Pro/Forums

Pro/Forums (http://forums.procooling.com/vbb/index.php)
-   General Liquid/Water Cooling Discussion (http://forums.procooling.com/vbb/forumdisplay.php?f=9)
-   -   Pumps and heat (http://forums.procooling.com/vbb/showthread.php?t=3756)

bigben2k 07-18-2002 12:02 PM

Quote:

Originally posted by myv65
The motor's inefficiency mainly goes off as convection for an inline pump. It obviously goes into the water for a submerged pump. The pump's inefficiency goes into thermal energy in the water. The pump's useful energy goes into circulating flow. All of the energy put into the pump's motor must eventually leave the box. Under steady-state conditions, energy leaving equals energy entering. For a while after starting up, energy leaving is less than entering with the excess showing up as thermal energy in the fluid (and kinetic energy of the moving fluid). After powering off, flow will slow to a halt and the fluid will cool to ambient. This is the release of stored energy and results in an energy output with no energy input.
I see what you mean, but...

I think that what we're talking about here involves 2 things:
1-The efficient part of the pump (say 70%), which moves the water, which heats up due to friction.
2-The inefficient part of the pump (the remaining 30%), which includes (but not limited to) heat, which heats up the water through the pumps housing. This is reduced if the pump is in-line.

myv65 07-18-2002 12:16 PM

The tubing portion is pretty easy to explain. It acts very much like a spring in a mechanical system. It will stretch (dilate) when a pressure gets applied. This stretch is very minor, but represents some finite change in system volume. The storage energy is simply delta-P (running versus stopped) times the system volume. It's not so easy to calculate, but is completely immaterial. Once you reach steady-state, stored energy has no bearing on the matter.

bigben2k 07-18-2002 12:35 PM

Quote:

Originally posted by myv65
The tubing portion is pretty easy to explain. It acts very much like a spring in a mechanical system. It will stretch (dilate) when a pressure gets applied. This stretch is very minor, but represents some finite change in system volume. The storage energy is simply delta-P (running versus stopped) times the system volume. It's not so easy to calculate, but is completely immaterial. Once you reach steady-state, stored energy has no bearing on the matter.
You're absolutely correct.

So that leaves what I was saying, which is that the energy (efficient part) supplied by the pump, fights friction, which turns into heat.

I'm at the end of the tunnel now... I think!

Let me try to apply this info:

Let's say that I set up a rig. It doesn't matter what it is, as long as it's the same.

In rig A, I use a 1000 gph pump, and I achieve a 200 gph effective flow rate. In rig B, I use a 400 gph pump, and achieve a 100gph flow rate. (pump sized accounting roughly for the increased restriction at the higher flow rate)

I get a higher flow rate in rig A than in rig B, but I'll have induced twice as much energy in the water.

Does that look about right?

bigben2k 07-18-2002 01:59 PM

Another bit, while I'm walking out of the tunnel...

Some people at OC (from the link posted in this thread) seem to think that an increased mass of water would make a difference. I disagree. Since we're inducing heat, in the form of a heat source (say 7 watts, for our little 1048), from its efficient part, the heat is "applied" to the water over time, regardless of the mass of the water, which can be translated into work.

This heat will be dissipated by the rad, as it achieves the balance point, between ambiant air, and the water temp. Since the rad dissipates the same heat, and reaches the same balance point, the increase in mass of the water would only delay reaching that balance point.

My other step-outside-the-tunnel is this: couldn't we calculate this heat, if we knew the exact (effective) flow rate?

Let's say that we have 100 gph. The work performed by moving 100 gallons of water over a time period of 1 hour, that can be calculated, can't it?

What if the density of the coolant was less (or more) than that of water? Would it make any difference? I'm guessing no, but why?

myv65 07-18-2002 02:20 PM

Yeah, I'd say you're reaching daylight now. :D

First let's look at your 1000 vs 400 gph pump example. Simply because one pumps 200 gph while the other does 100 gph doesn't tell us too much. What you need to know is how much energy the motor uses. We can assume that the 1000 producing 200 gph is running at a lower efficiency since it is pumping 20% of "rated" flow vs 25% for the other. This may be a bad assumption, however, as flow vs head (and efficiency) curves differ for different pumps.

If you have a chart of efficiency vs flow (commonly efficiency gets graphed right along with head vs flow), then you can answer the question. If not, you must measure power consumed by the motor.

