Lecture 4. Suspended growth treatment systems Secondary Treatment By Husam Al-Najar The Islamic University of Gaza- Civil Engineering Department Advance wastewater treatment and design (WTEC 9320)
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Lecture 4. Suspended growth treatment systems
Secondary Treatment
By
Husam Al-Najar
The Islamic University of Gaza- Civil Engineering Department
Advance wastewater treatment and design (WTEC 9320)
Suspended growth process: It is a biological treatment process in which
microorganisms are maintained in suspension while converting organic matter or
other constituents in the wastewater to gases and cell tissue.
.
Conventio
nal a
ctiv
ate
d s
ludge s
yste
m
Oxid
atio
n d
itches
Sequencin
g b
atc
h re
acto
r (SB
R).
Aera
ted la
goons.
Up flo
w s
ludge b
lanket re
acto
rs
Activated Sludge Principles
• Wastewater is aerated in a tank
• Bacteria are encouraged to grow by providing
• Oxygen
• Food
• Nutrients
• Correct temperature
• Time
• As bacteria consume BOD, they grow and multiply
• Treated wastewater flows into secondary clarifier
• Bacterial cells settle, removed from clarifier as sludge
• Part of sludge is recycled back to activated sludge tank, to maintain bacteria
population
• Remainder of sludge is wasted
Modeling Suspended Growth Treatment Processes
Biodegradable solution COD, Biomass and MLVSS versus SRT in
completely mixed activated sludge system
MLVSS equals the biomass concentration X plus the non-biodegradable
VSS concentration
Key Process Control Parameters
Mean Cell Resident Time (MCRT)
Food-to-Microorganism (F/M) ratio
Sludge Volume Index (SVI)
Specific Oxygen Uptake Rate (SOUR
Sludge (Solids) Wasting
Mean Cell Resident Time (MCRT)
– Is an average measure of how long the microorganisms remain in
contact with the substrate (food source) and is also known as solids
retention time (SRT).
– Used to control the mass of MLVSS in the aeration tank.
– The desired MCRT is achieved by adjusting the sludge wasting and
return rates.
– MCRTs ranging from 3 to 15 days are typical for conventional activated
sludge plants.
– MCRTs less than 3 days will produce a sludge that is young and slow
settling and produce a turbid effluent.
Food-to-Microorganism (F/M) ratio
– is a measure of the mass of food available in the primary effluent per
unit mass of MLVSS per unit time and has units of kg BOD or COD/kg
MLVSS-day.
– Food-to-Microorganism (F/M) ratio is calculated as follows:
F/M = Influent BOD (or COD) kg/day
MLVSS in aeration, kg/day
– The MLVSS represents the concentration of organisms in the aeration
tank.
– COD is often used instead of BOD because test results are available
four hours after sample collection instead of five days for BOD test
results.
– The F/M ratio can be used to control the concentration of MLVSS in the
aeration tank.
– To maintain a MLVSS concentration, the sludge wasting rate will need
to be adjusted.
Sludge Volume Index (SVI)
– Is the volume in mL occupied by one gram of MLSS after 30 minutes of
settling in a 1000 mL graduated cylinder and has units of mL/g.
– The SVI is a measure of the settleability of the activated sludge in a
secondary or final clarifier.
g
ml
MLSS
SVSVI 1000*
SVI = sludge volume index
SV = volume of settled solids in
one liter graduated cylinder
after 30 minutes settling,
MLSS in mg ss/l
Typical values of SVI:-
Typical range of SVI for activated sludge operating at concentrations of
MLSS of 2000 to 3500 mg ss/L is 50 to 150 mL/g.
Notice relation between SVI and MLSS:-
g
ml
MLSS
SVSVI 1000*
When MLSS is increased, SVI decrease, so if MLSS is increased
above 3500 mgSS/L to 5000 mgSS/L for example, SVI decrease
below 50, ml which means bad settling. If MLSS is decreased below
2000, then SVI increase above 150 g
mL leading to bad settling.
Specific Oxygen Uptake Rate (SOUR)
– SOUR is a measure of the quantity of oxygen consumed by
microorganisms and is a relative measure of the rate of biological
activity.
– As microorganisms become more active, the SOUR increases and vice
versa.
– Changes in the SOUR can be used to predict final effluent quality.
– SOUR is determined by taking a sample of mixed liquor, saturating it
with oxygen, and measuring the decrease in oxygen with a DO probe
with time.
– The results of that test, Oxygen Uptake Rate (OUR), measured in mg
O2/L-min, is divided by the MLVSS to yield the SOUR, measured in mg
O2/g MLVSS-hr.
Sludge (Solids)Wasting
– Solids in waste activated sludge (WAS) come from two sources.
