10 Sludge F12
10-Sludge_F12
Treatment and Disposal of Sludge (Processing Sludge) (Chap. 13, VH, p.569) Objectives 1. Reduction of sludge volume (dewatering) - to extract water from the solids and dispose of the dewatered residue - biological sludge = 10% solids + 90% water 2. Stabilization of sludge - to reduce obnoxious qualities of the sludge: a) organics, VSS b) odor c) appearance 3. Disinfection - to destroy pathogenic microorganisms - always an ultimate disposal problem Sludge Treatment Processes (Operations) – Units
Gravity thickening - by Gravity thickener Dissolved air floatation (DAF) Anaerobic digesting - by Anaerobic Digester Mechanical Dewatering - by Vacuum Filter with chemical conditioning (FeCl3, CaO) to improve filterability Passive Dewatering - by Lagoon dewatering by evaporation and seepage Disposal to farm land or landfill - ultimate disposal
Weight and Volume Relationships (VH, p.569; ME, p. 775) a. Sludge - solids from biological wastewater treatment b. Total solids - Total deposit remaining in a dish after evaporation of water from the sample and subsequent drying at 103ºC in oven. b. Volatile Solids (VS) - Loss of weight upon ignition by igniting the dried residue at 550ºC in a muffle furnace. - Volatile solids 70% (60 - 80 %) TS = (100%) SS = (30%) DS = (70%)
FS (30%) FSS
+ +
VS (70%) VSS
FDS
+
VDS
1
10-Sludge_F12
Specific Gravity (S.G.) of dry solids, Ss - The specific gravity of solid matter in sludge can be computed from the relationship: W Ws W f v Ss S f Sv
or
Wf Ws W v S s w S f w Sv w
where Ws = weight of dry solids, lb, kg Wf = weight of fixed solids (non volatile), lb, kg Wv = weight of volatile solids, lb, kg Ss = specific gravity of solids Sf = specific gravity of fixed solids Sv = specific gravity of volatile solids = unit weight of water, lb/ft3 (lb/gal), (62.4 lb/ft3, 8.34 lb/gal, 1000 kg/m3) w = density of water Weight of dry solids Ws W f Wv S .G. of dry solids S s S f Sv
The specific gravity (S.G.) = s / w where s = density of solids w = density of water S.G.
= 1.2 - 1.4 for organic matter = 1.5 - 2.5 for solids in chemical coagulation.
Specific Gravity of wet sludge (solid slurry), S Weight of Wet Sludge Ww Ws Ww Ws S .G. of Wet Sludge S Sw Ss
where Ww = weight of water, lb, kg Ws = weight of dry solids, lb, kg Ww + Ws = total weight of wet sludge, lb, kg S = S.G. of wet sludge Sw = S.G. of water = 1.00 Ss = S.G. of dry solids
2
10-Sludge_F12
The volume of wet sludge, V
V
Ws Ws s 100 P S S 100 100
where
V = volume of sludge, ft3, gal, m3 Ws = weight of dry solids, lb, kg s = solids content, % = unit weight of water (62.4 lb/ft3, 8.34 lb/gal, 1000 kg/m3) P = water content, % S = S.G. of wet sludge
Daily Sludge Volume, Qd Qd
Ws / t Ws / t V s 100 P t S S 100 100
where t = time (d) - The sludge volume is inversely proportional to the solids content. - If a waste is thickened from 2% to 4% solids, the waste volume is reduced by one-half. Example: In general, biological sludge consists of 10% dry solids and 90% water by weight. The dry solids consists of 70% volatile solids and 30% fixed solids. Assuming that the specific gravity (S.G) of volatile solids is 1.0 and S.G. of fixed solids is 2.5, determine S.G. of the biological sludge. Given: The dry solids (TS) consists of: 70% VS: Sv = 1.0 ..... S.G. of volatile solids 30% FS: Sf = 2.5 ..... S.G. of fixed solids thus
1.0 TS = 0.3 FS + 0.7 VS
1) Determine S.G. of dry solids, Ss Ws Wf Wv ------ = ------- + --------Ss Sf Sv
1.00 0.30 0.70 = ------- = -------- + -------- = 0.82 Ss 2.5 1.0
