Thermal Problem

Find a place that repairs resteraunt equipment and get the chiller from an orange juice or milk dispenser. Lots of machines get scrapped but they still have a good compressor and heat exchanger. The place I knew about moved, and I lost track of them, but they had over 100 spare chillers in their warehouse the last time I talked to them.

--
Service to my country? Been there, Done that, and I\'ve got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

Is the raised temperature caused by the lighting or is there another heat source?

Maybe some light source with less IR? Or do you need it for the plants?

Luhan

If the tank is hotter than room temperature, something has to be heating it - probably your lights. If you work out the wattage of all the lamps that you have got feeding visible and infra-red radiation into the tank, you'll probably get a reasonable idea of how much heat your chiller has to take out.

If you've got a circulating pump, the power that it dissipates is probably ending up in the tank as well, so add that in too.

You might save some money by illuminating your tank with efficient light sources - straignt or folded fluorescent tubes.

Incidentally the latent heat of evaporation of water is 2.26E6 joules per kilogram, and the latent heat of melting for ice is 3.33E5 joules per kilogram. The heat capacity of water is 4186 joules per kg. The American gallon is 8 lbs of water, or 3.632 kg.

Your two gallons of ice took 2,42E6 joules to melt, and should have cooled your tank by 0.66C or 1.2F. You need a better thermometer.

--=20 Bill Sloman, Nijmegen

It's a reef tank, the lighting is needed for the coral.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

Maybe you just need to put a big coil of hose in your pump line so it can heat exchange with the cooler room air.

Every gram of ice at 32° that melts absorbs 80 calories as it becomes water at 32°, which will lower the temperature of 1 gram of water 80°. Since it's all proportional, that means (all this is approximate, of course) that 2 gallons of ice at 32° upon melting will lower the temperature of 2 gallons of water 80°; it will lower the temperature of 20 gallons of water 8°, and it will lower the temperature of 200 gallons of water .8°--you get the picture. If the ice is at a temperature lower than

32° to start, then there will be a little additional cooling as the temperature of the ice comes up to 32° and melts. But, most of the cooling is from the change of phase of the ice.

If the tank is hotter than room temperature, something has to be heating it - probably your lights. If you work out the wattage of all the lamps that you have got feeding visible and infra-red radiation into the tank, you'll probably get a reasonable idea of how much heat your chiller has to take out.

If you've got a circulating pump, the power that it dissipates is probably ending up in the tank as well, so add that in too.

You might save some money by illuminating your tank with efficient light sources - straignt or folded fluorescent tubes.

Incidentally the latent heat of evaporation of water is 2.26E6 joules per kilogram, and the latent heat of melting for ice is 3.33E5 joules per kilogram. The heat capacity of water is 4186 joules per kg. The American gallon is 8 lbs of water, or 3.632 kg.

Your two gallons of ice took 2,42E6 joules to melt, and should have cooled your tank by 0.66C or 1.2F. You need a better thermometer.

--
Bill Sloman, Nijmegen


Pssst, Bill, you missed the fact that the heat source was STILL present, 
significantly reducing the 1.2F drop (balanced with heat loss to the outside 
world) and making it negligible in the full equation of all sources of 
heating and cooling. Not a "static" environment.

What is the total rating of the heat input? Lighting, pumps, etc. That'll be a starting point to sizing your chiller.

Next thing: What is the humidity of the environment where the tank is? If your AC keeps the area pretty dry, you are still going to lose some water even if you cool the tank down.

--
Paul Hovnanian     mailto:Paul@Hovnanian.com
------------------------------------------------------------------
100 buckets of bits on the bus
100 buckets of bits
   You take one down,
   and short it to ground
FF buckets of bits on the bus

Acrylic is a fairly good thermal insulator,

Most of the heat loss is from evaporation, conduction / convection, and some radiant heat. I would think that it would be reasonable to assume that the heat loss from convection /conduction from the surface of the water would be more then the heat loss through the acrylic. Also, how is the pump and filter setup done - can significant heat escape there? Unfortunately the only easy calculation is for evaporation. Another problem is the conduction / convection heat losses will disappear once the temperature differential disappears.

The easiest way to calculate the needed cooling is to figure out what the heat generation rate is for the aquarium.

Most of your heat is from lights, IR heat (sun?), the pumps mechanical work and possibly motor cooling. Some small amounts of heat may come from the fish and other bio heat.

What is the wattage and type of lighting? Qty of lights? What is the pump wattage, and how is the motor cooled? Does the sun shine on it? How much *food* energy is dumped in the aquarium a day?

Your tank has a thermal storage of: Size of tank = 243.8 cm x 60. 96 cm x 60.96 cm = 906 L water => ~906 kg water specific heat of water = 4.182 kj/kg K @ 25 deg C which gives: 3,788 kj / K

Assuming the 3% or so change in density from pure to salt water, the thermal storage of the tank, and the less then full tank sum to a neglectable amount.

