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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 11-12-2003, 04:10 PM   #1
jaydee
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Default Flow Rate and Pressure question.

Quote:
Originally posted by jlr.ii on 11-12-2003 at 12:59 PM

Jaydee his drawing shows the res b4 the pump...I would think at that point the system the pressure would be quite low if not negative. When i cracked mine it was actually drawing air into the system. I agree that the discharge would be a precarious location, which is why mine is in the return.
Quote:
Originally posted by jaydee116
That doesn't make since to me. Flow rate is causing the pressure, flow rate remains constant through the loop. Pressure should be pretty even through the entire loop.

I maybe wrong and I will double check and post back.
So is my statement above correct? I can't see how pressure would be different in one spot as opposed to another in a WC loop. maybe were volocity picks up and slams into a corner or something?
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Unread 11-12-2003, 04:24 PM   #2
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think of it in the same way you think of volts and amps...

volts being the pressure and amps being the flow...


you are correct in saying the flow is the same anywhere in a given loop.(series)

same as with an electric loop.... current is the same, no matter where you measure.... (in a series loop )

The pressure will differ through out the loop.

most do not realize, but electric cct's and water-loops are more similar in there reactions than they are different.

Funny thing, I used to use water-loops to better understand electricity, when I was still young and stupid..... nowa days it's the other way around.... I show someone a bunch of resistors in series, on pops the light-bulb.... [pun intended]
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Unread 11-13-2003, 01:02 AM   #3
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He he, it looks tricky all right, but it's not.

The thing is, there's pressure drop, then there's relative pressure (relative to atmosphere).

You'd think that if you have a leak anywhere in a loop, that water would spurt out of it, but that's not necessarily true: the pump intake can have a relative pressure that's below atmospheric, so if you have a leak/crack there, it'll suck air in.

Electric wise, think of a battery where the negative terminal is below ground.
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Unread 11-13-2003, 01:08 AM   #4
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Given the electricity - water flow thing (I don´t know what "liknelse" is called in english, OK I have other flaws as well, ask my boss), the pressure should continously fall from the pump, through the loop and finally drop to a minimum when the water is back at the pump intake.
First WB in the loop will get the highest pressure and the last WB will have the lowest pressure, right?
regards
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edit: Hm i see that Ben has posted a reply since I started writing this. Ben, go to bed, it´s far to late to stay up, it´s eight o´clock in Sweden, it should be night over at Your place.
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Unread 11-13-2003, 01:19 AM   #5
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Yeah, yeah... just doing some last minute catch up... (It's 1 AM here now)
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Unread 11-13-2003, 04:02 AM   #6
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jaydee

If you want to think of it in electronic circuit terms think about the pump, which provides a pressure difference, as a battery which provides a potential difference. The current flows through the circuit from the point of highest potential to the point of lowest potential. The same is true for water circuits where the water flows from the point of highest pressure, just after the pump, to the lowest, just before the pump.

you can also think of each component as a resistor, which in an electric circuit has a potential difference accross it. In a water circuit this is a pressure difference. Every time the flow encounters a resistance the pressure drops even when flowing through pipes (or wires in electric circuits). The larger the diameter tube and the shorter it is, the smaller the pressure drop will be. Bends in the tube create an even larger pressure drop. The smaller the radius of the bend the larger the pressure drop.

this is something that Bernoulli told us, which is that if the flow has constant energy (which isn't strictly true for water cooling but probably close enough) then:

Kinetic energy+energy due to pressure+energy due to height=constant

energy due to height(gravitational potential energy) can be neglected due to no net height difference through the circuit.

This means that if you speed up the flow by having a converging nozzle (increase kinetic energy) this will result in a decrease in pressure. The same is true vice versa.

Hope this helps some peeps out as people seem to have a bit of difficulty understanding how the flow thing works
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Unread 11-13-2003, 08:49 AM   #7
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Quote:
Originally posted by msv
First WB in the loop will get the highest pressure and the last WB will have the lowest pressure, right?
Not true, I don't think. (EDIT: I suppose it is true actually, though it doesn't affect performance as explained below)

Try thinking about it this way.

Think of a pressure vs flow rate curve for a pump. As the pressure drop of the system connected between the pump's outlet and inlet is increased the flow rate wil decrease, following the curve described above.

When you're thinking about the pressure drop of a system, it makes more sense to think of the pressure drop of the whole system as a sum of the pressure drop of the individual components.

The pressure drop vs flow rate curve for a restriction, eg block/rad, will be a non linear curve increasing from the origin (zero flow at zero pressure drop) as increased pressure drop results in increased flow rate.

