# Tank Retention Times Calculation Formula - PowerPoint by hjx11517

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CHEE 370
Waste Treatment
Processes

Last Lecture
Final Exam Review

1
Wastewater Constituents
   Suspended solids
   Micro-organisms
   Nutrients
   Refractory organics
   Heavy metals
   Dissolved inorganics
   Physical contaminants (temperature, colour, odour, solids)

2
Measurement of Solids
Evaporation
and drying at
105 ºC

Evaporation
and drying at
105 ºC

Ignition at 500 ºC

TS = TVS + TFS
TS = organic + inorganic   2.3 Metcalf & Eddy
The BOD Test
COHNS + O2 + bacteria (C5H7NO2)
 more C5H7NO2 + CO2 + H2O + NH3 + other

   An INDIRECT measure of the organic content
   Measures the amount of oxygen consumed as the
bacteria found in the WW use the organic material as a
carbon/energy source for the production of more bacteria

BODt=UBOD(1 - e-kt)
4
BODt=UBOD(1 - e-kt)
Chemical Oxygen Demand
(COD)
   An INDIRECT measure of the organic
content
   Mass of oxygen theoretically required to
completely oxidize an organic compound to
carbon dioxide
   Measured by mixing the WW with a very
strong chemical oxidant

CODin = CODout + O2consumed
6
Total Organic Carbon (TOC)
   A DIRECT measure of the organic content
   In theory - based on the chemical formula
   In practice - organic carbon is converted to
carbon dioxide, which can then be measured

Theoretical Oxygen Demand (ThOD)
   An INDIRECT theoretical measure of organic
content
   Calculated using stoichiometric equations
   Considers both carbonaceous and                7
nitrogenous oxygen demand
   Micro-organisms
   Total coliform, fecal coliform

   Toxicity
   Acute toxicity (LC50), Chronic toxicity

   Nutrients
   TKN, NH3, TP, ortho-phosphate

   Flowrates
   Hydraulic flowrates (peak and min), Loadings
8
WWTP for a Large Community

1 V 2  v 2 
hL             
C  2g 
   hL = headloss (m)
   C = empirical discharge coefficient to account for turbulence and
eddy losses (clean screen = 0.7; clogged screen = 0.6)
   V = velocity through the openings (m/s)
   v = approach velocity in upstream channel (m/s)

   g = acceleration due to gravity (9.81 m/s2)

10
Types of Settling

 Type I: Discrete Settling
 Type II: Flocculent Settling

 Type III: Hindered or Zone Settling

11
Type I Settling
Discrete Settling

   Settling of discrete, non-flocculating particles
   Found in grit removal tanks!
   Particles settle as individual entities at a
constant velocity
   Minimal interaction between particles
   Applies only to particles in a suspension with
a low solids concentration

12
Type I Settling
Critical Settling Velocity
   vo= critical settling velocity
   A particle starting at the top of the inlet zone with a
settling velocity of vo will just reach the bottom of the tank
at the beginning of the outlet zone

13
Type I Sedimentation
Analysis
   Use a batch settling column

   Withdraw samples from a fixed height “h” at time
intervals and measure the solids concentration

   Calculate the weight fraction remaining

   Calculate the settling velocity for the particles at each
of the time intervals (vs = h/t)

   Plot the weight fraction remaining versus velocity
(cumulative distribution curve)
14
fo
1
F  (1 f o ) 
vo
 v df
s
Total Removal
o
Type II Settling
Flocculant Particle Settling

   Particles coalesce as they settle

   Rate of settling (vs) changes with time
   Particles change in size, shape and weight as
they settle
   Larger particles have a higher vs

16
Type II Settling
Analysis
   Use a batch settling column with multiple sampling ports
   Withdraw samples from each port at time intervals and
measure the solids concentration
   Calculate the weight fraction removed
   Prepare a plot of sampling depth versus time and
indicate the weight fraction removed for each of the
samples
   Draw the “equal percent removal” lines at intervals of
10% on the plot

17
r3

r2

r1
Type I and Type II Settling
Clarifier Dimensions

   A = L x W = surface area of the basin (m2)

