VOLUME 2 Post-Treatment, Reuse, and Disposal
3.2 Mass Balance Analysis
The law of conservation of mass states that mass can neither be created nor destroyed. Mass balance analyses are routinely used in environmental engineering. To apply mass balance analysis it is necessary to establish a system boundary, which is an imaginary barrier drawn around the system. The system boun- dary may surround a node (junction), reactor, container, or a process diagram. Proper selection of system boundary is extremely important to identify all flows and masses into and out of the system.1,2The generalized mass balance is expressed by the following statement and equations (Equations 3.1 and 3.2a).
Accumulation rate
= Input
rate
− Output rate
+ Utilization or conversion rate
(3.1)
dmA
dt =VdCAi
dt =n
i=1
(QiCAi)−m
j=1
(QjCAj)+rAV (3.2a)
where
mA=mass of species A, mass
Qi =volumetric flow rate of the species entering the system through line i (i=0, 1, …, n), volume/time
CAi=concentration of species A entering the system through linei(i=0, 1,…,n), mass/volume Qj =volumetric flow rate of the species leaving the system through line j (j=0, 1, …, m),
volume/time
CAj=concentration of species A leaving the system through linej(j=0, 1,…,m), mass/volume rA =reaction rate of species A, mass/volume·time
V =volume of reactor, volume
3-1
Ai/ Equation 3.2a reduces to Equation 3.2b.
−rAV =n
i=1
(QiCAi)−m
j=1
(QjCAj) (3.2b)
3.2.1 Procedure for Mass Balance Analysis
Compounds with no chemical formation or loss within the control volume are termedconservative(i.e., mass is truly conserved). These compounds are not affected by chemical or biological reactions. Examples are chloride,fluoride, sodium, and tracer dyes.Nonconservativecompounds such as BOD, COD, TKN undergo consumption or generation. To apply mass balance analysis around the system boundary, the following steps must be followed:
1. Draw system boundary, and identify the volumetricflow rate into and out of the system by arrows.
All massflows that are known or to be calculated must cross the system boundary.
2. Assume that the liquid volume within the system does not change.
3. Assume that the control volume is well mixed.
4. Determine whether the compound being balanced is conservative (rA=0) or nonconservative (rAmust be determined based on reaction kinetics).
5. Determine whether the process is steady state (dC/dt=0) or nonsteady state (dC/dt≠0).
6. Solve the problem. This will require solution of a differential equation for nonsteady-state condi- tion, and algebraic solution for steady-state condition.
EXAMPLE 3.1: SIMPLIFIED MASS BALANCE EQUATION
Simplify Equation 3.2a for a single stream entering and leaving the reaction vessel. Assumefirst-order reaction rate (rA= −kC).
Solution
1. Draw the reactor and system boundary (Figure 3.1).
2. Write the mass balance (Equation 3.3).
VdC
dt =QC0−QC+V(−kC) (3.3)
wherek=reaction rate constant (first order), time−1.
Other variables are defined earlier, and are shown inFigure 3.1.
System boundary Conversion
–rAV C0
Q
C Q
FIGURE 3.1 Reactor and system boundary (Example 3.1).
3.2.2 Combining Flow Streams of a Single Material
A system boundary may receive several influent lines, and may have one or more effluent lines. Flow lines presenting one material may be combined in Equation 3.4.
Rate of flow change
= Rate of flow in
− Rate of flow out
+ Rate of volume increased
− Rate of volume reduced
(3.4)
0=Flow in−Flow out+0−0 Flow in=Flow out
EXAMPLE 3.2: FLUID-FLOW SYSTEM
Fluid enters and leaves a reactor. Assuming no accumulation offluid (dρV/dt)=0, andfluid is neither produced nor lost in the system (rm=0). Prove thatQin=Qout, if thefluid is incompressible.
