Asp_tf_rbc.docx

Bar Screen Raw sewage

Grit chamber

Primary clarifier

Primary sludge

Digestion supernatent Digestionsu pernatent

Primar y clarifier

Comminuter

Wet well

Aeration Tank  Activated Sludge Process Sludge Return

Secondary clarifier

Effluent

Secondar y clarifier

Waste sludge

Drying beds

Sludge digestion

Stabilized sludge

Typical layout of an activated sludge treatment plant

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1. Activated Sludge Process

𝐹𝑖 ,𝑆𝑖 ,𝑋𝑖

Aeration tank Va ,𝑋𝑎 ,𝑆𝑒

𝐹𝑖 (1 + 𝑅),𝑋𝑎 ,𝑆𝑒

(𝐹𝑖 − 𝐹𝑤 ),𝑋𝑒 ,𝑆𝑒

Secondary clarifier

𝐹𝑟 ,𝑋𝑟 ,𝑆𝑒

𝐹𝑤 ,𝑋𝑟

Typical flow scheme for a completely mixed activated sludge plant

𝐹𝑖 = rate of fresh wastewater inflow to the aeration tank (m3/day) 𝑆𝑖 = substrate concentration of BOD in raw sewage (mg/L) 𝑉𝑎 = volume of aeration tank (m3) 𝑋𝑎 = mixed liquor volatile suspended solids in the aeration tank (mg/L) 𝑋𝑒 = suspended solids in the effluent after treatment (mg/L) 𝑆𝑒 = steady-state substrate concentration after treatment (mg/L) 𝐹𝑟 = rate of sludge recycle = RFi 𝑅 = sludge recycle ratio = Fr/Fi 𝐹𝑤 = rate sludge wasting (m3/day) 𝑋𝑟 = mixed liquor volatile suspended solids in the recycled sludge from the secondary clarifier (mg/L)

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Inflow of microorganisms into the system boundary

Outflow of microorganisms from the system

Outflow of microorganisms from the system

Accumulation of microorganisms within the system boundary

Figure 1: Mass balances for the microorganisms and for the substrate flowing in the aeration tank.

𝐴𝑐𝑐𝑢𝑐𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 𝐼𝑛𝑓𝑙𝑜𝑤 − 𝑂𝑢𝑡𝑓𝑙𝑜𝑤 + 𝑁𝑒𝑡 𝑔𝑟𝑜𝑤𝑡ℎ

𝑑𝑋 𝑉 = 𝐹𝑖 𝑋𝑖 + 𝐹𝑟 𝑋𝑟 − (𝐹𝑖 + 𝐹𝑟 )𝑋𝑎 + 𝑉𝑎 𝑋𝑎 (𝜇 − 𝐾𝑑 ) 𝑑𝑡 𝑎 𝑑𝑋 𝑑𝑡

= rate of microorganisms growth measured in terms of mass of mixed liquor volatile suspended solids (MLVSS) per unit volume-time

𝑉𝑎 = reactor volume (aeration tank) 𝐹𝑖 = flow rate 𝑋𝑖 = concentration of microorganisms in influent 𝑋𝑎 = concentration of microorganisms in reactor µ = specific growth rate 𝐾𝑑 = decay rate

Specific growth rate µ can be substituted by equation below: 𝜇=

𝜇max(𝑆𝑒) 𝐾𝑠 + 𝑆𝑒

Substrate balance can be written similar to the microorganism mass balance as follow: 𝑑𝑆 𝑞𝑚𝑎𝑥 𝑆𝑒 𝑉𝑎 = 𝐹𝑖 𝑆𝑖 + 𝐹𝑟 𝑆𝑒 − (𝐹𝑖 + 𝐹𝑟 )𝑆𝑒 − 𝑉𝑎 𝑋𝑎 ( ) 𝑑𝑡 𝐾𝑒 + 𝑆𝑒 Where, 𝑞 max = µmax/Y Equations above can be solved if biokineticconstantµ𝑚𝑎𝑥 , 𝑌,𝐾𝑑 , and 𝐾𝑠 , were determined. Typical values for treatment of domestic wastewater and some industrial wastewater effluents treated biologically under aerobic or anaerobic environment are available. The most important parameters for the design and operation of activated sludge plants are: 3

1. The biological solids retention time (BSRT) 𝜃c, which is defined as the average number of days a unit of biomass remains in the treatment system. 2. The food-to-biomass ratio 𝐹/𝑀 which is the ratio of the ultimate BOD inflow per day in the aeration tank to the total biomass represented by the mixed liquor volatile suspended solids (MLVSS) in the aeration tank

