Hier zijn een aantal leuke quote's uit een
soortgelijk topic op overclockers.com van een aantal jaren terug... Have fun reading!
OK, specific heat (Cp) is the measure of how much energy it takes to raise the temperature of 1kg of material 1degree (C or K). The Cp for copper is 385 Joules/kg*K and for aluminum is 903 J/kg*K (at 300K = 23 deg. C). density for both is copper: 8933 kg/m^3 and alum: 2702 kg/m^3
So, if we have two same SIZED blocks of material, say 1m x 1m x 1m for simplicity, we have different masses, namely 2700kg of alum. and 8930 kg Cu. if we put 2.438 x 10^6 Joules of energy into each block, we can calculate the temp raise:
Q = m*Cp*deltaT --> deltT = Q/(m*Cp)
deltaT alum: = 2.4381 x 10^6 J/(2700 kg * 903 J/kg*K) =1 degree
deltaT Cu: = 2.4381 x 10^6 J /(8933 kg * 385 J/kg*K) = 0.71 degree
so for the same SIZE blocks, more heat is energy is required to heat the copper by the same delta T than the aluminum. In other words from the moment you turn your computer on it will take a longer time for a copper heatsink to reach steady state conditions.
Now since these blocks have the same size they have the same surface area (that's what we're comparing, right - same size/shape/etc... heatsinks)
Now, that is really transient heat transfer, where there is a buildup of heat in one of the components of the system (the heatsink). At steady state, BOTH Al. and Cu. heatsinks will stabilize at a certain temperature and the heat per unit time (J/s = Q_dot) will be exactly equal to the heat leaving the heatsink per unit time. It will just take a longer time for the Cu one to reach this point.
Now we're talking steady state:
2 processes going on:
(1) is conduction of heat through the heatsink
(2) is convection of heat from surface of heatsink to air
for conduction: Q_dot = (kAc/L)*(Tdie - Tsurface) <-- equation (1)here Q_dot is joules/second or Watts, k is the thermal conductivity of the material in Watts/meter*K, Ac is cross sectional area through the heatsink in meters^2, L is the lenth between Tdie and Tsurface in meters, Tdie is temperature at the die or temperature at the bottom of the heatsink (here we'll consider them equal) and Tsurface is the temp. at the surface of the heatsink, where the air contacts it. (temps in degrees C or K)
for convection: Q_dot = hA(Tsurface - Tair) <-- equation (2)
where A is the area of the surface of the heatsink (giving heat to the air) and h is the convection coefficient and has units Watts/meters^2*K.
from equation (1): Tdie = Q-dot * L/(k*Ac) + Tsurface
from equation (2): Tsurface = Q_dot/(h*A) + Tair
combining: Tdie = Q_dot*L/(k*Ac) + Q_dot/(h*A) + Tair
Taking the Tair, Across section of the heatsinks, A surface of the heatsinks, L length of the heat must travel between the die and surface, and Q_dot to be constant and the same in both cases (Ac, A, L wil be same because we are comparing the same heatsinks just of different materials, but the physical dimensions are the same, Q_dot is constant because the cpu puts out a fixed amount (ideally) say xxxWatts under load, and Tair is constant for it to be a fair comparison)
k for alum = 237 W/m^2*K
k for Cu = 401 W /m^2*K
--->You can see from the resulting equation that to make Tdie lower the only thing you can change is to make k, the thermal conductivity bigger - which is why COPPER WILL ALWAYS YIELD LOWER TEMPS FOR THE STEADY STATE HEAT TRANSFER.
Now, one more thing - after you turn your computer off, there is no more heat being put into the heatsink, and it is a possibility that the aluminum might cool down faster, but I'm not convinced of that yet. But who really cares then - the computer is off. All we really care about is steady state, which has been shown that copper will always yield lower die temps that for aluminum. (in the same situation, same mounting, same design heatsinks, etc....)
If we increase the amount of air blowing of the heatsink, it will lower the (convective) thermal resistance, that is it will lower the resistance of the heat to go from the metal into the air. Now suppose we blow infinite CFM's over the heatsinks, this would be equivalent to saying that the convective resistance is zero.
Now the limiting factor is the resistance of the heatsinks themselves, and of course copper has a lower conductive resistance (for those into equations, conduction resistance R = L/kA where L is the one dimensional length the heat is traveling through, A is the constant cross sectional area available, and k is the thermal conductivity; also convective resistance is R = 1/hA where h is the convection coefficient and A is the area exposed to the convection medium).
No. It has been established that at STEADY STATE heat transfer, which is ideally (and practically pretty close) what we are dealing with in CPU cooling, copper is ALWAYS better.
The ONLY time aluminum would "dissipate" heat faster is under transient (non-steady) conditions.
Conclusie:
Aluminium koelt wel sneller af dan koper, maar dat is voor de echte high-end niet interessant. Die willen dat de temperatuur onder 100% constante load zo stabiel mogelijk blijft. Echter is het in onze gevallen wél zo interessant om te gebruiken aangezien onze pc's in de meeste gevallen niet constant aan het 'crunchen' zijn. Dat wil zeggen, onze koeler mag wel wat sneller wat stiller worden als de pc gewoon teruggaat naar idle (vanuit een game).
Koper zou theoretisch dus de stabielere temperatuur kunnen leveren, maar is voor ons niet interessant. De redenen voor fabrikanten om daar niet / minder mee te werken zijn gewoon:
- Aluminium is goedkoper, en dus goedkoperen koelers voor ons
- Aluminium is lichter, en da's een belangrijke factor in gebruik
- Koper heeft een veel hoger smeltpunt, wat ook bewerking ervan moeilijker / duurder maakt voor de fabrikanten.
Aangezien ze wel weten dat koelblokken die 90 euro kosten en 2 kg wegen niet gaan verkopen, wordt koper achterwege gelaten. Voor de wat meer high-end workstations en crunchbakken is het dan wel weer interessanter en kunnen de kosten makkelijker gedrukt worden. Maar da's een vrij niche markt, dus sowieso zullen we dat niet veel zien
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