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 4067312 for flyrotary@lancaironline.net; Wed, 06 Jan 2010 16:57:38 -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 D9E44173C9E for ; Thu, 7 Jan 2010 05:57:02 +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 5551ABEC001 for ; Thu, 7 Jan 2010 05:56:58 +0800 (WST) Message-ID: <15DBD2C92F7D43D6BE32ADD6F2C1DDFD@ownerf1fc517b8> From: "George Lendich" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Waste Heat was [Fly Rotary] Re: Air Flow Question Date: Thu, 7 Jan 2010 07:56:59 +1000 MIME-Version: 1.0 Content-Type: multipart/mixed; boundary="----=_NextPart_000_000B_01CA8F6E.FDAEDB30" 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 100106-1, 01/06/2010), Outbound message X-Antivirus-Status: Clean This is a multi-part message in MIME format. ------=_NextPart_000_000B_01CA8F6E.FDAEDB30 Content-Type: multipart/alternative; boundary="----=_NextPart_001_000C_01CA8F6E.FDAEDB30" ------=_NextPart_001_000C_01CA8F6E.FDAEDB30 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Ed, Taking your idea of looking at BTU of fuel further, I'm wondering if we = shouldn't be removing from the calculations, the loss of BTU through the = exhaust and heat transfer from the engine to the air in the cowl. Looks = like 20% of energy going by the exhaust valve, some of which is removed = by the oil, but still a fair bit going out the exhaust. What's your take on that idea. Perhaps it something that could be looked = at in the take-off and climb figures, to compare to cruise figures, in = an endeavour to get a balanced view of overall need. George ( down under) Ok, George, you got it. I just want to emphasize that converting the = HP of your engine to its equivalent to BTU of heat energy really does = not tell you how much WASTE heat you need to get rid of - at least not = directly. You can generally assume if you are producing more HP you = will also have more waste heat to get rid of. But if you want to = really get a handle on how much, then you have to figure how much of the = total energy in the gasoline is going to work, exhaust and cooling uses = of that energy. =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 George Lendich Sent: Wednesday, January 06, 2010 4:27 AM To: Rotary motors in aircraft Subject: [FlyRotary] Re: Waste Heat was [Fly Rotary] Re: Air Flow = Question =20 Ed, That's an interesting, and obviously valid approach, didn't think of = 'the old BTU's in fuel used approach' -that goes in my notes as well. Nothing much is wasted on me, provided I can understand it, of course. I like the way you think Ed, I realized where the 18 knots came from - = silly me, 10% of cruise.=20 =20 I am currently rehashing my figures, like you Ed, I like to validate = the proximity of need using the maths, at least it gives a valid = starting point and a core understanding. =20 George (down under) =20 Hi George, =20 Like many results of using equations - what you get out depends on = what you put in {:>). =20 1st there is 42.41 BTU/min per HP. So 42.41 *270 =3D 11450.7 = BTU/Min - as you calculated - BUT (big but) that is the heat energy = needed to produce 270 HP - that is NOT the waste heat energy you need to = get rid of. The amount you calculated is basically the heat energy to = produce 270 HP of mechanical work (turning the prop). What you want to = figure out is how much "waste" heat you need to get rid of (that is - = the heat energy not used to rotate the prop). That waste heat is the = heat that has not/can not be used to produce mechanical work. Waste = heat is generally gotten rid of two ways - out the exhaust stack and = through convection (coolant and oil) system in our engines. So you are = primarily interested in how much heat you must get rid of by convection = (coolant and oil) systems. =20 =20 Here is my method and how I got the 8288 BTU/min . Again, your = calculation does gives you the amount of heat energy in BTU that = produces 270 HP - that heat has been used doing work by your engine, = however, that is NOT the heat you are getting rid of through your = coolers -. For cooling purposes, you want to find the BTUs of heat = energy you need to get rid of that is NOT producing power - or in other = words - the waste heat (I think I repeated myself).