Q (m^{3} / h) |
Flow - rate |

Q_{o} (m^{3} / h) |
Influent flow - rate |

Q_{e} (m^{3} / h) |
Effluent flow - rate |

Q_{r} (m^{3} / h) |
Recycled sludge flow - rate |

Q_{w} (m^{3} / h) |
Wasted sludge flow - rate |

BOD (mg / L) |
Biochemical oxygen demand |

BOD_{o} (mg / L) |
Influent biochemical oxygen demand |

V_{a} (m^{3}) |
Aeration tank volume |

SS (mg / L) |
Suspended solids (SS) |

SS_{e} (mg / L) |
Effluent SS |

SS_{r , w} (mg / L) |
Recycled and wasted sludge SS |

A (m^{3} / h) |
Air flow - rate |

MLSS (mg / L) |
Mixed liquor suspended solids |

t (h) |
Hydraulic retention time |

OL (kg BOD / m^{3} . day) |
Organic loading |

F / M (kg BOD / kg MLSS . day) |
Food to microorganism ratio |

r |
Recycle ratio |

SA (day) |
Sludge age |

ASR (m^{3} / kg BOD) |
Air supply rate |

E (%) |
BOD removal efficiency |

(2) Aeration time or hydralic retention time = Volume / Flow - rate

(3) BOD or organic loading = Mass of BOD applied per day / Tank volume

(4) Food to microorganism ratio = Mass of BOD applied per day / Mass of suspended solids in aeration tank

(5) Recycle ratio = Recycle flow - rate / Influent flow - rate

(6) Mean cell residence time or sludge age = Mass of suspended solids in aeration tank / Mass of solids leaving the system

(7) Air supply rate = Volume of air per unit time / Influent flow

or

(8) BOD removal efficiency ;

(BOD removal efficiency is a function of MLSS, sludge age, or F/M ; there is usually an optimum range of MLSS, sludge age, and F/M)

Primary effluent BOD

Compute BOD

(a) Final effluent BOD

(b) Overall efficiency = (200 mg / L - 13 mg / L) / (200 mg / L) = 93.5 %

(Q) (0) + (R) (Q) (8,000) = [Q + (R) (Q)] (2,800)

(R) (Q) (8,000) = [Q + (R) (Q)] (2,800)

(R) (8,000) = 2,800 + (2,800) (R)

(5,200) (R) = 2,800

R = 0.54

(b) Volumetric BOD load = Quantity of influent BOD / Aeration volume = [(173 mg / L) (8.34 lb / mil.gal / mg / L) (7.7 mil.gal / day)] / 300,000 ft

(c) F to M = Quantity of influent BOD / Mass of MLSS in Aeration Basin

F to M = [(173 mg / L) (8.34 lb / mil.gal / mg / L) (7.7 mil.gal / day)] / [(2,500 mg / L) (8.34 lb / mil.gal / mg /L) (2.24 mil.gal)] = 0.24 lb BOD / day.lb MLSS

(d) TS removal efficiency = (600 - 500) / 600 = 16.7 %

(e) SS removal efficiency = (120 - 22) / 120 = 81.7 %

(f) BOD removal efficiency = (173 - 20) / 173 = 88.4 %

(g) Sludge age = [(MLSS) (Volume)] / [(SS

Sludge age = [(2,500 mg / L) (2.24 mil.gal) (8.34 lb / mil.gal / mg / L)] / [(22 mg / L) (7.7 mil.gal / day) (8.34 lb / mil.gal / mg / L) + (9,800 mg / L) (0.054 mil.gal / day) (8.34 lb / mil.gal / mg / L)] = 8.0 day

(h) R = Q

Q = (25,000 cap) (120 gpcd) = 3 mgd

Influent BOD load = (0.20 lb / cap.day) (25,000 cap) = 5,000 lb / day

Assume 35% of the BOD

BOD load aeration basin = (0.65) (5,000 lb / day) = 3,250 lb / day

Compute the volume based on the volumetric loading = (3250 lb / day) / [(40 lb) / (1,000 day.ft

Compute the volume based on the aeration period = [(3x10

Use a volume equal to 100,300 ft

(b) Typically the average overflow rate for activated sludge plants with a flow greater than 1 mgd is 800 gpd / ft

Total surface area = (3x10

Surface area per clarifier = 3,750 ft

[(n) (D

D = 48.9 ft

Use a side water depth of 11 ft

Detention time = [(1,875 ft

Volume = (100,000 gal / day) (24 hr) (1 / 24 hr / day) = 13,368 ft

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actiexam.pdf |
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