𝜃𝑐 =

𝑋 𝑇𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑒𝑙𝑙𝑠 𝑖𝑛 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 𝑋𝑎 𝑉𝑎 = = ∆𝑋/∆𝑡 𝑀𝑎𝑠𝑠 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐𝑒𝑙𝑙𝑠 𝑟𝑒𝑚𝑜𝑣𝑒𝑑 𝑝𝑒𝑟 𝑑𝑎𝑦 𝐹𝑤 𝑋𝑟 + (𝐹𝑖 + 𝐹𝑤 )𝑋𝐸

∆𝑋/∆𝑡 = total active biomass withdrawn from the system (kg of MLVSS/day) X = total active biomass in aeration tank (kg of MLVSS)

𝐹 𝐹𝑖 𝑆𝑖 = 𝑀 𝑋𝑎 𝑉𝑎

The food-to-microorganism ratio (𝐹/𝑀) may be known as organic loading. Also, the specific substrate utilization rate 𝑞. 𝑞=

𝐹𝑖 (𝑆𝑖 − 𝑆𝑒 ) 𝑋𝑎 𝑉𝑎

The recycled solids 𝑋𝑟 to the aeration tank is a function of the settling characteristics of the solids in the secondary clarifier. When secondary clarifiers are operating properly, solids captured should approximate 100%, and the maximum solids concentration in the sludge return line (𝑋𝑟 )𝑚𝑎𝑥 can be estimated by (𝑋𝑟 )𝑚𝑎𝑥 =

106 𝑆𝑉𝐼

SVI = sludge volume index (mL/g), or volume or sludge occupied by 1g of solids when 1L of the activated sludge from the aeration tank is kept under quiescent conditions in a graduated cylinder for 30 minutes.

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Example – Activated Sludge An aeration tank of 30×10×5 meters treats 5000 cubic meters per day of primary wastewater effluent containing 150mg/L of ultimate BOD. Under steady-state conditions, the aeration tank contains 2000mg/L of MLVSS. The sludge recycle ratio is 0.3 and wasted sludge is 500kg/day. Calculate 1. Hydraulic detention time in aeration tank. 2. Concentration of return sludge. 3. Biological solids retention time if 90% substrate removal efficiency is desired, and the effluent suspended solids are not to exceed 20mg/L. 4. Specific substrate utilization rate.

Solution: Draw a schematic to describe the problem statement as shown below. The removal efficiency is 90%, therefore most of the variables can be quantified: 𝐹𝑖 = 5000 m3/day 𝑆𝑖 = 150 mg/L

30 × 10 × 5 m3

Secondary clarifier

𝑋𝑎 = 2000 mg/L

𝐹𝑖 = 5000 m3/day 𝑆𝑒 = 15 mg/L 𝑋𝑟 = 20 mg/L

𝐹𝑟 = 0.3 𝐹𝑖 m3/day conc = 𝑋𝑟 mg/L Return sludge

𝐹𝑤 ,𝑋𝑟 Waste sludge = 500 kg/day

1. Hydraulic detention time = Tank volume Flow =𝜃𝑏 . The flow is considered the incoming fresh feed flow 𝐹𝑖 plus the return sludge from 𝐹𝑟 . However, some investigators use only incoming fresh feed flow to determine the fresh feed hydraulic detention time.

𝐹𝑖 = 5000 𝑚3 / 𝑑𝑎𝑦 𝐹𝑟 = (0.3)(5000) = 1500 𝑚3 /𝑑𝑎𝑦 Tank volume = V = 30 × 10 × 5 = 1500 m3 𝜃ℎ =

1500 𝑚3 = 0.23 𝑑𝑎𝑦 = 5.54 ℎ𝑜𝑢𝑟𝑠 (5000 + 1500)𝑚3 /𝑑𝑎𝑦

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2. Concentration of return sludge: For this steady-state system, a mixed liquor suspended solids(MLVSS) concentration of 2000 mg/L is maintained I the aeration tank. These solids are supplied by return sludge and biological growth as shown from a material balance: Solids inflow in fresh feed

Solids inflow in return sludge

Biological growth

Solids outflow from tank

The concentration of solids inflow in fresh feed is considered insignificant or zero, and the biological growth is equivalent to solids lost from the system. 𝐵𝑖𝑜𝑙𝑜𝑔𝑖𝑐𝑎𝑙 𝑔𝑟𝑜𝑤𝑡ℎ = 𝑆𝑜𝑙𝑖𝑑𝑠 𝑖𝑛 𝑤𝑎𝑠𝑡𝑒 𝑠𝑙𝑢𝑑𝑔𝑒 + 𝐸𝑓𝑓𝑙𝑢𝑒𝑛𝑡 𝑠𝑜𝑙𝑖𝑑𝑠 Δ𝑋𝑔 Δ𝑋𝑤 Δ𝑋𝑒 = + Δ𝑡 Δ𝑡 Δt 1000 𝐿 ∕ 𝑚3 3 ⁄ (5000 = 500 𝑘𝑔 𝑑𝑎𝑦 + 𝑚 ∕ 𝑑𝑎𝑦)(20 𝑚𝑔 ∕ 𝐿) × ( 6 ) 10 𝑚𝑔 ∕ 𝑘𝑔 = 500 + 100 = 600 𝑘𝑔 ∕ 𝑑𝑎𝑦

