Wastewater Engineering - file · Web viewWhen wastewater enters a treatment plant it flows through screens, where large objects like stones are held back. This is called...

Kristof Van Hoye (449096) Wastewater Engineering

Wastewater EngineeringEG-M09

Course Assessment

Kristof Van Hoye449096

- 1 -


CONTENTS

1. Primary Treatment: BOD-removal 3

Introduction 3 BOD removal efficiency 3 Design parameters 5 Chemical coagulation and flocculants 7 Tube and plate settlers 8 Causes of low removal efficiencies 8 Sedimentation and flotation: a mathematical point of view 9 Conclusion 11 References 12

2. Secondary Treatment: Configuration 13

Batch 13 Continuous: Multiple Tanks 16 Continuous: Recycling 19 References 22 Appendices 23

3. Sludge Treatment & Disposal 25 Introduction 25 Incineration & Bio-Con 26 Incineration 26 Bio-Con 29 References 33

- 2 -


Primary Treatment: BOD-removal

1. Introduction

In Wastewater treatment one considers 3 stages, the first stage or the primary treatment is predominantly a physical or chemical process. This is in contrast with the secondary treatment which is mainly a biological process. The third stage includes de-nitrification, sand filtration and the destruction of viruses, bacteria et cetera (e.g. the use of UV-lamps in the Swansea Waste Water Treatment Works)

When wastewater enters a treatment plant it flows through screens, where large objects like stones are held back. This is called ‘Screening’.In a grit chamber the flow will slow down in order to allow the grit to fall out. In these two processes the larger parts are removed from the wastewater, some authors refer to it as being the pre-treatment others consider it as being a part of the primary treatment. A fact is that the most important stage in the primary treatment is the settling of dirt in so called primary sedimentation tanks. Primary sedimentation tanks can be rectangular or circular. To increase the sedimentation area one can use tube settlers. In the Swansea Waste Water Treatment Works one uses parallel plate separators (PPS). A disadvantage of the PPS is that, due to the great area, it costs a lot of time and money to clean them [Rob Jenkins, engineer at the SWWTW] .The basic idea of a settler is that particles will sediment due to gravity. The bigger the particles are the faster they will sediment (Stokes) Due to hindering and compaction the velocity decreases. If the particles form aggregates they become heavier and the velocity increases.

Primary treatment may also serve as a form of flow equalization or dilution when e.g. water with toxic components enters the treatment plant

The purpose of primary treatment of waste water is to decrease the amount of organic compounds (expressed as BOD: Biochemical Oxygen Demand), the Suspended Solids (SS), phosphor, etc., present in polluted water. It also reduces the amount of waste activated sludge in the activated sludge plant. (Kiely G., 1997)

2. BOD removal efficiency

- 3 -


To calculate the BOD removal efficiency one needs the BOD concentrations of the effluent (BODe) and influent (BOD0) The efficiency is expressed as:

E(%) = [(BOD0-BODe)/BOD0]/100

The degree of BOD removal depends on the design of the primary settling tanks (chapter 3): an efficiently designed and operated clarifier can remove 50 to 70 percent of the suspended solids. During the removal of the suspended solids 25 to 40 percent of the BOD5 is removed. (Metcalf & Eddy, 1972) The removal of the organic material (BOD) by primary treatment is not the result of bacteria who use the organic compounds as nutrients (which is the case in secondary treatment) but is caused by the gravitational settlement of the solids.

Industrial Wastewater : An Example from Flanders (Belgium)

One considers 3 companies:1. A thinning-factory with wastewater containing 700 mg/l suspended solids,5 000 mg/l COD, 3 000 mg/l BOD, 150 mg/l Ntot and 30 mg/l Ptot

2. A potato processing company with wastewater containing 700 mg/l suspended solids,10 000 mg/l COD,3 000 mg/l BOD,150 mg/l Ntot en 200 mg/l Ptot

3. A potato-peeling factory with wastewater containing 1 100 mg/l suspended solids,6 000 mg/l COD, 2 500 mg/l BOD, 200 mg/l Ntot en 30 mg/l Ptot

(Source: VITO, Flemish Institute for Technological Investigation)

Only considering the BOD removal we get:

Table 2.1 : an example on efficiency calculations(VITO)influent BOD

effluent BOD

BOD removed efficiency efficiency

Company mg/l mg/l mg/l %1 3000 1800 1200 0,4 402 3000 1800 1200 0,4 403 2500 1500 1000 0,4 40

The primary treatment of Company 1 exists of sieves, sedimentation and flotation. Measuring the BOD concentration of the water after each treatment step gives us the opportunity to calculate the efficiencies of each step. It is important to mention that the efficiencies of the considered steps will be different when used separately.

Table 2.2 : efficiency calculations : Company 1(VITO)

BODRemoved

BOD

Removed BOD

cumulativeRemoval

Efficiency

Removal Efficiencycumulative

mg/l mg/l mg/l % %influent 3000

Sieves 2850 150 150 5,0 5,0effluent Sedimentation 2565 285 435 10,0 14,5

Flotation 1795 770 1205 30,0 40,2

- 4 -


In this case the highest degree of BOD removal is the result of flotation. Flotation is a process in which solids (and BOD) attach to small air bubbles that will rise and so form froth on the top of the tank .Later this froth will be removed.

Municipal Waste Water

Typical influent data for MWW are

Table 2.3: typical influent data (MetCalf & Eddy,1972)BOD 220 mg/l Total Solids 720 mg/l

The composition of municipal waste is less predictable than industrial waste

3. Design Parameters

To design a primary settler following parameters are considered

(Equation 1)

(Equation 2)

(Equation 3)

(Equation 4)

A longer retention time and a lower surface loading rate (or surface overflow rate) will increase the removal efficiency (figure 3.1). But a too long retention time or a too low surface loading rate may cause septic conditions.Thus, In order to obtain satisfactory BOD-removal, a good design of the primary settler is indispensable.

