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Descripción: bakance de masa manjar blanco
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Descripción: determinacion dqo
DQO: Balance de masa
Curso Internacional:
Diseño, operación y mantenimiento de reactores anaerobios de flujo ascendente Mérida, México, 10-13 noviembre 2008
What is Chemical Oxygen Demand (COD) ?? The “theoretical COD” COD” calculation of organic compound
CnHaObNd is based on a complete oxidation. The amount of required O 2 depends on the oxidation state of C:
CnHaObNd Oxidation state:
+a
(2b + 3d - a)/n a)/n
-2b
-3d
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Chemical Oxygen Demand (COD) The number of electrons made free per atom C in the complete oxidation of CnHaObNd amounts to:
4 – (2b+3d-a)/n (2b+3d-a)/n
(or 4n 4n + a - 2b - 3d 3d))
as +4 is the most oxidized form of C (CO 2)
The required number of O2 molecules for the complete oxidation is: CnHaObNd + ¼ (4n+a-2b-3d) (4n+a-2b-3d) O2 → n CO2 + ½(a-3d) H2O + d NH3 1 O2 accepts max. 4 electrons Lettinga Associates Foundation
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Chemical Oxygen Demand (COD) In the standardized COD test with bichromate as oxidizer (150°C) almost all organic compounds
CnHaObNd are completely converted in CO 2 and H2O But: organic N stays reduced and is converted in NH 3 (similar to ‘O’)
The required number of O2 molecules for the complete oxidation is (no NOx produced): CnHaObNd + ¼ (4n+a-2b-3d) O2 → n CO2 + ½(a-3d) H2O + d NH3 Lettinga Associates Foundation
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That is Chemical Oxygen Demand (COD) !! 1 “mol” of organic matter demands: ¼ (4n+a-2b-3d) moles O2 or 8(4n+a-2b-3d) g O2
CnHaObNd + ¼ (4n+a-2b-3d) O2 → n CO2 + ½(a-3d) H2O + d NH3
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What is Total Organic Carbon (TOC) ?? Organic matter measured as CO 2 after incineration (corrections needed for inorganic carbon in waste sample) TOCt = 12n / (12n + a + 16b + 14d) g TOC/gC nHaObNd
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g COD and g TOC per g organic compound (no ‘N’) Ratio COD / TOC: 8(4n+a-2b-3d)/(12n) = 8/3 + 2(a-2b-3d)/(3n) Compound
Calculating the Theoretical Methane Production Theoretical methane production depends on: - the biodegradability of the substrate - the oxidation state of the organic compound Oxidation state of biodegradable compound:
CnHaObNd Oxidation state: +a
2b + 3d - a n
- 2b - 3d
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Theoretical CH4/CO2 production of various organic compounds
100 CH4 0 CO2
S A G O I B E H T F O CH4 N 50 O I CO2 T I S O P M O C
MEAN OXIDATION STATE OF CARBON Lettinga Associates Foundation
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Methane production from C nHaObNd Methane Production from C n HaOb Nd
Basic principles: - part of C will be completely oxidized - the other part of the C will be completely reduced - N and O stay completely reduced - the average oxidation state of C stays the same
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Methane production from C nHaObNd Assumption: - fraction X of C goes to CH4 (oxidation state C = -4 ) - fraction (1-X) of C goes to CO2 (oxidation state C = +4)
Since the oxidation state does not change:
-4X + 4(1-X) =
2b + 3d - a n
x = (a - 2b - 3d)/8n + 4/8
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Methane production from C nHaObNd so CnHaObNd is converted in: n.x.CH4 n. (1-x).CO2 d NH3 substitution of x gives: …...
