Calculating water reservoir heat-load (warning, math)

[Nightbird]

Hey Jorick, I thought about your problem for a bit and had some ideas.

First, and the simplest, is to keep your office doors open and put a high quality quiet, high airflow fan either inside your office blowing out (more effective), or outside blowing in (less noise). This will circulate the air in your house and keep your office no more than 2degrees warmer, though a light breeze from the fan will make you feel cooler. Rearrange your office if necessary to point the exhaust heat from your PC towards the door and away from you.

To go the large water reservoir route, I'd imagine that a closed reservoir is better as you wouldn't need to refill it often, which can get annoying. The problem is the size, so I looked online and the cheapest I found is this:
http://www.horsemansdepot.com/productdetails.aspx?part=FEP-50WT&ref=base
which is still 80$, but after you drop your radiator in and seal the top around the tubes with a material like cling-wrap, there won't be any need to refill it. The lid is only 5" so you have to be willing to cut it open once and then reseal it with some waterproof glue.

 

[Jorick]

Thanks for the ideas everyone.. Sorry I wasn't able to get back to this today; some friends of mine wanted to go out and try out some other water cooling... with some yeast and hops... yeah.


Anyways... My office right now is kind of an odd room. The wall my PC is on is a shared concrete block wall with the neighbor's townhouse, so drilling into it to do something would probably be a bad idea.

Going clockwise, the next wall holds a long closet, and then the door into the room. The other side of that closet is a bathroom, but the pipes are on the opposite side of the bathroom from the closet.

The third wall is common with my living room. It's got a spare love seat. I could do something with that wall, but I wouldn't want to move my PC there because...

The fourth wall is an external wall where the office tv/entertainment center is. It also has a floor to ceiling window (go go 70s construction).

If I moved the PC to the wall common with the living room, I wouldn't really be able to watch the tv, as my head would have to crank around Poltergeist style. I could move my pc along the external wall, move the TV to the living room wall, and move the love seat to the wall common with the neighbors.

However, it's kind of useless. My HOA is death on any external modifications, and there aren't any bushes along that wall to hide a vent.

As to your main idea, I don't see any problem with it. As long as you understand it will eventually reach an equilibrium then you're on the right track. I might add you'll need a pair of pumps to push through all the copper tubing you'll need inside your water container. While tubing in general is a small part of loop resistance, once you get into tens of feet of it it starts to add up - as does the bill at the hardware store.

Thanks for the pressure drop link.

Yeah, copper tubing isn't cheap. I don't think I'll need tens of tens of feet for a 30-40 gallon container, though.. In my head, I'm picturing myself now using something like http://www.walmart.com/ip/Sterilite-35-Gallon-Latch-Tote-Set-of-4/10401055 . I'm thinking of making a smallish coil, not something that loops around the entire bucket in a big spiral; then, assuming that's not good enough (Because it probably won't be), I'll drop something like http://www.homedepot.com/h_d1/N-5yc...splay?langId=-1&storeId=10051&catalogId=10053 in the bottom of the bucket to get some movement in the water to distribute the heat around. In my mind, at least (and my mind is a scary place) it's a decent trade-off between complexity/points-of-failure and price.

 

[Jorick]

Where will this bucket of water stand ??
Water + pc = condensation
You could always use mix some antifreeze in the reservoir of your pc and submerge your rad into water, i think?
Anyways I think I didn't got the idea at all, will read the topic again

The idea is pretty much the same concept as BobbyBubblehead's Most Professional-Looking Watercooling Setup Ever. (Sorry, couldn't resist; you really should call it the MPLWCSE ).

The large reservoir is filled with water. The loop from the PC runs through the water, and the heat from the loop will transfer to the water (because the water is cooler than the PC loop until it gets to equilibrium). I'm talking about using copper because ... Well... I never thought about using anything else. I know a few guys in the area who are big into fish tanks, they may have something like BB's stainless steel thing laying around I could have.

I don't know how WC rads take to being submerged. Good point, though; that may be the most cost/time effective for the intercooler (Get a cheap 1x120 or something for submerging; see my previous post with the water circulation pump too)

I'm not worried about the Bucket-O'-Water freezing. It'll be inside, in the same room as the computer, and I live in Arizona in the US. During the summer time, if it doesn't hit 100+ degrees F, we think we're in a deep cold-snap. I've never seen snow here, either, during the winter time. Of course, I keep the AC on so it's not 100 inside. You get the idea, though; it's not a cold place to live.

