aceprod
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Water cooling tubing size and impact on heat transfer... - 2007/08/03 21:01
I've been trying to find a way to show my appreciation for the help I've been receiving from these forums, so I decided to prepare this post hoping to help fellow enthusiasts. Many times people ask why use 1/2" ID tubing for their water cooling system and not use a 3/8" ID tube which is easier to handle and less bulky. Well, apart from the difference in the head losses of the system there is also an impact on the heat load and temperature of the cooling fluid.
Cross-Sectional Tube Difference...
--- 1/2" (13mm) ID Tube
A1 = (π x d^2)/4 = (π x 13^2)/4 = 132.665 mm^2
--- 3/8" (9mm) ID Tube
A2 = (π x d^2)/4 = (π x 9^2)/4 = 63.585 mm^2
Therefore a 1/2" ID tube has twice the cross-sectional area than a 3/8" ID tube. One very important property of a fluid, as heat transfer examination is concerned is its specific heat capacity. Specific heat capacity (c or C) can be described as the measure of the heat energy (Q) needed to increase the temperature of a unit quantity of a fluid by a certain temperature interval.
A larger amount of heat energy is required to increase the temperature of a high specific heat capacity fluid than a low specific heat capacity one. The reason why water cooling is more effective than air cooling is simply because water has a specific heat capacity of 4.18 Joule/gram.°C while air’s specific heat is 1.012 Joule/gram.°C.
Energy Theory Statements...
--- Q = C x m x ΔT
- Q = Heat energy [joules]
- C = Specific heat capacity of fluid [joules/gram.°C]
- m = Fluid mass [grams]
--- ΔT = T2 – T1
T2 = Final fluid temperature [°C]
T1 = Initial fluid temperature [°C]
Simply by replacing and rearranging, this will give the following understandings around the two formulas around the heat dissipation and energy. Just combining both and making the correct relations and equations.
--- Q = C x m(T2 – T1)
--- T2 = (Q/(C x m)) + T1
Let's try with an example... Assume that the total heat output of the water cooled parts of a PC is 500W and a total length of 1.5 metres of tube is used. The initial temperature of the cooling fluid (ionised water) is 20°C. Calculations for both 1/2" and 3/8" tubes are going to be carried so the difference in the final temperature are established.
Calculations Fluid Volume...
--- 1/2" ID Tube Fluid Volume
- V1 = A1 x L = 0.0001989975 m^3
--- 3.8" ID Tube Fluid Volume
V2 = A2 x L = 0.0000953775 m^3
- Q = 500 W
- T1 = 20°C
- L = 1.5 m
- A1 = 132.665 mm^2 = 0.000132665 m^2
- A2 = 63.585 mm^2 = 0.000063585 m^2
Calculation Final Temperatures...
--- 1/2" ID Tube Temperature
- T2 = (500/(4.18 x 198.9975)) + 20
- T2 = 20.6°C
--- 3/8" ID Tube Temperature
- T2 = (500/(4.18 x 95.3775)) + 20
- T2 = 21.25°C
Considering that the specific weight of water is 1000 Kg/m^3:
- m1 = 0.1989975 Kg = 198.9975 grams
- m2 = 0.0953775 Kg = 95.3775 grams
For some this difference might not be that important but I hope I could help those control freaks like me that like do take in consideration everything and achieve the max possible! Thanks a lot for reading...
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