0

10

20

30

40

50

60

70

80

0 20 40 60 80 100

Surface overflow rate(m³/d/m²)

Rem

oval

(%)

BOD

SS

Figure 3.1: removal in function of surface overflow rate(McGhee,1991)

- 5 -


Good BOD-removal-efficiencies are achieved when the Detention Time, Surface Loading Rate, the Weir Loading Rate and the flow through velocity have following values

Table 3.1: typical design criteria for primary settlers(Karia et al.)Unit Range Typical

Detention Time h 1 to 4 2Surface Loading Rate m³/m²/h 1.5 to 2.5 1.6

Weir Loading Rate m³/m/d125 to 500 250

Flow trough velocity m/min 0.6 to 3.6 0.9

Obeying these values results in following dimensional-features

Table 3.2: typical features of primary settlers (Metcalf & Eddy,1971)range typical

RectangularDepth, m 3 to 5 3.6

Length, m 15 to 9025 to

40Width, m 3 to24 6 to 10CircularDepth, m 3 to 5 4.5

Diameter, m 3.6 to 6012 to

45Bottom slope, mm/m 60 to 160 80

Calculations:

Indeed a rectangular tank with a typical surface loading rate of 1.6 m³/m²/hour or 38.4 m³/m²/day and a depth of 3.5 m (typical value = 3.6 m) has a detention time of 2.28 hours (using linear interpolation) this is slightly higher then the typical value of Karia et al.

Table 3.3: detention time, surface loading rate and depth ( …) DETENTION TIME, h

depthsurface-loading rate, m³/m².d 3 m 3.5 m 4 m 5 m

24 3 3.5 4 532 2.3 2.6 3 3.848 1.5 1.8 2 2.560 1.2 1.4 1.6 280 0.9 1.1 1.2 1.5

100 0.7 0.8 1 1.2120 0.6 0.7 0.8 1

Using a surface overflow rate of 38.4 m³/m²/d gives a BOD-removal of about 37 % (figure 1). Let us consider an average flow of 5000 m³/d . (Kiely G.,1997) The surface area required is

- 6 -


(Equation 2) 130 m² . If the depth of the rectangular tank is 3.5 m (table 3) the tanks has a volume of 456 m³. In that case the detention time is 2.19 hour (Equation 1)

4. Chemical coagulation and flocculants

Chemical coagulation treatment can be effectively adopted to a treatment plant to improve the suspended solids and BOD removal. It can be used intermittently during periods of high flows and heavy loadings. It also can help the plant to stay in compliance if it is consistently at or beyond its design flow capacity. Coagulation may also help to remove industrial pollutants that could inhibit biological organisms and interfere with secondary treatment processes.

When one adds chemicals to the primary settling tank one speaks of CEPT or Chemical Enhanced Primary Treatment. Due to the fact that the chemicals cause the suspended solids to clump to getter, the settling velocity will increase enhancing the treatment efficiency(Table 4.1). The increased removal efficiency allows to operate at greater overflow rates or using smaller tanks, still maintaining a high removal of suspended solids and BOD. (University of Leeds, EFM)

Table 4.1 : Removal Efficiency for CEPT (National Research Council USA,1992)Suspended Solids BOD

% %Conventional Primary treatment 55 35Chemically Enhanced Primary Treatment 85 57Conventional Primary Treatment + Biological Secondary Treatment 91 85

As on can see in table 1 the removal of suspended solids using chemicals will be similar to the removal efficiency of conventional Primary treatment followed by secondary treatment. The BOD-removal will increase significantly to near 60%

Table 4.2 : Treatment Costs for CEPT (National Research Council USA,1992)

Capital Costs

Operation andMaintanance

CostsTotal Costs

£/ 10³ m³ /d £/ 10³ m³ /d

£/ 10³ m³ /d

Conventional Primary treatment1.80 - 2.50 0.45 - 0.55 1.00 - 1.25

Chemically Enhanced Primary Treatment

2.50 – 3.15 0.50 - 0.65 1.20 - 1.55

Conventional Primary Treatment + Biological Secondary Treatment

5.35 - 5.80 0.70 - 0.95 2.00 - 2.50

- 7 -


The costs are higher using a flocculant or a coagulant. This is quite logical: One has to pay for the chemicals and the conventional tanks have to be adapted: e.g. injection equipment. Also important is the presence of a zone in which the chemicals are heavily mixed with the wastewater. The purpose of this is to create a homogenous liquid in less then 5 seconds (rapid mix)

4.1 Examples (Baeyens J. et al, 1995)

CoagulantsCoagulants are salts of high-valented metalsOn uses FeCl2, Fe2(So4)3 or copperas and Al2(SO4)3 or alum

Flocculants Flocculants are high molecular cationic poly-electrolytes

5. Tube and plate settlers

To improve the solids and organic material removal efficiency one can use plate or tube settlers. They consist of several parallel plates of tubes made of PVC or metal. These tubes are inserted in the primary tank at a 30 to 60° incline. Wastewater flows through the tubes and solids settle upwards the walls. (Tilman M., 1982)

Figure 5.1: Tube settler(brentwood industries)

6. Causes of low removal efficiencies

The removal of solids and BOD can be affected by following phenomena:

Hydraulic Short CircuitingHydraulic Short Circuiting is caused by currents .These currents can be induced by inlets or effluent weir plates that are not level. Another cause is the difference between the influent and clarifier water temperature. The wind can also cause problems.

The most important factors to consider in controlling short-circuiting are dissipation of inlet velocity, protection of tanks from wind sweep and uneven heating and reduction of density currents associated with high inlet suspended solids concentrations.

