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10GL + pump + W.block, does it work ? - Page 2

post #11 of 50
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if a 100$ fall from the sky around this month i will try it, an post the "lulz pictures" for you.

i honestly don't have anything better to try, or would you guys be so kind to tell me something to keep my mind/hand bussy for the moment.
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post #12 of 50
Quote:
Originally Posted by urgrandpasdog View Post
Not true, at load that PC is going to be putting out at least a couple hundred watts of heat. 10 gallons won't last long with that much.

The only way this "large res as radiator" would work is with huge amounts of water (like a swimming pool, or a fountain or something).
i agree

if you have MASSIVE amount of water, you will have so much surface area that the water reservoir can act as a rad (even though it will be super inefficient).

what you can do is make a shallow reservoir. try to maximize surface area, and 10 gal might work.

you'll have a LOT of evaporation, and that will cool your pc well. (see getto pic attached)

Edited by garbageacc3 - 4/10/10 at 5:40pm
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post #13 of 50
Quote:
Originally Posted by urgrandpasdog View Post
Not true, at load that PC is going to be putting out at least a couple hundred watts of heat. 10 gallons won't last long with that much.

The only way this "large res as radiator" would work is with huge amounts of water (like a swimming pool, or a fountain or something).
this man speaks the truth 5gal rez will go up 15c in under an hour
and after that it will get hotter faster

a big pool would work cause the concreete sides would absorb some heat for awhile

fountine would work as well cause the falling water would get cooled by the passing air as the water is falling kinda like a bong cooler
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post #14 of 50
Here comes some math, beware !!!

H = delta T * mCp
(heat) = (the change in temperature) * (the amount of energy required to change the temperature, per degree C, per unit of mass) * (mass)


10 Gallons = ~ 40 L
40 L = 40 KG (mass)
specific heatcapacity of water (Cp)= 4.186 kJ·kg−1·K−1
H = T * 40 KG * 4.18 KJ/(KG*k) = H = ~ 16.8 KJ/ T
T = (16.8KJ) / H
T is really delta T, I dropped the notation because of typing ^_^
that is to say, your 10 gallon reservoir requires 16.8 KJ to increase its temperature by one degree.
let us say you load 150 W from cpu + 48 W from pump,
heat dump (power)= 198W = 198 J/s
198 J / 16800J = 0.01178 S
that is to say, the increase in temperature will be 0.01178 K per second, if we assume no loss of heat to the outside world.
to continue, over one hour, your water will increase by 39.6 Degrees C after the addition of 712.8 KJ of heat.

This is all innaccurate as you will be losing heat to your surroundings, I have no way of estimating it though...If you use a steel resevoir and place some heat-pipes through it, to the top. I expect it could handle it
Edited by wcdolphin - 4/10/10 at 5:50pm
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post #15 of 50
Quote:
Originally Posted by shurik06_83 View Post
this man speaks the truth 5gal rez will go up 15c in under an hour
and after that it will get hotter faster

a big pool would work cause the concreete sides would absorb some heat for awhile
actually, it will heat up slower as temp of water starts to equal temp of steady state...

heat dissipation increases as the difference between the object temp and the ambient temp increases.

eg. look at temps when you run furmark, the temp rises sharply initially, then it increases slower as you reach steady state temp.

water is a better heat sink than concrete. and using concrete as a heat sink is no different than water.

the ONLY thing that matters for steady state temp is, at what temperature will the rate of heat into the water = heat dissipated by the water (via convection, conduction or irradiation).
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post #16 of 50
Ira-K should should check in and let you know why he runs a 5Gl res.

I run a DIY 3 Liter res that runs two systems, #1 single CPU and #2 CPU plus NB/Mosfet, both @ 2 degrees above ambient @ idle.

So, don't be afraid to experiment! It's a learning experience you shouldn't miss out on.

You'll get a lot of stated theory in this thread, but actual practice will get you farther than most book learning.
Edited by Dryadsoul - 4/10/10 at 5:55pm
  
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post #17 of 50
Quote:
Originally Posted by garbageacc3 View Post
actually, it will heat up slower as temp of water starts to equal temp of steady state...

heat dissipation increases as the difference between the object temp and the ambient temp increases.

eg. look at temps when you run furmark, the temp rises sharply initially, then it increases slower as you reach steady state temp.

water is a better heat sink than concrete. and using concrete as a heat sink is no different than water.

the ONLY thing that matters for steady state temp is, at what temperature will the rate of heat into the water = heat dissipated by the water (via convection, conduction or irradiation).
Actually, that is not too important. The temperature different just makes dQ/dTime larger, dQ is still the same though, it just increases the rate of transfer. The same amount of heat needs to be removed But yes, you are correct in correcting him, dTemp/dTime = k while the substance is in the same state.
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post #18 of 50
Quote:
Originally Posted by cdolphin View Post
Here comes some math, beware !!!

10 GL = ~ 40 L
40 L = 40 KG
specific heatcapacity of water = 4.186 kJ·kg−1·K−1
40 KG * 4.18 KJ/(KG*k) = ~ 16.8 KJ/K
K = 16.8KJ
i agree with your results but please keep track of units!!!

the result should just be 16.8 KJ/K

K= 16.8KJ

is like saying kelvin (a measure of temperature) = 16.8 kilojoule (a measure of energy)

i do agree with the "meaning, 10 gal will soak up 16.8kj for every 1C increase."



Quote:
Originally Posted by cdolphin View Post
that is to say, your 10 gallon resevoir requires 16.8 KJ to increase its temperature by one degree.
let us say you load 150 W from cpu + 48 W from pump,
heat = 198W = 198 J/s
198 J / 16800J = 0.01178
again, a s-1 disappeared

Quote:
Originally Posted by cdolphin View Post
This is all innaccurate as you will be losing heat to your surroundings, I have no way of estimating it though...If you use a steel resevoir and place some heat-pipes through it, to the top. I expect it could handle it
nonono, this is accurate assuming he is running it as a closed system (thermodynamically).

but the point is to run it as a open system so that it can dissipate heat via evaporation


if you add heat pipes and fins, then you're just running it as a isolated system, aka you're not using a res anymore, you're just running water through a rad = not an experiment, but a normal loop without a res (lol)
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post #19 of 50
Quote:
Originally Posted by cdolphin View Post
Actually, that is not too important. The temperature different just makes dQ/dTime larger, dQ is still the same though, it just increases the rate of transfer. The same amount of heat needs to be removed But yes, you are correct in correcting him, dTemp/dTime = k while the substance is in the same state.
uh, i dunno why you're talking about change in time as we're assuming t is infinity since we're talking stead state cause hes gonna be running the comp 24/7


change in temperature (units of K) dividied by time (units of, let say s-1) does not equal K (units of temperature)...
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post #20 of 50
Quote:
Originally Posted by garbageacc3 View Post
uh, i dunno why you're talking about change in time as we're assuming t is infinity since we're talking stead state cause hes gonna be running the comp 24/7


change in temperature (units of K) dividied by time (units of, let say s-1) does not equal K (units of temperature)...
note the lowercase k, constant.
Yes, I dropped units, unintentionally...doing this on an iPod is lame.
I believe the resevoir is closed?
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