   Default aspect geometry:
   L = 4W
   A = 4W2

19
Type I and Type II Settling
Scouring Velocity
   Re-suspension of particles due to large horizontal velocities
(u)

Q
u
   Where:
HW
   u = horizontal velocity (m/s)
   Q = water flowrate (m3/s)
   HW = cross-sectional area (entry area) in the direction of flow
(m2)

      To prevent scouring, u < (9 * vo)

20
Type III Settling
Hindered or Zone Settling
   Occurs in solutions with a very high solids
concentration

   Strong cohesive forces between the particles
cause them to settle collectively as a zone

   Distinct interface between the settled
particles and the clarified effluent

21
Type III Settling
Clarifier Design

   Secondary clarifiers need to be designed for
two purposes:

 Clarification
 Thickening

22
Type III Settling
Secondary Clarifier Design
1.       Calculate the area required for clarification

Qe
vo 
Ac
        Where:
    vo = initial zone settling velocity at the feed concentration (X),
[m/h], (function of X)
    Ac = surface area for clarification [m2]


    Qe = overflow rate of clarified liquid [m3/h]

23
Type III Settling
Secondary Clarifier Design
2.       Calculate the area required for thickening
•        Find the gravity mass flux

Gg  v i X i
        Where:
     Gg = gravity flux [M/L2•T] (kg/(m2•h))
     vi = settling velocity at solids concentration Xi [L/T] (m/h)
     Xi = local concentration of solids [M/L3] (kg/m3)


24
Type III Settling
Secondary Clarifier Design
2.       Calculate the area required for thickening
•        Find the bulk mass flux due to underflow pumping

Qu
Gu  ub Xi                      ub 
A
        Where:
     ub = bulk downward velocity of the solids [L/T] (m/h)
     Qu = underflow flowrate [L3/h] (m3/h)
     A = surface area of settling tank [L2] (m2)
                                        25
Type III Settling
Secondary Clarifier Design
2.   Calculate the area required for thickening
•    Find the total mass flux
    Plot G, Gg, Gu

G  Gg  Gu
G  X i v i  X i ub
26
QX o
Limiting Flux   AT 
GL



27
Type III Settling
Secondary Clarifier Design
   Identify which area is greater:
   Area for clarification
   Area for thickening

   Use the larger area to size the clarifier
   Adesign = 1.75*Acalculated
   For an ideal clarifier, L = 4W

28
Designing for a Specific Underflow Solids Concentration (Not Given ub)
Batch Bacterial Growth Curve
1.       Lag Phase
    Acclimation to environment

2.       Exponential Growth Phase
    Multiplication at max rate
    Rapid utilization of S

3.       Stationary Phase
    Growth is offset by death

4.       Death Phase
    Depletion of S
    Decrease in X due to cell
death                     30
Metcalf and Eddy; Figure 7-10
WW Treatment
Bacterial Growth Rate

dX
 X  kd X
dt
   Where:
   X = biomass concentration (mass/volume)
 = specific growth rate (time-1)



   kd = endogenous decay coefficient (time-1)

31
Bacterial Growth in Biological WW Treatment
Monod Kinetics
   Specific growth rate
increases as the
concentration of the
limiting substrate S
increases
    = specific growth rate
(time-1)
   max = maximum specific
growth rate (time-1)

m axS
   S = concentration of the
growth limiting substrate
                           (M/V)

S  KS                Ks = half saturation
constant (M/V)
32
Michaelis-Menten Continued
   Derivation carried out in class!
[S ]
vo  vm ax
S   K M
   Now analogously, if our “product” are cells
the above specific rate of cell formation is
as described as:
m axS

S  KS                               33
Estimation of Kinetic
Parameters
   It is possible to estimate the kinetic
parameters (Ks, kd, max, Y) from bench-scale
CSTR process data in order to design
biological waste treatment facilities

   Perform tests starting with a known limiting
substrate concentration So
   Measure X and S at various residence times ()

34
CSTR With No Recycle
Problems
   If the kinetic parameters (Ks, kd, max, Y) are known, and
you are given So, Q, and one additional variable (i.e. S,
U, V), then you can solve for the rest