- 8 -


High Sludge BedDue to a high water forward velocity the sludge may be scoured and re-suspended .It is also possible that, due to a high bed, there is a high drive torque. This is mostly the case for light organic solids.

Forward velocities should be from 9 to 15 times the settling velocity of critical size solids in order not to cause scour

Increase in Influent Suspended Solids

Increase in Influent Suspended Solids can cause particles to settle as a mass rather than discretely. Increase in Influent Suspended Solids can cause sudden increase in sludge bed height. If chemicals are used, a decrease in chemical/pound of solids is needed.

High residence time of the sludge

If Sludge is held too long in the clarifier it can create gasification by anaerobic decomposition. Gas bubbles can be seen breaking water surface. Re-suspension can occur of sludge solids. Floating black sludge can be seen. A strong hydrogen sulphide odour can be present in severe cases. This is a common problem in pulp & paper and food industries.

(www.environmentalleverage.com)

7. Sedimentation and flotation: a mathematical point of view

7.1 Sedimentation

One considers 4 types of settling (Sincero A.,1996) : 1. the settling of discrete particles2. the settling of flocculent particles3. hindered settling4. compression and compaction

In the case of type 1 settling the settling velocity is given by:

When one has laminar flow into the settler, the drag-coefficient will be 24/Re (Re: Reynolds number) and the settling velocity is given by:

g = 9.81 m/s²p= the density of the particlew = the density of waterd = diameter of the particleµ = dynamic viscosity of water

- 9 -


Thus heavy particles will settle faster (i.e.: bigger particles and/or particles with a high density)

It is important to mention that most of the particles are not spherical. In this case one can use the diameter of the equivalent spherical particle.

The settling velocity determines the fractional removal (fx)

V0 = the critical velocity = total volumetric flow/surface areaIn case Vs>V0 there will be complete removal

f0 is de fraction with settling velocities less than v0

The overall removal for all particles (fA) is the sum of the completely removed particles and the partial removed particles:

Thus fA can be found by an x-y graph, in which x = settling velocity and y = fraction of particles less than stated velocity

Figure 7.1.1: cumulative settling curve (Jones, M. ,course notes 2006-2007)

A = V0, here V0 = 1.36 (example in the course notes, Jones, M.)B = f0

C=

7.2 Flotation

- 10 -


Principle: gas bubbles with a diameter between 10 and 100 µm collide with a particle and attach to it. The density of the particle becomes lower and moves it upwards.

Two main phenomena occur:

1. Collision between the bubbles and the particle2. Permanent adhesion of bubbles to flocks thus bringing about their upward movement. The separated solids behave as a solid layer, which is scraped off.

In case of an ideal plug flow the efficiency (E) of the flotation process is given by:

WithNdi = particle density of influent (m-3)Nde= particle density of effluent (m-3) = contact timea = adhesion coefficient between flock particles and bubble (varies between 0.2 and 1.0)I = collision factor, should be 1db = average diameter of gas bubble(m)Qb = flow of a gas bubbleg = 9.81 m/s²= kinematic viscosity (1.0*10-6 m²/s , decreases with increasing temperature)

There is an optimal db of about 40 µm. In case of smaller bubbles the velocity is restrictive. In case of larger bubbles the likelihood of collision between particle and bubble is restrictive.

(Verstraete, W., 2006)

8. ConclusionThe statement ‘ Up to 60% of the BOD load of wastewater stream can be removed by Primary Treatment’ is true in case the primary settling tanks are good designed and when you add a chemical coagulants or flocculants . The removal efficiency can be increased using plate or tube settlers. It is also important to monitor the process in order to avoid problems like ‘Hydraulic Short Circuiting’ Also: combining several primary treatment methods is advisable.

- 11 -


References

BOOKS & ARTICLES

BAEYENS, J . HOSTEN, L. & VAERENBERGH, E. (1998)AfvalwaterzuiveringKluwer Editorial

JONES, M. , Wastewater Engineering , Course Notes (2006-2007)

KARIA,G.L. & CHRISTIAN,R.A. Waste Water Treatment: Concepts and Design Approach Prentice hall of India

KIELY, G. (1997)Environmental EngineeringLondon, McGraw-Hill Publishing

METCALF & EDDY (1972)Wastewater EngineeringNew York, McGraw-Hill Book Company

MCGHEE, T.J. (1991)Water Supply and SewerageNew York, McGraw-Hill

SINCERO, A.P. & SINCERO, G.A. (1996)Environmental Engineering : A Design ApproachNew Jersey,Prentice-Hall, Inc.

TILMAN,M. ,primary treatment at wastewater treatment plants (1998)

VERSTRAETE ,W., Biotechnological processes in environmental sanitation ,Course Notes (2006-2007)

WEBSITES

University of Leeds – EFM : Engineering Fluid Mechanics – An introduction of Chemically

Enhanced upgrading and primary treatment http://www.efm.leeds.ac.uk/CIVE/Sewerage/articles/Introduction_to_CEPT.pdf

www.environmentalleverage.com

www.vito.be

- 12 -

http://www.vito.be/


Secondary Treatment: Configuration

In this section we want to investigate which configuration, what kind of bacteria, … give the most satisfactory results in the biological treatment of wastewater.

I examined, using the program ‘Scientist’, what are the best values for the bio kinetic parameters. Therefore I used a batch set-up : bacteria and substrate are all put into a fermenter without the presence of a continuous flow of the wastewater.

I also investigated the benefit of recycling in case of a continuous set up and the influence of the number and volume of the aeration tanks.