To combine pressure drop for a number of components, the pressure drops at a given flow rate are simply cumulative. In other words, the pressure drop of a whole system with a flow rate of x gpm, is the sum of the pressure drops for all of the components at x gpm. In other words, if you imagine all of the pressure drop vs flow rate curves, (pd on vertical axis) they can be stacked up one on top of another, resulting in a steeper curve. This makes sense as more blocks means more pressure is required to maintain the same flow rate.

So you now have a pressure vs flow rate curve for the pump, and the same for the system connected between the outlet and the inlet. From this, we can predict the flow rate through the system.

The important thing to realise here is that the flow rate is the same through the whole system. So consider a water cooled dual processor rig, with two identical cpu blocks in series. These blocks "should" have the same pressure drop vs flow rate curves. Now, since the flow rate through each of the blocks is the same, the pressure drop across the blocks must also be the same, provided they are the same blocks. The key point here is that the PRESSURE DROP is the same, NOT the absolute pressure. But it is the pressure DIFFERENCE across the block which results in flow, and not the actual value of pressure. So while the value of pressure decreases from a maximum absolute pressure at the pump outlet to a minimum absolute pressure at the pump inlet, it is not true that the second of two identical blocks in series will suffer due to the presence of the first block before it. In terms of pressure that is. It is obvious and well known that the coolant entering the second block will be at a marginally higher temperature than the coolant entering the first block, due to the thermal energy picked up while flowing through said first block.

If you want to use the electrical analogy again, think about it this way, consider an electrical component connected between +5V and GRND, and then consider the same component connected between GRND and -5V. Is there any difference, NO, because they still have the same potential difference across them, resulting in the same current flow. It is exactly the same with water blocks, except the relationship between pressure drop and flow rate is often more complicated than the relationship between potential difference and current, (often linear, whereas flow vs pressure tends not to be).

The key point to remember is that liquid flows down a pressure gradient, much like current flows down a potential gradient.

So infact, it is quite possible for a leak sprung in any part of the loop where the pressure has dropped to below that of atmospheric, for air to be drawn in rather than water leak out, since the pressure difference between the atmosphere and the loop, will cause air to flow down the gradient, ie into the loop.

This then gets complicated as you are adding more matter into a closed loop, so in theory the pressure in the loop will rise until the pressure at the point of the leak will is the same as atmospheric. What happens then, I don't know. It is possible there will be an alternation of air bubbles leaking in followed by water leaking out. But that last bit is just guess work.

Hope this helps

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Unread 11-13-2003, 09:39 AM   #8
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Thanks guys!
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Unread 11-13-2003, 01:31 PM   #9
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8-Ball, thanx, though it took a while to get a grasp of it.
Analogy is a good translation to "liknelse".
OK, the flow rate / pressure drop relation is logaritmic. So if I meassure all my components separately
I can´t assume that the sum of all these flow restriction equals the total flow resistance in my system, when
assembled.
But: If one WB has twice the flow resistance of another WB, meassured separately, will there still be a 1:2
resistance ratio between the two blocks when in the system?
Any ideas how I cen determine the flow resistance in two parallell loops, after a Y-splitter? If there was a
linear relation this would be peanuts (oops, is that a kick approaching my but?), but so it isn´t.
regards
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Unread 11-14-2003, 06:13 AM   #10
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I think what 8 ball was saying (I hope I have this right) is that if you know the flow rate through your system and measure the pressure drop for each component at this flow rate then you can simply add them all up.

I would have thought (but I could be completely wrong) that the pressure drop through the whole system will be the pressure difference that the pump can create at that flow rate which you could read off the p/q charts.

If one water block has twice the pressure drop as another water block at a particular flow rate which is the same flow rate acheived in a system, then one water block will still have twice the pressure drop when installed in that system.

Not sure what you meant about the parallel flow bit
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Unread 11-14-2003, 06:42 AM   #11
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just a quick question. Wouldn't just be easier to put a pressure sensor at pump outlet and a pressure sensor at pump inlet saves adding up the pressure drop accross each component?

Alternatively you could measure the flow rate and then see what pressure difference that corresponds to on the p/q chart for that pump
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Unread 11-14-2003, 07:11 AM   #12
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Based on what u guys have written, the decision on where I place my blocks should not determined by flow or pressure drops, but instead based on temperature of water in the loop??

Water temps would reach sort of an equilibrum after some time. In that case, we can simply arrange our components any way we like?? Since flow,pressure and temps are negligable factors??

Quote "block #2 won't suffer from block #1 being there except for temperature changes" may only be true if both blocks are being feed with the desired flow rates that allows the blocks to attain optimum performance. More often than not, it is unattainable.