Condensation could be a factor in humid areas, I guess, but I'm not really worried about it here... It may get above 105-110 degrees F on a regular basis during the summer, but it's a *dry* heat... (Everyone says that who hasn't lived here... when it's triple-digits for a couple of months, it doesn't matter ) Here's a random graph I found on the Intertubes of humidity here : http://www.cityrating.com/cityhumidity.asp?City=Phoenix

 

[Jorick]

Hey Jorick, I thought about your problem for a bit and had some ideas.

First, and the simplest, is to keep your office doors open and put a high quality quiet, high airflow fan either inside your office blowing out (more effective), or outside blowing in (less noise). This will circulate the air in your house and keep your office no more than 2degrees warmer, though a light breeze from the fan will make you feel cooler. Rearrange your office if necessary to point the exhaust heat from your PC towards the door and away from you.

Yep, I already have a fan blowing... It's not a very good fan, and the bearings are almost shot, but it's a fan.

To go the large water reservoir route, I'd imagine that a closed reservoir is better as you wouldn't need to refill it often, which can get annoying. The problem is the size, so I looked online and the cheapest I found is this:
http://www.horsemansdepot.com/productdetails.aspx?part=FEP-50WT&ref=base
which is still 80$, but after you drop your radiator in and seal the top around the tubes with a material like cling-wrap, there won't be any need to refill it. The lid is only 5" so you have to be willing to cut it open once and then reseal it with some waterproof glue.

Heh, interesting find. The problem I see with a closed system is the solution it presents; no evaporation, so it'll take longer to cool back off. Still, worth a thought.

 

[BobbyBubblehead]

jorick - pffft MPLWCSE my only regret was forgetting the gaffer tape effect, ghetto fix essential

thought the irratic bands instead of a case were almost genius... its not caught on though

 

[Jorick]

And now that I've responded to people, the post I meant to do when I woke up this morning...

Why use water in the reservoir? Yes, water is what everyone in this forum works with, or wants to work with, so it's familiar... But.....

I had a thought (and as we've seen, my thoughts are weird). Why not use wax? According to http://en.wikipedia.org/wiki/Paraffin the specific heat of paraffin wax is 2.14–2.9 J g−1 K−1 (joule per gram per kelvin). Split the difference in the range and call it 2.52 J per gram per kelvin.

272.15 kelvins per C, so 685.818 J per gram per C... Much better than water.

Picking 5 pounds of wax for no good reason other than I had 5 beers last night.. A pound is 453.6 grams, so that's 2265 grams of wax.

Plug it into the equation, you get:

Joules Required = 2265 * 1degC * 685.818

1,553,377 Joules required to heat that 5 pound lump of wax up 1 degree C. Converting to time, using the 550watt from earlier, you get ... 2824 seconds (~47 minutes) to heat it up one degree C.

It seems you can get a pound of wax for ~$2-$3 US. Plus, you could add fragrance to it after it gets melted the first time

For volume, according to http://www.onestopcandle.com/candle/canmeasures.php, 1 pound of wax is ~96 teaspoons, which is 1/8 of a gallon, so a little over half a gallon for my 5-pound example. That's nothing, as far as taking up space in the room. No evaporation to worry about, etc.

The fly in the ointment of the idea, though, is the heat transfer coefficient. According to http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html it looks like water is roughly double paraffin wax. That can probably be counteracted, somewhat, by a bigger intercooler design. Physics, as noted above, is not my strong suit.

I'd see the build for this as a glass tray of some sort (baking pan, whatever); you'd melt the wax the first time in an oven, then sink the intercooler into the liquid wax; add any fragrances you want . Let it harden back up for easy carrying, then attach it to your PC loop.

This seems too nutty to work.

 

[QuietIce]

The idea with the wax doesn't seem feasible. The reason water might work is because; a) as you noted it transfers heat very well, and b) it will circulate itself by convection because the warmer water near the coil (inter-cooler) will tend to move upward and in that process will be replaced by cooler water. With wax you won't have any mixing of warm and cool material so the wax near the coil will get warmer, faster, and slow down your heat transfer.


By using the heat transfer rate of copper you can calculate now many inch² of copper you'll need to get rid of your heat. In turn that should give you a reasonable idea of how many feet of tubing you'll need ...

 

[Jorick]

The idea with the wax doesn't seem feasible. The reason water might work is because; a) as you noted it transfers heat very well, and b) it will circulate itself by convection because the warmer water near the coil (inter-cooler) will tend to move upward and in that process will be replaced by cooler water. With wax you won't have any mixing of warm and cool material so the wax near the coil will get warmer, faster, and slow down your heat transfer.

