FixedFilm Systems Trickling Filter Kinetics CEE 3650 Spring

Predicting Performance Great Deal of Discussion in Text on Kinetics of TF Systems n

Equation 12. 48, Page 537 n S -KD/Q L = e So n n

Notes on Using Equation 12. 48 K must be normalized to Operating Temperature of

Predicting Performance Additional Equations in Form of NRC on Page 531 n Note the

NRC Equations n Stage 1 - Equation 12. 42 English Units w/V = Volumetric

NRC Equations n Equations Predict Removal Efficiency through Each Stage of TF Based on

NRC Equations (continued) n Again Remember If No Intermediate Clarifier, E 1 = 50%

State of Utah TF Design Equations R 317 -7 j n First Stage n

State of Utah Equations n W/V = Raw BOD Loading!, lb BOD/1000 ft 3

Remember Example Problem n City of 5, 000 People to be Served by High-Rate

Problem (continued) Qin = 5, 000 people (95 gal/cap-d) = 475, 000 gal/d n

Problem (continued) TF SA = 1. 425 MGD/(25 MG/Ac-d) = 0. 057 Ac =

Check Performance Using Equation 12. 48 n n n Straight Block Media, Assume K

S Determination from Mass Balance Calculations Mass Balance Equation Q (BOD 1｡ Eff) +

Check Performance Using NRC Equation n w/V = (535 lb BOD/d)/(5 ft x 2.

Check Performance Using Utah DEQ Equation n n Wraw/V = (792 lb BOD/d)/(14. 135

Slides: 18

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Fixed-Film Systems - Trickling Filter Kinetics CEE 3650 Spring 2019

Predicting Performance Great Deal of Discussion in Text on Kinetics of TF Systems n All Equations have General Form Shown in Equation 12. 48, Page 537, for all Media Types Treating Municipal Wastewater n Equation 12. 48 is Particularly Useful for Evaluating Upgrade Possibilities for Old, Stone Media Filters n

Equation 12. 48, Page 537 n S -KD/Q L = e So n n n QL = Hydraulic Loading Rate, gpm/ft 2 K = BOD Removal Rate Constant D = Filter Depth, ft n = Hydraulic Retention Constant S = BOD effluent, mg/L So = BOD influent, mg/L

Notes on Using Equation 12. 48 K must be normalized to Operating Temperature of WW Using theta (1. 035) Temperature Correction Equation n So = BOD into TF after Dilution n n Use Mass Balance Equation to Calculate So Q (BOD 1｡ Eff) + Qr (BOD 2｡ Eff) = (Q +Qr ) So n Both K and n are Functions of the Media in the TF

TF Media Properties

Predicting Performance Additional Equations in Form of NRC on Page 531 n Note the Date of the NRC Reference Here (1940 s!) n Data & Equation Generated from TFs During WWII n Stone Media Filters Treating Domestic WW n Two-Stage TF w/Intermediate Clarifiers n If No Intermediate Clarifier, E 1 = 50% n

NRC Equations n Stage 1 - Equation 12. 42 English Units w/V = Volumetric BOD Loading n Stage 2 - Equation 12. 44 English Units n Recirculation Factor - Equation 12. 43

NRC Equations n Equations Predict Removal Efficiency through Each Stage of TF Based on n n Volumetric Organic Loading Rate w 1/V, lb BOD/1, 000 ft 3/d Recycle Factor - Equation 12. 43 • Modification of Recycle Ratio • Based on empirical data fit n Efficiency of First Stage Predicted (Equation 12. 42) and Used to Reduce Loading to Second Stage (w 2/V and E 1) to Allow Prediction of BOD Removal There (Equation 12. 44)

NRC Equations (continued) n Again Remember If No Intermediate Clarifier, E 1 = 50% n Must Reduce w/V in Second Stage for Removal Provided by the First Stage n n If Have Other than Rock Media or Domestic Wastewater, then Must Use Alternative Predictive Equation 12. 48

State of Utah TF Design Equations R 317 -7 j n First Stage n Second Stage

State of Utah Equations n W/V = Raw BOD Loading!, lb BOD/1000 ft 3 -d n BOD Used in this Equation is Before Primary Treatment! R/Q = Recirculation Ratio n Equation Used to n Determine Required V for Given E, OR n Determine E from Given V n

Remember Example Problem n City of 5, 000 People to be Served by High-Rate TF Q = 95 gal/cap-d n BOD to TF = 135 mg/L n R/Q = 2 n Bed Depth to be 5 to 6 ft n Design Based on Engineering Design Criteria Given in Table 17. 7 n

Problem (continued) Qin = 5, 000 people (95 gal/cap-d) = 475, 000 gal/d n Raw BOD Loading = 0. 475 MGD (200 mg/L) (8. 34) = 792 lb BOD/d n BOD Loading to TF= 0. 475 MGD (135 mg/L) (8. 34) = 535 lb BOD/d n Q + Qr = 3 * Q = 3 * 475, 000 gal/d = 1, 425, 000 gal/d = 1. 425 MGD n TF Volume = (535 lb BOD/d)/(40 lb/1000 ft 3 d) = 13. 375 1000 ft 3 = 13, 375 ft 3 n

Problem (continued) TF SA = 1. 425 MGD/(25 MG/Ac-d) = 0. 057 Ac = 2, 483 ft 2 n TF Diameter = ((2, 483 ft 2)*4/π)1/2 = 56. 3 ft n Design Diameter = 60 ft n TF SA = π *(602)/4 = 2, 827 ft 2 n Depth of TF from this SA is n Depth = Vol/SA = 13, 375 ft 3/2, 827 ft 2 = 4. 7 ft n SO Go with Minimum 5 ft Depth w/60 ft Diameter TF n n Final TF Vol = 5 ft x 2, 827 ft 2 = 14, 135 ft 3

Check Performance Using Equation 12. 48 n n n Straight Block Media, Assume K = 0. 2, n = 0. 345 QL = (1, 425, 000 gpd)/(24 hr/d)/(60 min/hr)/(2, 827 ft 2) = 0. 35 gpm/ft 2 S/So is: S -(0. 2) 5 ft /0. 35 0. 345 - 1. 44 = e = 0. 238 S 0 n S = Effluent from TF and is Found from Mass Balance Calculations as Follows

S Determination from Mass Balance Calculations Mass Balance Equation Q (BOD 1｡ Eff) + Qr (BOD 2｡ Eff) = (Q +Qr ) So n Q (135) + (2 Q)S = 3 Q (So) n 135 + 2 S = 3 So; and S = 0. 238 So from Previous Slide n 135 + 0. 476 So = 3 So n 2. 524 So = 135; So = 53. 5 mg/L n S = 0. 238 (53. 5) = 12. 7 mg/L n %Removal in TF = (135 -12. 7)/135 = 90. 6% n

Check Performance Using NRC Equation n w/V = (535 lb BOD/d)/(5 ft x 2. 827 1000 ft 2) = 37. 8 lb. BOD/1, 000 ft 3 -d F is: n Efficiency from TF n 1 + R/ Q 1 + 2 3 F = = 2. 08 2 = (1 + 0. 1 R/ Q) (1 + 0. 1 (2)) 1. 44

Check Performance Using Utah DEQ Equation n n Wraw/V = (792 lb BOD/d)/(14. 135 1000 ft 3) E is:

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