X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from poplet2.per.eftel.com ([203.24.100.45] verified) by logan.com (CommuniGate Pro SMTP 5.3.0) with ESMTP id 4065701 for flyrotary@lancaironline.net; Tue, 05 Jan 2010 18:02:04 -0500 Received-SPF: none receiver=logan.com; client-ip=203.24.100.45; envelope-from=lendich@aanet.com.au Received: from sv1-1.aanet.com.au (mail.aanet.com.au [203.24.100.34]) by poplet2.per.eftel.com (Postfix) with ESMTP id 032A31738F6 for ; Wed, 6 Jan 2010 07:01:26 +0800 (WST) Received: from ownerf1fc517b8 (203.171.92.134.static.rev.aanet.com.au [203.171.92.134]) by sv1-1.aanet.com.au (Postfix) with SMTP id 9D5B5BEC010 for ; Wed, 6 Jan 2010 07:01:24 +0800 (WST) Message-ID: From: "George Lendich" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Air Flow Question Date: Wed, 6 Jan 2010 09:01:25 +1000 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0011_01CA8EAE.D39CE910" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.5843 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5579 X-Antivirus: avast! (VPS 100105-0, 01/05/2010), Outbound message X-Antivirus-Status: Clean This is a multi-part message in MIME format. ------=_NextPart_000_0011_01CA8EAE.D39CE910 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Ed, I have been terribly busy, but wanted to follow this thread and finally = got around to it, but can't understand where you got 8288 BTU/min. =20 If 1hp =3D 2545BTU/hr /60 =3D42.41BTU/min x 270hp=3D1145.7 btu/min of = which we are needing only 2/3 of water cooling, as 1/3 is done by the = oil . Therefore 2/3 of 1145.7=3D7633.8 btu/min. Did I go wrong somewhere = ? Also I didn't know we were looking for 18knots as optimum air speed flow = through the rad. I did a quick copy of your calculations and got 28 = knots for 125 hp, so I will have to redefine my cooling set-up. Ed, I like the maths approach. Is Tracy following this line of thought = i.e. calculating for cooling, as an Engineer I assumed he would be doing = something following this line of approach? George ( down under) 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. =20 =20 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). =20 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. =20 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. =20 =20 So if your cruise speed is 180 knots then you would want the airflow = through your core to be ideally around 18knots. =20 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 =20 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. =20 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. =20 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.. =20 =20 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. =20 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. =20 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.=20 =20 Where I would Start: =20 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. =20 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. =20 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. =20 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 =20 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. =20 =20 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 =20 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. =20 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. =20 Sorry - got carried away. =20 Ed =20 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 =20 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? =20 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_0011_01CA8EAE.D39CE910 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Ed,
I have been terribly busy, but wanted = to follow=20 this thread and finally got around to it, but can't understand where you = got=20 8288 BTU/min. 
 
If 1hp =3D 2545BTU/hr /60 = =3D42.41BTU/min x=20 270hp=3D1145.7 btu/min of which we are needing only 2/3 of water = cooling, as 1/3=20 is done by the oil . Therefore 2/3 of 1145.7=3D7633.8 btu/min. Did = I go wrong=20 somewhere ?
 
Also I didn't know we were looking for = 18knots as=20 optimum air speed flow through the rad. I did a quick copy of your = calculations=20 and got 28 knots for 125 hp, so I will have to redefine my cooling=20 set-up.
 
Ed, I like the maths approach. Is Tracy = following=20 this line of thought i.e. calculating for cooling, as an Engineer I = assumed=20 he would be doing something following this line of = approach?
George ( down under)

Thomas, it = doesn=92t=20 work quite that way.  Old Bernoulli=92s law is (simplified by = removing=20 density) is A1V1 =3D A2V2 meaning the product=20 of the area and velocity anywhere in your duct is = equal. =20 Something to do with conservation of mass (can=92t created/destroy = it). So it=20 based on area expansion rather than volume. =20

 

So to = determine (more=20 or less) the air velocity you need to know the velocity of the air = entering=20 your duct (and the inlet area) and the area down stream that you are=20 expanding.  However, determining the velocity of air entering = your duct=20 may not be as simple as it first seems due to a condition known as = external=20 diffusion.  This is the air streaming being slowed down in front = of the=20 inlet  due to a pressure gradient extending out of your duct = opening (you=20 can sort of think of it as air molecules piling up before your core = and=20 increasing the pressure back out your = duct).

 

Where=20 to Start?  = Either find an=20 installation very similar to yours that is cooling adequate and copy = that OR=20 you can do some figuring on the back of an=20 envelope.

 

You gotta = start=20 somewhere and Mr. Horner indicated that you need to have the airflow = through=20 your core either 10% of your cruise speed or 30% of your climb = speed. =20

 

So if your = cruise=20 speed is 180 knots then you would want the airflow through your core = to be=20 ideally around 18knots.

 

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

 

A1V1 =3D = A2V2 so=20 solving for A2 =3D A1V1/V2 =3D 0.277 * 180/18 =3D 2.777 sq ft or 2.777 = * 144 =3D 400=20 square inches or an appox 10:1 difference between opening and expanded = area.

 

However, = another=20 couple of wizards (Kuchumman and Weber) indicated that for good = diffusion,=20 your ratio of inlet area to area before your core should be between = 0.25 and=20 0.40 - going beyond that you start to go = bad.

 

So lets say = you need=20 400 sq inches to accommodate your radiator core, then according to = K&W=20 your inlet would need to be between 0.25 and 0.40 X 400  =3D 100 = =96 160 sq=20 inches  The larger inlet would also tend to diminish the external = diffusion effect but not slow the air velocity as much as the smaller=20 opening.. 