For materials balance of MLVSS: 𝐹𝑖 𝑋𝑖 + 𝐹𝑟 𝑋𝑟 + 𝐺𝑟𝑜𝑤𝑡ℎ = (𝐹𝑖 + 𝐹𝑟 )𝑋𝑎 1000 𝐿⁄𝑚3 0 + (1500𝑚3 ∕ 𝑑𝑎𝑦)𝑋𝑟 𝑚𝑔⁄𝐿 ( 6 ) + 600 𝑘𝑔 ∕ 𝑑𝑎𝑦 10 𝑚𝑔⁄𝐿 1000 𝐿 ∕ 𝑚3 = (5000 + 1500)𝑚 ∕ 𝑑𝑎𝑦 (2000 𝑚𝑔 ∕ 𝐿 × 6 ) 10 𝑚𝑔 ∕ 𝑘𝑔 3

1.5 𝑋𝑟 = 13,000 − 600 = 12,400 𝑋𝑟 = 8267 𝑚𝑔 ∕ 𝐿

3. Biological solids retention time,𝜃𝑐 : 𝜃𝑐 =

𝑋 𝑇𝑜𝑡𝑎𝑙 𝑀𝐿𝑉𝑆𝑆 𝑖𝑛 𝑎𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑡𝑎𝑛𝑘 = ∆𝑋 ∕ ∆𝑡 𝑀𝐿𝑉𝑆𝑆 𝑟𝑒𝑚𝑜𝑣𝑒𝑑 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚

=

(1500𝑚3 )(2000𝑚𝑔 ∕ 𝐿)(1000𝐿 ∕ 𝑚3 )(𝑘𝑔 ∕ 106 𝑚𝑔) 600 𝑘𝑔 ∕ 𝑑𝑎𝑦

=

3000𝑘𝑔 = 5 𝑑𝑎𝑦𝑠 600 𝑘𝑔 ∕ 𝑑𝑎𝑦

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Notice from the flow balance of the whole system: 𝐹𝑖 + 𝐹𝑟 = 𝐹𝑒 + 𝐹𝑟 + 𝐹𝑤 𝐹𝑤 is generally very small, and 𝐹𝑖 approximates 𝐹𝑒 .

4. Specific substrate utilization 𝑞: 𝒒= =

𝐹𝑖 (𝑆𝑖 − 𝑆𝑒 ) 𝑋𝑎 𝑉𝑎

(5000 𝑚3 ∕ 𝑑𝑎𝑦)(150 − 15)𝑚𝑔 ∕ 𝐿 (2000𝑚𝑔 ∕ 𝐿)(1500𝑚3 )

= 0.225⁄𝑑𝑎𝑦

𝒒 represents the mass of BOD or substrate utilized per unit mass of microorganisms per unit time, as stated earlier. From the previous example, it is shown that some of the biokinetic parameters can be estimated by monitoring the operational parameters of a full-scale plant.

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2. Trickling Filters / Biotowers Bar Screen Raw sewage

Grit chamber

Primary clarifier

Primary sludge

Digestion supernatent Digestionsu pernatent

Primar y clarifier

Comminuter

Wet well

Trickling Filters / Biotowers

Sludge Return

Secondary clarifier

Effluent

Secondar y clarifier

Waste sludge

Drying beds

Sludge digestion

Stabilized sludge

Typical Biotower System

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Tricking Filter/Biotower with 40mm perforated spherical hollow celluloid medium in HCP arrangement

Trickling Filter/Biotower with 40mm perforated spherical hollow celluloid medium in HCP arrangement

Attached –Cultured Processes 9

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Example: Biotower A biotower composed of a modular plastic medium is to be used as the secondary-treatment component in a municipal wastewater treatment plant. Flow from the primary clarifier is 20,000m3/d with a BOD5 of 150mg/L. Pilot-plant analysis has established a treatability constant of 0.0055min-1 for the system at 20oC, and the n factor (modular plastic medium characteristics coefficient) can be taken as 0.5. Two biotowers are to be used, each with a square surface and separated by a common wall. The medium is to have a depth of 6.5m, and the recirculation ratio is to be 2 to 1 during average flow periods. Determine the dimensions of the units required to produce an effluent with soluble BOD5 of 10mg/L. The Mininum temperature is expected to be 25oC (use Equation kT = k20(1.035)T-20)