• The primary source of WAS is from the growth of new bacterial cells
in the aeration tank.
• The second source is from organic and inorganic solids in the raw
wastewater that pass through the primary clarifiers.
– Sludge is wasted to maintain the desired mass of microorganisms in
the aeration tank. It’s typically wasted when the actual MCRT is
higher than the target value.
– Typical secondary clarifiers thicken the activated sludge to three to
four times the concentration in the aeration tank.
– WAS (and return activated sludge, RAS) MLSS concentrations may
range from 2,000 to 10,000 mg/l (0.2 to 1.0 percent).
– Waste sludge on a continuous basis, changing the WAS rate as needed
by no more than 10 to 15 percent from one day to the next.
Record Keeping
– There are several process parameters that should be monitored daily
and recorded. They include:
• TSS and VSS
• BOD, COD or TOC
• DO
• settleable solids/SVI
• temperature
• pH
• clarity
• chlorine demand
• coliform group bacteria
It is important to understand the way by which the bacterial growth can be quantified.
The most famous equation used to describe the rate of bacterial growth is the Monod
equation:
SK
S
s
m
……………. (1)
m = maximum growth rate, T-1
S = concentration of the limiting
substrate, mg/L
Ks = half saturation constant, mg/L
Bacterial growth kinetics (Monod equation)
2
m
m
Ks Limiting Substrate (mg/l) S
It was also found that the rate of change of biomass (microorganisms) concentration
is governed by the following expression:-
xdt
dx ……………. (2)
) gor r growth= biomass growth rate, mg/L.t (it is also called r dx
dt = growth rate constant, t-1
X = concentration of biomass, mg/L
Substitute from (1) into (2):
SK
Sx
dt
dx
s
m
……………. (3) ( gr
dt
dx )
equation (3) accounts for growth only. To account for death or decay of biomass
another term is subtracted as follows:-
SK
Sx
dt
dx
s
m
- Kd X ……………. (4)
The rate of substrate utilization by the biomass is expressed using the following relation:-
- ds = 1 dx ……………. (5)
dt Y dt
where, Y= biomass yield, g biomass produced
g S consumed
Kd = endogenous decay rate constant, t-1
Substitute equ. 5 In to equ 4:-
XK
SK
SX
Ydt
dsd
s
m1……………. (6)
Note:-
ds/dt is also given the term: rsu rate of substrate utilization.
rsu=
XK
SK
SX
Yd
s
m1 ……………. (7)
if the decay term is neglected, then:-
dt
ds
SKY
SXr
s
msu
……………. (8)
• During reaction no
flow is allowed in
or out of the
reactor.
• The contents of the
reactor are mixed
completely, so the
concentration of
microorganism and
pollutants
Types of reactors used for wastewater treatment
Batch reactor Completely – mixed reactor Plug – flow reactor
(or tubular reactor)
• Wastewater flows
continuously in and out of
the reactor.
• The content of the reactor is
completely mixed and the
concentration of
microorganisms (biomass)
and pollutants are the same
every where inside the
reactor.
• Wastewater flows
continuously in and out
of the reactor.
• No mixing in the reactor,
fluid particles pass
through the tank and are
discharged in the same
sequence they inter.
• The concentration of
biomass and pollutants
is high at the inlet of the
reactor and low at the
outlet
Modeling of biological treatment kinetics
Modeling batch reactor
Qin = 0 Qout = 0
V
Accumulation = inflow mass – outflow mass + generation of mass or substrate
V = Q in So – Q out S + rsu V dt
ds
So from eq. (8) → surdt
ds
SKy
XSr
s
msu
SKy
XS
dt
ds
s
m
, by integration:-
tY
XSSS
SK m
t
t
s
0
0ln ……………. (9)
So = initial substrate concentration at t = 0.0
St = substrate concentration at time t, mg/L
t = time, days.
Equation (9) is used for the design of batch reactors
Modeling completely mixed reactors
Biomass – mass balance:-
dt
dXV = QXO - Q X +V rg
XKSK
XSr d
S
mg
(from eq. 4)
Thus
XK
SK
XSVQXQXV
dt
dxd
s
m0
we assume that Xo = 0.0, and this equation is simplified to
d
s
m KSK
S
V
Q
or →
d
s
m KSK
S
1……………. (10)
Q = flow rate, volume / time.
V = volume of the reactor,
= hydraulic detention time = V/Q,
X = concentration of biomass in the reactor, mass/volume.
So = concentration of substrate in the influent mass/volume.
S = concentration of substrate in the tank and in the effluent.
