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Unread 09-12-2005, 01:06 PM   #7
bobo5195
Cooling Savant
 
Join Date: Aug 2005
Location: uk
Posts: 400
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I guess we’ll see what the differences when your proggie can produce the graphs. It depends on the relative magnitude of the performance change. I think the plumbing side might be counteracted by ease of placement. Either way it would be something for the “cooling nutters” because of the extra cost of the second rad.

Good to here that you’ve nearly got the temps working. That makes things 10x more useful and might help me since ive always like low flow but nobody believes me! (I know is should keep that abit quiet around here, else ppl will start hammering me with facts). I would like to see how much temperature in the loop is steady state (ie never gets removed but is needed for heat transfer stake). Fieled for cost would be nice so that things get all added up (possibly iterate to find $$/w for cheapest cooling. Might earn some money off referrals there , although it’s a moon on a stick style idea at the moment.

The concept of a good approximation is less important than I imagined given how general your program is (a[i]*x^N ; equation input). This should make fitting any approximation even if it is not decimal polynomial easy. Fitting the “correct” graph from test data is important however. This is better left to being discussed elsewhere but I am concerned that testing data that can be obtained by “the man in his garden shed “ with an accurate interpolation would be less suitable than just saying the curve is of the form a.x^2 and trying to fit the best approximation through these points.
This is immediately obvious when talking about radiators. Im going to have a look over my heat transfer notes but im sure theres an analytical or empirical derivation of the change in C/W for a radiator under different conditions. Either way it’s a difficult task.

I think I got a method of evaluating a pump flow in parallel but I would like to read up on it. Two pumps (not necessarily similar but operating in the same direction) would work with your formula but pump / resistor calculation is impossible and would not work for any cases. So use with caution, should change to use never.

Quote:
Quote:
Originally Posted by bobo5195
If you would like pressure drop calcs for a pipe I could send you a subroutine with ease.
I can deduce a formulae using an app off the web, but the issue is it doesn't map to testing I found elsewhere. Perhaps silicon tubing is "sticky" to water flow vs the material properties of the apps?
Im not sure what the problem is with your pipe friction. The best method for friction would be http://www.lmnoeng.com/DarcyWeisbach.htm ) or perhaps a little worse but less of computational problem (http://www.lmnoeng.com/HazenWilliamsDesign.htm) . A little calculation of the Reynolds number (see below). Reveals that flow in water cooling pipes is near borderline for lamina/turbulent transition so that is perhaps where the problems are coming from, for a good model to be used some work will need to be carried out so perhaps hazen Williams is useful.

Re=Vl/v

Assume charcteistic length of pipe diameter and characteristic velocity of average fluid velocity V. Dynamic viscosity is 1*10^-6 for water. Using Q in lpm to calculate the velocity gives

Re=21.2*Q/D

For typical water cooling values of 1lpm and 12.5mm it becomes obvious that flow is in the transition range
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