Thermal Problem

Only if you use your own suspect thermal time constant as you admit. You have it estimated unreasonably high. Using something more reasonable, and that 1.2F gets noticeably smaller, something unlikely to be seen on a fish tank thermometer.

How small of a fridge can be had? I"ve seen little tiny boxes, presumably Peltier (or some kind of electric/heat transfer thing), about the size of a six-pack. Why not just wind a length of hose through one of them and hide it nicely behind the background plants with all the rest of the equipment?

Good Luck! Rich

Crushed ice should melt in a few minutes. To get the time constant of the fish tank down from 50 hours to just one hour, Jim would have to have 5 kilowatts worth of lights illuminating his fish tank, drawing 42 amps from a 120V AC mains, which is barely credible.

He certainly couldn't do any good tracking that sort of temperature excursion with a fish tank thermometer, but measuring the temperature by following the resistance of an interchangeable thermistor with a

4-decimal digit multimeter offering 0.01% resolution would divide the 4% per degree Kelvin response of the thermistor into 0.0025K steps, 0.0045 degrees F.

The last time I did something like that we used a Thurlby-Thandar Instruments 1906 5.5 digit logging multimeter, which can store up to

100 measurements at equal (preselectable) intervals. Very handy. The curve fittng program estimated the error on the time constant that we fitted to the data as +/-2 sec in 400 seconds.

--
Bill Sloman, Nijmegen

"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...

Odd.. If the room temp is 76°F, why is the tank @ 85°F? I would guess that the heat is probably coming from the lighting. It would be more energy efficient to reduce the heat input into the tank. That being said, a tiny amount of energy will enter the tank from fluid friction (fish swimming around, pumping action, etc.), but for all practical purposes this can probably be completely neglected.

****

I assume that your heater is turned off or unplugged at this point, or is not malfunctioning in some way?

****

I played around with an excel program to calculate some of the thermodynamic properties of water in an academic setting for a cooling tower and swamp cooler application. I don't know if it will help much, but I think that it has most of the calculations needed for the water / air energy transfer. You can download it from here:

. Look for the "Cooling Tower" tab.

Basically you need to know the energy input into your tank. This can be calculated from the volume rate 2 gal / day @ 85°F to ambient temperature and pressure @ some specific humidity. 2 Gal / Day = 16.69 Lbm H2O / Day. Specific heat of vaporization for water @ 1 atm = 970.4 Btu / lbm. 16.69 Lbm

• 970.4 Btu/Lbm = 16,195 Btu / Day = 0.197 kW = 197 Watts. ie: Btu/sec =

1.055 kW.

Now for the refrigerator, assuming that you cannot remove the 200 watts of power input into the tank you will need to remove it with the refrigerator that you want to buy. The size depends on the (c)oefficient (o)f (p)erformance of the chiller / (r)efrigerator. COPR = QL/Wnet,in = QL/(QH-QL) = 1/(QH/QL - 1), where W is the pump work in Watts or Btu/sec, and QH and QL is the high and low temperature energy transfer of the chiller in the same units. For an "ideal" refregerator QH and QL can be replaced with TH and TL to find the temperature in degrees Rankine of the low and high sides of the refrigerator. °F+460.67 = °R.

So now if you know the temperature of the refrigerator's heat transfer areas, you can find the COPR, if it is not given on the specifications. If you then know the power required to drive the compressor / pump, you can find out how much heat can be extracted from the tank.

Example. Low Temp = -40°F = 420.67°R High Temp = 140°F = 600.67°R COPR = 1/(600/420 - 1) = 2.337

85 Watts (input to the pump / compressor ) * 2.337 = 198.645 Watts extracted from the tank.

This is all kind of simplified, but my suggestion would be to replace any incandesent / halogen bulbs with the lowest wattage flourescent bulbs available.

But that's because I'm a green kinda guy. Why spend 85 Watts to take 200 Watts out when you can just not put the 200 Watts in in the first place?

Chris. T.

[snip]

They ARE fluorescents, but are simulating sunlight... it's a reef tank.

...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.

