Design Cal_2300PE.xls
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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 MAKMUR UNTUK TETUAN DOYENVEST (M) SDN BHD BASIC DATA: This is an outline design to produce 10:20 BOD5:SS effluent.
Population, Dry Weather Flow,
PE DWF
= =
2300 P.E 0.225 m3/c/d
Suspended Solid, BOD5, Influent Ammonia
SS BOD NH3-N
= = =
300 mg/l 250 mg/l 30 mg/l
Effluent BOD5, Effluent SS Effluent Ammonia
EBOD ESS ENH3-N
= = =
10 mg/l 20 mg/l 10 mg/l
LOADING a.
Hydraulic flow,
Qavg
= =
PE * DWF 517.50 m3/d
b.
Peak Flow,
Qpeak
=
Peak Factor * Qavg
Pff
= =
4.7(P) 4.29
Qpeak
= = =
Pff x Qavg 2219.31 m3/d 1.54 m3/min
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
-0.11
Peak Flow Factor
Therefore,
PRIMARY BAR SCREEN DESIGN Design Population,
PE
Design Flow, Peaking Factor, Peak Flow,
Qavg PF Qpeak
2300 517.50 m3/day 4.29 2219.31 m3/day
0.0060 l/s 25.81 l/s
Guidelines : Quantity of screenings = 30 m3 screening / 10^6 m3 wastewater Number of Channels Number of Back-up Channel
= =
1 1
Qavg
=
517.50 m3/day
Quantity of Screenings
= =
Qavg x 30 / 1000000 0.0155 m3/day
Quantity of Screenings
= =
7 x Quantity of Screenings per day 0.109 m3
Number of Storage Units Quantity per Unit
= =
1 0.109 m3
Dimension of screenings trough L W D
= = =
0.50 m 0.30 m 0.30 m
=
0.045 m3
Provide storage for 7 days
Volume
>
0.109 m3
Screen Design Guidelines: Max flow through velocity at Qpeak Guidelines: Min approach velocity at Qpeak
Vmax Vmin
Bar Size Clear Opening
Bs Co
= =
Efficiency Coefficient
Eff
=
10 mm 25 mm Clear Opening Clear Opening + Bar size
=
Clear area through each screen at Qpe AQpeak
Total cross sectional area of channel
AC
0.71
=
Qpeak Vmax x 24 x 60 x 60
=
0.026 m2
=
AQpeak Eff
=
Assume, Depth of Flow at Qpeak
D
= =
0.036 m2
=
0.050 m .
Required Width of Clear Opening @ Q
Wclr
= =
Number of Openings
No
AC / D 0.72 m
=
Wclr x 1000 Co
=
29
1.0 m/ses 0.3 m/sec
error
Number of Bars
Nbars
Gross Width of Screen
Wch
Set
= =
No - 1
=
Wclr + (Nbars x Bs / 1000)
28
=
1.00 m
Wch
=
0.50 m
Qpeak
=
2219.31 m3/day
Vapp
= =
Qpeak / (Wch x D) 1.03 m/sec
>=
0.30 m/sec
ok
= =
Qpeak / (Wclr x D) 0.71 m/sec
=
25.69 l/s
ok
Pump cycle time at Qavg Guideline: 6 min, 15 max @ Qavg Volume required for pump sump,
Assume number of start / stop where
Therefore,
times per hr (required 6 - 15 start/hour) Required Volume (m3) Cycle Time (minute) Pumping Rate (L/sec) Pumping Rate (m3/min) 1.56 m3
<
Pump Headloss Calculations
VALVES & FITTINGS
K VALUE
EXIT CHECK VALVE GATE VALVE TEE THRU SIDE ELBOW - 90 DEG
1.00 2.50 0.20 1.80 0.30
QUANTITY
DISCHARGE TOTAL
1 1 1 1 6
1.00 2.50 0.20 1.80 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 = (K x V2 / 2g)
0.25
m
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
79.60 - 76.00 3.45
m
3.75
m
S = STATIC HEAD LOSS (Worst Case Scenerio)
TOTAL DYNAMIC HEAD
1.68 m3
ok
Size pump for
26.00 L/sec
@
3.75 m
TDH
The submersible pump selected is as follows:Make = TSURUMI Model = 100B42.2 Capacity = 18.20 L/sec Total Head = 6.20 m Power = 2.20 kW Discharge Size = 100 mm No. of Units = 2 (1 duty; 1 standby)
CHANNEL INVERT
EL
76.75
HIGH LEVEL ALARM
EL
76.75
STANDBY PUMP CUT IN
EL
76.60
DUTY PUMP CUT IN
EL
76.45
ALL PUMPS CUT OFF
EL
75.80
INVERT ELEVATION
EL
75.40
Time to fill sump
=
0.0
m
0.15
m
0.15
m
0.65
m
0.4
m
Pump cycle time at Qpeak Guideline: 6 min, 15 max @ Qavg
= Time to empty sump
= =
Cycle time base on actual operating point
=
V Qpeak 1.09
min
V Qpump - Qpeak -3.75
min
=
60 Time fill + Time empty -22.6
min per hour min per cycle cycle per hour
=
[ 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
6 - 15 cycle / hr error
Determine Size of Force Main Guideline: 1 - 2.5m/s velocity in pipe Required size of raw sewage pipe
= = Actual velocity in selected pipe
=
0.115 m 100 mm Qpump / cross section area of pipe 2.32 m/s <
>
115 mm
error
2.5 m/s
ok
Determine Total Dynamic Head ( TDH ) TDH
=
hst + hf + hm
= = =
Static head (m) Losses through the pipe ( Hazen - William Formula ) Losses through fittings
Where ;
hst hf hm
A
Thus, static head of pump, hst
B
Hazen - William Formula hf
Where ;
79.60 - 76.00 m 3.45 m
=
6.82
V C
V
=
Velocity m/sec
V = Q/A
=
11.50 L/sec 1000 x 3.142 x 0.10 2 / 4
=
C
= =
1.85
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
Thus, losses through the pipe, h f
=
0.050 m
Losses through fittings hm
=
Where ;
K
=
Head loss coefficeint
7.30
V
=
Velocity m/sec
0.83
