X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from cdptpa-omtalb.mail.rr.com ([75.180.132.121] verified) by logan.com (CommuniGate Pro SMTP 5.2c1) with ESMTP id 2499946 for flyrotary@lancaironline.net; Wed, 21 Nov 2007 18:03:30 -0500 Received-SPF: pass receiver=logan.com; client-ip=75.180.132.121; envelope-from=eanderson@carolina.rr.com Received: from edward2 ([24.74.103.61]) by cdptpa-omta05.mail.rr.com with SMTP id <20071121230252.ICBH8432.cdptpa-omta05.mail.rr.com@edward2> for ; Wed, 21 Nov 2007 23:02:52 +0000 Message-ID: <000c01c82c92$ff9b3bf0$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Naca Report on Radiator Thickness Date: Wed, 21 Nov 2007 18:05:23 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0009_01C82C69.167677B0" 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 This is a multi-part message in MIME format. ------=_NextPart_000_0009_01C82C69.167677B0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable =20 > Ed, >=20 > The most the rejected heat could change in this case > is 300%. You made a 300% change in thickness which > results in a 300% change in wetted surface area. Hi Ron, The heat rejection was 283%, so a bit less than the 300% thickness - = how valid? Well, that is why I posted this to the list. What should the = heat rejection be?? That is what I am after. If there is a well-known = correlation between radiator characteristics and heat rejection - what = are they. If I look at K&W (page 266) they show the heat transfer Coefficient Kh = for turbulent flow in a core as Kh=3D 1/ 2*L/D*0.316*(Re)^(-1/4). Re = being proportional to Mass flow velocity. So If I look at this equation = it would appear that heat transfer goes up proportional to the thickness = (L) and decreases as the diameter of the tubes hydraulic diameter (D) = gets larger. It appears that as the velocity (mass flow) decreases the = Reynolds number goes down, as it goes down the 4th root of it decreases = it even more and then in the denominator it makes Kh coefficient higher = all implying more heat transfer per unit thickness of the core. But, then I could be misinterpreting the factors in the coefficient, but = I do not see any factor in the equation that would indicate this heat = transfer coefficient changes as the air flows through the core. But = clearly the decreasing deltaT causing the last part of the thicker's = core surface area to be less effective in transferring heat to the air - = but, how less? Does it decrease other than linearly? > Your number will be lower for a couple reasons: >=20 > 1) Decreased mass flow (5% in your case) I guess I did not make it clear. But, in calculating the heat removal, = I used the mass flow for the thicker radiator which was down by 5% - in = other words the mass flow used to calculate the heat removal for the 4" = thick rad was decreased by 5%. So I think that kept the apples with the = apples, at least in this case {:>). > 2) Boundary layer thickness grows as you go deeper > into the radiator. The last 1" of radiator will never > be as effective as the first 1". I agree, the boundary layer thickness grows and the deltaT decreases = both which decrease the heat transfer - but, I'm uncertain as to what = extent. But, There is still metal transferring heat to the air even if = it is less effective. As I mentioned, I have looked for = reports/experiments that would provide a better handle on what thickness = does or does not do. There do not seem to be a whole lot around = explicitedly addressing that question, although almost all the Naca = reports on radiators are referring to radiators that are quick thick by = our standards. If you look at their charts in the back of the report, they are plotting = effectiveness with radiators from 4" to 16". The equations they/I used = are based on imperical data so I don't have an answer other than they = did an experiment and reported the results. Perhaps the answer is = buried in the report and I simply don't have sufficient understand to = recognize it or interpret it correctly. =20 > I know your equations accounted for (1) but not sure > about (2). I haven't checked your reference. Certainly needs to be check by someone. I am not interested in leading = anybody astray - I don't get as many beers when I do that {:>). I'm the = first to admit a severely limited understanding of this stuff, but I = think there are questions regarding these areas (certainly is in my = mind) that folks like you can bring clarity to. Besides that, I could = easily have screwed up these equations in the spreadsheet - but, I did = check against their examples and think they are OK. > The drag increase of 58% sounds way too low. You > increased surface area by 300%. Unless mass flow > decreased a lot (it didn't) or drag coefficient > dropped a lot (it shouldn't), then this can't be > right. Well, there is no change in frontal area between the radiators, so the = old 1/2pV^2*A drag factor remains essentially the same for all - = discounting the small 5% decrease in mass flow which would (by itself) = help decrease the frontal drag some. So the question is would the = increase in skin friction be proportional to the increased internal = surface area ( I would presume it is)? And if it is? Then what is the = absolute amount of drag per square inch based on. Is the internal core = drag a small part or a large part of the overall core drag.???? I know = - it probably depends........ {:>) I know the pressure drop across a radiator is composed of 4 parts. The = initial decrease in flow area (open area/frontal area), the Ventura = Contracta effect as the airflow contracts inside the tube entrance, the = friction drag caused by the viscocity of the air and the tube skin (also = a function of hydraulic diameter), the pressure change as it exits the = core tubes . How does it compare to frontal drag? =20 I imagine it depends on the type of core. Cores with a few large = diameter tubes would probably have less skin drag than one with a number = of smaller diameter tubes. =20 I can send you a copy of the report if you have any trouble finding it = on the web. Thanks again, Ron, for your insight and comments. Ed > Ron >=20 > --- Ed Anderson wrote: >=20 >> But in any case, assuming I did not screw up >> someplace, here are the results I got when I changed >> only the thickness of the radiator on drag, mass >> flow and heat removal: >> 1. The Drag (R) went from 4.28 lbf/ft2 to 6.77 >> lbf/ft2 or a 58% increase presumably due to wall >> friction >>=20 >> 2. The Mass flow (M) decreased from 12.33 >> lbm/ft^2/sec to 11.705 or a 5.13 % decrease =20 >>=20 >> 3. The Heat Rejected(Q) increased from 6.1707 >> HP/100F to 23.66 HP/100F for a 283% increase >=20 >=20 > -- > Homepage: http://www.flyrotary.com/ > Archive and UnSub: = http://mail.lancaironline.net:81/lists/flyrotary/List.html > ------=_NextPart_000_0009_01C82C69.167677B0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
 