x = (a - 2b - 3d)/8n + 4/8 Lettinga Associates Foundation
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Methane production from C nHaObNd CnHaObNd
n • X CH4 + n • (1-X) CO2 + d NH 3
(n/2 + a/8 – b/4 – 3d/8) CH4
(n/2 – a/8 + b/4 + 3d/8) CO 2
Buswell’s formula
x = (a - 2b - 3d)/8n + 4/8 Lettinga Associates Foundation
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Buswell’s formula Theoretical CH 4-yield for a compound C nHaObNd follows from: CnHaObNd + (n – a/4 – b/2 + 3d/4) H 2O → (n/2 + a/8 – b/4 – 3d/8) CH 4 + (n/2 – a/8 + b/4 + 3d/8) CO 2 + d NH 3
Actual methane production differs owing to: - Limited biodegradability of compounds - Part of organic matter is used for cell growth (bacterial yield) - Possible presence of alternative electron acceptors - High solubility of CO 2 / HCO3- in the water fraction
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COD ‘consumed’ by alternative electron acceptors Required COD to be calculated from complete reduction reaction 1. COD oxidation with sulphate
COD + SO4
2−
→ H 2 S + CO2
8e + S 6+ → S 2− 2. COD oxidation with sulphite
3. COD oxidation with nitrate Stochiometric calc.:
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COD + SO3 4+
6e + S
2−
→ S
COD + 2 NO3 5+
5e + N
→ H 2 S + CO2 2−
−
→ N 2 + CO2
→ N
0
1 mol SO 42~2 mol O2 1g SO42=> 0.67 g COD 1 g SO32=> 0.6 g COD 1 g NO3 — N => 20/7 g COD = 2.8 6 g COD (1 g NO3 => 0.65 g COD) 14
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Ratio of unionised VFA to total VFA as a function of pH 1.0
+ CH CH33COOH COOH↔ CH33COO COO-++HH+ ↔CH
0.9
At neutral pH:
0.8 0.7
CH3COO- + H2O ↔ CH4 + HCO3-
Propionate
0.6
Butyrate Acetate
0.5 0.4 0.3
At low pH:
0.2
CH3COOH ↔ CH4 + CO2
0.1 0
4
5
6
7
8
9
pH
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Metabolism generated alkalinity Metabolism generated alkalinity from protein degradation: decrease in biogas CO2 CnHaO b Nd + (n -a/4 - b/2 + 3d/4) H2O (n/2 + a/8 - b/4 - 3d/8) CH4 + (n/2 - a/8 + b/4 + 3d/8) CO2 + d NH3
+ H2O ↔ OH- + H+
↓↑ d NH4+ CO2 + H2O ↔ H2CO3 ↔ H+ + HCO3Lettinga Associates Foundation
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Theoretical CH4/CO2 production of various organic compounds Actual production depends on overall wastewater characteristics ! 100 CH4 0 CO2
S A G O I B E H T F O CH4 N 50 O I CO2 T I S O P M O C
Conversion CH4 production - COD The oxidation of CH 4 requires 2 moles of O 2: CH4 + 2 O2 → CO2 + 2H2O x = (a - 2b - 3d)/8n + 4/8 Since “the fraction X” of the compound C nHaObNd will go to CH 4, The COD of compound C nHaObNd is 2 • “fraction X” mol O2/l or: 2 • (n/2 + a/8 – b/4 – 3d/8) mol O 2/l The “overall” oxidation state of organic compounds will not change during anaerobic conversion: A COD balance can be made: COD-in = COD-out Lettinga Associates Foundation
Working with the COD Balance COD equivalents in produced gas: 1 mol CH 4 = 2 mol O2 22.4 l (STP) CH 4 = 64 g O2 or 64 g COD 1 l CH4 (STP) = 64/22.4 = 2.86 g COD Or: 1 g COD = 0.35 l CH 4 (STP) COD equivalents in sludge: 1 g sludge - VSS = 1.42 g COD (based on heterotrophic biomass: C5H7O2N ⇒ 113 g VSS per mol X)
Question: what is the biogas and sludge production in a UASB treating sugar mill wastewater: - Q = 500 m3/day - COD = 3.5 kg/m3 - Sludge Yield = 10% - Efficiency = 90% Lettinga Associates Foundation