wet air oxidation for sampling - kexhu.people.ust.hkkexhu.people.ust.hk/ceng576/576-03.pdf · wet...

CENG 5760 Advanced Physico-Chemical Treatment Processes Professor Xijun Hu

88

Wet Air Oxidation Wet air oxidation (WAO) is a well-established technique for wastewater treatment particularly toxic and high concentration organic wastewater. WAO involves the liquid phase oxidation of organics or oxidizable inorganic components at elevated temperatures (125-320 oC) and pressures (0.5 – 20 MPa) using a gaseous source of oxygen (usually air). Enhanced solubility of oxygen in aqueous solutions at elevated temperature and pressure provides a strong driving force for oxidation. The elevated pressures are required to keep water in the liquid state. Water also acts as a moderant by providing a medium for heat transfer and removing excess heat by evaporation. In WAO Carbon CO2 H H2O N NH3, NO3 or N2 Halogen and sulfur inorganic halides and

sulfates The degree of oxidation depends on Temperature Oxygen partial pressure Residence time Oxidizability of the pollutants


89

for sampling

Schematic diagram of a batch wet air oxidation reactor The operating costs are almost entirely for power to compress air and high pressure liquid pumping. WAO becomes self-sustaining with no auxiliary fuel requirement when the COD (chemical oxygen demand) is above 20,000 mg/L. Incineration (combustion) becomes self-sustaining when the COD is in the range of 300,000 – 400,000 mg/L. Adding a catalyst can achieve the same or better oxidation efficiency at lower reaction temperatures and pressures so reducing the operation cost. When a catalyst is used, the process is called catalytic wet air oxidation (CWAO).


90

Schematic diagram of a continuous WAO reactor. In most applications, WAO is not used as a complete treatment method, but only as a pretreatment step where the wastewater is rendered nontoxic and the COD is reduced sufficiently, so that biological treatment becomes applicable for the final treatment. For industrial wastewater treatment, COD or TOC (total organic carbon) is often used to characterize the wastewater and to test the efficiency of the WAO process.


91

Measurement of Organic Content determines the approximate quantity of oxygen required to biologically stabilize the organic matter determines the size of waste treatment facilities measures the efficiency of some treatment processes determines compliance with wastewater discharge limits Chemical Oxygen Demand (COD) The oxygen equivalent of the organic matter that can be oxidized is measured by using a strong chemical oxidizing agent in an acidic medium. Potassium dichromate is excellent for this purpose. Test is performed at elevated temperature Catalyst (silver sulfate) is required in some cases The principle reaction is

Organic matter (C

aH O Cr O Hcatalyst

heatCr CO H O

b c )

2 72

32 2

Some inorganic compounds may interfere with the test


92

Total Organic Carbon (TOC) A known quantity of sample is injected into a high temperature furnace or chemically- oxidizing environment The organic carbon is oxidized to carbon dioxide in the presence of a catalyst The carbon dioxide produced is measured by an infrared analyser The test is very quick so it becomes popular Theoretical Oxygen Demand (ThOD) Organic compounds in wastewater is a combination of carbon, hydrogen, oxygen, and nitrogen The ThOD can be computed if the chemical formula of the organic matter is known Its usage is limited because the organic matter in wastewater is usually a mixture of many unknown substances


93

Example: Determine the ThOD for glycine (CH NH COOH2 2( ) ) using the following assumptions: 1. In the first step, the organic carbon and nitrogen are converted to carbon dioxide and ammonia. 2. In the second and third steps, the ammonia is oxidized sequentially to nitrite and nitrate. 3. The ThOD is the sum of the oxygen required for all three steps. Solution 1. Write the balanced reaction for the carbonaceous oxygen demand:

CH NH COOH O NH CO H O2 2 2 3 2 232

2( )

2. Write the balanced reaction for the nitrogenous oxygen demand: (a) NH O HNO H O3 2 2 2

32

(b) HNO O HNO2 2 312

------------------------------------- NH O HNO H O3 2 3 22 3. Determine the ThOD: ThOD = (3/2+2) mol O2/mol glycine = 3.5 mol O2/mol glycine 32 g/mol O2 = 112 g O2/mol glycine