G
=
Gravity, m/s²
9.81
hm
=
0.25 m
KV² 2g
Therefore , Total Dynamic Head ( TDH )
= =
hst + hf + hm 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)
0 3.33 6.67 10.00 13.33 16.67 20.00 23.00
0.00 0.11 0.21 0.32 0.42 0.53 0.64 0.73
3.45 3.45 3.45 3.45 3.45 3.45 3.45 3.45
0.000 0.001 0.004 0.009 0.015 0.022 0.031 0.040
* refer to Chart attached
x
hm (m) TDH (m) 0.000 0.004 0.017 0.038 0.067 0.105 0.151 0.199
3.45 3.46 3.47 3.50 3.53 3.58 3.63 3.69
Pump head (m) 16.0 14.0 12.6 10.8 9.0 7.0 5.2 3
…………A …………B …………C
L D1.167
INFLUENT RAW SEWAGE PUMP (TSURUMI MODEL 100 B42.2) 20.00
System Curve
Head (m)
15.00
Pump Operating Point 18.2L/sec @ 5.5 m TDH
10.00
5.00 Pump Curve
0.00 0
2
4
6
8
10
12
14
16
Flowrate, Q (L/sec)
18
20
22
24
26
28
30
FINE SCREEN DESIGN Parameters Population
PE
=
Peak Flow
Qpeak
=
2219.31 m3/day
=
0.03 m3/sec
Qavg
= =
517.50 m3/day 0.01 m3/sec
Guidelines: Max flow through velocity at Qpeak
Vmax
=
1.00 m/sec
Guidelines: Min approach velocity at Qpeak
Vmin
=
1.00 m/sec
Hd
=
0.00 m
Bar Size Clear Opening
Bs Co
= =
10.00 mm 12.00 mm
Efficiency Coefficient
Eff
=
Design Flow
Set Elevation of Datum
2,300
Design
Clear Opening Clear Opening + Bar size
= Clear area through each screen at Qpeak,
AQpeak
=
= Total cross sectional area of channel required
=
0.55 Qpeak Vmax 0.0257 m2 AQpeak Eff
= Assume Depth of Flow Width of Clear Openings
Number of Openings
D
=
Wclr
=
No
0.05 m2 0.055 m
Width of Frame
Nbars
=
0.47 m
=
Wclr x 1000 Co
= =
1
AQpeak / D
= Number of Bars
-----
39 No - 1 38
Wfr
=
0.10 m
Gross Width of Screen
W
= =
Set width of screen
W
=
0.50 m
Qpeak
=
1.71 B (H^1.5)
Qpeak
=
0.026 m3/s
Set depth of flow
H
=
0.055 m from - -
Determine width of throat
B
=
?
Therefore
B
= =
Qpeak / 1.71 / (H^1.5) 1.16 m
=
Throat width of grit chamber
Wclr + Wfr + (Nbars x Bs / 1000) 0.95 m
Determine Width of Downstream Throat Total Peak Flow,
m
1
OK
Energy equation between upstream and downstream of clean screen Velocity at Qpeak through clear openings of screen, Vs Vs
Headloss through clean screen
=
Qpeak Wclr x d1
=
0.65 m/sec
hLs
= =
(Vs^2 - v1^2) / 2g 0.0039 m
X2 X3
E2 = + d2 + (v2^2 /2g) = E3 + d3 + (v3^2 /2g) + hLs
X2
=
E2 d1 v1 E3 d2 v2
= = = = = =
0.000 0.09 0.60 0.000 0.055 0.93
X2 X3
= =
0.10 0.10
x 1/0.7
X3
Where, Trial and Error
Therefore,
m m m/sec m m m/sec
Height above datum Upstream Depth Upstream Velocity Height above datum Downstream Depth @ Qpeak Channel Velocity
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
=
Velocity through clogged screen (Assume 50% clogging)
Vcs
=
hLcs
(Vcs^2 - v2^2) / 2g x 1/0.7 Qpeak Wclr x 0.5 x Upstream Depth d"
=
0.92 m/sec
=
0.05
Therefore, X2 X3
= 2 + d" + (v"^2 /2g) E = E3 + d3 + (v3^2 /2g) + hLcs
X2
=
E2 d" v" E3 d3 v3
= = = = = =
0.00 0.120 0.43 0.00 0.055 0.93
X2 X3
= =
0.13 0.15
X3
Where, Trial and Error
m m m/sec m m m/sec
Check:
-0.018 Elevation of Channel Invert
=
79.00 m
Upstream Water Elevation
=
79.12 m
Downstream Water Elevation
=
79.06 m
ok
Height above datum Upstream Depth Upstream Velocity Height above datum Downstream Depth @ Qpeak Channel Velocity
DESIGN OF A CONSTANT VELOCITY GRIT CHANNEL Guidelines = 0.2 m/s flow through velocity Number of Channels Provided Number of Back-up Channels Provided
= =
1 1
Constant Velocity Channel Formula Q
=
1.71 B (H1.5)
Q B H
= = =
Flow Width of Throat Depth of Flow
=
( Q / 1.71 / B )^0.67
A W H
= = =
2/3 W H Width of Parabola Depth of Flow
V
=
0.2 m/s
Q
= =
A x V 2/3 W H x V
W
=
Therefore, Area of Parabola,
m3/s m m
m m
Assume:
Therefore
Assume:
B Qavg per Channel
= =
(Per guidelines)
1.5 Q H x V
0.33 m 517.50 m3/day
FLOW FACTOR ( x QAVG)
Q AT DIFFERENT FACTOR m3/sec
Depth H(m)
Width W(m)
0.10 0.25 0.50 1.00 2.00 3.00 4.00 4.29 5.00 6.00 7.00
0.001 0.001 0.003 0.006 0.012 0.018 0.024 0.026 0.030 0.036 0.042
0.01 0.02 0.03 0.05 0.08 0.10 0.12 0.13 0.14 0.16 0.18
0.43 0.58 0.73 0.92 1.16 1.33 1.46 1.50 1.57 1.67 1.76
0.0060 m3/s
@ Qavg
@ Qpeak
Determine Length of Channel Say settlement vel of particle
Vs
=
0.02 m/s 0.2 m/s
Say velocity of flow
V
=
Depth of channel at peak flow, Width of channel,
H W
= =
Length of Channel
L
=
L
=
1.29 m
L
=
3.00 m
Surface Area of Channel
SA
=
4.49 m2
Surface Overflow Rate
SR
=
Provided Length
=
0.13 m 1.50 m V x H Vs
Qpeak / SA 494.50 m3/m2/day
< 1500
m3/m2/day
Quantity of grit Guidelines = 0.03 m3 / 1000 m3 of wastewater Grit Quantity per Channel
Provide area for 30 days storage Grit quantity