=20 =20 Again, the HP of work your engine is producing is NOT the heat = energy that you have to get rid of through your two cooling systems = (radiators and oil coolers) (oops repeated myself again - but the = distinction is crucial!)=20 =20 Here is how I arrived at my figures for the waste heat. There is = more than one way to do this for sure and the results depends a lot on = your heat allocation (more on that in a bit). =20 Using a air/flow and air density and a formula for a 3 rotor, I = calculated that at a 12.65:1 air/fuel ratio (best power - so not cruise = {:>)) that you would burn around 25.2 GPH at 270 HP. =20 Using the old and simply (but good) approx power formula, I = calculate HP =3D 25.2 * 6 /0.55 =3D 274.9 HP which is pretty close to = the results (270 HP) of my more complex calculation=20 =20 which is based on engine air/flow and fuel mass consumed at that = ratio. So 25.2 GPH is a pretty valid figure for our fuel consumption at = that power. =20 =20 However, I calculate engine HP from a different approach (got to = make it more complex, you know {:>)). My approach is based on the = total number of BTU in the amount of fuel=20 =20 ingested by the engine to meet the power requirement (270 HP) with = following heat energy allocation: If your percentage of energy = allocation is different, then you will get a different results: =20 24% =3D HP (useful mechanical work) - some folks may think this is = too conservative, but I tend to be that way when calculating HP. 50%=3D out the Exhaust 26% =3D Cooling Waste (coolant + oil) =20 =20 We know coolant =3D 2/3 of 26% Cooling Waste and the oil =3D 1/3. =20 =20 There is approx 19000-20000 BTU/lbm in gasoline (I use 19000 = BTU/pound to be a bit conservative - besides 100LL has less energy than = 87 Mogas {:>) =20 So with 25.2 GPH * 6 lbm/gallon =3D 151 lbm/hour/60 =3D 2.52 = lbm/min of gasoline is being burned to produce the energy so out engine = produces 270 HP. =20 So 2.52 *19000 BTU/Lbm =3D 47880 BTU/min total energy of the fuel = consumed per minute at the 25.2 GPH rate. This total includes the work = energy (HP) and the waste energy. =20 So to find out the heat used to produce work 24% *47880 BTU/min =3D = 11491.2 BTU/min - Now that is not much different from what you got = 11450.7 BTU/Min. But, as I stated this is=20 =20 NOT the waste heat you need to reject.=20 =20 =20 So Allocation of heat energy we need to get rid of through the = cooling system, we have =20 26% =3D cooling waste =3D 0.26 * 47880 =3D 12448.8 BTU/Min of = which 2/3 is rejected through radiator , therefore=20 =20 0.6666 * 12448.8 =3D 8298 BTU/Min of waste heat through the = radiators. (does not include waste heat through the oil which is = 12448.8 - 8298 =3D 4150.8 BTU/min). =20 So altogether the radiator and oil cooler has to get rid of around = 12500 BTU/min at that power level - so you can see that if you do not = have adequate cooling at 270 HP power production you are going to fry = your engine fairly quickly. =20 =20 This is how it appears to me, George. Hope it helps. =20 =20 =20 Ed =20 =20 =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 George Lendich Sent: Tuesday, January 05, 2010 6:01 PM To: Rotary motors in aircraft Subject: [FlyRotary] Re: Air Flow Question =20 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 =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 = ? =20 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. =20 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 __________ 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_001_000C_01CA8F6E.FDAEDB30 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Ed,
Taking your idea of looking at BTU of=20 fuel further, I'm wondering if we shouldn't be removing from = the=20 calculations, the loss of BTU through the exhaust and heat transfer = from=20 the engine to the air in the cowl. Looks like 20% of energy going by the = exhaust=20 valve, some of which is removed by the oil, but still a fair bit going = out the=20 exhaust.
What's your take on that idea. Perhaps = it something=20 that could be looked at in the take-off and climb figures, to compare to = cruise=20 figures, in an endeavour to get a balanced view of overall = need.
George ( down under)
 