Solution
1. Draw the reactor and system boundary (Figure 3.2).
2. Write the mass balance equation.
d(ρV)
dt =Qinρ1−Qoutρ2+rmV where
ρ =mean density offluid in the control reactor, kg/m3
ρ1andρ2 =density of thefluid entering and leaving the control reactor, kg/m3 QinandQout=volumetricflow rate in and out of the control volume, m3/s V =volume of control reactor, m3
rm =mass rate of generation, g/m3s 3. Givend(ρV)/dt=0 andrw=0; write simplified equation.
Qinρ1=Qoutρ2
Since thefluid is incompressible,ρ1=ρ2
Qin=Qout
System boundary Qout
Qin –rwV
FIGURE 3.2 Reactor and system boundary (Example 3.2).
EXAMPLE 3.3: SEWERS FLOW AND JUNCTION BOXES
Intercepting sewers receiveflows from several laterals and then discharges into a trunk line. Theflow lines are shown inFigure 3.3. Determine theflow in thefinal trunk line.
Solution Method 1.
Draw three system boundaries.
1. Conductflow balance at Manhole A andfindflow in outgoing intercepting sewer.
Flow out into the interceptor=(10,000+15,000) L/min=25,000 L/min 2. Conductflow balance at Manhole B andfindflow in outgoing intercepting sewer.
Flow out into the interceptor=(30,800+12,900) L/min=43,700 L/min 3. Conductflow balance at Manhole C and determineflow in thefinal trunk sewer.
Flow in trunk line=(25,000+43,000) L/min=68,700 L/min Method 2.
Draw one system boundary around three manholes.
Flow in thefinal trunk line=(10,000 +15,000+30,800+12,900) L/min=68,700 L/min
EXAMPLE 3.4: HYDROLOGICAL CYCLE AND GROUNDWATER RECHARGE A 400 ha farm receives 100 cm precipitation per year. It is estimated that 40% precipitation returns into the atmosphere by evaporation, and 20% reaches the nearest watercourse as runoff. The remaining pre- cipitation is percolated into the aquifer. The water is withdrawn from the aquifer throughout the year for irrigation purposes. Approximately 80% of the withdrawn groundwater is eventually lost as evapotrans- piration. Calculate the amount of groundwater recharge in m3that can be withdrawn annually for
Final trunk line System boundary around Manhole A System boundary around Manhole C
System boundary around Manhole B 15,000 L/min
10,000 L/min
68,700 L/min A
C B
(a)
30,800 L/min
15,000 L/min 10,000 L/min
30,800 L/min A
C (b)
12,900 L/min
12,900 L/min B
Final trunk line System boundary around all three manholes
68,700 L/min
FIGURE 3.3 Sewer layout and system boundary: (a) Method 1 and (b) Method 2 (Example 3.3).
irrigation without depleting the groundwater reservoir. Also, draw the hydrological cycle with water components.
Solution
1. Determine the annual precipitation.
Precipitation volume =100 cm/year
100 cm/m ×400 ha×10,000 m2/ha=4×106m3/year 2. Determine the components of water budget.
Runoff ¼0.2(4106m3=year)¼0.8106m3=year Evaporation of precipitation ¼0.4(4106m3=year)¼1.6106m3=year Assume pumping rate for irrigation¼Qm3=year
Evaporation of irrigation water ¼0.8Qm3=year 3. Conduct the volume balance.
[Accumulation] = [Precipitation] − [(Evaporation of precipitation)+(Runoff )]
− [Evaporation of irrigation water]
=4×106m3/year−(1.6×106m3/year+0.8×106m3/year)−0.8Q Since annual withdrawal is without depletion of groundwater reservoir, net accumulation=0.
0 ¼1.6106m3=year0.8Q Q ¼2106m3=year
4. Draw the hydrological cycle showing water components inFigure 3.4.
Groundwater aquifier Cone of depression
Groundwater table
Irrigation water Q = 2 × 106 m3/year
Evapotranspiration of irrigation water 0.8 Q = 1.6 × 106 m3/year 1.6 × 106 m3/year
Evapotranspiration of precipitation
Pumping rate (or amount of groundwater available
for irrigation from recharge) Q = 2 × 106 m3/year
Repercolation of irrigation water 0.4 × 106 m3/year Runoff 0.8 × 106 m3/year
Percolation of precipitation 1.6 × 106 m3/year Precipitation
4 × 106 m3/year
Well for irrigation
FIGURE 3.4 Hydrological cycle showing water budget components (Example 3.4).
