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water loop, how to organize it.

T

thecarguy

Limp Gawd
Joined
Feb 17, 2004
Messages
206
Well, I have a water cooling question. Recently, I purchased an Eheim 1250 water pump, a BlackIce Xtreme2 (2x120), DangerDen RBX water bloc, and a BayRes. I also have hose clamps, 15 feet of Tygon, etc.

Anyway, where should I run the tubing, in what order? Pump to rad to CPU to res is what I was thinking, but tell me if I can do it otherwise. The pump is mounted under my HDD, the Rad is located on the top of the PC, the res is in a drive bay, block is on CPU (obviously).

Thanks for your help,

-thecarguy
 
For that setup I would recommend going:

pump->CPU->Radiator->bayres->pump

Another option would be Pump->Radiator->CPU->bayres->pump.

Either way would be great. You just have to decide what works best as far as tubing clutter and layout. I would tend towards the first one, but the second one would work in a few situations too depending on the dimensions of your case and the placement of the bayres in relation to the RBX and the Radiator.
 
Yep. OMP has the usual routes displayed. Either will work just fine. Use whichever is easiest to rig.
 
I would always have the radiator right before cpu so its gauranteed the coldest water...

I'd have it pump-->Rad-->Cpu-->res-->pump

and if you have a gpu block: pump-->Rad-->T-fitting-->cpu/gpu-->res-->pump *this is only recommended if you're using atleast 1/2" tubing and have a sufficient pump.. in your case u do.*
 
blargh! t-fittings add so much resistance. For overall better cooling it would be better to go pump->rad->cpu->gpu->res->Pump
 
the only thing I would say to someone doing their first setup, is that I had mine set up nicely, but it turns out that I imagined that the Eheim pump would pump out the side, but it pumps out the top. Just thought it might save you like 5 minutes, if you are crazy like me.
 
I would always have the radiator right before cpu so its gauranteed the coldest water
Click to expand...

it doesnt matter where it is in the loop


Thermal Equilibrium...


It is observed that a higher temperature object which is in contact with a lower temperature object will transfer heat to the lower temperature object. The objects will approach the same temperature, and in the absence of loss to other objects, they will then maintain a constant temperature. They are then said to be in thermal equilibrium.

-Zeroth Law of Thermodynamics


didnt type it it was cut and pasted but ya what ever:)
 
Originally posted by thebro
it doesnt matter where it is in the loop


Thermal Equilibrium...


It is observed that a higher temperature object which is in contact with a lower temperature object will transfer heat to the lower temperature object. The objects will approach the same temperature, and in the absence of loss to other objects, they will then maintain a constant temperature. They are then said to be in thermal equilibrium.

-Zeroth Law of Thermodynamics


didnt type it it was cut and pasted but ya what ever:)
Click to expand...

Good quote.. as I'm a physics man myself. But there are times when you wonder why application does not match theory.. these times often reflect things which are overlooked.. and in this case would be where these units are located. If the radiator is mounted low, It might not matter.. but if the radiator is mounted on the top of the case (where heat rises to), then I'd have it going straight to cpu.
 
Originally posted by thebro
it doesnt matter where it is in the loop


Thermal Equilibrium...


It is observed that a higher temperature object which is in contact with a lower temperature object will transfer heat to the lower temperature object. The objects will approach the same temperature, and in the absence of loss to other objects, they will then maintain a constant temperature. They are then said to be in thermal equilibrium.

-Zeroth Law of Thermodynamics


didnt type it it was cut and pasted but ya what ever:)
Click to expand...

Yup:D The "Zeroth" law sounds like it was made up, but it actually is called that, and thats exactly what it says and means. I'm glad to see someone else who has obviously studied some physics posting on these forums...

Every part of your loop will have the same water temperature, so in the end it doesnt matter how you orginize your loop before going reservoir->pump

w00t! Physics majors unite!
 
You've got a dangerden RBX block, which i believe has some kind of jet plate? (correct me if im wrong, i cant be arsed looking)

If thats the case, go pump > cpu because you want the pressure to go through those wholes unrestricted. I think. lol :p
 
Originally posted by Etacovda
You've got a dangerden RBX block, which i believe has some kind of jet plate? (correct me if im wrong, i cant be arsed looking)

If thats the case, go pump > cpu because you want the pressure to go through those wholes unrestricted. I think. lol :p
Click to expand...

It doesn't matter where it is in the loop, flow/pressure will be the same regardless. Imagine a parade running in a circle, no matter where you stand you'll always eventually see the same people go by and at the same speed. All parts of the parade have to move at the same speed or else someone will end up getting trampled.
Hmm, OK that's a pretty stupid example.... but try to see what I'm trying to explain...
 
Originally posted by Etacovda
You've got a dangerden RBX block, which i believe has some kind of jet plate? (correct me if im wrong, i cant be arsed looking)

If thats the case, go pump > cpu because you want the pressure to go through those wholes unrestricted. I think. lol :p
Click to expand...

it doesnt matter WHERE it is, pressure is going to be uniform.

go pump->rad->cpu->other stuff->pump, because you want the coldest possible water hitting the CPU, and the pump does add heat.
 