What then should we use as a guage when it comes to mapping out our watercooling system??
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Unread 11-14-2003, 07:45 AM   #13
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a good guage for mapping out your system is to have nice short tubing with as few bends as possible. Although the water temp dosn't change that much imo it would be best to have the radiator just before the cpu block then the radiator will be getting the water at its warmest and the cpu block will be getting the water at its coolest.
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Unread 11-14-2003, 08:05 AM   #14
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WAJ_UK, my misunderstanding.
Let´s see if I can get this straight, pressure drop adds in a linear fashion, and when pressure drop increases water flow drops in a logaritmic fashion, right?
Please clarify this, I begin to feel stupid.
regards
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Unread 11-14-2003, 10:02 AM   #15
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I believe that you can simply add up the pressure drops from each component in your system (as long as they are in series)

have a look at
http://forums.procooling.com/vbb/sho...&threadid=5068

It will show the relationship between pressure an flow for many pumps

If you have an eheim 1048 and the total pressure drop through your system is 5 feet then the flow will be zero.

If the pressure drop in your system is about 3 feet then your flow will be 100 GPM

(at least that is what seems to make sense to me)

I hope this helps
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Unread 11-14-2003, 08:27 PM   #16
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Since we are on the subject of flow rates.

How much pressure loss/drop or flow rate loss/drop are we expected running in #1 inline with res and #2 submerged in res as opposed to #3 pure inline (closed loop)??
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Unread 11-17-2003, 07:47 AM   #17
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Quote:
Analogy is a good translation to "liknelse".
Similarity is another.
Quote:
OK, the flow rate / pressure drop relation is logaritmic. So if I meassure all my components separately I can´t assume that the sum of all these flow restriction equals the total flow resistance in my system, when
assembled.
The relationship bewteen pressure and flow is a square law if that's what you mean by logaritmic. And you CAN pretty much add individual resistances to get a system resistance as long as you assume that the flow is a the same through each resistance.
Quote:
Any ideas how I cen determine the flow resistance in two parallell loops, after a Y-splitter?
Yes. By dividing the flow into two you instead have two half flows with both flows at half speed. Given that pressure drop is a function of the square of flowrate, running a 1/2 flow will give you a 1/4 pressure drop. And the fact that you have two paths agains halfs pressure drop. So you will have 1/8th of the pressure drop of running the flow through 1 WB. This is assuming that flowrate is held constant.
In reality the lower resistance will be compensated by an increase in flow provided by the pump (and assuming that your WB's are the biggest resistance in your loop). This will mean that in practice the flowrate through each WB will be only slightly less in the parallel configuration than if you just had the 1 block, subject to the pump being able to supply the flow.
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Unread 11-17-2003, 11:17 AM   #18
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Quote:
Originally posted by Ewan

Yes. By dividing the flow into two you instead have two half flows with both flows at half speed. Given that pressure drop is a function of the square of flowrate, running a 1/2 flow will give you a 1/4 pressure drop. And the fact that you have two paths agains halfs pressure drop. So you will have 1/8th of the pressure drop of running the flow through 1 WB. This is assuming that flowrate is held constant.
You've halved too many times.

Suppose you have two 'flow resistors' in series and the PQ relationship is defined by:

P = Q^2 * (R1 + R2)

Then if you put those two 'flow resistors' in parallel, (and assume that the necessary 'wyes' add no flow resistance) the equation becomes:

P = Q^2 / (1/R1 + 1/R2)

So assuming that:

R1 = R2 = 1

and therefore

Rs = R1 + R2 = 2

then:

Rp = 1/(1/R1 +1/R2) = 1/2

So:

Rp = 1/4 * Rs

Or, in words - the 'flow resistance' of the 'resistors' in parallel is a quarter of the flow resistance of the 'resistors' in series IF the two resistors are equal and the added resistance of 'wyes' is neglected.

Edit: This is incorrect. See below.
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Unread 11-17-2003, 01:41 PM   #19
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Yeah, That's it. What was I thinking? I must have been getting my fluid dynamics confused with electronics and using tricks from both to balls it up.

Sorry about that.
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Unread 11-17-2003, 06:31 PM   #20
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Thks for the info ... understood it much better with the formulae.
Blinding similarities between water and electricity.

Next step-up for me is to parallel my block and see what happens....
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Unread 11-17-2003, 07:28 PM   #21
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Quote:
Originally posted by georgeteo
Based on what u guys have written, the decision on where I place my blocks should not determined by flow or pressure drops, but instead based on temperature of water in the loop??

