End Semester Examination
Date: 18-05-2014
UCE-601: Sewerage and Sewage Treatment B.E. – Civil Engineering Instructor: Dr. Akepati S. Reddy Time: 3 hours (2-00 PM to 5-00 PM)
Max. Marks: 100
Roll No.: Note: Please assume if any requisite data is not given. Q.1 1.1
Wastewater being treated in an activated sludge process has the following characteristics: Wastewater flow rate
750 m3/hr
BOD5 at 20°C
210 mg/L
TSS
120 mg/L
VSS
80 mg/L
Biodegradable VSS
65 mg/L
Values of ASP kinetics parameters (qmax., Ks, Y and kd) are 6/day, 40 mg/L, 0.4 and 0.1/day respectively. TSS level in the treated is 40 mg/L. HRT and SRT of the activated sludge process are 8 hours and 9 days respectively. Estimate the secondary sludge wastage rate and the observed SRT of the activated sludge process? Treated effluent BODu
Se
K s 1 k d ( SRT ) SRT qm ax. .Y k d 1 Ks is 40 mg/L Kd is 0.1/day SRT is 9 days Qmax is 6/day Y is 0.4 Se = 3.858 mg/L or 0.004 kg/m3
Secondary sludge wastage rate = secondary sludge generation rate – sludge washout rate Sludge generation rate = net sludge synthesis rate + cell debris generation rate + non-biodegradable organic solids addition rate + inorganic solids addition rate Net sludge synthesis rate
NBSR
Y .Q ( S i S e ) 1 k d ( SRT )
Sewage flow rate is 750 m3/hr Si value is 1.6 x 210 mg/L = 0.336 kg/m3 Net biomass synthesis rate = 1258.1 kg/day or 52.421 kg/hr
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Cell debris generation rate
1 CDGR f d Y Q( Si Se ) 1 1 kd .SRT Fd is 0.15 Cell debris generation rate = 169.8 kg/day 7.077 kg/hr Non-biodegradable suspended solids addition NBVSS = QxVSSnb VSSnb is 80-65 = 15 mg/L or 0.015 kg/m3 NBVSS = 270 kg/day or 11.25 kg/hr Inorganic suspended solids addition INSS = QxTSSin TSSin is (120-80) = 40 mg/L or 0.04 kg/m3 TSSIN = 720 kg/day or 30 kg/hr Secondary sludge generation rate = 1258 + 169.8 + 270 + 720 = 2418 kg/day or 100.75 kg/hr Sludge washout rate Sludge washout rate = flow rate x TSSe TSSe is 40 mg/L or 0.04 kg/m3 Sludge washout rate = 720 kg/day or 30 kg/hr Sludge wastage rate = 1698 kg/day or 70.75 kg/hr
Observed SRT = sludge present in aeration tank / sludge wastage rate Sludge present in the aeration tank = aeration tank volume x MLSS MLSS estimation Active biomass concentration is
xa
SRT
S i Se Y
1 k d ( SRT )
HRT or τ is 8 hours or 0.333 days Active biomass concentration = 1.887 kg/m3 or 1887 mg/L Xa/MLSS = net biomass generation rate / total secondary sludge generation rate MLSS = 1.887 / 1258 x 2418 = 3.627 kg/m3 Observed SRT = (Q x τ x MLSS)/(sludge wastage rate) = 12.816 days
1.2
Present Peak Flow and Design Peak Flow to be conveyed by a sewer are 50 m3/hr and 160 m3/hr respectively. Design a sewer for the conveyance of this sewage? Find depth of flow and flow velocity in the sewer during the present peak flow and the design peak flow conditions? Hydraulic elements graph is given in the attachment for use if needed.
