Ramblings on guessing total hydraulic head for a system...

[pHaestus]

I just bought a (relatively) cheapo pump today, and happened to notice something interesting in the manual:

"Choosing the right output for the application

Laguna Statuary Pumps can be used in a wide variety of applications, and their wide range of output flows at various heights will accommodate almost every circumstance. What is most important is to define the needs of the application and ensure the pump will be adequate to the job. The factors that influence pump capacity include Head Height, hose and fitting resistance:

Head height: Vertical distance between water surface and water discharge.

Hose Resistance: For every ten feet (3 m) of hose, add one foot (30 cm) to the head height.

Fitting Resistance: For every elbow connector, add one foot (30 cm) to the head height.

Determine the head height first, the vertical distance from water surface to water outlet. To this, add the amount of hose needed to transport the water from pump exhaust to the water outlet. When pipe fittings are used to go around corners, the resistance to water flow created by a 90o elbow is equivalent to adding one foot (30 cm) per unit. If the installation requires a 90 cm (3 ft) lift from surface to output, requires 3 m (10 ft) of hose and one elbow, then the actual head is 150 cm (5 ft)."

Obviously, this doesn't give you enough data to actually calculate the head of your system, but it is basically the same approach used in Crane tech doc. 410 (with much more detail on fittings). I just found it interesting to have it laid out so clearly in a cheapo pump manual. You can quickly understand my aversion to 90 degree elbows (or Ts that block flow) now I would hope...

[gone_fishin]

So does this equate to a labyrinth design with three 180 degree turns and inlet and outlet at 90 degrees to the block top plate to be at least 5 foot of head if not more because 180 is obviously more restrictive than 90 degrees?

[pHaestus]

Not only that, the block's channel diameter is usually less than that of the tubing. This would further increase the head...

[Skulemate]

One important thing to remember is that the resistance is not simply a function of the shape of the fitting (or the roughness of tubing)... it also depends on the velocity of the fluid (i.e. the flowrate).

[pHaestus]

Yes but regardless you can still think of the resistance in terms of equivalent feet of pipe (or tubing). My main point was that I was happy to see that Laguna pumps was explaining hydraulic head with some eye to such concepts in their manual. May be more practical for overclockers to measure flow rate and then estimate hydraulic head from the pump's P-Q curve.

[Skulemate]

Originally posted by pHaestus
May be more practical for overclockers to measure flow rate and then estimate hydraulic head from the pump's P-Q curve.

That's exactly what I suggested in another thread when someone pointed out that it would be very hard to measure total head. The only problem is that most of the pumps commonly used in watercooling systems (Eheim, etc.) do not publish good performance curves that I know of. However, the manufacturer of my pump does publish such a curve, and I was thinking about doing some quick tests to determine the resistance through my new rad and block when I get them.

And I agree with you that it's nice to see the theory behind pumps and head loss in Laguna's manual (I actually looked at one of their pumps for my current system... the 3000 I think). Now all we have to do is get Dangerden or DTEK to publish flow-resistance curves for their blocks

[pHaestus]

From BillA's testing

[Skulemate]

Thanks... that's awsome. You're my new hero . Did you do those tests yourself? And if so, are there plans in the works to test the Maze 3 and the new Spir@l?

[pHaestus]

No way. BillA did that testing and is I believe working on an article able waterblock design parameters for overclockers. I pulled the image from this thread:

http://forum.oc-forums.com/vb/showth...hreadid=101864

[BillA]

easiest thing in the world to measure the total resistance of a WCing system
(Owenator did this)

put Ts in the pump intake and discharge lines (at the pump) with open vertical hoses
measure the difference between the two fluid levels
(think it out beforehand to understand how high the standpipe will need to be)

thats it, in in.H2O

be cool

[Skulemate]

Yeah, I was thinking of doing that, but my pump has a max head of 21', so I am not sure that's a good idea for me... I need to find a better way than by using simple water manometers.

[myv65]

The problem with such a simplification is that you may quickly add up total head losses higher than the dead head pressure of the pump. As skulemate stated, head loss through a fitting or hose is related to the flow rate. ie, a 90° elbow will have a head loss of 1 foot H2O only at a specific flow rate (not to mention elbow diameter). Fittings tend to have "equivalent lengths" that are proportional to their diameter. Since flow velocity is related to area (diameter^2 for round fittings), a given flow rate will have much lower head loss in a 1/2" fitting than in a 3/8" fitting. This becomes part of the balancing act where larger fittings result in higher flow, yet somewhat lower flow velocity and head loss.

It will provide an approximation, but could easily be off by a factor of two or more.

What is more accurate, though not much more difficult is to use tables relating head loss to flow rate and diameter. Even this has limits, however, as you'll be hard pressed to identify a head loss for radiators or blocks without actual testing.

