Text problems 14–17

14.

r_h = μ_h [ (X_S / X_B) / (K_x + (X_S / X_B)) ] X_B

difference is that r_h does not depend on

( X_S / (K_x + X_S) ) directly although X_S is the substrate

dependence on ratio says that when

X_B large relative to X_S, r_h will be

1st order (X_S-limited)

When X_S large relative to X_B, r_h is zero order

ratio reflects surface interaction between

particulate substrate & biomass (exo-enzymes)

15. processes

ammonification: biomass-N → NH3-N + debris-N

traditional

i_NXB biomass COD → NH3-N + i_NXD debris COD

r_SN4 = (i_NXB - i_NXD f_D) b X_B

lysis-regrowth

r_SN4 = - r_XNB

soluble organic N from X_S

r_XNB = -k_c S_NB X_BH

syntheses

r_SN4 = - μ_NXB X_B

15. nitrification (NH4-N → NO3-N)

e donor only

r_NH = μ X_BA = r_XA

r_XA / X_A = μ_A = μ̂_A (S_NH / (K_NH + S_NH)) (S_O / (K_O + S_O))

denitrification = biomass COD

- S_S / Y_H - (1 - Y_H) S_NO / 2.86(Y_H) = biomass COD

r_SNO / [ -((1 - Y_H) / 2.86 Y_H) ] = r_XBa = r_S / (1 / Y_H)

r_XBa = η_a r_XB = η_a μ_H X_BH

= η_a μ_H (S_S / (K_S + S_S)) (S_NO / (K_NO + S_NO)) (K_DO / (K_DO + S_O))

r_SNO = -((1 - Y_H) / 2.86 Y_H) r_XBa

16. PAO = PO4 removal

anaerobic (no growth): fermentation

- S_S + S_A = 0

r_SS = -r_SA

PHB formed (-S_A-COD + X_PHB-COD = 0)

r_XPHB = -r_SA

(X_PP) Poly-P destroyed, PO4 released (S_P)

r_SP = -r_XPP

energy from PP used to make PHB

- X_PP + Y_P X_PHB-COD = 0

Y_P = g P used / g PHB COD formed

-r_XPP = Y_P r_XPHB

-r_SA = r_XPHB = -r_XPP / Y_P = r_SP / Y_P

-r_SA = ĝ_A (S_A / (K_A + S_A)) (X_PPH / X_BP) / (K_PPH + (X_PPH / X_BP)) X_BP

rate depends on X_PP in PAO’s

PAO’s

max rate of acetate consumption is ref. since PAO’s NOT growing

16. aerobic PP synthesis & growth of PAO’s

growth

-(1 / Y_PAO) X_PHB-COD - ((1 - Y_PAO) / -Y_PAO) O2 + X_PAO-COD = 0

r_XPHB / (1 / Y_PAO) = r_SO / [(-1)(1 - Y_PAO)/Y_PAO] = r_XBP / 1

(growth at expense of X_PHB)

Y_PAO = X_PAO-COD formed / X_PHB-COD consumed

ref is

r_XPAO = μ̂ [ (X_PHB / X_BP) / (K_PHB + X_PHB / X_BP) ] (S_P / (K_P + S_P)) (S_O / (K_O + S_O)) X_BP

storage of P:P stoichiometry: (X_PP is reference)

- r_SP + r_XPP = 0

also P-stoichiometry:

- Y_XPHB-COD / S_B - ( -1 / Y_PHB ) S_O + X_PP = 0

Y_PHB = g P stored / g PHB consumed

5.

r_XPP = -r_SP = -r_XPHB = r_SO / [(-1)(-Y_PHB)]

r_XPP = ĝ_PP [ (X_PPH / X_BP) / (K_PPH + X_PPH / X_BP) ] (S_P / (K_P + S_P)) (S_O / (K_O + S_O)) × …

× (K_PPmax - (X_PP / X_BP)) / (K_PP + (K_PPmax - X_PP / X_BP))

note driving force term

only so much of cell mass can be

PP (K_PPmax ≈ 0.3–0.35 g P/g PHAO)

when X_PP approaches K_PPmax,

PP synthesis slows down

17. a. Ŷ_40 = 0.7 g X-COD / g S-COD

“observed” used considering both lysis & decay

oxygen requirement = (1 - 0.7) g O2 / g S-COD

= 0.3 g O2 / g S_S

Assume that i_NXB = 0.087 g N / g X-COD

Nitrogen requirement = 0.087(0.7) = 0.061 g N / g S_S

Assume phosphorus requirement = 0.2 × n-requirement

P requirement = 0.012 g P / g S_S

K

Ca

Mg

S

Na

Cl

Fe

Zn

Mn

70 mg/gss

70 mg/gss

5 mg/gss

4 mg/gss

2 mg/gss

2 mg/gss

1. mg/gss

0.1 mg/gss

0.05 mg/gss

Cu = 0.01 mg/gss

Mo = 0.003 mg/gss

Co = < 0.0003 mg/gss

Same procedure for b & c

(requirements will be lower

by 18.6% and 49%, resp.)