wide and 10" high or 27.7 inches wide by 6" high or what ever combination - again likely constrained by what the manufactures build. But let's say you chose 16 x10 x3 radiator. The next thing you need to know is how much airflow you must have through it to dissipate the heat (coolant only in this example). For a three rotor producing 270 hp the coolant needs to get rid of approx 8288 BTU/Minute. Assuming we can add heat to the cooling air increasing its temperature by 80F (might get 100), Then the air mass required can be found from Q (BTU) = M(mass)*Dt*Cp rearranging the formula M = Q/ Dt*Cp = 8288/(80*0.25) = 414.4 lbm/min of air. One Cubic foot of air at sea level = 0.0765 lbm So air flow in CFM = 414.4/0.0765 = 5416 CFM of cooling air. We need to pass that through the frontal area of our core (166.6 sq inches /144 = 1.1569 sq ft). 5416 / 1.1569 = 4681 ft/min of air velocity or dividing by 60 = 78.01 ft/sec = 46 knots air velocity through your core (166.6 sq inch). Not quite the 18 kts Horner wanted but at least a start. So what does this tell us. That if we want to get by with the ideal (slower) airflow through the core (18 kts) then our core frontal area needs to be larger than 166 sq inches. Back to A1V1 = A2V2 if we need 46 knots through 1.1569 sq ft of core frontal area and V1 (assuming no external diffusion) = 180 kts then the inlet A1 = 46 * 1.569/180 = 0.4 sq ft inlet = 0.4 * 144 = 57.7 sq inch inlet opening. Now remember this is all looking at cooling a cruise - where things are best. Conditions during take off and climbout are going to be worst case. So you need to do all of this for those airspeeds as well Note there are a whole bunch of assumptions made to simplify things which may not hold true in all cases. The first major one is the rule of thumb 1.8 cubic inch /Hp. IF your core is fabricated similar to the core from which this rule of thumb was drawn then you are probably OK. But, if substantive different then this rule of thumb may not be valid and then your basic assumption is flawed and need I add all following that is now garbage. The alternative to all of this stuff - is to find an installation as close as possible to yours that is cooling adequately and use that as you cooling system design basis. Sorry - got carried away. Ed Ed Anderson Rv-6A N494BW Rotary Powered Matthews, NC eanderson@carolina.rr.com http://www.andersonee.com http://www.dmack.net/mazda/index.html http://www.flyrotary.com/ http://members.cox.net/rogersda/rotary/configs.htm#N494BW http://www.rotaryaviation.com/Rotorhead%20Truth.htm _____ From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Thomas Mann Sent: Monday, December 21, 2009 2:04 PM To: Rotary motors in aircraft Subject: [FlyRotary] Air Flow Question If I have a volume of air entering my scoop at 180 kts and expand the volume of the chamber by 400% can I expect the speed of the airflow to drop to 45 kts at that point? T Mann __________ Information from ESET NOD32 Antivirus, version of virus signature database 3267 (20080714) __________ The message was checked by ESET NOD32 Antivirus. http://www.eset.com ------=_NextPart_000_000C_01CA8252.ADC309D0 Content-Type: text/html; charset="us-ascii" Content-Transfer-Encoding: quoted-printable

Thomas, it doesn’t work quite = that way.  Old Bernoulli’s law is (simplified by removing density) = is A1V1 =3D A2V2 meaning the product = of the area and velocity anywhere in your duct is equal.  Something to = do with conservation of mass (can’t created/destroy it). So it based = on area expansion rather than volume. 

 

So to determine (more or less) the = air velocity you need to know the velocity of the air entering your duct = (and the inlet area) and the area down stream that you are expanding.  = However, determining the velocity of air entering your duct may not be as simple = as it first seems due to a condition known as external diffusion.  This = is the air streaming being slowed down in front of the inlet  due to a = pressure gradient extending out of your duct opening (you can sort of think of it = as air molecules piling up before your core and increasing the pressure back = out your duct).

=  

= Where to Start?  Either = find an installation very similar to yours that is cooling adequate and copy = that OR you can do some figuring on the back of an = envelope.

 

You gotta start somewhere and Mr. = Horner indicated that you need to have the airflow through your core either 10% = of your cruise speed or 30% of your climb speed.  =

 

So if your cruise speed is 180 = knots then you would want the airflow through your core to be ideally around = 18knots.

 

Just as an example (disregarding = external diffusion) lets say your inlet opening A1  was 20 x2 =3D 40 sq = inches =3D 40/144 =3D  0.277 sq ft then if you want 18.4 knots the area in the = duct at A2 then you solve for A2

 

A1V1 =3D A2V2 so solving for A2 =3D = A1V1/V2 =3D 0.277 * 180/18 =3D 2.777 sq ft or 2.777 * 144 =3D 400 square inches or = an appox 10:1 difference between opening and expanded = area.

 

However, another couple of wizards (Kuchumman and Weber) indicated that for good diffusion, your ratio of = inlet area to area before your core should be between 0.25 and 0.40 - going = beyond that you start to go bad.

