Solucin a Problemas Sugeridos Wastewater Characteristics (P.61) 1. BODu = 292.7 mg/L k(30C) = k(20C)1.047(30-20) = 0.36/d BOD5(30C) = 245.3 mg/L 2. Graficando (t/yt)1/3 vs t = 0.0198 = 0.2007 k = 6B/A = 0.592/d L0 = 1/(kA3) = 209 mg/L 4. y5 = 160 mg/L Primary Treatment (P.69-70) 1. Q = 10.61 ft3/s

2. At effluent: BOD5 = 149.5 mg/L & SS = 122.5 mg/L

For Q = 50,000 m3/d: BOD5 removed = 4,025 kg/d & SS removed = 11,375 kg/d

7. Q = 11.3 ft3/s Activated Sludge Process (P.78-79) 1. V = 5943 m3

; Presumiendo un BOD5 a la salida (efluente) = 10 mg/L: Nota: Ignoren la ltima oracin del problema, ya que es redundante el cmputo que se les pide hacer.

( ) tLkkL

yt

t

+=

31

0

3231

0

31

6

= 31

0

32

6LkB ( ) 310 = kLA

dtdX

Xc = XkV

SSQYdtdX

d

=)( 0

dc 26.12=

2. Assuming 80% of biological solids: Sludge production = 1,515 kg/d O2 demand = 5,335 kg/d Air = 254,564 m3/d 4. V = 1,857 m3 12. O2 demand = 697 kg/d 13. Xr = 10,000 mg/L Qr = 2,560 m3/d Qw = 80 m3/d Trickling Filters (P.91) 1. Overall efficiency = 80.5%

2. a. BOD load = 463.7 g/d-m3

b. hyd load = 17.8 m3/d-m2 c. using NRC formula, filter efficiency = 82%

4. A = 815 m2

Oxidation Ponds 2. Vtotal = 9,065 m3

Atotal = 6,043 m2 Area 1st pond = 3,022 m2 Area 2nd and 3rd ponds = 1511 m2