1. Batch(The ‘Scientist’ input is given in Appendix 1)1.1 Equations

Starting from a material balance for Biomass:

Biomass in with bulk flow – Biomass out with bulk flow + Biomass formed by growth = Accumulation

F*X0-F*X+µ*X*V = V*dX/dt

We use following simplification: X=X+X0

In case of steady state dX/dt = 0

Knowing the yield coefficient it is possible to determine how much substrate has been used:

Y=

The equations are:

With

S: the substrate concentration X: the biomass concentrationµm: maximum specific growth rate

Y: yield coefficientKs: Monod coefficient

- 13 -


1.2 Ks

I used four values for Ks: 30, 60, 90 and 120 . After a time of 50 units I compared for each case the concentration of the substrate and biomass and also the time after which the biomass stays constant (steady state : dX/dt = 0)

Table 1.2.1 : influence of KsXeff at t=50

Seff at t=50

Time until dX/dt =0

KS = 30 185 8.37E-124 6KS =60 185 9.64E-61 8.5KS = 90 185 1.01E-39 11KS = 120 185 3.27E-29 13.5

The best result is achieved using a low Ks: after a time of 50 units most of the substrate is disappeared at a Ks of 30, in that case the time until steady state is obtained, is the lowest.

Explanation: Ks is the substrate-concentration at which µ = µmax /2 . A low Ks means that the affinity of the bacteria for the substrate is high (figure 1.2.1) whit the logical outcome that the substrate will diminish rapidly. When the biomass is at steady state there will be still substrate consumption due to maintenance.

0

10

20

30

40

50

60

0 100 200 300

Substrate

spec

ific

grow

th ra

te

Ks = low

Ks = high

Figure 1.2.1: Monod graphs 1.3 µm

Analogue to 1.2

Table 1.3.1: influence of µmXeff at t=50

Seff at t =50

Time until dX/dt =0

MUM =0.35 185 7.29E-46 10.5MUM = 0.40 185 2.65E-53 9.5MUM = 0.45 185 9.64E-61 8.5MUM =0.50 185 3.51E-68 7.5

The highest µm gives the best results: The bacteria will grow faster and thereby use more substrate.

- 14 -


1.4 The Yield coefficient

Analogue to 1.2

Table 1.4.1: influence of Y

The best result is obtained when using bacteria with a lower yield. A low yield implies that little substrate is used to create cell biomass (e.g. compare Xeff at t=50 for a yield of 0.30 and 0.45) .Due to a low yield there will be a higher substrate demand for energy-purposes. I also investigated what would happen if one used a bacteria with a Y of 1 (only theoretical possible) In that case the X is multiplied by seven (X0=50) in a time of 50 units.A consequence of a high yield is also that the sludge production (X) will be to high , which may cause severe problems : congestion, sludge spill,…

1.5 Starting values for S and X

Table 1.5:influence of X and S at t=0

S at t=0Xeff at t=50

Seff at t=50

Time until dX/dt =0

S=200 140 2.50E-46 9S=300 185 9.64E-61 8,5S=400 230 4.68E-75 8S=500 275 2.84E-89 8

X at t =0X=40 175 8.66E-57 9X=50 185 9.64E-61 8.5X=60 195 1.30E-64 8X=70 205 1.97E-68 7.5

The more substrate there is present in the beginning, the more will be removed after a while. This is quite astonishing! A possible explanation is: In this experiment the yield is kept constant. Thus when there is more substrate available the biomass of the bacteria will increase more. When there are more bacteria in the fermenter the demand for substrate will be higher and logically the substrate will diminish much faster.

One can derive the residence time at which failure will occur

Using :

Ks = 60 mg/Lµm = 0.45 h-1

- 15 -

Xeff at t= 50

Seff at t= 50

Time until dX/dt =0

Y = 0.30 140 2.51E-70 7Y=0.35 155 2.95E-66 7.5Y = 0.40 170 3.60E-63 8Y = 0.45 185 9.64E-61 8.5Y = 1 350 1.78E-49 11.5


Tabel 1.5.2:Critical theta calculationsS0(mg/L) Critical theta100 3.56 h200 2.89 h300 2.67 h400 2.56 h

Let us assume there is a of 2.7 h, in that case S0 has to be more than 300 mg/L, otherwise there will be no removal (S=S0) . On can conclude that the more substrate there is in the beginning the more likely it is that some substrate will be removed (S ≠ S0)

1.6 conclusions

If one uses bacteria in a fermenter one should choose bacteria with a low Ks, a low Yield coefficient and a high µm . High starting values for X and S will have a positive effect on the process, it is although important to keep the mechanical and physical borders of the fermenter in consideration.

2. Continuous We consider 2 configurations: A. Multiple tanksB. Recycling

2.A Continuous : multiple tanks (The Scientist input is given in Appendix 2)

I wanted to examine the influence of the number and the volume the of aeration tanks : How many tanks are needed to give the best results? Is it better to use 1 big tank or several smaller ones?

2.A.1 equationsAs an example we consider 2 tanks in series. The equations for 3, 4 and more tanks are analogue

Starting from a material balance: Material in with feed + Material formed by reaction – Material out with flow =Accumulation (Jones M., course notes)

This material balance is applicable for each tank.

