CIVL 1112 Water Treatment - Sedimentation 1/7

Treatment Processes Sedimentation

Sedimentation is the downwards movement of an object relative to its surrounding medium, due to the force of gravity.

Screening Aeration Prechlorination

Sedimentation Coagulation

Sedimentation Sedimentation

Dissolved-air flotation (DAF) is a method whereby bubbles are The purpose of sedimentation is to remove preexisting solids, produced by the reduction of pressure in a water stream as well as the precipitates formed in coagulation and saturated with air. flocculation.

Sedimentation Sedimentation

The purpose of sedimentation is to remove preexisting solids, Model of a circular settlement tank with sludge scrapers was as well as the precipitates formed in coagulation and used to estimate the distribution of particulate concentration flocculation. over time CIVL 1112 Water Treatment - Sedimentation 2/7

Sedimentation Sedimentation

Sedimentation Sedimentation

Click HERE for animations about sedimentation  Sedimentation is the accumulation through gravity of particulate matter at the bottom of a fluid. http://techalive.mtu.edu/meec/module03/WastewaterandWildlife.htm  This natural process is frequently used to separate contaminants from air, water, and wastewater.

 There are four types of settling:  discrete  flocculant  hindered  compression

Sedimentation Sedimentation

 Discrete - Individual particles settle independently, neither  Hindered - Particle concentration is sufficient that agglomerating nor interfering with the settling of the other particles interfere with the settling of other particles. particles present. This occurs in water with a low concentration of particles.  Compression - In the lower reaches of clarifiers where particle concentrations are highest, particles can settle  Flocculant - Particle concentrations are high enough that only by compressing the mass of particles below. agglomeration occurs. This reduces the number of particles and increases average particle mass. The heavier particles sink faster. CIVL 1112 Water Treatment - Sedimentation 3/7

Sedimentation Sedimentation

If the Vp > Vh then settling can occur

w Vh Vh

Vp Path of smallest Q consistently settled Q Q Q z particle Vp

sludge layer sludge layer

L L

Sedimentation Sedimentation

If the Vp < Vh then “short-circuiting can occur The horizontal velocity, Vh, of a particle can be approximated by considering the flowrate, Q, and the cross-sectional flow area of the tank. V h Q QVA V Vp h h  Q Q A Q sludge layer Vh  L wz

Sedimentation Sedimentation

The residence time of water in the sedimentation tank can be Estimate of the residence time of water in a small approximated as: sedimentation tank where Q = 1 liter/min, L = 6 in., Q w = 6 in., and z = 10 in. (dimensions of a tank in the lab). L Lwz 6in.(6in.)10in. Vh  Vh  t   wz t Q 1, 0 0 0 ml min Lwz 360in.3 min 16.39ml t   5.9 min t  3 Q 1000ml in. CIVL 1112 Water Treatment - Sedimentation 4/7

Sedimentation Sedimentation

 Discrete settling, can be analyzed by calculating the settling The forces acting on a settling particle are: velocity of the individual particles contained within the water. Fb Fd  The forces acting on a particle are:  gravity in the downward direction,  acting in the upward direction as the particle settles  upward buoyancy due the water displaces by the particle

Fg is the force due to gravity Fg Fd is the drag force

Fb is the buoyant force Fg = Fd + Fb

Sedimentation Sedimentation

The gravitational force can be expressed as: The drag on the particle can be calculated by the from fluid mechanics Fmggp 1 FCAv  2 Using the density and volume of the particle yields: ddw2

FVggpp  where Cd is the , dimensionless, A is the particle cross-sectional area, ft.2, 3 3 where: p is the density of the particle, lb-mass/ft. , w is the density of water, lb-mass/ft. , 3 Vp is the volume of the particle, ft. , and v is the velocity, ft./sec. g is the gravitational constant, ft./s2

Sedimentation Sedimentation

The buoyant force acting on the particle is: By balancing the forces acting on a settling particle and using

the relationships for Fg the force due to gravity, Fd the drag Fmg force, and Fb the buoyant force, the following relationship bw can be developed:

Substituting the particle volume and density of water, yields: 1 2 pVgpdwwp CA v Vg FVgbwp  2

3 where: w is the density of water, lb-mass/ft , CIVL 1112 Water Treatment - Sedimentation 5/7

Sedimentation Sedimentation

Solving for the settling velocity, v, results in: At low Reynolds numbers (for NRe, < 1) Cd, can be approximated by: 24 C  2(pwp )Vg d N v  Re For Reynolds Numbers is transition flow, 1 < N < 10,000, CAdw Re the drag coefficient for spheres is: If the particle is assumed to round and the formulas for area 24 3 and volume of a sphere are used: Cd  0.34 NRe NRe 4( )dg where d is the pwp p For turbulent flow, NRe > 10,000, the relationship for the drag v  diameter of the coefficient for spheres is: 3Cdw particle Cd  0.4

