Bang Tinh Can Bang Vat Chat

8/12/2019 Bang Tinh Can Bang Vat Chat

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Enviro-Mastered Calculation Program (ENMAS) 1

Wastewater characteristics:

Parameters Unit Value

1 Flow rate m3.day 4,000.0

2 Influent BOD g/m 3 120.0

3 Influent sBOD (Assumed 70% BOD) g/m3

84.0 4 Influent COD g/m 3 216.0

5 Influent sCOD (Assume 70% COD) g/m 3 151.2

6 Influent TSS g/m 3 135.0

7 Influent VSS (Assumed 85% TSS) g/m 3 114.8

8 Nitrogen g/m 3 96.0

9 NH 4-N g/m3

80.0

10 Total P g/m 3 39.0 12 pH 6.8

13 Alk (Assumed) g/m 3 as CaCO 3 150.0

15 Summer wastewater temperature oC 25.0

16 Autumn wastewater temperature oC 20.0 17 BOD:N:P ratio 100/80/32.5

I. CALCULATION OF EQUALIZATION BASIN, B-101

Conditions and assumption for design:

1 Daily flow rate Q 4,000 (m3 /d)2 Working time per day Td 24 (h)

3 Irregular factor (Assume) F i 2

4 Average hourly flow rate Q(aver) 166.7 (m3 /h)

5 Peak flow rate Q(peak) 333.3 (m3 /h)

6 Hydraulic retention time with average hourly flow rate 8 (h)7 Type of turbulane by mixing

Turbulancing intensity (not less than ) 6 (W/m 3 wastewater)8 Expected performance for BOD removal 5 %9 Expected performance for COD removal 3 %

Solution:a. Determine the volume of equalization basin, V

(B-101)V(B-101) = Q(aver) (m

3 /h) * (h)

= 166.7 (m3 /h) * 8 (h)

= 1,333.3 (m3)

Designed actual volume of equalization basin = 1,800.0 (m3)

1,800.0 (m3) versus 1,333.3 (m3), there is about 35.00 %bigger than volume as required.

Assesssing the designed value: Satisfiedb. Determine the concentration of organics in effluent of equalization basin

Effluent BOD concentration, S (BOD) = 114 (g/m3)

Effluent COD concentration, S (COD) = 210 (g/m 3)c. Determine the mixing power for equalization basin

Mixing power = 6 (W/m3wastewater) * V(B-102) (m3)= 8,000 (W)

Designed values of mixer:

MASS BALANCE CALCULATION FOR WWTP OF FOCOCEV QUANG NAM

Comparing

Project: WWTP of FOCOCEV Quangnam Copyright2014 WASEN. All rights reserved


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Mixing volume : 1,000 (m3)Motor capacity : 5.6 (kW)Quantity : 2 (pieces)Operation : 1 duty, 1 stand-by

II. CALCULATION OF CMAS PROCESS (Complete Mixed Activated Sludge)Concentration of pollutants go into CMAS process are as follows:Parameters

1 Flow rate 4,000 m3.day 4000 m3.day

2 BOD 5 120.0 (g/m 3) 50.0 (g/m 3)

3 sBOD (Assume 70% BOD 84.0 (g/m 3) (g/m 3)

4 COD 216 (g/m 3) 150.0 (g/m 3)5 bCOD/BOD ratio 1.6 Unitless Unitless

6 sCOD (Assume 70%COD 151.2 (g/m 3) (g/m 3)

7 rbCOD (30% sCOD) 45.4 (g/m 3) (g/m 3)

8 TSS 135 (g/m 3) 20 (g/m 3)

9 VSS (85%TSS) 115 (g/m 3) (g/m 3)

10 TKN 96.0 (g/m 3) 60 (g/m 3)11 NH 4-N 80.0 (g/m

3) 10 (g/m3)

12 Total P 39.0 (g/m 3) 6 (g/m 3)13 pH 7-7.5 7-7.5

14 Alk (Assumed) 150 (g/m 3 as CaCO 3) 80 (g/m 3 as CaCO 3)

15 Summer wastewater temperature oC 25.0

16 Autumn wastewater temperature oC 20.0 17 BOD:N:P ratio = 100/80/32

Kinetic coefficients for design:

For heterotrophic bacteria at 20 o C m, 20oC = 6.00 (g VSS/g VSS.d)

K s, 20oC = 20.00 (g bCOD/m3)

Y = 0.40 (g VSS/g bCOD)k d, 20oC = 0.12 (g VSS/g VSS.d)

f d = 0.15

For nitrifying bacteria at 20 o C nm, 20oC = 0.55 (g VSS/g VSS.d)

Kn, 20oC = 0.74 (g NH 4-N/m3)

