design cal_2300pe.xls
TRANSCRIPT
DESIGN OF AN EXTENDED AERATION TREATMENT PLANT 2300 P.E. CADANGAN PEMBANGUNAN BERCAMPUR DI ATAS TANAH KERAJAAN SELUAS 30.0EKAR, MUKIM PEDAH, DAERAH JERANTUT, PAHANG DARUL MAKMURUNTUK TETUAN DOYENVEST (M) SDN BHD
BASIC DATA:This is an outline design to produce 10:20 BOD5:SS effluent.
Population, PE = 2300 P.E
Dry Weather Flow, DWF = 0.225
Suspended Solid, SS = 300 mg/lBOD5, BOD = 250 mg/lInfluent Ammonia NH3-N = 30 mg/l
Effluent BOD5, EBOD = 10 mg/lEffluent SS ESS = 20 mg/lEffluent Ammonia ENH3-N = 10 mg/l
LOADING
a. Hydraulic flow, Qavg = PE * DWF
= 517.50
b. Peak Flow, Qpeak = Peak Factor * Qavg
-0.11
Peak Flow Factor Pff = 4.7(P) = 4.29
Therefore, Qpeak = Pff x Qavg
= 2219.31
= 1.54
c. Suspended Solids Loading Rate = Qavg * SS= 155.25 kg/d
d. BOD5 Loading Rate = Qavg * BOD = 129.38 kg/d
e. NH3-N Loading Rate = Qavg * NH3-N 15.53 kg/d
m3/c/d
m3/d
m3/d
m3/min
PRIMARY BAR SCREEN DESIGN
Design Population, PE 2300
Design Flow, Qavg 517.50 0.0060 l/sPeaking Factor, PF 4.29
Peak Flow, Qpeak 2219.31 25.81 l/s
Guidelines : Quantity of screenings = 30 m3 screening / 10^6 m3 wastewater
Number of Channels = 1Number of Back-up Channel = 1
Qavg = 517.50
Quantity of Screenings = Qavg x 30 / 1000000
= 0.0155
Provide storage for 7 days
Quantity of Screenings = 7 x Quantity of Screenings per day
= 0.109
Number of Storage Units = 1
Quantity per Unit = 0.109
Dimension of screenings troughL = 0.50 mW = 0.30 mD = 0.30 m
Volume = 0.045 > 0.109 error
Screen Design
Guidelines: Max flow through velocity at Qpeak Vmax = 1.0 m/sesGuidelines: Min approach velocity at Qpeak Vmin = 0.3 m/sec
Bar Size Bs = 10 mmClear Opening Co = 25 mm
Efficiency Coefficient Eff = Clear OpeningClear Opening + Bar size
= 0.71
Clear area through each screen at Qpe AQpeak = QpeakVmax x 24 x 60 x 60
= 0.026
Total cross sectional area of channel AC = AQpeakEff
= 0.036
Assume, Depth of Flow at Qpeak D = 0.050 m.
Required Width of Clear Opening @ Q Wclr = AC / D
= 0.72 m
Number of Openings No = Wclr x 1000Co
= 29
m3/day
m3/day
m3/day
m3/day
m3
m3
m3 m3
m2
m2
Number of Bars Nbars = No - 1= 28
Gross Width of Screen Wch = Wclr + (Nbars x Bs / 1000)
= 1.00 m
Set Wch = 0.50 m
Check velocities
Qpeak = 2219.31 0.02569 m3/sec
Velocity of approach Vapp = Qpeak / (Wch x D)= 1.03 m/sec > = 0.30 m/sec ok
Velocity through screen Vscr = Qpeak / (Wclr x D)= 0.71 m/sec < = 1.00 m/sec ok
Hydraulic Profile Through Screen
Headloss Through Clean Screen
Formula: HLc = .0729 (V^2 - v^2)
where HLc = ? Clean screen max headloss (m)V = 0.71 Velocity through screen (m/s)v = 1.03 Approach velocity (m/s)
Therefore HLc = -0.04 m
Elevation of Channel Invert = 76.75 m
Upstream Water Elevation = 76.76 m
Downstream Water Elevation = 76.80 m
Headloss Through Half Clogged Screen
Assume velocity through screen is doubled
Formula: HLc = .0729 (V^2 - v^2)
where HLc = ? Clogged screen max headloss (m)V = 1.429 Velocity through screen (m/s)v = 1.03 Approach velocity (m/s)
Therefore HLc = 0.07 m
Elevation of Channel Invert = 76.75 m
Upstream Water Elevation = 76.87 m
Downstream Water Elevation = 76.80 m
m3/day
PUMP SUMP DESIGN
Influent at Qpeak Influent = 2219.31 25.69 l/s
Influent at Qavg Influent = 517.50 5.99 l/s
Dimension of pump station L = 1.80 mW = 1.50 m
Depth between start and stop D = 0.65 mChamfer volume v = 1/2 x ( 2.05 x 0.2 x 0.4 )
= 0.072Therefore effective volume to fill /drain
V = L x W x D - v
= 1.68
Pumping rate @ Qpeak P = 26.00 l/s >= 25.69 l/s ok
Pump cycle time at QavgGuideline: 6 min, 15 max @ Qavg
Volume required for pump sump, V = T x q4
Assume number of start / stop = 15 times per hr (required 6 - 15 start/hour)
where V = ? Required Volume (m3)T = 4.0 Cycle Time (minute)q = 26.00 Pumping Rate (L/sec)
= 1.56 Pumping Rate (m3/min)
Therefore, V = 1.56 1.68 ok
Pump Headloss Calculations
VALVES & FITTINGS K QUANTITY DISCHARGEVALUE TOTAL
EXIT 1.00 1 1.00 CHECK VALVE 2.50 1 2.50 GATE VALVE 0.20 1 0.20 TEE THRU SIDE 1.80 1 1.80 ELBOW - 90 DEG 0.30 6 1.80
SUM OF K's FOR FITTINGS 7.30
PIPE DIAMETER 0.20 m
PUMPING RATE 26.00 L/sec
V = VELOCITY 0.83 m/sec
F = TOTAL FITTING HEAD LOSS 0.25 m = (K x V2 / 2g)
L = LINEAR METER OF STRAIGHT PIPE 8.00 m
C = HAZEN-WILLIAMS ROUGHNESS COEFFICIENT 100.00
P = HEAD LOSS USING HAZEN-WILLIAMS EQATION 0.05 m
TOTAL LOSS = F + P 0.30 m
S = STATIC HEAD LOSS (Worst Case Scenerio) 79.60 - 76.00 m3.45
TOTAL DYNAMIC HEAD 3.75 m
m3/day or
m3/day or
m3
m3
m3 < m3
Size pump for 26.00 L/sec @ 3.75 m TDH
The submersible pump selected is as follows:-Make = TSURUMIModel = 100B42.2Capacity = 18.20 L/secTotal Head = 6.20 mPower = 2.20 kWDischarge Size = 100 mmNo. of Units = 2 (1 duty; 1 standby)