Qppf = 50 m3/hr or 0.0139 m3/hr Qdpf = 160 m3/hr or 0.0444 m3/hr
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Finding slope of the sewer for the present peak flow using the following equation:
1 i K a K r 2 n
6
16
13
g
13
q
6
13
‘n’ is taken as 0.015 Ka and Kr corresponding to the d/D = 0.2 are taken as 0.1118 and 0.1206 ‘τ' is taken as 1.5 pascals ‘ρ' is taken as 1000 kg/m3 ‘g’ is taken as 9.81 ‘q’ is present peak flow = 50 m3/hr Slope ‘i’ = 2.5 x 10-3 or 1 in 387.5 Finding the sewer diameter for the design peak flow using the following equation:
q D n 8 K a 8 K r 4 1 i 2 3
3
1
3
8
‘n’ is taken as 0.015 Ka and Kr corresponding to the d/D = 0.8 are taken as 0.6736 and 0.3042 ‘q’ is design peak flow = 160 m3/hr ‘i’ is 2.5 x 10-3 or 1 in 387.5 Sewer diameter = 0.307 m or 307 mm (sewer diameter is taken as 300 mm) Finding flow depth and flow velocity for the present peak flow Hydraulic elements graph is used. The plots for constant ‘n’ value are used. d/D and V/Vfull are obtained corresponding to the Qppf/Qfull from the graph Qfull is obtained from the Manning’s equation
Q
AR 2 / 3 S 1/ 2 n A is cross sectional area of the sewer of 0.3 m diameter R is hydraulic radius of the circular sewer flowing full (D/4) ‘S’ is slope (i) = 2.5 x 10-3 or 1 in 387.5 ‘n’ is Manning’s ‘n’ taken as 0.015 Qfull = 0.0426 m3/sec or 153.27 m3/hr
Qppf/Qfull = 0.326 Qdpf/Qfull = 1.044 Finding Vfull from the Manning equation
V
R 2 / 3 S 1/ 2 n
Vfull = 0.602 Reading of dppf/D and Vppf/Vfull corresponding to the Qppf/Qfull and finding flow depth and flow velocity dppf/D = 0.4 Vppf/Vfull = 0.87 dppf = 0.12 m Vppf = 0.523 m/sec.
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Reading of dDpf/D and VDpf/Vfull corresponding to the QDpf/Qfull and finding flow depth and flow velocity dDpf/D = 0.80 VDpf/Vfull = 1.16 dDpf = 0.24 m VDpf = 0.698 m Vppf = 0.523 m/sec is below the self-cleansing velocity of 0.6 m/sec. For increasing the velocity, slope ‘S’ or ‘I’ and/or sewer diameter can be increased. Assume the increase of slope to 1 in 250 or 4.0x10-3, and find out flow depth and flow velocity for the present peak flow and for the design peak flow conditions. Qfull is obtained from the Manning’s equation
Q
AR 2 / 3 S 1/ 2 n A is cross sectional area of the sewer of 0.3 m diameter R is hydraulic radius of the circular sewer flowing full (D/4) ‘S’ is slope (i) = 4x10-3 or 1 in 250 ‘n’ is Manning’s ‘n’ taken as 0.015 Qfull = 0.053 m3/sec or 191 m3/hr
Qppf/Qfull = 0.262 Qdpf/Qfull = 0.839 Finding Vfull from the Manning equation
V
R 2 / 3 S 1/ 2 n
Vfull = 0.75 m/sec. Reading of dppf/D and Vppf/Vfull corresponding to the Qppf/Qfull and finding flow depth and flow velocity dppf/D = 0.355 Vppf/Vfull = 0.83 dppf = 0.107 m Vppf = 0.623 m/sec. Reading of dDpf/D and VDpf/Vfull corresponding to the QDpf/Qfull and finding flow depth and flow velocity dDpf/D = 0.69 VDpf/Vfull = 1.13 dDpf = 0.143 m VDpf = 0.848 m Sewer diameter = 0.3 m Sewer slope = 1 in 250 or 4x10-3
Marks: 2 x 10 = 20 Q.2 2.1
Estimate peak runoff (for a design storm event with 2 years return period) for a 5000 m2 size catchment basin (discharging the runoff into a storm sewer inlet)? Hydraulic flow path length of the catchment basin is 100 m, and length of sheet flow is 30 m. Slope, roughness coefficient and intercept coefficient values for the catchment basin are 0.3%, 0.11 and 0.490 respectively. Storm intensity, duration and frequency Page 4 of 16
information is given in the attached table for use if needed. Rational method is used for estimating the peak runoff
Q
CIA Ku Q peak flow in m3/sec. C runoff coefficient (is not given – taken as 0.7) I Rainfall intensity in mm/hr (not known – obtained from the storm density, duration and frequency table given for the 2 years return period – for this, time of concentration (tc) is required A drainage area in hectares (given and its value is 0.5 hectares) Ku units conversion factor (360)
Time of concentration = time of travel for sheet flow + time of travel for shallow concentrated flow Here channel flow is considered to occur in the sewer and it does not occur before the storm sewer inlet. Time of travel for sheet flow
Tti
K u nL I 0.4 S Tti is sheet flow travel time in minutes Ku is empirical coefficient (its value is 6.92) I is rainfall intensity in mm/hr (depends on the tc to be calculated) – taken as 40 mm/hr L is flow length in meters – given and its value is 30 m S is slope (catchment slope) – given as 0.3% or 3 in 1000 N is roughness coefficient – given and its value is 0.11 Time of travel for the sheet flow (Tti) = 95 min
Time of travel for shallow concentrated flow
V Ku KS p0.5 Tti
L 60V Ku is taken as 1.0 K is intercept coefficient (depends on land cover/ flow regime) –given - value is 0.49 Sp is slope percent (given – value is 0.3%) L is flow length (70 m) V is velocity in m/sec. = 0.268 m/sec. Tti is shallow concentrated flow travel time (in minutes.) = 4.35 min.
