Chapter_3.pdf

Chapter 3 Suspended growth treatment systems (Aerobic & Anaerobic) 3.1 Introduction:Many treatment systems based on suspended microorganisms have been developed and still used until now. Some of these systems are aerobic and other are anaerobic. Some of the most commonly used systems are:* Activated sludge systems:• Conventional activated sludge system. • Oxidation ditches. • Sequencing batch reactor (SBR) • Aerated lagoons. • Waste stabilization ponds. • Up flow anaerobic sludge blanket (UASB) 3.2 conventional activated sludge system:* The first version of activated sludge systems is called the conventional activated sludge system. This system is composed of two parts:a. Aeration tank. b. Final sedimentation tank. * The aeration tank in this system can be designed either as a complete mixed flow reactor (CMFR) or as a plug flow reactor (PFR).

Influent

Treated Aeration tank TTTTTTTTTTTTTT

flow

Waste Sludge

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Plug-flow aeration tank equipped with dome aeration devices.

A. Design of activated sludge system as a (CMFR). a. design of the aeration tank:• To design the aeration tank we need to find:1. Tank volume. 2. Recycle ratio. 3. Sludge wasting. 4. Oxygen requirements. 5. Check for some parameters such as θc ,θ, and S. • The aeration tank can be used to:1. Remove BOD only. (one sludge) or (separate stage) 2. Nitrify only (convert NH4+ to NO3-) (one sludge or separate stage) 3. Remove BOD + nitrify (two sludge or single stage).

Example 3.1:A completely mixed activated sludge system is to be used for organic matter removal only (one sludge system). Design this system knowing the following:-

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* microorganisms growth constants are:µ m = 2.5d-1 , Kd = 0.05d-1 , Y = 0.5 mg VSS/mg BOD5 removed KS = 100 mg BOD5 / L * Flow = 0.15 m3/s = 12960 m3/d BOD5 = 84 mg/L (Soluble) * required effluent → (BOD5)total = 30 mg/L Suspended solids (SS) = 30 mg/L Solution:1. Since we always deal with soluble substrate, first we need to find the effluent soluble BOD5 :(BOD5) soluble = S = (BOD5)total - BOD5 in suspended solids or paticulate (BOD5) in suspended solids = 63% * SS (BOD5)particulate = 0.63 * 30 = 18.9 mg BOD5 /L (BOD5)soluble in effluent = 30 – 18.9 = 11.1 mg BOD5 /L *Note: (BOD5)soluble in effluent = S = (BOD5) soluble in the aeration tank. Follow Example 3.1:2. Calculate θC:K s (1 + K d θ c ) S= (this is equ. (18) for CMFR) θ c (µ m − Kd ) − 1 100[1 + 0.05 ∗ θ c ] 1.11 = ⇒ Solve for θC θ c [2.5 − 0.05] − 1 θC = 5 days • Check for minimum sludge retention time θCm:= µ m - Kd = 2.5 – 0.05 = 2.45 θ cm = 0.408 d θc 5 = 12.25 d So S.F = m = 0.408 θc

(2<12.25<20) OK

• Check for smin:-

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Smin = K s

Kd  0.05  = 100  =2.04 mg BOD5/L µ m − Kd  2.5 − 0.05 

S > Smin →o.k (11.1 > 2.04) So use θc = 5 days, S = 11.1 mg BOD5/L 3. Calculate the aeration tank volume:• Assume the concentration of biomass (X) is equal to 3000mg MLVSS /L:θ Y(S0 − S ) X= c. (equ. 17) θ (1 + K d θ c ) 5 ∗ 0.5[84 − 11.1] 3000 = (solve for Ө) θ [1 + 0.05 ∗ 5] θ = 0.0486 day θ = 1.17 hours • Calculate the volume:V = θ Q = 0.0486*12960 ≈ 630 m3 • Check F/M ratio:12960m3 1 84mg 10 3 L 1m3 L F = . . . . . M 630m3 3000mg L 1m3 10 3 L d F

= 0.576 mg BOD5 /mg MLVSS .d (O.K) M Typical range for conventional activated sludge system is 0.1 – 0.6 mg BOD5 /mg MLVSS .d. this F/M is accepted. In case that we need to change F/M we can change the assumed X. 4. Calculate the amount of sludge to be wasted: Px = Yobs Q (So – S)

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Yobs =

Y 0.5 = = 0.4mgVss / mgBoD5 1 + K dθc 1 + 0.05 ∗ 5

Px = 0.4

3 mg biomass m3 ∗12960 [84 − 11.1] mg.BOD5 .10 3L . 6 1Kg mgBoD5 d L m 10 mg biomass

Px ≅ 378 Kg / d Another way to find PX:XV mg 630m3 103 L kg Px = Qw Xr + Qe Xe = = 3000 ∗ . . θc 1m3 5d 106 mg L Px = 378kg / d • 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) P L 378 ∗ 10 6 mg / d Px = Qw X r ⇒ Qw = x = = 37800 Xr d 10.000mg / l 3 QW = 37.8 m /d 5. Calculate the recycle flow Qr:Qr X = = R (Sometimes called ∝ or recycle ratio) Q Xr − X 3000 = 0.43 R= 10000 − 3000 Qr = 0.43Q ≅ 5573m3 / d 6. Calculate oxygen requirements:R0 = Q(S0 − S ) − 1.42 Px 3 m3 [84 − 11.1] mgBoD5 .10 3L . 16kg − 1.42 ∗ 378 kg d L d 1m 10 mg R0 = 408 KgO2 / d