CHANNEL INVERT EL 76.750.0 m
HIGH LEVEL ALARM EL 76.750.15 m
STANDBY PUMP CUT IN EL 76.600.15 m
DUTY PUMP CUT IN EL 76.450.65 m
ALL PUMPS CUT OFF EL 75.800.4 m
INVERT ELEVATION EL 75.40
Pump cycle time at QpeakGuideline: 6 min, 15 max @ Qavg
Time to fill sump = VQpeak
= 1.09 min
Time to empty sump = VQpump - Qpeak
= -3.75 min
Cycle time base on actual operating point = 60 min per hourTime fill + Time empty min per cycle
= -22.6 cycle per hour 6 - 15 cycle / hr error
Determine Size of Force Main Guideline: 1 - 2.5m/s velocity in pipe
Required size of raw sewage pipe =
=
= 0.115 m
= 100 mm > 115 mm error
Actual velocity in selected pipe = Qpump / cross section area of pipe2.32 m/s < 2.5 m/s ok
[ 4 x Qraw / (2.5 m/s x 3.14] 1/2
[ 4 x 0.0182 / (2.5 m/s x 3.14] 1/2
Determine Total Dynamic Head ( TDH )
TDH =
Where ; = Static head (m) …………A= Losses through the pipe ( Hazen - William Formula ) …………B= Losses through fittings …………C
A = 79.60 - 76.00 m= 3.45 m
B Hazen - William Formula= 6.82 1.85 x L
C
Where ; V = Velocity m/sec
V = Q / A = 11.50 L/sec
= 0.83 m/sec
C = Coefficient of roughnes = 100.00
L = Length of pipe, m = 8.00 m
D = Diameter of pipe, m = 0.20 m
= 0.050 m
C Losses through fittings
= KV² 2g
Where ; K = Head loss coefficeint 7.30
V = Velocity m/sec 0.83
G = Gravity, m/s² 9.81
= 0.25 m
Therefore ,
Total Dynamic Head ( TDH ) == 3.76 m
To plot chart for the System Curve & Pump Curve from the above equitition:-
Q (L/s) V (m/s) hst (m) hf (m) hm (m) TDH (m)
0 0.00 3.45 0.000 0.000 3.45 16.03.33 0.11 3.45 0.001 0.004 3.46 14.06.67 0.21 3.45 0.004 0.017 3.47 12.6
10.00 0.32 3.45 0.009 0.038 3.50 10.813.33 0.42 3.45 0.015 0.067 3.53 9.016.67 0.53 3.45 0.022 0.105 3.58 7.020.00 0.64 3.45 0.031 0.151 3.63 5.223.00 0.73 3.45 0.040 0.199 3.69 3
* refer to Chart attached
hst + hf + hm
hst
hf
hm
Thus, static head of pump, hst
hf V
D1.167
1000 x 3.142 x 0.10 2 / 4
Thus, losses through the pipe, hf
hm
hm
hst + hf + hm
Pump head (m)
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 300.00
5.00
10.00
15.00
20.00
INFLUENT RAW SEWAGE PUMP(TSURUMI MODEL 100 B42.2)
Flowrate, Q (L/sec)
Hea
d (m
)
System Curve
Pump Curve
Pump Operating Point18.2L/sec @ 5.5 m TDH
FINE SCREEN DESIGN
ParametersPopulation PE = 2,300
Peak Flow Qpeak = 2219.31
= 0.03
Design Flow Qavg = 517.50
= 0.01
Guidelines: Max flow through velocity at Qpeak Vmax = 1.00 m/sec
Guidelines: Min approach velocity at Qpeak Vmin = 1.00 m/sec
Set Elevation of Datum Hd = 0.00 m
Design
Bar Size Bs = 10.00 mmClear Opening Co = 12.00 mm
Efficiency Coefficient Eff = Clear OpeningClear Opening + Bar size
= 0.55
Clear area through each screen at Qpeak, AQpeak = QpeakVmax
= 0.0257
Total cross sectional area of channel required = AQpeakEff
= 0.05
Assume Depth of Flow D = 0.055 m - - - - - 1
Width of Clear Openings Wclr = AQpeak / D
= 0.47 m
Number of Openings No = Wclr x 1000Co
= 39
Number of Bars Nbars = No - 1= 38
Width of Frame Wfr = 0.10 m
Gross Width of Screen W = Wclr + Wfr + (Nbars x Bs / 1000)= 0.95 m
Set width of screen W = 0.50 m
Determine Width of Downstream Throat
Total Peak Flow, Qpeak = 1.71 B (H^1.5)
Qpeak = 0.026 m3/s
Set depth of flow H = 0.055 m from - - 1
Determine width of throat B = ? m
Therefore B = Qpeak / 1.71 / (H^1.5)= 1.16 m
= Throat width of grit chamber
m3/day
m3/sec
m3/day
m3/sec
m2
m2
OK
Energy equation between upstream and downstream of clean screen
Velocity at Qpeak through clear openings of screen, Vs
Vs = QpeakWclr x d1
= 0.65 m/sec
Headloss through clean screen hLs = (Vs^2 - v1^2) / 2g x 1/0.7= 0.0039 m
X2 =E2 + d2 + (v2^2 /2g) X3 = E3 + d3 + (v3^2 /2g) + hLs
X2 = X3
Where,E2 = 0.000 m Height above datum
Trial and Error d1 = 0.09 m Upstream Depthv1 = 0.60 m/sec Upstream VelocityE3 = 0.000 m Height above datumd2 = 0.055 m Downstream Depth @ Qpeakv2 = 0.93 m/sec Channel Velocity
Therefore, X2 = 0.10X3 = 0.10 ok
0.000
Elevation of Channel Invert = 79.00 m
Upstream Water Elevation = 79.09 m
Downstream Water Elevation = 79.06 m
Energy equation between upstream and downstream of 50% clogged screen
Headloss through clogged screen hLcs = (Vcs^2 - v2^2) / 2g x 1/0.7
Velocity through clogged screen Vcs = Qpeak(Assume 50% clogging) Wclr x 0.5 x Upstream Depth d"
= 0.92 m/sec
hLcs = 0.05Therefore,
X2 =E2 + d" + (v"^2 /2g) X3 = E3 + d3 + (v3^2 /2g) + hLcs
X2 = X3
Where,E2 = 0.00 m Height above datum
Trial and Error d" = 0.120 m Upstream Depthv" = 0.43 m/sec Upstream VelocityE3 = 0.00 m Height above datumd3 = 0.055 m Downstream Depth @ Qpeakv3 = 0.93 m/sec Channel Velocity
Check:X2 = 0.13X3 = 0.15 ok
-0.018
Elevation of Channel Invert = 79.00 m