Water temps would reach sort of an equilibrum after some time. In that case, we can simply arrange our components any way we like?? Since flow,pressure and temps are negligable factors??

Quote "block #2 won't suffer from block #1 being there except for temperature changes" may only be true if both blocks are being feed with the desired flow rates that allows the blocks to attain optimum performance. More often than not, it is unattainable.

What then should we use as a guage when it comes to mapping out our watercooling system??
The point I'm making is that the order of the blocks makes little difference to the delta T between coolant and heat source, provided the constituents of the loop have not been changed, only the order. However, the temperature of the coolant may vary, and since the delta T across each block should remain constant, the resulting temperature of the cooled objects may vary, though the magnitude of this effect is minimal.

Bearing that in mind, it "should" be more beneficial, given a set of components, to arrange the components in such a way that the total length of tubing, as well as the number of sharp bends is minimised, thus giving the best "overall" performance, of the whole system.

I hope this makes sense

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Unread 11-17-2003, 11:54 PM   #22
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Quote:
Since87:
So assuming that: R1 = R2 = 1 and therefore Rs = R1 + R2 = 2 then: Rp = 1/(1/R1 +1/R2) = 1/2 So: Rp = 1/4 * Rs
Excellent, but what happens in what I suspect might be a more common case of having a parallel setup where the two blocks are NOT equal? How would one figure out the amount of flow that would go through each block?

Example, with totally imaginary numbers...

Branch 1 CPU block, has a resistance of 2ft pressure
Branch 2 GPU block, with a resistance of 4ft pressure

My intuition would say that the system resistance would be around 1.5ft, and that Branch one would get about 2/3 of the total flow, and Branch 2 would get about 1/3. Is this correct? How would I show it by calculation?

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Unread 11-18-2003, 01:20 AM   #23
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Gooserider, quick (maybe too quick) calculations gave me that the Rp in Your example sums up to 1.333... ft, and the shares of that through each branch are as You suggest.

Ewan, thanx for the translation, "similarity" is even closer than "analogy".

regards
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Unread 11-18-2003, 05:30 AM   #24
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Lets state out the co-relation of electricity with hydraulics

Voltage = Pressure that causes water to flow (a.k.a piston)
V source = Pump Pressure/head (a.k.a battery)
V1 or V2 or V3 = Pressure drop across components
Current = Flow Rate
Resistance = Flow Resistance (tubing, blocks, rad etc..)

All scenarios assuming same components, length of tubing, etc.. Calculations based on the following:
Pump pressure = 10
Radiator #1 and #2 flow resistance = 2
Super restrictive Radiator #3 flow resistance = 6
CPU block flow resistance = 6
GPU block flow resistance = 4

Conditions :
1) CPU block has a more restrictive design as compared to the GPU block, which is commonly the case.
2) Pump Resistance / Y splitter resistance / Tubing Resistance(no kinks, no elbows, shortest length possible) are all Negligible. This is to make calculations easier.
3) Based on the assumption that the CPU block is 3 times more restrictive than the rad and 2 times more restrictive than the GPU block ... and vice versa.
4) Pump head is sustained at a rating of 10 throughout the scenarios. It is obvious that the performance would suffer if the pump can provide 10points of pressure and the components require 20points. But this is not the basis of discussion here. is calculation is based on a perfect situation where the Pump pressure is enough to cater for all the components.
-----------------------------------------------------------
Scenario #1a
Entire setup in series

Findings :
Flow rate is 0.83 across all components
-----------------------------------------------------------
Scenario #1b
1 CPU and 1 GPU block in parallel, rest of circuit in series.

Findings :
Flow rate across components is as follows :
Rad = 2.27
CPU = 0.91
GPU = 1.36
Block with higher resistance will receive lesser flow. CPU block is receiving 8%, GPU block is receiving 53% and Rad is receiving 144% more flow as opposed to a pure series setup.
Increasing the pump pressure to twice the value (20) has no effects on the flow distribution ratio between the 2 blocks
-----------------------------------------------------------
Scenario #2a
Addition of another radiator into the circuit in series. Assuming rad#1 is same as rad #2.

Findings :
Flow rate is 0.714 across all components
Overall Flow rate drops by 11.6% when compared to scenario 1a (only 1 rad present)
-----------------------------------------------------------
Scenario #2b
Addition of another radiator into the circuit. Rad#1 and Rad#2 now working in parallel. Rest of the circuit in series.

