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1/θ = μ = μ̂ Ss / (Ks + Ss)

Ks + Ss = μ̂ Ss

Ss(μ̂θ - 1) = Ks

Ss = Ks / (μ̂θ - 1)

@ Cmin, Ss → Ss0

Ss0 = Ks / (μ̂ Cmin - 1)

μ̂ Cmin - 1 = Ks / Ss0

Cmin = (1/μ̂)(Ks / Ss0 + 1) = (Ks + Ss0) / (μ̂ Ss0) = (50 + 250) / (2(250))

a) Cmin = 0.6 days @ washout

b) Ss = 0.1 Ss0 = 25 mg/L

τ = (Ks + Ss) / (μ̂ Ss) = (50 + 25) / (2(25)) = 1.5 d

V = Qτ = 10^4 m^3/d (1.5 d) = 15,000 m^3

X = Y(Ss0 - Ss) = (250 - 25)0.6 = 135 mgCOD/L

And X = 135 mgCOD/L 1 cells / 1.42 gCOD = 95 mg cells/L

d. PFR

dS/dτ = rS = -μ̂X(Ss / (Ks + Ss))

∫ from Ss0 to Ss (Ks/Ss + 1) dS = -μ̂X/Y ∫ from 0 to τ dτ

Ks ln Ss + Ss | from Ss0 to Ss = -μ̂Xτ/Y

Ks ln(Ss/Ss0) + Ss - Ss0 = -μ̂Xτ/Y

τ = Y/(μ̂X) [Ks ln(Ss0/Ss) + (Ss0 - Ss)]

0 = Y/(μ̂X) ( ) - 0

Effluent COD vs τ for ideal CSTR and PFR with Monod kinetics

Problem 1. PFR with Monod kinetics compared with CSTR

τ (d^-1)

0 CSTR PFR

0.2 250 4.6687E-05

0.4 250 6.6925E-06

0.6 250 0.00014206

0.8 83.3 0.000599214

1 50.0 0.0024484

1.02 48.1 0.00881E-05

1.1 41.7 8.0179E-05

1.2 35.7 0.00005643

1.4 27.8 0.0000243

1.5 25.0 0.00000571

1.6 22.7 0.00005136

1.8 19.2 -0.00002967

2 16.7 -

dτ for 90% COD removal in ideal PFR = 1.02d

VPFR = 10,000 m3/d x 1.02d = 10,200 m3

VCSTR for 90% removal = 15,000 m3, 50% larger than PFR

1 = (1/4) ln(S/S0)

Comparison of t for 85 and 99% COD removal

1st-order rate, k = 5 d^-1

0.99 1.13 0.20 0.49 PFR

0.85 1980 3.60 1.73 0.92

CSTR and cascade

n = 1, 2, and 4

CSTR and cascade

n = (1/k)(S0/S) / (1/n) - 1

d. 1 mb on COD, S

1 = μ - b = μ̂Ss/(Ks + Ss) - b

μ̂Ss/(Ks + Ss) = 1/θ + b = (1 + bθ)/θ

θμ̂Ss = (Ks + Ss)(1 + bθ)

Ss(θμ̂ - (1 + bθ)) = Ks(1 + bθ)

Ss = Ks(1 + bθ) / (θμ̂ - (1 + bθ)) (1)

"mb on COD, S"

QSo - (Q - Qw)Ss - QwSs + VX = 0

Q(So - Ss) + VX = 0

-Ss = (So - Ss)/θ = μX/Y

X = Y(So - Ss) / (μτ)

X = Y(So - Ss) / (τ(1/θ + b))

X = (θ/τ) Y(So - Ss) / (1 + bθ)

θμ - (1 + bθ) = Ks(1 + bθ)/Ss

θμ - bθ - (Ks/Ss)bθ = Ks + 1? [illegible]

θ = (Ks + Ss) / [Ss(μ̂ - b(1 + Ks/Ss))]

θ(Ss = 25) = (50 + 25) / (25(2 - 0.1((50 + 25)/25)))

θ = 1.76 d

V = Qwθ = 300 m3/d (1.76 d) = 528 m3

τ = 528 m3 / 10^4 m3/d = 0.053 d or 1.3 hours

X = 0.6(250 - 25) (1.76 / 0.053) = 3800 mgCOD/L

= 2700 mg VSS/L

V recycle = 0.035 / V no recycle = 28

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