Mailing List flyrotary@lancaironline.net Message #4900
From: Ed Anderson <eanderson@carolina.rr.com>
Subject: Re: [FlyRotary] Re: K&W Cooling factors.
Date: Mon, 22 Dec 2003 21:19:56 -0500
To: Rotary motors in aircraft <flyrotary@lancaironline.net>
 
----- Original Message -----
From: Al Gietzen
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.

 

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 = 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. 

 

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 ¼ 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= (1/s)*(L/D)*(0.316/(Re)^(1/4)) and Kh = 1/2*(L/D)*(0.316/(Re)^(1/4)).    With Re = 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 =larger),  the less pressure drop.

 

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

 

Q = 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

 

 .

Subscribe (FEED) Subscribe (DIGEST) Subscribe (INDEX) Unsubscribe Mail to Listmaster