 

However we = still have=20 A1V1 =3D A2V2  so with A1 at 100 sq inch (0.694 sq ft) and = assuming inlet=20 air velocity is 180 knots and now having A2 fixed at 400 sq inch or = 2.777 sq=20 ft we have V2 =3D A1V1/A2 =3D 0.694 * 180/2.777 =3D 45=20 kts.

 

So our air = velocity=20 at the core is a bit higher than we would like (according to Horner) = so while=20 it will cool, we may be encountering more cooling drag due to the = higher=20 velocity air through the core than we would = like.

 

But, this = is just a=20 back of the envelope calculation.  So many things can affect = cooling, we=20 should be so lucky that it would be just one major thing.=20

 

Where=20 I would Start:

 

I = personally think=20 the place to start is to 1st size your radiator core based = on the=20 heat you want to get rid of in  your worst case situation = (probably take=20 off/climb).  Since few of us have wind tunnels, starting with a = rule of=20 thumb for core volume to HP would probably be a good place to=20 start.

 

Then = looking at your=20 space constraints to determine your radiator size.  I would not = go much=20 thicker than 3=94 .  NASCAR car radiators are typically around = 3=94 in=20 thickness with some going up to 7=94 thick for the higher speed long = tracks at=20 speeds comparable to ours.  Also whatever the radiator builders = have sort=20 of mandates what you use.

 

So I think = you=20 mentioned a rule of thumb of 1.8 cubic inch of core/HP ( I personally = feel=20 this may be a little on the low side).  Assuming 270 HP max = engine power=20 then that would indicate a core volume of approx 1.8 * 270 =3D =  486 cubic=20 inches (lets round it up to an =93even=94 500 cubic inches).  = Assuming you=20 find a core 3=94 thick then its front area would need to be 500/3 =3D = 166.6 sq=20 inches.  So that could be 16 wide and 10=94 high or  27.7 = inches wide=20 by 6=94 high or what ever combination =96 again likely constrained by = what the=20 manufactures build.

 

But let=92s = say you=20 chose 16 x10 x3 radiator.  The next thing you need to know is how = much=20 airflow you must have through it to dissipate the heat (coolant only = in this=20 example).  For a three rotor producing 270 hp the coolant needs = to get=20 rid of approx 8288 = BTU/Minute.  =20 Assuming we can add heat to the cooling air increasing its temperature = by 80F=20 (might get 100),

Then the air mass = required can be=20 found from Q (BTU) =3D M(mass)*Dt*Cp=20 rearranging the formula

 

M =3D = Q/=20 Dt*Cp =3D=20 8288/(80*0.25) =3D 414.4 lbm/min of air.  One Cubic foot of air = at sea=20 level =3D 0.0765 lbm

So air flow in CFM =3D = 414.4/0.0765=20 =3D  5416 CFM of cooling air.  We need to pass that through = the=20 frontal area of our core (166.6 sq inches /144 =3D 1.1569 sq = ft).  =20 5416 / 1.1569 =3D 4681 ft/min of air velocity or dividing by 60 =3D = 78.01 ft/sec =3D=20 46 knots air velocity through your core (166.6 sq inch).  Not = quite the=20 18 kts Horner wanted but at least a start.  So what does this = tell=20 us.  That if we want to get by with the ideal (slower) =  airflow=20 through the core (18 kts) then our core frontal area needs to be = larger than=20 166 sq inches. 

 

Back to A1V1 =3D A2V2 if = we need 46=20 knots through 1.1569 sq ft of core frontal area and V1 (assuming no = external=20 diffusion) =3D 180 kts then the inlet A1 =3D 46 * 1.569/180 =3D 0.4 sq = ft inlet =3D=20 0.4 * 144 =3D 57.7 sq inch inlet opening.  Now remember this is = all looking=20 at cooling a cruise =96 where things are best.   Conditions = during=20 take off and climbout are going to be worst case.  So you need to = do all=20 of this for those airspeeds as well

 

Note there are a whole = bunch of=20 assumptions made to simplify things which may not hold true in all=20 cases.

The first major one is = the rule of=20 thumb 1.8 cubic inch /Hp.  IF your core is fabricated similar to = the core=20 from which this rule of thumb was drawn then you are probably = OK.  But,=20 if substantive different then this rule of thumb may not be valid and = then=20 your basic assumption is flawed and need I add all following that=20   is now garbage.

 

The alternative to all = of this=20 stuff =96 is to find an installation as close as possible to yours = that is=20 cooling adequately and use that as you cooling system design=20 basis.

 

Sorry =96 got carried=20 away.

 

Ed

 

Ed=20 Anderson

Rv-6A = N494BW Rotary=20 Powered

Matthews,=20 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.r= otaryaviation.com/Rotorhead%20Truth.htm

From:=20 Rotary motors in aircraft = [mailto:flyrotary@lancaironline.net] On=20 Behalf Of Thomas Mann
Sent: Monday, December 21, 2009 = 2:04=20 PM
To: = Rotary motors in aircraft
Subject: [FlyRotary] Air Flow=20 Question

 

If I = have a=20 volume of air entering my scoop at 180 kts and expand the volume of = the=20 chamber by 400% can I expect the speed of the airflow to drop to 45 = kts at=20 that point?

 

T=20 Mann



__________ Information from ESET = NOD32=20 Antivirus, version of virus signature database 3267 (20080714)=20 __________

The message was checked by ESET NOD32 = Antivirus.

http://www.eset.com

------=_NextPart_000_0011_01CA8EAE.D39CE910--