Ok, George, = you got=20 it.  I just want to emphasize that  converting the HP of = your engine=20 to its equivalent to BTU of heat energy really does not tell you how = much=20 WASTE heat you need to get rid of =96 at least not directly.  You = can=20 generally assume if you are producing more HP you will also have more = waste=20 heat to get rid of.   But if you want to really get a handle = on how=20 much, then you have to figure how much of the total energy in the = gasoline is=20 going to work, exhaust and cooling uses of that=20 energy.

 

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 George Lendich
Sent: Wednesday, January 06, = 2010 4:27=20 AM
To: = Rotary motors in aircraft
Subject: [FlyRotary] Re: Waste = Heat was=20 [Fly Rotary] Re: Air Flow Question

 

Ed,

That's an interesting, = and=20 obviously valid approach, didn't think of 'the old BTU's in fuel used=20 approach' -that goes in my notes as = well.

Nothing much is wasted = on me,=20 provided I can understand it, of = course.

I like the way you think = Ed, I=20 realized where the 18 knots came from - silly me, 10% of=20 cruise. 

 

I am currently rehashing = my=20 figures, like you Ed, I like to validate the proximity of need using = the=20 maths, at least it gives a valid starting point and a core=20 understanding.

 

George (down=20 under)

 

Hi=20 George,

 

Like many = results=20 of using equations =96 what you get out depends on what you put in=20 {:>).

 

1st=20 there is 42.41 BTU/min per HP.  So 42.41 *270 =3D 11450.7 = BTU/Min - as=20 you calculated  =96 BUT=20 (big but) that is the heat energy needed to produce 270 HP =96 that = is=20 NOT the waste heat energy you need to = get rid=20 of.   The  amount you calculated is basically the = heat energy=20 to produce 270 HP of mechanical work (turning the prop).  What = you want=20 to figure out is how much =93waste=94 heat you need to get = rid of=20 (that is - the heat energy  not used to rotate the = prop).=20   That waste heat is=20 the heat that has not/can = not=20 be used to produce mechanical work.  Waste heat is generally = gotten rid=20 of two ways =96 out the exhaust stack and through convection = (coolant and oil)=20 system in our engines.  So you are primarily interested in how = much=20 heat you must get rid of by convection (coolant and oil)=20 systems.

 

 

Here is = my method=20 and how I got the 8288 BTU/min .  Again,  your calculation = does=20 gives you the amount of heat energy in  BTU that produces 270 = HP - that=20 heat has been used doing work by your engine,  however, that is = NOT  the heat = you are=20 getting rid of through your coolers =96.   For cooling = purposes, you=20 want to find the BTUs of heat energy you need to get rid of that is = NOT=20 producing power  - or in other words - the waste heat (I think = I=20 repeated myself).

 

Again,=20   the HP of work your engine is producing is NOT the heat energy that you = have to=20 get rid of through your two cooling systems (radiators and oil = coolers)=20 (oops repeated myself again =96 but the distinction is crucial!)=20

 

Here is = how I=20 arrived at my figures for the waste=20 heat.  There is more than one way to do this for = sure and=20 the results depends a lot on your heat allocation (more on that in a = bit).

 

Using a = air/flow=20 and air density and a formula for a 3 rotor,  I calculated that = at a=20 12.65:1 air/fuel ratio (best power =96 so not cruise {:>)) that = you would=20 burn around 25.2 GPH at 270 HP.

 

Using the = old and=20 simply (but good)  approx power formula, I calculate HP =3D = 25.2 * 6=20 /0.55 =3D  274.9 HP which is pretty close to the results (270 = HP) of my=20 more complex calculation

 

which is = based on=20 engine air/flow and fuel mass consumed at that ratio. So 25.2 GPH is = a=20 pretty valid figure for our fuel consumption at that=20 power.

 

 

However, =  I=20 calculate engine HP from a different approach (got to make it more = complex,=20 you know  {:>)).  My approach is   based on = the total=20  number of BTU in the amount of fuel =

 

ingested = by the=20 engine to meet the power requirement (270 HP)  with following = heat=20 energy allocation:  If your percentage of energy allocation is=20 different, then you will get a different=20 results:

 

24% =3D = HP (useful=20 mechanical work) =96 some folks may think this is too conservative, = but I tend=20 to be that way when calculating HP.

50%=3D = out the=20 Exhaust

26% =3D = Cooling Waste=20 (coolant + oil)

 

 

We know = coolant =3D=20 2/3 of  26% Cooling Waste and the oil =3D=20 1/3.

 

 

There is = approx=20 19000-20000 BTU/lbm in gasoline (I use 19000 BTU/pound  to be a = bit=20 conservative =96 besides 100LL has less energy than 87 Mogas=20 {:>)

 

So with = 25.2 GPH *=20 6 lbm/gallon  =3D 151 lbm/hour/60 =3D 2.52 lbm/min of gasoline =  is=20 being burned to produce the energy so out engine produces 270=20 HP.

 

So 2.52 = *19000=20 BTU/Lbm  =3D  47880 BTU/min total=20 energy of the fuel consumed per minute at the 25.2 GPH=20 rate.  This total includes the work energy (HP)  and the = waste=20 energy.