Qin Qout
Substrate mass balance:- dt
dsV = Q So – QS + Vrus
SK
XS
Y
VQSQSV
dt
ds
SKy
XSr
S
m
s
sum
0
At steady state
dt
ds= 0.0, thus the above equation becomes:-
Q
Vbut
SK
XS
QY
VSS
S
m ,0.00
Thus 0.00
SK
XS
YSS
S
m……………. (11)
If equation (10) is rearranged in the following form:-
SK
SK
S
d
m
11, and this term is substituted in equation(11)
then eq.(11) Becomes:-
d
m
m KXY
SS
11
0Rearranging:-
dK
SSYX
1
0……………. (12)
Y obs is the observed biomass yield, which is the actual increase rate of
biomass. If Kd is assumed to be 0.0, then Yobs = Y. But usually Kd has a
Value > 0.0, and Yobs is <Y.
The endogenous rate
KdT= Kd20 (1.029) T-20 d-1
Kd20 =0.24/d
(valid between 12 and 30oC)
Modeling completely mixed reactors with solids recycle: This process is
called the activated sludge system.
Biomass – mass balance:-
geerw VrXQXQQXVdt
dx 0
dt
dx= rate of change of microorganisms in the reactor,
V = reactor volume,
Q = WW flow rate,
Xo = concentration of microorganisms in the influent,
Qw = Waste flow rate,
Xr = concentration of biomass in the return line,
Qe = effluent flow rate,
Xe = concentration of biomass in effluent,
rg = net rate of microorganisms growth (mass/unit volume time)
Assuming steady state conditions then → dt
dx= 0.0, and assume Xo= 0.0
and substitute for rg from equation (4), the above equation simplifies to:-
XK
SK
SXVXQXQ d
s
m
eerw
d
s
meerw KSK
S
VX
XQXQ
……………. (14)
The left hand side of equation (14) is the inverse of the mean cell residence time:-
eerw
cXQXQ
VX
so equation (14) becomes:-
d
s
m
c
KSK
S
1……………. (15)
* Substrate-mass balance:-
suew VrSQQQSVdt
ds 0 where?
)( SKY
SXr
s
msu
At steady state dt
dS = 0.0
SQSQSKY
SXVQS ew
s
m
)(0
where? we QQQ
QS
SKY
SXVQS
s
m
0 rearrange
1
)(0
V
Q
SKY
SXSS
V
Q
s
m
then ?
SK
S
X
SSY
s
m
0 ……………. (16)
x
BOD
Diffused Aeration System
Modeling of plug – flow reactor with solids recycle:-
Biological Nitrification
Nitrification is the conversion (by oxidation) of Ammonia (NH4-N) to nitrite (NO2-N) and then to nitrate (NO3-N).
The need for nitrification arises from water quality concerns:
Effect of ammonia on receiving water; DO demand, toxicity.
Need to provide nitrogen removal for eutrophication control.
Need to provide nitrogen removal for reuse applications.
The current drinking water MCL for nitrate is 45 mg/l as nitrate or 10 mg/l as nitrogen.
The total concentration of organic and ammonia nitrogen in municipal wastewater is typically in the range of 25-45 mg/l as nitrogen.
Primary Clarifier
Combined BOD removal and nitrification
Secondary Clarifier
Return Activated Sludge
Influent Effluent
Prim
ary Slud
ge
Waste Slu
dge
Organic matter and nitrification removal in the same bioreactor
Primary Clarifier
BOD removal
Secondary Clarifier
Return Activated Sludge
Prim
ary Slud
ge
Waste Slu
dge
Return Activated Sludge
Waste Slu
dge
Nitrification
Nitrification
Clarifier
Organic matter and nitrification removal in separate bioreactor
Ammonia oxidizing activity
NH4++ 3/2O2 → NO2
-+ H2O + 2H+ + Energy
Nitrite oxidizing organisms
NO2-+ 1/2O2 → NO3
-+ Energy
Nitrafication Catabolism
The biological oxidation of free and saline (NH3 and NH4+) ammonia by two
obligate aerobic activities:
Nitrification kinetics
SK
S
s
m
n n
n
N
N
µmn = Max. growth rate
µn= Growth rate at any concentration
N = concentration of the Ammonia nitrogen, mg/L
Kn = half saturation constant, mg/L
Effluent Ammonia- Nitrogen versus SRT
SRTmin= minimum sludge age
for nitrification
• Above certain SRT
Effluent Ammonia is very low.
• As SRT decreases
Effluent Ammonia increases
• When SRTMIN ≈ SRT,
Effluent = Influent Ammonia
SRT = SRTm
SRT > SRTm
Sludge age (SRT) is the most important design parameter for systems required
to nitrify
Temperature effects
Nitrifier kinetic constants μnM20, Kdn20 and Kn20 all dependent on temperature:
μnMT = μNM20 (Θn) (T-20) Θn=1.123
KnT = Kn20 (Θn) (T-20) Θn=1.123
KdnT = Kdn20 (Θd) (T-20) Θd =1.029
μNM20 values range between 0.3 -0.75 /d.