= 12960

Example 3.2:For example 3.1, we need to design the CMFR system for both organic matter removal and nitrification. The microorganisms growth constants

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for hetrotrophs are the same as in example 3.1, and for nitrifies (i.e. autotrophs) are:μmax = 0.25d-1, Yn = 0.2 mg Vss/mg NH4-N, Kd = 0.04d-1 Kn = 0.4 mg/L It is also given that:TKN = 40 mg/L (in the influent of the reactor) TKN = 1 mg/L (effluent nitrogen goal). Solution:1. It was calculated in example 3.1 that θc required for BOD5 removal was = 5 days. • We need to check if this θc is enough for to achieve complete nitrification. •

Find Smin for nitrogen:(kd ) n 0.04 Smin = Kn = 0.4 = 0.035 mg N/L <1 mg N/L (µ m )n − (kd ) n 0.5 − 0.04 (OK) • Calculate θc for complete nitrification:N=

K n [1 + (K d )n θ c ] θ c µ m n − Kd −1

[(

) ( )] n

0.4(1 + 0.04 * θ c ) θ c (0.25 − 0.04) − 1 Solve for θc ⇒ θc = 7.2 days So (θc )n > (θc )BOD5 ⇒ so take θc = 7.2 days for design purposes. 1=

2. Cheek for (θcmin)n:1 = µ m n − (K d )n = 0.25 − 0.04 ⇒ θ cmin = 4.76days m θc 7.2 S.F = = 1 . 5 < 2 not OK 4 .76 So take S.F = 2.1 ⇒ θc = 2.1 X 4.76 ≅ 10 days So take θc = 10 days

( )

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3. Calculate the actual S and N in the effluent:0.4(1 + 0.04 *10) 0.56 N= = = 0.51 mg N / L > 0.035mgN / L OK 10(0.25 − 0.04) − 1 1.1 100 (1 + 0.05 *10) 150 S= = = 6.38 mg BOD5 / L > 2.04 mg BOD5 / L OK 10 (2.5 − 0.05) − 1 23.5 4. Calculate θ:X nitrifiers

= 0.10 (this ratio is called nitrifies fraction ƒn ) XTotal So Xnitrifiers = 0.1 x 3000 = 300 mg Vss/L Xheterotrophs = 0.9 X 3000 = 2700 mg Vss/L * θ for heterotrophs:θ Y ( So − S ) 10 * 0.5(84 − 6.38) 388.1 θ= c = = = 0.103 d X (1 + K d θ c ) 2700(1 + 0.04 *10) 3780 *θ for nitrifies:10 * 0.2 (40 − 0.51) 78.98 θ= = = 0.188 d 300 (1 + 0.04 * 10) 420 θ for nitrifiers > θ for heterotrophs, so take θ = 0.188 d = 4.5 hours Assume thjat

5. Calculate the volume of the reactor:V = Q θ = 12960 * 0.188 = 2436 m3 Compare this volume with the 630 m3 needed for BOD5 removal only. Note:Xnitrifiers

= ƒn, use the following equation: Xtotal 0.16(N 0 − N ) fn = , where N0 = TKN in the influent, and N 0.6(S0 − S ) + 0.16(N0 − N ) = TKN in the effluent. To find

6. Calculate the sludge to be wasted:* for hetaotrophs:10 6 L 1kg Px = Yobs Q ( S o − S ) = 0.4 *12960 (84 − 6.38) • 3 • 6 m 10 mg

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P x = 402 kg Vss/d * for nitrifiers:Y 0.2 Yobs = = ≅ 0.143 mg Vss / mgN 1 + kd θ c 1 + 0.04 *10 Px = 0.143 *12960 (40 − 0.51)

103 l 1kg . m3 10 6 mg

p x = 37.8 kg Vss / d PX 37.8*10 6 mg / d (QW ) N = = = 37800 L / d ( X r ) N (10.000) * 0.1 mg / l 402 *10 6 mg / d (Qw ) BOD 5 = = 44667 L / d 10.000 * 0.9 mg / l Total Qw = 37800 + 44667 = 82467 L/d ≅ 83 m3/d 6. Calculate oxygen requirement:Ro = Q ( SO − S ) − 1.42 PX + 4.57 Q ( NO − N ) = 12960 (84 − 6.38) *

1kg 10 3 l 103 l 1kg . − 1 . 42 ( 402 + 37 . 8 ) + 4 . 57 * 12960 ( 40 − 0 . 51 ) . 10 6 mg 1m3 m3 10 6 mg

RO =1005.96 − 624.5 + 2338.9 ≅ 2720 Kg O2 / d

Example 3.3:For example 3.2, we need to design a separate stage CMFR for nitrification only. The Data for the nitrifies and TKN are the same as in Example 3.2. Q = 12922 m3/d Q = 12960 m3/d