94

Biochemical oxygen demand (BOD) measurement of the dissolved oxygen used by microorganisms in the biochemical oxidation of organic matter 5-day or 10-day BOD (BOD5 or BOD10) is most used Determination of BOD dilution to cover different ranges of BOD “seeding” with a bacterial culture (saprophytic and other micro-organisms and some autotrophic bacteria) that has acclimated to the organic matter or other materials in the wastewater incubation period of five days at the constant temperature of 20 oC measure dissolved oxygen before and after incubation Non-seeded BOD (mg / l) = D

P1 D2

Seeded BOD (mg / l) = D - B

P1 1 D B f2 2

D1 = dissolved oxygen of diluted sample immediately after preparation, mg/l D2 = dissolved oxygen of diluted sample after 5 days incubation at 20 oC, mg/l P = decimal volumetric fraction of sample used B1 = dissolved oxygen of seed control before incubation, mg/l


95

B2 = dissolved oxygen of seed control after incubation, mg/l f = ratio of seed in sample to seed in control = (% seed in D1)/ (% seed in B1)


96


97

Example: A wastewater sample is diluted by a factor of 10 using seeded dilution water. If the following results are obtained, determine the 5-day and 6-day BOD. The ratio of seed in sample to seed in control, f = 1. Dissolved oxygen, mg/L Time (day) Diluted sample Seeded sample 0 1 2 3 4 5 6

8.55 4.35 4.02 3.35 2.75 2.40 2.10

8.75 8.70 8.66 8.61 8.57 8.53 8.49

1,1.0101

fP

5-day BOD:

)/(3.591.0

)1)(53.875.8()4.255.8(

)()()/(

5

2121

LmgBOD

PfBBDDLmgBOD

6-day BOD:

)/(9.611.0

)1)(49.875.8()10.255.8(

)()()/(

6

2121

LmgBOD

PfBBDDLmgBOD


98

The kinetics of the BOD reaction can be formulated with first-order reaction kinetics, for practical purposes:

tt kL

dtdL

where Lt is the amount of the first-stage BOD remaining in the water at time t and k is the reaction rate constant. This equation can be integrated as

ln L kttt0 or L

Let kt

where L or BODL is the BOD remaining at time t=0 (i.e., the total or ultimate first-stage BOD initially present). The amount of BOD remaining at time t is

L L etkt ( )

and y, the amount of BOD that has been exerted at any time t, is

y L L L et tkt ( )1

Note that the 5-day BOD equals y L L L e k

5 551 ( )

For polluted water and wastewater, k (base e) is around 0.2 (0.05 - 0.3) day-1. Example: Determine the 1-day BOD and ultimate first-stage BOD for a wastewater whose 5-day, 20 oC


99

BOD is 200 mg/L. The reaction constant k (base e) = 0.23 day-1. Solution: 1. Determine ultimate BOD

L L etkt ( )

)1( 555

keLLLy

L=293 mg/L 2. Determine 1-day BOD

)1(11keLLLy

= 293(1-e-0.23) = 60 mg/L

23.051200 eL


100

3. Nitrogenous Biochemical Oxygen Demand (NBOD) - the oxygen demand associated with the oxidation of ammonia to nitrate. Carbonaceous Biochemical Oxygen Demand (CBOD) - suppressed BOD. Elimination of the interference of nitrifying bacteria by pretreatment or by the use of inhibitory agents. Limitation in the BOD Test A high concentration of active, acclimated seed bacteria is required Pretreatment is needed when dealing with toxic waste, and the effects of nitrifying organisms must be reduced Only biodegradable organics are measured An arbitrary, long period of time is required to obtain results Inhibition and Toxicity An organic substance that is biodegradable at one concentration can become persistent at higher concentrations by inhibiting the growth of the microbial culture. At even higher concentrations, the substance can become toxic to the culture.


101

WAO of cotton desizing wastewater at 290 oC.

0 20 40 60 80 100 120 140 16

CO

D R

educ

tion

(%

0

10

20

30

40

50

60

Reaction Time (min)0 20 40 60 80 100 120 140 16

TOC

Rem

oval

(%)

0

10

20

30

40

50

60

70

80

without O2

with O2

without O2

with O2

The organics in wastewater are stable to heating but oxidizable by oxygen


102

Effect of reaction temperature on the WAO of cotton desizing wastewater at 1.5 MPa partial oxygen pressure.