Kdn20 is accepted to stay constant (0.04/d).
Oxygen demand = 4.57 x Q (Ni-N)
The effect of Temperature on Minimum Sludge Retention Time (SRT)
Oxidation ditches (OD)
Oxidation ditches are type of suspended growth systems. It is a modification of the
conventional activated sludge system.
Characteristics of oxidation ditches:
A. Configurations:-
The oxidation ditch consists of a ring or oval – shaped channel. It is some times called
closed loop Reactor (CLR), and some times called Racetrack channel. The oxidation ditch
may have a trapezoidal or rectangular cross section.
The wastewater is re-circulated in the "CLR" using brush rotors (Kessners brush), which is
also used for aeration.
There are many configurations of oxidation ditches as shown in the figures.
The velocity of flow in the OD is maintained at 0.25 – 0.3 m/s to keep the biomass
in suspension. At this velocity, the mixed liquor completes a tank circulation in 5 –
15 min, leading to the dilution of the influent by 20-30 times.
The influent of raw sewage is introduced just upstream of the aerator (Rotor).
The effluent weir is located just upstream of the influent
effluent
Aerobic
Rotor (aerator)
influent
Anoxic
Difference between OD and conventional Activated sludge:-
Oxidation ditches were developed to minimize the net sludge
production compared to the conventional activated sludge system.
Net sludge production minimization is achieved by using law F/M
ratio (0.02 – 0.15dmgvss
mgBOD
5 ). In this case the active biomass is
forced to feed on the decaying biomass due to the shortage of
food. This leads to lower sludge production, and the sludge to be
wasted will be less and has lower organic content (i.e. more
stabilized)
OD are operated at high c (15-30 days) and at high (15-36 hrs).
It is theoretically possible in OD to minimize the net sludge
production to zero. This can be achieved by making the produced
biomass equal to the degraded biomass by endogenous decay (i.e.
biomass feeding on dead biomass).
)( 0 SSQK
YXV
d
)( SSK
YX o
d
This is presented mathematically as follows:-
Sludge produced = YQ (So – S)
Sludge decaying = Kd XV
Net production Px = YQ (So – S) – KdXV
Let net production (Px ) = 0.0
So YQ (So – S) = Kd XV
And or
This equation can be used to find X and V that can be used to a chive
zero net sludge production.
Processes that can be achived in oxidation ditches:-
Three processes can be achieved in oxidation ditches:-
Organic matter removal (BOD removal) in the aerobic zone.
Nitrification (in the aerobic zone).
Denitrification (in the anoxic zone).
At the influent to the OD, we have organic matter in addition to
nitrate (
3NO ) coming from the aerobic zone and the dissolved O2
is almost zero. This is called anoxic condition where
denitrification occurs.
At the end of the anoxic zone and the beginning of the aerobic
zone, we have the remaining organic matter that was not used for
denitrification in addition to ammonium (
4NH ) coming in the
influent in addition to O2 introduced by the aerator.
In this condition both BOD removal and nitrification occurs. At
the end of the aerobic zone the dissolved oxygen becomes almost
zero.
Advantages of oxidation ditches:
* low sludge production can be achieved due to using low M
F ratio
* The produced sludge, if any is stable and needs no further treatment. This
means that no sludge treatment installations are needed.
* no need for primary sedimentation, because the high c in the oxidation
ditches is enough to digest the solids that is usually separated in the primary
sedimentation tank.
* easy to operate and the operation and maintenance cost is much less than
conventional activated sludge.
* Ability to nitrify and denitrify in one tank
Sequencing Batch Reactors (SBR):
Sequencing Batch reactors are suspended growth activated sludge system.
The main difference between SBR and conventional sludge system is that in
the later process in continuous (CMFR) while in the SBR it is interment.
Hydraulic model of SBR:
SBR are designed as batch reactors. The reactor is filled, then time is
allowed for reaction to occur. During the reaction the reactor is completely
mixed. The design equation of this system is presented in Unit 2.
For SBR the following equation is applied:
tY
XSSS
SK m
to
t
o
s
ln
Sequencing Batch Reactors (SBR) Process
SBR process:
The SBR process is a fill and draw process. This process has five steps
as shown in the figure:
Fill
React (Aeration)
Settle (sedimentation)
Draw (decant)
Idle ( )
The following is a description of the five steps:
1. Fill:
It is the process of adding raw sewage to the SBR tank.