Time of concentration = 95 + 4 = 99 min. Intensity of rainfall corresponding to the 99 min time of concentration is not 40 mm – it is (from the table) <30 mm/hr For 30 mm rainfall intensity the time of travel for the sheet flow (Tti) is 107 min And, Time of concentration is 107 + 4 = 111 min. According to rational method the peak runoff will be
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Q
CIA Ku Q = 0.7x30x0.5/360 = 0.029 m3/sec
2.2
Sewage with the following characteristics is treated in an anaerobic pond of 3.5 m liquid depth. BOD
240 mg/L
TSS
320 mg/L
VSS
80% of the TSS
Biodegradable VSS
85% of the VSS
Treated effluent of the anaerobic pond is having 50 mg/L of suspended solids. If Consistency of the settled sludge in the anaerobic pond is 9%, and if the anaerobic pond is cleaned once the sludge reaches 1/3rd of the depth, find the frequency of cleaning of the anaerobic pond? Assume total hydrolysis of the biodegradable volatile suspended solids and conservation of all other suspended solids. Consider 0.1/day as the net yield coefficient for the anaerobic pond. TSS of the influent = 320 mg/L Non-biodegradable TSS = 320 (1– 0.8 x 0.85) = 102.4 mg/l TSS escaping in the treated effluent = 50 mg/L Volumetric loading rate for the anaerobic pond Volumetric organic loading rate (g/m3.day) = 20T-100 T is temperature (not given and take it is 16C) Volumetric organic loading rate (g/m3) = 20T-100 = 20x16-100 = 220 g/m3.day Surface hydraulic loading rate – it is given by Surface organic loading rate/BOD of the influent Surface organic loading rate – it is given by Volumetric organic loading rate x pond depth Pond depth is given as 3.5 0.22x3.5 = 0.77 kg/m2.day Surface hydraulic loading rate = 0.770/0.240 = 3.208 m3/m2.day Here 240 g/m3 or 240 mg/L is organic matter concentration as BOD Total solids accumulation rate (kg/m2.day) Surface hydraulic loading rate x (non-biodegradable solids + net biomass synthesis – TSS level in the effluent) Net biomass synthesis Efficiency of treatment = 2T+20 = 2x16+20 = 52% BOD removed = 240x1.6x52% = 199.7 g/m3 Net biosynthesis = 199.7x0.1 = 19.97 mg/L or g/m3 3.208 x (0.1024 + 0.01997 – 0.0 50) = 0.232 kg/m2.day Total sludge accumulation rate (in m3/m2.day) Consistency of the accumulated sludge = 9% given 0.232/90 = 2.579x10-3 m3/m2.day Volume available for sludge storage = 1x3.5/3 = 1.167 m3/m2 Time taken to fill this volume = 1.167/(2.579x10-3) =452 days
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Anaerobic pond cleaning frequency = once in 452 days
2.3
Design a facultative pond for the treatment of effluent from an anaerobic pond. Characteristics of this effluent are as given below: Flow rate
75 m3/hour
BOD
130 mg/L
TSS
70 mg/L
VSS
60% of TSS
Ammonical nitrogen
25 mg/L as N
Coliform count (MPN)
2.7x106
Alkalinity
160 mg/L as CaCO3
Find BOD, total coliform count (MPN) and ammonical nitrogen content of the treated effluent from the facultative pond? Take average ambient air temperature of the coldest month of the year as 17°C. Surface loading rate of the organic matter (kg/hectare.day)
s 3501.107 0.002T T 25 T is design temperature –given as 17C Surface loading rate = 199.19 kg/hectare Surface area of the facultative pond = flow rate x BOD /surface loading rate Flow rate = 75 m3/hr or 1800 m3/day BOD = 130 mg/L 75x24x0.13/199.9x10000=11747 m2 Hydraulic retention time = surface area x depth / flow rate Depth of the pond = taken as 1.5 m 11747x1.5/(75x24) = 9.789 days Finding BOD of the treated effluent
r 0.725 s 10 .75 0.725 x 199.19 + 10.75 = 155.16 kg/hectare BOD of the treated effluent = 130 x (199.19-155.16) / 199.19 = 28.74 mg/L Finding MPN count of the treated effluent
N fa
N an 1 K B (T ) fc
K B(T ) 2.61.19
T 20
T is temperature and its value is 17C Nan is MPN of the influent (given and its value is 2.7x106/100 mL) ϴfc is HRT of the facultative pond and its value is 9.789 days KB(T) = 1.543 Nfa = 1.68x105 Finding ammonical nitrogen level in the treated effluent
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Ce
