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General Liquid/Water Cooling Discussion For discussion about Full Cooling System kits, or general cooling topics. Keep specific cooling items like pumps, radiators, etc... in their specific forums.

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Unread 07-18-2002, 06:14 PM   #76
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Quote:
Originally posted by myv65
*sigh* Nothing is ever as cut and dried as you would like. There are two measures of viscosity, dynamic (absolute) and kinematic. The difference between the two is inclusion or exclusion of the material's density. In dynamic terms, the viscosity of mercury and water isn't so different and if water didn't freeze they would be equal at ~ -5 to -10°C. In kinematic terms, mercury's density makes its viscosity much, much lower than water's.

IIRC, its the absolute viscosity that goes into calculating pump flow rates, but don't quote me on that one.
Yeah... I remember seeing an experiment on PBS where they mixed something like corn starch, or some other kitchen product, and they showed that it looked really liquid, almost like water, but then the guy puts his hand in it, and quickly lifts the plate (an alu pie plate) and them slams it right back down. The plate followed the movement perfectly.

He was demonstrating that the viscosity changes, so I guess that's an example of dynamic (absolute) viscosity, versus kinematic viscosity. Sound about right, myv65?

(In this case though, I think the guy was demonstrating that some kind of starch has a low dynamic viscosity, but a high kinematic viscosity)
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Unread 07-18-2002, 06:19 PM   #77
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In one of the science links (above) it says this:
"Kinematic viscosity is defined as the ratio of the viscosity to the density."

Does that make any sense?
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Unread 07-18-2002, 07:52 PM   #78
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The most important reason for not using mercury vs. water is not visc. its Specific gravity. Mercury is by weight to volume inefficient. On the other hand if you take water and add a gylcol mixture to a 20% ratio you enable a better transfer of heat from the medium to the radiator. While inhibiting thermal gains from the pump.

To further improve Delta T across a radiator you should have the pump after the radiator with a larger inlet than discharge.

I do this for a living and have just started to look at trying my hand at cooling my PC this way. I'll post some Ideas soon.
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Unread 07-18-2002, 08:02 PM   #79
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Quote:
Originally posted by JimS
Actually my statement was based on experience as well as common sense. I have run waterblocks all nite with just a pump and no radiator and there is a noticeable amount of heat in the water after some time. On the other hand, by running the same test setup with a radiator, the difference between water temps. and ambient is not even noticeable.

Like I said, a good radiator makes this whole debate really a moot point. Unless you are approaching or at the limits of your radiator, pump heat is nothing to be concerned about.

bigben, I love how you keep your systems updated with the latest technology. I have a clawhammer CPU I will send you when I am done.
Looking forward to it!

I got into this thread becasue it's a topic of interest. For me (maybe just a few of us), I have a need to understand the physics behind it all.

Otherwise I agree, everything is going to come down to the rad.

So far, I think I've explored every component individually (or at least tried!) so that eventually, I'm able to understand how it all comes together.

It's clear to me though, that the rad is going to be the most difficult of all, since there really isn't any kind of standard for them. Fans have standards, pumps have standards, but rads are a whole different story.

So far, I've established that a higher pressure (higher velocity) can dissipate more heat. Good for the block, good for the rad. So I'm thinking of trying to build a rig with an effective flow rate of about 200 gph.

I've also spent a good deal of time looking into phase change, and it seems that the easiest thing to do, is to replace the rad with the bucket inside a humidifier, where you turn the cold coils inside the bucket.

Again, shooting for 200 gph (effective), it would be a fair bit easier to achieve, since there's no restriction from a rad, but as this thread indicates, I'd hit a wall because my pump would induce a fair amount of heat.

At this point, the block design matters a lot more, because I want a high pressure, but not so much that a pump will induce a lot more heat.

So I'm down to a cross-drilled design (which I believe to be of the best), but I'm still looking...

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Unread 07-18-2002, 08:11 PM   #80
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Quote:
Originally posted by Myrd
The most important reason for not using mercury vs. water is not visc. its Specific gravity. Mercury is by weight to volume inefficient. On the other hand if you take water and add a gylcol mixture to a 20% ratio you enable a better transfer of heat from the medium to the radiator. While inhibiting thermal gains from the pump.

To further improve Delta T across a radiator you should have the pump after the radiator with a larger inlet than discharge.

I do this for a living and have just started to look at trying my hand at cooling my PC this way. I'll post some Ideas soon.
Interesting... but even if mercury is "by weight" not as efficient as water, isn't it more efficient, for the same volume?

