3rd Revision DUAL LOOP ..Dark KNIGHT.. + Liquid COOLED! + 56k WaRn - Page 4 - Overclock.net

oblivion.sky · 01-28-09

Quote:

Originally Posted by procpuarie

hmm. you would be better to do:

res -> pump -> rad -> cpu -> nb -> res

no stupid 90's and split things. they all add unnecessary restriction and lower temps.

that's how i have it...

res -> pump -> 2 rads -> cpu -> nb -> res

and yeah i'll ditch the 90 degs

oblivion.sky · 02-03-09

shout out to mikenjill.

XxG3nexX · 02-03-09

i wish i wasnt such a woman,.. then i'd wc my setup

ddanko12 · 02-03-09

When i first read the title i thought you meant Soldam Dark Night....

Still very impressive though.

Cdpkook132 · 02-03-09

One question with the loops shown.

Pump>Rad>CPU>F fitting>nb/sb>F fiitting>Rad>pump

Are you sure water is circulating through the NB/SB Semi loop becuase it is in parrallel with just the Vertical fill line thus path of least resistance would say it would come out of the cpu up towards the Filling port and then to the Rad. I would imagine the water in the nb/sb portion of the loop is stagnant, or close to it. I WOuld suggest getting rid of the Tubing connecting the 2 F fittings. Correct me if I am wrong.

e-machine · 02-03-09

I was impressed till I saw the last pic with the dell mouse.

Cdpkook132 · 02-03-09

In response to my last post.

You pretty much have a single loop system in series with a portion of it (the nb/sb block) being in parallel.

We will call each of these loop 1 and loop 2

Q=Velocity * Cross Sectional Area
Qin=Q1+Q2=Qout

Since the Cross sectional area of the tubes is the same so velocities must be the same resulting in no problems right? WRONG!

Since there is a waterblock contained in loop 2, there is resistance present due to the narrowing into the water block which is when the water will gain velocity to make up for the fact that the cross sectional area has decreased. This is the resistance present in that loop. The water when thought about as energies ill try to remain constant and not want to go through that loop because its got the super tunnel with no restiriction running from one side of the loop to the other this is the preferred travel path by the streamline running through this parallel portion of the loop. Put flow gauges on both loop 1 and loop2 and you should see that the loop running from F fitting to F fitting is flowing more. Hope this helps.

oblivion.sky · 02-04-09

Quote:

Originally Posted by Cdpkook132

In response to my last post.

You pretty much have a single loop system in series with a portion of it (the nb/sb block) being in parallel.

We will call each of these loop 1 and loop 2

Q=Velocity * Cross Sectional Area
Qin=Q1+Q2=Qout

Since the Cross sectional area of the tubes is the same so velocities must be the same resulting in no problems right? WRONG!

Since there is a waterblock contained in loop 2, there is resistance present due to the narrowing into the water block which is when the water will gain velocity to make up for the fact that the cross sectional area has decreased. This is the resistance present in that loop. The water when thought about as energies ill try to remain constant and not want to go through that loop because its got the super tunnel with no restiriction running from one side of the loop to the other this is the preferred travel path by the streamline running through this parallel portion of the loop. Put flow gauges on both loop 1 and loop2 and you should see that the loop running from F fitting to F fitting is flowing more. Hope this helps.

very good point, but as far as i can see (in person) it flows in/out fast through waterblocks, but since i'm going to install loop (2) with independent pump/rad and 2 gpus blocks then i'm getting rid of the splits.

oblivion.sky · 02-04-09

Quote:

Originally Posted by Cdpkook132

In response to my last post.

You pretty much have a single loop system in series with a portion of it (the nb/sb block) being in parallel.

We will call each of these loop 1 and loop 2

Q=Velocity * Cross Sectional Area
Qin=Q1+Q2=Qout

Since the Cross sectional area of the tubes is the same so velocities must be the same resulting in no problems right? WRONG!

Since there is a waterblock contained in loop 2, there is resistance present due to the narrowing into the water block which is when the water will gain velocity to make up for the fact that the cross sectional area has decreased. This is the resistance present in that loop. The water when thought about as energies ill try to remain constant and not want to go through that loop because its got the super tunnel with no restiriction running from one side of the loop to the other this is the preferred travel path by the streamline running through this parallel portion of the loop. Put flow gauges on both loop 1 and loop2 and you should see that the loop running from F fitting to F fitting is flowing more. Hope this helps.

very good point, I'm no liquid flow specialist but as far as i can see (in person) it flows in/out fast through waterblocks It's ALMOST undeniable when you see it in person

, but since i'm going to install loop (2) with independent pump/rad and 2 gpus blocks then i'm getting rid of the splits.

oblivion.sky · 02-04-09

Quote:

Originally Posted by e-machine

I was impressed till I saw the last pic with the dell mouse.

rather beautify my rig that put money into keys/mice

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3rd Revision DUAL LOOP ..Dark KNIGHT.. + Liquid COOLED! + 56k WaRn

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e-machine New to Overclock.net Join Date: Jan 2009 Posts: 212 Rep: 23 Unique Rep: 22 Trader Rating: 0	I was impressed till I saw the last pic with the dell mouse.

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