432 Project
Transcript of 432 Project
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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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