X-Virus-Scanned: clean according to Sophos on Logan.com X-SpamCatcher-Score: 64 [XX] (100%) BODY: contains text similar to "low payment" Return-Path: Received: from ispmxmta09-srv.windstream.net ([166.102.165.170] verified) by logan.com (CommuniGate Pro SMTP 5.1.4) with ESMTP id 1733571 for flyrotary@lancaironline.net; Wed, 03 Jan 2007 14:26:25 -0500 Received-SPF: pass receiver=logan.com; client-ip=166.102.165.170; envelope-from=montyr2157@alltel.net Received: from ispmxaamta04-gx.windstream.net ([162.40.142.190]) by ispmxmta09-srv.windstream.net with ESMTP id <20070103192528.CRWZ9039.ispmxmta09-srv.windstream.net@ispmxaamta04-gx.windstream.net> for ; Wed, 3 Jan 2007 13:25:28 -0600 Received: from Thorstwin ([162.40.142.190]) by ispmxaamta04-gx.windstream.net with SMTP id <20070103192527.PEBS7955.ispmxaamta04-gx.windstream.net@Thorstwin> for ; Wed, 3 Jan 2007 13:25:27 -0600 Message-ID: <006301c72f6c$fd1e8e00$01fea8c0@Thorstwin> From: "M Roberts" To: "Rotary motors in aircraft" Subject: Ideal cooling Date: Wed, 3 Jan 2007 13:25:55 -0600 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0060_01C72F3A.B2679550" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.3028 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3028 This is a multi-part message in MIME format. ------=_NextPart_000_0060_01C72F3A.B2679550 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Just so that we are all on the same page here. First the disclaimer: The following assumes that you are flying a subsonic airplane where = compressibility effects can be neglected, and that you are not putting = out 1000+hp. I think these are fairly safe assumptions. The other thing = is it must be steady state with stabilized readings relative to time. = The math to look at this case is really very simple. If you can pass = algebra, you can do the math for this problem. Just keep the units = consistent and use absolute temps for calculating DT (deg Kelvin or = Rankine depending on metric or english units) The job of the cooling system is to remove enough waste heat to keep the = engine below its maximum temp in all operating ranges. This must be done = while keeping the weight and drag to a minimum.=20 The conflict is always between hot day climb and cruise efficiency.=20 The airplane spends most of its time in cruise, therefore cruise = efficiency is the most important design point.=20 Hot day climb only happens once in a while. It is the secondary concern. = Adding a spray bar or some other band aid to get past this one design = point is a perfectly logical and acceptable solution. It is not = "cheating".=20 In steady state cruise there is some heat flux Q from the engine that = must be disposed of to keep the engine at its desired operating temp. = There is no "optimum" coolant delta T (DT)across the radiator for all = conditions to accomplish this. There is only ONE coolant DT for any = given steady state operating point that will achieve this and it is = found by the equation Q =3D Mdot x Cp x DT Q =3D The heat flux that must be removed Mdot =3D mass flow rate of the water per unit time Cp =3D The specific heat of the coolant (how much heat each unit of mass = of the coolant absorbs per unit of temp increase)=20 DT =3D T1-T2 The change in temp of the water across the radiator (must = be equal to the water DT across the engine block for steady state = operation) Now for the optimum part: The absolute best possible performance would be a system that used just = enough air Mdot to heat the cooling air to the same temperature as the = water radiator exit temp ( water going back to the engine from the = radiator). This condition is impossible to achieve in practice. So you = try to get as close as you can. How close you get is called the heat = exchanger effectiveness. The closer the air and water exit temps are, = the better the effectiveness. The air side is also governed by the same equation:=20 Q =3D Mdot x Cp x DT =20 This is because Qwater =3D Qair for steady state operation The Cp for air is much less than for water. The density of water is also = much greater than air. Because of this the water DT will always be much = less than the Air DT =20 The optimum condition is to minimize Mdot air (thus drag) for a given = heat exchanger effectiveness to get the required Q. This results in the = smallest cooling system and the smallest drag/weight in cruise. The catch is you must oversize the system just enough so that a cowl = flap can increase Mdot air enough in hot day climb to adequately cool = the engine. In this regime there will be more drag and less Air DT. Who = cares! Cruise is where its at. Transient operation can only be analyzed using differential equations, = or piecewise analysis and a computer code. Climb is transient in nature, = that is the problem with trying to analyze it. The transient nature of = climb is why the thermal mass of the engine and coolant help us. It = absorbs heat as it comes up to temp, This is all Q that the rad does not = have to reject to the air. Taking off with the engine already at red = line temp is a different matter entirely. In this case you have no big = heat sink to help you. So use a lower rate of climb, a higher speed, = open cowl flaps and a spray bar. Or better yet, let the engine cool = before you take off then do all those things. For my design I prefer a light low drag installation that is optimized = for cruise with some Band-Aids to get me past the once in a while = condition that I rarely see.=20 If you are designing a glider tug or a STOL plane for use in North = Africa, you will have to adjust your design accordingly. Monty ------=_NextPart_000_0060_01C72F3A.B2679550 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Just so that we are all on the same = page=20 here.
 