Rad>pump huh? I've been recommending rad>block (with the pump anywhere) so that the coolant entering the block is at its lowest temp. Am I wrong?

"...with a larger inlet than discharge." That makes good sense, but I think that most of us are stuck with what the pump manufacturer dictates. I take it that the reason for this is to minimize the pressure at the pump inlet, so that the pump can be a little more effective? (I'm stretching here, I can almost see it, but I'm not there yet!) Which would mean that my rad>block suggestion is still good, but the tubing size to the pump inlet should be bigger (biggest of all tubing/channels)?

(That would be consistent with a lot of high power pump setups I've seen)

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Unread 07-18-2002, 09:03 PM   #81
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I agree that it is a very interesting thread. I enjoy learning from those who know a lot more than I do about the physics of heat and water.

Very interesting to see the science of PC watercooling mature.

Myrd, I am confused by your statement. Pump adds heat to water, water flows into radiator where heat is dissipated. The only thing I can think of is that you are indicating that once the water has gainied its maximum amount of heat(after WB and pump) that the radiator will be at its maximum effectiveness. Do you agree?

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Unread 07-18-2002, 10:16 PM   #82
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Here is a link to a resource for liquid weights and densities. This page shows the effective increase in work required by a pump to move a liquid.

Specific gravity chart

A normal impeller pump is limited in its ability to move liquid beyond a certain head pressure. As specific gravity increases losses in the impeller due to 'Slip' cause the effective losses to entropy to compound. Water at a specific gravity of 1 is roughly 16 times easier to move. If you were to replace the impeller type pump with a positive displacement pump you would gain a higher head capacity. But this would still not allow you to achieve the same efective cooling as with a chilled water medium.
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Unread 07-18-2002, 10:27 PM   #83
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Quote:
Myrd, I am confused by your statement. Pump adds heat to water, water flows into radiator where heat is dissipated. The only thing I can think of is that you are indicating that once the water has gainied its maximum amount of heat(after WB and pump) that the radiator will be at its maximum effectiveness. Do you agree?
This is the direct heat gain profile of most water cooled systems
[list=1][*]System pump imparts energy to water in the forms of Flow, Heat, entropy[*]Water passes to tubing which will remove heat from the enclosure to due Entropy. Heat always migrates to cold.[*]Water now slightly warmer 'Calories in gain from the tubing more dependent of Flow rate' moves to cooling block of component.[*]Based on temp. difference between the water and the cooling block and again the flow rate. Heat migrates to the liquid medium.[*]Water enters tubing where gains from the enclosure are now almost nonexistant.[*]Water enters radiator where in an ideal system it should reduce in velocity to enable a greater thermal transfer. [*]Water returns to the pump where velocity increases again.[/list=1]

The heat gain from the pump is not worth the reduced capacity caused by it preceeding the radiator. You want to have the least possible loss to entropy at the cooling block. Entropy caused by the tubing and fittings is from 'Laminar Flow'.

For once my job actually is fun!
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Unread 07-18-2002, 11:33 PM   #84
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Quote:
Originally posted by Myrd


This is the direct heat gain profile of most water cooled systems
[list=1][*]System pump imparts energy to water in the forms of Flow, Heat, entropy[*]Water passes to tubing which will remove heat from the enclosure to due Entropy. Heat always migrates to cold.[*]Water now slightly warmer 'Calories in gain from the tubing more dependent of Flow rate' moves to cooling block of component.[*]Based on temp. difference between the water and the cooling block and again the flow rate. Heat migrates to the liquid medium.[*]Water enters tubing where gains from the enclosure are now almost nonexistant.[*]Water enters radiator where in an ideal system it should reduce in velocity to enable a greater thermal transfer. [*]Water returns to the pump where velocity increases again.[/list=1]

The heat gain from the pump is not worth the reduced capacity caused by it preceeding the radiator. You want to have the least possible loss to entropy at the cooling block. Entropy caused by the tubing and fittings is from 'Laminar Flow'.

For once my job actually is fun!
Myrd,

I appreciate your enthusiasm, but a couple of your statements are either misleading or altogether incorrect. The two that particularly stick out regard heat moving from the enclosure through the tubing to the fluid and the second regards slowing flow in the radiator.

