Return-Path: Received: from [24.25.9.102] (HELO ms-smtp-03-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 4.1.8) with ESMTP id 2901596 for flyrotary@lancaironline.net; Mon, 22 Dec 2003 21:22:56 -0500 Received: from o7y6b5 (clt78-020.carolina.rr.com [24.93.78.20]) by ms-smtp-03-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id hBN2Miow021799 for ; Mon, 22 Dec 2003 21:22:53 -0500 (EST) Message-ID: <002601c3c8fb$4689fa60$1702a8c0@WorkGroup> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: K&W Cooling factors. Date: Mon, 22 Dec 2003 21:19:56 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0023_01C3C8D1.595A0E20" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2800.1106 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2800.1106 X-Virus-Scanned: Symantec AntiVirus Scan Engine This is a multi-part message in MIME format. ------=_NextPart_000_0023_01C3C8D1.595A0E20 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable =20 ----- Original Message -----=20 From: Al Gietzen=20 To: Rotary motors in aircraft=20 Sent: Monday, December 22, 2003 8:43 PM Subject: [FlyRotary] Re: K&W Cooling factors. Its fairly clear that air flow and core area are two of the most = significant factors as together they determine the basic Mass flow = possible through the cooler. If that is not sufficient to carry away = the heat you are trying to reject then all else is moot.=20 In K&W there are two coefficients they focus on, Kh the heat = transfer coefficient which is dependent on the characteristics of the = core, Kp the pressure drop coefficient - again based on core = characteristics. They end up with a combined coefficient that rates how = well the core transfers heat based on the pressure drop this is Kv =3D = Kh/Kp. These coefficients are dependent on such things as the openness = ratio, the hydraulic diameter of the individual passage in the core and = the thickenss of the core. Both Kh and Kp are also weakly dependent on = the Reynolds number of the air flow throught the core. I say weakly = because they relate to the inverse fourth power of the Reynolds number = or 1/(Re)^(1/4). Interestingly enough because it is an inverse relationship the = heat transfer is better to the air with lower velocity through the core. = Not quite certain that I fully understand why, except it appears to do = with the shear force and friction between the air flow in the core = passage and the the passage walls. However, while heat is apparently = passed from the core walls to the air better at the lower Reynolds = numbers, the downside is that low Reynold number also means lower air = velocity throught the core which in turn means Low mass flow which is = not good for cooling. =20 So from the equations in K&W that there is an optimum balance = between the heat transfer due to the lower Reynolds effect and the heat = removed due to higher mass flow through the core. Too much of one and = not enough of the other and your cooling is suboptimal. FWIW The Reynolds number (a dimensionless parameter) includes = "characteristic length" of the channel, so although the heat transfer = coefficient is increasing with increased velocity, the "residence time" = of the air is decreasing, meaning less time for the air to heat up. But = the =BC power of the Re in the demoninator is only one piece of the = equation (isn't it?). Somewhere in the numerator is the mass flow = (velocity) which wins out every time. For a given core, higher velocity = removes more heat. But of course, there are other limits on velocity. Or am I missing something. It's been a long time since I worked on = details of convective heat transfer. Al According to K&W the heat-transfer coefficient Kh and pressure-drop = coefficient Kp both decrease with increase in air velocity through the = core. However, even though these coefficients decrease, the overall = heat removal may increase (up to a point) due to the higher mass flow = through the core because of the higher air velocity. So as I understand = it lower velocity throught the core increases the coefficients, but = decrease the mass flow while higher velocity decreases the coefficients = but increase the mass flow. So from what I think I understand it = appears that there is a theoretical optimum balance between keeping the = value of these important coefficients sufficient high by limiting the = velocity through the core, but keeping the mass flow sufficiently high = by sizing the area of the core so that you removed the desired quantity = of heat. Again, a lot of compromise. The equation given for Kp=3D (1/s)*(L/D)*(0.316/(Re)^(1/4)) and Kh = =3D 1/2*(L/D)*(0.316/(Re)^(1/4)). With Re =3D VcD/sv with s being = the openness ratio, L (thickness of Core) and D the Hydraulic Diameter, = Vc being velocity at the face of the core. So as you mention the = pressure drop coefficent goes up with increased thickenss (L) but down = as the diameter (D) of the passage is increase and also down as the = velocity increases (Re increases). Also the more open (s =3Dlarger), = the less pressure drop. The equation given for heat transfer (the thing we are really = interested in)=20 Q =3D pVcAcCp(Tw-Ti)(1 - e^(-Kh). Which has the all important mass = flow factor pVcAc, the temp difference between wall and air and the = anitlog of heat transference coefficience Kh. Makes my head hurt. Ed=20 . ------=_NextPart_000_0023_01C3C8D1.595A0E20 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
 
----- Original Message -----
From:=20 Al = Gietzen=20
Sent: Monday, December 22, 2003 = 8:43=20 PM
Subject: [FlyRotary] Re: = K&W Cooling=20 factors.

