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# Water requirements of Aquaculture re-use systems

Water flow to satisfy oxygen requirements of the fish, Water flow to dilute waste products to acceptable concentrations.

## Theoretical models for construction of re-use systems

Water requirements of the system

The necessary amount of flowing water in the system, and the separation of new and re-used water depends on a number of factors including:

·  Amount of water to satisfy the oxygen requirements of the fish

·  Amount of water to dilute and remove waste products to acceptable levels

·  Amount of water to ensure self-cleaning of the tanks

·  Degree of re-use

·  Effectiveness of the water treatment system.

## Water flow to satisfy oxygen requirements of the fish

The addition of water to a fish tank to satisfy the oxygen requirements depends on the oxygen consumption of the fish, the oxygen concentration in the inlet water and the lowest acceptable concentration in the outlet water to achieve optimal growth for the fish species. The specific water requirements can be calculated from:

Qin= Mf/(Cin Co)

where:

Qin=specific water flow per kg fish

Mf=specific oxygen consumption of the fish (mgO2/(min kg fish))

Ci=concentration of oxygen in the inlet water to the tank (mg/l)

Co=concentration of oxygen in the outlet waterfrom the tank (mg/l).

Example

The fish size is 2000 g, the water temperature 12°C and the specific oxygen consumption is 3.63 mg O2/(min kg fish). The oxygen concentration in fully saturated water is 10.8 mg/l (from tables). The acceptable concentration in the outlet is set to 7 mg/l. Calculate Qin.

Qin= Mf/(Cin Co)

=3.63/(10.8  7)

=0.96 l/(min kg fish)

Here is it also important to remember the require-ments to dilute for other substances and those for self-cleaning.

By supersaturating the inlet water with pure oxygen the water requirements can, of course, be reduced. This means that the Cin is increased.

Example

The same numbers as in the previous example are used, but the inlet water has a supersaturation of oxygen of 150%, meaning that the concentration is 16.2 mg/l. Calculate the new Qin.

Qin= Mf/(Cin Co)

=3.63/(16.2  7)

=0.39 l/(min kg fish)

## Water flow to dilute waste products to acceptable concentrations

The amount of water required to dilute and remove substances produced by the fish (suspended solids (SS), CO2 and total ammonia nitrogen (TAN)) to acceptable concentrations can be calculated based on mass balance equations for the single substances:

Min+ Mro+ Mf= Mo

where:

Min=mass of substances in new incoming water

Mro=mass of substances from water entering fromthe re-use circuit

Mf=mass of substances produced by the fish inthe tank

Mo= Mass of substances in the outlet from the tank.

If Mo and the water flow out of the tank Qo are known, the concentration of a substance in the outlet from the tank (Co) can be calculated; this must not exceed a value that is acceptable for the fish. Based on this, can the lowest acceptable outlet water flow can be calculated from the following equation:

Qo  Mo/Co-acc

where:

Co-acc=acceptable concentration of the substance in the outlet to avoid reduction in growth.

Example

A fish of size 50 g has a specific growth during one day of 31 g resulting in waste production measured as SS = 6.2 g. The acceptable level of SS in the outlet is set to 25 mg/l. Calculate the necessary water flow out (Qo).

produced per minute = 6200 mg/(24 h × 60)

=4.3 mg/min

Qo=4.3 mg/min/25 mg/l= Mo/Co-acc0.17 l/min

This means that the water flow into the tank must be higher than 0.17 l/min to dilute the concentration of SS to acceptable levels or less.

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Aquaculture Engineering : Recirculationand Water Re-use Systems : Water requirements of Aquaculture re-use systems |