BOD5 removal

NH3 removal

Qr

Qr Qw

40 3

Qw = 37.8 m /d PDF created with pdfFactory Pro trial version www.pdffactory.com

Solution:1.Calculations for θ cmin , Smin for nitrifies:From example 3.2 it was found that:θ cmin = 4.76 days, Nmin = 0.035 mg N/L N = 1.0 mg-N/L. 2. The flow interring the nitrification CMER is: Q\ = Q – Qw = Qe from the BOD5 removal CMFR Q\ = 12960 – 37.8 ≅ 12922 m3/d 3.Calculate for the nitrification CMFR:Since no BOD5 removal occurs in this CMFR, only nitrifies are active in this reactor, this can be understood from this equation:0.16 ( NO − N ) fn = 0.6 (SO − S ) + 0.16 ( NO − N ) ⇒ but ( SO − S ) = 0.0 (no BOD5 removal ) So ⇒ fn =1.0 F mg TKN ⇒ take = 0.3 (Typical range 0.04 − 0.3) M mg Vss.d

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QNo F QN0 = ⇒ VX = F /M M VX

VX =

N m3 10 3 L * 3 * 40mg L = 1.723 *10 9 mgvss d m 0.3mgN / mgvss.d

12922 *

take X = 1500 mgvss / d V=

1.723 *10 9 ≅ 1149 *10 3 L = 1149 m3 1500

V 1149 = = 0.089 d = 2.13 hrs Q 12922 4. find θC:θ Y( N0 − N ) θ= c X (1 + K d θ c ) θ=

0.089 =

θ c * 0.2[40 − 1] ⇒ θ c = 54 days 1500(1 + 0.04 *θ c )

(typical range is 10 − 100 days )

5. Calculate the sludge to be wasted:Px = Yobs Q ( NO − N ) Yobs =

Y 0.2 = = 0.06mgvss / mg − N 1 + K d θ c 1 + 0.04 * 54

10 3 L 1kg Px = 0.06 *12922[40 − 1] 3 . 6 = 30 kgvss / d m 10 mg Px Qw = ⇒ assume X r = 10000 mg / L ( Xr )Nitrifiers Qw =

30 *10 6 mgvss / d = 3000 L / d = 3m3 / d 10.000mgvss / L

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6. Calculate oxygen requirements:Ro = 4.57 Q (NO + N) (Note:- this is the oxygen needed for nitrification only) L 1kg = 4.57 *12922 *10 3 (40 − 1) * 6 d 10 mg 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: R0 2303 kgO2 / d 3 Qair = /d = = 8377 mair ρ air * [O2 %] 1.185 kg / m3 * 0.232 •

Assume 8% oxygen transfer efficiency:8377 3 ≅ 104713 mair Qair = /d 0.08 • If pure oxygen to be used:R 2303 Qoxysen = 0 = = 1943 mO3 2 / d ρ air 1.185 Assume 8% oxygen transfer efficiency:1943 Qoxysen = = 24288 mO3 2 / d 0.08 •

So the separate stage nitrification system will look as the following:-

V = 630 m3

V = 1149 m3

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Example 3.4:- Denitrification For the system designed in Examples 3.1,3.2, and 3.3, design a separate stage denitrification completely mixed flow reactor (CMFR). The denitrifying bacteria have the following growth constants:−1 mgvss µ mD = 0.04d −1 , YDn = 0.9 , K d = 0.04d mgNo3 K Dn = 0.16mgNO3− − N / L • Required → No3- - N in the effluent = l mg No3- - N/L (So = (No3- - N)o = Do = 39 mg No3- - N/L) Solution:• The procedure is the same as that followed in example 3.1, except that we do not need oxygen for identification. • We need to add organic matter, be cause denitrifies are heterotrophic bacteria. 1. Calculate θcmin Smin (or Dmin) Do 1 39 = µ mDn − Kd = 0.4 * − 0.04 ≅ 0.36 min K Dn + Do 0.16 + 39 θc θ cmin = 2.78 d Smin = Dmin = K Dn

Kd 0.04 = 0.4 * ≅ 0.04mgNo3− − N µ − Kd 0.4 − 0.04 Dn m

2. Calculate θc:K Dn (1 + Kd θ c ) D= θ c ( µ mDn − Kd ) − 1 Follow Example 3.4:0.16[1 + 0.04θ c ] 1= ⇒ θ c = 3.28d θ c [0.4 − 0.04] − 1 θc > θcmin O.K Cheek factor of safty:-

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θc 3.28 = = 1.18 < 2 not o.k min 2.78 θc Take S.F = 2.1 θc = 2.1 θcmin = 2.1*2.78 ≅ 5.84 days o.k S.F =

3. Calculate θ:Assume X = 3000 mg MLVSS/L θ Y [D − D ] 5.84 0.9[39 − 1] . θ = c . Dn o = 0.54d X 1 + K d θ c 3000 (1 + 0.04 * 5.84 ) θ ≅ 1.3 hrs o.k

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4. Calculate V:θ = 12922 – 3 = 12919 m3/d V = Qθ = 12919 * 0.054 = 697.6 ≅ 698 m3 *Find Px , QW, Qr by the same way as in example 3.1. Q =12922 Q =12960

Q =12919 NHn+

BOD5

NO3-

Qr Qr Qw

Qr Qw

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Qw

b. Design of the final (Secondary) clarifier:The final sedimentation tank or clarifier, is an essential part in the activated sludge system. It is needed for gathering (by settling) the sludge and returning part of it to the system. The following parameters are used to design this tank:1. The overflow rate:- (or hydraulic loading) it is the amount of flow in m3/d applied to the unit area (m2) of the sedimentation tank and it's units are:Qo − Qw (As = Surface area) As Q O / F = e (note: Qr interring the settler is pumped from the bottom) As • Some times we ignore Qw

O/F =

•

The typical range of O/F is (20-34)

the unit is •

m3 m2 • d

m m3 or 2 d m •d

In this range we expect good separation of solids from liquid in the final sedimentation tank.

