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5.2 Design Criteria and Parameter

For general design purpose, kinetic expressions used for design were summarized in Table 5.2.

Because reported organic removal and nitrification kinetics cover a wide range, bench scale or

in-plant testing should be undertaken to evaluate site-specific organic removal and nitrification

kinetics value. Higher maximum specific growth rates than indicated in Table 5.2 could be found

or testing could reveal reaction rates and a significant nitrification inhibition problem. Aeration

tank volume and sludge residence time values are directl related to nitrification µm values. All

kinetics parameter in the table are adapted from !chobanoglous et al. "#$$%&.

Table 5.2. Activated sludge nitrification kinetic coefficients

Coefficient Unit Range Typical value

µmn g '(()g '((.d $.#$-$.*$ $.+K n g NH4-N/m

3 $.-.$ $.+

Y n g VSS/g NH4-N $.$-$. $.#k dn g '(()g '((.d $.$-$. $.$/K o g/m

3 $.$-$.0$ $.$

Ɵvalue µn unitless .$0-.#% .$+K n unitless .$%-.#% .$%k dn unitless .$%-.$/ .$

!he computation approach used in the design of the activated sludge process for B12 removal

and nitrification "ammonia-nitrogen& are as follows "!chobanoglous et al., #$$%&3

a& 1btain influent wastewater characterization data

b& 2etermine the effluent re4uirements in terms of 5H-5, !(( and B12 concentrations

c& (elect an appropriate nitrification safet factor for the design (6! based on expected

peak)average !75 loading. (afet factor ma var from .% to #.$

d& (elect the minimum 21 concentration for the aeration basin mixed li4uor. A minimum

21 concentration of #.$ mg)l is recommended for nitrification

e& 2etermine the nitrification maximum specific growth rate " µm& based on the aeration

basin temperature and 21 concentration and determine K n

f& 2etermine the net specific growth rate 8 and (6! at this growth rate, to meet the effluent

5H-5 concentration


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g& 1btain the design (6! b appling the safet factor to step f

h& 2etermine the biomass production

i& 9erform a nitrogen balance to determine 51x, the concentration of 5H-5 oxidized

:& ;alculate the '(( mass and !(( mass for the aeration basin

k& (elect a design


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%& !he point of air release for the diffusers is $. m above the tank bottom

& 21 in aeration basin ? #.$ mg)l

& Aeration ? $.$ for B12 removal onl and $.0 for nitrification C ? $.* for both

conditions, and diffuser fouling factor F ? $.*$

0& Dse kinetics coefficients given in !able .

+& 2esign


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μn=(1.28 g

g.day .10g /m3

1.12 g

m3+10g /m3 )( 2g /m30.50 gm3 +2g /m3 )−0.1095

8n ? $.%+# g)g.da

'tep 2.2etermine the theoritical and design (6!

a& Find theoritical (6!

(6! ?

1

μn=

1

0.372 g

g . day

=2.7day

b& 2etermine the design (6!

F( ? !75 peak)!75 average ? .

2esign (6! ? "F(& "theoretical (6!& ? . "#.+ da& ? da

'tep $. 2etermine biomass production

P x , bio=QY (So−S)1+( k d ) SRT

+ ( f d )( k d )Q (Y ) (So−S ) SRT 1+(k d ) SRT

+ Q Y n( NO x)1+( k dn ) SRT

............................"e4. ii&

a& 2etermine input data for "e4. ii&

Q ? $$$ m%)da

Y ? $.$ '(()g b;12

S o ? .0 "B12& ? .0 "#$$ mg)l& ? %#$ g)m

%

K d ? k #$.E!-#$ ? $.# ".$/-#$ ? $.0 g)g.da

8m! ? 8m.E!-#$ ? 0.$ g)g.da ".$+/-#$ ? $.% g)g.da


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S= K s [1+( k d ) SRT ]SRT ( μm−k d )−1

?

20 g/m3 [1+(0.164 gg .day )4day ]4day (10.31 gg .day−0.164 gg .day )−1

=0.836 g bCOD

m3

Y n? $.# g '(()g 51x

K dn? $.$* g)g.da

51x ? $./ "!75& ? $./ "%$$ mg)l& ? #$ g)m%

b& (ubstitute the value and solve for P x,bio

P x , bio=

(1000m3

d )(0.4

gVSS

gbCOD )(320

g

m3−0.836

g

m3)

1+(0.164 gg . d)4 d +(0.15 ) (0.164 )1000

m3

d (0.4 gVSSgbCOD )(320 m1+(0.164 gg . d)4 d

'tep (. 2etermine the amount of nitrogen oxidized to nitrate.

!he amount of nitrogen oxidized to nitrate can be found b performing a nitrogen balance

using "e4. iii&.

NO x ? !75 N e $.# P x,bio)G ...........................................................................e4. iii

NO x=300 g

m3−10

g

m3−0.12( 99000

g

day

1000 m3

day )=278.12 gm3

'tep 5. 2etermine the concentration and mass of '(( and !(( in the aeration basin


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P x ,TSS= P x , bio0.85

+Q (nbVSS )+Q (TSS0−VSS0 )=99kg

VSS

day

0.85+1000

m 3

day (16.8 gm3 )+1000 m3day (10 gm3 b& ;alculate the mass of '(( and !(( in the aeration basin


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? %*.0 kg)da

a& 1bserved ield based on !((

Y obs,TSS= P

x , TSSbCOD "'mo('d=

143.27 kg

day319.16

kg

day

=0.45 kgTSSkgbCOD=0.45 gTSSgbCOD x (1.6 g bCODg%OD )

=0.

b& 1bserved ield based on '((

Y obs,VSS= P x ,TSS

bCOD "'mo('d=

143.27 kg

day

319.16 kg

day

=0.45 kgTSS

kgbCOD=0.45

gTSS

gbCOD x(0.8 gVSSgTSS )=0.36

'tep ,. ;alculate the oxgen demand using "e4. iv&

R0=Q ( S0−S )−1.42 P x , bio+4.33Q ( NO x ) .........................................................."e4. iv&

R0=1000

m3

day (320−0.836 )

g

m3−1.42(99 kgday )+4.33 (1000 m 3day )( 278.12 gm3 )(10−3 kgg )=57.618 ko

'tep 1-. 2etermine air flowrate at average design flowrate

a& 2etermine the (1!6

SOTR= )OTR [ C s , 20* # ( + C ́s ,T , −C ) ] (1.02420−T )=57.618 [ 9.08

g

m3

0.65(0.9)(0.95 .(11.93 gm3 )−2.0 gm b& 2etermine the air flowrate

0.270 kg O2m3

ai"

(0.35)(60 min

o!")¿

¿¿

)i" f-o"at' , m3

min=

SOTR¿


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'tep 11. 2esign review summar

Table 5.(. 2esign review summar

Design Parameter Unit alue

>astewater flow m3/d $$$

Average B12 load kg)d #$$

Aerobic (6! 2

Aeration tank volume m3 $

Hdraulic detention time H $./

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