3.2.3 Combining a Conservative Substance and Flow
If a system boundary receivesflow streams that also contain a conservative material, a mass balance anal- ysis will includeflows and concentrations.
EXAMPLE 3.5: DISCHARGING A CONSERVATIVE SUBSTANCE IN A NATURAL STREAM
An industry is discharging waste brine into a stream. The concentration of total dissolved solids (TDS) in the industrial brine is 15,600 mg/L. The allowable concentration of TDS in the stream is 500 mg/L. The discharge of the stream under drought conditions is 8500 m3/d and background TDS concentration in the stream is 210 mg/L. Calculate the permissible discharge of industrial brine into the stream.
Solution
1. Draw the system boundary (Figure 3.5).
2. Apply the mass balance equation (Equation 3.2a).
VdC dt =
QCin−
QCout+rAV
There is no accumulationV(dC/dt)=0 and there is no conversion (rV=0).
QstreamCstream+QbrineCbrine−QmixtureCmixture=0
8500 m3/d×210 g/m3+Qbrine×15,600 g/m3−(8500 m3/d+Qbrine)×500 g/m3=0 17.85×105g/d+15,600 g/m3×Qbrine−42.5×105g/d−500 g/m3×Qbrine=0 15,100 g/m3×Qbrine=24.65×105g/d
Qbrine=163 m3/d
The permissible brine discharge=163 m3/d Qstream = 8500 m3/d
Cstream = 210 mg/L
Qmixture
Cmixture = 500 mg/L Qbrine
Cbrine = 15,600 mg/L Brine discharge
Stream System boundary
FIGURE 3.5 System boundary (Example 3.5).
EXAMPLE 3.6: MLSS CONCENTRATION IN AN ACTIVATED SLUDGE PLANT The biomass concentration in an activated sludge process is maintained by returning the sludge. The mixed liquor suspended solid (MLSS) concentration in the aeration basin is 2500 mg/L and TSS concen- tration (TSSras) in return activated sludge (RAS) is 10,000 mg/L. TSS concentration (TSSinf) in the influ- ent is small and ignored. The process diagram is shown below. Calculate (a) the return rate of sludge (Qras), and (b) theflow ratio of RAS to influent.
Solution
1. Draw the process diagram and system boundary.
The system boundary is drawn around Point A (Figure 3.6).
2. Conduct a mass balance at Point A.
VdCTSS
dt =QinfTSSinf−QrasTSSras+rTSSV
There is no accumulationV(dCTSS/dt)=0, and mass conversion in the connecting system is small and assumed zero (rTSSV=0).
0=(4500 m3/d×0+Qras×10,000 g/m3)−(4500 m3/d+Qras)×2500 g/m3+0 Qras= (2500×4500) g/d
(10,000−2500) g/m3=1500 m3/d 3. Determine the returnflow ratio (Rras) ofQrastoQinf.
Rras=Qras
Qinf =1500 m3/d 4500 m3/d=0.33
The RASflow is one-third of the influentflow.
Aeration basin MLSS = 2500 mg/L Qinf = 4,500 m3/d
TSSinf = 0 System boundary
Final clarifier A
TSSras = 10,000 mg/L Qras
FIGURE 3.6 Process diagram and system boundary (Example 3.6).
EXAMPLE 3.7: INDUSTRIAL SEWER SURVEY
An industry uses electroplating process in several shops at the plant site. Wastewater from each shop is collected in an equalization basin, and discharged in individual sewer lines over a 24-h period. The char- acteristics of waste stream from each shop are given below inFigure 3.7. Determine theflow and pollutant concentrations in the combined waste stream.
