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heatwave · 07-05-2004, 08:04 PM

Lporc - i'm not so sure how you would connect the radiators in parallel?

The reason I opt for series was the water would subsequently get cooler and cooler as it passes through each radiator. Its pretty much the same effect as having one large radiator but with smaller radiators in series, you'll also have entrance and exit losses - but since I am assuming the pump is able to overcome all flow restrictions.

Angry - I think you are using an incorrect equation there. When we design heat exchangers, we don't use thermal resistance, but overall heat transfer coefficient - U, which has the SI units W/m2.C. The typical heat exchanger is the counterflow of 2 liquids within 2 tubes and these having typical U values of 110-350 for a water-oil system. The heat transfer that occurs in the WC system is convection, not conduction.

For the watercooling system, we are interested in water-air and the typical U values for a "finned-tube heat exchanger, water in tubes, air across tubes" is between 25-55.

I got the value of 2kW but that was an error, it is actually larger as it depended on the:
* water flowrate
* temperature difference

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. But do remember that is assuming it takes 3L/min of water and 30C temperature difference. Now this does not show what material is being used - but these equations are used to design large commercial heat exchangers in power plants where heat removal is in the excess of hundreds of kilowatts.

Lets speculate what materials they could have used:

Diamond = 2300
Silver (pure) = 410
Copper (pure) = 385
Aluminum (pure) = 202
Iron (pure) = 73
Carbon steel (1% C) = 43

Those values are the "thermal conductivities" which is in W/m.C - the higher the number the better the material is to be used in a heat exchanger. For example, it takes 2300W of heat to bring 1m of the material up one degree C.

Diamond and silver are obviously cost prohibitive, iron and carbon steel are cheap but sizewise they would have to be built 3 to 10x the size of copper and aluminum. We all know that copper heatsinks are better than its aluminum counterparts so we can assume that these equations hold for our mini watercooling system.

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.

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.

Albeit I am attempting to solve for only 350W heat dissipation, the surface area would be considerably smaller than above.

Here's an interesting article on a undergraduate project from the chemical engineering department of University of Brunswick in Canada - http://www.unb.ca/che/Undergrad/proposed/radiator.pdf

I have the same book "Heat Transfer by J.P. Hollman" which is an excellent book on heat transfer. I will be investigating this PDF file more closely.....

07-05-2004, 08:04 PM	#7
heatwave Cooling Neophyte Join Date: Jul 2004 Location: Down Under Posts: 11	Lporc - i'm not so sure how you would connect the radiators in parallel? The reason I opt for series was the water would subsequently get cooler and cooler as it passes through each radiator. Its pretty much the same effect as having one large radiator but with smaller radiators in series, you'll also have entrance and exit losses - but since I am assuming the pump is able to overcome all flow restrictions. Angry - I think you are using an incorrect equation there. When we design heat exchangers, we don't use thermal resistance, but overall heat transfer coefficient - U, which has the SI units W/m2.C. The typical heat exchanger is the counterflow of 2 liquids within 2 tubes and these having typical U values of 110-350 for a water-oil system. The heat transfer that occurs in the WC system is convection, not conduction. For the watercooling system, we are interested in water-air and the typical U values for a "finned-tube heat exchanger, water in tubes, air across tubes" is between 25-55. I got the value of 2kW but that was an error, it is actually larger as it depended on the: * water flowrate * temperature difference 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. But do remember that is assuming it takes 3L/min of water and 30C temperature difference. Now this does not show what material is being used - but these equations are used to design large commercial heat exchangers in power plants where heat removal is in the excess of hundreds of kilowatts. Lets speculate what materials they could have used: Diamond = 2300 Silver (pure) = 410 Copper (pure) = 385 Aluminum (pure) = 202 Iron (pure) = 73 Carbon steel (1% C) = 43 Those values are the "thermal conductivities" which is in W/m.C - the higher the number the better the material is to be used in a heat exchanger. For example, it takes 2300W of heat to bring 1m of the material up one degree C. Diamond and silver are obviously cost prohibitive, iron and carbon steel are cheap but sizewise they would have to be built 3 to 10x the size of copper and aluminum. We all know that copper heatsinks are better than its aluminum counterparts so we can assume that these equations hold for our mini watercooling system. 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. 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. Albeit I am attempting to solve for only 350W heat dissipation, the surface area would be considerably smaller than above. Here's an interesting article on a undergraduate project from the chemical engineering department of University of Brunswick in Canada - http://www.unb.ca/che/Undergrad/proposed/radiator.pdf I have the same book "Heat Transfer by J.P. Hollman" which is an excellent book on heat transfer. I will be investigating this PDF file more closely..... Last edited by heatwave; 07-05-2004 at 09:09 PM.