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CE 432 Wastewater Treatment Plant Design

Sewer Design, edit bolded values: S, n, conduit fill ratio e.g. .75, Qavg (MGD) and the factors

Given:S = .005 ft/ft, n=.003, conduit 3/4 full BOD 5 250

S(ft/ft) = 0.005 SS 220

n = 0.013 Total P 9

conduit fill ratio = 0.75 NH3-N as 30

Q avg, MDG = 15

Q peak hourly, factor 2.5

Q min, factor 0.4

Find: D, Vmin, Vmax

D, inches = 40.28 Use next larger commerically available size, subsequent d's and V's bas

Vmax, fps = 8.15

davg, inches 28.56

Vavg, fps = 3.46

dmin, fps = 20.26

Vmin, fps = 2.08

Waste Flows

Design Flow Factor MGD CFS GPM

Q avg 1 15.00 23.21 10417

Q peak 2.5 37.50 58.01 26042

Q min 0.4 6.00 9.28 4167

Calculate the pipe diameter

d/D= 0.75

q = 2cos-1

(1-(2d/D))= 240.00

q, radians 4.19A=(D

2/8)(q-sin(q))

R=(D/4)(1 - sin(q)/qr)

AR2/3

=(D8/3

/ (8*42/3)

)(q-sin(q)(1 - sin(q)/qr)2/3

(q-sin(q)(1 - sin(q)/qr)2/3

= Qn(8*42/3

)/ [ (1.486)*D8/3

*S1/2

]

D based on peak flow = [ Qn(8*42/3

)/ (1.486)S1/2

(q-sin(q))(1 - sin(q)/qr)2/3

]3/8

D, feet = 3.36

D, inches = 40.28 Use next larger commercially available size

Calculate the maximum velocity based on the peak hourly flow

A=(D2/8)(q-sin(q))

d for max flow, ft 30.21

d/D= 0.75q = 2cos

-1(1-(2d/D))= 240.00

q, radians 4.19

A, ft2

= 7.12 Based on calculated D, not next larger commerically available size

Vpeak, fps = 8.15

Calculate the average velocity based on the average flow

d for avg flow, ft = 2.38 Based on calculated D, not next larger commerically available size

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d for avg flow, inches 28.56

d/D= 0.71

q = 2cos-1

(1-(2d/D))= 229.47

q, radians 4.00

A, ft2

= 6.71

Vavg, fps = 3.46

Calculate the minimum velocity based on the minimum flow

d for min flow, ft = 1.69 Based on calculated D, not next larger commerically available size

d for min flow, inches 20.26

d/D= 0.50

q = 2cos-1

(1-(2d/D))= 180.68

q, radians 3.15

A, ft2

= 4.46

Vmin, fps = 2.08

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for Qpeak hourly and Qmin.

ed on the calculated value shown.

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CE 432 Wastewater Treatment Plant Design

Bar Screens

Given: Provide 2 (1+1) mechanical screens, q=75, bars space = 1.0 inches, Vmax = 3.0 fps, Vavg = 2fps, V

q, degrees = 0.005Vmax, fps = 0.013

Vavg, fps = 0.75

Vmin, fps =

Find:

1. Bar rack dimensions

2. Chamber dimensions

d in the rack at peak flow

headloss through the rack

draw a dimensional plan and profile of the bar screen structure

Waste Flows

Design Flow Factor MGD CFS

Q avg 1 15.0 23.21Q peak 2.5 37.5 58.01

Q min 0.4 6.0 9.28

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in =1.3 fps

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CE 432 Waste water treatment plant design

Grit

Chamber

Waste characteristics BOD 5 250

Design Flow MGD CFS SS 220

Ave Q 12 18.60 Total P 9

Peak Q 25 38.75 NH3-N as N 30

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D. Primary

CE 432 Waste water treatment plant design

mg/l

Waste characteristics BOD 5 250

Design Flow MGD CFS SS 220

Ave Q 12 18.60 Total P 9

Peak Q 25 38.75 NH3-N as N 30

Plant Effluent Qualities (mg/l) L/W ratio 4

BOD 5 20

SS 20

VSS/SS ratio 0.65

Module D-- Primary Treatment

BOD 5 SS

Percent Removal 34 63

Effluent (mg/l) 165 81.4

Design Criteria

No. of Tanks 2

Detention Time 1.5 hours min.

