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General Liquid/Water Cooling Discussion For discussion about Full Cooling System kits, or general cooling topics. Keep specific cooling items like pumps, radiators, etc... in their specific forums.

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Unread 11-18-2003, 11:09 AM   #26
Since87
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Quote:
Originally posted by georgeteo
All scenarios assuming same components, length of tubing, etc.. Calculations based on the following:
Pump pressure = 10
Calculating comparative numbers based on a fixed dP across the system isn't too relevant to systems using centrifugal pumps.

And see my last post about incorrect parallel resistance equation.
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Unread 11-18-2003, 01:17 PM   #27
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All these electrical analogies are making this thread harder to follow for me. Why don't you all just express things as equivalent lengths so that I can join in on the fun?
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Unread 11-18-2003, 01:45 PM   #28
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Quote:
Originally posted by Skulemate
All these electrical analogies are making this thread harder to follow for me. Why don't you all just express things as equivalent lengths so that I can join in on the fun?
"Equivalent length"? What's that?

Seriously. How would you calculate characteristic for parallel pipes using equivalent lengths?

N parallel pipes of equivalent length X = 1 pipe of equivalent length X/N^2 ?

A gross simplification due to neglecting the splitting component(s), I know.
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Unread 11-18-2003, 01:57 PM   #29
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You first characterize each leg of the flow as a length of pipe or tubing that has the same resistance curve as the fittings, blocks or whatever else you are dealing with. Once this is done you can represent both legs together as an equivalent length of tubing in order to solve for the flow rate (and Dp) for your given pump. After this is done you can back-calculate to find the flow through each leg of the setup, and for the resistances of each component.

If this brief explanation doesn't help you understand, I can work out a brief sample tonight if you like.
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Unread 11-18-2003, 02:16 PM   #30
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I think equivalent length is a tool used by process engineers to do quick calculations of the pressure drop of a pipe system.
For example, you can make basic assumptions about the pressure drop of a meter of 2 inch pipe since this is a common diameter in industry and there is tons of empirical data. Bends, T-junctions etc would all have a resistance that can loosly be equated to the pressure drop a certain length of pipe would. A bend might for example have the same resistance as 5 meters of pipe.
I think it might a good system for quick estimation, since all you need to do is look at how many meters of pipe you have, add up the various bends and you can quickly get a rough pressure drop value.
This would realy be all you need to size a centrifugal pump.

This thread is beginning to look like the confused fluid dynamists convention, however Since87 has cleared everything up with his last posts. Shame it took a while to get there though because it's probably caused unnecessary confusion.

However to add to it all, fluid dynamics is very similar to electrics with some very important differences. The biggest is that flow resistance is proportional to the square of the flow, whereas resistance effects electrical current in a linear relationship (directly proportional).
This is because the fluid is moving and has a kinetic energy which is expressed as mass * velocity^2/2, which is recognisable from basic physics. I.e water hitting a pipe bend twice as fast will have a 4x impact energy on the bend, which naturally manifests itself as a resistance to flow which is 4 times greater.

Last edited by Ewan; 11-18-2003 at 02:43 PM.
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Unread 11-18-2003, 02:21 PM   #31
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You are correct Ewan, using equivalent lengths typically yields a rough solution, though for what I was taught to do it is typically good enough. Keep in mind that I am not an expert on fluids by any means (I am a student civil engineer specializing in concrete materials) though I have had some formal training in the subject. In any case, there are more exact methods than using equivalent lengths, but for what we are doing I would question whether or not the more complex solutions are warranted.
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Unread 11-18-2003, 05:23 PM   #32
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Quote:
Originally posted by Skulemate
You first characterize each leg of the flow as a length of pipe or tubing that has the same resistance curve as the fittings, blocks or whatever else you are dealing with. Once this is done you can represent both legs together as an equivalent length of tubing in order to solve for the flow rate (and Dp) for your given pump. After this is done you can back-calculate to find the flow through each leg of the setup, and for the resistances of each component.

If this brief explanation doesn't help you understand, I can work out a brief sample tonight if you like.
If you don't mind, I would like to see an example for multiple devices in parallel. Particularly, dissimilar devices in parallel.

Thanks.
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Unread 11-18-2003, 05:31 PM   #33
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How many legs would you like? Two or three (or more?)?
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Unread 11-18-2003, 05:59 PM   #34
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Quote:
Originally posted by Skulemate
How many legs would you like? Two or three (or more?)?
Three legs of with three different length equivalents would be great.
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Unread 11-18-2003, 06:01 PM   #35
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Now I truely understand the crudeness and errors in my calculations ... always good to learn something new eh?!?
<--Says here that I'm a noob
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Unread 11-19-2003, 01:19 PM   #36
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Another question. Would it be accurate in saying the most amount of pressure in the system is right out of the outlet of the pump?

If so, would it not give you a small advantage with center inlet blocks to go from the pump directly to the block? Instead of having the pressure drop from the rad first then the block?

:shrug:

Last edited by jaydee116; 11-19-2003 at 01:35 PM.
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Unread 11-19-2003, 01:34 PM   #37
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to your first question. Yes, the highest pressure in the circuit is at pump outlet.

to your second and third questions. No, it is the flow rate that determines the performance of the components. The higher the flow rate the better the performance.

It is the total pressure drop in your circiut that determines the flowrate (for a given pump).
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Unread 11-19-2003, 01:38 PM   #38
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Quote:
Originally posted by WAJ_UK
to your first question. Yes, the highest pressure in the circuit is at pump outlet.

to your second and third questions. No, it is the flow rate that determines the performance of the components. The higher the flow rate the better the performance.