Q0 + Qr

Qe = Q0 - Qw

Qr

Qw

2. The weir loading rate:It is the amount of flow in m2/d applied to the unit length (m) of the effluent weir. The weir is the circumference of the sedimentation tank from which the wastewater leaves the tank. • The typical range of weir loading (WL) is:-

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WL →125 − 250

m3 or m• d

(

Qe m2 ) , (WL = ) d weir length

3. The solids loading rate (SL):It is the amount of solids in (Kg) applied to the unit area of the settling (Q + Qr ) • X) tank per day. ( SL = o As • Typical range of SL is 130 – 300 Kg/ d.m2. If SL is higher than Kg 300 the suspended solids will increase in the effluent of the d .m2 settling tank. Final settling basin side water depth Tank diameter, m <12 12 to 20 20 to 30 30 to 42 >42

Minimum 3.0 3.4 3.7 4.0 4.3

Side water depth, m Recommended 3.4 3.7 4.0 4.3 4.6

Source: Joint Task Force of the Water Environment Federation and the American Society of Civil Engineers, Design of Municipal Wastewater Treatment Plants Vol. I, Manual of Practice No. 8, Chapters 1-12, Alexandria, VA, 1992

Photograph of sludge collector for circular sedimentation basins.

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Schematic diagram of sludge collector for circular sedimentation basins.

Effluent channel of circular sedimentation basins.

Example 3.5:Design the secondary clarifier (final settling tank) for the CMFR in Example 3.1. Solution:(Qo = 12960 m3/d) , ( Qw = 37.8 m3/d) * find As:Take O / F = 33

m d

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Qe As Q m3 1 As = e =12922 . ≅ 392 m2 O/F d 33 m / d (note: Qe = Qo – Qw = 12960 – 37.8 = 12922 m3/d)

O/F =

* find the diameter:π d2 As = 4 4 * 392 d= = 22.3 m II ww surface

weir

Channel

outlet pipe

SWD

Qe d H Inlet pipe

Qr+Qw

Qo+Qr

Sludge outlet pipe

weir d= Efflueat channel

sludge pipe

ww inlet pipe

treated water pipe

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* Select a side wall depth (SWD): From the final setting basin side water depth:For (d) in the range 20-30 m → SWD = 4.0 m * find (H), the depth of the inclined bottom depth:The typical slope of the bottom is 1:12, So:H 1 = ⇒ H = 0.93 m 11.15m 11.15m 11.15 12

H

Take H = 1.0 m * Check for the solids loading rate SL:[Q + Qr ]* X = (12960 + 5573)m3 . 3000mgvss . 103 L . 1kg SL = o As L m3 10 6 mg 392m2 SL ≅ 142kg / d .m2 (in the range130 − 300) o.k. *Check for the weir loading rate (WL):Qe 12422 m3 / d m3 WL = = ≅ 184 weir length π (22.3)m m• d 3 Typical range of WL is 125-250 m /m.d

OK

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* Sludge volume index SVI:* This parameter is used to check the quality of settling and the efficiency of the secondary settling tank in activated sludge systems. SVI is also used to control the concentration of the biomass in the biological reactor (X) and the concentration of the biomass in the return sludge (Xr). *SVI is the volume in milliliters (ml) occupied by 1g of activated sludge after the aerated liquor has settled 30 minutes and calculated as follows:SV mg SVI = * 1000 X g

1000 ml

1000 ml

30

Min SV Sludge Volume: ml Where, mL g SV = volume of settled solids in one liter graduated cylinder after 30 mL minuts settling, L X = biomass concentration in the biological reactor such as (CMFR), mgSS/L→ (MLSS). SVI is related to Xr by the following relation:106 Xr = mgss / L SVI SVI is related to Qr (recycle flow) as follow:X Q R= = r Xr − X Q

SVI = sludge volume index,

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 X  Q = Q   Xr − X    X   Qr = Q 6 10 − X SVI   * Typical values of SVI:• Typical range of SVI for activated sludge operating at concentrations of MLSS (X) of 2000 to 3500 mg ss/L is 50 to 150 mL/g. • Notice relation between SVI and X:SV mg SVI = * 1000 X g When X is increased, SVI decrease, so if X is increased above 3500 mgSS/L to 5000 mgSS/L for example, SVI decrease below 50, ml which means bad settling. If X is decreased below 2000, then SVI mL increase above 150 leading to bad settling. g Example 3.6:For example 3.1 find the SVI, SV. MLVSS Given that MLVSS = 0.8 MLSS ⇒ MLSS = 0.8 Solution:X r =10.000 mg vss / L (or 10.000 mg MLVSS / L) from example 3.1 mg Vss 1mg ss X r =10.000 * = 12500 mg ss / L L 0.8 mg vss =12500 mg MLss 10 6 mg MLSS / L SVI 10 6 mL mL SVI = 80 > 50 o.k , good settling g g 12500 Xr =

* find sludge volume SV:-

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SV mg *1000 X g X.SVI SV = 1000

SVI =

X = 3000 mg MLVSS / L = SV =

3000 = 3750 mg MLSS / L 0.8

3750 *80 mL (this is the volume of sludge in one liter of ww after 30 min settli = 300 1000 L

B. Design of activated sludge system as a plug flow (PFR):The conventional activated sludge system can be designed as a PFR. The following is an example to illustrate the procedure used.