 

> Ed,
>
> The most the rejected = heat could=20 change in this case
> is 300%. You made a 300% change in thickness = which
> results in a 300% change in wetted surface = area.
Hi Ron,

The  heat rejection = was  283%,=20 so a bit less than the 300% thickness - how valid? Well, that is = why I=20 posted this to the list.  What should the heat rejection be??  = That is=20 what I am after.  If there is a well-known correlation between = radiator=20 characteristics and heat rejection - what are they.
 
If I look at K&W (page 266) =  they=20 show the heat transfer Coefficient Kh for turbulent flow in a core = as Kh=3D=20 1/ 2*L/D*0.316*(Re)^(-1/4).  Re being proportional to Mass flow=20 velocity.  So If I look at this equation it would appear that heat = transfer=20 goes up proportional to the thickness (L) and decreases  as the = diameter of=20 the tubes hydraulic diameter (D) gets larger.  It appears that = as the=20 velocity (mass flow) decreases the Reynolds number goes down, as it goes = down=20 the 4th root of it decreases it even more and then in the denominator it = makes=20 Kh coefficient higher all implying more heat transfer per unit thickness = of the=20 core.
 
But, then I could be = misinterpreting the=20 factors in the coefficient, but I do not see any factor in the equation = that=20 would indicate this heat transfer coefficient changes as the air flows = through=20 the core.  But clearly  the decreasing deltaT = causing the=20 last part of the thicker's core surface area to be less effective in=20 transferring heat to the air - but, how less?   Does it = decrease other=20 than linearly?
 
> Your number will be lower for a couple=20 reasons:
>
> 1) Decreased mass flow (5% in your=20 case)
I guess I did not make it = clear.  But,=20 in calculating the heat removal, I used the mass flow for the thicker = radiator=20 which was down by 5% - in other words the mass flow used to calculate = the heat=20 removal for the 4" thick rad was decreased by 5%.  So I think that = kept the=20 apples with the apples, at least  in this case = {:>).