35
AS Design Equations
V                   VX                 Ks  S
      c  c  Q X  Q X  c  S(   k ) k K
Qo               E E   w R          max   d   d s

 c Y(So  S)        K s(1 c kd )             So  S
X                S                        U
 1 kd c       c (  m ax  kd ) 1           X
 1  S                       
 max          1                      (K s  So )(1 R  R)
          kd         YU  kd  w 
 c Ks  S          c                      So (  max  kd ) kd K s


 maxXSV              Qr    XR
BiomassWasted  QE X E  QwX R           kd XV R        
Ks  S                Qo    X

 m axSX
                                  m axS
RXR  (1 R)X              kd X  0 R  (1 R)             kd  0
Ks  S                              Ks  S
F Qo  So   S                            
So  S
         o               
Efficiency        100%
M V  X  X                              So
COD Mass Balance
Oxygen Consumed
    When working in COD units, you can always perform a mass
balance

COD substrate in = COD substrate out + COD biomass out + O2
consumed

 O2 consumed = COD substrate in - COD substrate out -
COD biomass out

 O2 consumed = COD substrate consumed - COD biomass
out

37
Conversion Factor (f)
   Required to determine the oxygen requirements if
the substrate concentration is expressed in terms of
BOD5
BOD5 converted
f 
UBODconverted

   If you are given the influent substrate concentration
in terms of BOD5 and UBOD, you can calculate “f” to

determine the effluent substrate concentration in
UBOD units
38
AS - Aeration Requirements
   Air supply requirements can be expressed in a
variety of units
   kg O2/day, kmol O2/day, m3 O2/day, m3 air/day
   Conversion factor: 22.4 m3 gas/kmol (@ STP)
   Air contains ~21 % O2
   If the oxygen transfer efficiency of the aeration
system is known or can be estimated, the air
requirements may be determined

ALWAYS DESIGN AERATION SYSTEMS WITH A
SAFETY FACTOR OF 2
39
SVI Determination
   Take a sample of MLSS from the aeration basin
   Settle for 30 min (usually in a 2 L container with a
diameter larger than a graduated cylinder)
   Determine the volume and mass of the settled solids

Y         Y, settled volume of sludge (mL)
SVI            X, mass of settled solids (g)
X         Typical range: 50 - 150 mL/g

6
10
Xr              Units of mg/L
SVI                              40
TF Design Equation
Se                  n 
A 
 exp  z    
K
Si      
       Q  

   z = depth of the packing media/bed [ft]
   Q = applied flow (Qo + Qr) [MG/D]
   A = filter bed cross-sectional area = π•r2 [Acres]

   n, K = constants; f(packing media);
   See table 6.11

41
   Design based on the rate-limiting step - breakdown of
volatile fatty acids (VFAs)
   Non-biodegradable fractions of COD remain unchanged
by the digestion process
   Heterotrophic bacteria only decays and the COD
associated with decay will be accumulated as VFAs
available to the methanogens
   Complete hydrolysis and fermentation of biodegradable
organic matter -> fully available to methanogens
   Use the kinetics for the growth of the methanogens to
determine the minimum SRT, then use this value with a
safety factor to determine the operating conditions   43
Minimum SRT Calculation

K vfa  Svfa,available
m in 
Svfa,available  (m ax,m  k d ,m )  K vfa  k d ,m

   Where:
    umax,m = maximum specific growth rate for the
methanogens
    Kd,m = decay rate for the methanogens

44
Factor of Safety for Growth
   It is necessary to provide a factor of safety for
methanogen growth (prevent “stuck” digester) and
   Use a factor of safety of at least 2.5

design  2.5  min
   The ministry of the environment requires at least 15
days SRT at 35 C
   Compare with your calculation and select the larger value

                                                                  45
Heterotroph Mass Balance
   Assume there is no growth - only decay
   Perform a mass balance on the digester for the
heterotrophic bacteria:

X H ,o
XH 
1 kd , H   c
   As the SRT increases, the amount of active
heterotrophic biomass in the effluent decreases

                                                    46
Debris Mass Balance
   Debris (XD) can enter the digester in the influent
(XDo) stream and is also generated during biomass
decay
   Perform a debris mass balance on the digester :
 k   
X D  X Do  f d  X H,o   d ,H   c

 kd , H  c 
1
   Where fd = debris fraction of the degraded biomass
(fd ranges from ~ 0.08 - 0.20)