Qi Si , Xi

V Q0 – Qw

Q0 S0 X0

S Xe

Q0 + Qr

Qr,S Qw Xr S

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Example 3.7:Solve example 3.1 using a PFR. a. find θc:θ cmin was calculated as 0.408 d , for PFR we find θc from equation 20:µ m ( So − S) 1 = − kd θc  Si  ( So − S ) + (1 + α ) K s In  s  Qo So + Qr S Qi

Si =

α = R=

R=

X 3000 = ≅ 0.43 X r − X 10000 − 3000

Qr ⇒ Qr = 0.43Qo = 0.43 *12960 = 5573m3 / d Qo

Qi = Qo + Qr = 18533m3 / d Si =

12960 * 84 * 5573 *11.1 = 62mgBoDS / L 18533

1 = θc

2.5[84 − 11.1]

[84 − 11.1] + [1 + 0.43]*100In 62  11.1 θc = 1.92 days

0.05 = 0.522d −1

* Check for S.F:θ 1.92 S.F = mc = = 4.71 > 2 O.K θc 0.408 b. Calculate θ from equation 19:θ Y(So − S) X= c , let X = 3000 mg / l θ 1 + Kd θ c

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1.92 * 0.5 (84 −11.1) = 0.0213 d = 0.5 hrs. 3000 (1 + 0.05 *1.92) * it is typically preferred to have a minimum θ of 1.0 hr. To increase θ we can either decrease X or increase θc , or do both things. So, assume θ=

θ = 1 hr = 0.042 d , X = 3000 mg / l , and solve for θc . 3000 =

θc 0.5)84 − 11.1) . ⇒ θc = 4.18 days 0.042 (1 + 0.05 *θc)

* for PFR

θC Should be > 5 θ

4 . 18 = 99 . 5 >> 5 o.k 0 . 042 C. Calculate the reactor volume:V = Qθ =12960 * 0.042 = 544 m3 d. Calculate Px, Qw, Ro the same as in example 3.1. 3.3 Oxidation ditches:- (OD) Oxidation ditches are type of suspended growth systems. It is a modification of the conventional activated sludge system. 3.3.1 . 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 recirculated 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

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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 pipe.

Oxidation Ditch

Oxidation Ditch (empty)

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Oxidation Ditch: carrousel configurations

Typical Oxidation Ditch layout

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B. Hydraulic model:Oxidation ditehs combine features of both PFR and CMFR models: C. CMFR similarity:The rapid flow in the OD results in 20-30 dilutions which gives a considerable amount of mixing. The influent ww is mixed with the rotating ww at the inlet. D. PFR similarity:The OD are long reactors, and thus they have some similarity with PFR a long the reactor. E. Which model is used for OD design? CMFR or PFR?:Since PFR assumes no mixing, this case is not found in OD. So OD is designed as a CMFR. The error in this assumption leads to higher hydraulic detention time, which gives a safty factor in the design. Moreover, OD are designed at high (θ) anyway to achieve sludge stabilization so assuming that OD is CMFR is accepted. 3.3.2. 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 mgBOD5 ratio (0.02 – 0.15 ). In this case the active biomass is mgvss • d 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 θc (15-30 days) and at high θ (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). 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

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So ⇒ YQ (So – S) = Kd XV Y Y And ⇒ XV = Q( S0 − S ) or X = ( S0 − S ) Kd Kdθ This equation can be used to find X and V that can be used to a chive zero net sludge production. 3.3.3 Processes that can be achived in oxidation ditches:Three processes can be achieved in oxidation ditches:− Organic matter removal (BOD5 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 ( NO3− ) 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 ( NH 4+ ) coming in the influent in addition to O2 introduced by the aerator. In this condition both BOD5 removal and nitrification occurs. At the end of the aerobic zone the dissolved oxygen becomes almost zero. influent effluent

Aerobic Anoxic

Rotor (aerator)

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Example 3.8:Design an oxidation ditch for BoD5 removal only. The following Data are given: • influent BOD5= 300 mg BOD5/L ( soluble) • effluent BOD5=15 mg BOD/L (soluble) 3 • Q0 = 20.000 m /d • Y= 0.5 mgvss/mgBOD5 , Kd = 0.03d-1,Ks = 30 mgBOD5, µ m = 2.5d −1 Assume that we want to operate the OD at Zero net sludge production solution: 1- Calculate X to achieve zero net production: Y ( S0 − S) Xθ = Kd mgBOD5 mgvss d mgvss Xθ = 0.5 . .[300 − 15] = 4750 .d mgBODs 0.03 L L 2- check for

