Pressure Loss Or Flowrate

[Titan]

I want to ask if any pressure will be lost or will the flow rate suffer if a small reservoir(16mm * 16mm * 2mm) is made in between the inlet of the hose barb and the base plate.
For example,this. When water enters the block, it enters a small reservoir (which is bigger than the ID of the Hose barb ), before it passes through the jet arrays to the base plate.Or will there be none?

[myv65]

Any time you change water velocity you will have a loss of pressure. Upon entering the reservoir, the water will "disperse", losing its concentrated velocity profile that existed in the tube. It must re-establish this when leaving the reservoir. As a conservative estimate, take the water velocity, square it, and divide by 2 * gravitational constant. With the right units, this will yield the head loss for the combined entrance and exit loss measured in height of a water column (feet of water, meters of water, etc.).

It will be a fairly low number, but real.

[Titan]

so i could say that a strong pump could minimise this head loss.
As it could fill up this reservoir fast enough and still maintain a flow rate close to the inlet flow when it passes through the array to the base plate. (As in the picture attached)
If so, does this mean only an area slightly bigger than ID of the hose barb could be made or could it be bigger. I want to make a small reservoir about half the size of the P4 IHS, with the ID of the hose barb ~11mm , im concentrating on cooling this portion of the IHS. I know there are some blocks out there with this reservoir, i just want to know whether is it almost as "efficient" as those with "direct" inlet to the baseplate.

[bigben2k]

Don't confuse "flow rate" with "flow velocity": they are quite different. Re-read myv65's post!

A strong pump will only allow a higher flow rate, which is going to increase the pressure drop across that restriction. The pump doesn't "compensate" for nothing: you need to select a pump that's going to fit your targetted flow rate.

[myv65]

The "head loss" will be entirely a function of the volumetric flow rate. "Strong" pumps provide more flow at a given head compared to "weak" pumps.

You asked whether there would be a loss. Yes. For a given pump what you suggest will reduce flow. How much? Depends on the pump and the rest of the setup. I gave you a means to estimate the head loss for a given flow rate. I can't do much more beyond that.

[Gooserider]

I agree with Myv65 and friends, but I'm not entirely sure how useful their answers are in regard to what you're asking. I would add that ANY time you have a change in the diameter or direction of flow you will have a measurable amount of turbulence, and consequential loss of pressure and flow rate. Heck, you even get such a loss (albeit a smaller one) in a straight length of pipe.

The amount of these losses is a function of how much pressure and volume is passing through. The functions are inter-related, thus; higher pressure <-> greater volume <-> greater losses. A 'stronger' pump will give more pressure, thus a larger flow volume, but with a greater restriction loss. However in a given system, the higher pressure pump will move a greater volume per time unit, how much greater depends on the system.

The flow volume is a function of the TOTAL amount of pressure losses in the system, which can be plotted as a curve of pressure vs. flow. A pump's output can be plotted on the same curve, and where the two intersect is the amount of flow you will get for that setup.

However, what you probably want to keep in mind is that given a simple, non branching, system with a known flow rate, the same amount of flow will take place at EVERY point in the system. It works like current in a simple electrical circuit - the amount of current is a function of the voltage and amount of resistance, but is constant at every point in the circuit.

Thus, what you are proposing would work as a design, and I think is what you will find in some form on any jet impingement type design. There always has to be a transition between the tube and the jets of some sort, which would include changes in the area of the tube to the size of the cooled area of the block.

The only caution I would mention is that IF you make that transition area to short in relation to the diameter change then there will be a significant restriction and pressure drop between the central jets and the ones on the outside... Also, right angle bends are more restrictive / turblulence causing than gentle bends, so a conical transition is theoretically better than a cylindrical one. I.E. in ASCII art:

 Note; in each example, input barb and jet plate are the same size...

       | |         ____| |____              | |
       | |         |          |            /   \
       | |         |          |           /     \
   ____| |____     |          |          /       \
   |__________|    |__________|         /_________\
    | | | | | |     | | | | | |          | | | | | |

     BAD!            BETTER!              BEST!

I hope this intuitive type explanation is helpful...

Gooserider

[Titan]

I see what u are implying....thanks for the nice ascii art i appreciate it, it really helps me to think better now.... If anyone has any comments about pls speak up. my waterblock is currently stuck in this portion. Thanks