VOLUME 2 Post-Treatment, Reuse, and Disposal
5.2 Physical Quality
The physical quality of municipal wastewater is generally reported in terms of temperature, color, turbid- ity, odor, and suspended (settleable and nonsettleable) and dissolved solids. The significance of these parameters is briefly summarized below. Measurement procedures may be found in References 1–3.
5.2.1 Temperature
The temperature of municipal wastewater is slightly higher than that of the water supply, and stays in the range of 10–21◦C. The average temperature varies slightly with the season, and is higher than the average air temperature most of the year except during the hot summer months. The wastewater temperature has significant effect upon the microbiological activity, treatability, solubility of gases, density, and viscosity.
5.2.2 Color, Turbidity, and Odor
The color of fresh wastewater is slightly gray. Stale or septic wastewater is dark gray or black. Turbidity is due to suspended and colloidal particles. In general, stronger wastewaters have higher turbidity.
Fresh wastewater has soapy or oily odor that is somewhat disagreeable. Stale wastewater has putrid odor due to hydrogen sulfide, indol, and skatol, and other decomposition products. The intensity of odor is 5-1
at which the odor is just perceptible.TONis expressed by Equation 5.1:
TON=A+B
A (5.1)
where
TON=threshold odor number A =volume of sample, mL
B =volume of odor-free water, mL. The recommended volume of A+B is 200 mL
A series of dilutions and blanks (no sample) are made and maintained at 60◦C. Panel members are requested to give their response. Geometric mean of TON is developed from the response of panel mem- bers. Larger the TON value, more odorous is the sample.
EXAMPLE 5.1: TON TEST
TON study was conducted on a wastewater sample. A series of dilutions were made in odor-free water such that the total volume of diluted sample in each case was 200 mL. The results of a three-member panel are summarized below. Determine TON.
Panel Member
Panel Response to Sample Volume Diluted to 200 mL
0.5 1.0 B 1.5 2.0 2.5 B 3.0 3.5 4.0
1 þ þ þ þ þ
2 þ þ þ þ
3 þ þ þ þ þ þ
B¼blank (dilution water only) þ ¼odor detected
¼No odor perception
Solution
1. Determine the TON values based on results of each panel member.
Panel member 1, TON1¼200 2 ¼100 Panel member 2, TON2¼200
2:5 ¼80 Panel member 3, TON3¼200
1:5 ¼133 2. Determine the mean TON values.
a. Arithmetic mean: TON¼1
n(TON1þTON2þ. . .þTONn)
¼1
3(100þ80þ133)¼104 b. Geometric mean
Method 1 TON¼(TON1TON2. . .TONn)1n¼(10080133)13¼102
Method 2 m¼1
n½log (TON1)þlog (TON2)þ. . .log (TONn)
¼1
3(log 100þlog 80þlog 133)¼1
3(2:00þ1:90þ2:12)¼2:01 TON¼10m¼102:01¼102
5.2.3 Settleable and Suspended (Nonfilterable), Dissolved (Filterable), Volatile, and Fixed Solids
The settleable solids are organic and inorganic. They settle under low velocity, and may block the channels and pipes. Organic content of settleable solids will undergo decomposition and cause odors.
Solids in municipal wastewater contain 50–80% volatile, and 20–50%fixed solids, although this ratio may vary greatly. The solids settleability test is conducted by settling the wastewater in anImhoffcone for 1 h. It is also obtained from the results of total suspended solids (mg/L) minus nonsettleable solids (mg/L). The suspended or nonfilterable solids are determined byfiltration of sample through a glassfiber filter with a nominal pore size of about 1.2μm (1μm=10−6m). The dissolved orfilterable solids are determined by evaporation of afiltered sample in a steam bath. The volatile solids in each category are determined by ignition of dry solids in a muffle furnace at 550+50◦C. Procedures for chemical analysis may be found in the Standard Methods.1
EXAMPLE 5.2: SETTLEABLE SOLIDS IN IMHOFF CONE
A 1-L sample of raw municipal wastewater is settled for 1 h in anImhoffcone (Figure 5.1). The volume of settled solids in the bottom of the cone is 15 mL. Express the result in mL/L and mg/L. Assume bulk density of settled solids in the cone is 1015 kg/m3.