F M

F S mgBODs 300 = 0 = = 0.063 M Xθ 4750 mgvss.d 3- Assume θ in the typical range ( 15-36 hrs), take θ = 1 day (24 hrs) mgvss Xθ = 4750 .d L X = 4750 mgvss/L , typical range for OD is (2500 – 6000), O.K. 4- In this example we do not need to check for θ c because we assumed that no sludge wasting will take place, and theoriticaly θ c →∞ 5- find the volume of the oxidation ditch

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V = Qθ = 20000

m3 *1d = 20000m3 d

Notice that the volume is very high due to the high θ

6- Calculate R=

R=

Qr:

X Xr − X 4750 10.000 − 4750

, assume Xr = 10000 mgvss/L

= 0.9

Qr = QR = 20.000*0.9 = 18000 m3/d 7- find the oxygen requirements: R0 = Q0 ( S0 − S) − 1.42 Px , (note that Px = 0.0) R0 = Q0 ( S0 − S) = 20000

m3 10 3 L mgBOD kg • 3 [300 − 15] • 6 d L m 10 mg

R0 = 5700kgO2 / d

Example 3.9: Repeat the Design in example 3.8, assuming that we want to allow for some sludge waste, by using a sludge age ( θ 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 (θ ,θ c , X , F ) M a- Assume θ c , = 30 days, assume θ = 15 hrs ( 0.625 days). (note: in oxidation ditches we allow S.F above 20)

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b- Calculate S: 30(1 + 0.03 * 30) = 0.78mgBODs / L S= 30(2.5 − 0.03) − 1 Kd 0.03 = 30 * = 0.36mgBOD5 / L < 0.78 Smin = K s 2.5 − 0.03 µ m − Kd c- calculate X: θ Y( S0 − S ) X= c θ (1 + Kdθ c )

[300 − 0.78] = 3780mgvss / L 30 .0.5 0.625 1 + 0.03 * 30 The typical range of X is 2500 to 6000 mgVSS/L, O.k X=

d- check for

F M

300mgBOD5 / L mgBOD5 F S0 = = = 0.121 within M θx 0.625 * 3780mgvss / L mgvss.d (0.02 – 0.15) O.K

the

e- calculate V: V=Q = 20000 * 0.625 = 12500 m3 F- calculate sludge production: Y 0.5 mgvss Px = Yobs Q(S0 − S ), Yobs = = = 0.26 1 + K d θ c 1 + 0.03 * 30 mgBOD5 Px = 0.26 * 20000 Qw =

10 3 L Kg m3 (300 − 0.78) 3 . 6 = 1556kg / d d m 10 mg

Px 1556 *10 6 L = = 155600 = 155.6m3 / d Xr 10.000 d

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range

3.3.4 Advantages of oxidation ditches: * low sludge production can be achieved due to using low F

ratio M * 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 θ 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. 3.3.5 OD process flow sheet: O.D Screening

Secondary clarifier Effluent

Grit removal

return sludge

Waste sludge

3.4 Aerated lagoons (AL): Aerated lagoons are suspended growth waste water treatment system. They are not considered as an activated sludge system because no solids recycle is applied. This system ( i.e AL) is a low cost low efficiency treatment system compaired to Activated sludge systems. 3.4.1 Configuration of "AL" Aerated lagoons consist mainly of an earthen basin that has a large surface area and a shallow depth ( 1-3m). The sides slopes are generally 1:3 ( some times (1:2)). The face area is usually square to achieve the best power transfer applied by the mechanical aerators. Surface mechanical aerators are used for both oxygen transfer and complete mixing of the lagoons.

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Mechanical aerator influent

Final clarifier

Effluent Sludge to disposal

Aerated lagoon

Floating mechanical aerator Influent Effluent sludge 3.4.2 hydraulic model of Aerated lagoons: Aerated lagoons are designed as completely mixed reactor without solids recycle. The derivations of the equations of such a system are presented in chapter 2. Qo So

Qe S

S,X Qw S

For "AL" the following design equations apply: Ks (1 + θKd ) Y( S0 − S) ,X = S= θ (µ m − Kd ) − 1 1 + Kdθ these equtions were derived previously ( Ch-2) 3.4.3 Differences between Aerated lagoons and Activated sludge: * Since no solids are recycled in AL, the biomass concentration X is in the range of 100 – 400 mgvss/L , which is too low compared to 1500-

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6000 mgvss/L in activated sludge, As a result, we need much more reactor volume (V) to achieve similar treatment efficiency to that of Activated sludge systems. * Only BOD removal is achieved in Aerated lagoons because the oxygen supply and sludge age θ c , is not enough to achieve nitrification ( θ c in AL is equal to θ and have a typical range of 3-10 days). Note that θ c dose not appear in the equations above , but θ c is equal to θ in this system.