For starters, our plastic tubing is a pretty good insulator. Even if the fluid was 10°C cooler than the air within the case (extreme delta T) there would be little energy exchange across the tubing. As to the radiator, one only needs to consider heat flux between the tubing within it and the fluid. Heat flux integrated over the total surface area equals the total heat transfer. Heat flux depends on many factors, one of which is velocity. Higher velocity equals higher heat transfer coefficient. Higher heat transfer coefficient equals greater heat transfer for a given delta T or lower required delta T for a given amount of heat transfer. The differences are pretty minor and overall results are impacted a lot more by air flow, but nonetheless it is a fact that heat dissipation in a radiator will be more efficient at higher velocity. Only when the energy required to drive the higher flow rate exceeds the incremental gain in convection coeffient will reducing flow improve heat exchanger efficiency.

If you want the coolest fluid to strike the block, there can be no doubt that the radiator should preceed the block. Again, for "typical" flow rates, the differences are all but immeasurable, but it does not change the fact about which location is "best". On a more practical note, what really ought to dictate radiator position is which option yields the coolest air to the fans. A 2°C drop in air temperature will have a larger effect than moving the radiator ahead of vs after the pump.
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Unread 07-19-2002, 07:00 PM   #85
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I have a made a few changes from this photo. I have added a new TC-4 and I have removed all the PCI case covers. You will see in the photo air is pushed over the pump and out the back of the case. Now for heat being added to the water from the pump is entirely true.

I shut down the computer, not one fan was running in the entire system. Only the Ehiem 1250pump was running. I ran the pump over night. The ambient air temp inside the house was 27c and the water temp was a cool 46c.
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Unread 09-23-2002, 01:17 PM   #86
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Bump.

I've got 2 follow up questions, as I'm selecting my pump.

1- If the pump induces heat (and we've shown that it does), WHERE does this heat appear? I mean, if the heat comes from water friction on itself/tube walls, wouldn't the heat appear there? Wouldn't that mean that the heat appears mostly where the restrictions are, i.e. the rad and waterblock?


2-I'm shooting for a high flow rate, but I'm not willing to spend $100 on a pump, so I'm thinking about using two Via 1300 pumps in series. I realize that they probably won't be running in a very efficient range, but I'll get the higher head. Comments?
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Unread 09-23-2002, 03:53 PM   #87
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Hey Ben,

It's all about energy. You get heat as a by-product of energy dissipation in the flow. Energy dissipation in the flow is nothing more than pressure drop. So yeah, you get some small heat generation simply flowing through tubing, but the majority comes where the big pressure drops are. This will be any restrictive fitting, the block, and to varying degrees within the radiator.

And to answer #2, yes, you'll see higher total head with two pumps in series. This will result in higher total system flow, but probably lower overall pumping efficiency unless everything in your system is very low resistance.

Last edited by myv65; 09-23-2002 at 03:58 PM.
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Unread 09-23-2002, 03:56 PM   #88
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That's what I thought. Thanks Dave!

Anyone care to comment on using two pumps in series for more head?
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Unread 09-23-2002, 07:47 PM   #89
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Before i got the last pieces to my system (CPU and MB), i made a little (and somewhat unscientific) test of the amount of heat, the pump puts into the water.
test setup:
Ambient: 22,6
Eheim 1048 (600Ltrs/h @head)
radiator: BlackIce (classic i think)
block: DD Maze2
Senfu temp-diode (which i think i pretty accurate)
hooked it up like this:
pump, radiator, WB, res, pump.

filled and bleed the system, and put the probe between the fins of the rad. I let it sit still overnight, so the watertemp would reach ambient. startingtemp was 22,6C measured between the fins in the rad. fired up the pump, and had it run overnight: the following morning, the temp was at 28,4C, and ambient still 22.6C
made the same test again, but measured the temp on the WB instead - watertemp this time (measured on WB) 29,4C.
So, although not all that much, the pump does add somewhat heat to the loop, but nothing significant (at least for the Eheim 1048).
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Unread 09-23-2002, 10:12 PM   #90
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Hum... I was considering 2 Via Aqua's in series, until I came across the Rio 2500HP:

flow @ 0head: 782 gph
max head: 10 ft
Power: 55 Watts
3/4 connections
found it at $34.

The Via pump
flow @0: 370 gph
max head: 6 ft.
Power: 20.5 Watts
1/2 connections
Can be found at $18.

What to do, what to do...
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Unread 09-24-2002, 05:02 PM   #91
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Well I see you boys (and girls) finally have sorted out the question on where all the enegy goes . But I'll post a link anyway:
http://forums.procooling.com/vbb/sho...=&threadid=533

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Unread 09-24-2002, 05:15 PM   #92
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Geesh! You could have saved us a lot of trouble, a long time ago!