Upstream Water Elevation = 79.12 m
Downstream Water Elevation = 79.06 m
DESIGN OF A CONSTANT VELOCITY GRIT CHANNEL
Guidelines = 0.2 m/s flow through velocity
Number of Channels Provided = 1Number of Back-up Channels Provided = 1
Constant Velocity Channel Formula
Q =
Q = Flow B = Width of Throat mH = Depth of Flow m
Therefore, = ( Q / 1.71 / B )^0.67
Area of Parabola, A = 2/3 W HW = Width of Parabola mH = Depth of Flow m
Assume:V = 0.2 m/s (Per guidelines)
Q = A x V= 2/3 W H x V
Therefore W = 1.5 QH x V
Assume: B = 0.33 m
Qavg per Channel = 517.50 0.0060
FLOW FACTOR Depth Width( x QAVG) m3/sec H(m) W(m)
0.10 0.001 0.01 0.430.25 0.001 0.02 0.580.50 0.003 0.03 0.731.00 0.006 0.05 0.92 @ Qavg2.00 0.012 0.08 1.163.00 0.018 0.10 1.334.00 0.024 0.12 1.464.29 0.026 0.13 1.50 @ Qpeak5.00 0.030 0.14 1.576.00 0.036 0.16 1.677.00 0.042 0.18 1.76
Determine Length of Channel
Say settlement vel of particle Vs = 0.02 m/s
Say velocity of flow V = 0.2 m/s
Depth of channel at peak flow, H = 0.13 mWidth of channel, W = 1.50 m
Length of Channel L = V x HVs
L = 1.29 m
Provided Length L = 3.00 m
Surface Area of Channel SA = 4.49
Surface Overflow Rate SR = Qpeak / SA
= 494.50 < 1500
1.71 B (H1.5)
m3/s
m3/day m3/s
Q AT DIFFERENT FACTOR
m2
m3/m2/day m3/m2/day
Quantity of grit
Guidelines = 0.03 m3 / 1000 m3 of wastewater
Grit Quantity per Channel = Qavg x 0.03/1000
= 0.016
Provide area for 30 days storage
Grit quantity = 0.47
Dimension of hopper at bottom of grit channelL : W ration 2:1 as per guidelines
L = 3.00 mW = 0.50 mD = 0.15 m
Volume = 0.225 >= 0.466 ok
1500 mm
700 mm
50 mm
450 mm
1000 mm
Check ratio of grit tank dimension
Ratio W: D = 1 : 2 W D2.00 1
Ratio W: L = 1 : 2 W L1 2
Hydraulic Retension Time, Tr
Guidelines = 3 min @ Qpeak < 5,000 pe
Area of retangular = 1.05
Area of trapezium = 0.06
Area of grit storage = 0.45
Total Area = 1.56L = 3.00 m
Volume, V = 4.69
= 2219.31
= 1.54
Tr == 3.04 min >= 3 min ok
m3/day
m3
m2
m2
m2
m2
m3
Qpeak m3/day
m3/min
V / Qpeak
X-Axis Y-Axis-0.837 0.161-0.787 0.143-0.748 0.129-0.731 0.123-0.664 0.101-0.580 0.077-0.460 0.049-0.365 0.031-0.290 0.019-0.214 0.011
0 00.214 0.0110.290 0.0190.365 0.0310.460 0.0490.580 0.0770.664 0.1010.731 0.1230.748 0.1290.787 0.1430.837 0.161
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
CONSTANT VELOCITY FLOW CHANNELX - SECTION
Width (m)
Dep
th (
m)
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
CONSTANT VELOCITY FLOW CHANNELX - SECTION
Width (m)
Dep
th (
m)
GRIT STORAGE
Width DepthX - Axis Y - Axis
0.427 0.0110.580 0.0190.731 0.0310.921 0.0491.160 0.0771.328 0.1011.462 0.1231.496 0.1291.575 0.1431.673 0.161
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
CONSTANT VELOCITY FLOW CHANNELX - SECTION
Width (m)
Dep
th (
m)
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
CONSTANT VELOCITY FLOW CHANNELX - SECTION
Width (m)
Dep
th (
m)
GRIT STORAGE
-0.650 -0.450 -0.250 -0.050 0.150 0.350 0.550-0.150
-0.050
0.050
0.150
0.250
0.350
0.450
0.550
0.650
0.750
0.850
0.950
1.050
1.150
1.250
CONSTANT VELOCITY FLOW CHANNELX - SECTION
Channel Width (m)
Dep
th (
m)
GRIT STORAGE
DESIGN OF A GRIT / GREASE CHAMBER
Guidelines : Detension time T = 360 sec @ Qpeak
Number of Channels Provided = 1Number of Back-up Channels Provided = 1
Peak Flow Qpeak = 2219.31 m3/day= 0.026 m3/sec
Detension Time T = 6 min= 360 sec
Volume Required V = Qpeak x T= 9.25 m3
Provide:Length L = 2.90 mWidth W = 1.60 mDepth D = 0.85 m
Volume Provided V = 3.94 m3 > 9.25 m3 error
Ratio L : W = 1.81 : 1.0W : D = 1.88 : 1.0 ok
Determine Length of Channel
Provided Length L = 2.90 mWidth of channel, W = 1.60 mSurface Area of Channel SA = 4.64 m2
Surface Overflow Rate SR = Qpeak / SA= 478.30 m3/m2/day< 1500 m3/m2/day ok
Quantity of grit
Guidelines = 0.03 m3 / 1000 m3 of wastewater
Grit Quantity = Qavg x 0.03/1000= 0.016 m3/day
Allow for storage of 30 days = 0.47 m3
Quantity of grease
Average quantity of grease S = 8 kg / 1000 m3Specific gravity of grease sg = 0.95
Therefore quantity of grease Qs = Qavg x S= 4.14 kg/day
Flowrate of Grease Fs = Qs / 1000 / sg= 0.0044 m3/day
Allow for storage of 30 days Vs = 30 x Fs= 0.131 m3
Grit/Grease Drying Bed Sizing
Accumulated grease for 30 days Vs = 0.131 m3Accumulated grit for 30 days = 0.466 m3
Depth of Feed D = 0.25 m
Therefore, Area required A1 = 2.39 m2 Grease
Dimension of Grit/Grease Drying Bed
Length = 1.50 mWidth = 0.75 m
Area = 1.13 m2 > 2.39 m2 error
DESIGN OF A GREASE CHAMBER
Guidelines : Detension time T = 180 sec @ Qpeak
Number of Chamber Provided = 1Number of Back-up Chamber Provided = 1
Peak Flow Qpeak = 2219.31
= 0.026
Detension Time T = 180 sec