Note: since you are operating in an environment that is cooler then the tank, not hotter, which is when most people need chillers, you won't need as large a chiller. Also going by HP numbers is useless, as all that states is motor size. You need to know the actual cooling rate of the chiller, which is what to base the requirements on.

Have you done any more experiments with vents in the tank lid, and maybe a fan?

Thanks, Rich

Lets assume a 95% power factor since they are electronic (a switcher) as poor power factor ballasts have been obsolete for decades, and the lamps are running on 120V, this gives 240W per bulb set - hummm, looking up actinic aquarium bulbs, the only ones I found with that power handling in a 4 or 8 foot length was 110W in a 4 foot bulb. Power factor must be a little closer

90%, even so, Those are big fluoro lamps!

So you likely have about 880W of heat from lights. I'd assume about 1/2 of that makes it to the aquarium

Count on a least a few hundred watts from that.

This gives you a electrical heat load of 440 + 300 W

Lets assume 50W from this.

Also the bio load - lets assume you put in the equivalent of 100 to 150 g of oily food a day. This is an energy input of about 4,000 kj per day!

This gives you a heat load of: 440 + 300 + 50 W = 0.790kW, which is 2844 kj per hour + 166 kj per hour from food = 3011 kj per hour, during the day

You have 712 kj per hour of evaporative cooling (from previous post calculation)

And I'll assume a few hundred kj per hour lost through convection and IR radiation.

This gives a heat gain somewhere between 2000 kj per hour, and 2299 kj per hour if assuming the convection and IR nears zero as the temp differential diminishes.

To be on the safe side, I would buy a unit rated for 3000 to 4000 BTU, to cool the aquarium with. Too large of a unit and it will cycle on and off a lot, killing efficiency, but with such a large thermal mass with the water, this may not be a problem. A big problem may come and bite you with:

IMPORTANT!!!!! you are drawing at least 1200W, and 11.4 A (2.1 A x 4 ballasts, 3 A from the pump (assuming) ) from that circuit. For a household / commercial circuit, you should not go over about 80 - 85% continuous loading of the wirings ampicity, as heat accumulation in the wiring can cause the insulation to degrade or worse. NEC and the Canadian electric code state that electric heaters can only be installed up to 80% of the circuits capacity for this reason. The max ampicity of 14 AWG is 15 A, and is generally safe for 12 to 13A without problems (this is often another reason why appliances are limited to about 13A, and it's rare that appliances will sustain a maximum continuous load for any substantial length of time). Newer electrical code required wiring to be done with 12 AWG minimum in some states - this gives a 20A max ampicity, which is safe for 16 to 17A continuous.

If you don't have an outlet on another circuit near by, you should limit the load to say 13 A continuous if you have 14 AWG wiring, and you could go as high as 15A for 12 AWG wiring, but that might be too close to the breaker's max (cause false tripping), so you may have to change out the breaker and outlet for 20A units, which would allow you to easily go a little more then

2000W at 17A.

If you are limited to 1500W, 13A, then the chiller will be limited to about

1.6A, which is a pretty small chiller. Also note the power factor for small refrigeration compressors may be quite poor, substantially increasing the current of the circuit.

Better venting for the lights (forced convection with computer fans as suggested by another poster) and a stainless steel heat exchanger in a nice box with a few pancake fans blowing through it might start looking more attractive if you have power problems.

How long are the tubes? Four foot tubes seem to be mostly 32W, and two foot tubes 20W.

Stick an AC current measuring multimeter into the power supply lead - you will porobably have to dismantled the plug to do it, or make a "break-out" extension cord.

0=2E7F is 0.4K. Heat capacity of the tank is 3.65E6 joule/degree Kelvin, so this is 1.42E6 joules lost over eight hours, which imples that the lamps are putting in at least 50W when they are on..

The latent heat of evaporation of water is 2.260E6 joule/killogram. Evaporating two US gallons of water (7.264 kg) soaks up 16.4 MJ of heat per day, 190 joules per second, which is 190W of free cooling.

This is only weakly dependent on water temperature, so we have to balance the heat input from the lamps - at least 50W - and the pump motor - (unknown) with the 190W lost by evaportion and the conduction.convection transfer out of the tank, which is probably linearly proportional to temperature.

