COD and Nitrogen Stoichiometric and Kinetic Matrix Notes

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Concept

NH4+-N → Cells

NH4+-N → NO3-

H2O

a. O2 only used for nitrification

COD/O2 equivalent = 4.57 g O2 / g NH4+-N

(from NH4+ + 2 O2 → NO3- + H2O + 2H+)

mass O2/d = 0.8(40 - 2) g NH4+-N/m3 (4.57 g O2/g NH4+-N) 70 m3 1440 min/d

≅ 14,000 kg/d

b. cells: C5H7NO2

uHXB = 14 gN / 113 g cells = 0.124 gN/g cells

mass cell formation rate = 0.2(40 - 2) g N/m3 (70 m3) (1440 min/d) 365 / 0.124

= 2,255 MT/yr

(4 points)

2. CH2O + O2 → CO2 + H2O

(3 points) CH2O + 0.8NO3- + 0.8H+ → 0.4N2 + CO2 + 1.4H2O

compare

(1(32)O2 / 1(30)CH2O) / (0.8(14)N / 1(30)CH2O) = 32 g O2 / 11.2 g NO3--N = 2.86 g O2 / g NO3--N

Purposes:

a. Compare stoichiometries for 2 e- acceptors

b. get familiar with writing redox stoichiometry

3. (3 points)

COD-S in

Y = 0.4

growth 1 → cells

assume Y = 0.4

growth 2 → cells

decay → debris

COD-S recycled

a. Yobs = growth1 - decay + growth2

= 0.4 - 0.25(0.4) + 0.4(0.5)(0.25)(0.4)

= 0.4(1 - 0.25 + 0.5(0.25)(0.4)) = 0.32 g cells / g CODS

introduce component non-degradable organic matter

4. a. 0.45 g alls x 1.42 g allOD / g alls = 0.64 g all-OD / g COD

(4 points)

b. reference COD-S

(1) -1 COD-S + (1-Y) COD-O2 + Y COD-cells = 0

CR for O2

(2) -1 COD-S - (1-Y) O2 + Y COD-cells = 0

reference COD-cells

(3) -(1/Y) COD-S - ((1-Y)/Y) O2 + 1 COD-cells = 0

from (2)

r = rs/(-1) = rO/-(1-Y) = rX/Y

rs = -300 g 50,000 m3/d assuming 100% removal

rO2 = -(1 - 0.64)(300)(50000)0 kg/d = 5,400 kg/d

reference rate

5. 1 CH3COO- + 0.03O2 + 0.02NH4+ + 0.92H2O → 0.95CH4 + 0.02C5H7NO2 + 0.78HCO3-

a.) ref CH3COO-, mw = 59

ψ1 = 1, ψ2 = 0.03(44) / 1(59) = 0.022, ψ3 = 0.02(8) / 1(59) = 0.006

ψ4 = 0.92(18) / 1(59) = 0.281, ψ5 = 0.95(16) / 1(59) = 0.258

ψ6 = 0.02(13) / 1(59) = 0.038, ψ7 = 0.78(61) / 1(59) = 1.01

1 CH3COO- + 0.022 CO2 + 0.006 NH4+ + 0.281 H2O → 0.258 CH4 + 0.038 C5H7NO2 + 1.01 HCO3-

1/30? = 1.306 ✓

b. ψ = 1, COD-Ac = 1.085 g COD/g Ac

COD-O2 (acceptor) = -1.45 g COD/g O2

COD-CH4 = 4 g COD/g CH4 + 2 O2 → CO2 + 2 H2

COD-cells = 1.42

COD-everything else = 0

ψ2 = -1.45(0.022) / 1.085 = -0.029

ψ3 = 0.258(14) / 1.085 = 0.95

ψ4 = 0.038(1.42) / 1.085 = 0.05

b) 1 COD-Ac - 0.029 COD-O2 → 0.95 COD-CH4 + 0.05 COD-cells

1 - 0.029 = 0.971, 0.95 + 0.05 = 1.0 (close)

transform rates using matrix

method for multiple phases in one

reactor.

