fish tank water cooling

[E-Licious]

Hey, this reminded me of an article I saw in maximum PC about a girl making an Aquarium/PC case. Check it out at her site. And no, she didn't mix her water-cooling system with her aquarium system.

 

[Korndog]

He does have a large body of water though. And the water will also be dissapating some of the heat to the air, so it may not rise that quickly. Does anyone wanna do the calculations for heat transfer?

I have a 50gal aquarium and a 100watt heater, even with the heater off and nights being as low as 60F (my fish tank has to be over 75F) i'll usually find it with a 3-4c drop in the morning.
there's only one exposed side and ratio of water to exposed surface area is very very large.

all i'm worried about are the fish, some fish are very very sensitive to heat changes, even if its between their recommened living conditions, its the change over a short period of time.

 

[Korndog]

Hey, this reminded me of an article I saw in maximum PC about a girl making an Aquarium/PC case. Check it out at her site. And no, she didn't mix her water-cooling system with her aquarium system.

wow.. shes got skillz..

oh yea, as for the calculated heat, i believe cather used this same method, i remember doing it in physics too.
say he has a 300gph pump, with no rad, just waterblock, u'll probably get 200, 150gph? lets say 200gph or ~12 LPM, which is ~.20lps which is .20KG of water passing through. to get the thermal capacity is 4186, so 4186 times .2 is ~870 watts per celcius. so if 100watts is being dumped it, 100/870, units cancel out, u get ~.115c increase. so of course, after ur pump runs all 30gal of water through ur system, it'll gain .115c. with a 200gph pump, it will pump it 6 2/3 times so every hour, it'll gain .8c granted that ur system will produce 100watts of heat.

 

[squeakygeek]

I have a 50gal aquarium and a 100watt heater, even with the heater off and nights being as low as 60F (my fish tank has to be over 75F) i'll usually find it with a 3-4c drop in the morning.
there's only one exposed side and ratio of water to exposed surface area is very very large.

all i'm worried about are the fish, some fish are very very sensitive to heat changes, even if its between their recommened living conditions, its the change over a short period of time.

Most of your heat loss is probably not through your exposed side, but through evaporation.

 

[sean uk]

re:-

thanks for that link E-Licious (what a cool case )


korndog some good points
i think the copper base idea is right maybe a copper insert over the die area to give a even heat spread is the way to go as the plastics tend to have hot spots

the fish would have to be aclimatised

btw my job is to achieve the impossible

i'm currently working on another project at the moment
(i was also told this was impossible by everybody inc the polymer manufacturers but after 2yrs and about £???k it finally works field trials soon)

so watch this space

 

[zabomb4163]

He does have a large body of water though. And the water will also be dissapating some of the heat to the air, so it may not rise that quickly. Does anyone wanna do the calculations for heat transfer?

blah.

30 gallons of water = 113.7 liters
thermal capacity of water - 1C rise over 1gram of water = 1calorie
113.7 liters = ~113,700 grams or 113,700 calories of capacity.

-that is 113,700 calories per 1C rise in our system

1 calories = 4.18400 Joules
1 Watt hour = 3 600 Joules

Heat = 3600joules * 100watt hours
Heat = 360,000 Joules per hour or 86,042.065 calories per hour

Temp rise per hour = Capacity/Heat
= 113,700 / 86,042
=1.32144767C per hour not including evaporation or heat loss through the surface.


---------------------------------------------------------------------

 

[Korndog]

Most of your heat loss is probably not through your exposed side, but through evaporation.

i'm not far enough in physics to determine that or i might have been asleep while we were learning that, but i'm sure it'll be fine if he was to try it, my only concern are the fish.
plus i'm sure u have a filter pump, and the UV sterilizers, all that stuff adds more watts.
if u leave the fish out, i'm sure it'll be fine.

 

[squeakygeek]

i'm not far enough in physics to determine that or i might have been asleep while we were learning that, but i'm sure it'll be fine if he was to try it, my only concern are the fish.
plus i'm sure u have a filter pump, and the UV sterilizers, all that stuff adds more watts.
if u leave the fish out, i'm sure it'll be fine.

I wasn't saying his idea would or wouldn't work, I just pointed that out because he didn't seem to realize this (posted that his tank had only one exposed surface).

 

[Korndog]

blah.

30 gallons of water = 113.7 liters
thermal capacity of water - 1C rise over 1gram of water = 1calorie
113.7 liters = ~113,700 grams or 113,700 calories of capacity.

-that is 113,700 calories per 1C rise in our system

1 calories = 4.18400 Joules
1 Watt hour = 3 600 Joules

Heat = 3600joules * 100watt hours
Heat = 360,000 Joules per hour or 86,042.065 calories per hour

Temp rise per hour = Capacity/Heat
= 113,700 / 86,042
=1.32144767C per hour not including evaporation or heat loss through the surface.

wow.. i screwed up..
how the hell did i pass physics??