0.5 MPa1 MPa1.5 MPa2 MPa3 MPa

PO2


PO2

0 50 100 1

CO

D R

educ

tion

(%)

0

10

20

30

40

50

60

70

Reaction time (min)0 50 100 1

TOC

Rem

oval

(%)

0

10

20

30

40

50

60

70

80

150 oC200 oC240 oC270 oC290 oC

150 oC200 oC240 oC270 oC290 oC

WAO is better at a higher temperature Near 80% COD and TOC removals at 290°C


103

Effect of reaction pressure on the WAO of cotton desizing wastewater at 240oC.

0 50 100 15

CO

D R

educ

tion

(%

0

10

20

30

40

50

60


TOC

Rem

oval

(%)

0

10

20

30

40

50

60

70

0.375 MPa0.75 MPa1.125 MPa1.5 MPa2.25 MPa

PO2


PO2

WAO is better at a higher oxygen partial pressure


104

WAO of chemical fibre desizing wastewater at 2 MPa partial oxygen pressure at various temperature.

0 2 0 4 0 6 0 8 0 1 0 0 1 2

CO

D R

educ

tion

(%)

01 02 03 04 05 06 07 08 09 0

R e a c tio n tim e (m in )

0 2 0 4 0 6 0 8 0 1 0 0 1 2

TOC

Rem

oval

(%)

01 02 03 04 05 06 07 08 0

1 5 0 o C2 0 0 o C2 4 0 o C2 7 0 o C

1 5 0 o C2 0 0 o C2 4 0 o C2 7 0 o C

0 2 0 4 0 6 0 8 0 1 0 0 1 2

Bio

degr

adab

ility

(%)

01 02 03 04 05 06 07 08 09 0

1 5 0 o C2 0 0 o C2 4 0 o C2 7 0 o C

WAO is better at a higher temperature 90% COD & 80% TOC removals at 270° Biodegradability = BOD/COD BOD = biochemical oxygen demand


105

Possible Reaction Kinetics COD as reactant (C) Reaction mechanisms:

Wastewater CO & H Ok

Intermediate organicproducts (COD)

(COD)k

fast2 2

slowkfast

Rate data modeled by first order kinetics

dCdt

kC

where t is reaction time, and k is the specific reaction rate constant which has the following temperature dependency: RTEkk /exp0 where k0 is a pre-exponential factor, E is the activation energy, R is the universal gas constant and T is the temperature in Kelvin. Integration gives

ktC

C

0ln

where C0 is the initial COD value. By plotting ln(C0/C) versus time, the slope is the specific reaction rate constant k. A typical plot of the WAO treatment of cotton desizing wastewater at a fixed partial oxygen pressure of 1.5 MPa and four


106

different reaction temperatures is shown in the following figure. The data fit well into two straight lines for a given temperature, indicating that oxidation proceeds in two distinct steps: a fast initial reaction of large molecules decomposed into intermediate products, followed by a slow reaction of further oxidizing the intermediate products into end products of low molecular weight organic acids, carbon dioxide, and water.

Reaction time (min)0 20 40 60 80 100 120 140 160

LnC

0/C

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

200 oC240 oC270 oC290 oC

WAO of cotton desizing wastewater at 1.5 MPa partial oxygen pressure (theoretical oxygen requirement).


107

The specific rate constant k is a function of temperature:

k k ERT

0 exp or ln lnk k ERT0

1/T (K-1)00.0018 00.0020 00.0022

- ln(

K)

3

4

5

6

7

KfastKslow

Effect of temperature on rate constants of cotton desizing wastewater at 1.5 MPa partial oxygen pressure.

The activation energies are: Efast = 30 kJ/mol; Eslow = 9 kJ/mol


108

If oxygen is not in excess, then k k POn

'2

Reaction time (min)0 20 40 60 80 100 120 140 160

InC

0/C

0.0

0.2

0.4

0.6

0.8

1.0

1.2


PO2

WAO of cotton desizing wastewater at 240 oC.


