Designing a New Sand Filter for a Small Operation: Sizing, Media, and Filtration Methods
The estimated flow rate for this site is 2000 m3/day. The surface overflow rate should exceed 125 m3/m2/d. What will be the surface area of the filter? You can use some leftover Ottawa sand (d10 = 0.00055m) from your last filtration project. How deep should the filter be?
Let’s make the filter 4 m x 4m.
Depth (L) should be: 1000
So, let L = 1 m; gives us: L/d10 = 1m/0.00055m = 1818 so choose 1m depth.
* Removal Efficiency = 1 – 0.29 = 0.71 = 71%
Determine the matching sizes (d10) of anthracite (S.G. = 1.63 and uniformity coefficient = 1.5) and granular activated carbon (GAC) (S.G. = 1.38 and uniformity coefficient of 1.7). When the sand size (d10) is 0.45 mm, S.G. = 2.63, and the uniformity coefficient = 1.5. Uniformity coefficient (cu) = d60/d10 – Sand d60 = d10 × cu = 0.45 × 1.5 = 0.675 mm – Matching anthracite size – From the graph, d60 = 1.27 mm – d10 = 1.27 / 1.5 = 0.85 mm – Matching GAC size – From the graph d60 = 1.73 mm – d10 = 1.73 / 1.7 = 1 mm Types of Filtration Methods? Slow Sand Filters Contains only Sand–Cleaning occurs every few weeks or months.Rapid Filters Sand or multimedia filters that have sand, anthracite (hard coal), garnet, ilmenite or granular activated carbon (GAC).-Much more common in municipal treatment.
Filter type Size -Number of filters -Filtration rate -Control of the filter flow rate -Characteristics of the filter bed -Type of filter wash system-Filter underdrains
-Auxiliary scouringTypes of filter media(s):Single media (usually sand) -Dual media (usually sand and anthracite) -Multi-media (sand, anthracite, garnet.Backwashing removes accumulated solid matter.Backwash occurs when:Headloss through the filter reaches a set value-Effluent criteria reaches a set value.
Taypes of Underdrains:Header and lateral:Backwash water enters through a pipe called a header
Plenum/nozzle floor :False floor penetrated by nozzles or strainers-Dual parallel lateral Relatively new.
Conventional Filtration Chemical coagulation, rapid mixing.Direct Filtration Chemical coagulation
Membrane Filtration Passage of the water through a thin synthetic organic polymer film in a straining filtration step.
Cartridge/Bag Filtration:Filters made from fiber, have a broad range of pore/opening
Slow Sand FiltrationAdsorption and biological flocculation take place in microbial growth in upper sand layer.
Diatomaceous Earth Filtration Water passes through layer of diatomite media on a fine metal screen,
Coagulation: It is a rapid dispersion of coagulant (chemical) in water to destabilize particles taypes?
AluminiumSulphate-Ferric Chloride-Polyaluminium Chloride-Cationic Polymers.
Determine the setting on a dry alum feeder in kg/d when the flow is 450 m3/d (for a sticker, how many people will this serve?). Jar tests indicate that the best alum dose is 8 mg/L. Feeder Setting (kg/d) = Flow (m3/d) x Alum (mg/L) x 103 (L/m3) x 10-6 (kg/mg)= 450 m3/d x 8 mg/L x 103 L/m3 x 10-6 kg/mg= 3.6 kg/d.
2. A water treatment plant uses alum for coagulation. The optimum dose of alum is determined to be 35 mg/L as dry alum. The rated plant capacity is 5300 m3/d. Alum is dosed as a liquid, which is 48.5% dry alum. The density of liquid alum is 1.35 g/mL.?Liquid alum (mg/mL) = 1.35 (g/mL) x 103 (mg/g) x 0.485 = 655 mg/mL– Dosage rate (mL/min) = 
= 197 mL/min–c. What quantity of liquid alum will they use per day at an average day flow rate of 3200 m3/d?Liquid alum (L/d) =197(mL/min)x1440(min/day)x
x10-3(L/mL)= 171 L/d-d. What quantity of liquid alum do they use in 1 month?Alum (L) = 171 L/d x 30 d = 5130 L. Types of Aerators? The water-in-air method is designed to produce small drops of water that fall through the air. The air-in-water method creates small bubbles of air that are injected into the water stream.
You must design a weir used for mixing at various flow conditions. You want to achieve a G = 1000 s-1 and the channel must have a minimum velocity of 0.5 m/s downstream. You have measured the flow over the course of one year and you know you will need to design for a maximum flow rate Q=28,000m3/d. The downstream channel should be rectangular. Assume water at 20◦C. Ignore velocity heads upstream and downstream of the weir. Ensure that the channel section satisfies the relation w=2d, where w is the width and d is the depth of the channel.
Sedimentation is the physical separation of suspended material from a water by the action of gravity.Flocculation: gentle agitation of water to promote contact between particles and coalescence to form flocs.
water contains calcium and magnesium compounds it becomes Hard Water difficult to wash, because hard water contains dissolved substances.Taypes? Temporary hard water : (which is removed by boiling) Permanent hard water: Permanent hardness (which is not removed by boiling).
The soap reacts with the magnesium and calcium ions in the water, forming SCUM -Scum is not the only problem hard water can potentially cause. Pipes can suffer from scale(also limescale).
Soft water does not contain the dissolved salts that produce scum and scale.Compute the quantities of Lime and Soda required to soften raw water containing
- Alkalinity of 220 mg/L (as CaCO3)
- Hardness as CaCl2 = 40 mg/L and as MgSO4 = 60 mg/L. Water required to be treated is 1 million liters.
- Lime required for Carbonate Hardness of 220 mg/L (as CaCO3)
100 mg/L of CaCO3 needs = 56 mg/L of CaO
220 mg/L of CaCO3 needs = 56/100 * 220 = 123.2 mg/L of CaO
b) Lime required for 60 mg/L of MgSO4
120 mg/L of MgSO4 needs = 56 mg/L of CaO
60 mg/L of MgSO4 needs = 56/120 * 60 mg/L of of CaO = 28 of CaO
Total pure lime (CaO) required = 123.2 + 28 = 151.2 mg/L
Aeration brings water and air in close contact in order to remove dissolved gases (such as carbon dioxide) and oxidizes
Chemicals removed or oxidized by aerationAmmonia- Chlorine- Carbon dioxide- Hydrogen sulfide-Methane-Iron and Manganese