> 2) Boundary layer thickness grows as = you go=20 deeper
> into the radiator. The last 1" of radiator will = never
> be=20 as effective as the first 1".
I agree, the boundary layer = thickness grows=20 and the deltaT decreases both which decrease the heat transfer  - = but, I'm=20 uncertain as to what extent.  But, There is still metal = transferring heat=20 to the air even if it is less effective.  As I mentioned, I have = looked for=20 reports/experiments that would provide a better handle on what thickness = does or=20 does not do.  There do not seem to be a whole lot around = explicitedly=20 addressing that question,  although almost all the Naca reports on=20 radiators are referring to radiators that are quick thick by our=20 standards.
 
If you look at their = charts in the=20 back of the report, they are plotting effectiveness with radiators from = 4" to=20 16".  The equations they/I used are based on imperical data so = I don't=20 have an answer other than they did an experiment and reported the = results. =20 Perhaps the answer is buried in the report and I simply don't have = sufficient=20 understand to recognize it or interpret it correctly.  =

> I know your equations accounted for (1) = but not=20 sure
> about (2). I haven't checked your = reference.

Certainly needs to be check  by = someone.  I am not interested in leading anybody astray - I don't = get as=20 many beers when I do that {:>).  I'm the first to admit a = severely=20 limited understanding of this stuff, but I think there are questions = regarding=20 these areas (certainly is in my mind) that folks like you can bring = clarity=20 to.  Besides that,  I could = easily have=20 screwed up these equations in the spreadsheet - but, I did check against = their=20 examples and think they are OK.
 

> The drag increase of 58% sounds way too low. You
> = increased=20 surface area by 300%. Unless mass flow
> decreased a lot (it = didn't) or=20 drag coefficient
> dropped a lot (it shouldn't), then this can't=20 be
> right.
Well, there is no change in frontal area = between the=20 radiators, so the old 1/2pV^2*A drag factor remains essentially the same = for=20 all  - discounting the small 5% decrease in mass flow which would = (by=20 itself) help decrease the frontal drag some.  So the question is = would the=20 increase in skin friction be proportional to the increased internal = surface area=20 ( I would presume it is)?  And if it is? Then what is the absolute = amount=20 of drag per square inch based on.  Is the  internal core=20 drag  a small part or a large part of the overall core = drag.???? =20 I know - it probably depends........ {:>)
 
 
I know the pressure drop across a radiator is = composed=20 of 4 parts.  The initial decrease in flow area (open area/frontal = area),=20 the Ventura Contracta effect as the airflow contracts inside the tube = entrance,=20 the friction drag caused by the viscocity of the air and the tube skin = (also a=20 function of hydraulic diameter), the pressure change as it exits the = core tubes=20 .  How does it compare to frontal drag?  
 
  I imagine it depends on the type of = core. =20 Cores with a few large diameter tubes would probably have less skin drag = than=20 one with a number of smaller diameter tubes.  
 
I can send you a copy of the report if you = have any=20 trouble finding it on the web.
 
Thanks again, Ron,  for your insight and = comments.
 
Ed
 

> Ron
>
> --- Ed Anderson <eanderson@carolina.rr.com> = wrote:
>
>> But in any case, assuming I did not screw=20 up
>> someplace, here are the results I got when I = changed
>>=20 only the thickness of the radiator on drag, mass
>> flow and = heat=20 removal:
>> 1. The Drag (R) went from 4.28 lbf/ft2 to = 6.77
>>=20 lbf/ft2 or a 58% increase presumably due to wall
>>=20 friction
>>
>> 2. The Mass flow (M) decreased from=20 12.33
>> lbm/ft^2/sec to 11.705 or a 5.13 % decrease  =
>>=20
>> 3. The Heat Rejected(Q) increased from 6.1707
>> = HP/100F=20 to 23.66 HP/100F for a 283% increase
>
>
> = --
>=20 Homepage:  http://www.flyrotary.com/
> Archive=20 and UnSub:   http://mail.lancaironline.net:81/lists/flyrotary/List.html
>
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