Regarding your "volume of water present" bit, you are entirely correct. Ultimately, volume has zero bearing if we run the system long enough to reach steady-state. All it does is factor into how long it takes to reach steady-state. If you want to pick nits, you can argue that more volume requires more surface area (tubing, reservoir, etc) and this added surface area aids the radiator in dispelling heat. Yeah, whatever. Unless the reservoir is relatively large, its convective heat loss will be peanuts next to a good radiator with sufficient flow of ambient air.

Finally, for your question of calculting "the power of moving 100 gph". Sure, power is nothing more than flow rate multiplied by delta-P. If you can measure delta-P (ala BillA's manometer explaination a few days back), then you can determine the useful work of the pump.

You also asked about density. Density alone doesn't have much bearing on flow power. In an open system where suction and discharge are at different elevations, sure, but not in a closed system. Viscosity really determines flow for a given pump and piping system. Lower viscosity = higher flowrate. As viscosity increases, pressure drop versus flowrate decreases. Total pressure drop tends should remain reasonably constant as the lower viscosity gets offset by higher flow.

bigben2k 07-18-2002 02:20 PM

For those with a knack for physics, check this out.

Let's see... Watts can be calculated as Joules/seconds.

I'm moving 100 gallons of water, over 1 hour, or 60 minutes, or 3'600 seconds. How do I get the Joules?

myv65 07-18-2002 02:36 PM

As the old commercial says, "Sorry Charlie". You can't get there from here. You need delta-P.

It may help to take a gander at the units in their base form. Energy has units of force * length. Power has units of force * length / time. Flow has units of length^3 / time.

In order to get power based on flow, the units need to come out right. What are the missing units? Well, divide power by flow and you have (force * length / time) / (length^3 / time) = force / length^2. Recognize that? What if I said lbf / in^2 (psi)?

To calculate flow power you require flow rate and pressure rise. No two ways around it.

JimS 07-18-2002 02:41 PM

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. :) :)

bigben2k 07-18-2002 02:45 PM

Quote:

Originally posted by myv65
Yeah, I'd say you're reaching daylight now. :D

Thanks, I needed that!

000 07-18-2002 02:50 PM

I'm pretty sure that if your density is greater and your volume is the same your mass will be greater which means you will be doing more work. (well at least most of the time it does, in some cases mass can get canceled out ( I've learned this after many tricky engineering physics ))


It seems that density is very closely tied to viscousity also, I'd be interested in a liquid that is more dense than another but yet less viscous.

-Sidney

000 07-18-2002 02:52 PM

Also the point about pump heat being fairly moot with a radiator may be true for ambient cooling, but it's not for sub ambient cooling, you want the least amount of heat into your system as possible so you can have the coldest temps going to your components.

bigben2k 07-18-2002 02:57 PM

Quote:

Originally posted by myv65
As the old commercial says, "Sorry Charlie". You can't get there from here. You need delta-P.

It may help to take a gander at the units in their base form. Energy has units of force * length. Power has units of force * length / time. Flow has units of length^3 / time.

In order to get power based on flow, the units need to come out right. What are the missing units? Well, divide power by flow and you have (force * length / time) / (length^3 / time) = force / length^2. Recognize that? What if I said lbf / in^2 (psi)?

To calculate flow power you require flow rate and pressure rise. No two ways around it.

Of course, and I've even said that myself, in another thread, where I wrote something along the lines of 100 gph at 10 psi, isn't the same as 100 gph at 100 psi.

So I need to know the pressure drop between the pump inlet and outlet, as well as the effective flow rate.

bigben2k 07-18-2002 03:04 PM

Quote:

Originally posted by 000
I'm pretty sure that if your density is greater and your volume is the same your mass will be greater which means you will be doing more work. (well at least most of the time it does, in some cases mass can get canceled out ( I've learned this after many tricky engineering physics ))


It seems that density is very closely tied to viscousity also, I'd be interested in a liquid that is more dense than another but yet less viscous.

-Sidney

For the same flow rate, yes, but you won't achieve the same flow rate with the same pump, wether its because of the density, or the viscosity. although, I remember reading somewhere though, that most pumps reach the same flow rate, regardless of fluid density, but since a heavier fluid will probably be thicker too, we just can't picture it.

I need a viscosity table! I believe that mercury, although lots denser, might have a similar viscosity as water. Does anyone know?