Volume Required V = Qpeak x T
= 4.62
Provide:Length L = 3.80 mWidth W = 1.20 mDepth D = 1.10 m
Volume Provided V = 5.02 4.62 ok
Grease Quantities
Average quantity of grease S = 8Specific gravity of grease sg = 0.95
Therefore quantity of grease Qs = Qavg x S= 4.14 kg/day
Flowrate of Grease Fs = Qs / 1000 / sg
= 0.0044
Allow for storage of 30 days Vs = 30 x Fs
= 0.13
Depth of grease baffle required to contain scum for 30 daysLength L = 3.80 mWidth W = 1.20 mDepth of water below baffle D = 0.02 m
Volume of grease storage provide V = 0.09 0.13 ok
m3/day
m3/sec
m3
m3 > m3
kg / 1000 m3
m3/day
m3
m3 > m3
WEIR BEFORE AERATION BASINS not use
Design Population, PE 2300Design Flow, Qavg 517.50 m3/day 0.01 m3/secPeaking Factor, PF 4.29Peak Flow, Qpeak 2219.31 m3/day 0.03 m3/sec
Number of Aeration Basins, N = 1
Peak Flow to Each Aeration Basin = Qpeak / N= 0.026 m3/sec
Formula Q(m3/sec) = Cw x Le x H^1.5
where Cw = 1.84 Weir CoefficientLe = 0.3704 m LengthH = 0.148 m Height
With end contractionLe = L - (0.1)nH
where L = 0.40 m Lengthn = 2 Number of side contractionsH = 0.148 m Height of Flow
Therefore Le = 0.3704 m
Therefore Q = 0.039 m3/sec >= 0.026 m3/sec ok
Height of water above weir = 102 mm Qpeak if both tanks open
Height of weir = 170 mm Qpeak if one tank open
Therefore, size opening for 400mm x 250 mm H = 272 mm
RECTANGULAR WEIR AT OUTLET BOX
Design Population, PE 2300Design Flow, Qavg 517.50 m3/day 0.01 m3/secPeaking Factor, PF 4.29Peak Flow, Qpeak 2219.31 m3/day 0.03 m3/sec
Formula Q(m3/sec) = Cw x Le x H^1.5
where Cw = 1.84 Weir CoefficientLe = 0.578 m LengthH = 0.11 m Height
With end contractionLe = L - (0.1)nH
where L = 0.60 m Lengthn = 2 Number of side contractionsH = 0.110 m Height of Flow
Therefore Le = 0.578 m
Therefore Q = 0.039 m3/sec >= 0.026 m3/sec ok
Height of water above weir = 110 mm @ Qpeak
90 DEG V-NOTCH WEIR AT OUTLET BOX
Design Population, PE 2000
Design Flow, Qavg 450.00 m3/day
Peaking Factor, PF 4.35
Peak Flow, Qpeak 1959.73 m3/day 0.023 m3/sec
90 deg V-Notch Weir Formula5/2
q (m3/sec) = 1.417 H
H (m) = 192 mm or 0.192 m at Qpeak
Therefore, q = 0.023 m3/sec >= 0.023 m3/sec ok
Set depth of weir @ 250mm ok
DESIGN OF ANOXIC CHAMBER
Population PE = 2300
Peak Flow = 2,219.31
= 92.47
= 1.54
= 0.026= 25.69 l/s
Design Flow = 517.50
= 21.56
= 0.36
= 0.01= 5.99 l/s
Peaking Factor PF = 4.29
Using the Lodification Ludzack-Ettinger Process
Assume
Influent Ammonia-N Ni = 30 mg/lEffluent Ammonia-N Ne = 10 mg/lMixed liquor Suspended Solids MLSS = 3600 mg/lMixed Liquor Volatile Suspended Solids MLVSS = 2880 mg/l (MVLSS = 0.8 x MLSSTemperature Temp = 28 deg CDissolved Oxigen Do = 0.1 mg/lSpecific Denitrification Rate U = 0.11 per dayOverall Denitrification rate Ua = ? per dayResidence Time Temp = ? hrs
Calculation Overall Denitrification Rate
Formula : Ua =0.20 per day
Calculation Residence Time
Formula : T = Ni - NeUa x MLVSS x 24 hrs/day
= 0.81 hrs required
Size anoxic zone for a residence time of Anox = 2 hrs
Determine size of anoxic tank
Anoxic zone volume required, Vol = Anox x Qavg24
= 43.13
Number of Anoxic Tank, N = 2 nos
Volume required per Tank, Vreq = 21.56
Tank DimensionLength = 2.30 mWidth = 1.50 mDepth = 4.50 m
=
Anoxic tank volume = 31.05 > 21.56 ok
Actual hydraulic retention time provided = Anoxic tank volumeQavg
= 2.88 hrs
Qpeak m3/day
m3/hr
m3/min
m3/s
Qavg m3/day
m3/hr
m3/min
m3/s
U x 1.09(Temp-20)X(1-DO)
m3
m3
m3 m3
EXTENDED AERATION TANK DESIGN
Sizing
Number of Aeration Tank, No. = 2
Minimum Hydraulic Retention Time, HRT = 19.4 hours
Volume Required, V = 418.31
Volume Required Per Tank, Vol = 209.16
Dimension of each tank: Ratio
Area A = 46.48 L : WLength L = 10.60 m 10.60 : 4.40Width W = 4.40 m 2.41 : 1Depth D = 4.50 mFreeboard FB = 0.65 m
Volume Vol = 209.88
Total Volume Provided Vp = 419.76 418.31
Therefore, HRT Provided = 19.47 hrs > 19.4 hrs ok
Sludge Age (MCRT)
Guidelines > 20 daysAssume, MLSS = 3600 mg/l
SA = Total solids in aeration tankExcess sludge wasting / day + Solids in effluent
Total solids in aeration tank = MLSS x Vp / 1000= 1511.14 kg
Solids in effluent = 10 mg/l= 5.18 kg/day
Sludge Yield, = 0.40 @ 24 hours HRT= 0.60 @ 18 hours HRT
Determine Sy @ Actual HRTVia Interpolation Therefore actua = (24 -21.9) x (0.6 - 0.4) + 0.4
(24 - 18)
Sy = 0.55 kg / kg BOD removed based on HRT interpolation
Excess sludge wasting / day Sd = QAVG x Sy (BOD5-EBOD5)(Sludge accumulation per day) = 68.45 kg/day
SA = 20.53 days ok
Waste Activated Sludge (WAS)
Excess Sludge Wasting / day WAS = 68.45 kg/day
Assume underflow concentration = 1% or 10,000 mg/l or 10 kg/m3
Volume of WAS = 6.84= 6844.61 l/day
m3
m3
m2
