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Originally Posted by heatwave
[b]
I assumed:
* all units are in SI
* flowrate of water = 3 Litres/min = 0.05 kg/s (m)
* specific heat of water = 4180 J/kg.K (Cp)
* delta T = 75-45 = 30C (delta T)
The formula to calculate heat, q = m * Cp * delta T (where q is in Watts)
Therefore with the above values:
q = (0.05) * (4180) * (30)
= 6,270W
= 6.27kW!
That is a very large heat loss.
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Indeed it is. I believe the delta T in this equation should be the difference between inlet and outlet water temperatures, not the delta between starting temp and final temp. 30C for any radiator of the size we talk about, at this flow rate, in air, is not possible. Replace the 30 with 0.3 and you'd be nearer the mark I'd say.
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Originally Posted by heatwave
... they would have to be built 3 to 10x the size of copper and aluminum.
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I'd say not. The proportion of the radiators thermal resistance due to conduction is minimal. As you say, convection dominates.
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To solve for the required cooling surface area the equation q = U * A * Tm is used.
That q is the same as before, the A is the surface area in m2, U is the heat transfer coefficient (see above) and I chose 25 as the worst case scenario, Tm is the log mean temperature difference which is:
Tm = [ (Tw1-Tw2) - (Ta2-Ta1) ] / ln [ (Tw1-Tw2) / (Ta2-Ta1) ]
where:
* Tw1 - initial water temperature (75C)
* Tw2 - final water temperature (45C)
* Ta1 - initial air temperature (35C)
* Ta2 - final water temperature (45C)
solve for Tm = 18.2C
I chose an initial air temperature at 35C because this is pretty much as hot a typical room gets in the summer period. The two 45C shows the temperature I expect the CPU to be operating at under maximum load.
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An operating CPU is never at water temperature. It is at a temperature determined by the thermal resistance of the block in C/W, the power output of the CPU, the water temperature and a host of other variables. Think 15C above water T. Typically-ish for a good block.
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Rearranging the equation to solve for A:
A = q / (U * Tm)
= 6270 / (25 * 18.2)
= 13.8m2
Now this example shows how the equations work and relate to one another.
NOTE:
Don't be mistaken that this WC system assumes the CPU is at 75C! Its not. I assumed that in a stagnant system, the CPU at such high speed will be heating the water to 75C. Thus at 3L/min water flowrate and with a 14m2 surface area, the water temperature will drop to 45C and stay so in a steady state operation.
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If are saying that ALL the water in the system was pre-heated to 75C in a stagnant system then OK. Maybe I'm not reading this correctly but I think your assumptions are not quite right. Aren't you just plugging random numbers into formulae?