Excuse me, we are a carbon removal process, COD1500-2000mg/L, how much ammonia nitrogen and total phosphorus are maintained, buy urea and potassium dihydrogen phosphate, how to calculate the dosage?

    The determination of nitrogen and phosphorus nutrients is the prerequisite for reasonable nutrient dosing. When confirming the amount of nitrogen and phosphorus nutrients added during the carbon removal process, the following empirical ratio is usually used, that is, organic matter: nitrogen: phosphorus = 100:5:1. In the proportional formula, organic matter can be represented by C, nitrogen can be represented by N, and phosphorus can be represented by P, and the expression can be transformed to: C:N:P=100:5:1. This ratio can be understood as the corresponding consumption of 5g of nitrogen and 1g of phosphorus for every 100g of organic matter decomposed to ensure that the demand for nutrients in the decomposition of activated sludge is balanced.

   In practical engineering applications, it is often found that the amount of N and P added by theory is often larger than the actual demand, and the main reason for the analysis is that the influent sewage and wastewater still contain nutrients to a greater or lesser extent. Therefore, it is necessary to pay enough attention to the nitrogen and phosphorus values in the influent sewage and wastewater, calculate this part of the nitrogen and phosphorus content, and deduct it from the theoretical calculation value, so that the nitrogen and phosphorus content added will not be excessive.

1. Calculation method of nitrogen source addition

    At present, there is little controversy about N, generally recognized as TKN, except for specific industrial sewage, there is very little organic nitrogen in the actual influent, so the dosage of N source is: N=V*G/Y:

　　N-N source dosage

　　V—The amount of water in the pool

　　G—The difference in N needs to be supplemented

　　The amount of N converted from Y-N sources

　　1. Urea as an added N source (CH4N2O molecular weight: 60.06 g/mol)

　　Urea contains 46.7% N, and if 1g N source needs to be added, urea Y=1/0.467=2.14 g needs to be added

　　2. Ammonium sulfate was added as the N source ((NH4)2· SO4 molecular weight: 132.14)

　　Ammonium sulfate contains 21.2% N, and if 1g N source needs to be added, ammonium sulfate Y=1/0.212=4.72 g needs to be added

　　3. Ammonium nitrate is added as an N source (NH4NO3 molecular weight 80g/mol)

　　Ammonium nitrate contains 35% N, if 1g N source needs to be added, ammonium nitrate Y=1/0.35= 2.86 g needs to be added

　　2. Calculation of phosphate addition

　　Ordinary sludge culture is generally calculated according to the CNP ratio of 100:5:1, and there is no dispute about TP at present.

　　P=V*G/Z

　　Where:

　　P—P source input increase

　　V—The amount of water in the pool

　　G—The difference in N needs to be supplemented

　　The amount of phosphate converted from the Z-P source

　　1. Sodium dihydrogen phosphate as the added P source (Na2HPO4.7H2O, molecular weight 268.07 g/mol)

　　Sodium dihydrogen phosphate contains 11.57% P, and if 1g P source needs to be added, sodium dihydrogen phosphate Z=1/ 0.1157= 8.64 g should be added

　　2. Dipotassium hydrogen phosphate was added as the P source (K2HPO4-3H2O, molecular weight 228.22g/mol)

　　The P content of dipotassium hydrogen phosphate is 13.6%, and if 1g P source is added, dipotassium hydrogen phosphate Z=1/ 0.136 = 7.35 g needs to be added

　　3. Phosphate fertilizer superphosphate is used as a P source

　　The available phosphorus in phosphorus fertilizer is soluble phosphorus pentoxide (P₂O5, molecular weight 141.94g/mol)

　　The available phosphorus content in phosphorus fertilizer was 12%, and the P content of P₂O5 was 43.66%, and if 1g P source was added, phosphate fertilizer Z=1/(0.12×0.4366)=19.09 g was added

　　3. Instance calculation

　　Let's take the community post as an example, the carbon source COD that the system can use is calculated as 2000mg/L, and the CNP ratio is 100:5:1, then ammonia nitrogen: 100mg/L, TP: 20mg/L! 1 ton of water requires 100g of ammonia nitrogen and 20g of TP20g!

　　Urea as an added N source, if you need to add 1g N source, you need to add 2.14 g of urea, so 1 ton of water needs 200*2.14=428g!

　　Potassium dihydrogen phosphate KH2PO4 is used as the added P source, if 1g P source is added, 4.38 g of potassium dihydrogen phosphate needs to be added, so 1 ton of water needs 20*4.38=87.6g!

　　Then subtract the nitrogen and phosphorus content that comes with the system, and it is dosed!