SOR 800 gpd/sf

Calculations

1 Tank Dimensions

Tank Area = Q/ SOR= 15000 sf total or 7500 sf each tank

Tank Vol = Q * DT 750000 gal total or 375000 gal ea tank

Tank Depth = V/A = 6.68 ft plus 2 ft freeboard

A = Wx L = W x 4 W W= 43.3 use 40 ft

L = A/W= 187.5 ft use 190 ft

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E. AS

Q Xo + Qr Xr = ( Q+Qr) X

Q Xo X Xr Qr Qr/Q

Qr=Q X / ( Xr-X) 12 0 2800 10000 4.67 0.39

MG

E5 Waste Activated Sludge Rate

Method A. Calculated from Cell Residence time

C V X Xr Xe

C = V X/(QwXr+QeXe) 10 2.53 2800 10000 13

Qw =(VX/C-QeXe)/Xr= 0.0553301 MG

WAS--VSS (lbs/day)= 4615

WAS--SS (lbs/day) = 5768

Method B. Calculate from BOD removal Px = Yobs Q ( So-Se) 8.34

Y Kd C Yobs

Yobs = Y / ( 1+kd C) 0.6 0.06 10 0.375

Q So Se Px

Biomass 12 165 7.4 5916

SS 7394

WAS --SS (lbs/day) = Px - Q Xe 8.34 = 5393

Use the larger of two methods WAS--SS = 5768 lbs/day

E-6 RAS and WAS Pumps Tot Q Q/pump Qp/Pump

MG gpm gpm

no. of RAS pumps 3 4.67 1080 2251

no of WAS pumps 2 0.055 58 120

(WAS pumps 8 hr/d)

E-7 Oxygen Requirements O2 = C BODu - BODu in Px + N BODu

Q So Se f C BODu

C BOD = Q ( So-Se) 8.34 / f 12 165 7.4 0.684 23070

Px (SS lb/d) Biodg/SS BOD in WAS

BOD in WAS = 1.42 Px (biod) 5916 0.65 5460

O2 (lbs/d) = 17610 For ave flow

36688 For peak flow

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E. AS

E-8 Compressor Requiement

Air requirement at STP ( 14.7 psi and 68 F)

lbs/cf O2/Air Vol O2/Air Wt air wt(lb/d) air (lb/sec) air (scfm)

Air 0.0752 0.21 0.23 76566 0.886 707

Air requirement= air /transfer efficiency transfer eff = 0.128 6.923 5524

Air requirement at Qp 14.42 11508

Compressor HP = [WRT/(550 n e) ][(P2/P1)**n-1] Sized at Qp

W (lb/sec) n e P2 (psia) P1(psia) R T (R)

14.42 0.283 0.8 24.7 13.7 53.3 570

Total Each compressor

Av disch P 19.7

no of compressor

HP max 639 6 106

Check MIxing scfm V (1000 cf) scfm/1000cfm T tank P ave tank

Ave Q (scfm/1000 cf aeration vol) 5524 339 16.3 560 17.9

At Qp 34.0

cfm/scfm* cfm/1000cfm *Vt=(Pstp/Pt)x (Tt/Tstp) Vstp

Ave flow 0.87 14.2 Range 20-30

At peak flow 29.5

E-9 Calculate numuber of diffusers (Ts/Ts) Vs

From M&E p. 562 T 10-7, typical values 3.3-10 cfm per diffuser

Use diffuser with 5 cfm/diffuser at Qp Ceramic Disc Grids

number of diffusers = 2302

number of diffuser per tank= 575

L n Spacing(ft) spacing(in)

Coarse bubble spiral Diffuser spacing = L/n

Grid Spacing =sqrt( tank area/ no of diffusers) = 3

E-10 F/M Ratio So Q F (lbs/d)

F (lb/d) = Sox Qx 8.34 165 12 16513

MLVSS V M

M(lbs) = MLVSS V 8.34 2800 2.53 59155.7

F/M ratio = 0.28 Typical range 0.2-0.4

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E. AS

E-11 Organic Loading ( #BOD/d per 1000 cf aetation volume)

F ( lbs/d) 16513

V ( 1000 cf) 339

F/V 48.8 Typical range 50-120

Alternate I. Trickling Filter

Es = 100/(1+0.0085 sqrt(W/VF)) V = Volume in Acre-Ft W = BOD applied lbs/d

V = W/F [(100/Es-1)/0.0085]^(-2) Es= BOD removal Efficiency

Es=100*(So-Se)/So

F = (1+R)/(1 + 0.1 R)**2

R = Recirculaiton ratio

Es R F W V (AC-Ft) V ( cf)

Calculate V 88 2.0 2.08 16513 30.10 1,311,216

# filters Depth (ft) SF each Diam ( ft)

Filter Dimensions 6 6 36423 215

Alternate II. RBC See Text Figure 10-39

Hydraulic loading depends in soluble BOD in influent and total BOD in effluent.