IN tank 1 – OUT tank 1 = Accumulation in tank 1 – Material formed 1IN tank 2 – OUT tank 2 = Accumulation in tank 2 – Material formed 2

OUT tank 1 = IN tank 2

So:For biomass:

- 16 -


In case of steady state : = 0

For substrate

= dS/dt * V

= dS/dt * V

And Rsi = In case of steady state : dS/dt = 0

To derive S1,S2,X1 en X2 we use the program ‘Scientist’The typical values of MetCaff & Eddy for Ks, µ1,µ2,Y were used

Considering 2 tanks in series :

WithX0 = the influent biomass concentration of tank 1X1 = the effluent biomass concentration of tank 1 or the influent biomass concentration of tank 2X2 = the effluent biomass concentration of tank 2S0 = the influent substrate concentration of tank 1S1 = the effluent substrate concentration of tank 1 or the influent substrate concentration of tank 2S2= the effluent substrate concentration of tank 2µ1 = the specific growth rate of the bacteria in tank 1µ2 = the specific growth rate of the bacteria in tank 2Y= the yield coefficient= the residence time

- 17 -


2.A.2 number of tanks

In this experiment I used 1,2,3,4,5 and 6 aeration tanks , all of the same volume (10 units) After a time of 200 units I analysed the effluent concentration of the biomass & substrate, the removal efficiency and the times at which the substrate and biomass stayed constant (steady state)

Tabel 1.A.2.1: 1-6 tanks with a volume of 10number of tanks (each one V= 10)

Xeff at St.state

Seff at St.State

% removal

Time until dX/dt =0

Time until dS/dt =0

1 149.400 48.000 87.36842 80 802 169.498 3.338 99.12162 100 1003 170.901 0.220 99.94217 100 904 170.994 0,014 99.99621 110 905 170.999 0.001 99.99975 130 806 171.000 6.198E-05 99.99998 140 70

Using more tanks gives higher removal efficiencies. This is because the detention time (DT) is higher when the volume is bigger: DT = V/Q, a larger DT implies that the bacteria have more time to consume the substrate. The steady state bacteria biomass will be a bit higher and logically it will take more time to achieve steady state.

-20

0

20

40

60

80

100

120

140

160

180

0 50 100 150 200 250

Time

Con

cent

ratio

n

2 tanks of 10, Biomass2 tanks of 10, Substrate6 tanks of 10, Biomass6 tanks of 10, Substrate

Figure 2.A.2.1: 2 tanks of 10 and 6 tanks of 10

There are indeed no big differences between the Xeff at steady state, but the differences between Seff in case of 2,3,4,5 and 6 tanks are also not that big. The most significant result is found between 1 and 2 tanks.

2.A.3 constant volume

- 18 -


Table 2.A.3.1: 1 to 6 tanks with total volume of 20

total volume =20

Xeff at St.state(or at t=200)

Seff at St.State(or at t=200)

% removal

Time until dX/dt =0

Time until dS/dt =0

1 tank of 20 163.286 17.143 95.48872 140 1502 tanks of 10 169.498 3.338 99.12162 100 1003 tanks of 6.666 170.371 1.397 99.63237 80 804 tanks of 5 165.151 12.999 96.57933 +200 +2005 tanks of 4 0.038 379.916 0.021988 +200 +2006 tanks of 3.333 5.337E-09 380.000 0

To explain these data one has to consider two principals:1. The principal of detention time, see 2.A.12. e.g.: using 1 tank has the disadvantage that after a while the bacteria will become saturated with food and thus the substrate-removal-rate will flatten. Dividing this tank into 2 tanks will create a zone (the 2nd tank) in which ‘fresh’ bacteria will consume the effluent substrate of tank 1

The second principal is the most import in case of 1, 2 and 3 tanks . The first one in case of 4,5 and 6 tanks : the detention time is to small, mostly in case of 5 and 6 tanks were the process will fail.

These conclusions are only applicable in this particular case. When the aeration tanks have other volumes, or there is another flow another number of tanks may be favourable: e.g. in the Swansea Waste Water Treatment Works one uses 4 aeration tanks.

2.B. Continuous : recycling(The ‘Scientist’ input is given in Appendix 3)

In almost all the wastewater treatment plants one uses aeration tanks with recycling. I wanted to investigate what are the advantages of recycling and when it is advisable.

2.B.1 Equations

Starting from a material balance: Material in with feed + Material in with recycle + Material formed by reaction – Material out with flow = accumulation (Jones M., course notes)

For biomass :

F0X0+FRXR+µXV-(F0+FR)*X = V*dX/dt

In case of steady state , dX/dt =0 this equation becomes:

F0X0+FRXR+µXV-(F0+FR)*X = 0

For substrate

- 19 -


F0S0 + FRS - RsV- FS = V*dS/dt

with

Rs =

In case of steady state, dS/dt = 0

We get : F0S0 + FRS - RsV- FS = 0Using both balances you become following equations :

With S : the effluent substrate concentrationX: the effluent biomass concentrationS0 : the influent substrate concentrationX0 : the influent biomass concentration

R : the recycle ratio , R =

Fr : the recycle flowF0: the incoming flow

: the hydraulic residence time,

V : the volume of the tankY: the yield coefficientKs: the Monod coefficient

the settler performance,

Xr : the recycle biomass concentrationµm: the maximum growth rate

In a first experiment I used several recycle ratios : 0, 0.15, 0.20, 0.30 ,0.45, 0.60, 0.75, 0.90.For every recycle ratio I analysed the biomass and substrate concentration at steady state (or when there was no steady state within 200 time units the concentration at a time of 200 units), the time in which steady state occurred and the removal efficiency

Table 2.B.1.1: influence of R

R

Xeff at St.state(or at t=200)

Seff at St.State(or at t=200)

% removal

Time until dX/dt =0

Time until dS/dt =0

0.00 1337.019 28.846 99.03846 120 1400.15 1916.933 18.103 99.39655 170 1800.20 2238.749 15.000 99.5 200 200

- 20 -


0.30 3364.264 9.376 99.68748 +200 +2000.45 12322.784 2,352 99.9216 +200 +2000.60 899590.814 0.031 99.99896 +200 +2000.75 569638912 4.937E-05 100 +200 +2000.90 5.668E+11 4.962E-08 100 +200 +200

A high recycle ratio results in high removal efficiencies, this is quite logical : the more sludge you return , the more substrate will be used by the bacteria. Looking at Xeff , it seems that a high R results in high biomass production. In case of R=0.2 the biomassconcentration will be higher, but the process will not fail, this is not the case using a ratio of e.g. 0.6

