Description
Technical Parameters
Oxidation Ditch Design Calculator
The following is Aquasust for you to sort out the correct way to calculate the amount of aeration：
- Blue block is the design datameter : be filled in
- Brown: calculate process data
- greener : last result for your proces
| 1.Aquasust--Oxidation Ditch Process Design parameters: | ||||||||||
| Design treatment water volume Q = | 300 | m3/d= | 12.50 | m3/h | ||||||
| Intake water quality: | Water quality of effluent: | |||||||||
| Inlet water CODCr= | 1620 | mg/L | CODCr= | 324 | mg/L | |||||
| BOD5=S0= | 840 | mg/L | BOD5=Sz= | 126 | mg/L | |||||
| TN= | 250 | mg/L | TN= | 30 | mg/L | |||||
| NH4+-N= | 180 | mg/L | NH4+-N= | 18 | mg/L | |||||
| Alkalinity SALK= | 280 | mg/L | pH＝ | 7.2 | ||||||
| SS= | 180 | mg/L | SS=Ce= | 20 | mg/L | |||||
| f=MLVSS/MLSS= | 0.7 | Mixture concentration X = | 4000 | mgMLSS/L | ||||||
| Minimum sludge age is used | 30 | d | Dissolved oxygen concentration of aeration tank effluent | 2 | mg/L | |||||
| Attenuation factor Kd= | 0.05 | d-1 | Activated sludge yield coefficient Y = | 0.5 | mgMLSS/mgBOD5 | |||||
| Average summer temperature T1= | 25 | ℃ | Denitrification rate constant qdn,20 at 20℃ = | 0.07 | kgNO3--N/kgMLVSS | |||||
| Average winter temperature T2= | 15 | ℃ | Denitrification temperature correction factor = | 1.09 | ||||||
| Residual alkalinity | 100 | mg/L | Nitrification reaction safety factor K = | 2.5 | ||||||
| Required alkalinity | 7.14 | mg alkalinity/mgNH4-N oxidation | Oxygen required for nitrification | 4.6 | mgO2/mgNH4-N | |||||
| Output alkalinity | 3.57 | mg alkalinity/mgNO3+-N reduction | Oxygen available for denitrification | 2.6 | mgO2/mgNO3+-N | |||||
| Dissolved oxygen concentration during denitrification |
0.2 | mg/L | If the biological sludge contains about | 12.40% | Nitrogen for cellular synthesis |
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| 2.Design Calculation | ||||||||||
| (1) Aerobic zone volume calculation | ||||||||||
| 459 | m3 | Aerobic tank hydraulic retention time t1= | 1.53 | d = | 36.72 | h | ||||
| 3.Volume calculation of hypoxic zone | ||||||||||
| (1) Oxidation ditch biological sludge production | ||||||||||
| 42.84 | kg/d | |||||||||
| (2)For cell synthesisTKN= | 5.31 | kg/d | That is, there are TKN x 1000/300 in TKN = | 17.71 | mg/L | |||||
| Therefore, the [NH4-N] to be oxidized = | 144.29 | mg/L | Reduction required [NO3+-N]= | 43.29 | mg/L | |||||
| (3) Denitrification rate | ||||||||||
 |
0.036 | kg/(kg.d) | ||||||||
| (4) Volume of hypoxic zone V2 | ||||||||||
 |
425 | m3 | ||||||||
| Anoxic tank hydraulic retention timet2= | 1.42 | d= | 33.98 | h | ||||||
| 4.Total tank volume of oxidation ditch | ||||||||||
| V=V1+V2= | 884 | m3 | Total hydraulic retention time t= |
2.95 | d | |||||
| The design takes V= | 900 | m3 | ||||||||
| Design effective water depth h= | 3.5 | m | Design width b = | 5.5 | m | |||||
| Then the total length of the required ditch L = |
46.75 | m | Take the length of the linear trench section = |
22.5 | m | |||||
| Actual effective volume = | 1198.87 | m3 | Actual dwell time t'= | 4.00 | d | |||||
| 5.Alkalinity balance calculation | ||||||||||
| (1) Nitrification consumes alkalinity = | 1030.25 | mg/L | ||||||||
| (2) Denitrification produces alkalinity= | 154.54 | mg/L | ||||||||
| (3) Removal of BOD5 produces alkalinity= | 71.4 | mg/L | ||||||||
| (4) Residual alkalinity = | 175.69 | mg/L | ||||||||
| 6.Calculation of actual oxygen demand | 7.Calculation of standard oxygen demand | |||||||||
| (1) Carbonation Oxygen Demand (COD) | Press set conditions α= | 0.85 | β= | 0.95 | ||||||
| 254.17 | kg/d | CS(20)= | 9.17 | θ= | 1.024 | |||||
| CS(25)= | 8.38 | |||||||||
| 678.83 | kg/d | |||||||||
| (2) Nitrification Oxygen Demand (NOD) | ||||||||||
| D2=4.5×Q(N0-Ne)= | 218.7 | kg/d | ||||||||
| (3) Oxygen production by denitrification | 8. Sludge Return Flow Calculation | |||||||||
| D3=2.6×Q×NT= | 33.76 | kg/d | According to the set condition X0= | 250 | mg/L | Xr= | 10000 | mg/L | ||
| From QX0+Qr=(Q+Qr)X we get | ||||||||||
| (4) Nitrifying residual sludge NH4-N oxygen demand | 187.5 | m3/d | ||||||||
| D4=0.56×WV×f= | 16.79 | kg/d | ||||||||
| (5) Total Oxygen | ||||||||||
| D=D1+D2-D3-D4= | 422.31 | kg/d | 9.Residual sludge volume | |||||||
| W=WV+X1Q-XeQ= | 27.54 | m3/d | ||||||||
| Take the sludge water content P = | 99.20% | |||||||||
| 3.44 | m3/d | |||||||||
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