Ws = wt of dry solids (TS) = 1.0 Wf = wt of fixed solids (FS) = 0.3 Wv = wt of volatile solids (VS) = 0.7 thus, Ss = 1.22 ... S.G. of dry solids 3
10-Sludge_F12
2) Determine S.G. of wet solids, S W w + Ws Ww Ws -------------- = {-------- + -------} S Sw Ss Since 10% dry solids and 90% water by weight in general, Ww = weight of water = 0.9 Ws = weight of dry solids = 0.1 Sw = S.G. of water = 1.00 0.9 + 0.1 0.9 0.1 ------------- = -------- + -------- = 0.98197 S 1.00 1.22 thus, S = 1.018 ≈ 1.02 ... S.G. of the wet sludge
These calculations demonstrate that: - For organic sludge of less than 10% solids, the specific gravity may be assumed to be 1.00 without introducing significant error. If dry solids in the sludge is 1000 lb/d and solid concentration is 10%, the daily sludge volume, Qd, is V Ws / d 1000 lb / d Qd = ---- = ------------- = -------------------------------- = 157 ft3 /d d s 10 ----- γ S ------ (62.4 lb/ft3) (1.02) 100 100
Example 13.1 Water Treatment (VH, p.571): Coagulation of surface water using alum produces 10,000 lb (4540 kg) of dry solids/day; of which 20% are volatile. i) Both the settled sludge following coagulation and filter backwash water are concentrated in clarifier-thickeners to a solids concentration of 2.5%. ii) Centrifugation can be used to increase the concentration to 20%, a consistency similar to soft wet clay, or iii) clarifier-thickener underflow can be dewatered to a 40% cake by pressure filtration. (a) Estimate the specific gravities (S.G.) of the thickened sludge, concentrate from centrifugation, and filter cake. (b) Calculate the daily sludge volume from each process: i) Clarifier-Thickner; ii) Centrifugation; and iii) Pressure filtration
4
10-Sludge_F12
(Solution) Assume that the solids consist of: 20% VS: Sv = 1.0 ... S.G. of volatile solids 80% FS: Sf = 2.5 ... S.G. of fixed solids thus, 1.00 TS = 0.8 FS + 0.2 VS 1) Determine S.G. of dry solids, Ss Ws Wf Wv ------ = ------- + -------- = Ss Sf Sv
1.00 0.80 0.20 -------- = -------- + --------- = 0.52 Ss 2.5 1.0
Thus, Ss = 1.9 2) Determine S.G. of wet sludge, S W w + Ws Ww Ws --------------- = {------- + -------} S Sw Ss S.G. of water, Sw = 1.00 i) Thickened sludge (2.5% solids): S.G. of the sludge is 0.975 + 0.025 0.975 0.025 ------------------ = --------- + ---------S 1.00 1.9 S = thickened sludge = 1.0 ... S.G. of the wet sludge
ii) Centrifuge discharge (20% solids): S.G. of the sludge is 0.8 + 0.2 0.8 0.2 ------------- = --------- + --------S 1.00 1.9 S = 1.1
5
10-Sludge_F12
iii) Filter cake (40% solids): S.G. of the sludge is 0.6 + 0.4 0.6 0.4 ----------- = ------- + -------S 1.00 1.9 S = 1.2
b. The volume of waste sludge V
ws ws s 100 P S S 100 100
where V = volume of sludge, ft3 , gal, m3 Ws = weight of dry solids, lb, kg s = solids content, % = S.G. of wet sludge (8.34 lb/gal, 62.4 lb/ft3, 1000 kg/m3) P = water content, % S = S.G. of wet sludge Thickened sludge i) 10,000 lb/d Qd = ---------------------- = 48,000 gpd 2.5 (-----)(8.34)(1.0) 100 or 4540 kg/d Qd = ------------------------------- = 182 m3/d 2.5 (-----)(1000 kg/m3)(1.0) 100 ii) Centrifuge discharge 10,000 lb/d Qd = ----------------------- = 54,000 gpd 20 (-------)(8.34)(1.1) 100 or 6
10-Sludge_F12
4540 kg/d Qd = ------------------------------- = 20.6 m3/d 20 (-----)(1000 kg/m3)(1.1) 100
iii) Filter cake 10,000 lb/d Qd = --------------------- = 2500 gpd 40 (------)(8.34)(1.2) 100 or 4540 kg/d Qd = ------------------------------- = 9.5 m3/d 40 (-----)(1000 kg/m3)(1.2) 100 The daily sludge volume was reduced from 182 m3/d, 20.6 m3/d to 9.5 m3/d.
