Activated Sludge Processes MK Unit Proses Teknik Lingkungan

Activated Sludge Processes MK Unit Proses Teknik Lingkungan Program Studi Teknik Lingkungan FTSP ITS

Suspended Growth • Organisms are suspended in the treatment basin fluid. • This fluid is commonly called the “mixed liquor”. Suspended Growth: • • • Activated sludge Oxidation ditch/pond Aerated lagoon, stabilization pond

Activated Sludge • • Process in which a mixture of wastewater & microorganisms is agitated & aerated Leads to oxidation of dissolved organics After oxidation, sludge is separated from wastewater Detention time = approximately 6 - 8 hours

Activated Sludge • Designed based on loading (the amount of organic matter added relative to the microorganisms available) • Commonly called the food -to-microorganisms ratio, F/M • F measured as BOD. M measured as volatile suspended solids concentration • F/M is the pounds of BOD/day per pound of MLSS in the aeration tank

Activated sludge process

Activated sludge flocs Note filamentous bacteria Note Vorticella and other protozoa

Activated sludge model

Primary aeration tank

Oxygenated systems Cryogenic air separation facility, Hyperion, Playa del Rey, CA)

Settling tanks Secondary settling tank, Hyperion, Playa del Rey, CA)

Nitrogen removal q Nitrification (Nitrosomonas and Nitrobacter) NH 3 + O 2 NO 2 - NO 3 - q Denitrification NO 3 - + organics CO 2 + N 2 q Process adaptations Air Anoxic Aerobic

Phosphate removal q BNR plants q Discarding phosphate anaerobically q Luxury aerobic uptake of P in aerobic stage q Process adaptations for N and P removal Wastewater Anaerobi c Air Anoxic Aerobic

Excess biomass disposal q Production q Separation q Further biological treatment – (an)aerobic q Dewatering q Drying – solar or gas heated q Disposal/ beneficial use – soil amender/fertilizer or fuel The cost of biomass disposal amount to about half the cost of wastewater treatment. Aeration, if used, almost up to half of the rest of the cost. If no aeration, the capital cost , including the cost of land, could be very high.

Typical steps in modern wastewater treatment

Type of Activated Sludge • Activated sludge without cell recycle • Activated sludge with cell recycle

Design of Activated Sludge • Influent organic compounds provide the food for the microorganisms and is called substrate (S) • The substrate is used by the microorganisms for growth, to produce energy and new cell material. • The rate of new cell production as a result of the use of substrate may be written mathematically as: • Y is called the yield and is the mass of cells produced per mass of substrate used (g SS/g BOD)

Monod Model for Substrate Utilization

ACTIVATED SLUDGE WITHOUT CELL RECYCLE

Mean Cell Residence Time, θc Mean cell residence time (MCRT, θc) is the mass of cells in the system divided by the mass of cells wasted per day. Consider the system: For no recycle systems, θc = θh At steady state, the amount of solids wasted per day must equal the amount produced per day:

Mass Balance on Microorganisms: Accumulation = input – output – process In steady state condition (d. X/dt) V = 0, and QX 0 = 0 dibagi VX Jika kd diabaikan

Example A CSTR without cell recycle receives an influent with 600 mg/L BOD at a rate of 3 m 3/day. The BOD in the effluent must be 10 mg/L. The kinetic constants are: Ks = 500 mg/L and μm = 4 days-1. How large should the reactor be? Solve for θc: V = θc Q = 12. 75 (3) = 38. 25 m 3

Given the conditions in the previous example, What would the percent reduction in substrate be if the reactor volume was 24 m 3? Reduction = [(600 – 16. 1)/600] x 100 = 97. 3%

ACTIVATED SLUDGE WITH CELL RECYCLE

• Aerobic suspended systems – activated sludge

Now consider a CSTR with cell recycle: Since Xe = 0: Removal of substrate often expressed in terms of substrate removal velocity, q:

Mass balance on microorganisms: X 0 = Xe = 0 The substrate removal velocity, q, can also be expressed as: q = μ/Y By substitution: But q is also equal to:

If we equate these two equations for q and solve for S 0 – S: Hydraulic retention time: Since q = μ / Y Reactor volume:

Solids Separation The success of the activated sludge process depends on the efficiency of the secondary clarifier, which depends on the settling characteristics of the sludge (biosolids). Some system conditions result in sludge that is very difficult to settle. In this case the return activated sludge becomes thin (low MLSS) and the concentration of organisms in the aeration tank goes down. This produces a higher F/M ratio (same food input, but fewer organisms) and a reduced BOD removal efficiency. One condition that commonly causes this problem is called bulking sludge. Bulking sludge occurs when a type of bacteria called filamentous bacteria grow in large numbers in the system. This produces a very billowy floc structure with poor settling characteristics.

