# Thread: Hydrology or hydro dynamics question

1. ## Hydrology or hydro dynamics question

I am thinking about the design of water blocks after reading some of Cathers tenents of water block design, namely the part about high velocity creating high turbulance over the die area. And in my dads company, a steel and industrial supply and fabrication house we use water to drill in addition to clean and fire prevention. This cutting water is piped though a series of smaller pipes to gain some serious pressure and though a small diamond outlet..enough to cut/drill steel in some cases. It works on the simple principle of the potential energy (large pipe) turning into kenetic energy (small pipe small hole) thus creating massive water speeds.

Anyway I was wondering why we don't use 1" inlets piped into very small designs, like the innvoteck blocks for example or even smaller, to create this massive turbulence over the die area for maxium cooling?

2. I think (and I can be pretty dumb) its because thats overkill. The core isn't 1 inch and if there is a heatspreader the heat doesn't spread very far, so it would be cooling areas that dont need to be cooled.

3. Becuase 1" tubing is a pain in the neck to deal with.

Simply making the inlet bigger isn't going to increase turbulence. You have to increase the overall flowrate, and increasing the tubing size will do that a bit... but really the only effective way is to buy a bigger pump.

4. its actually because of the power that it takes to make that kind of pressure is outrageous. The power consumption and power dissipation into the water would by far outweigh what it cools.

Jon

5. Originally posted by 9mmCensor
I think (and I can be pretty dumb) its because thats overkill. The core isn't 1 inch and if there is a heatspreader the heat doesn't spread very far, so it would be cooling areas that dont need to be cooled.
No no, the 1" would be mearly the hose size, we would drop that down to say 1/8" which is indeed smaller inlet than current designs even, the block *could* even shrink in size under this scenario and would create greater turbulance.

I think the next poster after you stated the obvious though...stupid me...OverKill and pump needed would be to the point of diminishing returns.

6. Interesting concept none the less.

7. Greater turbulance typicaly = greater total resistance = way out of the total head range pumps we use can provide.

Mr. Bernoulli says that pressure and flow will be equal at all points in a closed loop. The only variable is velocity. Velocity is inversely proportional to the diameter of the pipe. Halve the pipe diameter and velocity will quadruple, and so will the energy required maintain the same flow rate of a pipe twice as big.

Increasing velocity is a great way to increase turbulence, but there is a limit to how much energy a centrifugal pump can provide without becoming horribly inefficent.

So unless you can do what pump engineers have failed to do and dramaticaly increase the efficiency of centrifugal pumps, or feel like dropping some major bucks on a positive displacement pump, the jet velocity of a Cascade is about as good as it gets.

To generate 15lpm of flow at 10 meters of head will require 18.34 watts. A damn good centrifugal pump is at best 20% efficient at its point of best efficiency (PBE). 20% efficiency means that 91.7 watts total of energy will have to input into the pump by the pump motor to generate 15lpm @ 10M. The other 73.36 watts is turned to heat. Wander a little bit to the left or right of the PBE and it gets REALLY crappy quick.

8. It pretty much requires a pump like an Iwaki MD40-RZ at the least to give enough pressure to get to the super-turbulent velocities for which jet impingement can continue to show good improvement beyond what the Cascade is doing. This would require a slight redesign of the Cascade block to enable that to work.

Problem is, the MD40-RZ is a 110/140W (50/60Hz) pump. Also the resultant jet velocities (15-20m/s or 35-45mph) would be high enough that metal erosion would become an issue over time.

Having said that, I won't say that I won't try it once.