Yn = 0.12 (g VSS/g NO x)k dn, 20oC = 0.08 (g VSS/g VSS.d)

Ko, 20oC = 0.50 (g/m3)

Solution:1. Develop wastewater characteristics needed for process design

a. Determine the S o = bCOD (biodegradable COD) in influent of CMAS) by the following Equation:

bCOD = 1.6 BOD = 192 (g/m 3)b. Find nbCOD using Equation:

COD = bCOD + nbCOD nbCOD = COD - bCOD

= 216 - 192 = 24 (g/m 3)

bCOD = sbCOD + rbCOD

30 % sCOD = 45.4 (g/m 3)Assume that, rbCOD =

OutletInlet



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sbCOD = bCOD - rbCOD = 147 (g/m 3)c. Determine the nbVSS by the following Equation:

nbVSS = [1-(bpCOD/pCOD)]VSSDetermine the bpCOD/pCOD ratio by the following Equation:

bpCOD (bCOD/BOD) * (BOD-sBOD)pCOD

bpCOD 1.6 * 36 pCOD 65

nbVSS = 12.75 (g/m 3)d. Determine iTSS (Inert Total Suspended Solid)

iTSS = TSS o - VSS o

= 135 - 114.8 = 20.3 (g/m 3)

2. Determine the Maximum specific growth rate of nitrifying bacteria

mn*N DO

(Kn + N) (K o + DO)

Substitute the above and given values and solve for n

10 (g/m3)

DO = 3 (g/m 3) n = 0.51 * 0.86 - 0.08

n = 0.36 (g/g.d)

3. Determine the theoretical and design SRTa. Find theoretical SRT by using Equation:

1 n

b. Determine the design SRT by the following Equation:SF = TKN peak/TKN average = 2 (Safety factor)Design SRT = (SF) * (theoretical SRT) = 5.6 (d)

4. Determine biomass production by the following Equation, Parts A, B, C

Determine S from the following Equation:

33.4

31.76 Assume NO x = 80% (TKN) as nitrogen balance can not be done yet. The eeror in assuming that NO x = 80% (TKN)is small as the nitrifier VSS yield is a small fraction of total MLVSS concentration.

NO x = 76.8 (g/m3) or 0.077 (kg/m

3)

1 + (k dn)SRT

Part A

N = (NH 4-N) e =

n = * -

+

(d)

QY n(NO x)

=(COD-sCOD)

=

S =Ks[1 + (k d)SRT]

1 + (k d)SRT(f d)(k d)Q(Y)(S o-S)SRT

=

QY(S o-S)

0.89

k dn

=

1 + (k d)SRTPart B

+

S =

Ks[1 + (k d)SRT]

=

SRT( m - k d) - 1SRT(Yk - k d) - 1

SRT = = 2.79

Px,bio =

Part C

(g/m 3)1.05



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Px,bio = Part A + Part B + Part C = 226.96 (kg/d)

5. Determine the amount of nitrogen oxidized to nitrate. The amount of nitrogen oxidized to nitratecan be found by performing a nitrogen balance, using the folowing Equation:

NO x = TKN o - Ne -

NO x = 96.0 - 10 -

NO x = 79.19 (g/m3) or = 0.079 (kg/m

3)

Where: NO x: Nitrogen oxidized, g/m3

TKN o: Influent TKN concentration = 96.00 (g/m3)

Ne: Effluent NH 4-N concentration = 10 (g/m3)

6. Determine the concentration and mass of VSS and TSS in the aeration basin

Mass = PX (SRT)

a. Calculate the concentration of VSS and TSS in the aerarion basin(i). Px,vss can be determined by the following Equation. Part A, B, and C have already been calculatedabove as P x,bio . Part D must be added to determine P x,vss

PX,VSS = P x,bio + Q(nbVSS) = 278 (kg/d)

(ii). P X,TSS can be determined by the following Equation, with the term E added to account for inertinfluent TSS

Px,bio0.85

Parts (A+B+C) Part D

PX,TSS = 267.01 + 51.00 + 81PX,TSS = 399.01 (kg/d)

b. Calculate the mass of VSS & TSS in the aeration basin by the folllowing Equations:(i). Mass of MLVSS

= 1,548.74 (kg)

(ii). Mass of MLSS

= 2,223.22 (kg)

7. Select a design MLSS mass concentration and determine the aeration basin volume and detention time,using the TSS mass computerd in Step 6

Part C =1 + (k dn)SRT

Part A =QY(S o-S)

Part E

PX,TSS =

1 + (k d)SRT

(f d)(k d)Q(Y)(S o-S)SRT

+

Part B =1 + (k d)SRT

(kg/d)

Q(TSS o - VSS o)

(kg/d)25.5

6.81

Q(nbVSS)

