water supply calculation
TRANSCRIPT
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Characteristics of Wastewater
Treatment Sludge
By
Ricardo B. Jacquez
Professor
CAGE Department
NMSUAssistant: Maung Myint
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Important Relations for Sludge Calculations
Ws = weight of the dry solids
Wf = weight of the fixed solids
Wv = weight of the volatile solidsSs = specific gravity of dry solids
Sf = specific gravity of fixed solids (2.5)
Sv = specific gravity of volatile solids (1.2)
v
v
f
f
s
s
S
W
S
W
S
W
+=
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Sludge volume = V
Ws= weight of dry solid [lb or gm]r = unit weight of water = 62.4 lb/ft3 = 1000 kg/m3
Swet.sludge= specific gravity of wet sludge
sludgewet
Srsolid
WsV
.100
%
=
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Sources of wastewater sludge
Primary clarifier
Secondary clarifiers:
activated sludgebio-towers
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Typical domestic wastewater sludge,
Ws=1
Wf=0.3 Wv=0.7
Sf=2.5 Sv=1.2
42.12.1
7.0
5.2
3.01=+=
+=
s
s
v
v
f
f
s
s
SS
S
W
S
W
S
W
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Typical sludge has 96.5% water, 3.5% solid,
and specific gravity of solid=1.42.
Determine the specific gravity of typical wet
sludge?
( ) ( )
( ) ( ) 0.101.142.1/5.31/5.96
5.35.96
/%/%
%%
.
.
==++=
++=
sludgewet
sludgewet
S
SssludgeinsolidSwwater
sludgeinsolidwaterS
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Primary clarifier solids
Mp=E Q [MGD] SS [mg/L] 8.34 [lb/MG(mg/L)]
Mp = mass of primary solids, [lb/day]
E = efficiency of primary clarifier (see Fig)SS = suspended solid concentration in influent
[mg/L]
Q = flow rate [MGD]
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(from Peavy, Rowe, and Tchobanoglous, 1985, p. 228)
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Secondary clarifier solids
Ms=YQ[MGD] BODR [mg/L] 8.34[lb/MG mg/L]
Ms=mass of secondary solids [lb/day]Y =yield coefficient (see Fig. 5-37)
BODR=BOD removed [mg/L]
Q = flow rate [MGD]
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Volume of sludge
sludgewetofgravityspecificS
gallbwaterofweightunitr
daygalsludgeofvolumeV
SrwaterMMV
Srsolid
MMV
sludgewet
sludgewet
SP
sludgewet
SP
=
=
=
+=
+=
.
.
.
]/34.8[
]/[
)100/%100(
)100/(%
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Typical solids content of sludge
Type of sludge Sludge concentration %
Un-thickened ThickenedSeparate
Primary sludge 2.5 - 5.5 8 - 10
Trickling-filter sludge 4 - 7 7 - 9
Activated sludge 0.5 - 1.2 2.5 - 3.3
Pure O2
sludge 0.8 - 3 2.5 - 9
Combined
Primary and trickling-filter sludge 3 - 6 7 - 9
Primary and modified-aeration 3 - 4 8.3 - 11.6
Primary and air-- -
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Ex. Reduction of volume by sludge thickening
A wastewater-treatment plant consists ofprimary treatment units followed by an
activated-sludge secondary system. Theprimary and secondary sludges are mixed,thickened in a gravity thickener, and sent to
further treatment. A schematic of the system isshown in the next slide.
Influent SS=200mg/L Primary sludge =5%
Influent BOD=225mg/LEffluent BOD=20mg/L Thickened sludge=4%
Flow Q = 5 MGD Secondary sludge=0.75%
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Primary clarifier diameter = 82 ft
Aeration volume = 102,400 cft = 766,180 gal
Influent Primary Aeration Secondary
clarifier clarifier
Sludge return
LiquidReturn Thickener Thickened sludge
to sludge-treatment facilities
MLSS in aerator 3,500 mg/L
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1. Mass and volume of primary sludge
a) area of primary clarifier = A = 3.14 D2/4
A = 3.14 (82 ft)2
/ 4 = 5,283 ft2
b) overflow rate = Q/A = 5.0 x 106/5283 ft2
=950 gpd/ft2 (42 m/s)
c) from Fig. 5-13, the efficiency of clarifier isEss = 58% EBOD= 32%
d) mass of primary solids removed is Mp
Mp=Ess Q SS
=0.58(5 MGD)(200 mg/L)(8.34lb/MG
mg/L)
=
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e) volume of primary sludge
( )
daygalV
gallb
dlbV
Srsolid
MV
sludgewet
P
/600,11
)0.1(/34.8)100/5(
/840,4
)100/(% .
=
=
=
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2. Mass of secondary solids
b) from Fig. 13.1 (text, pg 651), Y=0.48
c) mass of secondary solid Ms
Ms=Y BOD5
Q
=0.48(5MGD)133mg/L(8.34lb/MG mg/L)
Ms = 2,660 lb/day
( ) ( )
( )29.0
/3500766.0
/22532.015
/
=
=
=
LmgMG
LmgMGD
M
F
masssludge
dayappliedBOD
M
F
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d) volume of secondary sludge
3a. Total mass of solids MT=Mp +Ms
MT= 4,840 + 2,660 = 7,500 lb/day
3b. Total volume of sluge VT=Vp +Vs
VT= 11,600+ 42,600 = 54,200 gal/d
( )
daygalV
gallb
dlbV
Srsolid
MV
sludgewet
s
/600,42
)0.1(/34.8)100/75.0(
/660,2)100/(% .
=
=
=
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4a) mass of solids sent to sludge treatment
Neglect the solids in the thickener
supernatant, the mass of solids in thethickened sludge is equal to 7,500 lb/day.
4b) volume of thickened sludge
( )
daygalV
gallb
dlbV
Srsolid
MV
sludgethickened
thickened
/500,22
)0.1(/34.8)100/4(
/500,7
)100/(% .
=
=
=
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5. Volume reduction (%) by thickening
%58100200,54
500,22200,54
100
==
=
reductionvolume
V
VVreductionvolume
T
thickenedT
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Thank you
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