I was thinking about, in the wax situation, instead of having a coil, have a simple loop of copper, but braising/attaching some thin copper fins (or maybe just a plate, like you can get at arts'n'crafts stores) to the copper tube to pump up the area of copper exposed to wax. It has the added side benefit of spreading the heat out from the pipe, as well.

By using the heat transfer rate of copper you can calculate now many inch² of copper you'll need to get rid of your heat. In turn that should give you a reasonable idea of how many feet of tubing you'll need ...

I've got no idea how to do that... Off to google I go.

 

[Jorick]

... and my brain exploded...

I know I have to somehow use the Fourier equation. But, I've got no idea what the numbers I have to plug in are.

Rather, I do know some... From http://www.engineeringtoolbox.com/conductive-heat-transfer-d_428.html :

 Fourier's Law express conductive heat transfer as
    q = k A dT / s         (1)
    where
    q = heat transferred per unit time (W, Btu/hr)
    A = heat transfer area (m2, ft2)
    k = thermal conductivity of the material (W/m.K or W/m [sup]o[/sup]C, Btu/(hr oF ft2/ft))
    dT = temperature difference across the material (K or oC, oF)
    s = material thickness (m, ft)

q = 550 W
A = what we'd try to solve for.
k = 401 W/(m^o C) --> This doesn't make sense to me. I found this at http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html I assume it's supposed to be 401 W/m ^oC
dT = 1 deg C for the sake of argument.
s = Type L copper tubing at 1/2" ID has a thickness of .001 m

550W = 401 W/(m^o C) * A * 1 ^oC/.001m
multiple both sides by (.001m / 1 ^oC))

(550 W * (.001m / 1 ^oC)) = 401 W/(m^o C) * A
Finish the multiplication on the left side

(.55 W(meter) / 1 ^oC) = 401 W/(m^o C) * A
divide both sides by 401 W

(.55 W(meter) / ^oC) * (1 / 401 W ) = (1 /(m^o C)) * A
Finish unit cancellation

(.55 m / 401 ^oC) = (1 /(m^o C)) * A
Multiple both sides by 1 m ^oC to get A by itself

(.55 m / 401 ^oC) * (1 m^o C) = A
Cancel out the degrees

(.55 m / 401) * (1m) = A

(.55 / 401) * 1 m² = A

.001 m² = A

1 square mm of copper can conduct 550w of heat? This seems awfully... convenient and round.

 

[markp1989]

If you want to keep that room cool, you only have one option. Get the heat out. Vent the PC exhaust out of the room.

getting the heat is what i have been trying to do my self, im planning on putting a fan in the window to remove the heat soon.

could keep the 50gal drum out side, but in winter you have to wory about it getting frozen, i dont know how a submerged rad would cope under frozen water?

or have the rad out side, but i already mentioned somthing similar to that before, and the ops layout stops this 1 from happening. maybe he could run holes in the ceiling, and put the radbox and fans in the loft . (asuming there is aloft and not anohter person living up top)

Or.... If you are handy at all as a plumber and have plumbing nearby, eliminate your PC pump, hook up your cold water line (with a pressure/volume regulator) to your PC, then have the other end of the loop go to a drain line or, better yet, outside to your lawn/garden.

im sure the water board wont be pleased about wasting this much water, and if your waters on a meter it would cost crap loads.

 

[Nightbird]

1 square mm of copper can conduct 550w of heat? This seems awfully... convenient and round.

Yes and no.
http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html

Copper can pull an amazing amount of heat quickly, but how fast heat is removed by your setup isn't determined by the fastest heat conductor in the loop, but the slowest. Normally, this is air. In your case, it is water.

The thermal conductivity of air is 0.024, whereas as it is 0.58 for water. This means that a 1x120mm passive radiator in water performs as well as a 24x120mm passive radiator in air, when the ambient temperatures are equal.

To me, 24x120mm is enough to passively cool your system, but hey, doubling to 48x120mm would just mean dunking in a 2x120mm radiator. I also don't see the need to use expensive copper coils, as copper radiators are readily available and cheap, and provide what you need most, lots of surface area, for very little copper in weight.