Solution
1. Draw the overall system boundary (Figure 3.7).
2. Conductflow balance.
0=(QA+QB+QC+QD) −Qcombined
Qcombined=(3500+800+18,900+21,000) L/d=44,200 L/d 3. Conduct a mass balance for each contaminant.
Since there is no accumulation and all heavy metals are conservative substances, simplified mass balance analysis will give the result.
a. Material balance for Cr.
0=(QB×CrB+QC×CrC+QD×CrD)−(Qcombined×Crcombined)
Crcombined=800 L/d×160 mg/L+18,900 L/d×4 mg/L+21,000 L/d×8 mg/L 44,200 L/d
=371,600 mg/d
44,200 L/d =8.4 mg/L b. Material balance for Cu.
0=(QA×CuA+QB×CuB+QD×CuD)−(Qcombined×Cucombined)
Cucombined=3500 L/d×12 mg/L+800 L/d×215 mg/L+21,000 L/d×4 mg/L 44,200 L/d
=298,000 mg/d
44,200 L/d =6.7 mg/L CuA = 12 m/L
ZnA = 10 mg/L
Crc = 4 mg/L Znc = 7 mg/L
Crcombined Cucombined Zncombined
CrB = 160 mg/L CrD = 8 mg/L CuB = 215 mg/L CuD = 4 mg/L ZnD = 9 mg/L
Shop B Shop D
System boundary
Shop A Shop C
Qcombined Qc = 18,9000 L/d
QA = 3500 L/d
QB = 800 L/d QD = 21,000 L/d
FIGURE 3.7 Industrial sewer plan (Example 3.7).
EXAMPLE 3.8: SPLIT TREATMENT OF INDUSTRIAL WATER
An industrial water treatment plant produces 100 m3/minfinished water. The hardness in raw water is 15 mg/L as CaCO3. The plant uses split treatment. Partialflow is treated by zeolite softener. Softened water has hardness of 0.02 mg/L as CaCO3. A small stream is bypassed around the softener and then mixed with the softened water. The upper limit of hardness in treated water is 0.9 mg/L as CaCO3. Calculate (a) hardness capture rate in the softener, (b) partialflow to the softener, and (c)flow bypassed around the softener.
Solution
1. Draw the process diagram and the system boundaries around the water treatment plant, softener, and flow splitter (Figure 3.8).
2. Conduct a mass balance around the water treatment plant to determine hardness capture rateZin the softener.
Z¼15 g=m3as CaCO3100 m3=min0.9 g=m3as CaCO3100 m3=min Z¼1410 g=min as CaCO3
3. Conduct a mass balance around the softener to determinefiltration rateQ2. 15 g/m3as CaCO3×Q2=1410 g/min as CaCO3+0.02 g/m3as CaCO3×Q2
Q2=1410 g/min as CaCO3
14.98 g/m3as CaCO3=94.1 m3/min c. Material balance for Zn.
0=(QA×ZnA+QC×ZnC+QD×ZnD)−(Qcombined×Zncombined)
Zncombined=3500 L/d×10 mg/L+18,900 L/d×7 mg/L+21,000 L/d×9 mg/L 44,200 L/d
=356,300 g/d
44,200 L/d =8.1 mg/L
Q1 (m3/min) Hardness = 15 mg/L
as CaCO3
Capture rate Z (g/min as CaCO3) Q2 (m3/min)
Hardness = 15 mg/L as CaCO3
Q2 (m3/min) Hardness = 0.02 mg/L
as CaCO3
System boundary around water treatment plant
Q = 100 m3/min Hardness = 15 mg/L
as CaCO3
Q = 100 m3/min Hardness = 0.9 mg/L
as CaCO3 System boundary around flow
splitter
System boundary around softener Zeolite
softener Flow
splitter
FIGURE 3.8 Process diagram of water treatment plant and system boundaries (Example 3.8).