Influence of Recycling R=0 & R=0.2

0

500

1000

1500

2000

2500

0 50 100 150 200 250

Time

Biom

ass

conc

entra

tion

R=0R=0.2

Figure 2.B.1.1 : influence of recyling R = 0 , R=0.2 and R=0.6

I also investigated what would happen if the F jumps from 2 to 4 after a time of 120

Table 2.B.1.2: recycle and increased flowno jump jump Xjump/Xno

jumpSjump/Sno jumpXeff 200 Seff 200 Xeff 200 Seff 200

R = 0 1337.019 28.846 1307.812 93.75 0.978 3.25R =0.2 2238.749 15.000 2221.875 37.50 0.992 2.50R = 0.3 3364.264 9.376 3350.891 21.43 0.996 2.29R = 0.45 12322.78 2.352 13029.22 4.57 1.057 1.94R= 0.6 899590.8 0.031 1.11E+08 0.0005 122.96 0.016

As one can see the Xeff 200 (the effluent concentration after a time of 200) decreases due to the jump in case of a recycle ratio of 0(no recycle), 0.2 and 0.3; but increases in case of a recycle ratio of 0.45 and more. This is because when F0 increases the hydraulic residence time decreases , this phenomenon is less import in case R is e.g. 0.6 : a high amount of recycling combined with a high input flow will cause an increase of input of organic material ,present in the influent water (F0)

- 21 -

Influence of Recycling R=0.6

0100000200000300000400000500000600000700000800000900000

1000000

0 50 100 150 200 250

Time

biom

ass

conc

entr

atio

n


Seff increases when there is a jump (in case of lower recycle ratio’s), but the higher R is, the less there will be a difference between the effluent concentration with or without a jump. In case of R=0.6 Xeff is 1.11E+08, this is too high and the process will fail.

The data also gives us the opportunity to estimate when Xjump will be equal to Xnojump and when Sjump is the same as Snojump:Xjump = Xnojump when 0.3<R<0.45Sjump = Snojump when 0.45<R<0.6

We can conclude that when 0<R<0.5 the process will become more robust (less effect of an increased flow)

References

JONES, M. , Wastewater Engineering , Course Notes (2006-2007)

METCALF & EDDY (1972)Wastewater EngineeringNew York, McGraw-Hill Book Company

APPENDICES 1- 22 -


BATCH

// MicroMath Scientist Model FileIndVars: TDepVars: X, SParams: KS, MUM, Y//define paramKS=75MUM=0.5Y=0.45//define functionsS'=-MUM*S*X/(Y*(KS+S))X'=MUM*S*X/(KS+S)//starting valuesS=300T=0X=50

2CONTINUOUS , MULTIPLE TANKS

e.g : 2 tanks

/ MicroMath Scientist Model FileIndVars: TDepVars: X1,S1,X2,S2Params: KS,MUM,Y,V,F,jumpMU1=MUM*S1/(KS+S1)MU2= MUM*S2/(KS+S2)KS=60MUM=0.45Y=0.45V=10S0=380X0=0jump=0F=ifgezero(T-jump,2,0)THETA=V/FX1'=(X0-X1)/THETA+MU1*X1S1'=(S0-S1)/THETA-MU1*X1/YX2'=(X1-X2)/THETA+MU2*X2S2'=(S1-S2)/THETA-MU2*X2/YX1=100S1=0X2=0S2=0T=0

- 23 -


3CONTINUOUS , RECYCLING

// MicroMath Scientist Model FileIndVars: TDepVars: X, SParams: KS, MUM, Y, theta, S0,X0,R, beta//define paramKS=75MUM=0.45Y=0.45theta=8S0=3000X0=0R=0.3beta=3//define functionsS'=S0/theta+R*S/theta-MUM*S*X/(Y*(KS+S))-(1+R)*S/thetaX'=X0/theta+R*beta*X/theta+MUM*S*X/(KS+S)-(1+R)*X/theta//starting valuesS=0T=0X=10***

Sludge stabilization and disposal

1. Introduction

- 24 -


Sludge stabilization

The use of sludge is crucial in most of the wastewater treatment systems. After being used the sludge will be disposed. It may be necessary, before disposing the sludge, to stabilize it in order to reduce pathogens, eliminate offensive odours and to control the potential for putrefaction of organic matter (Qasim, S.R.) Sludge stabilization can be accomplished by physical, chemical or biological means. The sludge disposal method will determine which sludge stabilization method will be used.

Some common methods are - Anaerobic digestion- Aerobic digestion- Chemical stabilization- Thermal conditioning

The key factor for the selection of a sludge stabilization process is the ultimate sludge disposal. But there is also a juridical factor: Regulation agencies have regulations that directly impact the process selection, design criteria, redundancy requirements and the ultimate disposal, e.g. odour restrictions.

The costs for the treatment and disposal of sludge are estimated to be almost 40% of the total cost in water purification (Tebbutt,T.,1998)

(table 1.1)

Sludge disposal

After the sludge is stabilised one can choose several methods to get rid of the redundant sludge:- conversion processes - incineration - wet oxidation - pyrolysis - composting

-land disposal - land application - landfilling

- …

Table 1.1: Summary of Conversion processes for Disposal of Municipal Sludges(Qasim, S.)

Conversion ProcessRecommendedPretreatment

Incineration thickening and dewatering

- 25 -


wet oxidation thickeningPyrolysis thickeningRecalcining thickening and dewateringcomposting by heat drying thickening and dewateringcomposting by microbial action thickening,digestion and dewatering

Conversion Process Additional Processing Requirementsincineration Ash landfilling

wet oxidationseparation of ash, treatment of returned liquid, landfilling of ash

Pyrolysisutilization of by-products(gas,liquid,carbon,etc) disposal of residues

recalcining recovery of lime, landfilling of ashcomposting by heat drying utilization or sale of compostcomposting by microbial action utilization or sale of compost

As can see in Table 1.1 the use of conversion processes requires pre-treatment and additional processing what makes them quite costly : they are either energy or labor-intensive.