(X TSS)(V) = (P X,TSS )SRT

(X VSS) (V) = (P X,VSS )SRT

=

0.12 P x,bio /Q

= 18.36

183.10 (kg/d)

=QY n(NO x)

+



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= Oxygen transfer correction factor for waste (Aquaturbo surface aerator): = 0.95 = Salinity-surface tension correction factor, typically = 0.95F = Fouling factor, typically = 0.65-0.9. = 1

T = Operating temperature, oC. 25 oC (Summer time)

H = Atmosphere (standard): = 1 at = 1.0133 * 10 2 (kPa)= 101.33 (kPa)

Cs,20 = Dissolve oxygen saturation concentration in clean water at 20oC and 1 atm, mg/L

= 9.08 (mg/L)Cs,T,H = Average dissolved oxygen saturation concentration in clean water in aeration tank

at temperature T and altitude H, mg/LCs,T,H = Oxygen saturation concentration in clean water at temperature T and altitude H, mg/LCs,25 = 8.24 (mg/L)

CL = Operating oxygen concentration, mg/L. Chosen value: C L = 2 (mg/L)

Solving the above Equation for AOR:AOR = 58.93 * 1.640 * 0.822 AOR = 79.44 (kg/h)

Choosing the technical data of floating aeratorType: Screw impeller floating aeratorModel: AER-AS-7500-24Motor power: 75 kWOxygen capacity: 82.5 kg O 2 /h

Quantity: 1 piece

In the case of using air blowers and tube diffusers: = Oxygen transfer correction factor for waste. For diffuser: = 0.65 = Salinity-surface tension correction factor, typically = 0.95

F = Fouling factor, typically = 0.65-0.9. = 0.9

Solving the above Equation for AOR:AOR = 58.93 * 2.66 * 0.822 AOR = 129.01 (kg/h)

c. Determine the air flow rateAir flow rate can be determine by the following equation

Where: E = Fine bubble diffuser with an aeration clean water oxygen transfer efficiency

E = 25 % at a 4.5 m depth and 15 m 3 /h/tube air flow rateda = density of air at 30

oC0.21 = percentage of oxygen in the air

(i). Determine the density of air d a at 30oC

PMRT

Where: P = Atmospheric pressure = 1.01325*10 5 (N/m 2)

M = Mole of air = 28.97 (kg/kg-mole)R = Universal gas constant = 8314 (N*m/kg-mole*K)

T = Air temperature, K = (273.15 + oC)Solving the above equation for d a

1.01325 * 100,000 * 28.97 8314 * 303.15

da =

Note: For diffuser, F

da =

AOR (kg/h)

[(E)(60 min/h)(0.21*d a (kg/m3)]

Air flow rate, m 3 /h =

Note: For surface aerator, F



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= 1.165 (kg/m3)(ii). Determine the air flow rate

= 35.16 (m3 /min)Choosing the technical data of air blowersType: Roots air blower

Model: BE-200H (ANLET manufacturer)Motor power: 45 kW

Air flow rate: 35.6 m 3 /minQuantity: 2 piecesOperation: 1 duty, 1 stand-by

11. Check alkalinitya. Prepare an alkalinity balance

Alklinity to maintain pH 7 Influent Alk - Alk used +

Influent alkalinity = 150 (g/m 3 as CaCO 3)

Amount of nitrogen converted to nitrate: NO x = 79.19 (g/m3)

Alkalinity used for nitrification = 7.14 (g CaCO 3 /g NH 4-N) x NO x (g/m3)

= 565.43 (g/m 3 used as CaCO 3)b. Substitute known values and solve for alkalinity needed.

Residual alkalinity concentration needed to maintain pH in the range of 6.8-7.0 = 70-80 (g/m 3 as CaCO 3)

Select : 80 (g/m 3)

80 (g/m 3) = Influent Alk - Alk used + Alk to be added

80 (g/m 3) = 150 - 565.4 + Alk to be added

495.43 (g/m 3 as CaCO 3)

1,981.70 (kg/d as CaCO 3)c. Determine the alkalinity needed as NaOH

= 50 (g/equivalent)= 40 (g/equivalent)

1,981.70 (kg/d CaCO 3) * 40 (g NaOH/eq)

50 (g CaCO 3 /eq)

1,585.36 (kg/d NaOH)

12. Estimate effluent soluble BOD (sBOD e)

(g BOD) (0.85 g VSS)(1.42 g VSS) (g TSS)

(g BOD) (0.85 g VSS)(1.42 g VSS) (g TSS)

sBOD e = 50.0 - 0.7 * 0.85 *sBOD e = 38.03 (g/m

3)

20 (g/m3)