109

Oxygen partial pressure (MPa)0 1 2 3

K

0.000

0.005

0.010

0.015

Kfast

Kslow

Effect of oxygen concentration on rate constants of cotton desizing wastewater at 240 oC. The slow reaction is independent of oxygen partial pressure. The fast reaction strongly depends on the oxygen supply when it is less than the theoretical oxygen requirement (1.5 MPa), with excess oxygen, even the fast reaction becomes independent of oxygen partial pressure.


110

For most WAO operations, the reaction is assumed

to consist of two steps: the decomposition of large

molecules into intermediate products and the further

oxidation of the intermediates into the end products

of carbon dioxide and water. If starch is assumed

the major content of the wastewater, which can be

hydrolyzed into glucose at first, and glucose is

oxidized into carbon dioxide and water thereafter.

Furthermore, it is assumed that a portion of the

organic compound is very difficult to be oxidized.

Therefore, the following reaction routes were

assumed as:

S K1 G k2 CO2 + H2O

k3

N

Where S is the substrate organic (starch) of

wastewater, G is glucose, and N is a non-oxidizable

organic. Reactions 1 and 3 do not change the COD

or TOC value of the solution.


111

To simplify the analysis, it is further assumed that

the reactions are kinetics controlled and the

dissolved oxygen concentration is a constant since

enough oxygen gas is supplied. The conversion

between the substrate organic and glucose is a fast

reversible reaction and reaches equilibrium very

quickly, represented by the equilibrium constant K1.

Reactions 2 and 3 are assumed to follow first order

kinetics. Let the total organic in the solution during

reaction be X, its removal rate is then

G k = dtdX 2 (1)

where [G] stands for the concentration of glucose.

There exists equilibrium between the substrate and

glucose concentrations:

[G]=K1 [S] (2)


112

where [S] is the concentration of starch. By

substituting Equation (2) into Equation (1), one

obtains

S k K = dtdX 21 (3)

The total organic in the wastewater comprises S, G

and N:

X = [G] + [S] + [N] (4)

where [N] represents the concentration of non-

oxidizable product. Elimination of [G] by

substituting Equation (2) into Equation (4), we can

get

[N]XK+11 = [S]

1 (5)

Thus, Equation (3) becomes

[N] - X K+1kK =

dtdX

1

21 (6)

Meanwhile

S k = dt

d[N]3 (7)


113

Combining Equation (5) with Equation (7) we obtain

[N] - X K+1

k = dt

d[N] 1

3 (8)

Dividing Equation (8) by Equation (6) yields

k K

k =

dXd[N]

21

3 (9)

with the initial condition

t=0 [N]=0 (10)

The solution for Equations (9) and (10) is

X) - (X k K

k = [N] 0

21

3 (11)

Now we substitute Equation (11) into Equation (6)

to give

X) - (X

k Kk

-X K+1k K

= dtdX 0

21

3

1

21 (12)

with the initial condition

t=0 X = X0 (13)


114

where X0 is the total organic concentration in the

wastewater at time zero.

The solution for Equations (12) and (13) is

e k + k K

kK + k + k K

k = XX t

K + 1k + k K

-

321

21

321

3

0

1

321

(14)

If we assume the TOC value in the solution is

proportional to the total organic concentration in the

wastewater, X, i.e.

00 X

X

TOC

TOC (15)

then the removal of TOC, TOC, would become

e k + k K

kK + k + k K

k TOCT - 1 =

TOCTOC

TOCTOC1

TOCTOC1

t K + 1k + k K

-

321

21

321

3

i

0

0i

0

iTOC

1321

OC

(16)


115

where TOCi is the initial TOC value of the fresh

wastewater, which is different from the TOC value

at the reaction time t=0, TOC0, by a factor due to the

thermal decomposition.

Equation (16) is applied to simulate the WAO

treatment of natural fiber desizing wastewater at

different temperatures.


PO2


TOC

Rem

oval

(%)

0

10

20

30

40

50

60

70

80

150 oC

200 oC

240 oC270 oC

290 oC

The model (lines) is in good agreement with the

experimental data. The kinetic parameters are


116

optimized from the experimental data by a least

square method, and listed in the following table.