Here's a quote of the relevant part:
Quote:
If i place my pump inline only the "flowheat" is added to the water and if i place it submerged all consumed energy is added to the water.

The question everybody wants to know is:
How much of the total amount of consumed energy is transformed into flow and how much into heatloss in pump.
----------------------------------------
Here comes a calculation=answer of that question:
(I've used data from my Eheim 1250 as input)

E_tot=28W (total consumed power)
F_max=1200l/h=0.33l/s (maximum waterflow)
h_max=2m (maximum height)
E_flow=? power transformed to flow
E_loss=? powerlosses in pump
g=9.80665N/kg (gravityconstant)
rho=0.998kg/l (density of water in 20C)

(note that F_max and h_max don't occur simultaneusly but using them together gives the "best case senario")

E_flow=F_max*rho*g*h_max=
=0.33(l/s)*1(kg/l)*9.81(N/kg)*2(m)=6.5(Nm/s)=
=6.5W

E_loss=E_tot-E_flow=21.5W

Conclusion:
If the pump was able to pump 1200l/h to a height of 2m the powerinput into the water would be 6.5W and the powerloss in the pump would be 20.5W.
So by placing the pump under water you add 20.5W of unnecessery heat to the water.
Now let's run the numbers for a typical flow rate, say 100 gph, for the same Eheim 1250.

1250 curve here!

At 100 gph, the head is about 1.7 meters, or 2.42 psi.

E_tot=28W (total consumed power)
F=100 gph, or 0.10 L/sec
h=1.7m
E_flow=? power transformed to flow
E_loss=? powerlosses in pump
g=9.80665N/kg (gravityconstant)
rho=0.998kg/l (density of water in 20C)

E_flow=F*rho*g*h=
=0.10(l/s)*1(kg/l)*9.81(N/kg)*1.7(m)=1.67(Nm/s)=
=1.67W

E_loss=E_tot-E_flow=26.33W


That doesn't feel right...

Last edited by bigben2k; 09-24-2002 at 05:29 PM.
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Unread 09-24-2002, 06:16 PM   #93
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In a word, wrong. Yes, we have debated this to death, but that does not mean that everyone reached a consensus.

The referenced post makes at least two incorrect assumptions. One, that the maximum power consumed by the motor occurs at a mythical point of both maximum flow and maximum head. These two conditions are mutually exclusive of one another. The second is that the "flow energy" or whatever they heck they called it, is the only energy put into the fluid.

You've got motor inefficiency and pump inefficiency. For all practical purposes, all of the pump inefficiency goes into the water. Whether or not the motor inefficiency goes into the water depends on the particular situation. Submerged? No question it all goes into the water. Open-air? Most will convect off the motor casing into air with very little going into water.

A "typical" motor, however, will operate at >75% efficiency. Not so the typical pump. These tend to max out around 50% (among the pumps we use) or less. The "lost energy" is due to churning of the water that doesn't produce flow. It's the eddies and shearing around the impeller and between the impeller and volute.

If someone wanted to be really geeky about this, all you need to do is measure torque on the impeller shaft. Torque times rpm equals input power to the water. The fact that only a portion of this power is useful, namely flow rate times pressure rise, is irrelevant. The power still goes into the water.
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Unread 09-24-2002, 06:42 PM   #94
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I want to be geeky!!!

We're so near, closing the gap between the theoretical, and the practical...

The thing is, it would be just as easy to measure the torque of a pump impeller, as it would to measure a fan's... I have this mental picture of a belt driven pump: it would be easier to measure that, I think!

I also have another picture: a pump mod where the impeller is replaced with one that is a bit thicker, to minimize the loss of energy at that point.
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Unread 09-25-2002, 02:16 AM   #95
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myv65
Do you dissagree that it takes 6.5W of power to move 1200l/h to a height of 2m with a 100% effective pump?

What i find interesting is that my theoretical approach gives a power input of (6.5/28)=23% and skulemate got power input of 25% with his testing (see first page in this thread).