13. Determine nutrient removal

Equivalent weight of CaCO 3Equivalent weight of NaOH

Alk to be added =

Equivalent to

NaOH needed =

(BOD 5)e

NaOH needed =

(BOD 5)e =

sBOD e = *

sBOD e *

-

+

(m3 /h)=AOR (kg/h)

2,109.85

TSS e (g/m3)

TSS e (g/m3)

20

Alk to be added

*

Assume TSS e =

*

Air flow rate, m 3 /h =[(E)(60 min/h)(0.21*d a (kg/m

3)]



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Biomass produced (P x,bio ) in the CMAS process = 226.96 (kg VSS/d)a. Determine the amount of Nitogen removed by biomass

Nitrogen requirement to synthesis the biomass equals 12.0 % of biomass producedPx,bio * 0.12 (kg/kg)

= 27.2 (kg/d)b. Determine the amount of Phosphorous removed by biomass

Conditions for design:1 Phosphorous content of Phosphorous Accumulating Organisms (PAOs) 0.3 (kg P/kg VSS)2 Phosphorous content of other bacteria, 0.024 (kg P/kg VSS)

b1. Determine phosphorous removed by PAOs due to the fermentation of t 0.045 (kg rbCOD/m 3) in the wastewater influent.(i). Determine biomass produced by the following Equation, neglecting cell debris

QY1 + (k d)SRT

1,600.00 1.67

(ii). Determine the phosphorous removed by PAOsP removed = 0.3 (kg P/kg VSS) * 43.49 (kgVSS/d)

= 13.0 (kg P/d)b2. Determine phosphorous removed by other bacteria

(i). Determine biomass produced by the following Equation, neglecting cell debrisQY

1 + (k d)SRT1,600.00

1.67 (ii). Determine the phosphorous removed by other bacteria

P removed = 0.024 (kg P/kg VSS) * 140.61 (kgVSS/d)= 3.4 (kg P/d)

b3. Determine total phosphorous removed by bacteria13.0 (kg/d) + 3.4 (kg/d)

= 16.42 (kg/d)Influent nutrients:

TKN = 96.00 (g/m 3) * Q (m3 /d)

= 96.00 (g/m 3) * 4,000 (m3 /d)= 384,000 (g/d) or 384 (kg/d)

TP = 39.0 (g/m 3) * Q (m3 /d)

= 39.0 (g/m 3) * 4,000 (m3 /d)= 156,000 (g/d) or 156 (kg/d)

TKN influent (kg/d) - TKN used (kg/d)= 384.0 (kg/d) - 27.2 (kg/d)

= 356.77 (kg/d), equivalent to 89.2 (g/m 3)TP influent (kg/d) - TP used (kg/d)

= 156.0 (kg/d) - 16.42 (kg/d)

= 139.58 (kg/d), equivalent to 34.9 (g/m 3)

It means that, there is sufficient both nitrogen and phosphorous for growing of biomass.

15. Clarifier design

a. Determine the returned sludge flow rateThe returned sludge flow rate can be determine by following Equatiion:

QR = RQ (m3 /d)

=

(kg/d)

N requirement =

*Biomas produced = rbCOD

43.49 (kgVSS/d)

(kg/d)

(kgVSS/d)

=0.045

0.15

*

TP in effluent =

sbCODBiomas produced =

*

* 140.61

=

TKN in effluent =

=

Total P removed =



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Where: R = Return sludge recycle ratio

Q = Influent flow rate = 4,000 (m3 /d)(i). Determine the return sludge recycle ratio, R

XXR - X

Where: X = Concentration of MLSS in aeration basin = 2.5 (kg/m 3)

XR = Concentration of MLSS in recirculation activated sludge = 8 (kg/m3

)(ii). Determine the returned sludge flow rate, Q R

QR = RQ (m3 /d) = 1,818.2 (m3 /d) or 75.76 (m3 /h)b. Determine the wasted sludge flow rate per day, Q w

Wasted sludge flow rate per day can be determined by the following Equatiion:

VX - (SRT)Q eXe

2,223.22 - 445.7

= 1.66 (m3 /h)

Where: Qw: Wasted sludge flow rate (m3 /d)

SRT: Solid Retention Time = 5.57 (d)

V: Volume of aeration basin = 889.29 (m3)

X: Concentration of MLSS in aeration basin = 2.5 (kg/m 3)

Q: Flow rate of influent wastewater = 4,000 (m3 /d)XR: Concentration of MLSS in recirculation activated sludge = 8 (kg/m

3)

Xe: Concentration of MLSS in effluent = 0.02 (kg/m3)

c. Determine size of clarifier

16 (m3 /m 2.d) at average flow rate

QSHR

d. Check solids loading rate (SLR)