The fill volume is determined so that the added “Q” rises
the volume from 25% VT to 100%VT.
Typical time needed for the fill step is 25% of the cycle
time.
The volume addition is controlled by automatic valves or
timers.
Interment aeration is needed in this step is needed to prevent an aerobic
conditions
2. React:
The purpose of this step is to start the aerobic reactions by
applying oxygen and complete mixing. In this step both
organic matter removed and nitrifications achieved.
The volume of the tank during this step is100% full with
wastewater .
The time needed for this step is typically 35% of the cycle
time. This time should be checked using the batched reactor
design equation .
3. Settle:-
The purpose of settle step is to allow solids separation to
occur providing a clarified supernatant to be discharged
as treated effluent . It is a sedimentation step .
The settle step is controlled by [using automatic timers] ,
it takes CycleThrtohr %2012
1 .
During this step no mixing or aeration is applied.
3. Draw:
The purpose of this step is to remove the clarified treated
wastewater from the reactor.
Draw ( is achieved by floating decants or ( غ
automatic adjustable weirs.
Draw time is 15 % T cycle ( typically 45 minutes ).
The volume is reduced to 35% TV
No aeration or mixing is applied during this step.
To prevent solids from leaving with the effluent, it is usually preferred to
add an extra volume above the sludge blanket.
4. Idle:
The purpose of idle step in a multi-tank system( i.e. more than
2 tanks)is to provide time for one reactor to complete its fill
cycle before switching to another tank. Idle is not a necessary
step, and can be eliminated.
Aeration and mixing can be applied to prevent anaerobic
conditions, depending on the idle time.
Idle time is%5of cycle time or longer in some cases. For example if the
flow “Qin” is minimum and the other tank is in the fill phase is not
receiving it’s design “Qin”, then the tank in the idle step has to wait until
the first tank completes the fill step.
Sludge Wasting:
Sludge wasting is not a separate step, it can be done in the idle step, or
during the react step if the idle step is eliminated.
Sludge Recycle:
No sludge recycle is needed since sedimentation occurs in the biological
reactor, so sludge is already there.
Cycle time in SBR:
The cycle time is the total time needed to complete the five steps
mentioned above:
T cycle = tƒ + tr + ts + td + ti
tƒ = fill step time Note: there is a relation between
tr = react step time tƒ and tr, ts and td:
ts = settle step time 1
n
tttt dsr
f
td = draw step time
n= number of SBR tanks used.
ti = idle step time
Typical cycle time is 4-8hrs.
At least two SBR tanks are needed
No final sedimentation is needed
No sludge recycle is needed
If c is <20 days, no primary sedimentation tank is needed
SBR tanks are square tanks in which mL 305
Typical depth = 5m.
Advantages of SBR:
Biological reactions and final sedimentation is achieved in one
tank, so we do not need final sedimentation tank.
No need for sludge recycle pumping station.
If c is <20 we do not need primary sedimentation tank, and
the wasted sludge is stable.
( Note: M
F ratio is similar to that of oxidation ditches, i.e. 0.02-
0.15 dmgvssmgBOD ./5 )
Appendix
Example:-
A completely mixed activated sludge system is to be used for organic matter and
nitrogen removal. Design this system knowing the following:-
Flow = 0.15 m3/s = 12960 m3/d
BOD influent = 84 mg/L (Soluble)
required effluent → (total BOD) = 11.1 mg/L
TKN = 40 mg/L (in the influent of the reactor)
TKN = 1 mg/L (effluent nitrogen goal).
Required → NO3- - N in the effluent = l mg NO3
- - N/L and Do = 39 mg NO3- - N/L)
Heterotrophic microorganisms growth constants are:
= 2.5d-1 Kd = 0.05d-1 Y = 0.5 mg VSS/mg BOD removed KS=100 mg BOD/L
Autotrophs for nitrifies microorganisms growth constants are:
μmax = 0.25d-1 Yn = 0.2 mg Vss/mg NH4-N K d = 0.04d-1 Kn = 0.4 mg/L
The denitrifying bacteria have the following growth constants:
m
LNmgNOK
dKmgNo
mgvssYd
Dn
dDn
D
m
/16.0
04.0,9.0,4.0
3
1
3
1
1)(
)1(
dmc
cds
K
KKS
105.05.2
05.011001.11
c
c
m
m
c
m
c
c
408.0
5
For organic matter removal
1. Calculate C:-
(this is equ. (18) for CMFR)
Solve for C C = 5 days (mean cell residence time)
- Kd = 2.5 – 0.05 = 2.45
= 0.408 d (minimum sludge retention time)
= 12.25 d (2<12.25<20) OK
mc
1=
m - Kd
So S.F = =
dm
sK
KK
d
05.05.2
05.0100
cd
c
K
SSYX
1. 0
505.01
1.11845.053000
2. Check for smin:-
S > Smin → o.k (11.1 > 2.04) So use c = 5 days, S = 11.1 mg BOD/L
3. Calculate the aeration tank volume:-
Assume the concentration of biomass (X) is equal to 3000 mg MLVSS (Mixed liquor
volatile suspended solids)/L:-
(equ. 17)
(solve for Ө) = 0.0486 day = 1.17 hours
S min = = 2.04 mg BOD/L
Calculate the volume:- V = Q = 0.0486*12960 ≈ 630 m3
Check F/M ratio:-
= 0.576 mg BOD/mg MLVSS .d (O.K)
Typical range for conventional activated sludge system is 0.1 – 0.6 mg BOD/mg
MLVSS .d. (Mixed liquor volatile suspended solids)
This F/M is accepted. In case that we need to change F/M we can change the
assumed X.