[ matrix diagram ]

multiply 2 vectors

for ith component, n reactions

for ith component / jth component

r_i = Σ(ψ_i,j r_j)

will be used extensively in BioWin

have already used in simple CSTR

rX = (μm - bH) XBH

7/32 = 2.38 g O2/COD / g phenol

a. mass added = 500 g phenol/m3 2.38 g O2/g phenol 1000 m3 10^-? /d

= 1,191.5 kg phenol COD/d

regular ww mass/d = 450 g COD/m3 (25000 m3) 10^-3 kg/g = 11,250 kg/d

~10% increase in COD loading

b. additional O2 = (1 - 0.35) g O2/g CODphenol (1191.5 g CODphenol/d)

= 774.5 kg O2 additional/d

c. toxicity of phenol to WWTP bacteria

discharge of unbiodegraded phenol

volatilization of phenol during aeration

Purpose: be able to assess impact of new

waste/industrial waste on WWTP

CH3COO- + 2O + H+ → 2CO2 + 2H2O

2(32) / 1(44) = 1.45 g O2 / g CO2

5 c. 500 kg-CH3COO- consumed (0.258 g CH4 / g CH3COO-) = 129 kg CH4/d

rAc = rCH4 / -1 = 500 kg Ac/d

V = nRT / P = 129 kmol CH4/d (8.314 kPa·m3/kmol·K) (357.3 K) / 101 kPa

V = 204 m3 CH4/d

d) rVSS / ψ6 = rAc / -1 = 500 kg Ac/d VSS = C5H7NO2

rVSS = 500 kg Ac/d (0.038 kg VSS / kg Ac)

rVSS = 19 kg VSS/d

5. purpose

Practice converting stoichiometries

and related rates

COD AND NITROGEN STOICHIOMETRIC AND KINETIC MATRIX FOR GROWTH AND D...

a. (4 points)

Aerobic Heterotrophic Growth

rH = -μH XBH

= -1.82

-4XBH = -0.08

Aerobic Growth of Autotrophs (Nitrification)

- FA* =

-8.33

Decay and Lysis of Heterotrophs

1-fa = 0.98

μWB = 0.079

[equations handwritten along right side, partly unreadable]

And rates are [3 points]

μH, XBH = growth rate for heterotrophs (d-1)

μA, XBA = growth rate for autotrophs (d-1)

bH, XBH = decay rate for heterotrophs (d-1)

* could use all active biomass in the?

r = -1/4

= 0.171? something unclear

Concept

NH4+-N → Cells

NH4+-N → NO3-

H2O

a. O2 only used for nitrification

COD/O2 equivalent = 4.57 g O2 / g NH4+-N

(from NH4+ + 2 O2 → NO3- + H2O + 2H+)

mass O2/d = 0.8(40 - 2) g NH4+-N/m3 (4.57 g O2/g NH4+-N) 70 m3 1440 min/d

≅ 14,000 kg/d

b. cells: C5H7NO2

uHXB = 14 gN / 113 g cells = 0.124 gN/g cells

mass cell formation rate = 0.2(40 - 2) g N/m3 (70 m3) (1440 min/d) 365 / 0.124

= 2,255 MT/yr

(4 points)

2. CH2O + O2 → CO2 + H2O

(3 points) CH2O + 0.8NO3- + 0.8H+ → 0.4N2 + CO2 + 1.4H2O

compare

(1(32)O2 / 1(30)CH2O) / (0.8(14)N / 1(30)CH2O) = 32 g O2 / 11.2 g NO3--N = 2.86 g O2 / g NO3--N

Purposes:

a. Compare stoichiometries for 2 e- acceptors

b. get familiar with writing redox stoichiometry

3. (3 points)

COD-S in

Y = 0.4

growth 1 → cells

assume Y = 0.4

growth 2 → cells

decay → debris

COD-S recycled

a. Yobs = growth1 - decay + growth2

= 0.4 - 0.25(0.4) + 0.4(0.5)(0.25)(0.4)