= 2.42 (kg MLSS/m 2.h)(Within acceptable range of solids loading of 4-6 kg/m2.h)

16. Summary of CMAS design

Design parameters Unit BOD removal & nitrification

Average wastewater flow m3

/d 4,000 Average BOD load kg/d 480 Average TKN load kg/d 384.0 Aerobic SRT d 5.57 Aeration basin Nos 1Aeration basin volume m3 889.29

ASLR =

Clarifier area, A =

44.57

(1 + R)Q(MLSS)(Q + Q R)(MLSS) =A

(m2)=

= 39.88

SRT = QwXR + Q eXe

(SRT)X R

0.45

Qw =

=

250

(m3 /d)

R =

Qw =

VX

Assume a Surface Hydraulic Rate (SHR) of clarifier =



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Hydraulic detention time, h 5.34 MLSS kg/m 3 2.5MLVSS kg/m 3 1.7F/M ratio g BOD/g VSS.d 0.31 BOD loading kg BOD/m 3.d 0.54 Sludge production, P X,TSS kg/d 399.01 Observed yield kg TSS/kg bCOD 0.52

kg VSS/kg bCOD 0.36 Oxygen required kg/h 79.44 Air flow rate sm3 /minRAS ratio Unitless 0.455Clarifier hydraulic rate m3 /m 2.d 16Clarifier Number 1

Surface area, m 2 250Alk added as NaOH kg/d 1585.36Effluent BOD 5 g/m

350.00

TSS e g/m3

20

Effluent NH 4-N g/m3

10Qw m

3 /d 39.9

QR m3 /d 1,818

SLR kg MLSS/m 2.h 2.42

III. CALCULATION OF ANOXIC PROCESSAssumption:

1 Use the same coefficiency as the nitrification process design

2 Mixing energy for anoxic reactor = 10 (W/m 3)

Solution:1. Determine the active biomass concentration by the following Equation and substituing V/Q for

Q(SRT)

V

Xb = 25.06 *Xb = 1,147.19 (g/m 3)

2. Determine the IR ration by the following Equation:

NO xNO 3-N

a. Determine the NO 3-N concentration:

NO 3-N concentration in aeration basin = (NO 3-N) eNO 3-N concentration can be determine by the following Equation:

TKN e - Ne -

Where: TKN e: Effluent TKN concentration = 60.0 (g/m3)

Ne: Effluent NH 4-N concentration = 10 (g/m3)

Assume the effluent TSS e concencentratio = 20 (g/m 3)The effluent VSS e concencentration = TSS e (g/m 3) * 0.85

1

45.77

-=

(NO 3-N) e =

*Xb =Y(S o-S)

1 + (k d)SRT

IR - R

0.12(VSS e)



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SDNR 20 * (X b /X T)

0.030 * 0.46 0.01 (g/g.d)

The computed value is in the range of reported SDNR value (0.04-0.42)

8. Go to nitrification step in design and determine net oxygent required73.82 (kg/h)

(2.86 g O 2)

(g NO 3-N)

2.86 * 31.23 * 4,000 357.29 (kg/d) or 14.89 (kg/h) (Go back to Step II-10a)

-

73.82 (kg/h) - 14.89 (kg/h)58.93 (kg/h) (Go back to step II.10a)

Note the required aeration rate will decrease to proportion to the lower R o. Oxygen required can be reducedby 20 %

9. Check alkalinitya. Prepare an alkalinity mass balance

Alkalinity to be added to maintain pH ~ 7 =

Influent Alk - Alk used +

+

(i). Influent alkalinity = 150 (g/m 3 as CaCO 3)

(ii). Alkalinity used = 7.14 (NO x; g/m3) = 565.43 (g/m

3)

(iii). Alkalinity produced = 3.57 (NO x-NO 3-N); g/m3 = 111.50 (g/m

3)

(iv). Alkalinity needed to maintain neutral pH = 7 = 80 (g/m 3 as CaCO 3)b. Solve the above expression for Alkalinity to be added

Alk to be added = 383.93 (g/m 3 as CaCO 3)

Mass of alkalinity needed = 1,535,720 (g/d) or 1,535.72 (kg/d as CaCO 3)Equivalent to 1,228.58 (kg/d as NaOH)

c. Compare to alkalinity needed for nitrification onlyFor the nitrification only design, the alkalinity needed = 1585.36 (kg/d NaOH)Alkalinity savings = 356.79 (kg/d NaOH)

10. Determine anoxic zone mixing energy

= 10 (W/m 3)

= 2722.4 (m3)= 27.22 (kW)

11. Summary of anoxic design

Item Unit Value

Effluent (NO 3-N) e g/m3

48.0 Internal recycle ratio Unitless 0.20

Oxygen credit =

Ro (with nitrification) =

SDNR (MLSS) =

SDNR (MLSS) =

Net oxygen required = Ro (with nitrification)