L
m
m
L
L
mg
mg
L
md
m
MF
3
3
3
3
3
3
10
1.
1
10.
84.
3000.
630
1.
12960
4. Calculate the amount of sludge to be wasted:
P x = Y obs Q (So – S)
dKgP
mg
Kg
m
L
L
BODmg
d
m
mgBoD
mgP
mgBoDmgVsscK
YY
x
biomass
biomassx
d
obs
/378
10
1.
10.
.1.1184129604.0
/4.0505.01
5.0
1
63
3
5
3
5
RXX
X
Q
Q
r
r
(Sometimes called or recycle ratio)
R= 43.0300010000
3000
dmQQr /557343.0 3
dKgOR
d
kg
mg
kg
m
L
L
mgBoD
d
m
PSSQR x
/408
37842.110
1.
1
10.1.118412960
42.1
20
63
3
5
3
00
6. Calculate the Oxygen required:-
5. Calculate the recycle flow Q r:-
d
L
lmg
dmg
X
PQXQP
r
x
wrwx 37800/000.10
/10378 6
Calculate Qw (waste sludge flow):-
Assume Xr = 10000 mg VSS/L (Typical range: 8000 – 12000 mg VSS/L)
Px = QW Xr + Qe Xe, (neglect Xe compared to Xr)
QW = 37.8 m3/d
BOD5 removal NH3
removal
Q = 12960 m3/d
Qr Qr
Qw Qw = 37.8 m3/d
Q = 12922 m3/d
Continue for nitrification
Solution:
1. Calculate c for complete nitrification:-
Smin = Kn
073.0
04.025.0
04.04.0
)(
)(
ndnm
nd
k
k
mg N/L <1 mg N/L (OK)
2. The flow interring the nitrification CMER is:
Q\
= Q – Qw = Qe from the BOD removal CMFR
Q\ = 12960 – 37.8 12922 m
3/d
3.Calculate for the nitrification CMFR:-
Since no BOD removal occurs in this CMFR, only nitrifies are active in
this reactor, this can be understood from this equation:-
)(16.0)(6.0
)(16.0
NNSS
NNf
OO
On
=
)(0.0)( removalBODnoSSbut O
0.1nf
)3.004.0(.
3.0 rangeTypicaldVssmg
TKNmg
M
Ftake
hrsdQ
V
mLV
dmgvssXtake
mgvssdmgvssmgN
L
Nmg
m
L
d
m
VX
MF
QNoVX
VX
QN
M
F
13.2089.012922
1149
114910*11491500
10*723.1
/1500
10*723.1./3.0
40*10
**12922
/
339
93
33
0
daysisrangetypicaldays
KX
NNY
c
c
c
cd
c
1001054)*04.01(1500
1402.0*089.0
)1(
)( 0
)( NNQYP Oobsx
4. find C:-
5. Calculate the sludge to be wasted:-
dmdLLmgvss
dmgvssQ
LmgXassumeX
PQ
dkgvssmg
kg
m
LP
NmgmgvssK
YY
w
r
Nitrifiersr
xw
x
cd
obs
/3/3000/000.10
/10*30
/10000
/3010
1.
1014012922*06.0
/06.054*04.01
2.0
1
36
63
3
6. Calculate oxygen requirements:-
Ro = 4.57 Q (NO - N)
(Note:- this is the oxygen needed for nitrification only)
mg
kg
d
L6
3
10
1*)140(10*12922*57.4
Ro = 2303 kg O2/d
7. Calculate the volume of air to be supplied:-
At standard conditions i.e → T = 20 oC, pressure = l atm, air density =
1.185 kg/m3
% oxygen by mass in air = 23.2%.