Sludge Volume Index, SVI (volume of sludge after 30 min. settling, ml) x 1000 SVI = mg/L suspended solids A mixed liquor has 4000 mg/L suspended solids. After 30 minutes of settling in a 1 L cylinder, the sludge occupied 400 ml. SVI = (400 x 1000)/ 4000 = 100 Good settling if SVI < 100, if SVI > 200 …. problems

Problem An activated sludge system operates at a flow rate of 400 m 3/day and has an influent BOD of 300 mg/L. The kinetic constants for the system have been determined to be: Ks = 200 mg/L, Y = 0. 5 kg SS/kg BOD, μm = 2 day-1. The mixed liquor suspended solids concentration will be 4000 mg/L. IF the system must produce an effluent with 30 mg/L BOD, determine: A. The volume of the aeration tank B. The sludge age (MCRT) C. The quantity of sludge wasted per day The hydraulic retention time may be found from the following equation: θh = [0. 5(300 – 30)(200 + 30)] / [2 (30) (4000)] = 0. 129 days = 3. 1 hr

V = θh Q = 400 (0. 129) = 51. 6 m 3 θc = 1/ (q. Y) q = (S 0 – S) / (X θh ) = (300– 30) / [(4000)(0. 129)] = 0. 523 (kg BOD removed/day) / (kg SS in the reactor) θc = 1/ (q. Y) = 1 / (0. 523 x 0. 5) = 3. 8 days Also θc = (X V) / (Xr Qw) Xr Qw = (X V) / θc = [(4000)(51. 6)( 103 L/m 3)( 1/106 kg/mg)] / 3. 8 = 54. 3 kg/day

Using the same data what MLSS is necessary to produce an effluent concentration of 15 mg BOD/L? q = (μm S) /[Y(Ks + S)] = [2(15)] / [0. 5(200 + 15)] = 0. 28 day-1 X = (S 0 – S) / ( θh q ) = (300 – 15) / [0. 129(0. 28)] = 7890 mg/L θc = 1 / (q Y) = 1 / [0. 28(0. 5)] = 7. 2 days

Ringkasan Persamaan-persamaan yang digunakan dalam proses activated sludge: 1. TANPA RECYCLE:

• Mencari kd dan Y dari percobaan lab. : TANPA RECYCLE • Mencari Ks dan μm dari percobaan lab. :

2. DENGAN RECYCLE:

• Mencari ke dan. Y dari percobaan lab. : RECYCLE • Mencari Ks dan μm dari percobaan lab. :

Parameters Sludge age & hydraulic residence time Microorganism in system Substrate in system Determining kd & Y Determining Ks & m F-M ratio Recycle Non-Recycle

Recycle Ratio R=Qr/Qo

Mass Balance at Junction • Microorganism Xo = 0 • Substrate

• Sludge Production SR = (So - Se). Qo • Oxygen Requirement Y’ = oxygen coeff, mass oxygen/mass substrate utilized = 1 – 1, 42 Y kd‘ = endogenous respiration coeff, mass oxygen/ mass cell-day = 1, 42 kd On = oxygen for nitrification = mass N x 3. 84

Soal-soal • Suatu percobaan proses lumpur aktif dalam skala laboratorium yang dioperasikan secara Batch, dengan waktu aerasi 24 Jam dan diketahui nilai MLVSS = 70 % dari nilai MLSS. Hitunglah nilai Y dan kd dengan acuan masa MLVSS. Data hasil percobaan : No. Xo (MLSS) Xt (MLSS) So (BOD 5) St (BOD 5) Reaktor (mg/L) 1 450 790 725 95 2 860 1160 725 89 3 1650 1960 725 77 4 3670 3875 725 70

Suatu kawasan pemukiman dengan jumlah penduduk 8000 jiwa dilayani dengan sistem IPAL terpusat. Kapasitas air buangan sebesar 225 L/orang. hari dan rata-rata BOD sebesar 425 mg/L. Konsentrasi NH 3 -N sebesar 25 mg/L dan teroksidasi sebesar 95 %. Susunan sel C 5 H 7 NO 2. Removal BOD pada pengendap I diperkirakan sebesar 32 %. Jika diinginkan reaktor utama adalah proses Lumpur Aktif, dengan nilai : Y = 0, 81 kg VSS/ kg BOD; ke = 0, 07/hari ; Y’ = 0, 73 kg O 2/kg BOD; ke’ = 0, 16 kg O 2/kg VSS; X = 2500 mg MLSS/L ; Xr = 10. 000 mg/L MLVSS/MLSS = 0, 7, umur lumpur = 10 hari , waktu tinggal hidrolik = 8 jam. Hitunglah : • Konsentrasi BOD effluen. • Produksi lumpur • Kebutuhan Oksigen. • Debit resirkulasi yang diperlukan

Air limbah industri rumah tangga diolah dengan Proses lumpur aktif. Karakteristik air buangan dan Parameter perencanaan ditentukan sbb: • Q = 6. 000 m 3/hari. • BOD 5 influen (So) = 320 mg/ L • BOD 5 effluen (Se) = 20 mg/ L • Umur Lumpur = 7 hari, • Konsentrasi mikroorganisme dalam reaktor (X) = 2500 mg VSS/L • SVI = 85 • Y = 0, 6 kg VSS/kg BOD ke = 0, 04 • MLVSS = 0, 75 MLSS. 1. Hitung kebutuhan Oksigen untuk proses lumpur aktif tersebut 2. Hitung nilai ratio resirkulasi (R = Qr/Qo) 3. Hitung volume tangki aerasi 4. Hitung ratio F/M.