 

[sean uk]

re:-

korndog how about this
when i boil the fish on my first trial i'll have a tasty snack while the system carries on without any further probs

i'm off shopping now and to the aquarium shop as my powerhead siezed up
i have an idea for an easier solution i might try this evening if it works i'll post some pics

 

[sean uk]

squeekygeek

hi
i didn't post that my tank had one exposed side that was someone else
my tank is exposed to air both sides and one end plus currently with no actual tank heating evaporation caused by the lighting is a lot probably 5-7gal per week

 

[Korndog]

Re: re:-

korndog how about this
when i boil the fish on my first trial i'll have a tasty snack while the system carries on without any further probs

i'm off shopping now and to the aquarium shop as my powerhead siezed up
i have an idea for an easier solution i might try this evening if it works i'll post some pics

*sniff* *sniff* fish killer
i'll just sick my bala shark fish at you
hes only 4", but has teeth

btw, i'd get a maxijet 1200 or 900 from the shop, they run good and are cheap.

 

[zabomb4163]

I'm calculating the heat loss through evaporation. the reason i have not posted it yet is because its a little more complicated.

 

[MoreGooder]

Are there laws agains subjecting animals to dangerous environments in the UK? This idea needs some further thought. Why do it if it's going to have some many risks, especially to your fish?

I've been reluctant to post in this thread, but I just can't help myself. The whole idea of using a fish tank with live fish to cool a CPU is just plain silly. It's analogous to wearing your family dog as a winter coat just cause he has fur, and while he's still alive.

 

[sean uk]

moregooder

the dog analogy is good
already done that( though not in the uk there is a law against it)
i am not going into dog stories it may offend?

i wouldn't really subject my fish to my sometimes crazy experiments
i feel it's okay to endanger yourself through stupidity as long as others are either not at risk or can make there own judgements
fish & children cannot

 

[zabomb4163]

Done

what is our surface area and ambient temp? then i can plug them into my program to give you our heat loss through evaporation.

 

[sean uk]

Surface area is 0.333m*2 (370mm x 900mm)

ambient room temp is 21'C

 

[Hookj5]

There are so many flaws in your idea I can't even add them all up

This is the only thing I have read on this thread so far.... I can't stop laughing long enough to read any more.

It's analogous to wearing your family dog as a winter coat just cause he has fur, and while he's still alive.

Man...... just keeps getting better and better. This might be the most entertaining thread I have ever read.

 

[zabomb4163]

Evaporation

E = A * Km * (Mw * Pv)/(R * T) (kg/s)

E = evaporation rate, in kg s-1
A = area of the evaporating puddle
KM = mass transfer coefficient, in m s-1
MW = molecular weight
PV = vapor pressure,
R = the gas constant
T = ambient temperature, in K (21°C is equal to 294.15 K K).



Km = 0.0048 * U^(7/9) * Z^(-1/9) * Sc^(-2/3) (m/s)

U = wind speed m/s,
Z = the pool diameter in the along-wind direction (m)
Sc = the laminar Schmidt number for the selected chemical


Sc = (v/Dm)
v = the kinematic viscosity of the air, assumed to be 1.5 x 10^-5 m^2/s,
Dm = the molecular diffusivity of the selected liquid in air, in m^2/s.


----------------------------------------------------

E = A * Km * (Mw * Pv)/(R * T) (kg/s)
E= 1.45066E-05 Kg/s
E= 0.014506575 g/s
E = 0.8703945 g/m
E= 52.22367001 grams per hour


A = 0.333
Km = 2.528553666
Mw = 18.015
Pv = 2.3388 kPa @ 20C
R = 8314 J/(kmol K))
T = 294.15 K (21C)



v = 1.5 x 10^-5 m^2/s
Dm = 0.2178
http://www.fs.fed.us/rm/landscapes/Solutions/Mole.shtml
Sc = (1.5*(10^-5))/0.2178
Sc = 6.88705E-05
----------------------------------------------------------------
Heat loss =
2,260 joules @100C /g
2,500 joules @0C /g

Loss max = 52.22 grams per hour * 2,500 joules
= 130,550 Joules

Loss min = 52.22 gramas per hour *2260 Joules
= 118,017.2 Joules

 

[zabomb4163]

blah.

30 gallons of water = 113.7 liters
thermal capacity of water - 1C rise over 1gram of water = 1calorie
113.7 liters = ~113,700 grams or 113,700 calories of capacity.

-that is 113,700 calories per 1C rise in our system

1 calories = 4.18400 Joules
1 Watt hour = 3 600 Joules

Heat = 3600joules * 100watt hours
Heat = 360,000 Joules per hour or 86,042.065 calories per hour

Temp rise per hour = Capacity/Heat
= 113,700 / 86,042
=1.32144767C per hour not including evaporation or heat loss through the surface.


---------------------------------------------------------------------

Heat = 360,000 Joules per hour - Evaporation max
= 229,450 per hour

Heat = 360,000 Joules per hour - Evaporation min
= 241,983 per hour

-----------------------------------------------------------------------
Our evaporation/heat equilibrium requires 144 grams/hour

(before somone complains about airflow, I assumed 1m/s for airflow across the surface which equates to ~197cfm....600cfm would get you enough evaporation though)