I can't seem to find what I'm looking for. I'm hopeful that one of you might know of a tool/website/spreadsheet/formula.
Is there a way to *easily* calculate the impact of an aquarium on household humidity.
Eg. My home is 3300sq/ft, two story and is currently 54% relative humidity at 76 degrees F. If I add 200 gallons of water in an open aquarium and keep the aquarium at 79 degrees F evaporation will increase the household humidity by X?
I'm sure that I don't understand the complicated science behind evaporative cooling and humidity. I also know that temperature in my house can fluctuate a lot as I don't cool the house mid day and I open the house up at night to bring in cool air.
I'm asking this because I had a few hundred gallons in a previous, and smaller house, and condensation created mold problems around windows. I'm assuming that I may have similar problems again and am trying to quantify the problem.
I may be asking the wrong question. In which case I'm open to suggestions on how to figure out what to ask.
I took meteorology in undergrad 8 years ago. Here is what I remember.
Part 1: RH to mass per volume
Starting with the basics, RH is actual vapor density (AVD) over saturation vapor density (SVD). Saturation vapor density changes with temperature (
http://www.chemguide.co.uk/physical/phaseeqia/svpwaterplot.gif ). So, of course, as it gets warmer, the SVD increases, corresponding to a drop in RH. As it gets colder the SVD decreases and the RH rises. As an aside, the temperature for which the SVD equals the AVD, is the dew point (more on this in part 2)
Using the SVD, and atmospheric pressure (you can find a useful approx. given your elevation ASL), you can calculate the actual gH20/m3Air
Part 2: Add new mass per volume to new RH
Now, take the volume of your room. Determine how many gallons of water evaporate. Use the specific density of water to determine how much mass is added to your room's air. Calculate your new AVD (the prior AVD plus addition from evaporation), then divide that by the SVD (we are assuming your air temp is the same), to get your new RH
Note: This calculator will get you through Part 1. You'll have to backcalculate for Part 2.
http://www.cactus2000.de/uk/unit/masshum.shtml
The caveats:
Airflow--Of course, that room's air isn't isolated. You can estimate the exchange rate between that room and the rest of the house (and the outside for your HVAC) to get some RH reduction factor. Your HVAC system might actually have tags on it that specify the CFM air exchange rate. Simply use the RH of the rest of the house, the RH in the fish room, and the exchange rate to see how much the rest of the house relieved the fish room of humidity. The easiest way to do this is to figure out how much air is recycled within a day, how many gallons of water are evaporated in a day.
Here is a calculator that can help you incorporate the outside.
http://www.lenntech.com/calculators/humidity/relative-humidity.htm
Insulation--Ignoring window insolation for a moment, if it is cold outside, your inside RH might be 50%, but once that inside air hits the window, if the temperature outside is below dew point, your will have condensate. To figure out if your window might condensate, sans insulation, determine the new RH (that is, including tank evaporation), then determine the dew point. That will tell you the outside conditions for which your window will have condensate. Of course, you still have insulation. I don't know how to include conductivity but the easy way is to, on the coldest days of the year, measure the room temperature right next to the window. Consider that temperature your dew point threshold for condensation.
Evaporation rate--In my explanation, I assume you know your evaporation rate (gal/day). The other posters hit on some of the difficulties to calculate this.
Sorry.. no easy answers and I'm probably only half aware, but hopefully close enough for an estimate.
The EASIEST way is to pay me to give me your house and tank specs. I can run a computational fluid dynamics model to figure this out for you