Ci
A 1 0.0038 0.000134T exp1.041 0.044T pH 6.6 Q pH 7.3 exp 0.0005 Ai Ci is influent Ammonical-N concentration (mg/L) – given and its value is 25 mg/L A is area of the facultative pond = 11747 m2 Q is flow rate = 75x24 = 1800 m3/day T is temperature = 17C Ai is alkalinity of the influent (in mg/L as CaCO3) – given and its value is 160 mg/L pH = 7.91 Ce = 17.69 mg/L
Marks: 3 x 8 = 24
Q.3 3.1
Write note on sewer design computation table and suggest format for the computation table?
Sewer design computation table can be an excel worksheet. It is meant for carrying out the repetitive sewer design calculations and presenting the results. This computation table can include the following columns: • Columns identifying the sewers and summerizing the basic data • Sewer code, upstream and downstream manhole numbers and sewer length • Local (tributary) area, its present and design average and peak sewage flow including infiltration allowance • Columns showing cumulative present average flow and peak flow, and the average and peak flow at the end of design period • Surface elevation at the upstream and downstream sewer ends • Columns showing computed slope and diameter of the sewer and Qfull • Columns showing hydraulic elements for the present and the design peak flows - d/D corresponding to the qPFP and to the qPFD, and flow velocity at qPFP and at qPFD • Columns showing sewer layout data • Invert elevations at the upstream and at the downstream ends of the sewer and corrected invert elevations of the sewer • Crown cover, sewer depth, etc. 3.2
Write note on manholes, manhole components and manhole types?
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Manholes: Openings by which humans along with their machinery have access to sewers for inspection, cleaning, repair and other maintenance operations. Locations of manholes • At change of sewer diameter or slope or direction • At the upstream ends and at the sewer junctions • At regular intervals/distances along straight sewer stretches Components of a manhole are 1. Manhole frame and cover 2. Access shaft 3. Manhole chamber 4. The sewer and benching within the manhole Types of manholes Precast manholes and constructed onsite manholes RCC manholes and brick manholes Rectangular, arch type and circular manholes Types of manholes • Straight through manholes • Junction manholes • Side entrance manholes • Drop manholes • Scraper type manholes Manhole types Maintenance manholes Purge manholes Drop manholes Junction manholes
3.3
Draw a properly labeled schematic diagram of a UASB reactor used for sewage treatment and write note on the components of a UASB reactor? UASB reactor has a reactor zone and a settling zone. The reactor zone is divisible into sludge bed zone, sludge blanket zone and diffused sludge zone. There are provisions for the sampling the reactor contents from different elevations for the treatment process monitoring and control. There are provisions for the draining out of sludge from the sludge bed, sludge blanket and diffused sludge zones. Distribution tubes load sewage at the bottom of the reactor to flow upwards through the sludge bed and sludge blanket. The settling zone has a 3-phase separator comprising of deflectors, biogas collection tunnels and clarified effluent overflow weirs and collection troughs. The wastewater, together with the biogas bubbles generated, flows upward and get into the settling zone. Deflectors direct the biogas bubbles into the biogas collection tunnels. Wastewater free from biogas bubbles enter the settling zone and suspended solids of the wastewater settle on the inclined sufaces and slide down back into the reactor zone. Clarified water overflows into the clarified effluent trough and
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drained out from the UASB reactor, and scum baffle do not allow washout of the floating scum. A distribution box placed over the reactor with multitude of distribution tubes facilitate the application of sewage to the UASB reactor.