2.Incineration & Bio-Con

There is no ultimate way of disposing the waste-sludge. On has to find a compromise between economical and environmental factors. A good solution may be the Bio-Con . The Bio-con process is developed in Denmark and is a sludge incineration process, which integrates recovery of phosphorus, energy and precipitation chemicals.

Before we describe the Bio-Con process in detail we first consider the incineration process: 2.1 incineration (Metcalf & Eddy,1972 ; Tebbutt T,1998)

Incineration is the combustion of sludge and thereby converting the material into gases and residues. What remains is an inert ash. If the incineration process is preceded by adequate dewatering to ca. 30 percent solids there is no need to ad any supplemental fuel, because dewatered sludge has a high calorific value (16-20 MJ/kg dry solids) .I most cases although a small amount of fuel oil or gas is added to support self-sustaining and to achieve optimum performance .

The remaining ash amounts to 5 - 10 percent of the original solids. The inert material can disposed as fly ash, but environmental problems may be caused due to the heavy metal content of the ash. The emission of several constituents, formed during the combustion, is another environmental disadvantage (table 2.1.1)

Table 2.1.1 : emissions rates from sewage sludge incineration(Tebbutt T.,1998)Constituent Emission rate (kg/tonne dry solids)Particulates 16NOx 2.5

- 26 -


SOx 0.5Hydrocarbons 0.5

Stages of the incineration process(Skitt J.,1979)

1. Drying

One uses - fuel- radiation- convection- pre-heated air- hot waste gases

2. Combustion

The dried and heated refuse gives off volatiles which burn due to the presence of O2 in the gases in the combustion chamber. A good burn-out of these volatiles is very important in order to avoid bad smell and smoke(A temperature of at least 750 °C is necessary to ensure this) In case the temperature is above 1000 °C it is likely that ash fusion may take place, resulting in the creation of sticky particles who may create tensions and lead to slagging problems.

3. Burnout

This is the combustion of the slower burning elements until an organic free residue with a very low carbon content is left for disposal.

2.1.1 Heat Balance of incineration (Sincero P., 1996)

Heat losses: sensible heat losses due to the sensible heat content of the ash( 1047 J/kg °C) and unburned carbon remaining in the ash(heating value of carbon is 32,851 kJ/kg), heat loss due to radiation(from 0.003 to 0.005 kJ/kg of fuel) and heat loss due to the vaporization of water( latent heat of vaporization of water is 2420 kJ/kg)

Heat input: the total input of heating energy (9300-12 800 kJ/kg)

Available heat = Heat input – Heat losses

Example calculation

-inputHeat input per day = sludge mass/day * heating value

Heat input per day = 100 000 kg/day * 12 000 kJ/kg = 1.2 * 109 kJ/day (data : Sincero,P.,1996)

-output- Heat loss in ash = mass of the ash * T * heating value + mass of C * heating value

- 27 -


Heat loss in ash = 21.053 kg ash/day *(420-25) °C *1.047 kJ/kg °C + 1053 kg C/day * 32.851 kJ/kg = 4.33 *107 kJ/day

- Water losses = mass of water * heating value Water losses = 53.761 kg water/day * 2420 kJ/kg = 1.301 * 108 kJ/day - Radiation losses = mass of sludge / day * heating value Radiaton losses = 100 000 kg sludge per day * 0.004 kJ/day = 1.734 * 108 kJ/day

So the available heat is : input – sum of the outputs = 1.0266 * 109 kJ/day

(data : Sincero, P.,1996)

2.1.2 Incinerator technology

There are two primary technologies used to incinerate sludge: MHF (Multiple Hearth Furnaces) and Fluidized Bed Incinerators (FBI) (Ky, D. et al,2000)

2.1.2.1 MHF

MHF’s can be divided into 4 zones :The first, which consists of the upper hearths, is the drying zone where most of the water is evaporatedThe second, consisting the central hearths is the combusting zone. In this zone the combustibles are burned (temperature : 760°C – 930 °C) The third is the fixed carbon burning zone, where the remaining carbon is oxidized into carbon dioxide.The fourth is the cooling zone and includes the lowest hearths. In this zone ash is cooled by the incoming combustion air.

2.1.2.2 FBI

An FBI is a single chamber unit in which both drying and combusting occur in a fluidized sand bed.Ash is carried out the top of the furnace and is removed by air pollution control devices,e.g. Venturi Tray Scrubbers

There are two basic process configurations for the FBI :HWB or the hot wind box design: the fluidizing air passes through a heat exchanger, prior to injection into the combustion chamberCWB or the cold wind box design: the fluidizing air is injected directly into the furnace

- 28 -


Figure 2.1.2.2.3: Fluidized Bed Incinerator(http://www.eccj.or.jp/hicot/eng/pamph/2002/02_03.html)

2.2 Bio-Con(Lundin,M.,2004)

The Bio-Con method includes three processes1. drying : The sludge is dried to 90% DM2. combustion : The dried sludge is fed to an incinerator. Some of the energy generated is

used to dry the sludge, the rest is used for district heating3. recovery : After the combustion, ash and slag remain. They are dissolved in sulphuric

acid . The generated solution enters 4 ion exchangers.- In the first ( a cat ion exchanger) iron is collected- In the an ion exchanger, sulphate is collected as potassium sulphate- In the third ion exchanger phosphoric acid is recovered- In the last section heavy metals are collected and disposed off

2.2.1 Nutrients and Heavy Metals

Thus the Bio-Con process combines the advantages of incineration and agricultural usage of waste-sludge : The advantage of the land application of sludge is that it provides the soil with phosphorus and other essential nutrients. The advantage of co-incineration is that it destroys harmful and toxic components: e.g. heavy metals. In Belgium the land application of sludge is not allowed because of the too high concentration of heavy metals.