= 0.4(1 - 0.25 + 0.5(0.25)(0.4)) = 0.32 g cells / g CODS

introduce component non-degradable organic matter

4. a. 0.45 g alls x 1.42 g allOD / g alls = 0.64 g all-OD / g COD

(4 points)

b. reference COD-S

(1) -1 COD-S + (1-Y) COD-O2 + Y COD-cells = 0

CR for O2

(2) -1 COD-S - (1-Y) O2 + Y COD-cells = 0

reference COD-cells

(3) -(1/Y) COD-S - ((1-Y)/Y) O2 + 1 COD-cells = 0

from (2)

r = rs/(-1) = rO/-(1-Y) = rX/Y

rs = -300 g 50,000 m3/d assuming 100% removal

rO2 = -(1 - 0.64)(300)(50000)0 kg/d = 5,400 kg/d

reference rate

5. 1 CH3COO- + 0.03O2 + 0.02NH4+ + 0.92H2O → 0.95CH4 + 0.02C5H7NO2 + 0.78HCO3-

a.) ref CH3COO-, mw = 59

ψ1 = 1, ψ2 = 0.03(44) / 1(59) = 0.022, ψ3 = 0.02(8) / 1(59) = 0.006

ψ4 = 0.92(18) / 1(59) = 0.281, ψ5 = 0.95(16) / 1(59) = 0.258

ψ6 = 0.02(13) / 1(59) = 0.038, ψ7 = 0.78(61) / 1(59) = 1.01

1 CH3COO- + 0.022 CO2 + 0.006 NH4+ + 0.281 H2O → 0.258 CH4 + 0.038 C5H7NO2 + 1.01 HCO3-

1/30? = 1.306 ✓

b. ψ = 1, COD-Ac = 1.085 g COD/g Ac

COD-O2 (acceptor) = -1.45 g COD/g O2

COD-CH4 = 4 g COD/g CH4 + 2 O2 → CO2 + 2 H2

COD-cells = 1.42

COD-everything else = 0

ψ2 = -1.45(0.022) / 1.085 = -0.029

ψ3 = 0.258(14) / 1.085 = 0.95

ψ4 = 0.038(1.42) / 1.085 = 0.05

b) 1 COD-Ac - 0.029 COD-O2 → 0.95 COD-CH4 + 0.05 COD-cells

1 - 0.029 = 0.971, 0.95 + 0.05 = 1.0 (close)

transform rates using matrix

method for multiple phases in one

reactor.

[ matrix diagram ]

multiply 2 vectors

for ith component, n reactions

for ith component / jth component

r_i = Σ(ψ_i,j r_j)

will be used extensively in BioWin

have already used in simple CSTR

rX = (μm - bH) XBH

7/32 = 2.38 g O2/COD / g phenol

a. mass added = 500 g phenol/m3 2.38 g O2/g phenol 1000 m3 10^-? /d

= 1,191.5 kg phenol COD/d

regular ww mass/d = 450 g COD/m3 (25000 m3) 10^-3 kg/g = 11,250 kg/d

~10% increase in COD loading

b. additional O2 = (1 - 0.35) g O2/g CODphenol (1191.5 g CODphenol/d)

= 774.5 kg O2 additional/d

c. toxicity of phenol to WWTP bacteria

discharge of unbiodegraded phenol

volatilization of phenol during aeration

Purpose: be able to assess impact of new

waste/industrial waste on WWTP

CH3COO- + 2O + H+ → 2CO2 + 2H2O

2(32) / 1(44) = 1.45 g O2 / g CO2

5 c. 500 kg-CH3COO- consumed (0.258 g CH4 / g CH3COO-) = 129 kg CH4/d

rAc = rCH4 / -1 = 500 kg Ac/d

V = nRT / P = 129 kmol CH4/d (8.314 kPa·m3/kmol·K) (357.3 K) / 101 kPa

V = 204 m3 CH4/d

d) rVSS / ψ6 = rAc / -1 = 500 kg Ac/d VSS = C5H7NO2

rVSS = 500 kg Ac/d (0.038 kg VSS / kg Ac)

rVSS = 19 kg VSS/d

5. purpose

Practice converting stoichiometries

and related rates

COD AND NITROGEN STOICHIOMETRIC AND KINETIC MATRIX FOR GROWTH AND D...

a. (4 points)

Aerobic Heterotrophic Growth

rH = -μH XBH

= -1.82

-4XBH = -0.08

Aerobic Growth of Autotrophs (Nitrification)

- FA* =

-8.33

Decay and Lysis of Heterotrophs

1-fa = 0.98

μWB = 0.079

[equations handwritten along right side, partly unreadable]

And rates are [3 points]

μH, XBH = growth rate for heterotrophs (d-1)

μA, XBA = growth rate for autotrophs (d-1)

bH, XBH = decay rate for heterotrophs (d-1)

* could use all active biomass in the?

r = -1/4

= 0.171? something unclear

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