3.4
Through a schematic process flow diagram show what happens to the biodegradable organic matter of a sample in a BOD bottle when the sample is tested for BOD? With the help of this diagram, differentiate carbonaceous BOD from nitrogenous BOD, and show why BOD is lower than ThOD and COD?
Oxygen is demanded for the following three different purposes: 1. Bio-oxidation of organic matter 2. Auto-oxidation of microbial biomass 3. Nitrification of the ammonical nitrogen into nitrite and nitrate nitrogen Oxygen demanded for the nitrification is considered as nitrogenous BOD, while the oxygen demanded for both bio-oxidation of organic matter and for the auto-oxidation of microbial biomass put together is known as carbonaceous BOD. COD testing does not differentiate biodegradable organic matter from non-biodegradable organic matter, and hence COD measures both the biodegradable and non-biodegradable organic matter. In the BOD test on the other hand only biodegradable organic matter is bio-oxidized and measured as oxygen demanding. Hence BOD is usually lower than the COD. In the BOD testing, some of the organic matter is left unutilized by the microorganisms, some of the produced microbial biomass is left un-auto-oxidized and the auto-oxidation leaves behind cell debris. Further, BOD test requires infinite time duration for the complete oxidation of the organic matter into inorganic end products. Because of these reasons BOD is always lower than the ThOD (a stoichiometric oxygen demand of the organic matter present in the sample).
Marks: 4 x 6 = 24 Q.4 4.1
State the objectives of tertiary treatment of sewage, and indicate the pollutants removed and the Page 10 of 16
treatment technologies typically used in the tertiary treatment of sewage? Tertiary treatment is required when the treated effluent is to be reused as reclaimed water. Tertiary treatment usually involve polishing, nutrient removal and pathogen reduction from the secondary effluent. Maturation ponds, filtration and chlorination units, vegetated ponds and constructed wetland systems are usually used for the tertiary treatment. Maturation ponds are designed for the MPN reduction. Filtrations units are usually designed for the TSS removal and marginal removal of BOD and nutrients also occurs mainly through denitrification. Vegetation ponds and constructed wetlands can result in nutrient removal and BOD reduction and even MPN count is reduced. 4.2
Write note on interceptor tanks and state how these differ from septic tanks?
Interceptor tanks are used in the small bore (solids free) sewerage systems for clarifying the sewage prior to its entry into the sewer. These are provided closer to the points of sewage generation and these remove both floating and settlable materials from the sewage. These are very similar septic tanks (that are usually meant for the clarification of the sewage and stabilization of the settled sludge). Interceptor tanks are smaller in size and have lesser HRT. Only coarse materials will be removed. These require frequent maintenance and cleaning (removal of the accumulated floating and settled materials). HRT of septic tanks vary from 24 hours to 72 hours. These are more efficient as clarifiers. Solids or sludge separated gets stabilized. Cleaning (removal of stabilized sludge) of the septic tanks is less frequent (once a year or so).
4.3
Draw a properly labeled schematic process and material flow diagram for the sewage treatment plants of Ludhiana?
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Sewage treatment plants at Ludhiana are UASB based systems and include the following units: Bar screens Sewage pumping Grit channels/chambers UASB reactors Pre-aeration tank Polishing pond Chlorination unit There are division boxes and distribution boxes for dividing the incoming flow into n number of streams and loading the UASB reactors. There is a biogas collection handling system comprising of gas metering, gas scrubbing, gas holding and gas flaring units. There is ETP sludge handling system comprising of sludge sumps and pumping facilities and sludge drying beds.
