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4 min readTo determine the digestion of aquaponic sludge treatment in aerobic and anaerobic bioreactors, a specific methodology needs to be followed. A methodology adapted for aquaponic sludge treatment purposes is presented in this chapter. Specific equations have been developed to precisely quantify their performance (Delaide et al. 2018), and these should be used to evaluate the performance of the treatment applied in a specific aquaponic plant.

In order to evaluate the treatment's performance, a mass balance approach needs to be achieved. It requires that TSS, COD and nutrient masses are determined for the all reactor inputs (i.e. fresh sludge) and outputs (i.e. effluents). The reactor content also needs to be sampled at the beginning and at the end of the studied period. The input, output and content of the reactors have to be perfectly mixed for sampling. Reactor input and output should basically be sampled every time the reactors are fed with fresh sludge.

Then, reactor sludge reduction performance (η) can be formulated as follows:

$\eta__s = 100\%(1-( \Delta S + S__{out})/S_{in})$ (10.6)

where ΔS is the sludge inside the reactor at the end of the studied period minus the one at the beginning of the period, Ssubout/sub is the total sludge that left the reactor in the outflow, and Ssubin/sub is the total sludge that entered the reactor via inflow.

For organic reduction, the sludge (i.e. the term S) can be characterised by the dry mass of sludge (i.e. TSS) or the mass of oxygen needed to oxidise the sludge (i.e. COD). Thus, for COD and TSS reduction performances, the smaller the accumulation and the smaller the quantity in the outflow, the higher the reduction performance (i.e. high percentage) and so the less solids discharged out of the loop.

Based on the same mass balance, the nutrient mineralisation performance of the treatment (ζ), i.e. the conversion into soluble ions of the macro- and micronutrients present in the sludge under undissolved forms, the following formula can be used:

$\zeta__N = 100\%((DN__{out}-DN__{in})/(TN__{in}-DN_{in}))$ (10.7)

where ζ is the recovery of N nutrient at the end of the studied period in percent, DNsubout/sub is the total mass of dissolved nutrient in the outflow, DNsubin/sub is the total mass of dissolved nutrient in the inflow and TNsubin/sub is the total mass of dissolved plus undissolved nutrients in the inflow.

Thus, similar to the organic reduction performances, the smaller the accumulation inside the reactor and undissolved nutrient content in the outflow, the higher the mineralisation performance (i.e. high percentage) and so the dissolved nutrient recovered in the effluent (or outflow) for aquaponic crop fertilisation (see Example 10.1). The presented mass balance equations are used in the example box.

Example 10.1The digestion performance of a 250-L anaerobic bioreactor has been evaluated for an 8-week period. It was fed once a day with 25 L of fresh sludge coming from a

TilapiaRAS system, and the equivalent supernatant volume (or output) was removed from the bioreactor. The fresh sludge (input) had a TSS of 10 g dry mass (DM) per litre or 1%, and the supernatant (output) had a TSS of 1 gDM/L or 0.1%. The TSS inside the bioreactor at the beginning and at the end of the period was 20 gDM/L. Consequently, the total DM inputs, outputs and inside the bioreactor during the evaluated period are calculated as follows:DM in = 0.01 kg/LD $\times $ 25 L $\times $ 7 days $\times$ 8 weeks = 14 kg

DM out = 0.001 kg/Ld $\times$ 25 L $\times $ 7 days $\times$ 8 weeks = 1.4 kg

DM to = DM tf = 250 L $\times$ 0.02 kg/L = 5kg

The TSS reduction performance (

ηTSS) of the bioreactor can then be calculated as follows:$\bold{\eta}_{TSS}=100\%(1-((5-5)+1.4)/14) = 90\%$

The bioreactor P mineralisation performance can be evaluated knowing that the fresh sludge (input) had a concentration of dissolved P of 15 mg/L and a total P content of 90 mg/L. The concentration of dissolved P in the supernatant (output) was 20 mg/L. Consequently, the total P content in the input, the total dissolved P in the inputs and outputs during the evaluated period are calculated as follows:

TP in = 0.090 g/LD $\times $ 25 L $\times $ 7 days $\times$ 8 weeks = 126 kg

DP in = 0.015 g/Ld $\times$ 25 L $\times $ 7 days $\times$ 8 weeks = 21 kg

DP out = 0.020 g/Ld $\times$ 25 L $\times $ 7 days $\times$ 8 weeks = 28 kg

The P mineralisation performance (ζsubP/sub) of the bioreactor can then be calculated as follows:

$ζ_P = 100\%((28 - 21)/(126 - 21)) = 6.67\%$

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