It is the total pressure drop in your circiut that determines the flowrate (for a given pump).
Yeah I am totaly lost still I see....

If the inlet of the block is getting the highest amount of pressure then doesn't that mean the water will "press" harder against the block and lower the boundry layer creating better heat transfer?
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Unread 11-19-2003, 02:31 PM   #39
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I'm afraid it doesn't work like that. You can reduce the boundary layer by increasing velocity and turbulance which are linked by what the fluid is flowing around and the reynolds number.

The pressure drop across the block is an indication of the turbulance created hence the heat transfer coefficient acheived. This pressure drop will be the same no matter where the wb is positioned in the circuit for a given flow rate


sorry if I'm not very good at explaining it
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Unread 11-19-2003, 02:41 PM   #40
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Quote:
Originally posted by WAJ_UK


sorry if I'm not very good at explaining it
I am just not recieving it well. I understand now though. Thanks!
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Unread 11-19-2003, 08:27 PM   #41
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It's quite a hard thing to visualise, but I think you're getting it.

Consider it this way.

The pressure drop across a block at any point in a loop of identical components will be the same, since the flow rate will be the same, yes.

So while there may be a greater pressure at the block inlet by situating the block right next to the inlet, but the pressure at the outlet of the block will also be considerably higher than if the block were at the tail end of the loop. The relative difference between these two values should not change regardless of positioning, and it is the pressure difference, essentially a pressure GRADIENT which causes flow, much like a temperature gradient causes flow of thermal energy.

Hope this helps

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Unread 11-20-2003, 07:48 AM   #42
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Quote:
Another question. Would it be accurate in saying the most amount of pressure in the system is right out of the outlet of the pump?

If so, would it not give you a small advantage with center inlet blocks to go from the pump directly to the block? Instead of having the pressure drop from the rad first then the block?
Just for accuracy one may wish to remember that the head of water adds to the absolute pressure at any given point. Therefore in addition to the pump's pressure there is pressure at the bottom of the loop which is equal to the height of the loop. So if the waterblock is mounted 20 cm above the pump, then the absolute pressure will be 20cm of water less than the pump outlet pressure. This information is completely useless though because the absolute pressure of any particular point in the loop is irrelevant. If you wished to raise the pressure of your system you could hook up a T connector to your water cooling loop and plug that into your water mains which will be anywhere between 2- 6 bar. However flowrate will remain the same since the pressure drop through the loop will remain unchanged. Thge high pressure will stress all your components though, so don't do this.

One could assume that having a higher pressure would mean a greater contact force against heat transfer surfaces which would be beneficial in much in the same way that heat will transfer better between a CPU and water block the harder you clamp them together. But it doesn't work this way.
When you increase force between two solids you are in effect increasing the contact surface area, since irregularities between the surfaces get squashed out. While the materials seem hard, on a molecular scale they will get better squashed together the harder you squeeze them. This squashing together means that more mocules from one surface will come into contact with molecules from the other, thereby increasing the heat transfer surface area.
This doesn't happen to the same extent with liquids since liquids find their way into gaps anyway. Increasing the pressure may help them into very tight gaps, but it's not an effect which I think would be noticable unless you had extremely high pressures. If you had pressures of 50 bar or thereabouts then the effect may be noticable but that's a far stretch from the 1.1 bar that one would find in a watercooling situation (1 bar being atmospheric pressure and the 0.1 bar provided by the pump).
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Unread 11-20-2003, 09:55 AM   #43
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Uh, yeah, what he said...


Pressure is tricky, and I have to really concentrate to explain it, because it was explained to me wrong the first time.

First, there's relative pressure (relative to atmosphere). Imagine that you have a tube, standing up, capped at the bottom, and full of water. There is some pressure inside the tube, from the water pressing down (thanks to gravity). That pressure is slightly higher than atmospheric. Now... at the water level itself (top), the difference between the pressure inside, and the pressure outside is zero: it's the same.


When you throw in a pump, everything goes awry: you create pressure where there might not be any, normally. If you can imagine a PC loop without a pump, you still have the relative pressures (relative to atmosphere), and the whole thing is predictable.

Add pump.

Now you've got an additional set of variables. If we stick to relative pressure (relative to atmosphere), the pump outlet will register the highest pressure point, and the pump inlet will register the lowest (in fact, the pump inlet pressure can be below atmospheric).

But none of that is a concern, for water cooling, except to note that if there's a leak in the loop, it'll either spill water, or suck in air, depending on where the leak is.

Obviously, the relative pressure drops, from the pump outlet, to the pump inlet, as the water goes through various components. It really doesn't matter in what order the components are, they'll all drop the same pressure.

The only thing that's different, is the relative pressure. The only effect it has, is on the joints. Example:

A heatercore drops 1 psi at a given flow rate. If the core is right after the pump, then those joints will have the highest pressure points, relative to atmosphere, but the core will still drop 1 psi.

If the core is at the pump inlet, then those joint will have the lowest pressure. If there's a leak there, it's even possible for the joint to suck air inside the core. (It's often the "invisible" leak that some people can't figure out).

The different relative pressure will have no measurable effect on performance: water is still a mostly incompressible liquid.
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Unread 11-20-2003, 06:43 PM   #44
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I'm sorry guys about not having those examples up... I am swamped with marking this evening, and will be lucky to get through it. I'll get them up this weekend though.
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