ENVIRONMENTAL ENGINEERING-I Prof. Rajesh Bhagat Asst. Professor Civil Engineering Department Yeshwantrao Chavan College Of Engineering Nagpur B. E. (Civil Engg.) M. Tech. (Enviro. Engg.) GCOE, Amravati VNIT, Nagpur Mobile No.:- 8483003474 / 8483002277 Email ID:- [email protected] Website:- www.rajeysh7bhagat.wordpress.com UNIT-IV 1) Sedimentation: Principles types of setting basins, inlet and outlet arrangements. 2) Clariflocculators: Principles and operation. 3) Filtration: Mechanism of filtration, types of filters RSF, SSF, pressure filters, elements of filters, UDS, design aspects of filter and operational problems in filtration. Sedimentation 1) Discrete or granular particle are those which do not change their size, shape & weight. 2) Settling- process by which particulates settle to the bottom of a liquid and form a sediment. 3) The other type of particles are those which change their size, shape & weight and thus loose their identity. 4) The basin in which the flow of water is retarded or storage is offered is called the settling tank or sedimentation tank or basin or clarifier. 5) The average time theoretically for which the water is detained in the tank is called detention period. 6) Sedimentation is process by which the suspended particles that are heavier than water are removed by gravitational settling. 7) Sedimentation process can remove 60% of SS & 75% of bacteria. 8) Floatation is a process by which lighter particles can remove. Factors Affecting Sedimentation: 1) Flow velocity 2) Viscosity of water 3) Temperature 4) Size of particle 5) Shape of particle 6) Specific gravity of particle Characteristics of the particles Size and shape Specific gravity Properties of the water Specific gravity Viscosity Physical environment of the particle Velocity of the water Inlet and outlet arrangements of the structure Advantages of Sedimentation:- Simplest technologies Little energy input Relatively inexpensive to install and operate No specialized operational skills Easily incorporated into new or existing facilities Disadvantages of Sedimentation:- Low hydraulic loading rates Poor removal of small suspended solids Large floor space requirements Re-suspension of solids. Types of Sedimentation Tank A. Depending upon shape: 1) Circular 2) Rectangular 3) Square B. Depending on process of operation: 1) Continuous Tank: Flow velocity of water is reduced by providing sufficient length of travel. This tank is designed such that the time taken by the water particle to travel from one end to another end is kept slightly more than the time required for settling of suspended particles in water. 2) Intermittent Tank: In this tank, water is completely brought to rest. This type of tank works intermittently. SEDIMENTATION BASIN ZONES 1) Inlet Zone: The inlet or influent zone should distribute flow uniformly across the inlet to the tank. The normal design includes baffles that gently spread the flow across the total inlet of the tank and prevent short circuiting in the tank. 2) Settling Zone: The settling zone is the largest portion of the sedimentation basin. This zone provides the calm area necessary for the suspended particles to settle. 3) Sludge Zone: The sludge zone, located at the bottom of the tank, provides a storage area for the sludge before it is removed for additional treatment or disposal. Sludge is removed for further treatment from the sludge zone by scraper or vacuum devices which move along the bottom. 4) Outlet Zone : The basin outlet zone (or launder) should provide a smooth transition from the sedimentation zone to the outlet from the tank. This area of the tank also controls the depth of water in the basin. Circular Basin Rectangular Basin Critical Settling Velocity and Overflow Rate Critical settling velocity (Vo)is the settling velocity of particles which are 100% removed in the basin Typical Dimensions of Sedimentation Tanks ______________________________________________________ Description Dimensions Range Typical ______________________________________________________ Rectangular Depth, m 3-5 3.5 Length, m 15-90 25-40 Width, m 3-24 6-10 Circular Diameter, m 4-60 12-45 Depth, m 3-5 4.5 Bottom Slope, mm/m 60-160 80 ______________________________________________________ 14 SEDIMENTATION Particles Size Discrete Flocculating Shape weight Suspension Dilute Concentrated 15 Types of Settling:- Depending upon the concentration of solids and the tendency of a particle to interact the following types of settling may occur Type I: Discrete Settling: This corresponds to the sedimentation of discrete particle in a suspension of low solids concentration. No interaction between particles Settling velocity is constant for individual particles Type II: Hindered Settling: This type of settling refers to dilute suspension of particles that flocculate during sedimentation process. 