4. Conduct a mass balance around the splitter to determine bypassflowQ1around the softener.
100 m3/min=94.1 m3/min+Q1
Q1=(100−94.1) m3/min=5.9 m3/min
EXAMPLE 3.9: PARTICULATE REMOVAL IN A BAGHOUSE
An air pollution facility is using a baghouse to remove dust from an air exhaust streamflowing at a rate of 200 m3/min. The dirty air contains 10 g/m3of particulate, while the clean air from the baghouse contain 0.02 g/m3particulate. The operating permit allows the exhaust stream to contain as much as 0.9 g/m3of particulate matter. The industry wishes to use split treatment by bypassing some of the dirty air around the baghouse and mixing it back into the clean air so that the total exhaust stream meets the permit limit.
Assume that there is no air leakage, and there is negligible change in pressure or temperature of air though the process. Calculate theflow rate of air through the baghouse and kilogram of dust collected per day at the baghouse.
Solution
1. Draw the process diagram of air pollution control facility.
The process diagram and system boundaries are shown inFigure 3.9.
2. Conduct aflow balance at system boundary A.
200 m3=min¼XþY
X ¼200 m3=minY
Baghouse Baghouse inflow
Flow = Y (m3/min) Dust = 10.0 g/m3
Baghouse exhaust Flow = Y (m3/min) Dust = 0.02 g/m3
Dust removed W (kg/d)
Exhaust air to atmosphere Flow = 200 m3/min
Dust = 0.9 g/m3 Bypass mixed with baghouse exhaust
Flow = X (m3/min) Dust = 10.0 g/m3 Exhaust steam
Flow = 200 m3/min Dust = 10.0 g/m3
Calculation results X = 18 m3/min Y = 182 m3/min W = 1816 g/min or 2615 kg/d
A B
C
FIGURE 3.9 Process diagram of air pollution control facility (Example 3.9).
3. Conduct a mass balance at system boundary B and calculate gasflow through baghouse.
X10.0 g=m3+Y0.02 g=m3 ¼200 m3=min0.9 g=m3 (200 m3=minY)10 g=m3þ0.02 g=m3Y ¼180 g=min
2000 g=min10 g=m3Yþ0.02 g=m3Y ¼180 g=min
9.98 g=m3Y ¼1820 g=min
Y ¼182 m3=min
4. Determine theflowXbypassed the baghouse at system boundary A.
X=(200−182) m3/min=18 m3/min
5. Conduct a mass balance at system boundary C, and calculate dust collected in the baghouse.
Y×10 g/m3−Y×0.02 g/m3 =W 182 m3/min×(10.0–0.02) g/m3=W W=182 m3/min×9.98 g/m3=1816 g/min
Dust collected in the baghouse=1816 g/min×60 min/h×24 h/d
1000 g/kg =2615 kg/d
EXAMPLE 3.10: SLUDGE SOLIDS CONCENTRATION IN THE FILTRATE
A sludge dewateringfilter press receives thickened sludge that has 3% solids and specific gravity of 1.02.
The dry solids in thickened sludge are 1800 kg/d. The solids capture efficiency of dewatering facility is 85%. The dewatered sludge cake has solids content of 30% and specific gravity of 1.04. Calculate (a) vol- umetricflow rate of sludge cake, m3/d and (b) volumetricflow rate and TSS in thefiltrate.
Solution
1. Draw the process diagram and system boundary (Figure 3.10).
2. Determine the volumetricflow rate of thickened sludge (Stream A).
Qsludge=1800 kg dry solids/d×100 kg wet sludge
3 kg dry solids × 1
1020kg/m3wet sludge=58.8m3/d 3. Calculate the solids in the sludge cake (Stream B) andfiltrate (Stream C).
Solids capture efficiency of dewatering facility is 85%.
Solids captured in sludge cake (Stream B)¼1800 kg=d dry solids0.85¼1530 kg=d dry solids Solids infiltrate (Stream C)¼1800 kg=d dry solids(10.85)¼270 kg=d dry solids 4. Calculate the volumetricflow rate of sludge cake (Stream B).
