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Growbeds design/sizing

Discussion in 'Commercial Systems' started by Anish, Jul 16, 2018.

  1. Anish

    Anish New Member

    Joined:
    Apr 18, 2018
    Messages:
    13
    Country:
    qatar
    State:
    Doha
    City:
    Doha
    Hey All,

    We have 12,000liter tank which we are planing to add some tilapia. The design is based on my ADC course using the CHOP system. I am just not sure how to size the growbed. As a rough estimate i've put 4 growbeds (2 IBC tanks cut into half) with 20mm rocks as media. The water volume in each is about 300 liters

    My current design has the following
    1. My swirl filter is a 3000 liter round tank
    2. 2 x 1000 liter tank as a sump
    3. 1 x 500 liter tank as distribution tank
    4. 250 liter mineralization tank
    5. 350 liter bio-filter with 20 liter K1 media
    6. 120 Hielia aerator aerating the fish tank, bio-filter and the mineralization tank.
    The main produce are the fish hence the Grow beds are here just to control the Nitrate levels in the water. That was the only way i understand to control Nitrate levels other than water changes which is not easy to get.
     
  2. Glenn Holguín

    Glenn Holguín New Member

    Joined:
    Mar 16, 2022
    Messages:
    1
    Country:
    Dominican Republic
    State:
    Hato Mayor
    City:
    Hato Mayor del Rey
    I'm new to the blog but if it's not too late and you follow up on the line you published, to size the grow beds you must be practical in some things, the recommended depth is 30 cm, the rest is basic mathematics to determine the width you must think about the size of the floating polystyrene raft that you are going to use, it can be 100x60cm or 120x60cm if you are going to use the 120x60 cm one then you say I will use 2 rafts it would be a total width of 2.4 m x 0.6 m you already have the width of the bed in 2.4 m and the depth in 0.3 m, now you consider the total volume of the water of the fish ponds you indicated 12000 liters, about 12 cubic meters of water, then 2.4mx 0.3m = 0.72 and we have 12/0.72=16.67 m this calculation is for determine the length an inverse way to determine a variable of the 3 that we need to calculate volume, then there must be a balance between the volume of water in the fish ponds and the volume of water in the culture beds, if we review new You will see that 2.4 x 0.3 x 16.67 (Bed size)= 12.00 m3, you can divide in two beds of (2.4 x 8 x 0.3 m) which is the same total volume indicated by you. Since you know the length and that they are 2 floating rafts wide, then you divide the length by the depth of the raft, which is 0.6m, and you will see that in total there will be 27.78 x 2 floating rafts, you will have about 54 floating rafts of 120x60cm, distribute the difference in fractions in the edges and ends so you have room to insert and remove the rafts.

    Something interesting is that Murray speaks of a balance in the volumes of water while other authors such as Dr. James Rakocy of the virgin islands uvi experimental station makes the calculations according to the amount of food supplied 2-3 oz/yd2/day (60- 100 g/m2/day) and for a commercial scale design I did the calculations with both principles and I must indicate that the difference in crop area was minimal.
    Regards from Dominican Republic.
     

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