Originally posted by zer0signal667
It doesn't matter where it is in the loop, flow/pressure will be the same regardless....
Click to expand...

I'm pretty sure I understand, but something just popped into mind: In most watercooling systems the components are set up in series. Using an electrical circuit as an example, when components are set up in series, each takes a little hit off of the voltage because of their resistance. Current stays the same, though. IIRC, pressure is analogous to voltage, and the flowrate is analogous to current. So (assuming the electrical circuit analogy is valid) wouldn't each component in a WC loop reduce the pressure initially delivered by the pump?

Eh. Time to go refresh my fluid mechanics memory. Fox & Mcdonald, here I come, again.:p

Edit: Couldn't lay my hands on Fox & Mcdonald. But I did manage to lay my hands on Marks' Standard Handbook for Mechanical Engineers. In the section "Flow in Pipes" (page 3-49) it presents the following example for calculating flow through pipes with rough surfaces, which I quote: "Case 1: 2000gal/min of 68 degree F (20 degree C) water flow through 500 ft of cast-iron pipe having an internal diameter of 10 in. At point 1 the pressure is 10psi and the elevation 150ft, and at point 2 the elevation is 100ft. Find P2 [the pressure at point 2]."

Further, Fluid Mechanics by Granger discusses pipe flow head loss and the Moody Diagram, presenting an example on p.503 where gasoline is passed through a rubber fuel line with certain roughness characteristics. The student is asked to find the flow rate, given the tubing diameter, length, and a pressure drop. This could easily have been rewritten to give the flow rate, and ask the student to find the pressure drop.
 
The electrical analogy is pretty good, except that pressure is equal to resistance times flow squared.

And, if you look at that Moody Diagram, you'll see that the resistance (labelled friction factor, or the Greek letter lambda) isn't actually constant. But with the flow rates and pipe sizes used in watercooling that isn't a big source of error.
 
Aggie, your electrical circuit analogy is a pretty good one, in fact it's what I usually compare a water loop to. You're right, the "current", or volumetric flowrate, will be the same throughout. And yes, a certain pressure drop is created by each "device" in the circuit such as waterblocks and even the tubing. But just like a circuit, the pressure drops created by the devices plus the pressure created by the pump sum up to zero. Regardless of the position of the device, it still creates the same pressure drop (this is in a simple simulation, I'm sure it gets much more complex when you consider fluid flow more extensively).

So, lets say you have a waterblock, a radiator and a pump. The pump creates a pressure of 5 psi, so the waterblock and radiator combined must drop pressure by 5 psi. The loop is continuous, so it's not that the pump "sees" the waterblock first or last, but sees the restriction of both devices combined at the same time because they're in series. The pressure drop of each individual device depends on the resistance that each has to water flow. No matter where you place them, they have the same characteristics, so the pressure drop won't change either.

I think I'm just rambling now, so I'll stop. Hopefully this made a little sense.

Oh, by the way, the same thing sort of applies to the problems you found in that book. The fuel line one- the fuel pump applies a pressure, the hose applies resistance. It gradually drops the pressure compared to that immediately after the pump, and by the time the fuel squirts out the end, the pressure is zero.
 
Originally posted by HeThatKnows
The electrical analogy is pretty good, except that pressure is equal to resistance times flow squared.
...
Click to expand...

So, by that equation, pressure = power?

In electricity: voltage, potential difference (V) V=IR, and power(P) P=VI. P = (I^2)R
 
Hydraulic power = pressure drop * volumetric flow rate.

If you want to be really cool....calculate how much your water heats up just by flowing through a waterblock. :)
 
Originally posted by AggieMEEN
So, by that equation, pressure = power?

In electricity: voltage, potential difference (V) V=IR, and power(P) P=VI. P = (I^2)R
Click to expand...

No, pressure is a force/area. Power is energy/time, or joules/sec in electrical terms. He was trying to state that the relationship between flowrate and resistance is non-linear due to turbulence effects, I think.

Check out this link , it may be helpful.
 
Originally posted by zer0signal667
Aggie, your electrical circuit analogy is a pretty good one, in fact it's what I usually compare a water loop to. You're right, the "current", or volumetric flowrate, will be the same throughout. And yes, a certain pressure drop is created by each "device" in the circuit such as waterblocks and even the tubing. But just like a circuit, the pressure drops created by the devices plus the pressure created by the pump sum up to zero. Regardless of the position of the device, it still creates the same pressure drop (this is in a simple simulation, I'm sure it gets much more complex when you consider fluid flow more extensively).
Click to expand...

OK, gotcha. Net pressure drop.

Oh, by the way, the same thing sort of applies to the problems you found in that book. The fuel line one- the fuel pump applies a pressure, the hose applies resistance. It gradually drops the pressure compared to that immediately after the pump, and by the time the fuel squirts out the end, the pressure is zero.
Click to expand...

Depends on where you take the final pressure measure, but yeah, I understand. Mass flow in = mass flow out.

EDIT: Yes, I know P = F/A. I was surprised that the equation he quoted was the one for power, and not potential. Thanks for the link, too.
 
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