My dad (he is an Chemistry graduate so understands it all a lot better than me) put together this system calculate the volume of water which would neet changing.
|a||Volume of water in aquarium (litre)|
|b||Volume of water removed in bucket (litre)|
|c||Concentration of contaminant (e.g. nitrate) in solution at the start|
|n||Number of steps|
If we siphon off b litres of liquid from the aquarium, and replace this volume with b litres of clean water, the concentration of material in solution will be: (a-b)c / a
e.g. If our aquarium holds 360 litres, and the bucket has a capacity of 10 litres, then replacing the first 10l of water will reduce the concentration of contaminant by a factor of (360 – 10)/360. i.e. 0.9722
After two steps, the contaminant will be reduced by 0.9722 x 0.9722. i.e. 0.9452
After three steps, the contaminant will be reduced by 0.9722 x 0.9722 x 0.9722. i.e. 0.97223 or 0.9189.
After n steps, the contaminant will be reduced by a factor 0.9722n.
To find out how many steps it will take to halve the concentration of contaminant, we need to solve the following mathematical equation:
0.9722n = 0.5
n log 0.9722 = log 0.5
n = log 0.5 / log 0.9722
n = 24.6
Rounding up to a whole number, we arrive at 25 steps.
To conclude, using a 10 litre bucket on a 360 litre aquarium, we must replenish the volume that we remove 25 times to halve the concentration of a contaminant, such as nitrate.
Fish are able to tolerate the slow reduction in concentration in a large number of small steps. Removing and replacing half the water in an aquarium in one step would be too much of a shock to them.
The table below shows what percentage of the original concentration of contaminant is left after each step to empty and replenish.