3.4.4 mixing power requirements: The power needed for mixing is usually more than power needed for aeration in aerated lagoons. So we always need to check for mixing requirements using the following equation: P= 0.004 X + 5 Where, Kw 103 m3 X= MLSS in the "AL" , mgss/L

P = power input ,

Example 3.8: Design an aerated lagoon to treat a domestic waste water with a total BOD5 of 400 mgBOD5/L , a TSS of 130 mg/L and a daily flow of 8000 m3/d. Heterotrophic bacteria growth constants are µ m = 2.8d −1 , Ks = 60 mgvss mgBODs/L, Kd= 0.03d-1, Y = 0.5 , assume θ = 5 days. mgBODs Assume that (BOD5)/ Tss=0.63, and MLVSS = 0.8 MLSS , T

Slope = 1:3

h =2.5m

h = 3h

B

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T B

Solution: 1- determine (S0) soluble: (S0)soluble=(BOD5)total – mgBOD5/L

(BOD5)particulate=

400

–

0.63*130 ≅

318

2- calculate S: Ks (1 + θKd ) S= θ (µ m − Kd ) − 1 S=

mgBOD5 60(1 + 5 • 0.03) = 5.37 L 5(2.8 − 0.03) − 1

3- calculate X: X=

0.5(318 − 5.37) mgvss ≅ 136 , typical range 100-400mgvss/L, OK. 1 + 0.03 • 5 L

4- Calculate the volume and surface area required: * V = Qθ = 5*8000 m3/d = 40000 m3/d * Assume the depth of the lagoon as 2.5m, and that the lagoon is square, find the surface area:  T 2 + B2  From the geometry of the AL→ V =   • h , and T = B + 6h  2 

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 ( B + 6 • 2.56)2 + B 2  40000 =   • 2.5 , solve for B→ B ≈ 118.8 m 2   Then, T= 118.8+6*2.5 = 133.8 m A surface or As = (133.8)2 ≈ 17902 m2 Note that this is a very large area. 5. Assume that only 9000 m2 available, what changes should we do? 9000 ≅ 0.50 , so the proposed lagoon 17902 volume (V) = 0.5*40000 = 20000 m3 We can reduce (V) by the ratio

* calculate θ in this case: V 20000 = = 2.5 days Q 8000 * calculate S in this case: mgBOD5 60(1 + 2.5 • 0.03) S= ≅ 10.9 L 2.5(2.8 − 0.03) − 1 * calculate X in this case: 0.5(318 − 10.9 ) mgvss X= = 143 , typical range 100-400, OK (1 + 0.03 • 2.5) L 6. Calculate the sludge production for the first case, when θ = 5 days : PX = Yobs Q( S0 − S) , Y 0.5 mgvss Yobs = = = 0.43 , 1 + kd θ 1 + 0.03 • 5 mgBOD5 θ=

m3 10 3 L 1Kg Kg • 3 [318 − 5.37] 6 = 1075 d d m 10 mg 7. Calculate the oxygen requirements: RO = Q (S0 − S ) − 1.42 PX PX = 0.43 • 8000

RO = 8000

KgO2 m3 10 3 L 1Kg Kg • 3 (318 − 5.3) 6 − 1.42 • 1075 = 975 d d d m 10 mg

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7. Calculate the power needed for oxygen transfer assuming that 1.8 KgO2 requires 1KWh: KgO2 1 1d Power = 975 • • ≅ 23 Kw d 1.8( KgO 2 / KWh ) 24h 8. Check power requirements for mixing: Xvss 136 mgss = = 170 P = 0.004 X ss + 5 , Xss = 0.80 0.80 L Kw P = 0.004 • 170 + 5 = 5.68 3 , but V= 40000 m3 10 m3 5.68 So the total power needed Ptotal = 40000 • 3 = 227 Kw 10 So mixing power controls the design.

3.5

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 (CHFR) while in the SBR it is interment.

3.5.1 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 chapter 2. For SBR the following equation is applied:

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K s ln

So µ  − (So − St ) = X  m t St  Y 

3.5.2 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)‫ﺗﻔﺮﯾﻖ اﻟﻤﺎء اﻟﺮاﺋﻖ‬ Idle (‫)ﻓﺘﺮة اﻧﺘﻈﺎر ﺑﻌﺪ اﻧﺘﮭﺎء دورة اﻟﻤﻌﺎﻟﺠﺔ‬

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 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 .

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•

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] , 1 it takes hr to 1 hr [≅ 20% T Cycle ]. 2 During this step no mixing or aeration is applied.

4. Draw: • • • • • •

The purpose of this step is to remove the clarified treated wastewater from the reactor. Draw (‫ ) ﺗﻔﺮﯾﻎ اﻟﻤﺎء‬is achieved by floating decants or automatic adjustable weirs. Draw time is 15 % T cycle ( typically 45 minutes ). The volume is reduced to 35% VT 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.

5. 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 ≅ 5 % of cycle time or longer in some cases. For example if the flow “Qin” is minimum and the other tank is in

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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.

3.5.3 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.

3.5.4 Sludge Recycle: No sludge recycle is needed since sedimentation occurs in the biological reactor, so sludge is already there. 1.5.5 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: t +t +t tf = r s d ts = settle step time n −1 td = draw step time n= number of SBR tanks used. ti = idle step time Typical cycle time is 4-8hrs. 4. Typical flow chart of SBR treatment plant: ………………………………………………………………………… ………………………………………………………………………… ……………………………… • • • •

At least two SBR tanks are needed No final sedimentation is needed No sludge recycle is needed If θ c is >20 days, no primary sedimentation tank is needed

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• •

SBR tanks are square tanks in which 5 ≤ L ≤ 30m Typical depth = 5m.