Kinetic parameters of WAO of cotton desizing

wastewater at different temperatures

T ( oC) 150 200 240 270 290

K1 0.0231 0.0667 0.0758 0.0807 0.161

k2

(min-1)

0.0428 0.131 0.171 0.329 0.574

k3

(min-1)

2.411x10-4 9.8510-3 8.2110-3 0.0154 0.0428

From the small value of hydrolization equilibrium

constant, K1, at 150oC, it can be seen that starch does

not hydrolyze easily at low temperature. Once the

temperature is above 200oC, however, the effect of

reaction temperature on the rate of hydrolysis

becomes less significant. The equilibrium constant is

within the range of 0.067 to 0.08 for temperatures of


117

200 to 270oC. The temperature dependence of k2

and k3 are assumed to follow the arrhenius form:

)RTE-

exp(k=k a0 (17)

where k0 is the pre-exponential factor, Ea is the

activation energy, and R is the gas constant.

1/T (K-1)

0.0018 0.0020 0.0022 0.0024

ln(k

)

-9

-8

-7

-6

-5

-4

-3

-2

-1

0

ln(k2)

ln(k3)

k2 and k3 are with the following equations


118

T

4106 - 1.87=)ln(k2 (18)

T

7728 - 2.35=)ln(k3 (19)

The activation energy for the oxidation of glucose,

34.1 kJ/mol, is much larger than 25 kJ/mol, a value

where mass transfer resistance can be ignored.

Therefore, the reactions here are indeed kinetics

controlled. On the other hand, the value of

activation energy obtained here is smaller than the

value reported in the literature for the oxidation of

glucose. This means that the fast-formed

intermediates of this kind of wastewater are easier to

oxidize than the pure glucose. The activation energy

for the conversion of the original organic to the non-

oxidizable product is 64.2 kJ/mol, much larger than

that for the oxidation of glucose. This implies that

oxidation is the major reaction.


119

Other possible WAO reaction mechanisms During WAO, the long molecules are oxidized to various intermediates products. Most of the initial intermediates formed (except the low molecular weight carboxylic acids) are unstable and further oxidized to end products (CO2, etc.) or to low molecular carboxylic acids (mainly acetic acid). The low molecular carboxylic acids are resistant to further oxidation. Thus, the organics in the effluent from a WAO system can be divided into three groups: A: all initial & relatively unstable intermediates B: refractory intermediates like acetic acid C: oxidation end products A + O2 ----k1------- C （CO2 + H2O） k2 k3 B + O2 Assume oxygen is in excess, we may have

AAA CkCk

dtdC

21

BAB CkCk

dtdC

32

12

1 /0,11

nO

RTE Cekk


120

22

2 /0,22

nO

RTE Cekk

32

3 /0,33

nO

RTE Cekk

tkkAA eCC )(

0,21

][ )(0,

321

2

0,

213

3

tkktkA

tkBB

eeCkkk

k

eCC

CB,0 can be assumed to be zero:

e k - k + k

k - k +

e k - k + k

k = C

C+C

tk + k -

321

31

k-

321

2

0,A,0

BA

21

3t

BC

The COD or TOC in wastewater should be CA + CB, so

e k - k + k

k - k + e k - k + k

k TOCTOC tk + k -

321

31 tk -

321

2

0

213


121

Improved WAO The efficiency of WAO can be improved by various means, such as adding a catalyst or using a stronger oxidant. Catalytic wet air oxidation (CWAO) The catalyst used may be metal salt solution, metal oxide powders, or porous solid supported metals. By using metal ion solutions and metal oxide powders as catalysts in the treatment of wastewater, The benefits: Higher COD and TOC removals Lower reaction temperature and total pressure The disadvantages: Cause secondary pollutants The solution: Immobilize metals onto granular porous solids Used catalysts can be recovered by filtration


122

0 20 40 60 80 100 120

CO

D R

emov

al (%

)

0

20

40

60

80

Reaction time (min)0 20 40 60 80 100 120

TOC

Rem

oval

(%)

0

10

20

30

40

50

60

Cu(NO3)2

FeSO4

Mn(NO3)2

CuSO4

No Catalyst

Cu(NO3)2

FeSO4

Mn(NO3)2

CuSO4

No Catalyst

Effect of catalysts on the CWAO of dyeing and printing wastewater at 200oC, pO2

=2.65 MPa. Use of catalysts greatly improves the oxidation The effectiveness of catalysts is Cu(NO3)2 > CuSO4 > Mn(NO)2 > FeSO4