Assuming 100% oxygen transfer efficiency:
dm
mkg
dkgO
O
RQ air
air
air /8377232.0*/185.1
/2303
%*
3
3
2
2
0
Assume 8% oxygen transfer efficiency:-
dmQ airair /10471308.0
8377 3
If pure oxygen to be used:-
dmR
Q O
air
oxysen /1943185.1
2303 3
20
Assume 8% oxygen transfer efficiency:-
dmQ Ooxysen /2428808.0
1943 3
2
So the separate stage nitrification system will look as the following
V = 630 m3
V = 1149 m3
37.8m3/d 3m3/d
Continue for Denitrification
The procedure is the same as that followed in BOD removal,
except that we do not need oxygen for denitrification.
We need to add organic matter, because denitrifies are
heterotrophic bacteria.
1. Calculate cmin
Smin (or Dmin)
NmgNoK
KKDS
d
KDK
D
d
Dn
m
dDn
c
d
oDn
oDn
m
c
3minmin
min
min
04.004.04.0
04.0*4.0
78.2
36.004.03916.0
39*4.0
1
2. Calculate c:-
1)(
)1(
d
Dn
mc
cdDn
K
KKD
dc
c
c 28.3104.04.0
04.0116.01
c > cmin
O.K Cheek factor of safety:-
218.178.2
28.3.
min
c
cFS
not o.k Take S.F = 2.1
c = 2.1 cmin
= 2.1*2.78 5.84 days o.k
3. Calculate :-
Assume X = 3000 mg MLVSS/L
dK
DDY
X cd
oDnc 054.084.5*04.01
1399.0.
3000
84.5
1.
1.3 hrs o.k
4. Calculate V:-
Q = 12922 – 3 = 12919 m3/d
V = Q = 12919 * 0.054 = 697.6 698 m3
Find Px , QW, Qr by the same way as in example 1
BOD5 NHn+ NO3- 12960Q =
Qr
Qr Qr
Qw Qw
Qw
12922Q = 12919Q =
Example Oxidation diches design:-
Design an oxidation ditch for BOD removal only. The following Data are
given:
influent BOD = 300 mg BOD/L ( soluble)
effluent BOD =15 mg BOD/L (soluble)
Q0 = 20.000 m3/d
Y= 0.5 mgvss/mgBOD , Kd = 0.03d-1
, Ks = 30 mgBOD, 15.2 dm
Assume that we want to operate the OD at Zero net sludge production
solution:
1- Calculate X to achieve zero net production:
)( 0 SSK
YX
d
dL
mgvss
L
mgBODd
mgBOD
mgvssX
s
.475015300.03.0
.5.0 5
2- check for M
F
dmgvss
mgBOD
X
S
M
F s
.063.0
4750
3000
3- Assume in the typical range ( 15-36 hrs),
take = 1 day (24 hrs)
dL
mgvssX .4750
X = 4750 mgvss/L , typical range for OD is (2500 – 6000), O.K.
4- In this example we do not need to check for c because we
assumed that no sludge wasting will take place, and theoriticaly c
5- find the volume of the oxidation ditch
= V 33
200001*20000 mdd
mQ
Notice that the volume is very high due to the high
6- Calculate Qr:
R= XX
X
r , assume Xr = 10000 mgvss/L
9.04750000.10
4750
R
Qr = QR = 20.000*0.9 = 18000 m3/d
7- find the oxygen requirements:
dkgOR
mg
kg
L
mgBOD
m
L
d
mSSQR
PxthatnotePxSSQR
/5700
1015300
1020000)(
0.0,42.1)(
20
63
33
000
000
Example Oxidation ditches :
Repeat the Design in previous example, assuming that we want to allow
for some sludge waste, by using a sludge age ( )c in the range 15 – 30
days.
Solution:
In this case design the oxidation ditch as a CMFR and use the equation of
CMFR. The difference between the conventional CMFR and OD is the
design parameters typical ranges ),,,(M
FXc
a- Assume ,c = 30 days, assume = 15 hrs ( 0.625 days).
(note: in oxidation ditches we allow S.F above 20)
b- Calculate S:
78.0/36.003.05.2
03.0*30
/78.01)03.05.2(30
)30*03.01(30
5min
LmgBODK
KKS
LmgBODS
dm
ds
s
1)(
)*1(
mc
c
d
ds
K
KKS
c- calculate X:
LmgvssX
K
SSYX
cd
c
/378030*03.01
78.03005.0.