Solution
1. Express the result as mL/L.
Since the volume of sample in theImhoffcone is 1 L, the volume of settleable solids=15 mL/L 2. Express the result as mg/L
Based on bulk density, the concentration of settled solids in theImhoffcone
=15 mL/L× m3
106mL×1015 kg/m3×106mg
kg =15,225 mg/L 1 L
Imhoff cone Settled solids Stand
FIGURE 5.1 Determination of settleable solids in anImhoffcone (Example 5.2).
EXAMPLE 5.3: SETTLEABLE, NONSETTLEABLE, AND DISSOLVED (FILTERABLE) SOLIDS
Twenty-five milliliters of raw wastewater sample wasfiltered through a dried and preweighed glass-fiber filter to determine the total suspended solids (TSS). One liter of raw wastewater sample was settled in an Imhoffcone for 1 h to determine the volume of settleable suspended solids (SS). Twenty-five milliliters of settled sample wasfiltered through another predried and weighed glass-fiberfilter to determine the non- settleable SS. Fifty milliliters offiltrate was evaporated in a predried and weighed petri dish over a steam bath to determine the total dissolved orfilterable solids. Calculate (1) TSS, (2) nonsettleable (or nonfil- terable) SS, (3) settleable SS, (4) total dissolved (filterable) solids (TDS), and (5) the concentration of solids in the settled sludge in theImhoffcone.
The results of the tests are given below:
TSS, including settleable and nonsettleable SS
Weight of driedfilter paper in an aluminum dish =12.3478 g
Weight of driedfilter paper in the aluminum dish and TSS from raw wastewater=12.3534 g Nonsettleable SS
Weight of driedfilter paper in an aluminum dish =12.3480 g
Weight of driedfilter paper in the aluminum dish, and nonsettleable suspended solids =12.3508 g Settleable SS
Volume of settled SS in theImhoffcone =10 mL/L
TDS
Weight of dried petri dish =65.4711 g
Weight of dried petri dish andfilterable residue =65.5080 g
Solution
1. Calculate the concentration of TSS in raw wastewater sample, including settleable and nonsettleable SS.
Concentration of TSS =(12.3534−12.3478) g
25 mL ×103mg
g ×103mL
L =244 mg/L 2. Calculate the concentration of nonsettleable SS.
Concentration of nonsettleable SS =(12.3508−12.3480) g
25 mL ×103mg
g ×103mL
L =112 mg/L 3. Determine the concentration of settleable SS.
Concentration of settleable SS =TSS−nonsettleable SS=(244−112) mg/L=132 mg/L 4. Calculate the concentration of TDS.
Concentration of TDS =(65.5080−65.4711) g
50 mL ×103mg
g ×103mL
L =738 mg/L 5. Calculate the concentration of settled solids in theImhoffcone.
Volume of settled SS in 1-LImhoffcone=10 mL/L Concentration of solids in the sludge =132 mg/L
10 mL/L ×103mL
L =13,200 mg/L
EXAMPLE 5.4: SLUDGE QUANTITY AND PUMPING RATE
Municipal wastewater is settled in a primary sedimentation basin. The settled solids are pumped out inter- mittently from the basin. The average wastewaterflow to the basin is 0.25 m3/s. A constant speed sludge pump runs 15 min per h of cycle time. The average pumping rate is 0.321 m3/min. The concentration of total solids in raw and settled wastewater is 250 and 150 mg/L, respectively. The average solids content in the sludge is 1.85%. Calculate the following: (a) total volume of wet sludge, (b) total quantity of settleable solids in the sludge, (c) density and specific gravity of liquid sludge, and (d) the volumetric concentration expected in theImhoffcone.
Solution
1. Calculate the volume of sludge pumped per day.
Pump running time =15 min
h ×24 h
d =360 min/d Total volume of wet sludge pumped =0.321 m3
min ×360 min
d =116 m3/d 2. Calculate the quantity of settleable SS.