3.5.6 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 θ c is >20 we do not need primary sedimentation tank, and the wasted sludge is stable. F ( Note: ratio is similar to that of oxidation ditches, i.e. 0.02M 0.15 mgBOD5 / mgvss.d ) Example 3.9: Design an SBR system to achieve both BOD5 removal and nitrification. The following Data is available: mg m3 mgss • ( BOD5 ) soluble = 150 , Q = 7500 , X = 3500 , L d L mgss mgN X s (i.e X r ) = 10,000 , K n = 0.5 L L mgvss Yn = 0.12 , µ mn = 0.44d −1 , K dn = 0.05d −1 mgBOD5 mgvss Y = 0.5 mgBOD5 K d = 0.05d −1 mgBOD5 K s = 50 L −1 µ m = 2.5d • Assume: ts= 0.5hrs (typical) tD= 0.5hs (typical) tr = 1.0hrs (typically 1-2hrs)→Should be checked by the batch reactor design equation.

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1. Determine SBR operating cycle: Tcycle = tƒ + tr + ts + tD (ti = 0.0, not needed) t +t +t tf = r s D assume n =2 SBR tanks, if the dimension of SBR n −1 are within 5 ≤ L ≤ 30 o.k., other wise more than 2 tanks are needed. 1.0 + 0.5 + 0.5 = 2hrs 2 −1 Tc = 2 + 1.0 + 0.5 + 0.5 = 4hrs tf =

2. Determine number of cycles per tank per day: 24hrs cycles No cycles = =6 4hrs tan k • d 3. Determine fill volume per cycle per tank: Q0 7500 m3 n 2 VF = = = 625 No cycles 6 fill V 4. Determine F fraction: VT VFill = fill volume Vs = settle volume VT = VF+Vs V X VT • X = Vs • X s → Vs = T Xs

Note:

VFill =

Vdecant 3500 • VT = 0.35 VT 10,000 treated VF = VT − 0.35VT = 0.65VT in each cycle. Since : VF = 625m3 V 625 VT = F = ≅ 962m3 0.65 0.65 Vs =

Vdecant = volume of wastewater disposed

5. Determine the surface area of each SBR: Assume the depth of each tank is 5m (typical depth)

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Asurface =

VT 962m3 = = 192m2 depth 5m

L = 192 ≅ 13.90m so two tanks are o.k. S ≤ 13.90 ≤ 30m 6. Determine the portion of heterotrophic microorganisms and θ c : X= Xnitrifiers + Xhetrotrophs = Xn+Xh mgVss X= 3500 • 0.8 = 2800 L QY( S − Se)θ c QYn ( N − Ne )θ c Xn= , Xh = [1 + Kdnθ c ] [1 + Kd θ c ]VT

and

nitrifiers

3750 • 10 3 • 0.12(40 − 50)θ c 3750 • 10 3 • 0.5(150 − 10)θ c + [1 + 0.05θ c ] 962 • 10 3 [1 + 0.05θ c ]962 • 10 3 18.5θ c 272.9θ c mgVss 2800 = + ≅ 178 [1 + 0.05θ c ] (1 + 0.05θ c ) L θ c ≅ 18.5 days 18.5θ c mgVss 18.5 • 18.5 Xn = + ≅ 178 L 1 + 0.05θ c 1 + 0.05 • 18.5 272.9θ c mgVss 272.9 • 18.5 Xh = + = 2623 L 1 + 0.05 • θ c 1 + 0.05 • 18.5 mgVss Or X h = 2800 − 178 ≅ 2622 L 2800 =

7. Check for the reaction time(tr): Use the batch reactor design equation: For nitrification: No µ mn K n ln + ( N o − N t ) = X n ( )t Nt Yn No = concentration of nitrogen in the SBR after dilution resulting from mixing VFill in VTotal: N • VFill 40 • 625 mgN N • VFill = No • VT → No = = ≅ 26 VT 962 L Nt = 0.5 mgN/L (the required nitrogen influent).

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2.6 0.44 + (26 − 0.5) = 178( )t 0.5 0.12 t = 0.042 day ≅ 1hr (so tr = 1.0 is o.k. as assumed) • Check for BOD5 removal: S µ K s ln o + ( So − St ) = X h ( m )t St Y S • VFill 150 • 625 mgBOD5 So = = ≅ 97.5 VT 962 L mgBOD5 St = 10 (the required BOD5 in the effluent) L 97.5  2.5  50 ln + (97.5 − 10) = 2623 t 10  0.5  t = 0.0154d ≅ 0.37hrs < 1.0hr (t r = 1.0hr ) * Note: tr for nitrification always control the design of SBR. 0.5 ln

8. Calculate sludge procedure: XV 2800 • 962 • 10 3 KgVss Px = ≅ 146 = 6 day θc 18.5 • 10 Px m3 146 • 10 6 18 ≅ = X s 8000 • 10 3 d If Qw is taken during the react step: P 146 • 10 6 m3 Qw = x = ≅ 52 X 2800 • 10 3 d Qw =

9. Calculate oxygen requirements: Ro = Q( Sin − Se ) − 1.42 Px + 4.57Q ( Nin − N e ) Ro = 3750 •

10 3 10 3 Kg o 2 ( 150 − 10 ) − 1 . 42 • 146 + 4 . 57 • 3750 • ( 40 − 0 . 5 ) = 995 per d 10 6 10 6

mgBOD5 F QSo 3750 • 40 3 • 150 10. Check = = = 0 . 21 M Vx mgVss • d 2800 • 962 • 10 3

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Typical range is(0.02-0.15), this is not in the range.

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