 

So lets say you need 400 sq inches = to accommodate your radiator core, then according to K&W your inlet = would need to be between 0.25 and 0.40 X 400  =3D 100 – 160 sq = inches  The larger inlet would also tend to diminish the external diffusion effect = but not slow the air velocity as much as the smaller opening..  =

 

However we still have A1V1 =3D = A2V2  so with A1 at 100 sq inch (0.694 sq ft) and assuming inlet air velocity is = 180 knots and now having A2 fixed at 400 sq inch or 2.777 sq ft we have V2 = =3D A1V1/A2 =3D 0.694 * 180/2.777 =3D 45 kts.

 

So our air velocity at the core is = a bit higher than we would like (according to Horner) so while it will cool, = we may be encountering more cooling drag due to the higher velocity air through = the core than we would like.

 

But, this is just a back of the = envelope calculation.  So many things can affect cooling, we should be so = lucky that it would be just one major thing.

 

= Where I would Start:

 

I personally think the place to = start is to 1st size your radiator core based on the heat you want to = get rid of in  your worst case situation (probably take off/climb).  = Since few of us have wind tunnels, starting with a rule of thumb for core = volume to HP would probably be a good place to start.

 

Then looking at your space = constraints to determine your radiator size.  I would not go much thicker than = 3” .  NASCAR car radiators are typically around 3” in thickness = with some going up to 7” thick for the higher speed long tracks at = speeds comparable to ours.  Also whatever the radiator builders have sort of mandates = what you use.

 

So I think you mentioned a rule of = thumb of 1.8 cubic inch of core/HP ( I personally feel this may be a little on = the low side).  Assuming 270 HP max engine power then that would = indicate a core volume of approx 1.8 * 270 =3D  486 cubic inches (lets round = it up to an “even” 500 cubic inches).  Assuming you find a core = 3” thick then its front area would need to be 500/3 =3D 166.6 sq = inches.  So that could be 16 wide and 10” high or  27.7 inches wide by = 6” high or what ever combination – again likely constrained by what = the manufactures build.

 

But let’s say you chose 16 = x10 x3 radiator.  The next thing you need to know is how much airflow you = must have through it to dissipate the heat (coolant only in this = example).  For a three rotor producing 270 hp the coolant needs to get rid of approx = 8288 BTU/Minute.   Assuming we can add heat to the cooling air = increasing its temperature by 80F (might get 100),

Then the air mass required can be found from Q (BTU) = =3D M(mass)*Dt*Cp = rearranging the formula

 

M =3D Q/ = Dt*Cp =3D 8288/(80*0.25) =3D 414.4 lbm/min of air.  One Cubic foot of air at = sea level =3D 0.0765 lbm

So air flow in CFM =3D 414.4/0.0765 =3D  5416 = CFM of cooling air.  We need to pass that through the frontal area of our = core (166.6 sq inches /144 =3D 1.1569 sq ft).   5416 / 1.1569 =3D = 4681 ft/min of air velocity or dividing by 60 =3D 78.01 ft/sec =3D 46 knots air = velocity through your core (166.6 sq inch).  Not quite the 18 kts Horner = wanted but at least a start.  So what does this tell us.  That if we want = to get by with the ideal (slower)  airflow through the core (18 kts) then = our core frontal area needs to be larger than 166 sq inches.  =

 

Back to A1V1 =3D A2V2 if we need 46 knots through = 1.1569 sq ft of core frontal area and V1 (assuming no external diffusion) =3D 180 kts = then the inlet A1 =3D 46 * 1.569/180 =3D 0.4 sq ft inlet =3D 0.4 * 144 =3D 57.7 = sq inch inlet opening.  Now remember this is all looking at cooling a cruise = – where things are best.   Conditions during take off and = climbout are going to be worst case.  So you need to do all of this for those = airspeeds as well

 

Note there are a whole bunch of assumptions made to = simplify things which may not hold true in all = cases.

The first major one is the rule of thumb 1.8 cubic = inch /Hp.  IF your core is fabricated similar to the core from which = this rule of thumb was drawn then you are probably OK.  But, if substantive different then this rule of thumb may not be valid and then your basic assumption is flawed and need I add all following that   is = now garbage.

 

The alternative to all of this stuff – is to = find an installation as close as possible to yours that is cooling adequately = and use that as you cooling system design basis.

 

Sorry – got carried = away.

 

Ed

 

From: = Rotary motors in aircraft = [mailto:flyrotary@lancaironline.net] On Behalf Of Thomas = Mann
Sent: Monday, December = 21, 2009 2:04 PM
To: Rotary motors in aircraft
Subject: [FlyRotary] Air = Flow Question

 

If I have a = volume of air entering my scoop at 180 kts and expand the volume of the chamber = by 400% can I expect the speed of the airflow to drop to 45 kts at that = point?

 <= /o:p>

T = Mann



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The message was checked by ESET NOD32 Antivirus.

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