Concentration of settleable SS in wastewater= (250−150)mg/L=100 mg/L
Wastewater flow rate =0.25 m3
s ×(60×60×24) s
d =21,600 m3/d Quantity of settleable dry sludge =100 mg/L× kg
106mg×21,600 m3/d×103L m3
=2160 kg/d 3. Calculate the density and specific gravity of wet sludge.
Quantity of wet sludge ¼2160 kg/d dry solids100 kg of wet sludge 1:85 kg dry solids
¼117,000 kg/d wet sludge
Density of wet sludge ¼117,000 kg/d
116 m3=d ¼1010 kg=m3
Specific gravity ¼1010 kg=m3sludge
1000 kg=m3water¼1:01 4. Calculate the expected volumetric concentration in theImhoffcone.
The volumetric concentration in the 1−LImhoffcone=116 m3/d×106mL m3 21,600 m3/d×103L
m3
=5.4 mL/L
5. Summarize thefinal results.
a. Total volume of wet sludge ¼116 m3=d
b. Total quantity of settleable solids in the sludge ¼2160 kg=d
c. Density and specific gravity of wet sludge ¼1010 kg=m3and 1.01 d. Volumetric concentration expected inImhoffcone ¼5.4 mL=L
EXAMPLE 5.5: DETERMINATION OF VOLATILE AND FIXED SOLIDS
Twenty-five milliliters of municipal wastewater sample wasfiltered through a dried preweighed glass- fiberfilter. Thefilter afterfiltration of solids was dried and weighed. Fifty milliliters of the sample was evaporated in a preweighed petri dish. The dish was dried and weighed. Both thefilter paper and the dish were ignited in a muffle furnace for 15–20 min, cooled, and weighed. Determine (a) nonsettleable SS, andfixed and volatile components, (b)filterable and nonfilterable solids, andfixed and volatile com- ponents, and (c) dissolved orfilterable solids,fixed and volatile components.
EXAMPLE 5.6: BLOCK PRESENTATION OF SUSPENDED, DISSOLVED, VOLATILE, AND FIXED SOLIDS
Draw a block diagram. Indicate the concentrations of volatile andfixed solids of suspended and dissolved solids in typical municipal wastewater.
Solution
The block representation of various components of suspended and dissolved solids in municipal waste- water are given inFigure 5.2. The typical values are given in parentheses.
The experimental values are given below:
Weight of driedfilter paper in the aluminum dish =12.3469 g
Weight of aluminum dish,filter paper, and suspended solids after drying =12.3540 g
Weight of aluminum dish and ash after ignition =12.3488 g
Weight of ignited petri dish =65.3821 g
Weight of petri dish, and total (filterable and nonfilterable) solids after drying=65.4221 g
Weight of petri dish and ash after ignition =65.3920 g
Solution
1. Determine the total suspended solids (TSS),fixed suspended solids (FSS), and volatile suspended solids (VSS).
TSS=(12.3540−12.3469) g
25 mL ×103mg
g ×103mL
L =284 mg/L
FSS=(12.3488−12.3469) g
25 mL ×103mg
g ×103mL
L =76 mg/L
VSS=(12.3540−12.3488) g
25 mL ×103mg
g ×103mL
L =208 mg/L
2. Determine the total solids (TS), totalfixed solids (TFS), and total volatile solids (TVS).
TS=(65.4221−65.3821) g
50 mL ×103mg
g ×103mL
L =800 mg/L
TFS=(65.3920−65.3821) g
50 mL ×103mg
g ×103mL
L =198 mg/L
TVS=(65.4221−65.3920) g
50 mL ×103mg
g ×103mL
L =602 mg/L
3. Determine the total dissolved orfilterable solids (TDS), fixed dissolved solids (FDS), and volatile dissolved solids (VDS).
TDS ¼TSTSS¼(800284) mg=L¼516 mg=L
Fixed dissolved solids (FDS) ¼TFSFSS¼(19876) mg=L¼122 mg=L Volatile dissolved solids (VDS)¼TVSVSS¼(602208) mg=L¼394 mg=L