Q*

Net oxygen required =

Oxygen credit =

SDNR (MLSS) =

(NO x-NO 3-N)

Oxygen credit

Alk to be added

*

Net oxygen required =

PowerAnoxic volume

Oxygen credit =

Alk produced

Mixing energy



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RAS recycle ratio Unitless 0.45

Anoxic volume m3 2722.41

MLSS kg/m 3 2.5Overall SDNR g NO 3-N/g MLSS.d 0.01 Detention time h 16.33Mixing power kW 27.22Alkalinity required kg/d as NaOH 1,228.6

Equivalent to kg/d as liquid NaOH 32% 3,839.30

IV. CALCULATION OF COAGULATION BASIN, B-106


1 Daily flow rate Q 4,000 (m3 /d)2 Operation time per day Td 24 (h)

3 Hourly average flow rate Q(aver) 166.7 (m3 /h)

4 Hydraulic retention time 10 (min.)5 Type of mixing Agitator

Solution:1. Determine the volume of coagulation basin, V (B-106)

V(B-106) = Q (aver) (m3 /h) * (h)

= 166.7 (m3 /h) * 0.17 (h)

= 27.78 (m3)

Designed actual volume of coagulation basin = 33.30 (m3)

Comparin 33.3 (m3) versus 27.8 (m3), there is about 19.9 %bigger than volume as required.

Assesssing the designed value: Satisfied

Designed values of agitator:Motor capacity : 1.5 (kW)Gear-speed : 140 (rpm)Quantity : 1 (piece)Operation : Continuously

2. Determine the Alum requirement for removal of Phosphorous

Parameters Inlet Outlet

Flow rate 4,000 (m3.day) 4000 (m3.day)

Total P 34.9 (g/m 3) 6 (g/m 3)

Conditions for design:1 Mole of Al required per mole of P remove = 1.2 (mole Al required/mole P removed)2 Fomular for liquid alum Al2(SO 4)3.18H 2O3 Alum strength = 48 %4 Density of liquid alum solution = 1.2 (kg/L)5 Alum storage capacity required = 5 (d)

Determine the amount of liquid alum required to precipitate phosphorous in a wastewater that contains:

+

2 x 78 3 x 136 6 x 44 18 x 182Al(OH) 3 + 3CaSO 4 + 6CO 2 + 18H 2O

3 x 100 (as CaCO 3) 666.5

Al 2(SO4) 3.18H 2OAluminum sulfate

(Soluble)Calcium bicarbonnate

(Soluble)

3Ca(HCO 3)2



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Aluminum Calcium sulfate Carbon dioxidehydroxide(Insoluble) (Soluble)

Al3+ + PO 43- AlPO 4 (1)

Al3+ + 3OH-

Al(OH) 3 (2)

Solution:a. Determine the weight of aluminium (Al) avalable per liter of liquid alum

(i) The weight of alum per liter is:Alum/L = 0.48 * 1.2 (kg/L) = 0.576 (kg/L)

(ii) The weight of aluminium per liter is:Molecular weight of Alum = 666.5

2 * 26.98

= 0.0466 (kg/L)b. Determine the weight of Al required per unit weight of P

(i) Theoretical dosage = 1 mole Al per 1 mole P(ii) 1 (kg) * (mw Al/mw P)

26.9830.97

= 0.87 (kg Al/kg P)c. Determine the amount of alum solution required per kg P

(kg Al)(kg P) 0.0466 (kg)

= 22.4 (L Alum solution/kg P)d. Determine the amount of alum solution required per day

Alum = Q(m 3 /d) * P (g/m 3) *

4000 * 28.9 *1,000

= 2,591.0 (L Alum solution/d) or 1,492.39 (kg Alum powder/d)= 108.0 (L/h)

Technical data of Alum dosing pumpsFlow rate : 130.0 (L/h)Motor capacity : 0.18 (kW)Pressure : 3 (bar)Quantity : 2 (pieces)

Operation : 1 duty, 1 stand-byModel :

e. Determine the required alum solution storage capacity based on average flow.2,591.0 (L Alum solution/d) or * 5 (d)

= 12.95 (m3)f. Determine the mass of sludge produced from alum used

1,492.39 (kg/d)Determine the Al(OH) 3 formed, M Al(OH)3

(2*78)666.5

= 349.3 (kg/d)

d. Determine the mass of Lime required to convert Al 2(SO 4)3.18H 2O to aluminium hydroxide

(3*100) as CaCO 3666.5

Alum dosage = 1.2

luminium/L =

Aluminium required =

= 1 (kg) *

*

*(L Alum solution)

Alum dose (L Alum solution/kg P)