If we assume a 10 hour thermal time constant for the tank - which is not unreasonable -we are iin fact assuming 500W of convection cooling at a 9.6F temperature difference, so the total heat going out has to be

690W (including evaporative cooling). The evaporative cooling should make the equilibrium temperature of the tank about 3.65F below ambient.

There is probably enough there for Jim to set up a Spice model ....

--=20 Bill Sloman, Nijmegen

That is exactly Jim's original question, however, that information seems to be unavailable.

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

Nah - Jim will never read about it. He's got everyone with common sense plonked. ;-)

Cheers! Rich

wihout evaporation, and assuming a constant room temperature, the experiment you describe would take forever.

--

Bye.
   Jasen

Two pieces of information are needed.

1> the rate at which the lamps are putting energy into the tank 2> the rate at which various models of chiller chiller can remove heat from the tank at your target temperature and in a room at your ambient temperature. the chiller's cooling capacity should be specified somewhere... the other you'll have to calculate,

Chuck Hariss has the right idea.

Heat (or cool) the tank to 2 degrees below the target temperature and then time how long you need to run the lamps until it's 2 degrees too hot. (you'll need a precise thermometer) from that, and the amount of water in the tank, you can figure out the rate of heat flow into the tank at around the target temperature, and thus the rate at which you need to extract that heat. Bye. Jasen

Two pieces of information are needed.

1> the rate at which the tank is gaining energy when it's at the desired temperature. 2> the rate at which various models of chiller chiller can remove heat from the tank at your target temperature and in a room at your ambient temperature. the chiller's cooling capacity should be specified somewhere... the other you'll have to calculate,

Chuck Harris has the right idea, time the temperature change.

Heat (or cool) the tank to 2 degrees below the target temperature and then time how long you need to run the lamps until it's 2 degrees too hot. (you'll need a precise thermometer) from that, and the amount of water in the tank, you can figure out the rate of heat flow into the tank at around the target temperature, and thus the rate at which you need to extract that heat. Bye. Jasen

True,

I didn't mean to say room temperature, I should have said heat the tank until it rises 1 F above its stable temperature, and then cut the heat off, and time how long it takes to drop back to the original temperature. From there you can do your calculations.

-Chuck

Here's the Btu/hr ratings...

Also, results from lifting hinged lid up about 4" (propped up with sardine tins :-), dropped temperature from 85.5°F to 81.1°F, rather significant. Looks like adding a fan would help a lot.

...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.

Bear in mind that a fan up in the hood will increase your water loss.

Since you can pretty well see that the heat rise in your tank is due to your lighting, (Water does a pretty good job of absorbing the energy from light that is aimed into it.) and you know how many watts of light you are running in your tank hood, you can easily find out how big a chiller you need to remove the heat added by the lights.

1 watt = 3.413 BTU/hr

If you have 200W of lights in your hood, you will need 682 BTU/hr of cooling just to eliminate their contribution to heating your tank. However, don't think that a 682 BTU/hr chiller will do the job for your situation, it will hold things steady, but has no capability of dropping the temperature of an already too hot tank... So, double the capacity, and you will be in great shape.

Now, run the numbers using the real wattage in your lamps, and submersible pumps, and you can select your chiller.

-Chuck Harris

temperature.

things

pumps,

Yep. I'm getting there ;-)

...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.

te of

te at

If you go back to the start of this thread, you will find that Jim tried to do exactly what you propose by dumping two gallons of crushed ice into the sump of his system.

It is pretty easy to calculate exactly what effect a specific mass of ice would have had - though Jim would have been well-advised to weigh his crushed ice rather than throw in a two gallon volume.

A little later in the thread Jim quoted the "overnight" temperature drop of the tank as from 86.5F to 84.9F (while the lights are off) which imples that he as access to a moderately sensitive thermomenter. If he had been thinking about what he was doing, he'd have recorded precisely how long the lights were off.

By now he has probably had enough advice to tackle the job a little more systematically.

--=20 Bill Sloman, Nijmegen