=
Qavg x 0.03/1000
=
0.016 m3/day
=
0.47 m3
= = = =
3.00 m 0.50 m 0.15 m 0.225
Dimension of hopper at bottom of grit channel L : W ration 2:1 as per guidelines L W D Volume
1500
>=
0.466
ok
mm
700 mm 50 mm
450 mm
1000
mm
Check ratio of grit tank dimension Ratio W: D = 1 : 2
W 2.00
D 1
Ratio W: L = 1 : 2
W 1
L 2
Hydraulic Retension Time, Tr Guidelines = 3 min @ Qpeak < 5,000 pe Area of retangular Area of trapezium Area of grit storage
= = =
1.05 m2 0.06 m2 0.45 m2
Total Area L
= =
1.56 m2 3.00 m
Volume, V
=
4.69 m3
Qpeak
= =
2219.31 m3/day 1.54 m3/min
Tr
= =
V / Qpeak 3.04 min
>= 3 min
ok
X-Axis -0.837 -0.787 -0.748 -0.731 -0.664 -0.580 -0.460 -0.365 -0.290 -0.214 0 0.214 0.290 0.365 0.460 0.580 0.664 0.731 0.748 0.787 0.837
Y-Axis 0.161 0.143 0.129 0.123 0.101 0.077 0.049 0.031 0.019 0.011 0 0.011 0.019 0.031 0.049 0.077 0.101 0.123 0.129 0.143 0.161
CONSTANT VELOCITY FLOW X - SECTION 0.700
)
0.600
0.700
Depth (m)
0.600 0.500 0.400 0.300 0.200 0.100
GRIT STORAG
0.000 -0.100 -0.500
-0.400
-0.300
-0.200
-0.100
0.000
Width (m)
0.10
ITY FLOW CHANNEL ECTION
Width Depth X - Axis Y - Axis 0.427 0.011 0.580 0.019 0.731 0.031 0.921 0.049 1.160 0.077 1.328 0.101 1.462 0.123 1.496 0.129 1.575 0.143 1.673 0.161
GRIT STORAGE
0.000
Width (m)
0.100
0.200
0.300
0.400
0.500
CONSTANT VELOCITY FLOW CHANNEL X - SECTION 1.250 1.150 1.050
Depth (m)
0.950 0.850 0.750 0.650 0.550 0.450 0.350 0.250 0.150 0.050
-0.650
-0.450
-0.250
-0.050 GRIT STORAGE -0.150 -0.050 Width (m) Channel
0.150
0.350
0.550
DESIGN OF A GRIT / GREASE CHAMBER Guidelines : Detension time T = 360 sec @ Qpeak Number of Channels Provided Number of Back-up Channels Provided
= =
1 1
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
L W D
= = =
2.90 m 1.60 m 0.85 m
V
=
3.94 m3
>
L:W W:D
= =
1.81 1.88
: :
Provided Length Width of channel, Surface Area of Channel
L W SA
= = =
2.90 m 1.60 m 4.64 m2
Surface Overflow Rate
SR
= = <
Qpeak / SA 478.30 m3/m2/day 1500 m3/m2/day
Grit Quantity
= =
Qavg x 0.03/1000 0.016 m3/day
Allow for storage of 30 days
=
0.47 m3
8 kg / 1000 m3 0.95
Peak Flow
Provide: Length Width Depth Volume Provided Ratio
9.25 m3 1.0 1.0
ok
Determine Length of Channel
ok
Quantity of grit Guidelines = 0.03 m3 / 1000 m3 of wastewater
Quantity of grease Average quantity of grease Specific gravity of grease
S sg
= =
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
Accumulated grease for 30 days Accumulated grit for 30 days
Vs
= =
0.131 m3 0.466 m3
Depth of Feed
D
=
0.25 m
Therefore,
A1
=
2.39 m2
= =
1.50 m 0.75 m
Grit/Grease Drying Bed Sizing
Area required
Dimension of Grit/Grease Drying Bed Length Width
Grease
error
Area
=
1.13 m2
>
2.39 m2
error
DESIGN OF A GREASE CHAMBER Guidelines : Detension time T = 180 sec @ Qpeak Number of Chamber Provided Number of Back-up Chamber Provided
= =
1 1
Peak Flow
= =
2219.31 m3/day 0.026 m3/sec 180 sec
Qpeak
Detension Time
T
=
Volume Required
V
=
Qpeak x T
=
4.62 m3
L W D
= = =
3.80 m 1.20 m 1.10 m
V
=
5.02 m3
Average quantity of grease Specific gravity of grease
S sg
= =
8 kg / 1000 m3 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 0.13 m3
Provide: Length Width Depth Volume Provided
>
4.62 m3
ok
0.13 m3
ok
Grease Quantities
Depth of grease baffle required to contain scum for 30 days Length L = 3.80 m Width W = 1.20 m Depth of water below baffle D = 0.02 m Volume of grease storage provided
V
=
0.09 m3
>
WEIR BEFORE AERATION BASINS Design Population, Design Flow, Peaking Factor, Peak Flow,
Number of Aeration Basins,
not use
PE Qavg PF Qpeak
N
=
Peak Flow to Each Aeration Basin
2300 517.50 m3/day 4.29 2219.31 m3/day
Qpeak / N 0.026 m3/sec
Q(m3/sec)
=
Cw x Le x H^1.5
Cw Le H
= = =
1.84 0.3704 m 0.148 m
Le
=
where
L n H
= = =
0.40 m 2 0.148 m
Therefore
Le
=
0.3704 m
Therefore
Q
=
where
0.03 m3/sec
1
= =
Formula
0.01 m3/sec
Weir Coefficient Length Height
With end contraction L - (0.1)nH
0.039 m3/sec
Length Number of side contractions Height of Flow
>=
0.026 m3/sec
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
ok
RECTANGULAR WEIR AT OUTLET BOX Design Population, Design Flow, Peaking Factor, Peak Flow,
Formula
PE Qavg PF Qpeak
2300 517.50 m3/day 4.29 2219.31 m3/day
Q(m3/sec)
=
Cw x Le x H^1.5
Cw Le H
= = =
1.84 0.578 m 0.11 m
Le
=
where
L n H
= = =
0.60 m 2 0.110 m
Therefore
Le
=
0.578 m
Therefore
Q
=
0.039 m3/sec
where
0.01 m3/sec 0.03 m3/sec
Weir Coefficient Length Height
With end contraction
Height of water above weir