123

PtO2

PtO2

0 20 40 60 80 100 120

CO

D R

emov

al (%

)

0

20

40

60

80

CuOFe2O3

MnO2

PtO2

TiO2

No catalyst


TOC

Rem

oval

(%)

0

20

40

60

CuOFe2O3

MnO2

PtO2

TiO2

No catalyst

Effect of metal oxide catalysts on the CWAO of dyeing and printing wastewater at 200oC, pO2

=2.65 MPa. Use of catalysts greatly improves the oxidation The efficiency of catalysts is CuO > Fe2O3 > TiO2 > MnO2 > PtO2


124

0 20 40 60 80 100 120

TOC

Rem

oval

(%)

0

20

40

60

0 20 40 60 80 100 120

CO

D R

emov

al (%

)

0

20

40

60

80


Col

or R

emov

al (%

)

0

20

40

60

80

Cu-Al2O3Cu(NO3)2

No catalyst

CuO

Cu-Al2O3Cu(NO3)2

No catalyst

CuO

Cu-Al2O3Cu(NO3)2

No catalyst

CuO

Effect of various copper catalysts on the CWAO of dyeing and printing wastewater at 200oC, pO2

=2.65 MPa.


125

Addition of H2O2 as promoter

0 20 40 60 80 100 120

TOC

Oxi

datio

n R

emov

al (%

)

0

15

30

45

Reaction Time (min)0 20 40 60 80 100 120C

olor

Oxi

datio

n R

emov

al (%

)

0

20

40

60

80WAOCWAOPCWAO

WAO of dyeing wastewater at 200oC, pO2

=2.65 MPa. CWAO & PCWAO: Cu/AC (copper supported on

activated carbon) catalyst was used. PCWAO: 10% H2O2 of the theoretical oxidation requirement was added in additional to oxygen.


126

Wet Peroxide Oxidation (WPO) Completely replace oxygen by H2O2.

0 15 30 45 60 75 90 105 120 135 150

TOC

Oxi

datio

n R

emov

al (%

)

0

20

40

60

80

Reaction Time (min)0 15 30 45 60 75 90 105 120 135 150C

olou

r Oxi

datio

n R

emov

al (%

)

0

20

40

60

80

100

70oC110oC130oC150oC

Reaction is very fast High TOC & color removal at 130oC H2O2 is expensive


127

0 20 40 60 80 100 120

TOC

Oxi

datio

n R

emov

al (%

)

0

20

40

60

80

Reaction Time (min)0 20 40 60 80 100 120

Col

our O

xida

tion

Rem

oval

(%)

0

20

40

60

80

100

50%Qth

100%Qth

200%Qth

Effect of hydrogen peroxide dosage on WPO of dyeing wastewater concentrate. Increasing H2O2 dosage accelerates the TOC

reduction when it is below its theoretical amount.

However, when the H2O2 dosage is above the

theoretical requirements it little affects the final TOC


128

and color reductions, although the initial reaction

rate increases as the H2O2 dosage increases. This

indicates the maximum final TOC removal

efficiency can not be improved by increasing the

H2O2 dosage. The reason for this might be that the

excess H2O2 reacts with the hydroxyl radical to form

water and HO2 radical which will further react with

H2O2 to form water and hydroxyl radical. Therefore,

H2O2 is self-consumed.

OHOOHOHHO

HOOHOHOH

22222

2222


129

Catalytic Wet Peroxide Oxidation (CWPO)

0 20 40 60 80 100 120

TOC

Oxi

datio

n R

emov

al (%

)

0

10

20

30

40

50

60

70

No CatalystFe++ 200mg/lCu++ 200mg/lAC-Cu 2g/l

Reaction Time (min)0 20 40 60 80 100 120C

olor

Oxi

datio

n R

emov

al (%

)

01020304050607080

Effect of catalyst on WPO of dyeing wastewater at 110oC.

wet air oxidation for sampling - kexhu.people.ust.hkkexhu.people.ust.hk/ceng576/576-03.pdf · wet...

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