Influent total BOD 165 mg/l Assume 70% soluble

Influent soluble BOD 115.5

Effluent total BOD 20

Hydraulic loading rate 1.6 gpd/sf From Fig 10-39 , P. 634

Calculate A required 7500000 SF

Check Organic Loading 2.20 lbs/d per 1000 SF Typical range 2- 3.5

Each shaft provides 140 MSF( average value) for shaft length 27 ft

No of shafts required 54

No of shafts per train 4

No.of shafts per train 13.4

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E. AS

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E. AS

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E. AS

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F. Final Clar

Module F. Final Clarifier

Qp SOR SL

Design Criteria (MGD) (gpd/sf) (#/d per sf)

at Qp 25 1500 50

at Q ave 12

Area required for clarification = Q/SOR = 16667 SF

Note: See P.588. Q excludes Qr becasue return sludge is drawn off the bottom, therefore,

does not contibute to overflow rate.

Area required for thickening = Total SS lbs/d / SL

Qr/Q Qp (Q+Qr) peak MLSS Total SS

Total solids (lb/d) 0.39 25 34.72 3500 1013541.7

Area required for thickening (SF) = 20271 Use the higher area for design

d DT (h) ave DT (h) pk

Detention time = A d / (Q+Qr) 10 2.2 1.05

No. of circular tanks= 4

Tank dimensions A Dia

Each Tank 5068 80

Module G. Gravity Thickener

# tanks SLR SOR Thicken S

Design Criteria lbs/d/sf gpd/sf % solids

2 10 800 8%

Sludge Quantity SS(mg/l) Q SS (lbs/d) VSS/SS VSS(lbs/d)

SS in WW 220 12 22018 0.7 15412

WAS ( From E-5) 5768 0.80 4615

Total Sludge 27786 20027

Total A Each A Diam (ft)

Circular Tank Dimensions 2779 1389.2881 42

SS (lb/d) SG Solid % V (cf/d) Depth (ft/d)

Volume of Thickend Sludge 27786 1.01 8% 5511 2.0

Blanket(ft) Rmvl(ft/d) SVI

Sludge Volume Ratio = V of sludge blanket/ V sludge removal 8 2.0 4.0

Typical range of SVI is 0.5-20. If SVI > 20, Bulking sludge.

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CE 432 Waste water treatment plant design

Gravity

Thickener

Waste characteristics BOD 5 250

Design Flow MGD CFS SS 220

Ave Q 12 18.60 Total P 9

Peak Q 25 38.75 NH3-N as N 30

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H. Ana Dig

Module H. Anaerobic Digester

Design Criteria

# Tank T digester T slud in T earth T air DT VSS load

oF days lb VSS/d/cf

3 95 50 40 40 15 0.2

SS (lbs/d) VSS/SS VSS(lbs/d) V (cf/d)

Primary sludge 22018 0.7 15412

Biosludge 5768 0.8 4615

Total sludge 27786 0 20027 5511

1 Tank Dimensions

Volume required for hyd loading = 82664 cf

Volume required for VSS loading= 100134 cf Use the higher value

Digester Dimensions

#Tank V (cf) each Depth(ft) A each Daim

Each Tank 3 33378 30 1113 38

2 Energy Production

Assumptions

15 cf gas per lb solids destroued

600 BTU per cf gas

50% of VSS applied are destroyed

TSS app Fix SS VSS appl VSS destr Gas prod E prod

#/d #/d #/d #/d cf/d MBTU/d

27786 7759 20027 10013 150201 90121

Digested Sludge Characteristics

VSS FS TSS ( lbs/d)

10013 7759 17772

3 Energy Requirements for Digester Heating (BTU/d)

Total Heat Req( Ht) = Heat req to raise temp of sludge ( Hs) + Heat req to compensate for loss ( Hl)

Hs = P ( #/d sludge dry solids) x ( 100/ps) ( Td-Ts) x (1/24 )x Cp

ps=% solids Cp = sp heat of sludge Td= temp of digester Ts = temp of sludge

P ps Td Ts Cp Hs (BTU/hr)

Cal Hs 27786 8% 95 50 1 651229

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H. Ana Dig

Hl= C ( Coef of heat flow) x A ( contact area) x ( T d- To) To is temp of outside

Cal Hl C Td To A Hl

Dry earth 0.08 95 40 3338 14686

Air ( Cover) 0.3 95 40 3338 55074

Air ( walls) 0.3 95 40 10642 175591

Wet earth 0.26 95

Total heat loss 245351

Total Heat requirements ( BTU/h) = Hs + Hl = 896580

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I. Sludge Dewater

Module I Sludge Dewatering

Disgested Sludge TSS (lbs/d) 17772

1 Solid Loading Rate for filter drum 4 #/hr per sf

Filter Drum Area (sf) = 185

2 Chemical Requirements

Dose lbs/d

Lime 8% 1422

Ferric Chloride 2% 355

3 Volume of Sludge Cake

Total Weight of sludge cake = 19550

Solid content of sludge cake = 18%

SG of dry solids is assumed to be 2

Calculate SG of sludge 1.18

Volume of sludge cake (cf/d) 1475

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