=
L - (0.1)nH
110 mm
Length Number of side contractions Height of Flow
>=
@ Qpeak
0.026 m3/sec
ok
90 DEG V-NOTCH WEIR AT OUTLET BOX Design Population, Design Flow,
PE Qavg
2000 450.00 m3/day
Peaking Factor,
PF
Peak Flow,
Qpeak
4.35 1959.73 m3/day
0.023 m3/sec
90 deg V-Notch Weir Formula 5/2
q (m3/sec)
=
H (m)
=
q
=
Therefore,
Set depth of weir @ 250mm
1.417 H 192 mm 0.023 m3/sec
ok
or
0.192 m
at Qpeak
>=
0.023 m3/sec
ok
DESIGN OF ANOXIC CHAMBER Population
PE
=
Peak Flow
Qpeak
= = =
2,219.31 m3/day 92.47 m3/hr 1.54 m3/min
= =
0.026 m3/s 25.69 l/s
= =
517.50 m3/day 21.56 m3/hr
= = =
0.36 m3/min 0.01 m3/s 5.99 l/s
=
4.29
= = = = = = = = =
30 10 3600 2880 28 0.1 0.11
Qavg
Design Flow
Peaking Factor
PF
2300
Using the Lodification Ludzack-Ettinger Process Assume Influent Ammonia-N Ni Effluent Ammonia-N Ne Mixed liquor Suspended Solids MLSS Mixed Liquor Volatile Suspended Solids MLVSS Temperature Temp Dissolved Oxigen Do Specific Denitrification Rate U Overall Denitrification rate Ua Residence Time Temp
? ?
mg/l mg/l mg/l mg/l (MVLSS deg C mg/l per day per day hrs
=
0.8
x MLSS
Calculation Overall Denitrification Rate Formula :
Ua
=
U x 1.09(Temp-20)X(1-DO) 0.20 per day
T
=
Ni - Ne Ua x MLVSS
Calculation Residence Time Formula :
=
Size anoxic zone for a residence time of Anox
x
24 hrs/day
0.81 hrs required
=
2
hrs
=
Anox 24
=
43.13
m3
N
=
2
nos
Vreq
=
21.56
m3
Length Width Depth
= = = =
2.30 1.50 4.50
m m m
Anoxic tank volume
=
31.05
m3
Actual hydraulic retention time provided
=
Determine size of anoxic tank Anoxic zone volume required,
Number of Anoxic Tank, Volume required per Tank,
Vol
x
Qavg
Tank Dimension
=
Anoxic tank volume Qavg 2.88 hrs
>
21.56 m3
ok
EXTENDED AERATION TANK DESIGN Sizing Number of Aeration Tank,
No.
=
Minimum Hydraulic Retention Time,
HRT
=
V
=
418.31 m3
Vol
=
209.16 m3
Area Length Width Depth Freeboard Volume
A L W D FB Vol
= = = = = =
46.48 10.60 4.40 4.50 0.65 209.88
Total Volume Provided
Vp
=
419.76 m3 >
HRT Provided
=
19.47 hrs >
=
3600 mg/l
Volume Required, Volume Required Per Tank,
2 19.4 hours
Dimension of each tank:
Ratio
Therefore,
m2 m m m m m3
L 10.60 2.41
: : :
W 4.40 1
418.31 m3 19.4 hrs
ok
Sludge Age (MCRT) Guidelines > 20 days Assume,
MLSS SA
=
Total solids in aeration tank Excess 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
Therefore actua
=
Sy
=
Sd
= =
QAVG x Sy (BOD5-EBOD5) 68.45 kg/day
SA
=
20.53 days
WAS
=
68.45 kg/day
Determine Sy @ Actual HRT Via Interpolation
Excess sludge wasting / day (Sludge accumulation per day)
(24 -21.9) (24 - 18)
x
Assume underflow concentration = 1% or 10,000 mg/l or 10 kg/m3 Volume of WAS
= =
+
0.4
0.55 kg / kg BOD removed based on HRT interpolation
Waste Activated Sludge (WAS) Excess Sludge Wasting / day
(0.6 - 0.4)
6.84 m3/day 6844.61 l/day
ok
Return Activated Sludge Flow (RAS) QRAS
=
MLSS Cu - MLSS
Cu
=
Underflow concentration assume at 0.8 % solid or 8,000 mg/l
QRAS
=
423.41 m3/day
4.92 l/s
Set QRAS
=
465.75 m3/day
5.42 l/s
Qavg
=
517.50 m3/day
Ratio of QRAS : QAVG
x
QAVG
=
0.90
ok
OR
=
2.00 kg O2 / kg BOD5
AOR
= =
QAVG Total
= =
517.50 m3/day 517.50 m3/day
= =
Total x 2 / 1000 1.04 kg/day
=
249.44 kg/day
Oxygen Requirements
Oxygen required per kg of BOD5 re Actual Oxygen Required
OR x BOD5 removed 248.40 kg/day
Maintain DO of 2 mg/l in aeration tank
Oxygen required in tank
Total AOR
Oxygen correction factor
AOR/SOR
AOR/SOR = Where
(((BETA * ACF * Csf @ 28 deg) - C ) / Css ) * ALPHA * 1.024^T-20 BETA ACF Csf C Css Alpha T
= = = = = = =
0.98 0.989 7.83 1.5 9.08 0.75 30
AOR/SOR
=
0.64
Standard Oxygen Required
SOR
=
391.23 kg/day
O2 transfer efficiency
O2eff
=
20 %
Actual O2 transfer efficiency
O2act
=
Therefore
@
6
m3/hr (See catalog)
SOR kg/day O2eff/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
Flowrate per diffuser
= =
0.10 m3/min 100.00 L/min
Quantity
=
Number of diffusers required
= Number of diffusers per tank
=
or
6
Total air required / Flowrate 48.75
Use
52
26 Nos in each Aeration tank
ok
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 F/M Ratio
= = =
419.76 m3 INBOD / (MLSS x Vp) 0.086
ok
Check Aeration Loading Guidelines :
0.1 - 0.4 kg/m3/day AL
=
INBOD / Vp
=
0.31 kg/m3/day
ok
Check Mixing Rate Per EPA guidelines,
0.010 to 0.025 m3/min. m3
Volume of air from Blower
Va
=
Volume of tank provided
Vp
=
Mixing Rate
=
Therefore,
=
#NAME? m3/min
(See Blower calculations)
419.76 m3
Va / Vp #NAME? m3/min. m3
#NAME?