If a water manometer is out, you could always use a mercury one.

[Skulemate]

Yeah right... I'll just steal some from the Mercury Intrusion Porosimetry machine in the lab I work in...

[bigben2k]

I think that a foot or 2 more than the rig's total height, should do the job, but not sure... It will measure the pressure drop between the pump intake and outlet, which is the total system pressure drop. The size of the tubing will be critical here, but the point is to be able to measure a height difference, then calculate the mass of the water, and convert it to psi.

[BillA]

jeez fellahs, its a super simple test yielding an approximation;
no need to try to 'over-engineer' it

"inches of H2O" are very common units used to measure low pressures
and incidentally quite handy if using a water manometer
using mercury will not yield any more accuracy, just accommodates higher pressure
i.e. 21ft.H2O = 18.5in.Hg

bigben2k
take your pump's deadhead pressure, convert the units to in.H2O (or ft.H2O); and rest assured, NO water will come out the top
- well, there is a surge when the pump starts up - so add a foot (or pinch it off)
-- the size of the risers is irrelevant as there is no flow through them

myv65
"The problem with such a simplification is that you may quickly add up total head losses higher than the dead head pressure of the pump."
nope, not if you are referring to the manometer test

"What is more accurate, though not much more difficult is to use tables relating head loss to flow rate and diameter."
this is what Crane No.410 is all about
- but w/o some experience it is indeed easy to have nonsensical calcs

be cool

[bigben2k]

Originally posted by unregistered
jeez fellahs, its a super simple test yielding an approximation;
no need to try to 'over-engineer' it

bigben2k
take your pump's deadhead pressure, convert the units to in.H2O (or ft.H2O); and rest assured, NO water will come out the top
- well, there is a surge when the pump starts up - so add a foot (or pinch it off)
-- the size of the risers is irrelevant as there is no flow through them

But over-engineering is what we do best!

Besides, if we didn't do that, we wouldn't have noticed that this simple test also involves taking the result of the weight of that water column, convert it to psi, and multiplying it by two, since the rig is still running, and the pump will split its load between the cooling loop and keeping that column up.

The part that was hard for me to understand is how this tube on the intake isn't going to suck air in: I have to remember that its the column at the pump output that's preventing that from happening, assuming that there's been a good enough quantity of water added. It's otherwise clear that both these tubes will have the same level of water, when the pump is off, as they balance each other out (through the pump, assuming that it's not an odd pump).

The result, a difference in height of coolant level that stabilizes at a certain height, will indicate the pressure drop between the outlet and inlet, but not the actual pressure readings. Few people know that the pump inlet's pressure is negative (i.e. below atmospheric).

I don't think that Skulemate has anything to worry about. I would expect pressure drops to be around 5, maybe 10 psi in the worst of cases, which translates into... what's the calculation again?

[myv65]

Originally posted by unregistered
"The problem with such a simplification is that you may quickly add up total head losses higher than the dead head pressure of the pump."
nope, not if you are referring to the manometer test

Hey BillA,

Sorry for any confusion. This was directed toward the "one foot per 10 feet of tubing and one foot per elbow" stuff. Adding the typical fittings, tubing, block and radiator of a standard setup would yield a result higher than many pump's dead-head pressures. IMHO, having something so simple yet inaccurate is no better than having something more accurate that one doesn't know how to use.

It isn't a question of smarts so much as it is of experience. Anyone using tables and canned equations (especially engineers such as myself) needs to have a handle on what's really going on in a given situation. Otherwise it's kinda like the computer clichè of "garbage in, garbage out". To this end, verification through testing ala your manometer comments is required.

To suggest it is as simple as applying a fixed rule as first quoted leads to a lot of misunderstandings. Folks that don't know why the relationships that exist between flow rate and head loss are what they are ask questions like, "Why doesn't everyone provide such a simple table for their {insert pump, fitting, tubing, block, radiator, etc. here}?" The answer is that it really isn't quite that simple.

Yeah maybe trying to be more accurate is over-engineering it. Funny thing is though, a lot of people seem to be curious about the finest aspects of their systems with only vague clues about how it all works together. Still others will look at a 1°C temperature difference on a graph and conclude a particular product is superior to another without regard for test conditions and (in)accuracies.

OK, I'll get off my soap box now.

As to the mercury manometer, I made the comment mainly in jest. As I'm sure you know, it's main benefit is allowing a manometer that's only 1/13.61 times as tall as a water manometer. Instead of needing a twelve foot tube to measure your pump's dead-head pressure you could use a one foot tube. Too bad it can be messy to work with those things.