This removes about: Loss of water = ~ 7.5 L a day -> which is about 0.31 L (kg) per hour Latent heat of fusion of water = 2260kj / kg Which gives a cooling rate of about 712 kj per hour, assuming a constant rate

Note, this cooling effect will decrease as the temp differential decreases.

No good - that's like specking a air conditioner on house square footage.

This would possible be equivalent to 1 gallon of ice due to the air content.

Lets assume 4 kg of ice: Latent heat of fusion of water = 333kj / kg This therefore removed about 1332 kj of heat. Considering the thermal mass of the tank is 3,788 kj / K, the temp would have dropped by about 0.35 deg C

I like this solution, however a coil won't likely do it, as several kj of heat need to be removed an hour at an almost zero temp differential. A liquid to air heat exchanger would be more ideal (think small radiator). Temp could be controlled by a diverter valve and / or fan, or just use the ambient air temp to keep things inline. Circulating cool water through a liquid to liquid heat exchanger would also work. Jim, what's your cold water source like (cheap, cold - if so, what temp?)? Due to the salt water, the heat exchangers would likely have to be stainless steel.

A reasonable guestimate would be that you need to remove about 2000 to 3000 kj per hour, and considering that a BTU is 1.055 kj, a 2000 to 3000 (which is a 1/4 ton unit!) BTU chiller would do if that guestimate is correct. If you want faster cooling and more compressor cycling, which reduces efficiency somewhat, a larger unit could be used.

Also putting lot's of bubbling air ornaments in the water will help cool it, mostly by aiding in substantial evaporation.

You haven't worked with thermostatted baths, have you - two gallons of cracked ice fed into a pumped system looks very like a Heavyside impulse.

240 US gallons of water is 3.65E6 joules/degree C of heat capacity. Jim's tank is 5C hotter than ambient, and let us say he has about 100W worth of lighting, so the thermal resistance from tank to ambient is 0.05C/W, then the thermal time constant of the tank would be about 50 hours. That seems a bit long - my very well insulated hot water tank in Cambridge, which held the same sort of volume of water, had that sort of time constant, so Jim has probably got bigger lights than the number I first thought of.

In any event, the ice will melt a lot faster than the lamps could have melted it.

If you can measure temperature accurately, at reasonalby regular intervals over something like one thermal time constant, you can ensure the thermal resistance to ambient surprisingly accurately - I've mesured a thermal time constant to +/-0.5% on at least one occasion.

--
Bill Sloman, Nijmegen

1 BTU will lower(raise) 1 pound of water 1 degree F. 1 gallon of water weighs 8 pounds.

You need to lower 240 gallons, 1920 pounds of water 9 degrees F. That takes 17280 BTU in a perfect no heat gain/loss system.

All you need to figure out is how quickly you want to make the temperature change. That is where the insulation factors of the acrylic, and the room air temperature come into play.

You can experimentally determine the heat gain by heating the tank to

1 degree above room temperature, and timing how long it takes to drop back down to room temperature. That will be the number of hours the system needs to lose 1920 BTU.

-Chuck

but what about the fish?

what temp do they need the water to be?

set up a dehumidifer in the room so that it drains back into the tank?

Mark

Heat sources....

4 Electronic ballasts driving fluorescent lighting, 4 tubes are "actinic" (blue), 4 tubes are white.

Each of the ballasts are labeled 2.1A, but I have no idea of the true heat production. The ballasts themselves are cool to the touch, so I presume some kind of switcher, but the heat emanating from the lamps is substantial.

Pump is submersible, so I have no idea of power.

Room temperature is 76°F, air conditioned, so not a source of heat TO the tank.

Windows are thermopane with metallic film, so I don't think there's significant heat there either.

Commercial aquarium websites are no help either... they just specify a chiller size based on tank volume.

Water temperature is 85.6°F when lighting is on, cools to 84.9°F overnight.

Evaporation loss is 2 gallons/day, out of a total of 240 gallons.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

If you have 76 degree air, why not just heat exchange the water to the air? I have a few copper fin radiators from old water cooled computers and I will send you one if you want to experiment.

(1) Saltwater (reef tank)

(2) This is in my Great Room... anything resembling a lab experiment will not pass a "War Department" review ;-)

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

Why not buy such a device? If it is not right, return it. Let them deal with your problem :-))

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Reply to nico@nctdevpuntnl (punt=.)
Bedrijven en winkels vindt U op www.adresboekje.nl

Ordering from the web gives by far the best price. But it also gives a considerable pain doing an exchange.

I just propped the lid up every so slightly so there's some cross ventilation. If that improves things I'll go looking for some appropriate vents.

But I'll probably go for overkill and buy the 1/3 HP, since it's trivially more expensive then the 1/4 HP...

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

Why not just try a little fan to move the air above the tank. Maybe a PC type deal on a wall wart.

Looks neat. I'm not sure about the shop though: 'Sm Animal Specials' ????

--
Reply to nico@nctdevpuntnl (punt=.)
Bedrijven en winkels vindt U op www.adresboekje.nl