m3
m3 > m3
m3/day
Return Activated Sludge Flow (RAS)
QRAS = MLSS x QAVGCu - MLSS
Cu = Underflow concentration assume at 0.8 % solid or 8,000 mg/l
QRAS = 423.41 4.92 l/s
Set QRAS = 465.75 5.42 l/s
Qavg = 517.50
Ratio of QRAS : QAVG = 0.90 ok
Oxygen Requirements
Oxygen required per kg of BOD5 re OR = 2.00 kg O2 / kg BOD5
Actual Oxygen Required AOR = OR x BOD5 removed= 248.40 kg/day
Maintain DO of 2 mg/l in aeration tank
QAVG = 517.50
Total = 517.50
Oxygen required in tank = Total x 2 / 1000= 1.04 kg/day
Total AOR = 249.44 kg/day
Oxygen correction factor AOR/SOR
AOR/SOR = (((BETA * ACF * Csf @ 28 deg) - C ) / Css ) * ALPHA * 1.024^T-20
Where BETA = 0.98ACF = 0.989Csf = 7.83C = 1.5
Css = 9.08Alpha = 0.75
T = 30
Therefore AOR/SOR = 0.64
Standard Oxygen Required SOR = 391.23 kg/day
O2 transfer efficiency O2eff = 20 % @ 6 (See catalog)
Actual O2 transfer efficiency O2act = SOR kg/dayO2eff/100 x 1.201 kg/m3 x 0.232 kg O2/kg air x 1440 min/day
Total amount of air required= 4.88 m3/min
Number of diffusers required
Flowrate per diffuser = 0.10 or 6= 100.00 L/min
Quantity = Total air required / Flowrate
= 48.75 Use 52 ok
Number of diffusers per tank = 26 Nos in each Aeration tank
m3/day
m3/day
m3/day
m3/day
m3/day
m3/hr
m3/min m3/hr
Check F/M Ratio
Guidelines for F/M ratio is between 0.05 to 0.10 kg BOD / kg MLSS
Influent BOD5, = 250 mg/l= 129.38 kg/day
MLSS = 3600 mg/l
Volume provided, Vp = 419.76
F/M Ratio = INBOD / (MLSS x Vp)= 0.086 ok
Check Aeration Loading
AL = INBOD / Vp
= 0.31 ok
Check Mixing Rate
Volume of air from Blower Va = 4.88 (See Blower calculations)
Volume of tank provided Vp = 419.76
Therefore, Mixing Rate = Va / Vp
= 0.012 ok
m3
Guidelines : 0.1 - 0.4 kg/m3/day
kg/m3/day
Per EPA guidelines, 0.010 to 0.025 m3/min. m3
m3/min
m3
m3/min. m3
CLARIFIER DESIGN
Guidelines : Hydraulic retension time (HRT) = 2 hours @ QpeakGuidelines : Surface overflow rate = 30 m3/m2/dayGuidelines : Side water depth (SWD) = 3 m
Surface Area =
= 73.98
Number of clarifiers = 2
Surface area required per clarifier = 36.99
Dimension of each clarifier
L = 6.20 mW = 6.20 m
Therefore, actual area provided = 38.44 36.99 ok
Provided Length, L = 6.20 mProvided Width, W = 6.20 m
Effective Volume = Volume of cone with slope of 60 deg.SWD in cone = 3.46 m (See attach volume cals)
Cone volume = 65.73 okVertical SWD = 1.04 m
Vertical SWD volume = 39.98
Total Volume = 105.70Total SWD = 4.50 m
HRTPer guidelines: < 2 hours
== 2.29 >= 2 hours ok
Overflow WeirPer guidelines: 150 - 180 m3/day/m
Max weir loading rate, Wr = 180Peak Flow, Qpeak = 2219.31
Min Weir Length Required per Clarifier = Qpeak / Wr / No. of Clarifiers= 6.16 m
Total length of weir provided per Clarifier, Tw = 3.00 m
Actual weir loading rate = 369.89 error
Solids Loading Rate (SLR)Guideliness : < 150 kg/m2/day at Qpeak
< 50 kg/m2/day at Qavg
MLSS = 3600 mg/l or 3.6
Total Q = Qpeak + QRAS(For QRAS look under Extended Aeration Tank Sizing)
= 2685.06
SLR (Qpeak) = (Total Q / Actual Surface Area) x MLSS kg/day
= 125.73 ok
SLR (Qavg) = (Qavg + Qras) / Actual Surface Area x MLSS kg/day
= 46.04 ok
Qpeak / 30 m3/m2/day
m2
m2
m2 > m2
m3
m3
m3
Total Volume m3 / Qpeak m3/day x 24 hours
m3/day/mm3/day
m3/day/m
kg/m3
m3/day
kg/m2/day
kg/m2/day
Launder Calculation
Overflow from weir to lauder = Qpeak / number of tank
= 1109.66
Launder sizeLength = 3.00 mWidth = 0.25 mFlow velocity = 0.80 m /s < 1.0 m/s okDepth of flow in launder = 21.41 mm < 0.3m as provided
m3 / launder
Determine Volume of Cone
L
W
L W
h h 60 y
BW 1 A BW2 B
L = 6.20 mW = 6.20 m
BW 1 = 2.20 mBW 2 = 2.20 m
A = (L - BW1) / 2= 2.00
B = (W - BW2) / 2= 2.00
h = height of cone= A x Tan 60
3.46 m
Tan y = h /By = Tan -1 (h/B)
= 60.00 degrees
Therefore, Volume of Cone V = [ (1/3 x 2A x 2B x h) +(1/2 x 2A x BW2 x h) + ( 1/2 x (W + BW2) x h) ]
V = 65.73 m3
L (m)
W (m)
h (m)
A (m) BW1 (m)
Scum Withdrawal Airlift Pipe Design
Percent submergence = Hs / (Hs + Hl) x 100
Hs = Depth (m) of air pipe below water surfaceHl = Height (m) of lift
Scum AirliftHs = 3.8 mHl = 0.7 m
Percent Submergence = 84.44 %
With reference to the attached chart, Discharge = 45 gpm 2.84 l/sVelocity = 2 ft/sec 0.61 m/sAir Requirement = 7 cfm 0.20 m3/min
AEROBIC SLUDGE HOLDING TANK DESIGN
Guidelines: Sludge yield = 0.6 kg/kg BOD5 (Standard "A")
Design Population, PE = 2300
Design Flow, QAVG = 517.5
Sludge Yield Ys = 0.55 kg/kgBOD5/day (Calculated earlier)
Sludge accumulation per day Sd = QAVG x Ys (BOD5-EBOD5)= 68.31 kg
Temperature of Wastewater T = 28 Deg C
Percent Volatile Suspended Solid (VSS) VSS = 75 %
Percent VSS Destruction VSSd = 55 %
Influent to Digester WAS = 68.31 kg/day