22.4

(kg/L)0.576

*

Al(OH) 3 formed =

Alum used =

Storage capacity =

0.87

1,492.39

666.5*

(kg/d)

*Lime required = 1,492

=

(kg/d)

DMX 226-130-3 (M)



15/18


= 671.7 (kg/d)e. Determine the mass of NaOH needed as the same alkalinity with Lime

= 50 (g/equivalent)= 40 (g/equivalent)

671.7 (kg/d CaCO 3) * 40 (g NaOH/eq)

50 (g CaCO 3 /eq)

537.4 (kg/d NaOH), equivalent to 134 (g/m 3)

V. CALCULATION OF FLOCCULATION BASIN, B-107




4 Hydraulic retention time 10 (min.)

5 Type of mixing Agitator

Solution:a. Determine the volume of flocculation basin, V (B-107)

V(B-107) = Q(aver) (m3 /h) * (h)

= 166.7 (m3 /h) * 0.17 (h)

= 27.8 (m3)

Designed actual volume of flocculation basin = 33.3 (m3)

Comparin 33.3 (m3) versus 27.8 (m3), there is about 19.88 %bigger volume as required.

Assesssing the designed value: SatisfiedDesigned values of agitator:

Motor capacity : 1.5 (kW)Gear-speed : 70 (rpm)Quantity : 1 (piece)Operation : Continuously

b. Determine the flocculation chemical requirementAssumptions:

1 Chemical for coagulation Anion polymer2 Fomular for Anion polymer3 Polymer strength = 0.2 %4 Density of liquid A. polymer solution = 1 (kg/L)5 A. polymer dosage (Assume) = 1 (g/m 3)6 Liquid A. polymer storage capacity requir = 1 (d)

(i). Determine the weight of Anion polymer needed per day:

Q (m3 /d) * (g/m 3)

= 4,000 (m3 /d) * (g/m 3)= 4,000 (g/d) or 4.0 (kg/d)

(ii). Determine the weight of liquid A. polymer needed per day :The weight of A. polymer per liter is:

0.002 * 1 (kg/L) = 0.002 (kg/L)

The weight of liquid A. polymer needed per day is:(kg/d)(kg/L)

= 2,000 (L/d) or 83.33 (L/h)

Weight of A. polymer needed = A. polymer dose

1

A. Polymer/L =

eight of liquid A. polymer needed per day =

NaOH needed =

Weight of A. polymer needed per day

Equivalent weight of CaCO 3

NaOH needed =

Weight of A. polymer per liter

Equivalent weight of NaOH



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Technical data of Polymer dosing pumpsFlow rate : 130.0 (L/h)Motor capacity : 0.18 (kW)Pressure : 3 (bar)Quantity : 2 (pieces)Operation : 1 duty, 1 stand-byModel :

(iv). Determine the liquid Sodium Hydroxide storage capacity required:2,000 (L/d) * 1 (d)

= 2,000 (L) or 2.00 (m3)

VI. CALCULATION OF SECONDARY CLARIFIER, B-108




4 Influent TSS concentration 20 (g/m 3)

5 Surface hydraulic loadings (in range) v 24-48 (m3 /m 2 /d)

6 Effluent weir hydraulic loadings (in range of (125-500)) 250 (m3 /m/d)7 Hydraulic retention time (in range of 1.5-2.5) 2 (h)8 Expected performance for TSS removal 50 %

9 Sludge concentration in bottom of secondary clarifier (Assumed) 4 (kg/m 3)10 Specific gravity of sludge (Assumed) 1.05

Solution:a. Determine the surface area needed of secondary clarifier, A

The surface area needed of secondary clarifier can be determined by the following Equation:Q (m3 /d)v (m3 /m 2 /d)

For secondary ary clarifier, the standard value of v is in range 24-48 (h), choosing v = 24 (m3 /m 2 /d)4,000 (m3 /d)

24 (m3 /m 2 /d)

b. Determine the volume of secondary clarifier, VThe volume needed of secondary clarifier can be determined by the following Equation:

V = Q(aver) (m3 /h) * (h)

For secondary clarifier, the standard value of is in range of 1.5-2.5 (h), choosing = 2 (h)

V = 166.7 (m3 /h) * 2 (h)

= 333.3 (m3)

Designed parameters of secondary clarifier:

115.59 (m2)

697.81 (m3)

4.19 (h)

34.60 (m3 /m 2 /d)

101.91 (m3 /m/d)Comparing the actual designed parameters vesus standard conditions:

The designed values: Satisfied

c. Determine the mass of physico-chemical sludge produced per day

(actual) =

v(actual) =

Effluent weir loading =

he A. polymer storage capacity required =

A =

A =

A(actual) =

V(actual) =

(m2)167 =

DMX 226-130-3 (M)