CLARIFIER DESIGN Guidelines : Hydraulic retension time (HRT) = 2 hours @ Qpeak Guidelines : Surface overflow rate = 30 m3/m2/day Guidelines : Side water depth (SWD) = 3 m Surface Area
= =
Qpeak / 30 m3/m2/day 73.98 m2
Number of clarifiers
=
2
Surface area required per clarifier
=
36.99 m2
L W
= =
6.20 m 6.20 m
Therefore, actual area provided
=
38.44 m2
Provided Length, L Provided Width, W
= =
6.20 m 6.20 m
Effective Volume SWD in cone
= =
Cone volume Vertical SWD
= =
65.73 m3 1.04 m
Vertical SWD volume
=
39.98 m3
Total Volume Total SWD
= =
105.70 m3 4.50 m
Dimension of each clarifier
>
36.99 m2
ok
Volume of cone with slope of 60 deg. 3.46 m (See attach volume cals) ok
HRT Per guidelines: < 2 hours = =
Total Volume m3 / Qpeak m3/day x 24 hours 2.29 >= 2 hours
ok
Overflow Weir Per guidelines: 150 - 180 m3/day/m Max weir loading rate, Wr Peak Flow, Qpeak
= =
Min Weir Length Required per Clarifier
= =
Total length of weir provided per Clarifier, Tw
=
Actual weir loading rate
=
Solids Loading Rate (SLR) Guideliness :
MLSS
180 m3/day/m 2219.31 m3/day Qpeak / Wr / No. of Clarifiers 6.16 m 3.00 m 369.89 m3/day/m
< 150 kg/m2/day at Qpeak < 50 kg/m2/day at Qavg =
3600 mg/l
Total Q = Qpeak + (For QRAS look under Extended Aeration Tank Sizing) = #NAME? m3/day SLR (Qpeak)
= =
SLR (Qavg)
error
= =
or
3.6 kg/m3
QRAS
(Total Q / Actual Surface Area) x MLSS kg/day #NAME? kg/m2/day
#NAME?
(Qavg + Qras) / Actual Surface Area x MLSS kg/day #NAME? kg/m2/day
#NAME?
Launder Calculation Overflow from weir to lauder
Launder size Length Width Flow velocity Depth of flow in launder
= =
= = = =
Qpeak 1109.66
3.00 0.25 0.80 21.41
/ number of tank m3 / launder
m m m /s mm
< 1.0 m/s ok < 0.3m as provided
Determine Volume of Cone L
W
L
W
h
h 60
Therefore, Volume of Cone
BW 1
A
L W BW 1 BW 2
= = = =
6.20 6.20 2.20 2.20
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 y
= = =
h /B Tan -1 (h/B) 60.00 degrees
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
=
L (m) W (m)
h (m)
A (m)
y
BW1 (m)
BW2
B
m m m m
65.73 m3
Scum Withdrawal Airlift Pipe Design Percent submergence Hs Hl
Scum Airlift Hs Hl Percent Submergence
=
Hs / (Hs + Hl)
= =
Depth (m) of air pipe below water surface Height (m) of lift
= = =
x 100
3.8 m 0.7 m 84.44 %
With reference to the attached chart, Discharge Velocity Air Requirement
= = =
45 gpm 2 ft/sec 7 cfm
2.84 l/s 0.61 m/s 0.20 m3/min
AEROBIC SLUDGE HOLDING TANK DESIGN Guidelines: Sludge yield = 0.6 kg/kg BOD5 (Standard "A")
Design Population,
PE
=
Design Flow,
QAVG
=
Sludge Yield
Ys
=
0.55 kg/kgBOD5/day
Sludge accumulation per day
Sd
= =
QAVG x Ys (BOD5-EBOD5) 68.31 kg
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
TSd
=
Nonvolatile Solid + VS Remaining
=
(WAS - TVS) + { (1 - VSSd/100) x
=
(26.73 - 20.05) + { (1 - 55/100) x 20.05)
Temperature of Wastewater
Therefore, Total Solids Remaining After Digestion
TSd
=
Density of water
ρ
=
Specific gravity of sludge
sg
=
2300 517.5 m3/day
1000.00 kg/m3 1.015 3.95 m3/day
T
=
30 days
Volume of Tank Required
Vtank
=
Number of Tank Provided
No.