Influent Solid Content Conc = 1 %
Total Volatile Solid TVS = VSS/100 x WAS= 51.23 kg/day
Total Volatile Solid Destruction TVSd = VSSd/100 x TVS= 28.18 kg/day
Therefore, Total Solids Remaining After Digestion TSd = Nonvolatile Solid + VS Remaining
= (WAS - TVS) + { (1 - VSSd/100) x TVS }
=
TSd = 40.13 kg/day
Density of water ρ = 1000.00
Specific gravity of sludge sg = 1.015
Therefore, TSd in term of volume = 3.95
Storage days provided T = 30 days
Volume of Tank Required Vtank = 118.62
Number of Tank Provided No. = 1
Volume of Tank Required V = 118.62
SIZE OF EACH TANK
Depth = 4.5 mWidth = 3.3 m
Length = 8.0 m
Volume = 118.80 118.62 ok
m3/day
(26.73 - 20.05) + { (1 - 55/100) x 20.05)
kg/m3
m3/day
m3
m3
m3 > m3
OXYGEN REQUIREMENT FOR DIGESTIONGuidelines: 1.5kgO2/kgBOD
Wt of sludge digested Ws = Sd - TSd
= 28.18 kg
Amount of oxygen required, (AOR) Ws x 1.5 = 42.27 kg
Oxygen correction factor AOR/SOR
AOR/SOR = (((BETA * ACF * Csf @ 28 deg) - C ) / Css ) * ALPHA * 1.024^T-20
Where BETA = 0.98ACF = 0.989Csf = 7.83C = 1.5
Css = 9.08Alpha = 0.75
T = 30
Therefore AOR/SOR = 0.64
Standard Oxygen Required SOR = 66.29 kg/day
Assume O2 transfer efficiency O2eff = 10 %
Total amount of air required = SOR kg/dayO2eff/100 x 1.201 kg/m3 x 0.232 kg O2/kg air x 1440 min/day
= 1.65
Number of diffusers required
Flowrate per diffuser = 200.00 L/min
0.200Quantity =
Total air required / Flowrate=
8.26 Use 10 ok
m3/min
m3/min
DRYING BED DESIGN
Digested Sludge From Aerobic Digester DS = 40.13 kg/day (See Aerobic Digester cals)
Concentration of Sludge C = 1 %
= 10
Specific gravity of sludge sg = 1.015
Volume of Digested Sludge Vds = 3.95
Per guidelines: Provide 4 week cycle for 450mm thick feed depth
Volume of Vds required V = Vds(m3/day) x 28days(Based on 21 days drying; 7 days feeding)
= 110.71
Depth of Feed D = 0.45 m
Therefore, Area required A = 246.02
Provide fully covered drying beds
Therefore actual area required A = 164.01
Number of Drying Beds provided = 4
Dimension of Drying Beds
Dimension of Bed Length = 16.50 mWidth = 2.50 m
Area = 41.25
Total area of drying bed provided = nos of bed x bed area
= 165.0 164.0 ok
Determine Volume of Dewatered Sludge (DS)
Assume final sludge concentration is 25%
DS = Vs x Si x SGi x n SGo x So
DS = ? Volume of Dewatered Sludge, (m3/day)Vs = 3.95 Volume of Influent Sludge (m3/day)Si = 0.01 Fractional Percent Solid Content of Influent SludgeSo = 0.25 Fractional Percent Solid Content of Dewatered SludgeSGi = 1.015 Specific Gravity of Sludge Before ThickeningSGo = 1.03 Specific Gravity of Sludge After Thickening
n = 0.95 Fractional Percent Capture
Therefore
DS = 0.15 @ 25% solid
Provide covered storage for 30 days
Volume required V = DS x 30 days
= 4.44
Dimension provided Depth = 0.50 mWidth = 1.00 mLength = 4.60 m
Volume = 2.30 4.44
kg/m3
m3/day
m3
m2
m2 ( 1/3 reduction in area)
m2
m2 > m2
m3/day
m3
m3 > m3 ok
Drying Bed Feed Pump Design
Pump Sizing
Pumping rate per pump P = 9.10 L/sec= 546.00 L/min
Pump Headloss Calculations
VALVES & FITTINGS K QUANTITY SUCTION QUANTITY DISCHARGEVALUE TOTAL TOTAL
ENTRY 0.50 1 0.50 0 0.00 EXIT 1.00 0 - 1 1.00 CHECK VALVE 2.50 0 - 0 0.00 SLUICE VALVE 0.20 0 - 1 0.20 REDUCER 0.30 0 - 1 0.30 TEE THRU SIDE 1.80 0 - 0 0.00 TEE THRU RUN 0.60 0 - 1 0.60 ELBOW - 45 DEG 0.23 0 - 0 0.00 ELBOW - 90 DEG 0.30 0 - 5 1.50
SUM OF K's FOR FITTINGS 0.50 3.60
PIPE DIAMETER 0.10 m 0.10
PUMPING RATE 9.10 L/sec 9.10
V = VELOCITY 1.16 m/sec 1.16
F = TOTAL FITTING HEAD LOSS 0.03 m 0.25 = (K x V2 / 2g)
L = LINEAR METER OF STRAIGHT PIPE 0.00 m 15.00
M = MULTIPLYING FACTOR 1.10 1.10
C = HAZEN-WILLIAMS ROUGHNESS COEFFICIENT 100.00 100.00
P = HEAD LOSS USING HAZEN-WILLIAMS EQATION 0.00 m 0.39
TOTAL LOSS = F + P 0.03 m 0.64
TOTAL SUCTION + DISCHARGE LOSS 0.73 m
S = STATIC HEAD LOSS (Worst Case Scenerio) 4.00 m
TOTAL DYNAMIC HEAD 4.73 m
Size pump for 9.10 L/sec @ 4.73 m TDH
The submersible pump selected is as follows:-Make = EbaraModel = 65DVS51.5Capacity = 9.10 L/secTotal Head = 4.73 mPower = 1.50 kWDischarge Size = 65 mm connect to 100 mmNo. of Units = 1 (1 duty)
Determine Size of Force Main Guideline: 1 - 2.5m/s velocity in pipe
Required size of sludge pump =
=
= 0.068 m
= 80 mm > 68 mm ok
Actual velocity in selected pipe = 1.81 m/s > 2.5 m/s ok
[ 4 x Qsludge / (2.5 m/s x 3.14] 1/2
[ 4 x 0.0091 / (2.5 m/s x 3.14] 1/2
Determine Total Dynamic Head ( TDH )
TDH =
Where ; = Static head (m) …………A= Losses through the pipe ( Hazen - William Formula ) …………B= Losses through fittings …………C
A = 4.00 m
B Hazen - William Formula= 6.82 V 1.85 X L
C
Where ; V = Velocity m/sec 1.16 m/s
V = Q / A = 8 L/sec
= 1.16 m/sec