Qcake=1530 kg dry solids/d×100 kg sludge cake 30 kg dry solids × 1
1040 kg/m3=4.9 m3/d
5. Calculate the volume and TSS concentration offiltrate (Stream C).
Qfiltrate=58.8 m3/d−4.9 m3/d=53.9 m3/d TSSfiltrate= 270 kg/d
53.9 m3/d×1000 g/kg=5009 g/m3or 5009 mg/L Sludge dewatering
filter press solids capture 85%
Sludge Qsludge 1800 kg/d 3% solids sp. gr. = 1.02 Filtrate
Qfiltrate
TSSfiltrate Stream C
Stream A
Stream B Dewatered sludge cake Qcake
30% solids sp. gr. = 1.04 FIGURE 3.10 Process diagram and system boundary (Example 3.10).
EXAMPLE 3.11: SOLIDS CONCENTRATION IN THICKENER OVERFLOW Total quantity of sludge collected in a secondary wastewater treatment plant is 8500 lb per day. Primary sludge is 60% of combined sludge by weight, and has solids concentration of 3% by weight, and sp. gr. of 1.02. The secondary sludge has solids concentration of 0.8% and sp. gr. is 1. The combined sludge is thick- ened in a gravity thickener, and the supernatant is returned to the head of the plant. The solids capture efficiency of thickener is 90%. The thickened sludge has solids content of 8% and sp. gr. of 1.04. Calculate (a) the volume of the thickened sludge and (b) TSS in thickener supernatant.
Solution
1. Draw the process diagram and system boundary (Figure 3.11) 2. Calculate the dry solids and volumetricflow rate of primary sludge.
TSS (dry solids) in primary sludge=8500 lb/d(dry solids)×0.6=5100 lb/d Qprimary=5100 lb dry solids/d×100 lb
3 lb × 1
62.4 lb/f t3×1.02=2671 f t3/d 3. Calculate the dry solids and volumetricflow rate in secondary sludge.
TSS (dry solids) in secondary sludge=8500 lb/d(drysolids)×0.4=3400 lb/d Qsecondary=3400 lb dry solids
d ×100 lb
0.8 lb× 1
62.4 lb/f t3×1.00=6811 f t3/d
3.2.4 Mass or Concentration of Nonconservative Substances in Reactors A nonconservative substance may undergo decay or generation; therefore, the mass balance analysis must also include conversion reactions.
4. Determine the volumetricflow rate of combined sludge.
TSS (dry solids) in combined sludge=8500 lb/d Qcombined=2671 f t3/d+6811 f t3/d=9482 f t3/d 5. Determine the volumetricflow rate of thickened sludge.
TSS (dry solids) in thickened sludge=8500 lb/d×0.9=7650 lb/d
Qthickened=7650 lb dry solids
d ×100 lb
8 lb × 1
62.4 lb/f t3×1.04=1474 f t3/d 6. Determine the TSS concentration in supernatant.
Qsupernatant=(9482−1474) ft3/d=8008 ft3/d
TSS (dry solids) in supernatant=(8500−7650) lb/d=850 lb/d
TSSsupernaqant= 850 lb/d
8008 f t3/d× 453.6 g/lb
0.0283 m3/f t3=1700 g/m3or 1700 mg/L Solids
capture 90%
Primary sludge Qprimary
60% of combined sludge by weight 3% dry solids
sp. gr. = 1.02
Secondary sludge Qsecondary
40% of combined sludge by weight 0.8% dry solids
sp. gr. = 1.00 Combined sludge
Qcombined
8500 lb/d (dry solids)
Thickened sludge Qthickened 8% dry solids sp. gr. = 1.04
Supernatant Qsupernatant
TSSsupernatant
FIGURE 3.11 Process diagram and system boundary (Example 3.11).