=
Volume of Tank Required
V
=
Storage days provided
TVS }
40.13 kg/day
=
Therefore, TSd in term of volume
(Calculated earlier)
118.62 m3 1 118.62 m3
SIZE OF EACH TANK Depth Width Length
= = =
Volume
=
4.5 m 3.3 m 8.0 m 118.80 m3
>
118.62 m3
ok
OXYGEN REQUIREMENT FOR DIGESTION Guidelines: 1.5kgO2/kgBOD Wt of sludge digested
Ws
Amount of oxygen required, (AOR) Ws x 1.5
Oxygen correction factor AOR/SOR Where
= =
Sd - TSd 28.18 kg
=
42.27 kg
AOR/SOR =
(((BETA * ACF * Csf @ 28 deg) - C ) / Css ) * ALPHA * 1.024^T-20
BETA ACF Csf C Css Alpha T
= = = = = = =
0.98 0.989 7.83 1.5 9.08 0.75 30
AOR/SOR
=
0.64
Standard Oxygen Required
SOR
=
Assume O2 transfer efficiency
O2eff
=
Therefore
Total amount of air required
=
66.29 kg/day 10 % SOR kg/day O2eff/100 x 1.201 kg/m3 x 0.232 kg O2/kg air x 1440 min/day
=
1.65 m3/min
Flowrate per diffuser
=
200.00 L/min 0.200 m3/min
Quantity
=
Number of diffusers required
Total air required / Flowrate = 8.26
Use
10
ok
DRYING BED DESIGN Digested Sludge From Aerobic Digester Concentration of Sludge
DS
=
C
=
#NAME? kg/day 1% 10 kg/m3
= Specific gravity of sludge Volume of Digested Sludge
sg
=
Vds
=
(See Aerobic Digester cals)
1.015 #NAME? m3/day
Per guidelines: Provide 4 week cycle for 450mm thick feed depth Volume of Vds required (Based on 21 days drying; 7 days feeding)
V
=
Vds(m3/day) x 28days
=
#NAME? m3
Depth of Feed
D
=
Therefore,
A
=
#NAME? m2
A
=
#NAME? m2 ( 1/3 reduction in area)
Area required
0.45 m
Provide fully covered drying beds Therefore actual area required Number of Drying Beds provided
=
4
= = =
16.50 m 2.50 m 41.25 m2
Dimension of Drying Beds Dimension of Bed Length Width Area Total area of drying bed provided
=
nos of bed x bed area 165.0 m2
=
>
#NAME? m2
###
Determine Volume of Dewatered Sludge (DS) Assume final sludge concentration is 25% DS
=
Vs x Si x SGi x n SGo x So
DS Vs Si So SGi SGo n
= = = = = = =
Therefore DS
=
### m3/day
V
=
DS x 30 days
=
#NAME? m3
= = = =
0.50 1.00 4.60 2.30
? ### 0.01 0.25 1.015 1.03 0.95
Volume of Dewatered Sludge, (m3/day) Volume of Influent Sludge (m3/day) Fractional Percent Solid Content of Influent Sludge Fractional Percent Solid Content of Dewatered Sludge Specific Gravity of Sludge Before Thickening Specific Gravity of Sludge After Thickening Fractional Percent Capture
@ 25% solid
Provide covered storage for 30 days Volume required
Dimension provided
Depth Width Length Volume
m m m m3
>
#NAME? 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
ENTRY EXIT CHECK VALVE SLUICE VALVE REDUCER TEE THRU SIDE TEE THRU RUN ELBOW - 45 DEG ELBOW - 90 DEG
K VALUE
QUANTITY
SUCTION TOTAL
QUANTITY
DISCHARGE TOTAL
0.50 1.00 2.50 0.20 0.30 1.80 0.60 0.23 0.30
1 0 0 0 0 0 0 0 0
0.50 -
0 1 0 1 1 0 1 0 5
0.00 1.00 0.00 0.20 0.30 0.00 0.60 0.00 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 = (K x V2 / 2g)
0.03 m
0.25
L = LINEAR METER OF STRAIGHT PIPE
0.00 m
15.00
M = MULTIPLYING FACTOR
1.10
1.10
100.00
100.00
C = HAZEN-WILLIAMS ROUGHNESS COEFFICIENT 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
The submersible pump selected is as follows:Make = Ebara Model = 65DVS51.5 Capacity = 9.10 L/sec Total Head = 4.73 m Power = 1.50 kW Discharge Size = 65 mm No. of Units = 1 (1 duty)
TDH
connect to
100
mm
Determine Size of Force Main Guideline: 1 - 2.5m/s velocity in pipe Required size of sludge pump
=
[ 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
=
Actual velocity in selected pipe
0.068 m
=
80 mm
=
1.81 m/s
>
>
68 mm
ok
2.5 m/s
ok
Determine Total Dynamic Head ( TDH ) TDH
=
hst + hf + hm
= = =
Static head (m) Losses through the pipe ( Hazen - William Formula ) Losses through fittings
Where ;
hst hf hm
A
Thus, static head of pump, h st
B
Hazen - William Formula hf = 6.82
Where ;
=
V C
1.85
X
=
Velocity m/sec
V = Q/A
=
8 L/sec 1000 x 3.142 x 0.10
1.16 m/s
2
/ 4
1.16 m/sec
C
=
Coefficient of roughness
L
=
Length of pipe, m
D
=
Diameter of pipe, m
Thus, losses through the pipe, hf
C
L D1.167
V
=
4.00 m
0.00 15.00 0.10
=
#DIV/0! m
Losses through fittings hm
=
KV² 2g
Where ;
K
=
Head loss coefficeint
4.10
V
=
Velocity m/sec
1.16
G
=
Gravity, m/s²
9.81
hm
=
0.29 m
Therefore , Total Dynamic Head ( TDH )
= =
hst + hf + hm ### m
To plot chart for the System Curve & Pump Curve from the above equitition:Q (L/s)
V (m/s)
TDH (m)
0 2 4 8 9.5
0.00 0.25 0.51 1.02 1.21
4 4.04 4.15 4.57 4.80
* refer to Chart attached
Pump head (m) 18.6 16.4 14.2 8.3 4.2
18.6 16.4 14.2 13 8.33 4.2
4 4.04 4.15 4.23 4.57 4.8
…………A …………B …………C
DRYING BED FEED PUMP Submersible Pump EBARA Model 65DVS51.5 20 Pump Curve
18
Head (m)
16 Pump Operating Point 9.2 L/sec @ 4.7 m
14 12 10 System Curve 8 6 4 2 0 0
2
4
6 Flowrate, Q (L/sec)
8
10
X-Axis 0 3.3 6.67 10 13.33 16.67 19.98 65DVS
X-Axis 0 2 4 8 9.5
Pump System 16.2 4 14.2 4.25 12.4 4.92 11 5.97 9 7.39 7.4 9.17 5.6 11.27 Pump System 18.6 4 16.4 4.04 14.2 4.15 8.33 4.57 4.2 4.8
3.580986
Pump 0 198 400.2 600 799.8 1000.2 1198.8
System
Blower- 1 Sizing Air Requirement for Aeration
=
#NAME? m3/min
Air Requirement for Aerobic Sludge Digestion
=
#NAME? m3/min
=
#NAME? m3/min
Qd
TOTAL AIR REQUIREMENT
Determine air flow under standard condition, Qs Qs
=
Qd x (1.0332 + Pd)
273 + St
x
273 + Sd
1.0332
Qs
=
?