C = Coefficient of roughness 0.00
L = Length of pipe, m 15.00
D = Diameter of pipe, m 0.10
= #DIV/0! m
C Losses through fittings
= KV² 2g
Where ; K = Head loss coefficeint 4.10
V = Velocity m/sec 1.16
G = Gravity, m/s² 9.81
= 0.29 m
Therefore ,
Total Dynamic Head ( TDH ) == ### m
To plot chart for the System Curve & Pump Curve from the above equitition:-
Q (L/s) V (m/s) TDH (m) Pump head (m)
18.6 40 0.00 4 18.6 16.4 4.042 0.25 4.04 16.4 14.2 4.154 0.51 4.15 14.2 13 4.238 1.02 4.57 8.3 8.33 4.57
9.5 1.21 4.80 4.2 4.2 4.8
* refer to Chart attached
hst + hf + hm
hst
hf
hm
Thus, static head of pump, hst
hf
D1.167
1000 x 3.142 x 0.10 2 / 4
Thus, losses through the pipe, hf
hm
hm
hst + hf + hm
0 2 4 6 8 100
2
4
6
8
10
12
14
16
18
20
DRYING BED FEED PUMPSubmersible Pump
EBARA Model 65DVS51.5
Flowrate, Q (L/sec)
Hea
d (m
)
System Curve
Pump Curve
Pump Operating Point9.2 L/sec @ 4.7 m
X-Axis Pump System Pump System0 16.2 4 0
3.3 14.2 4.25 1986.67 12.4 4.92 400.2
10 11 5.97 60013.33 9 7.39 799.816.67 7.4 9.17 1000.219.98 5.6 11.27 1198.8
65DVS X-Axis Pump System0 18.6 42 16.4 4.044 14.2 4.158 8.33 4.57
9.5 4.2 4.8
3.580986
Blower- 1 Sizing
Air Requirement for Aeration = 4.88
Air Requirement for Aerobic Sludge Digestion = 1.65
TOTAL AIR REQUIREMENT = 6.53
= Qd x (1.0332 + Pd) x
1.0332
= ? Air flow under standard condition
= 6.53 Air flow under discharge condition
= 0.41 Discharge static pressure
PS = -0.05 Suction static pressure
= 30 Suction temperature
= 38 Discharge temperature
Therefore,
= 8.88
= - 1 x 1.03321.0332 + PS
= 0.48
Size blower for 9.09 m3/min @ 0.48 kgf/cm2Provide 2 blowers : 1 duty, 1 standby
Model: Fu Tsu Model TSC 125, 920rpm, 15.0kW
m3/min
m3/min
Qd m3/min
Determine air flow under standard condition, Q s
Qs 273 + St
273 + Sd
Qs (m3/min)
Qd (m3/min)
Pd (kgf/cm2)
(kgf/cm2)
StoC
SdoC
Qs m3/min
Determine discharge pressure under standard condition, P s
Ps 1.0332 + Pd
kgf/cm2
Blower-3 Sizing
Air Requirement for Scum Airlift = 0.20
Air Requirement for RAS = 0.6091
Air Requirement for WAS = 0.0090
Air Requirement for MLSS = 2.10
TOTAL AIR REQUIREMENT = 2.91
= Qd x (1.0332 + Pd) x
1.0332
= ? Air flow under standard condition
= 2.91 Air flow under discharge condition
= 0.45 Discharge static pressure
PS = -0.05 Suction static pressure
= 30 Suction temperature
= 38 Discharge temperature
Therefore,
= 4.08
= - 1 x 1.03321.0332 + PS
= 0.53
Size blower for 1.76 m3/min @ 0.53 kgf/cm2Provide 1 blowers : 1 duty
Model: Fu-Tsu Model TSC 80, 760 rpm @ 5.5 Kw
m3/min
m3/min
m3/min
m3/min
Qd m3/min
Determine air flow under standard condition, Q s
Qs 273 + St
273 + Sd
Qs (m3/min)
Qd (m3/min)
Pd (kgf/cm2)
(kgf/cm2)
StoC
SdoC
Qs m3/min
Determine discharge pressure under standard condition, P s
Ps 1.0332 + Pd
kgf/cm2
Sludge Withdrawal Airlift Pipe Design
Calculate the submerged distance of the air inlet (S) by the following formula:
S = Basin SWD - ( 1 + (3 x Pump Dia) / 12 )
where
SWD (ft) = 14.76 ft 4.50 mPump Dia (in) for RAS = 3.00 in 0.075 m
Pump Dia (in) for MLSS & scum = 2.00 in 0.050 m
S = 13.01 ft 3.97 m
= h / { C x LOG [ (H + 10.4) / 10.4 ] }
where
h = 0.75 total lift required (m)H = 4.50 submergence (m)C = 10.20 constant for less than 15m lift
= 0.47 m3/min of air per m3 of water
= 25.00 air-lift efficiency (%)
SCUM = 75.0
= 0.0521
=
= 0.0981
RAS = 465.8 See Aeration Tank cals
= 0.3234
=
= 0.6091
WAS = 6.84 See Aeration Tank cals
= 0.0048
=
= 0.0090
MLSS == 2070.00
= 1604.25
MLSS Return = 1604.3
= 1.1141
=
= 2.0984
Calculate the Air Supply Volume (V air) required:
Vair
Vair
A eff
Determine Volume of Air Requirement for SCUM, V SCUM
m3/day
m3/min
Vair-SCUM Vair x RAS x Aeff
m3/min
Determine Volume of Air Requirement for RAS, V RAS
m3/day
m3/min
Vair-RAS Vair x RAS x Aeff
m3/min
Determine Volume of Air Requirement for WAS, V WAS
m3/day
m3/min
Vair-WAS Vair x WAS x Aeff
m3/min
Determine Volume of Air Requirement for MLSS Return, V MR
4 Qave - QRAS
- 182.3 m3/day
m3/day
m3/day
m3/min
Vair-MLSS Vair x MLSS x Aeff
m3/min
Aeration Piping Headloss To Aeration Tanks
CriteriaAmbient air temperature To = 30 deg C 303.2 deg K
Ambient barometric pressure Po = 1.00 atm
Air supply pressure P = 1.450 atm
Blower capacity Qb = 6.53 m3/min
Blower efficiency e = 75 %
Equations
Friction factor f = 0.029 x D^0.027 Q^0.148
Temperature in pipe (deg K) T = To x (P/Po)^0.283
Velocity head Hv = 9.82E-8 x TQ^2 PD^4
Headloss (mm) hL = f x (L/D) x Hv or K x Hv
A. Pipe Fittings Losses