Skulemate
I'm glad to see that someone actually tried to measure the way you did. I presented an idea of measuring C/W a while ago that is not totally different than what you did.
(But it got shot down by BillA off course ).
Lookie here:http://forums.procooling.com/vbb/sho...W&pagenumber=4

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Unread 09-25-2002, 08:09 AM   #96
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Quote:
Originally posted by Dix Dogfight
myv65
Do you dissagree that it takes 6.5W of power to move 1200l/h to a height of 2m with a 100% effective pump?
Actually, I do disagree. Your statement would be true only if the pressure on the suction side of the pump was atmospheric. If you were running the pump, say, 2m above a reservoir, then the true pump lift would be 4m. If you had a reservoir 10m above the pump, the pump would effectively be "in the way" of natural, gravity-fed flow. Useful flow energy is flow rate multiplied by pressure rise. In a closed system, vertical lift really has no meaning and you need to look at total head loss through the system. Note that I've also left out flow loss to get it up those two meters. What if you're going through a 1/4" ID line? What if you've got elbows or other restrictions in the way? All you are considering is the gravity lift. All the pump knows is total head, a number that includes suction as well as discharge resistance. How those resistances are generated make no difference to the pump.

Even so, all one needs to do is read a pump curve (one that include pump efficiency). The folks that make most pumps have already done this work for us. The pumps we use at the flow situations we have usually run less than 50% efficient and often below 30%.
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Unread 09-25-2002, 10:38 AM   #97
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Thanks Dave.

To get back to my practical approach, could we take a minute here and run some calculations on that Eheim 1250, at 100 gph?

Like I wrote earlier, at 100 gph, the pressure drop will be 2.4 psi. The pump uses a 28 Watt source. Where do I go from here?


DixDogfight: you wrote: "Do you dissagree that it takes 6.5W of power to move 1200l/h to a height of 2m with a 100% effective pump?". I don't quite see the relation with the Eheim 1250, since it can't achieve that. It's either 1200L/h at 0 head, or 0 L/h at 2meters. What's up?


To recap what we have so far (correct me as necessary!):

-A pump uses a fix amount of energy, supplied in electrical form. (*** correction: a variable amount of energy***)

1-In a centrifugal design, less than 50% of that power is converted to create a flow (outlet to intake). This energy is converted to heat, from friction of water-on-water within the loop.

2-The remaining energy appears in the form of: churning, and energy loss. The churning again, induces heat from the water-on-water friction.

2b-The energy loss includes:
-Motor inneficiency (i.e. not all of the EM field generated by the coils moves the impeller)
-Heat from the coils (i.e. the coil wires, not being superconductors, dissipate some power)
-Friction/noise (as the impeller rattles around the housing)
-Noise (as the coils rattle, if applicable)

Last edited by bigben2k; 09-25-2002 at 11:21 AM.
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Unread 09-25-2002, 11:05 AM   #98
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Ben,

The only part of yours I'd take exception to is the statement that "A pump uses a fix amount of energy, supplied in electrical form." Energy input to the motor will vary with flow and head.

Here's a "typical" graph of an industrial centrifugal pump. There's a lot more information here than you'll normally get from the Eheim's, Via's, Rio's, Little Giant's, etc. of the world.

Among the additional info is efficiency, required suction pressure, impeller diameter, motor power at the operting point, etc. Note that many industrial pumps allow you to specify the impeller diameter so that you can vary flow at a given head. No, you can't change the diameter dynamically, but you can specify it up front.
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Unread 09-25-2002, 11:17 AM   #99
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BB2k
I don't know what amount of flow a 1250 produces at a 1m lift. But using the maximum flow at the maximum lift results in the maximum kinetic energy added.
A sort of worst case scenario if you will.
It was only a theoretical exaple to figure out a rough estimate of the energy added to the water due to pumping.

I agree on the recap.

myv65
Ok so i didn't state the pressure on the suction side. I didn't state the gravity constant either. If you change the setup (like adding 2-10m of tubing below) then of course the figure/number (6,5W) is wrong.
I'm talking about figure A and you about B and C (or close to it anyway).
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Last edited by Dix Dogfight; 09-25-2002 at 11:24 AM.
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Unread 09-25-2002, 11:25 AM   #100
myv65
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Join Date: May 2002
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All-righty then. The only thing missing is the kinetic energy of getting that flow going. So you get a V^2 * rho / 2 factor that'll raise the input power slightly. I understand where you're trying to get, and practically speaking your diagram "A" matches your previous statement about a 100% efficient pump/motor.

I think more to the point is the actual motor power. Motor nameplates merely state the maximum power the motor will consume. This does not mean that they consume that amount of power irrespective of flow and/or head.

That's the problem with skulemate's data on page 1. He assumes that his motor is drawing 180 watts with ~46 watts showing up as thermal energy within the water. In the graph I posted, the operating point (as indicated by the red right-angle) calls for ~1.45 hp. At other flows/heads, it would list a different number. We use a 1.5 hp motor because it can handle all combinations of flow/head that we use. This does not mean that the motor draws 1.5 hp.
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