Table 2.2.1.1:Environmental impact from the Swedish society with a population of 8.9 millions(Lundin,M.,2004)

- 29 -


Unit/person and year Agricultural use Co-incineration Bio-Con

Cd to arable land Mg 33 0 0Pb to arable land Mg 1180 0 0Hg to arable land Mg 24 0 0Cu to arable land Mg 10400 0 0Hg to air Mg 0 2.7 13Cd to landfill Mg 0 35 X

plant available N Kg 0.34 0 0plant available P Kg 0.49 0 0.56

The impact of the heavy metals is high when one uses the waste sludge on land without pre-treatment : The amounts of Cd, Pb, Hg and Cu are high.

The impact is low if the sludge is incinerated (co-incineration and Bio-con): The amounts of Cd, Pb, Hg and Cu are zero.A disadvantage of the Bio-con method are the high mercury emissions into the air.

The plant available nitrogen and phosphorus are quite high for land-application. This means the sludge can serve as a fertilizer. Incineration destroys all the nitrogen and phosphorus, i.e the ashes are too contaminated when mixed with other waste. The phosphorus although can be recovered using the Bio-Con method and due to the process it will be in a more plant available form.

2.2.2 Energy Since the energy recovery from the incineration is of great importance co-incineration and Bio-con show an advantage compared to the agricultural option. Agricultural usage is an option which requires a lot of energy : energy is required for transportation, pasteurisation and spreading of sludge

- 30 -

Spreading Transport Pasteurisation Chemicals Incineration

Fertiliser prod.


A: agricultural application B: co-incineration C: Bio-Con

Figure 2.2.2.1 : Energy Bio-Con (Lundin,M.,2004)

Al 3 methods require energy for transport, especially the agriculture use where transport requires about 400 kWh/ton DM sludge.

The energy generated by incineration is the largest for common co-incineration. Bio-con also creates energy due to the production of fertiliser and the usage of chemicals

2.2.3 Environmental impact

Relative Environmental Inpact

-1,5

-1

-0,5

0

0,5

1

1,5

Resourcedepletion

excl.Sulphur

Resourcedepletion incl.

Sulphur

Acidification Eutrophication GWP

Agricultural use co-incineration Bio-Con

Figure 2.2.3.1:environmental impact of sludge disposal(Lundin,M.,2004)

This figure shows the results from the characterisation, using the categories resource depletion (excluding and including sulphur), acidification, eutrophication and global warming potential (GWP). The results have been normalised by dividing the values for each category with the highest absolute values.

Considering the resource depletion we conclude that the ‘agricultural use’ reduces the environmental impact (negative values) if you also include the Sulphur usage. This is due to the reduced depletion of phosphorus (minor effect), but the fuel-use is much higher here than the other two methods. The results for Bio-con are quite good, but the use of Sulphur (needed to recover phosphorus) makes it less good then co-incineration.

Compared to the other options agricultural application of sludge contributes to a higher degree to eutrophication and acidification. The cause of this phenomenon is the emission of ammiona (NH4). The results for Bio-con are satisfactory. Co-incineration has the best results, considering both acidification and eutrophication.

- 31 -


All 3 methods lower the effects on global warming (GWP = global warming potential), since the reduced use of mineral fertilisers decrease the emissions of CO2 and N2O

2.2.4 Economical

10,95

118,99

58,438,69

64,24

44,53

0

20

40

60

80

100

120

140

160

180

200

Agricultural use Co-incineration Bio-Con

Operational costCapital cost

Figure 2.4.1: Costs for sludge disposal (£/tonnes DM sludge) (Lundin,M.,2004)

The total net cost for Bio-Con is £103 / tonnes DM sludge. The operational costs account for 43% of the net costs; these costs are quite high due to the large use of chemicals. The costs can be lowered by the revenues from sale of the product chemicals in the process. To obtain this the quality of the product chemicals has to be high.The most expensive method is co-incineration due to the costs of expanding the waste incinerator. Reducing the cost for expansion can lower the total costs by £100 /tonnes DM sludge, in this case the Bio-con method will be more expensive then co-incineration. 2.2.5 Conclusion

Although the Bio-Con process is still in a start-up phase it seems to be an interesting method : It is a good method considering the usage of energy sources and the environment. A problem may be that, because it is a new method, the costs may be higher than predicted in the investigation

References

- 32 -


KY, D., MULLEN, J. & MAYROSE, D. (2000), current practices and future direction for biosolids incinerators, proceedings of the residuals and biosolids management conference, February)

LUNDIN, M. , OLOFSSON, M., PETTERSON, G.J. & ZETTERLUND, H. (2004)Environmental and economic assessment of sewage sludge handling options ,resources conservation and recycling 41,255-278

METCALF & EDDY (1972)Wastewater EngineeringNew York,McGraw-Hill Book Company

QASIM, S.R. ( ) Wastewater Treatment Plants: Planning, Design, and Operation, Second Edition

SINCERO, A.P. & SINCERO, G.A. (1996)Environmental Engineering : A Design ApproachNew Jersey, Prentice-Hall, Inc.

SKITT, J. (1979)Waste disposal management and practiceLondon, Knight

TEBBUTT, T.H.Y. (1998)Principles of Water Quality ControlButterworth Heinemann

- 33 -

Wastewater Engineering - file · Web viewWhen wastewater enters a treatment plant it flows through screens, where large objects like stones are held back. This is called...

Documents