No. Valves & Fittings Size (mm) Quantity K Value Q (m3/min) T (deg K) hL, headloss (mm)1 Check valve 80 1 2.50 6.53 336.77 59.322 Gate valve 80 1 0.80 6.53 336.77 18.989 Reducer 80 0.20 6.53 336.77 0.003 Tee thru side 80 1 1.80 6.53 336.77 42.714 Tee thru run 80 1 0.60 6.53 336.77 14.245 Elbow 90 deg 80 2 0.30 6.53 336.77 14.246 Elbow 90 deg 80 0.30 6.53 336.77 0.004 Tee thru run 50 2 0.60 4.35 336.77 82.939 Reducer 25 2 0.20 1.09 336.77 27.647 Tee thru run 25 0.60 1.09 336.77 0.004 Tee thru run 25 0.60 1.09 336.77 0.005 Elbow 90 deg 25 1 0.30 1.09 336.77 20.738 Elbow 90 deg 25 0.30 1.09 336.77 0.009 Reducer 25 0.20 1.09 336.77 0.0010 Gate valve 25 1 0.80 1.09 336.77 55.29
Subtotal 336.08
B. Straight Pipe Losses
No. Length (m) DIA (mm) Velocity (m/min) Q (m3/min) f, fric. factor T (deg K) hL, headloss (mm)1 8.00 100 1130.94 8.88 0.02000 336.77 28.802 4.00 50 3015.84 5.92 0.02100 336.77 215.033 7.00 50 1507.92 2.96 0.02300 336.77 103.044 4.00 25 3015.84 1.48 0.02500 336.77 511.98
Subtotal 858.84
C. Supply Pressure At The Blower
1 Losses in piping = 858.84 mm2 Losses in pipe fittings = 336.08 mm3 Losses in air filter = 50.00 mm3 Losses in silencer = 50.00 mm4 Losses in blower = 150.00 mm5 Losses in diffusers = 160.00 mm6 Static head = 4100.00 mm
Total 5704.92 mm @ 5.70 m
Therefore, the absolute supply = 1.55 atmpressure
D. Power Requirement of Blower, P (kw)
P = w RTo x [ (P/Po)^0.283 - 1 ]8.41 e
where R = 8.314 kJ/k mole deg K
w = air mass flow, kg/s
Therefore, P = 7 Kw or 9 HP
Select next bigger size motor = 5.5 Kw 7 HPok
Size blower for 9.09 m3/min @ 0.48 kgf/cm2
Fu Tsu Model TSC 125, 920rpm, 15.0kW
Provide 2 blowers; 1 running, 1 standby
CHLORINATION TANK DESIGN
= 2219.31
= 1.54
Detension Time at Qpeak t = 15.00 min
Volume of Tank V =
= 23.12
Number of Tanks N = 1
Number of Bays per Tank n = 4
Dimension of Tank ProvidedDepth H = 1.80 m ok max 3mWetted depth h = 1.50Width W = 0.75 mLength L = 4.50 mNumber of pass n = 4
Volume Provided Vp = 24.30 23.12
Check:
Ratio Ratio okWetted depth : Width
1.50 : 0.752 : 1
Length : Width4.50 : 0.75
6 : 1
Detension Time at Qpeak t == 15.77 min > 15 min ok
PARSHALL FLUME FOR DISCHARGE
Design Flow, Qavg = 517.5Peaking Factor PF = 4.29
Peak Flow Qpeak = 2219.31
Formula for flow calculation with diffrent throat width of Parshall Flume by Harlan Bengtson
Flow thru 1" PF, Q =where
Flow tru PF Q = flow in PF in cfs= 0.9071091 cfs = Qpeak
Head over flume H = in ft
water level in PF H = 1.8263509 ft556.7 mm
Qpeak m3/day
m3/min
Qpeak x t
m3
m3 >= m3
Vp / Qpeak
m3/day
m3/day
0.338 x H1.55
For Q > Qpeak,
90 DEGREE V-NOTCH WEIR AT OUTLET BOX
Design Population , PE = 2300Design Flow, Qavg = 517.5 m3/dayPeaking Factor PF = 4.29Peak Flow Qpeak = 2219.31 m3/day
90 Deg V-Notch Weir Formula
q(m3/sec) =
H (m) = 150 mm or 0.15 m at Qpeak
Therefore, q = 0.012348 m3/sec > = 0.02569 m3/sec ok
Set depth of weir @ 200 mm ok
1.417H 5/2
INFLUENT INFLUENT INFLUENT SECONDARY EFFLUENT
Flow (m3/d) 517.50 Flow (m3/d) 521.31 Flow (m3/d) 510.88 MLSS Flow (m3/d) 510.73
BOD (kg/day) 129.38 BOD (kg/day) 129.38 BOD (kg/day) 109.97 Flow (m3/d) 1604.25 BOD(kg/d) 5.11
TSS(kg/d) 155.25 TSS(kg/d) 157.26 TSS(kg/d) 133.67 TSS(kg/d) 10.21
INFLUENT SCREEN PUMP ANOXIC TANK AERATION SECONDARY FINALFLOW CHAMBER * STATION BASINS CLARIFIERS EFFLUENT
LAST 10 mg/L BODMANHOLE 20 mg/L TSS
RASRAS = 1%
Flow (m3/d) 423.41
TSS(kg/d) 4234.09
LIQUID FLOW
SOLID FLOW WAS
* Assume 15% of BOD/TSS is removed from screenings and grit* Assume 2% of flow is removed from screenings WAS = 1%
Flow (m3/d) 6.84
TSS(kg/d) 68.45DIGESTED SLUDGE DEWATERED SLUDGE
Flow (m3/d) 3.95 Flow (m3/d) 0.15
TSS(kg/d) 40.13 TSS(kg/d) 38.13
AEROBIC SLUDGESLUDGE HOLDING DRYING BEDS
TANK(75% VSS) (25-40% SOLIDS)
(55% VSS Destruction) (95% CAPTURE)
SUPERNATANT
Flow (m3/d) 3.81
TSS(kg/d) 2.01
SOLIDS BALANCE @ Qavg
INFLUENT INFLUENT INFLUENT SECONDARY EFFLUENT
Flow (m3/d) 689.63 Flow (m3/d) 697.82 Flow (m3/d) 683.87 Flow (m3/d) 683.82BOD (kg/day) 172.41 BOD (kg/day) 172.41 BOD (kg/day) 146.55 BOD(kg/d) 6.84
TSS(kg/d) 206.89 TSS(kg/d) 236.97 TSS(kg/d) 201.43 TSS(kg/d) 13.68
INFLUENT SCREEN AERATION SECONDARY FINALFLOW CHAMBER * BASINS CLARIFIERS EFFLUENT
10 mg/L BOD20 mg/L TSS
RAS
LIQUID FLOW
SOLID FLOW
* Assume 15% of BOD/TSS is removed from screenings and grit* Assume 2% of flow is removed from screenings
WAS
WAS = 1%
Flow (m3/d) 9.10
TSS(kg/d) 91.03THICKENED SLUDGE DIGESTED SLUDGE DEWATERED SLUDGE
Flow (m3/d) 2.09 Flow (m3/d) 1.23 Flow (m3/d) 0.05
TSS(kg/d) 62.79 TSS(kg/d) 36.89 TSS(kg/d) 35.05
GRAVITY AEROBIC SLUDGE SLUDGESLUDGE THICKENER SLUDGE HOLDING DRYING BEDS HOLDING
TANK TANK AREA(3% Solid) (80% VSS) (25-40% SOLIDS) (30 DAYS)
(55% VSS Destruction) (95% CAPTURE)
SUPERNATANT SUPERNATANT
Flow (m3/d) 7.01 Flow (m3/d) 1.18
TSS(kg/d) 28.24 TSS(kg/d) 1.84
SOLIDS BALANCE @ Qavg