Sedimentation Sedimentation

vd This relationship is known as Stokes' law, and the velocity is The is: NRe   known as the Stokes velocity. where u is the absolute viscosity of the water, lb-force- 2 sec./ft.2 (at 500F, μ = 2.73(10-5) lb.-sec./ft.2). () dg v  pw p 18 For NRe, < 1 the particle settling velocity can be estimated as a function of the properties of the particle and water, and the particle diameter, or The vertical velocity of water in a is often described as the overflow rate (OFR). 2 ()pw dg v  It is usually expressed as gal./ft.2-day (m3/m2-day). p 18

Sedimentation Sedimentation Example 1

The overflow rate is calculated in the following way:  Estimate the settling velocity of sand (p = 2,650 kg/m3) with a mean particle diameter of 0.21 mm. Q OFR  A  Assume the sand is approximately spherical.  Using a safety factor of 1.4 to account for inlet and where: OFR is the overflow rate, gal./ft.2-day, outlet losses, estimate the area required for a chamber Q is the flowrate, gal./day, and to remove the sand if the flowrate is 0.10 m3/sec A is the clarifier area, ft.2. (1,000 liters = 1 m3). CIVL 1112 Water Treatment - Sedimentation 6/7

Sedimentation Example 1 Sedimentation Example 1

The density of water at 200C is 998 kg/m3 and the viscosity of Knowing the overflow rate, the area required is: water at 200C is 1.01(10-3) N-s/m2 (Newton = kg-m/s2). The Stokes settling velocity is: 3 0.10m () dg2 Q s 2 vvOFR pw A  ()SF  (1.4)  3.6m ps OFR 0.039m 18 s

2 kg kg4 m  265033 998 2.1 10m 9.81 2 mm s   where SF is the safety factor, 1.4 3 kg 18 1.01 10 ms

= 0.039 m/s = 3.9 cm/s

Sedimentation Example 2 Sedimentation Example 2 Group Problem Group Problem  Estimate the settling velocity of the floc particles we have  What are “good” estimates of the particle density and seen in lab - especially the jar test results. diameter?

 Use Stokes' law to estimate the settling velocity.  Let’s assume the following values:

 What are “good” estimates of the particle density and  Particle density = 1,100 kg/m3 diameter?  Particle diameter = 10-4 m  How does your estimate compare to what you have seen in the lab?

Sedimentation Example 2 Sedimentation Example 2 Group Problem Group Problem

2 -4 ()pw dg OFR = 5.5 x 10 m/s = 0.055 cm/s vvOFRps  18 2 0.055cm cm2 86,400s  1 gal 30.48 cm OFR  23 kg kg2 m s cm day 3785.41cm 1ft 1,100 998 1 104 m 9.81 mm33 s 2   OFR  1,166.3 gpd 3 kg ft 2 18 1.01 10 ms For ferric chloride typical OFRs are in the 700 - 1,000 gpd/ft.2 = 5.5 x 10-4 m/s = 0.055 cm/s CIVL 1112 Water Treatment - Sedimentation 7/7

Sedimentation Example 3 Sedimentation Example 3

Group Problem Knowing the overflow rate and the minimum flowrate, the area required is:  If the settling velocity of the floc particles is 0.055 cm/s, ml cm3 Q 750 1 determine the area of the sedimentation tank. A  ()SF  (1.3) min ml  295.5 cm2 OFR 0.055cm 60s s min

 Assume a factor of safety of 1.3 2 2 1in. 2 Acm 295.5  45.8 in.  Assume the system flowrate can varying from 750 ml/min 2.54 cm to 1,250 ml/min In lab, each tank is 6 in. by 6 in. or 36 in.2.  How does your estimate compare to what you have seen in Therefore, for this estimate of particle velocity we need 1.27 tanks the lab? or 2 sedimentation tanks

Sedimentation Example 3 Sedimentation Example 3

Knowing the overflow rate and the maximum flowrate, Group Questions the area required is: ml cm3  What if the settling velocity of the floc particles is greater Q 1,250 1 A  ()SF  (1.3) min ml  492.4 cm2 than the computed 0.055 cm/s? OFR 0.05 5cm 60s s min

2  What if the settling velocity of the floc particles is less than 2 1in. 2 the computed 0.055 cm/s? Acm 492.4  76.3 in. 2.54 cm  How do these estimates compare to what you have seen in In lab, each tank is 6 in. by 6 in. or 36 in.2. the lab?

Therefore, for this estimate of particle velocity we need 2.1 tanks or 3 sedimentation tanks

Treatment Processes Any Questions?