As far as whether 50gallons is enough, I really believe it is. I don't think your system will be offloading 550watts constantly, and even in a close reservoir I recommended earlier, some heat will be radiated off the water container.
http://www.pugetsystems.com/submerged.php
Here, they submerged a pc in mineral oil with no radiator. With full stress, the temperature went up to 88c after 4 hours, with 6 gallons of mineral oil. Water can hold 2.5 times the heat of mineral oil (http://www.engineeringtoolbox.com/specific-heat-fluids-d_151.), and you have 50 gallons of it. This means roughly your 50 gallons of water is equivalent to 125 gallons of mineral oil. The amount of energy needed to increase 6 gallons of mineral oil from 20c to 88c would only increase 50 gallons of water by 3.3c.

 

[QuietIce]

... and my brain exploded...

I know I have to somehow use the Fourier equation. But, I've got no idea what the numbers I have to plug in are.

Rather, I do know some... From http://www.engineeringtoolbox.com/conductive-heat-transfer-d_428.html :

 Fourier's Law express conductive heat transfer as
    q = k A dT / s         (1)
    where
    q = heat transferred per unit time (W, Btu/hr)
    A = heat transfer area (m2, ft2)
    k = thermal conductivity of the material (W/m.K or W/m [sup]o[/sup]C, Btu/(hr oF ft2/ft))
    dT = temperature difference across the material (K or oC, oF)
    s = material thickness (m, ft)

q = 550 W
A = what we'd try to solve for.
k = 401 W/(m^o C) --> This doesn't make sense to me. I found this at http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html I assume it's supposed to be 401 W/m ^oC
dT = 1 deg C for the sake of argument.
s = Type L copper tubing at 1/2" ID has a thickness of .001 m

550W = 401 W/(m^o C) * A * 1 ^oC/.001m
multiple both sides by (.001m / 1 ^oC))

(550 W * (.001m / 1 ^oC)) = 401 W/(m^o C) * A
Finish the multiplication on the left side

(.55 W(meter) / 1 ^oC) = 401 W/(m^o C) * A
divide both sides by 401 W

(.55 W(meter) / ^oC) * (1 / 401 W ) = (1 /(m^o C)) * A
Finish unit cancellation

(.55 m / 401 ^oC) = (1 /(m^o C)) * A
Multiple both sides by 1 m ^oC to get A by itself

(.55 m / 401 ^oC) * (1 m^o C) = A
Cancel out the degrees

(.55 m / 401) * (1m) = A

(.55 / 401) * 1 m² = A

.001 m² = A

1 square mm of copper can conduct 550w of heat? This seems awfully... convenient and round.

I didn't check everything else in there but the last lines have a BIG oopsy! in them.

0.001 m² = 1000 mm², not 1 mm²


Quite frankly, that doesn't seem like nearly enough surface area. Have you thought about how much heat the water on the other side of the pipe can absorb? I mean, 1000 mm² of copper is enough but is 1000 mm² of water area enough ...?

 

[Jorick]

Yes and no.
http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html

Copper can pull an amazing amount of heat quickly, but how fast heat is removed by your setup isn't determined by the fastest heat conductor in the loop, but the slowest. Normally, this is air. In your case, it is water.

The thermal conductivity of air is 0.024, whereas as it is 0.58 for water. This means that a 1x120mm passive radiator in water performs as well as a 24x120mm passive radiator in air, when the ambient temperatures are equal.

To me, 24x120mm is enough to passively cool your system, but hey, doubling to 48x120mm would just mean dunking in a 2x120mm radiator. I also don't see the need to use expensive copper coils, as copper radiators are readily available and cheap, and provide what you need most, lots of surface area, for very little copper in weight.

As far as whether 50gallons is enough, I really believe it is. I don't think your system will be offloading 550watts constantly, and even in a close reservoir I recommended earlier, some heat will be radiated off the water container.
http://www.pugetsystems.com/submerged.php
Here, they submerged a pc in mineral oil with no radiator. With full stress, the temperature went up to 88c after 4 hours, with 6 gallons of mineral oil. Water can hold 2.5 times the heat of mineral oil (http://www.engineeringtoolbox.com/specific-heat-fluids-d_151.), and you have 50 gallons of it. This means roughly your 50 gallons of water is equivalent to 125 gallons of mineral oil. The amount of energy needed to increase 6 gallons of mineral oil from 20c to 88c would only increase 50 gallons of water by 3.3c.

Interesting.. I definitely see the point that it's the weakest link in the chain that has to be looked at, not the strongest. In this case, the copper is, by far, the strongest link.

I also see the point about radiators.. I can probably get a random copper heater core from craigslist, or even bought new at the store, for less hassle than doing it myself.

And getting a used aquarium is easy enough. Hmm.. Warm water fish? Maybe just some coral?