9. I thought of that after I posted. The cascade/RBX do provide great "impingement" relative due to thier dual outlets increasing thier kenetic energy right? Or maybe not since it's still a closed system and water does'nt compress and that's just for equalateral flow over the die. 4 outlets with a cross cut design for more surface area next i bet

10. With a higher velocity you increase your convection coefficient h. As water passes over surface there is a small area of water that doesn’t move at the same rate as the rest of the water called a boundary layer. This happens in the tubing and inside the water block where you don’t want it to happen. They try to make the block have a turbulent flow, which will help stop boundary layers from forming. The down side is with their designs you lose flow rate, which also plays a big roll in cooling. So you need two thing flow and velocity. They go hand in hand you can’t change one with out changing the other, unless you put more energy into the system like upgrading you pump. I would check out these two books “Liquid Cooling Of Electronic Devices By Single-Phase Convection” ISBN#0-471-15986-7 and “Heat Transfer in Electronic and Microelectronic Equipment” ISBN#0-89116-277-1 another good book is “Fundamentals of Heat and Mass Transfer” ISBN#0-471-38650-2

11. You can find the Reynolds number and if it's over 4000 you will have a turblent flow. You will find out your velocity has to be so high that your flow will to be small to cool the chip. Work is being done here at Iowa State Unv. to solve this problem. Those books will help you alot and cover everything your talking about. Testing has been done and the results are in the books.

12. Originally posted by extremecorvette
You can find the Reynolds number and if it's over 4000 you will have a turblent flow.
Re=4000 will only bring you to the *transition* zone of the Moody diagram, ie you will have partially turbulent flow. Besides, in a 1/2" pipe, you only need a volumetric flow of 2.4lpm to have a Re=4000 and since most of the high-end waterblock use impingment plate (smaller surface than with your 1/2 tubing), you can get a value of Re higher than 4000.
Maybe you had Re=40,000 in mind?

Originally posted by UberBlue
Mr. Bernoulli says that pressure and flow will be equal at all points in a closed loop. The only variable is velocity
False, only the mass flow is and this is in accord with the Conversation of mass principle.

13. Originally posted by Prandtl

False, only the mass flow is and this is in accord with the Conversation of mass principle.
Yep. I've come to realize that myself the past week from messing around with pump efficiency. Pressure(head) + flow = work, all of which is disapated by the loop. Therefore if work is to be done, pressure cannot remain constant.

14. Originally posted by UberBlue
Yep. I've come to realize that myself the past week from messing around with pump efficiency. Pressure(head) + flow = work, all of which is disapated by the loop. Therefore if work is to be done, pressure cannot remain constant.
yep, but keep in mind that there is no actual work (no moving part) done per se in a watercool system, only heat transfert. But anyway, even in a theorical frictionless incompressible flow close loop, pressure wouldn't stay constant if speed isn't. When you have a diameter (or suface) changes, both the velocity and pressure changes.

15. Originally posted by Prandtl

yep, but keep in mind that there is no actual work (no moving part) done per se in a watercool system, only heat transfert. But anyway, even in a theorical frictionless incompressible flow close loop, pressure wouldn't stay constant if speed isn't. When you have a diameter (or suface) changes, both the velocity and pressure changes.
No moving parts? No pump? The water isn't moving?

16. Originally posted by extremecorvette
With a higher velocity you increase your convection coefficient h. As water passes over surface there is a small area of water that doesn’t move at the same rate as the rest of the water called a boundary layer. This happens in the tubing and inside the water block where you don’t want it to happen. They try to make the block have a turbulent flow, which will help stop boundary layers from forming. The down side is with their designs you lose flow rate, which also plays a big roll in cooling. So you need two thing flow and velocity. They go hand in hand you can’t change one with out changing the other, unless you put more energy into the system like upgrading you pump. I would check out these two books “Liquid Cooling Of Electronic Devices By Single-Phase Convection” ISBN#0-471-15986-7 and “Heat Transfer in Electronic and Microelectronic Equipment” ISBN#0-89116-277-1 another good book is “Fundamentals of Heat and Mass Transfer” ISBN#0-471-38650-2
Thank you. I'm not so sure if it's to advanced for me but I'll check those out just because Id like to learn something. Next year I planned to major in finance eventually MBA at the university so everything I read is geared twards that but I have had AP chemistry, physics and calculus in High School. Worth a shot

17. Originally posted by squeakygeek
No moving parts? No pump? The water isn't moving?
I should have made myself clearer, there is no net work done by the system... better? :P

Beerhunter,
a good source of information on fluid dynamic (an alot of other related things to watercooling) can be found on the front page in an article called "Basic Principles Involved in Heat Transfer and Cooling Related to Processors".

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