 

Its = fairly clear=20 that air flow and core area are two of the most significant = factors as=20 together they determine the basic Mass flow possible through the=20 cooler.  If that is not sufficient to carry away the heat you = are=20 trying to reject then all else is moot.

 

In = K&W there=20 are two coefficients they focus on, Kh the = heat transfer=20 coefficient which is dependent on the characteristics of the core,=20 Kp the = pressure drop=20 coefficient - again based on core characteristics.  They end up = with a=20 combined coefficient that rates how well the core transfers heat = based on=20 the pressure drop this is Kv =3D = Kh/Kp. =20 These coefficients are dependent on such things as the openness = ratio, the=20 hydraulic diameter of the individual passage in the core and the = thickenss=20 of the core.  Both Kh and=20 Kp=20 are also weakly dependent on the Reynolds = number=20 of the air flow throught the core.  I say weakly because they = relate to=20 the inverse fourth power of the Reynolds number or=20 1/(Re)^(1/4).

 

  = Interestingly enough because it is an inverse relationship the heat = transfer=20 is better to the air with lower velocity through the core.  Not = quite=20 certain that I fully understand why, except it appears to do with = the shear=20 force and friction between the air flow in the core passage and the = the=20 passage walls.  However, while heat is apparently passed from = the core=20 walls to the air better at the lower Reynolds numbers, the downside = is that=20 low Reynold number also means lower air velocity throught the core = which in=20 turn means Low mass flow which is not good = for=20 cooling. 

 

So = from the=20 equations in K&W that there is an optimum balance between the = heat=20 transfer due to the lower Reynolds effect and the heat removed due = to higher=20 mass flow through the core. Too much of one and not enough of the = other and=20 your cooling is suboptimal.

 

FWIW

 

 

The = Reynolds=20 number (a dimensionless parameter) includes =93characteristic = length=94 of the=20 channel, so although the heat transfer coefficient is increasing = with=20 increased velocity, the =93residence time=94 of the air is = decreasing, meaning=20 less time for the air to heat up.  But the =BC power of the Re = in the=20 demoninator is only one piece of the equation (isn=92t it?).  = Somewhere=20 in the numerator is the mass flow (velocity) which wins out every=20 time.  For a given core, higher velocity removes more = heat.  But=20 of course, there are other limits on velocity.

 

Or am I = missing=20 something.  It=92s been a long time since I worked on details = of=20 convective heat transfer.

 

Al

 

According to=20 K&W the heat-transfer coefficient Kh and = pressure-drop=20 coefficient Kp both decrease = with=20 increase in air velocity through the core.  = However,=20 even though these coefficients = decrease,=20 the overall heat removal may increase (up to a = point) due=20 to the higher mass flow through the core because of the higher air = velocity.=20 So as I understand it lower velocity throught the core increases the = coefficients, but decrease the mass flow while higher velocity = decreases the=20 coefficients but increase the mass flow.  So from what I think = I=20 understand it appears that there is a theoretical optimum balance = between=20 keeping the value of these important coefficients sufficient high by = limiting the velocity through the core, but keeping the mass flow=20 sufficiently high by sizing the area of the core so that you removed = the=20 desired quantity of heat.  Again, a lot of=20 compromise.

 

The = equation=20 given for Kp=3D (1/s)*(L/D)*(0.316/(Re)^(1/4))=20 and Kh =3D 1/2*(L/D)*(0.316/(Re)^(1/4)).  =  =20 With Re =3D VcD/sv  = with s being the openness ratio, L = (thickness of=20 Core) and D the Hydraulic Diameter, Vc being velocity at the face of = the=20 core.  So as you mention the pressure drop coefficent goes up = with=20 increased thickenss (L) but down as the diameter (D) of = the=20 passage is increase and also down as the velocity increases (Re=20 increases).  Also the more open (s=20 =3Dlarger),  the less pressure = drop.

 

The = equation=20 given for heat transfer (the thing we are really interested in)=20

 

Q=20 =3D pVcAcCp(Tw-Ti)(1 - e^(-Kh).  = Which has the=20 all important mass flow factor pVcAc, the temp = difference=20 between wall and air and the anitlog of  heat transference = coefficience=20 Kh.

 

 

Makes = my head=20 hurt.

 

Ed=20

 

 .

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