Is it valid, physics-wise, to just look at the ratios in thermal conductivity and heat capacity, if you have a known baseline like that mineral oil video? It runs in my head that I read something about a larger volume having changing values as far as one of those numbers goes.

 

[Jorick]

I didn't check everything else in there but the last lines have a BIG oopsy! in them.

0.001 m² = 1000 mm², not 1 mm²

Quite frankly, that doesn't seem like nearly enough surface area. Have you thought about how much heat the water on the other side of the pipe can absorb? I mean, 1000 mm² of copper is enough but is 1000 mm² of water area enough ...?

Oops, I forgot the whole "squared" part... I stand corrected.

I tend to agree with you... 1000 mm² is only 1.5 square inches. This does not seem right.

Nightbird did make a good point, that copper is not the limiting factor here... Instead, the water is.

Running through that calculation, using .58 for water (listed in the link in my last equation run-through) gives .95 m² needed which is 1472.503 square inches.

Using the equation to calculate surface area of a cylinder (no top/bottom), assuming .5" inner diameter (.25" inner radius) copper tubing (since that's the worst part of this system) gives :

2 * pi * .25" * h = 1472.503

it'll need a tube 937 inches long, if we only went with a tube. Impractical and not cost-effective, I agree.

 

[QuietIce]

As someone else noted, if you're willing to go with submerged radiators a cheap alternative is a heater core. I love heater cores and use them almost exclusively for my WC loops. A 1977 Bonneville heater core runs a little over $20 at the autoparts store. Generally, one heater core per component being cooled is good enough.


I might consider a used heater core it if it had been used only for a WC loop but never if it had been on a car ...

 

[Jorick]

As someone else noted, if you're willing to go with submerged radiators a cheap alternative is a heater core. I love heater cores and use them almost exclusively for my WC loops. A 1977 Bonneville heater core runs a little over $20 at the autoparts store. Generally, one heater core per component being cooled is good enough.


I might consider a used heater core it if it had been used only for a WC loop but never if it had been on a car ...

Lol, I've got no problem with heater cores.. My first WC rig was a silver-painted heater core I bought as part of a kit from somewhere; the pump+res was a little aluminum cube with a little submersible pump inside; the cpu block was a slug of copper, and the mounting system was a metal bar with wingnuts to tighten it down on the cpu socket. Mmm... mixed metals... and I ended up tearing the plastic tab off the socket that the wingnut hooked on to.

Ahh, the bad old days.

Anyways, the issue I always had with the heater cores is that installing barbs always seemed ... troublesome. I've seen the walkthroughs, but I don't have any of the tools needed; it might be worth it to get it pre-modded from Danger Den, even if they're only 3/8" barbs.

 

[Bobnova]

What is under this room?
How bout a couple holes in the floor to put the radiator in the basement/downstairs?
EDIT:
Or above for that matter, dump the heat in the already hot attic.

 

[Jorick]

What is under this room?
How bout a couple holes in the floor to put the radiator in the basement/downstairs?
EDIT:
Or above for that matter, dump the heat in the already hot attic.

Under -- Concrete Foundation; no basement/first floor
Above -- Sloped ceramic tile roof; no attic.

 

[Bobnova]

Well now that's a bit of an issue.



Relating to a radiator in air vs a radiator in water:
In air there is forced fluid (air) movement over the fins.
Unless you have a water pump forcing water through the submerged radiators fins, you won't get anywhere near 24x the heat transfer.

Personally, i would leave the computer where it is and run pipes or hoses over to (and out the) window.

 

[QuietIce]

Anyways, the issue I always had with the heater cores is that installing barbs always seemed ... troublesome. I've seen the walkthroughs, but I don't have any of the tools needed; it might be worth it to get it pre-modded from Danger Den, even if they're only 3/8" barbs.

Why on Earth anybody puts barbs on a heater core is beyond me. It's not that I don't have the tools or can't do it, I've plumbed whole kitchens, but it's just not necessary. So far I've used a total of four heater cores and have as yet to add a barb to one. 1/2" tubing goes right over the end of 1/2" copper pipe - in fact I use a short piece of 1/2" copper pipe as a joiner for my tubing runs outside/inside the case. It's no harder to put 1/2" tubing over 1/2" copper pipe than it is to put it on an MCP-655 pump and, trust me, with a single screw clamp the tubing is NOT coming off the pipe. I doubt it would come off without the clamp but I'd rather be safe than sorry.

I should note you do need to cut off part of the pipe that comes with it but a $4 hacksaw can do that in ~5 minutes - if you take your time ...