First the disclaimer:
 
The following assumes that you are = flying a=20 subsonic airplane where compressibility effects can be neglected, and = that you=20 are not putting out 1000+hp. I think these are fairly safe = assumptions. The=20 other thing is it must be steady state with stabilized readings relative = to=20 time. The math to look at this case is really very simple. If you can = pass=20 algebra, you can do the math for this problem. Just keep the units = consistent=20 and use absolute temps for calculating DT (deg Kelvin or Rankine = depending on=20 metric or english units)
 
The job of the cooling system is to = remove enough=20 waste heat to keep the engine below its maximum temp in all operating = ranges.=20 This must be done while keeping the weight and drag to a minimum. =
 
The conflict is always between hot day = climb and=20 cruise efficiency.
 
The airplane spends most of its time in = cruise, therefore cruise efficiency is the most important design = point.=20
 
Hot day climb only happens once in a = while. It is=20 the secondary concern. Adding a spray bar or some other band aid to get = past=20 this one design point is a perfectly logical and acceptable solution. It = is=20 not "cheating". 
 
In steady state cruise there is some = heat flux Q=20 from the engine that must be disposed of to keep the engine at its = desired=20 operating temp. There is no "optimum" coolant delta T (DT)across the=20 radiator for all conditions to accomplish this. There is = only ONE=20 coolant DT for any given steady state operating point that will achieve = this and=20 it is found by the equation
 
Q =3D Mdot x Cp x DT
 
Q =3D The heat flux that must be = removed
 
Mdot =3D mass flow rate of = the water per=20 unit time
 
Cp =3D The specific heat of the coolant = (how much=20 heat each unit of mass of the coolant absorbs per unit of temp = increase)=20
 
DT =3D T1-T2 The change in temp of = the water=20 across the radiator (must be equal to the water DT across=20 the engine block for steady state operation)
 
Now for the optimum part:
 
The absolute best possible performance = would be a=20 system that used just enough air Mdot to heat the cooling = air to the same temperature as the water radiator exit temp ( = water=20 going back to the engine from the radiator). This condition is = impossible to=20 achieve in practice. So you try to get as close as you can. How close = you get is=20 called the heat exchanger effectiveness. The closer the air and = water exit=20 temps are, the better the effectiveness.
 
The air side is also governed by = the same=20 equation:
 
Q =3D Mdot x Cp x = DT  
 
This is because Qwater =3D Qair for = steady state=20 operation
 
The Cp for air is much less than for = water.=20 The density of water is also much greater than=20 air.  Because of this the water DT will always be = much less=20 than the Air DT
 
The optimum condition is to minimize = Mdot air (thus=20 drag)  for a given heat exchanger effectiveness to get the = required Q.=20 This results in the smallest cooling system and the smallest drag/weight = in=20 cruise.
 
The catch is you must oversize the = system just=20 enough so that a cowl flap can increase Mdot air enough in hot day climb = to=20 adequately cool the engine. In this regime there will be more drag and = less Air=20 DT. Who cares! Cruise is where its at.
 
Transient operation can only be = analyzed using=20 differential equations, or piecewise analysis and a computer code. Climb = is=20 transient in nature, that is the problem with trying to analyze it. The=20 transient nature of climb is why the thermal mass of the engine and = coolant=20 help us. It absorbs heat as it comes up to temp, This is all Q that = the rad=20 does not have to reject to the air. Taking off with the engine = already at=20 red line temp is a different matter entirely. In this case you have no = big heat=20 sink to help you. So use a lower rate of climb, a higher speed, open = cowl flaps=20 and a spray bar. Or better yet, let the engine cool before you take off = then do=20 all those things.
 
For my design I prefer a light low drag = installation that is optimized for cruise with some Band-Aids to get me = past the=20 once in a while condition that I rarely see.
 
If you are designing a glider tug or a = STOL plane=20 for use in North Africa, you will have to adjust your design=20 accordingly.
 
Monty
 
 
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