4.4
Write note on Vortex type grit separators?
Vortex type grit separators are compact type degritting facilities. Spiraling motion of sewage in the grit separator results in the separation of grit by the centrifugal force developed. Grit strikes the side walls and collects at the bottom. Spiraling motion of sewage in the grit separator is caused either by a mechanical device or by the flowing sewage. Based on how the spiraling motion is created two types of vortex grit separators (type-1 and type-2) are identified. Type-1: Wastewater enters and exits tangentially; Rotating turbine maintains constant flow velocity and promotes separation of organics from grit; and HRT is 20-30 seconds for average flow. Type-2: Vortex is generated by the flow entering tangentially at the top of the unit; Effluent exits the center of the top of the unit (axial exit); These are sized to handle peak flow rates upto 0.3 m3/sec. per unit
4.5
Write note on flow division or flow distribution box?
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Division boxes and distribution boxes are used for dividing the incoming flow into desired number of streams of equal flow. These are used for loading the sewage to the UASB reactors. Division boxes divide the incoming flow and supply to different UASB reactor cells. Each of the UASB reactor cells have one or more distribution boxes for dividing the incoming flow and loading at the bottom of the reactor through distribution tubes. The division boxes and the distribution boxes allow the incoming water to calm and flow over a properly leveled overflow weir into collection chambers. Through fixing the overflow weir length for each of the collection chambers the flow is divided. From each of the collection chamber a drain pipe is provided to the carry the divided flow. Hydraulic conditions within the division box or the distribution box are maintained such that no settling of solids occurs within.
Marks: 5 x 4 = 20 Q.5 5.1
Write note on sludge drying beds? Sludge drying beds have an under-drain system and a bed of graded gravel and sand supported over it. Sludge is loaded on the sand layer upto 300 mm depth for the sludge thickening, dewatering and drying. Water percolates downwards and comes out as filtrate, and the sludge is thickened and dewatered. Water evaporates from the dewatered sludge and sun drying of sludge occurs. Sludge drying beds are operated as batch units. After each batch of the sludge drying, the dried sludge is removed along with the top thin layer of sand. Then the bed is conditioned through addition fresh sand and kept ready for the next batch of treatment.
5.2
Write note on sequencing batch reactors? A sequence of steps of a treatment process are carried out usually in the same tank (not separated in space but separated in time) one step after the earlier step. SBR can be aerobic, anaerobic or a combination of both the treatment processes. In a sequencing batch reactor, filling, aeration, settling and decanting are carried out in the same sequence repeatedly. Frequent sludge wastage is also made the part of the batch processing. SBR is preferred for smaller flows.
5.3
Write note on small bore sewerage system? Page 13 of 16
It is sewerage system (also known as solids free sewerage) collecting and carrying clarified sewage. Interceptor tanks are used at the individual points of sewage generation for the sewage clarification prior to entry into the sewer. Sewers are smaller in diameter (minimum diameter is as small 38 mm (typically 50 to 75 mm diameter) than the conventional sewers (usually >100 or 150 mm). Used when density of population is relatively low and maintaining self-cleansing velocity is difficult.
5.4
Write note on catch basins?
These are masonry chambers provided below the storm sewer inlets for collecting and clarifying the storm water prior to its entry into the storm sewer. Both floating material and grit are separated from the storm water in these catch basins. Catch basins require frequent maintenance (degritting and cleaning).
5.5
Write note on Limiting Solids Flux?
Limiting solids flux (SFL) is the maximum surface loading rate of mixed liquor solids for a secondary clarifier for a specified underflow sludge concentration. SFL is found for the mixed liquor in the laboratory experimentation and used in sizing the secondary clarifier. Finding. The solids flux includes two components (gravity flux and mass flux) and increasing underflow sludge concentration decreases the SFL.
5.6
Write note proportional weir?
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Proportional weirs are used as control sections on the down-stream side of rectangular grit channels. Proportional weirs ensure constant horizontal flow velocity in the grit channels in the face varying flow rates and flow depths.
Marks: 6x2 = 12
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