Findings :
Flow rate across components is as follows :
Rad#1 = Rad#2 = 0.454
CPU = GPU = 0.91
Flow rate of CPU and GPU has increased by 19.6% over scenario 2a. CPU flow rate has improved as is now on par with that obtained from scenario 1b, the highest attained so far. GPU flow rate is also clocking in its 2nd highest scores so far. Rad flow rate is at an all time low, the lowest calculated so far.
------------------------------------------------------------
Scenario #3a
Rad#1 and CPU in one loop, Rad#2 and GPU in another. Both loops starting and ending at the same pump.

Findings :
Flow rate across components is as follows :
Loop#1 (Rad#1+CPU) = 1.25
Loop#2 (Rad#2+GPU) = 1.667
Total flow rate of both loops = 2.917
Highest flow rates calculated for both the CPU, GPU and rads in parallel. Only losing out to a rad in series as in scenario 1b
-----------------------------------------------------------
Scenario #3b
One of the rads have been removed and a new restrictive rad is introduced into the setup.
Rad#1 and CPU in loop#1, Rad#3 and GPU in loop#2. Both loops starting and ending at the same pump.

Findings :
Loop#1 (Rad#1+CPU) = 1.25
Loop#2 (Rad#3+GPU) = 1
Total flow rate of both loops = 2.25
Inclusion of the restrictive rad has no impact on the flow rates of loop#1. Flow rates of loop #2 dropped drastically due to the inclusion of the super restrictive rad#3
-----------------------------------------------------------

Conclusion:
Highest calculated flow rates are ....
CPU = 1.25 (3a,3b)
GPU = 1.25 (3a,3b)
RAD1# = 2.27 (1b)
RAD2# = 1.667 (3a)
RAD3# = 1 (3b)

This is meant to be a pureply mathamatical calculation of the various flow rates we can except from different setups using ohm's and kirchhoff's laws. If I'm wrong ... pls correct me.
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Unread 11-18-2003, 11:03 AM   #25
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Ewan, you gave up too easily. Although neither of us was really right. For constant flowrate, the pressure drop for two blocks in parallel would be 1/4 that of a single block, but 1/8th that of two blocks in series.

I did some equation handwaving and no one caught me.

The equation I listed earlier for calculating with flow resistances in parallel is wrong.

For resistance values relevant to the equation:

P = Q^2 * R

parallel resistances are calculated using the equation.

Rp = 1 / ( (1/R1^-2 + 1/R2^-2 + 1/R3^-2 ... ) ^2 )

A bit uglier than I'd suggested earlier, but I believe this equation actually works.

Quote:
Originally posted by Gooserider
Excellent, but what happens in what I suspect might be a more common case of having a parallel setup where the two blocks are NOT equal? How would one figure out the amount of flow that would go through each block?

Example, with totally imaginary numbers...

Branch 1 CPU block, has a resistance of 2ft pressure
Branch 2 GPU block, with a resistance of 4ft pressure
First of all; a waterblock has a dP vs Q curve. A waterblock doesn't just have a pressure number. It is reasonable to say:

Branch 1 CPU block, has a resistance of 2ft pressure @ 2 gpm

but, just listing a pressure doesn't really say anything.

Anyway, assuming a flowrate of 2 gpm to go along with these hypothetical pressures, we can calculate 'flow resistances' for the two devices. (Although this is only valid inasmuch as the PQ curve of the devices conforms to the equation P = Q^2 * R. Not everything does.)

2 = 2^2 * R1
4 = 2^2 * R2

R1 = 0.5 (ft-H2O/gpm^2)
R2 = 1 (ft-H2O/gpm^2)

Using my newly derived parallel resistance equation:

Rp = 1/(1/squareroot(R1) + 1/squareroot(R2))^2 = 0.172 (ft-H2O/gpm^2)

Now you can graph the curve:

P = Q^2 * 0.172

and the curve for a pump simultaneously, and find the point where the two curves intersect. This will give the dP and flowrate for the parallel system. Using the dP determined for the system you can then calculate the flowrate through each parallel leg using the known resistance value for each leg.

For example, if the pressure that the pump produces across the parallel system is 3 ft, then:

Q1 = squareroot( 3 / 0.5 ) = 2.45 gpm
Q2 = squareroot( 3 / 1 ) = 1.73 gpm
Qp = squareroot( 3 / 0.172 ) = 4.18 gpm

So block 1 gets about 59% of the flow,
and block 2 gets about 41% of the flow.

Of course for a real system you'd need to throw a radiator into the mix, as well as values for tubing, fittings, etc.

Now I've got to take a look at some previous work based on my incorrect assumption for calculating parallel resistance.

Edit: Corrected first paragraph.

2nd Edit: Corrected flow resistance units from ft-H2O/gpm to ft-H2O/gpm^2. Plus grammar.
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