=

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 =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.

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. =A0But the =BC power of the Re in the = demoninator is only one piece of the equation (isn’t it?).=A0 Somewhere in the numerator is the mass flow (velocity) which wins out every time.=A0 For = a given core, higher velocity removes more heat.=A0 But of course, there are = other limits on velocity.

Or am I = missing something.=A0 It’s been a long time since I worked on details of = convective heat transfer.

Al