(It doesn't mean that mercury is a better coolant, we can look into that some other time!)

000 07-18-2002 03:08 PM

Mercury is a much better coolant it is about 16 times better than water... just really dangerous and I think pretty hard to pump, but maybe not.

-Sidney

myv65 07-18-2002 03:10 PM

Quote:

Originally posted by 000
It seems that density is very closely tied to viscousity also, I'd be interested in a liquid that is more dense than another but yet less viscous.

-Sidney

OOOOO! I know the answer to that one!

Actually, there's plenty of examples. Water is more dense than oil, yet has far less viscosity (especially at low temperatures).

Water is actually pretty unique in its cooling abilities. Its heat transfer characteristics are better than darn near everything and its viscosity is also lower than darn near everything else. Too bad it doens't do well below 0°C. But that's where additives come into play.

bigben2k 07-18-2002 03:16 PM

Ok, fluid viscosity, measured in Pa*s for

Water at 0C=1.8 * 10^-3
Water at 20C=1.0 * 10^-3
Mercury = 1.55 * 10^-3

Note: Liquid viscosities tend to decrease with increasing temperature

From
http://www.phys.virginia.edu/classes...ds2/node2.html

and
http://www.mas.ncl.ac.uk/~sbrooks/bo...01/node13.html

000 07-18-2002 03:19 PM

Hmm yeah thats true I forgot about water and oil. Well I guess density doesn't necessarily affect viscosity. I have tried looking up better liquids and the only one that I could find was mercury it's thermal conductivty is 8.something and water is about 0.6

bigben2k 07-18-2002 03:24 PM

Yeah, mercury is nasty stuff!
http://danpatch.ecn.purdue.edu/~epad...src/poison.htm

http://www.healthyvermonters.info/hp.../mercury.shtml

myv65 07-18-2002 03:27 PM

*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.

000 07-18-2002 03:27 PM

Hmm well if mass doesn't affect the pump then I guess you can use mercury no problem (except for the toxicity problem). Maybe someone should try that, would look really neat in some tygon tubing, just make sure that your tubing won't absorb it (You sure don't want mercury vapor around you).

-Sidney

bigben2k 07-18-2002 03:30 PM

Quote:

Originally posted by 000
Hmm well if mass doesn't affect the pump then I guess you can use mercury no problem (except for the toxicity problem). Maybe someone should try that, would look really neat in some tygon tubing, just make sure that your tubing won't absorb it (You sure don't want mercury vapor around you).
NO!

Mercury is very, very nasty stuff! If you spill it, it will be released in the form of vapors, and your body can absorb a lot of it, very quickly!!!

You can't even pick it up with a paper towel, because if you touch it, you'll absorb even more!!!

000 07-18-2002 03:35 PM

Yeah I know, although I don't think it's as bad as they make it sound because I personally have played with the stuff and didn't get sick or anything (touching it pushing it around messing with it quite a bit), and that was when I was only about 7 or 8 so I should have been affected much quicker than an adult. (What can I say I was a curious kid and found a mercury filled thermometer).

Edit:
Although I don't suggest that anyone do what I did, or that anyone actually mess around with mercury.

-Sidney

bigben2k 07-18-2002 03:54 PM

It certainly would take care of a few problems though: corrosion, algae, ...

(Note: don't even think about it!)

000 07-18-2002 04:04 PM

Hehe, that's very true. And for the people using chillers they wouldn't have to worry about it freezing ;) And you could see really easily if you were leaking any.... or you could just do a self evaluation (am I going mad... am I going mad...) hmm might not work for some people though :D

I imagine it's fairly hard to get a large amount of the stuff though (probably a good thing ;) )

bigben2k 07-18-2002 04:15 PM

Well, I found an internet link, but I don't want to post it.

They sell it for over $260 per pound, and seeing that a cubic foot of it weighs about 850 (eight hundred and fifty) pounds, I'm sure that it's not a lot, in volume.

Also, since it is hazardous, it'd be VERY expensive to ship.


All times are GMT -5. The time now is 11:00 AM.

Powered by vBulletin® Version 3.7.4
Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
(C) 2005 ProCooling.com
If we in some way offend you, insult you or your people, screw your mom, beat up your dad, or poop on your porch... we're sorry... we were probably really drunk...
Oh and dont steal our content bitches! Don't give us a reason to pee in your open car window this summer...