Example 13.2 (VH, p. 572): Estimate the quantity of sludge produced by a trickling-filter plant treating 1.0 MGD of domestic wastewater. Assuming a suspended-solids concentration of 220 mg/L in the raw wastewater, a solids-content in the sludge equivalent to 90% removal, and a sludge of 5.0% concentration withdrawn from the settling tanks. (Solution) Solids in the sludge = (1.0 MGD)(220 mg/L)(8.34)(0.90) = 1651 lb/d 1650 lb/d Qd = ------------------------------ = 530 ft3/d 5 (------)(62.4 lg/ft3)(1.0) 100
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Treatment and Disposal of Sludge (Processing Sludge) (Chap. 13, VH, p.569) Objectives 1. Reduction of sludge volume (dewatering) - to extract water from the solids and dispose of the dewatered residue - biological sludge = 10% solids + 90% water 2. Stabilization of sludge - to reduce obnoxious qualities of the sludge: a) organics, VSS b) odor c) appearance 3. Disinfection - to destroy pathogenic microorganisms - always an ultimate disposal problem Sludge Treatment Processes (Operations) – Units
Gravity thickening - by Gravity thickener Dissolved air floatation (DAF) Anaerobic digesting - by Anaerobic Digester Mechanical Dewatering - by Vacuum Filter with chemical conditioning (FeCl3, CaO) to improve filterability Passive Dewatering - by Lagoon dewatering by evaporation and seepage Disposal to farm land or landfill - ultimate disposal
Weight and Volume Relationships (VH, p.569; ME, p. 775) a. Sludge - solids from biological wastewater treatment b. Total solids - Total deposit remaining in a dish after evaporation of water from the sample and subsequent drying at 103ºC in oven. b. Volatile Solids (VS) - Loss of weight upon ignition by igniting the dried residue at 550ºC in a muffle furnace. - Volatile solids 70% (60 - 80 %) TS = (100%) SS = (30%) DS = (70%)
FS (30%) FSS
+ +
VS (70%) VSS
FDS
+
VDS
1
10-Sludge_F12
Specific Gravity (S.G.) of dry solids, Ss - The specific gravity of solid matter in sludge can be computed from the relationship: W Ws W f v Ss S f Sv
or
Wf Ws W v S s w S f w Sv w
where Ws = weight of dry solids, lb, kg Wf = weight of fixed solids (non volatile), lb, kg Wv = weight of volatile solids, lb, kg Ss = specific gravity of solids Sf = specific gravity of fixed solids Sv = specific gravity of volatile solids = unit weight of water, lb/ft3 (lb/gal), (62.4 lb/ft3, 8.34 lb/gal, 1000 kg/m3) w = density of water Weight of dry solids Ws W f Wv S .G. of dry solids S s S f Sv
The specific gravity (S.G.) = s / w where s = density of solids w = density of water S.G.
= 1.2 - 1.4 for organic matter = 1.5 - 2.5 for solids in chemical coagulation.
Specific Gravity of wet sludge (solid slurry), S Weight of Wet Sludge Ww Ws Ww Ws S .G. of Wet Sludge S Sw Ss
where Ww = weight of water, lb, kg Ws = weight of dry solids, lb, kg Ww + Ws = total weight of wet sludge, lb, kg S = S.G. of wet sludge Sw = S.G. of water = 1.00 Ss = S.G. of dry solids
2
10-Sludge_F12
The volume of wet sludge, V
V
Ws Ws s 100 P S S 100 100
where
V = volume of sludge, ft3, gal, m3 Ws = weight of dry solids, lb, kg s = solids content, % = unit weight of water (62.4 lb/ft3, 8.34 lb/gal, 1000 kg/m3) P = water content, % S = S.G. of wet sludge
Daily Sludge Volume, Qd Qd
Ws / t Ws / t V s 100 P t S S 100 100
where t = time (d) - The sludge volume is inversely proportional to the solids content. - If a waste is thickened from 2% to 4% solids, the waste volume is reduced by one-half. Example: In general, biological sludge consists of 10% dry solids and 90% water by weight. The dry solids consists of 70% volatile solids and 30% fixed solids. Assuming that the specific gravity (S.G) of volatile solids is 1.0 and S.G. of fixed solids is 2.5, determine S.G. of the biological sludge. Given: The dry solids (TS) consists of: 70% VS: Sv = 1.0 ..... S.G. of volatile solids 30% FS: Sf = 2.5 ..... S.G. of fixed solids thus
1.0 TS = 0.3 FS + 0.7 VS
1) Determine S.G. of dry solids, Ss Ws Wf Wv ------ = ------- + --------Ss Sf Sv
1.00 0.30 0.70 = ------- = -------- + -------- = 0.82 Ss 2.5 1.0
