Pro/Forums - View Single Post - Good article on effect of radiator size in watercooling

myv65 · 07-22-2002, 08:35 AM

Just remember, you asked for this stuff BigBen2k,

O.K. Here's the quick run down on fin effectiveness and efficiency equations.

Individual fin effectiveness is defined as the amount of heat transferred off the fin versus the amount that would be transferred off the portion of the base connected to the fin if the fin didn't exist. If we call fin heat transfer "qf" and heat transfer potential from the base equals h * A * delta-T then fin effectiveness equals qf / (h * A * delta-T). "h" is the convection coefficient, A is the area at the fin base, and delta-T is the temperature differential from base to ambient air.

Fin efficiency is defined as the heat transfer off the fin versus the heat transfer that would leave the fin if the entire fin was the same temperature as the base material. This is like saying that the fin has a conduction coefficient of infinity.

There are different equations that apply to fins based upon how you wish to model them. These variations include convection from the fin tip, no heat transfer from the fin tip, a given specific temperature at the fin tip, and an infinitely long fin. The easiest equation is for the infinite fin where qf = (h * P * k * A)^(.5) * delta-T. Here the additional terms are "P" for fin perimeter (approximated as 2 * width for very thin, flat fins) and "k" is fin conduction coefficient. This approximation actually fits "real world" fins of finite length reasonably well.

A better approximation comes from assuming that the fin is of finite length with no heat transfer from the tip. The effective length of the pin (Leq) may be considered as actual fin length + 2 * fin thickness. The effective surface area (Ap) becomes Leq * fin thickness. In this case, fin efficiency boils down to:

Leq^(3/2) * (h / k * Ap)^(1/2)

Things begin getting really fun when you start to consider arrays of fins (ala radiators). You can define an overall efficiency to equal actual heat transfer divided by potential heat transfer if all the fins matched the temperature of the base material.

I'll skip ahead and spit out the equation for overall efficiency. It is:

1 - Af / At * (1 - fin efficiency)

Where:

"Af" is total fin area
"At" is total area of fins and exposed base material
"fin efficiency" is a single fin's efficiency based on the earlier equations.

Slapping all this stuff down in this form is only so much gibberish without all the supporting material that goes with it. If you really want to get into this sort of stuff, I'd suggest picking up a heat transfer book by Frank Incropera and David DeWitt. Along with a little calculus experience, you'd be well on your way.

07-22-2002, 08:35 AM	#9
myv65 Cooling Savant Join Date: May 2002 Location: home Posts: 365	Just remember, you asked for this stuff BigBen2k, O.K. Here's the quick run down on fin effectiveness and efficiency equations. Individual fin effectiveness is defined as the amount of heat transferred off the fin versus the amount that would be transferred off the portion of the base connected to the fin if the fin didn't exist. If we call fin heat transfer "qf" and heat transfer potential from the base equals h * A * delta-T then fin effectiveness equals qf / (h * A * delta-T). "h" is the convection coefficient, A is the area at the fin base, and delta-T is the temperature differential from base to ambient air. Fin efficiency is defined as the heat transfer off the fin versus the heat transfer that would leave the fin if the entire fin was the same temperature as the base material. This is like saying that the fin has a conduction coefficient of infinity. There are different equations that apply to fins based upon how you wish to model them. These variations include convection from the fin tip, no heat transfer from the fin tip, a given specific temperature at the fin tip, and an infinitely long fin. The easiest equation is for the infinite fin where qf = (h * P * k * A)^(.5) * delta-T. Here the additional terms are "P" for fin perimeter (approximated as 2 * width for very thin, flat fins) and "k" is fin conduction coefficient. This approximation actually fits "real world" fins of finite length reasonably well. A better approximation comes from assuming that the fin is of finite length with no heat transfer from the tip. The effective length of the pin (Leq) may be considered as actual fin length + 2 * fin thickness. The effective surface area (Ap) becomes Leq * fin thickness. In this case, fin efficiency boils down to: Leq^(3/2) * (h / k * Ap)^(1/2) Things begin getting really fun when you start to consider arrays of fins (ala radiators). You can define an overall efficiency to equal actual heat transfer divided by potential heat transfer if all the fins matched the temperature of the base material. I'll skip ahead and spit out the equation for overall efficiency. It is: 1 - Af / At * (1 - fin efficiency) Where: "Af" is total fin area "At" is total area of fins and exposed base material "fin efficiency" is a single fin's efficiency based on the earlier equations. Slapping all this stuff down in this form is only so much gibberish without all the supporting material that goes with it. If you really want to get into this sort of stuff, I'd suggest picking up a heat transfer book by Frank Incropera and David DeWitt. Along with a little calculus experience, you'd be well on your way.