 

So to = find out the=20 heat used to produce work 24% *47880 BTU/min =3D  11491.2 = BTU/min =96 Now=20 that is not much different from what you got 11450.7 BTU/Min.  = But, as=20 I stated this is

 

NOT the = waste heat you=20 need to reject.

 

 

So = Allocation of=20 heat energy we need to get rid of through the cooling system, we=20 have

 

26% =3D = cooling waste=20 =3D 0.26 * 47880 =3D  12448.8 BTU/Min  of which 2/3 is = rejected=20 through radiator , therefore

 

0.6666 * = 12448.8 =3D=20 8298 BTU/Min of waste = heat=20 through the radiators.  (does not include waste heat through = the oil=20 which is 12448.8 =96 8298 =3D 4150.8 = BTU/min).

 

So = altogether the=20 radiator and oil cooler has to get rid of around 12500 BTU/min at = that power=20 level =96 so you can see that if you do not have adequate cooling at = 270 HP=20 power production you are going to fry  your engine fairly=20 quickly. 

 

This is = how it=20 appears to me, George.  Hope it = helps.

 

 

 

Ed

 

 

 

Ed=20 Anderson

Rv-6A = N494BW Rotary=20 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.r= otaryaviation.com/Rotorhead%20Truth.htm

From:=20 Rotary motors in = aircraft=20 [mailto:flyrotary@lancaironline.net] On=20 Behalf Of George Lendich
Sent: Tuesday, January 05, = 2010 6:01=20 PM
To: = Rotary motors in aircraft
Subject: [FlyRotary] Re: Air = Flow=20 Question

 

Ed,

I have been terribly = busy, but=20 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=20 =3D42.41BTU/min x 270hp=3D1145.7 btu/min of which we are needing = only 2/3 of=20 water cooling, as 1/3 is done by the oil . Therefore 2/3 of=20 1145.7=3D7633.8 btu/min. Did I go wrong somewhere=20 ?

 

Also I didn't know we = were=20 looking for 18knots as optimum air speed flow through the rad. I did = a quick=20 copy of your calculations and got 28 knots for 125 hp, so I = will have=20 to redefine my cooling set-up.

 

Ed, I like the maths = approach.=20 Is Tracy=20 following this line of thought i.e. calculating for cooling, as = an=20 Engineer I assumed he would be doing something following this = line of=20 approach?

George ( down=20 under)

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

 

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

 

Where=20 to Start?  = Either=20 find an installation very similar to yours that is cooling = adequate and=20 copy that OR 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=20 through your core either 10% of your cruise speed or 30% of your = climb=20 speed. 

 

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=20 example (disregarding external diffusion) lets say your inlet = opening A1=20  was 20 x2 =3D 40 sq inches =3D 40/144 =3D  0.277 sq ft = then if you=20 want 18.4 knots the area in the duct at A2 then you solve for=20 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=20 400 square inches or an appox 10:1 difference between opening and = expanded=20 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=20 and 0.40 - going beyond that you start to go=20 bad.

 

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

 

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

 

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

 

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

 

Where=20 I would Start:

 

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

 

Then = looking at=20 your space constraints to determine your radiator size.  I = would not=20 go much thicker than 3=94 .  NASCAR car radiators are = typically around=20 3=94 in thickness with some going up to 7=94 thick for the higher = speed long=20 tracks at speeds comparable to ours.  Also whatever the = radiator=20 builders have sort 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=20 power then that would indicate a core volume of approx 1.8 * 270 = =3D=20  486 cubic inches (lets round it up to an =93even=94 500 = cubic=20 inches).  Assuming you find a core 3=94 thick then its front = area would=20 need to be 500/3 =3D 166.6 sq inches.  So that could be 16 = wide and 10=94=20 high or  27.7 inches wide by 6=94 high or what ever = combination =96 again=20 likely constrained by what the manufactures=20 build.

 

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

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

 

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

 

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

The first major one = is the=20 rule of thumb 1.8 cubic inch /Hp.  IF your core is fabricated = similar=20 to the core from which this rule of thumb was drawn then you are = probably=20 OK.  But, if substantive different then this rule of thumb = may not be=20 valid and then your basic assumption is flawed and need I add all=20 following that   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=20 Rotary 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=20 [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=20 a volume of air entering my scoop at 180 kts and expand the = volume of=20 the chamber by 400% can I expect the speed of the airflow to = drop to 45=20 kts at that point?

 

T=20 Mann



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