EXAMPLE 3.12: ESTIMATION OF MASS CONVERSION RATE
A reactor receives a chemical compound at a rate of 0.2 mole/L·h. The exit rate of the same compound from the reactor is 0.001 mole/L·h. Determine the rate of mass conversion for the following conditions:
1. There is no accumulation in the reactor, and 2. The accumulation is 0.08 mole/L·h.
Solution
1. Draw the reactor and system boundary (Figure 3.12).
2. Calculate the conversion rate.
a. Accumulation=0
0 ¼inflowoutflowþconversion
0 ¼0.2 mole=L·h0.001 mole=L·hConversion Conversion ¼(0.20.001) mole=L·h¼0.199 mole=L·h b. Accumulation=0.08 mole/L·h
0.08 mole=L·h¼0.2 mole=L·h0.001 mole=L·hConversion Conversion ¼(0.20.0010.08) mole=L·h¼0.119 mole=L
Input rate = 0.2 mole/L∙h Accumulation a. 0
b. 0.08 mole/L∙h
Exit rate = 0.001 mole/L∙h
FIGURE 3.12 Reactor and system boundary (Example 3.12).
EXAMPLE 3.13: SLUDGE STABILIZATION IN AN ANAEROBIC DIGESTER The thickened sludge is pumped into an anaerobic digester for stabilization. The volatile matter is partly converted into gaseous products (CH4and CO2). The digested sludge is pumped out daily and the super- natant from the digester is returned to the aeration basin. The information regarding the raw sludge and input rate, and digester performance is given below. Determine the quantity of solids (kg/d) andflow rate (m3/d) of digested sludge and supernatant.
Thickened wet sludge input rate,Qthickened sludge ¼132 m3=d Thickened dry solids input rate,Wthickened sludge ¼8180 kg=d VSS=TSS ratio in thickened sludge solids ¼71%
VS reduction (VSR) in anaerobic digester ¼52%
Solids in supernatant and density ¼0.4% and 1000 kg=m3 Digested sludge solids and density ¼5% and 1030 kg=m3 Solution
1. Draw the process diagram and system boundary (Figure 3.13).
2. Conduct theflow balance.
Assuming the loss of moisture is negligible, Equation 3.5a is developed.
Rate of flow accumulation
=Qthickened sludge− Qdigested sludge+Qsupernatant
+ Volume lost or consumed
(3.5a) Since both the rate of accumulation offlow, and volume lost or consumed=0, relationship between weight of supernatant and weight of digested sludge is established by Equation 3.5b.
Qthickened sludge=Qdigested sludge+Qsupernatant
132 m3/d= Wdigested sludge
0.05×1030 kg/m3+ Wsupernatant
0.004×1000 kg/m3
0.0194 m3/kg×Wdigested sludge+0.25 m3/kg×Wsupernatant=132 m3/d
Wsupernatant=528 kg/d−0.0776 ×Wdigested sludge (3.5b)
3. Conduct a mass balance of volatile solids in the digester.
From mass balance to develop Equation 3.6a.
Rate of solids accumulation
= Rate of thickened solids input
− Rate of digested solids withdrawn
+ Rate of solids lost in supernatant
+ − Rate of volatile solids stabilizaion
(3.6a)
Substitute the values of solids mass in Equation 3.6a to establish Equation 3.6b.
Volatile solids in input sludge=8180 kg/d×0.71=5808 kg/d Volatile solids reduced=5808 kg/d×0.52=3020 kg/d
0 kg/d=8180 kg/d−[Wdigested sludge]−[Wsupernatant]−3020 kg/d
Wsupernatant=5160 kg/d−Wdigested sludge (3.6b)
VSR = 52%
Digested sludge Qdigested sludge
Solids = 5%
Density = 1030 kg/m3 Supernatant Qsupernatant
Solids = 0.4%
Density = 1000 kg/m3 Thickened sludge
Qthickened sludge = 132 m3/d Wthickened sludge = 8180 kg/d VSS/TSS ratio = 71%
FIGURE 3.13 Mass balance around anaerobic digester (Example 3.13).