Determine influent concentration of BOD5  Eq. 5-28 for Sa 2. Adjust treatability constant for temperature kT  Eq. 5-26 for K25 3. Determine Hydraulic Loading Rate (m3/m2.min)  Eq. 5-27 for Q 4. Determine the surface area of each biotower (m2) 1.

kT = k20(1.035)T-20

Sa = mixture of raw and recycled mixture applied to the medium

Solution: 1. Determine the influent concentration of BOD5. Sa = Sa =

So + RSe 1+R 150+(2x10) 1+2

= 56.7 mg/L

2. Adjust treatability constant (k) for temperature. kT = k20(1.035)T-20 k25 = k20(1.035)25-20 = 0.055(1.035)5 = 0.065 min-1

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3. Determine/Solve Hydraulic Loading Rate, m3/m2.min

10 56.7 10 56.7 10 56.7

0.5

=

e−0.065x6.5/Q

0.5

(1+2)− (2)e−0.065x6.5/Q

(1 + 2) −

10 56.7

(2)e−0.42/Q

(1 + 2) = e−0.42/Q 0.529 = 1 e−0.42/Q

0.5

0.5

+

0.5

10 56.7

= e−0.42/Q

0.5

(2)e−0.42/Q

+ 0.353e−0.42/Q

0.5

0.5

0.5

0.529 = 1.353e−0.42/Q 0.529 1.353

= e−0.42/Q

0.5

0.39 = e−0.42/Q

0.5

0.39 = (e−0.42 )1/Q

0.5

0.5

0.39 = (0.655)1/Q

0.5

Log 0.39 = Log (0.655)1/Q

Log 0.39 = 1/Q0.5 Log (0.655) Log 0.39 Log 0.655

Q0.5 =

=

1 Q0.5

Log 0.39 Log 0.655

(Q0.5) = (0.45) Q = (0.45)2 Q = 0.20

m3 m2 −min

4. The surface area of each unit is determined as follows: (20,000 + 40,000) m3 41.67 min m3 (2)(0.20) 2 m −min

m3 d

1d

m3

x 1440 min = 41.67 min

= 104.17 m2

Each unit is square, so dimensions are L=W= √104.17 m2 = 10.21 m, say 10 m Each unit is 10m x 10m x 6.5m deep. 12

5. Rotating Biological Contactors Bar Screen Raw sewage

Grit chamber

Primary clarifier

Primary sludge

Digestion supernatent Digestionsu pernatent

Primar y clarifier

Comminuter

Wet well

Rotating Biological Contactors (RBCs) Sludge Return

Secondary clarifier

Effluent

Secondar y clarifier

Waste sludge

Drying beds

Sludge digestion

Stabilized sludge

Rotating Biological Contactor System

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Efficiency & Loading Rate Relationship for Bio-Surf Medium Treating Municipal WW

Temperature Correction for Loading Curves for Above Figure (Multiply Loading Rate by Correction Factors)

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Example: Rotating Biological Contactor Determine the surface area required for a Rotating Biological Contactor (RBC) system to treat the wastewater with following characteristics. Flow from the primary clarifier is 20,000m3/d with a BOD5 of 150mg/L. Pilot-plant analysis has established a treatability constant of 0.0055min-1 for the system at 20oC, and the n factor (modular plastic medium characteristics coefficient) can be taken as 0.5. Two biotowers are to be used, each with a square surface and separated by a common wall. The medium is to have a depth of 6.5m, and the recirculation ratio is to be 2 to 1 during average flow periods. Determine the dimensions of the units required to produce an effluent with soluble BOD5 of 10mg/L. The Mininum temperature is expected to be 25oC (use Equation kT = k20(1.035)T-20)

Solution: 1. Enter the figure - “Efficiency & Loading Rate Relationship for Bio-Surf Medium Treating Municipal WW” with: Influent BOD = 150mg/L Effluent BOD = 10mg/L The Design or Maximum Hydraulic Loading Rate is approximately 50L/m2.d = 0.05m3/m2.d 2. Disk area (Ad) required Flow from Primary clarifier = 20,000 m3/d Recirculation Ratio = 2:1, Thus, R=2, Recirculation Rate = 2(20,000) = 40,000 m3/d Total Flow to Biotowers = (20,000 + 40,000) = 60,000 m3/d 𝟔𝟎,𝟎𝟎𝟎 𝐦𝟑 /𝐝

Estimated Disk Area, Ad = 𝟎.𝟎𝟓 𝐦𝟑 /𝐦𝟐 .𝐝 = 1.2(106) m2 3. Assuming a 7.6-m shaft for a 3.7-m diameter disk with a total surface area of 1(104) m2, 1.2(106 ) m2 1(104 ) m2

= 120 modules in parallel will be required to provide single-stage treatment of

the wastewater. For nitrification, a maximum of 5 stages (200 modules) will be required.

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