Air flow under standard condition
(m3/min)
Qd
=
#NAME?
Air flow under discharge condition
(m3/min)
Pd
= =
0.41 -0.05
Discharge static pressure Suction static pressure
(kgf/cm2) (kgf/cm2)
=
30
Suction temperature
o
=
38
Discharge temperature
o
=
#NAME? m3/min
PS St Sd
C C
Therefore, Qs
Determine discharge pressure under standard condition, Ps Ps
=
=
1.0332 + Pd 1.0332 + PS
-
0.48 kgf/cm2
Size blower for 9.09 m3/min @ 0.48 kgf/cm2 Provide 2 blowers : 1 duty, 1 standby Model:
Fu Tsu Model TSC 125, 920rpm, 15.0kW
1
x
1.0332
Blower-3 Sizing Air Requirement for Scum Airlift
=
0.20 m3/min
Air Requirement for RAS
=
#NAME? m3/min
Air Requirement for WAS
=
Air Requirement for MLSS
=
#NAME? m3/min #NAME? m3/min
=
#NAME? m3/min
Qd
TOTAL AIR REQUIREMENT
Determine air flow under standard condition, Qs Qs
=
Qd x (1.0332 + Pd)
273 + St
x
273 + Sd
1.0332
Qs
=
?
Air flow under standard condition
(m3/min)
Qd
=
#NAME?
Air flow under discharge condition
(m3/min)
Pd
=
0.45
Discharge static pressure
(kgf/cm2)
PS St
=
-0.05
Suction static pressure
(kgf/cm2)
=
30
Suction temperature
o
Sd
=
38
Discharge temperature
o
=
#NAME? m3/min
C C
Therefore, Qs
Determine discharge pressure under standard condition, P s Ps
=
=
1.0332 + Pd 1.0332 + PS
-
0.53 kgf/cm2
Size blower for 1.76 m3/min @ 0.53 kgf/cm2 Provide 1 blowers : 1 duty Model:
Fu-Tsu Model TSC 80, 760 rpm @ 5.5 Kw
1
x
1.0332
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) Pump Dia (in) for RAS Pump Dia (in) for MLSS & scum
= = =
14.76 ft 3.00 in 2.00 in
4.50 m 0.075 m 0.050 m
S
=
13.01 ft
3.97 m
Calculate the Air Supply Volume (Vair) required: Vair
=
h / { C x LOG [ (H + 10.4) / 10.4 ] }
where h H C
= = =
Vair
=
A eff
=
0.75 total lift required (m) 4.50 submergence (m) 10.20 constant for less than 15m lift 0.47 m3/min of air per m3 of water 25.00 air-lift efficiency (%)
Determine Volume of Air Requirement for SCUM, VSCUM SCUM
= =
Vair-SCUM
= =
75.0 m3/day 0.0521 m3/min Vair x RAS x Aeff 0.0981 m3/min
Determine Volume of Air Requirement for RAS, VRAS RAS
= =
Vair-RAS
= =
#NAME? m3/day #NAME? m3/min
See Aeration Tank cals
Vair x RAS x Aeff #NAME? m3/min
Determine Volume of Air Requirement for WAS, VWAS WAS
= =
Vair-WAS
= =
#NAME? m3/day #NAME? m3/min
See Aeration Tank cals
Vair x WAS x Aeff #NAME? m3/min
Determine Volume of Air Requirement for MLSS Return, VMR MLSS
= = =
MLSS Return
= =
Vair-MLSS
= =
4 Qave - QRAS 2070.00 - 182.3 m3/day #NAME? m3/day #NAME? m3/day #NAME? m3/min Vair x MLSS x Aeff #NAME? m3/min
Aeration Piping Headloss To Aeration Tanks Criteria Ambient air temperature
To
=
Ambient barometric pressure
Po
=
1.00 atm
Air supply pressure
P
=
1.450 atm
Blower capacity
Qb
=
#NAME? m3/min
Blower efficiency
e
=
75 %
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
Headloss (mm)
hL
=
f x (L/D) x Hv
Size (mm) 80 80 80 80 80 80 80 50 25 25 25 25 25 25 25
Quantity 1 1
K Value 2.50 0.80 0.20 1.80 0.60 0.30 0.30 0.60 0.20 0.60 0.60 0.30 0.30 0.20 0.80
30 deg C
303.2 deg K
Equations
x
TQ^2 PD^4 or
K x Hv
A. Pipe Fittings Losses No. 1 2 9 3 4 5 6 4 9 7 4 5 8 9 10
Valves & Fittings Check valve Gate valve Reducer Tee thru side Tee thru run Elbow 90 deg Elbow 90 deg Tee thru run Reducer Tee thru run Tee thru run Elbow 90 deg Elbow 90 deg Reducer Gate valve
1 1 2 2 2
1
1
Q (m3/min) #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME?
T (deg K) 336.77 336.77 336.77 336.77 336.77 336.77 336.77 336.77 336.77 336.77 336.77 336.77 336.77 336.77 336.77 Subtotal
hL, headloss (mm) #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME?
B. Straight Pipe Losses No. 1 2 3 4
Length (m) 8.00 4.00 7.00 4.00
DIA (mm) 100 50 50 25
Velocity (m/min) #NAME? #NAME? #NAME? #NAME?