[myv65]

Originally posted by bigben2k

Besides, if we didn't do that, we wouldn't have noticed that this simple test also involves taking the result of the weight of that water column, convert it to psi, and multiplying it by two, since the rig is still running, and the pump will split its load between the cooling loop and keeping that column up.

Actually, the pump won't expend any energy keeping that column up that it isn't already expending. It "keeps that column up" simply because of the resistance provided by the cooling loop.

PS: 1 psi = 27.708 in H2O = 2.036 in Hg = 6894.8 Pa

[BillA]

myv65

I call Crane No.410 "The Piper's Bible", kinda 'old school' now
- though a $400 software version is available

I'm trying to get pHaestus to do an article on 'accuracy',
or 'measurement uncertainity' as it is now called

there is a whole bunch of smoke floating around

bigben2k

"we wouldn't have noticed that this simple test also involves taking the result of the weight of that water column, convert it to psi, and multiplying it by two, since the rig is still running, and the pump will split its load between the cooling loop and keeping that column up."

what on earth is this sentence all about ?

here is a converter:
http://www.convert-me.com/en/

be cool

[Skulemate]

My comments were also made in jest... if I were truely serious I would have also mentioned the thick gloves, spill kit and mineral oil needed to handle and dispose of the material safely.

Also, about the conversion confusion. Perhaps it would be easier for all if one of us simply explained the piezometric head and energy grade lines in the context of a system with a pump...

[bigben2k]

Originally posted by unregistered
"we wouldn't have noticed that this simple test also involves taking the result of the weight of that water column, convert it to psi, and multiplying it by two, since the rig is still running, and the pump will split its load between the cooling loop and keeping that column up."

what on earth is this sentence all about ?

Thanks for the link BillA, I've added to my favorites, so that I won't be caught short again!

I do mean it though, about doubling the value. The pump exercises an effort to push the coolant through the rig, and there is an effort involved in maintaining that column up. Since the column is maintained at a steady level, there is no flow restriction to calculate, since there's no flow in that column, but the pump does do some work there too, as well as creating a flow through the rig. I am assuming that the difference is equal, so I say, take the result, and multiply it by two. That way, when you restore your rig to normal (i.e. remove the T's and water columns), then you have the full pressure/flow going through the rig.

I also said that the tubing size (for the columns) is critical, because if the pressure is say 10 psi (worst case), the water column would be subjected to 5 psi of pressure, and at 27.7 in. of water per psi, that's 138 inches of water, or 11.5 ft, assuming that the column/tube has a diameter that accomodates 1 sq. in. of water. If you used a 2 in. diameter tube, the area is 3.14 sq in., so the water would rise 44 inches, or 3.8 ft.

Of course most of us would be lucky to get half of that.

BillA, maybe you can provide us with a link of what this fellow "Owenator" actually did, and what he got. Maybe I've got it all wrong, and I just can't see it.

[bigben2k]

Originally posted by unregistered
easiest thing in the world to measure the total resistance of a WCing system (Owenator did this)

put Ts in the pump intake and discharge lines (at the pump) with open vertical hoses
measure the difference between the two fluid levels
(think it out beforehand to understand how high the standpipe will need to be)

thats it, in in.H2O

be cool

If this involved adding a T only to either the outlet or intake, there would not be a need to multiply the result by 2.

...and right, now, as I'm writing this, is (finally!) when I realize what I was missing!

The other column is maintaned LOWER, which negates the effect of the other column, so there's no need to multiply anything by anything...

Phew!

[Skulemate]

Firstly, I don't think that the manometer diameter affects how much the water rises at all... that rise is totally dependant on the pressure that is present in the fluid at a given point in the system. The manometers we use in school are often less than 1/4" OD tubes. Also, the pump does not have to worry about dividing its work into two to maintain the water column... as I have said before, the pump imparts potential energy rather than kinetic energy onto the fluid (represented by the piezometric head line). This manometer is only demonstrating how much potential head the water has. A clear way to illustrate this piezometric head line is to run a whole series of manometers at various points through the system, and then connect the tops of the columns (though this makes a much nicer visual picture if everything is in a straight line).

[bigben2k]

Originally posted by Skulemate
Firstly, I don't think that the manometer diameter affects how much the water rises at all... that rise is totally dependant on the pressure that is present in the fluid at a given point in the system. The manometers we use in school are often less than 1/4" OD tubes. Also, the pump does not have to worry about dividing its work into two to maintain the water column... as I have said before, the pump imparts potential energy rather than kinetic energy onto the fluid (represented by the piezometric head line). This manometer is only demonstrating how much potential head the water has. A clear way to illustrate this piezometric head line is to run a whole series of manometers at various points through the system, and then connect the tops of the columns (though this makes a much nicer visual picture if everything is in a straight line).

So what did you use to measure with your manometer?

[Skulemate]

I'm sorry, I don't understand your question.