Ws = wt of dry solids (TS) = 1.0 Wf = wt of fixed solids (FS) = 0.3 Wv = wt of volatile solids (VS) = 0.7 thus, Ss = 1.22 ... S.G. of dry solids 3
10-Sludge_F12
2) Determine S.G. of wet solids, S W w + Ws Ww Ws -------------- = {-------- + -------} S Sw Ss Since 10% dry solids and 90% water by weight in general, Ww = weight of water = 0.9 Ws = weight of dry solids = 0.1 Sw = S.G. of water = 1.00 0.9 + 0.1 0.9 0.1 ------------- = -------- + -------- = 0.98197 S 1.00 1.22 thus, S = 1.018 ≈ 1.02 ... S.G. of the wet sludge
These calculations demonstrate that: - For organic sludge of less than 10% solids, the specific gravity may be assumed to be 1.00 without introducing significant error. If dry solids in the sludge is 1000 lb/d and solid concentration is 10%, the daily sludge volume, Qd, is V Ws / d 1000 lb / d Qd = ---- = ------------- = -------------------------------- = 157 ft3 /d d s 10 ----- γ S ------ (62.4 lb/ft3) (1.02) 100 100
Example 13.1 Water Treatment (VH, p.571): Coagulation of surface water using alum produces 10,000 lb (4540 kg) of dry solids/day; of which 20% are volatile. i) Both the settled sludge following coagulation and filter backwash water are concentrated in clarifier-thickeners to a solids concentration of 2.5%. ii) Centrifugation can be used to increase the concentration to 20%, a consistency similar to soft wet clay, or iii) clarifier-thickener underflow can be dewatered to a 40% cake by pressure filtration. (a) Estimate the specific gravities (S.G.) of the thickened sludge, concentrate from centrifugation, and filter cake. (b) Calculate the daily sludge volume from each process: i) Clarifier-Thickner; ii) Centrifugation; and iii) Pressure filtration
4
10-Sludge_F12
(Solution) Assume that the solids consist of: 20% VS: Sv = 1.0 ... S.G. of volatile solids 80% FS: Sf = 2.5 ... S.G. of fixed solids thus, 1.00 TS = 0.8 FS + 0.2 VS 1) Determine S.G. of dry solids, Ss Ws Wf Wv ------ = ------- + -------- = Ss Sf Sv
1.00 0.80 0.20 -------- = -------- + --------- = 0.52 Ss 2.5 1.0
Thus, Ss = 1.9 2) Determine S.G. of wet sludge, S W w + Ws Ww Ws --------------- = {------- + -------} S Sw Ss S.G. of water, Sw = 1.00 i) Thickened sludge (2.5% solids): S.G. of the sludge is 0.975 + 0.025 0.975 0.025 ------------------ = --------- + ---------S 1.00 1.9 S = thickened sludge = 1.0 ... S.G. of the wet sludge
ii) Centrifuge discharge (20% solids): S.G. of the sludge is 0.8 + 0.2 0.8 0.2 ------------- = --------- + --------S 1.00 1.9 S = 1.1
5
10-Sludge_F12
iii) Filter cake (40% solids): S.G. of the sludge is 0.6 + 0.4 0.6 0.4 ----------- = ------- + -------S 1.00 1.9 S = 1.2
b. The volume of waste sludge V
ws ws s 100 P S S 100 100
where V = volume of sludge, ft3 , gal, m3 Ws = weight of dry solids, lb, kg s = solids content, % = S.G. of wet sludge (8.34 lb/gal, 62.4 lb/ft3, 1000 kg/m3) P = water content, % S = S.G. of wet sludge Thickened sludge i) 10,000 lb/d Qd = ---------------------- = 48,000 gpd 2.5 (-----)(8.34)(1.0) 100 or 4540 kg/d Qd = ------------------------------- = 182 m3/d 2.5 (-----)(1000 kg/m3)(1.0) 100 ii) Centrifuge discharge 10,000 lb/d Qd = ----------------------- = 54,000 gpd 20 (-------)(8.34)(1.1) 100 or 6
10-Sludge_F12
4540 kg/d Qd = ------------------------------- = 20.6 m3/d 20 (-----)(1000 kg/m3)(1.1) 100
iii) Filter cake 10,000 lb/d Qd = --------------------- = 2500 gpd 40 (------)(8.34)(1.2) 100 or 4540 kg/d Qd = ------------------------------- = 9.5 m3/d 40 (-----)(1000 kg/m3)(1.2) 100 The daily sludge volume was reduced from 182 m3/d, 20.6 m3/d to 9.5 m3/d.
Example 13.2 (VH, p. 572): Estimate the quantity of sludge produced by a trickling-filter plant treating 1.0 MGD of domestic wastewater. Assuming a suspended-solids concentration of 220 mg/L in the raw wastewater, a solids-content in the sludge equivalent to 90% removal, and a sludge of 5.0% concentration withdrawn from the settling tanks. (Solution) Solids in the sludge = (1.0 MGD)(220 mg/L)(8.34)(0.90) = 1651 lb/d 1650 lb/d Qd = ------------------------------ = 530 ft3/d 5 (------)(62.4 lg/ft3)(1.0) 100
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