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Secondary Treatment Processes

Secondary Treatment Processes. Two Types. 1. Attached growth or Fixed Film. Organisms attached to some inert media like rocks or plastic. . 2. Suspended Growth. Organisms are suspended in the treatment basin fluid. This fluid is commonly called the “mixed liquor”.

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Secondary Treatment Processes

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  1. Secondary Treatment Processes

  2. Two Types 1. Attached growth or Fixed Film Organisms attached to some inert media like rocks or plastic. 2. Suspended Growth Organisms are suspended in the treatment basin fluid. This fluid is commonly called the “mixed liquor”.

  3. Attached Growth or Fixed Film Reactors Trickling Filters Rock Media Typically 4 – 8 feet deep.

  4. Trickling Filters With time, the “slime” layer becomes thicker and thicker until oxygen and organic matter can not penetrate to the organisms on the inside. The organisms on the inside then die and become detached from the media, causing a portion of the “slime” layer to “slough off”. This means the effluent from a trickling filter will have lots of solids (organisms) in it which must be removed by sedimentation

  5. Trickling Filters Single Stage Trickling Filter Two Stage Trickling Filter

  6. Bio-towers

  7. Rotating Biological Contactors (RBC’s) In trickling filters the moving wastewater passes over the stationary rock media. In an RBC, the moving media passes through the stationary wastewater. Commonly used out in series and parallel

  8. Suspended Growth Processes 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

  9. 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: dX/dt = - Y dS/dt Y is called the yield and is the mass of cells produced per mass of substrate used (g SS/g BOD)

  10. : = : S /(Ks + S) dX/dt = : X = (: S X) /(Ks + S) So: dS/dt = - dX/dt (1/Y) = (: S X) / [Y (Ks + S)] Monod Model for Substrate Utilization dX/dt = - Y dS/dt

  11. Mean Cell Residence Time, 2c Mean cell residence time (MCRT, 2c) is the mass of cells in the system divided by the mass of cells wasted per day. Consider the system: 2c = VX/QX = V/Q At SS the amount of solids wasted per day must equal the amount produced per day: 2c = XV/[Y(dS/dt)V] = X/[Y(dS/dt)

  12. So: dS/dt = X/Y (: S /(Ks + S) 1/2c = (: S /(Ks + S) S = Ks/(:2c – 1) Mass Balance on Microorganisms: V dX/dt = Q X0 – Q X + Y(dS/dt)V S.S.: (dX/dt) V = 0, and QX0 = 0

  13. A CSTR without cell recycle receives an influent with 600 mg/L BOD at a rate of 3 m3/day. The BOD in the effluent must be 10 mg/L. The kinetic constants are: Ks = 500 mg/L and : = 4 days-1. How large should the reactor be? S = Ks/(:2c – 1) Example Solve for 2c: 2c = (Ks + S)/(S :)) = (500+10)/(10 x 4) = 12.75 days 2c = V/Q V = 2c Q = 12.75 (3) = 38.25 m3

  14. S = Ks/(:2c – 1) = 500/[4(8) – 1] = 16.1 mg/L Given the conditions in the previous example, What would the percent reduction in substrate be if the reactor volume was 24 m3? 2c = V/Q = 24/3 = 8 days Reduction = [(600 – 16.1)/600] x 100 = 97.3%

  15. = [(S0 – S)/ t ] V /(XV) = (S0 – S)/(X t ) Now consider a CSTR with cell recycle: 2c = (X V) / [(QwXr) + (Q – Qw)Xc] Since Xc = 0: 2c = (XV)/(QwXr) Removal of substrate often expressed in terms of substrate removal velocity, q: q = (mass of substrate removed/time)/(mass of microorganisms under aeration)

  16. Since : = : S /(Ks + S) (: S) /[Y(Ks + S)] (: S X t ) /[Y(Ks + S)] By substitution: q = But q is also equal to: q = (S0 – S)/(X t ) S0 – S = Mass balance on microorganisms: V dX/dt = Q X0 – QwXr – (Q – Qw)Xc + : X V X0 = Xc = 0 : = (Xr Qw)/(X V) = 1/ 2c The substrate removal velocity, q, can also be expressed as: q = :/Y If we equate these two equations for q and solve for S0 – S:

  17. And: x = (S0 – S) / t q Since q = : / Y 2c = 1/ (q Y)

  18. An activated sludge system operates at a flow rate of 400 m3/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, : = 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 S0 – S = (: S X t ) /[Y(Ks + S)] t = [Y(S0 – S) ( Ks + S)] / ( : S X) Problem The hydraulic retention time may be found from the following equation: = [0.5(300 – 30)(200 + 30)] / [2 (30) (400)] = 0.129 days = 3.1 hr

  19. V = t Q = 400 (0.129) = 51.6 m3 2c = 1/ (qY) Q = (S0 – S) / (X t ) = (300– 30) / [(4000)(0.129)] = 0.523 (kg BOD removed/day) / (kg SS in the reactor) 2c = 1/ (qY) = 1 / (0.523 x 0.5) = 3.8 days Also 2c = (X V) / (Xr Qw) Xr Qw = (X V) / 2c = [(4000)(51.6)( 103 L/m3)( 1/106 kg/mg)] / 3.8 = 54.3 kg/day

  20. q = X = (S0 – S) / ( t q ) (: S) /[Y(Ks + S)] Using the same data what MLSS is necessary to produce an effluent concentration of 15 mg BOD/L? = [2(15)] / [0.5(200 + 15)] = 0.28 day-1 = (300 – 15) / [0.129(0.28)] = 7890 mg/L 2c = 1 / (q Y) = 1 / [0.28(0.5)] = 7.2 days

  21. 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.

  22. Aeration Diffused Aeration Coarse Bubble Fine Bubble

  23. Mechanical Aeration

  24. Modeling Gas Transfer Henry’s Law S = K P S = solubility of the gas, mg gas/L P = partial pressure of the gas K = solubility constant If a gas is 60% O2 and 40% N2 and the total pressure of the gas is 1 atm (101 KPa), the partial pressure of O2 = 0.6 x 101 = 60.6 Kpa. The total pressure is equal to the sum of the partial pressures (Dalton’s Law)

  25. Example At one atmosphere, the solubility of pure oxygen is 46 mg/L in water with no suspended solids. What would be the solubility if the gas were replaced by air? With Pure oxygen: S = K P 46 = K x 1 K = 46 mg/(L-atm) With air: S = K P Since air is 20% oxygen, P = 1 x 0.2 = 0.2 atm S = 46 x 0.2 = 9.2 mg/L

  26. Oxygen Transfer The rate of oxygen transfer is proportional to the difference in the oxygen concentration that exists in the system and the saturation concentration: dC/dt % (S – C) The constant of proportionality is called the gas transfer coefficient, KLa dC/dt = Kla (S – C) S – C = D, so: dD/dt = Kla D Integrating: Ln (D/D0) = -Kla t

  27. Example Two diffusers are to be tested for their oxygen transfer capability. Tests were conducted at 20oCusing the system shown below, with the following results:

  28. Kla = slope of the lines Air-Max: Kla = 2.37 min-1 Wonder: Kla = 2.69 min-1

  29. (volume of sludge after 30 min. settling, ml) x 1000 SVI = mg/L suspended solids Sludge Volume Index, SVI 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

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