16 Type III: Zone of Settling: Refers to flocculent suspension of Intermediate solids concentration Solids move as a block rather than individual particles Mass of particles subside as whole Type IV: Compression settling Flocculent suspension of high concentration. Volume of solids may decrease Inlet & Outlet Arrangement for Sedimentation Tank Design Concept of Sedimentation: 1) Flow Velocity, vH = (Q/(B x H)) 2) Over Flow Rate = Surface Loading, vO = (Q/(B x L)) 3) Settling Velocity, vs :- (vH / vS ) = (L / H) Particles with settling velocity ‘vS’ equal to or greater than vO will settle down. 4)Over flow rate = 500 to 750 lit/Hr/m2 ( Plain Sedimentation) 5) Over flow rate = 1000 to 1250 lit/Hr/m2 ( Sedimentation with coagulation) 6) Length of Tank, L greater than or equal to 4B (Width of tank, B should not exceed 12 m) 7) Flow velocity = 0.15 to 0.9 m/min. 8) Length of Tank = L = vH x DT (Detention Time = 4 to 8 Hrs) 9) Depth of Tank = 3 to 4.5 m 10)Provision for Sludge = 0.3 to 1.2 m & FB = 0.3 to 0.5m Design Concept of Sedimentation: 1) Stokes Law :- Vs = ((g/18)(Gs-1)(d2/ u)) g = acceleration due to gravity Vs = settling velocity D = diameter of particle, cm Gs = specific gravity u = kinematic viscosity = 0.01 cm2 /s Find the settling velocity of a particle in water for following condition: Diameter of particle = 0.06 mm, specific gravity of particle = 2.65, temperature of water = 200c, Kinematic Viscosity of water at 200c = 1.007 centistokes. G = 9.81 m/s2 = 9810 mm/s2, u = 1.007 centistokes = 1.007mm2/ s Design Concept of Sedimentation: 1) Stokes Law :- Vs = ((g/18)(Gs-1)(d2/ u)) g = acceleration due to gravity Vs = settling velocity D = diameter of particle, cm Gs = specific gravity u = kinematic viscosity = 0.01 cm2 /s Find the settling velocity of a particle in water for following condition: Diameter of particle = 0.06 mm, specific gravity of particle = 2.65, temperature of water = 200c, Kinematic Viscosity of water at 200c = 1.007 centistokes. G = 9.81 m/s2 = 9810 mm/s2, u = 1.007 centistokes = 1.007mm2/ s Vs = 3.21 mm/s Re = (Vs. Dp) / u = (3.21 x 0.06 )/1.007 = 0.195 < 1 Therefore stokes law is applicable. sedimentation A circular sedimentation tank is generally provided with its bottom cone shaped with a slope of 1 vertical to 12 horizontal. Under this condition, its diameter is given by: V = D2 (0.011 D + 0.785 H) Design a plain sedimentation tank for max. daily demand of water 9.5 x 106 lit / day. Assume the velocity of flow to be 20 cm / minute and detention time 8 hours. Ans: Quantity of water to be treated = 9.5 x 106 lit / day = 395833.33 lit / Hr Capacity of Sedimentation tank required = V = Q x DT = 395833.33 x 8 = 3.17 x 106 Lit = 3170 m3 Velocity of flow to be maintained through the tank, v = 20 cm / min = 0.2 m / min. The length of tank required, L = v x DT = 0.2 x 8 x 60 = 96 m The c/s area of the tank is required = ( Capacity of tank / Length of Tank ) = 3170 / 96 = 33 m2 Assuming the water depth in the tank is 4 m. The width of tank , B = C/s area / Depth = 33/4 = 8.25 m (< 12m) Design a plain sedimentation tank for max. daily demand of water 9.5 x 106 lit / day. Assume the velocity of flow to be 20 cm / minute and detention time 8 hours. Ans: Quantity of water to be treated = 9.5 x 106 lit / day = 395833.33 lit / Hr Capacity of Sedimentation tank required = V = Q x DT = 395833.33 x 8 = 3.17 x 106 Lit = 3170 m3 Velocity of flow to be maintained through the tank, v = 20 cm / min = 0.2 m / min. The length of tank required, L = v x DT = 0.2 x 8 x 60 = 96 m The c/s area of the tank is required = ( Capacity of tank / Length of Tank ) = 3170 / 96 = 33 m2 Assuming the water depth in the tank is 4 m. The width of tank , B = C/s area / Depth = 33/4 = 8.25 m (< 12m) FB = 0.5m Depth for sludge = 1m Overall Depth of Tank = 4 + 1 + 0.5 = 5.5 m Dimension of Tank = 96 x 8.25 x 5.5m Design a plain sedimentation tank for a population of 100000 with water supply rate 135 lpcd. Ans: Quantity of water to be treated = 100000 x 135 x 1.8 = 1012.5 m3/ hr Assuming Detention Time = 6 Hours Capacity of Sedimentation tank required = V = Q x DT = 1012.5 x 6 = 6075 m3 Assuming Velocity of flow through the tank, v = 0.6 m / min. The length of tank required, L = v x DT = 0.6 x 6 x 60 = 216 m The c/s area of the tank is required = ( Capacity of tank / Length of Tank ) = 6025 / 216 = 28.125 m2 Assuming the water depth in the tank is 4 m.