17/18


(i). Determine the mass of sludge produced from TSS removal, M TSSMTSS = 0.5 * TSS i (g/m

3)

= 0.5 * 20 (g/m 3)

= 10 (g/m 3) or 0.01 (kg/m 3)Daily M TSS = Q (m

3 /d) * 0.01 (kg/m3)

= 40 (kg/d)(iii). Determine the total mass of physico-chemical sludge produced per day, T M

TM = MTSS + MAl(OH)3= 389 (kg/d)

(iv). Determine the volume of physico-chemical sludge produced per dayTM (kg/d)

1.05 * 4 (kg/m 3)

= 92.69 (m3 /d)

Technical data of physico-chemical sludge pumps

Flow rate : 6.0 (m3 /h)Motor capacity : 0.25 (kW)Head : 5.5 (m)Quantity : 2 (pieces)Operation : 1 duty, 1 stand-by

d. Determine the concentration of pollutants in effluent of primary clarifierEffluent TSS concentration, X (TSS) = 10 (g/m

3)

VII. CALCULATION OF DISINFECTION BASIN, B-109




4 Hydraulic retention time 0.5 (h)

Effluent total coliform count before disinfection (N) = 1*10 6 /100 mLRequired effluent total coliform count (N o) = 4.5*10

3 /100 mL (B level)

Initial effluent chlorine demand = 4 (g/m 3)

Demand due to decay during chlorine contact = 2.5 (g/m 3)Required chlorine contact time = 30 (min)

Slope of inactivation curve (n) = 2.8Value of x-intercept (b) = 4

Solution:a. Determine the volume of disinfection basin, V (B-109)

V(B-109) = Q(aver) (m3 /h) * (h)

= 166.7 (m3 /h) * 0.5 (h)

= 83.3 (m3)

Designed actual volume of disinfection basin = 84.00 (m3)

Comparin 84.0 (m3) versus 83.3 (m3), there is about 0.8 %bigger than the volume as required.

Assesssing the designed value: Satisfiedb. Determine the required chlorine dosage

Assumptions:1 Chemical for disinfection Liquid Sodium Hypochlorite, 10% available chlorine2 Fomular for liquid Sodium Hypochlorite NaOCl3 Free available chlorine in liquid Sodium Hypochlorite = 10 %

Vsludge =

For coliform:



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4 Density of liquid Sodium Hypochlorite = 1.16 (kg/L)5 Liquid Sodium Hypochlorite storage capacity required = 10 (d)

(i). Estimate the required chlorine residual using Equation:

N 1No (CRt/b)

n

Where: CR = Chlorine residual remaining at the end of time tt = Contact timen = Slope of inactivation curveb = Value of x -intercept when N/No = 1

Solve the above Equation to determine the C R value:

30

= 0.92 (g/m3

)(ii). The required chlorine dosage is:

4 + 2.50 + 0.92

= 7.42 (g/m 3)

Equivalent to 74.2 (g liquid Sodium Hypochlorite / m 3)(iii). Determine the mass of Sodium Hypochlorite required

Dosage (g/m 3) * Q (m3 /d)

= 7.4 (g/m 3) * 4,000 (m3 /d)= 29,671 (g/d) or = 29.67 (kg/d)

(iv). Determine the mass of liquid Sodium Hypochlorite required

10010

= 296.7 (kg/d)(v). Determine the volume of liquid Sodium Hypochlorite required

296.7 (kg/d) : 1.16 (kg/L)= 255.8 (L/d) or = 10.66 (L/h)

Technical data of chemical dosing pumpsFlow rate : 17.0 (L/h)Motor capacity : 0.09 (kW)Pressure : 4 (bar)

Quantity : 2 (pieces)Operation : 1 duty, 1 stand-byModel :

(vi). Determine the liquid Sodium Hypochlorite storage capacity required:255.78 (L/d) * 10 (d)

= 2,558 (L) or 2.56 (m3)Summary of sludge from WWTPAerobic sludge: 399.0 (kg/d)Physyco-chemical sludge 389.3 (kg/d)Total daily sludge mass: 788.3 (kg/d)Summary of chemical used:Alum solution, 8% Al 2O3 777.3 (g/m

3)

Anion polymer 1.0 (g/m 3)

NaOH 441 49 (g/m 3)

=

=

CR =b

t (N/No) 1/2.8

CR4

Liquid NaOCl storage capacity required =

(4.5*10 3 /1*10 6)0.357

29.67 (kg/d) *

Volume of liquid Sodium hypochlorite required =

DMX 221 17-4

Chlorine dosage =

Volume of liquid Sodium hypochlorite required =

Mass of Sodium Hypochlorite required =

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