Q (m3/min) #NAME? #NAME? #NAME? #NAME?
f, fric. factor #NAME? #NAME? #NAME? #NAME?
T (deg K) 336.77 336.77 336.77 336.77 Subtotal
hL, headloss (mm) #NAME? #NAME? #NAME? #NAME? #NAME?
C. Supply Pressure At The Blower 1 2 3 3 4 5 6
Losses in piping Losses in pipe fittings Losses in air filter Losses in silencer Losses in blower Losses in diffusers Static head
= = = = = = =
Total
#NAME? #NAME? 50.00 50.00 150.00 160.00 4100.00
mm mm mm mm mm mm mm
#NAME? mm
Therefore, the absolute supply pressure
=
@
#NAME? m
#NAME? atm
D. Power Requirement of Blower, P (kw)
where
Therefore,
P
=
R
=
8.314 kJ/k mole deg K
w
=
air mass flow, kg/s
P
=
#NAME? Kw
=
5.5 Kw ok
Select next bigger size motor
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
w RTo 8.41 e
x [ (P/Po)^0.283 - 1 ]
or
#NAME? HP 7 HP
CHLORINATION TANK DESIGN Qpeak
=
2219.31 m3/day
=
1.54 m3/min
Detension Time at Qpeak
t
=
Volume of Tank
V
=
15.00 min Qpeak x t
=
23.12 m3
Number of Tanks
N
=
1
Number of Bays per Tank
n
=
4
Dimension of Tank Provided Depth Wetted depth Width Length Number of pass
H h W L n
= = = = =
1.80 m 1.50 0.75 m 4.50 m 4
Vp
=
Volume Provided
24.30 m3 >=
ok
max 3m
23.12 m3
Check: Ratio
Ratio Wetted depth 1.50 2 Length 4.50 6
Detension Time at Qpeak
t
= =
ok : : : : : :
Width 0.75 1 Width 0.75 1
Vp / Qpeak 15.77 min >
15 min ok
PARSHALL FLUME FOR DISCHARGE Design Flow, Peaking Factor Peak Flow
Qavg PF Qpeak
= = =
517.5 m3/day 4.29 2219.31 m3/day
Formula for flow calculation with diffrent throat width of Parshall Flume by Harlan Bengtson
Flow thru 1" PF,
Q
=
0.338 x H1.55
Flow tru PF
Q
= =
flow in PF in cfs 0.9071091 cfs = Qpeak
Head over flume
H
=
in ft
H
=
1.8263509 ft 556.7 mm
where
For Q > Qpeak, water level in PF
90 DEGREE V-NOTCH WEIR AT OUTLET BOX Design Population , Design Flow, Peaking Factor Peak Flow
PE Qavg PF Qpeak
= = = =
2300 517.5 4.29 2219.31
m3/day m3/day
90 Deg V-Notch Weir Formula
1.417H 5/2
q(m3/sec)
=
H (m)
=
150
Therefore, q
=
0.012348
Set depth of weir @
200
mm
ok
mm or m3/sec > =
0.15
m at Qpeak
0.02569
m3/sec ok
INFLUENT Flow (m3/d) BOD (kg/day) TSS(kg/d)
INFLUENT
517.50 129.38 155.25
Flow (m3/d) BOD (kg/day) TSS(kg/d)
INFLUENT FLOW
INFLUENT
#NAME? 129.38 #NAME?
Flow (m3/d) BOD (kg/day) TSS(kg/d)
SECONDARY EFFLUENT
#NAME? 109.97 #NAME?
SCREEN CHAMBER *
MLSS Flow (m3/d)
Flow (m3/d)
#NAME?
BOD(kg/d) TSS(kg/d)
PUMP STATION
ANOXIC TANK
AERATION BASINS
#NAME? #NAME? #NAME?
SECONDARY CLARIFIERS
FINAL EFFLUENT 10 mg/L BOD 20 mg/L TSS
LAST MANHOLE RAS RAS = 1% Flow (m3/d) TSS(kg/d)
423.41 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) TSS(kg/d)
#NAME? #NAME? DIGESTED SLUDGE Flow (m3/d) TSS(kg/d)
DEWATERED SLUDGE
3.95 #NAME?
Flow (m3/d) TSS(kg/d)
AEROBIC SLUDGE HOLDING TANK (75% VSS) (55% VSS Destruction)
SLUDGE DRYING BEDS (25-40% SOLIDS) (95% CAPTURE)
SUPERNATANT Flow (m3/d) TSS(kg/d)
SOLIDS BALANCE @ Qavg
#NAME? #NAME?
#NAME? #NAME?
INFLUENT Flow (m3/d) BOD (kg/day) TSS(kg/d)
INFLUENT FLOW
INFLUENT
689.63 172.41 206.89
Flow (m3/d) BOD (kg/day) TSS(kg/d)
INFLUENT
697.82 172.41 236.97
Flow (m3/d) BOD (kg/day) TSS(kg/d)
SECONDARY EFFLUENT
683.87 146.55 201.43
Flow (m3/d) BOD(kg/d) TSS(kg/d)
SCREEN CHAMBER *
AERATION BASINS
683.82 6.84 13.68
SECONDARY CLARIFIERS
FINAL EFFLUENT 10 mg/L BOD 20 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) TSS(kg/d)
9.10 91.03 THICKENED SLUDGE Flow (m3/d) TSS(kg/d)
GRAVITY SLUDGE THICKENER TANK (3% Solid)
SUPERNATANT Flow (m3/d) TSS(kg/d)
DIGESTED SLUDGE
2.09 62.79
Flow (m3/d) TSS(kg/d)
AEROBIC SLUDGE HOLDING TANK (80% VSS) (55% VSS Destruction)
SOLIDS BALANCE @ Qavg
Flow (m3/d) TSS(kg/d)
SLUDGE DRYING BEDS (25-40% SOLIDS) (95% CAPTURE)
SUPERNATANT
7.01 28.24
DEWATERED SLUDGE
1.23 36.89
Flow (m3/d) TSS(kg/d)
1.18 1.84
0.05 35.05
SLUDGE HOLDING AREA (30 DAYS)
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