                                                            47
VFAs for Methanogens
   Multiple Sources:
   Soluble biodegradable COD (Ss)
   Biodegradable particulate COD (Xs)
   Decay of heterotrophic biomass

 k   
Svfa,available    Ss  X S  (1 f d )  X H ,o  d ,H   c

 kd , H  c 
1

48
Effluent VFA and Formation of
Methanogenic Bacteria
   CSTR without recycle

K vfa  (1  c k d ,m )
Svfa   
 c (m ax,m  k d ,m ) 1

(Svfa,available  Svfa )
X m  Ym 
                      1  c  k d ,m
49
Methane Production
   COD balance can be performed in order to
determine the amount of methane produced
   CODin = Q(SSo + XSo + XHo + XDo)
   CODout = Q(Svfa + XH + Xm + XD)

CODin = CODout + CODmethane produced

CH4 + 2O2  CO2 + 2H2O
   64 g COD/mol CH4
50
Methane Production
   Use the ideal gas law to calculate the volume
produced per day (V=nRT/P)
   Textbook example 7-9, p. 633 - Effect Of Temp!
   Volume of methane produced per day:

mCH 4 RT
FCH 4 
64P
   Where mCH4 is mass-COD of CH4 produced/time
51
Breakpoint Chlorination

Dose

52
Estimating the Kill Efficiency
   A correlation often used to estimate the amount of
residual chlorine required to achieve a certain kill
efficiency is:

 1  0.23Ct 
Nt                3
   Collins Model
N0
   N = number of organisms
   [C] = total chlorine residual, mg/L; [t] = time, minutes

53
Chlorine as a Disinfectant
   Reliable
   Cheap
   Simple
   Provides a stable residual

   Limitations:
   Extremely toxic and corrosive
   Can influence water taste and odor
   Forms trihalomethanes by reacting with organic
matter in the WW - these compounds are known
carcinogens                                    54
Nitrification
   In a suspended growth (AS) process, low-rate
extended aeration conditions lead to nitrification
   Involves two types of autotrophic bacteria:
   Nitrosomonas oxidizes ammonia to nitrite
NH3 + 3/2 O2  HNO2 + H20
   Nitrobacter oxidizes nitrite to nitrate
HNO2 + 1/2 O2  HNO3
   Overall conversion:
NH3 + 2 O2  HNO3 + H20

55
Nitrifying Systems
   The AS design algorithm can be used to develop
strategies to remove ammonia from the effluent
through nitrification
   Ammonia is used as the substrate in the AS design
model (in place of the biodegradable organics in the
influent (BOD))
   The systems are designed to facilitate longer solids
retention times
   Accounts for the slower growth the the nitrifying bacteria

   In practice, typically a plug flow system is used
(results in coupled PDEs)                                        56
One-Step Process

Two-Step Process

57
Metcalf and Eddy; Figure 7-19
Summary of the BEPR Process
   In the anaerobic zone of the system:
   PAOs take up the VFAs from the liquid phase
   Phosphorus is released to the liquid phase due to polyphosphate
cleavage to provide energy for VFA transport
   Glycogen is utilized and transformed into PHAs to provide enough
reducing power to drive the transformation of VFAs into PHAs

   At the end of the anaerobic period:
   No VFAs left
   Large phosphorus concentration in the liquid phase
   PAOs will have low intracellular glycogen and high PHA contents
58
Summary of the BEPR Process
   In the aerobic zone of the system:
   PAOs use stored PHA as a substrate for growth and for replenishing the
glycogen pool
   PAOs use stored PHA for phosphorus uptake and replenishment of the
polyphosphate pool
   The amount of phosphorus taken up in the aerobic phase is higher than
the amount of phosphorus released in the anaerobic phase
   Net P removal from the liquid phase

   At the end of the aerobic period:
   More PAOs will be present
   Intracellular glycogen content is high, polyphosphate content is high,
PHA content is low
   Soluble phosphorus concentration is very low (can be zero)
59
The BEPR Process

60

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