625.0
30
)1(
)( 0
The typical range of X is 2500 to 6000 mgVSS/L, O.k
d- check for M
F
dmgvss
mgBOD
Lmgvss
LmgBOD
x
S
M
F
.121.0
/3780*625.0
/300 550
within the range
(0.02 – 0.15) O.K
e- calculate V:
V=Q = 20000 * 0.625 = 12500 m3
F- calculate sludge production:
dmd
L
X
PQ
dkgmg
Kg
m
L
d
mP
mgBOD
mgvss
K
YYSSQYP
r
xw
x
cd
obsobsx
/6.155155600000.10
10*1556
/155610
.10
)78.0300(20000*26.0
26.030*03.01
5.0
1,
36
63
33
5
0
Example SBR:
Design an SBR system to achieve both 5BOD removal and
nitrification. The following Data is available:
)(BOD soluble = ,3500,7500,1503
L
mgssX
d
mQ
L
mg
L
mgNK
L
mgssXeiX nrs 5.0,000,10).(
11
5
05.0,44.0,12.0 dKdmgBOD
mgvssY dn
n
mn
mgBOD
mgvssY 5.0
105.0 dK d
L
mgBODK s 50
15.2 dm
Assume:
ts= 0.5hrs (typical)
tD= 0.5hs (typical)
tr = 1.0hrs (typically 1-2hrs)→Should be checked by the
batch reactor design equation.
1. Determine SBR operating cycle:
Tcycle = tƒ + tr + ts + tD (ti = 0.0, not needed)
1
n
tttt Dsr
f assume n =2 SBR tanks, if the dimension of SBR
are within 305 L o.k., other wise more than 2 tanks are needed.
hrst f 212
5.05.00.1
hrsTc 45.05.00.12
2. Determine number of cycles per tank per day:
No cycles =dk
cycles
hrs
hrs
tan6
4
24
3. Determine fill volume per cycle per tank:
fill
m
cyclesNo
nQ
VF
30
6256
27500
4. DetermineT
F
V
V fraction:
VFill = fill volume
Vs = settle volume
VT = VF+Vs
ssT XVXV →s
Ts
X
XVV Note: VFill = Vdecant
TTs VVV 35.0000,10
3500 Vdecant = volume of treated
TTTF VVVV 65.035.0 wastewater disposed in each cycle.
Since : VF = 625m3
396265.0
625
65.0m
VV F
T
5. Determine the surface area of each SBR:
Assume the depth of each tank is 5m (typical depth)
23
1925
962m
m
m
depth
VA T
surface
mL 90.13192
m3090.135 so two tanks are o.k
6. Determine the portion of heterotrophic and nitrifiers
microorganisms and c :
X= Xnitrifiers + Xhetrotrophs = Xn+Xh
X=L
mgVss28008.03500
Xn= Tcd
ch
Tcdn
cn
VK
SeSQYX
VK
NeNQY
1
)(,
1
)(
3
3
1096205.01
)5.040(12.01037502800
c
c
3
3
1096205.01
)10150(5.0103750
c
c
L
mgVssc
c
c
c
c
178
)05.01(
9.272
05.01
5.182800
daysc 5.18
L
mgVssX
c
cn 178
5.1805.01
5.185.18
05.01
5.18
L
mgVssX
c
ch 2623
5.1805.01
5.189.272
05.01
9.272
Or L
mgVssX h 26221782800
7. Check for the reaction time(tr):
Use the batch reactor design equation:
For nitrification:
tY
XNNN
NK
n
n
m
nto
t
o
n )()(ln
No = concentration of nitrogen in the SBR after dilution resulting from
mixing VFill in VTotal:
ToFill VNVN →L
mgN
V
VNN
T
Fillo 26
962
62540
Nt = 0.5 mgN/L (the required nitrogen influent).
t)12.0
44.0(178)5.026(
5.0
26ln5.0
hrdayt 1042.0 (so tr = 1.0 is o.k. as assumed)
Check for BOD removal:
tY
XSSS
SK m
hto
t
o
s )()(ln
L
mgBOD
V
VSS
T
Fill
o
55.97962
625150
L
mgBODSt
510 (the required BOD in the effluent)
t
5.0
5.22623)105.97(
10
5.97ln50
t = )0.1(0.137.00154.0 hrthrhrsd r
* Note:
tr for nitrification always control the design of SBR.
8. Calculate sludge procedure:
day
KgVssXVP
c
x 146105.18
1096228006
3
d
m
X
PQ
s
x
w
3
3
6
18108000
10146
If Qw is taken during the react step:
d
m
X
PQ x
w
3
3
6
52102800
10146
9. Calculate oxygen requirements:
)(57.442.1)( einxeino NNQPSSQR
kperd
KgR
o
o tan995)5.040(10
10375057.414642.1)10150(
10
103750
2
6
3
6
3
10. Check
dmgVss
mgBOD
Vx
QS
M
F o
5
3
3
21.0109622800
150103750
Typical range is(0.02-0.15), this is not in the range.
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