Automatic (Drip/Overflow) Water Change Systems

[rkc772]

This is a good start, but you'll have to tweak your drip rates accordingly. http://www.angelfish.net/DripSystemcalc.php

i agree! i don't even mind changing "x" amount of water on my 480gal system. just drain one end and fill on the other side every week. i let it run for 3-4 hours and water my garden (garden looks really green!!!). i would conservatively estimate at 50 gal per hour exchange rate.

 

[CHOMPERS]

How does one calculate the 30% per week,

If you are asking what the math is that you would enter into a calculator, it is:
tank volume x .30 = volume changed

keeping in mind that when one drains 30% and replaces it, one has 30% new water, where as when new water is dripping in at a slow rate, each drop falls into a tank of "dirty" water???

Actually you asked it about the way I would explain it. So, yes, you are right

I gather you are viewing the big picture, that if you add 30% and then drain out 30% then you won't actually drain out 30% of the old water. Part would be new water. This isn't really how it works with a drip system.

The drip system should drip in after old water has drained out. In other words, if you drip into your sump, the emitter should be in the last chamber and the overflow should be in the first chamber. The reason for this is the new water should mix with the largest volume possible before being removed from the aquarium system. If the water is dripped into the first chamber and then drained out the last chamber, your calculations would have to be limited to the sump volume (then that solution used in dilution of the tank system). If dripped in properly, the new water will be forced to mix with the entire aquarium volume rather than just the sump volume before being allowed to exit the system.

Now with that said, when one drop enters the system one drop exits the system. Drip systems work one drop at a time. That one drop when mixed with the tank system exits the system as a ratio of the mixture of new water and old water.

My drip system drips about 20 drops per second with a rate of one gallon per minute. That works out that one drop is .000833 gallons. My system is 250 gallons (neglecting filter volume). When the one drop mixes with the tank volume, the ratio of water in the drop leaving is .000833:250 of new water to old water. Simplified that is for 1:300,120. There is very little new water that actually goes to waste.

 

[CHOMPERS]

By the way, I am aware of the built in fallicy. Feel free to point it out. Just be prepared to show the differential equation and give a laymans explanation for it. Keep in mind that there are about six people here that will actually understand it if you refer to calculus terms.

The 'one drop at a time' example is an effective model even though there are multiple drops in the system at a time.

 

[rkc772]

bring in the scientific calculators out! i'll just get a regulator like R1 has and open it up to change 300 gal in 6 hours or more. then drain on the other size. surely 1/2 of the 480 should have been