Question

Your task is to build a wastewater treatment plant to handle all of the domestic wastewater produced by a housing estate. Using population equivalent, PE = 65000, design, describe and calculate briefly the sizes of all the unit processes in the treatment plant that are going to be built. You may assume the value that is not given.

Answer 1

Design

Descripton

Wastewater collection is done in many steps, some common steps are the collection of wastewater that is mainly done by the municipal administration. It is done channel technique and water is made to flow in channels to one central point. This water is then directed to a treatment plant using underground systems. Another important factor is the odor control as wastewater contains a lot of dirty substances with foul smell over time. All the odor sources are contained and neutralized using chemicals.

DESIGN PARAMETERS

The overall design of the wastewater treatment plant consists of 4 stages: i) Primary treatment which consists of screening, grit removal and sedimentation ii) Secondary treatment consists of a bioreactor iii) Tertiary treatment consists of nitrogen removal, adsorption and pH control.

SCREENING

Screens typically consist of wedge wire. It is carried out in two phases. In the 1st phase also called coarse screening, the size of the opening is 20-30 mm. It captures the large objects. In the second phase called fine screening, the openings vary between 1.5 to 6.4 mm. Cross section area of the screens is typically 1 msq.

PRIMARY SEDIMENTATION

Sedimentation is the process of removing solid particles heavier than water by gravity settling. In wastewater treatment, sedimentation is used to remove both inorganic and organic materials which are settleable in continuous-flow conditions. The sedimentation tank consists of a tank with 2 settling funnels where solid waste settles down. Baffles are provided to enhance the settling process. The wastes removed in this stage 20 kg/h for a 1000 kg/h feed.

SAND BED BIOREACTOR

A bioreactor packed with sand of effective particle size 0.3-0.5 mm. The uniformity coefficient (a measure of similarity of shape and size in sand particles) of sand is 4. Assuming 30% BOD is removed by sedimentation, about 92% BOD removal takes place in this reactor.

SLUDGE DIGESTION

Sludge digestion involves the treatment of highly concentrated organic wastes in the absence of oxygen by anaerobic bacteria. The dried sludge is fed to a cylindrical RCC tank of 6m diameter and the height of sludge in the tank is 6m.

PH CONTROL UNIT

The water discharged to the streams should neither be highly acidic nor alkaline. The acceptable discharge pH of 6:5. The pH of water in the tank is measured, which serves as an input to a controller. If pH is not close to 6.5, the exit stream is shut off.

NITROGEN AND PHOSPHORUS REMOVAL

For nitrogen removal, a Step Feed Anoxic Aerobic Process is used. In this process, wastewater is introduced at several feed points. Phosphate release and denitrification take place in the anoxic zone. In the aerobic zone, nitrification and BOD removal takes place.

MEASURMENTS

2.2.1 Grit chamber: N = 5 Silt particles diameter = 0.017*(1+N*0.1) = 0.0255 cm ρs = 3*(1-N*0.1) = 1500 kg/m3 Note that the product of the specific gravity of the particle (1.5) and the density of water is the density of the particles (ρs) Before we can calculate the terminal settling velocity of the particle, we should gather some information from Table A-1 in Appendix A from book (introduction to environmental engineering fourth edition) at wastewater temperature of 22oC, we find the water density to be 997.774 kg/m3 . We will use 1000 kg/m3 as a sufficiently close approximation. Form the same table, we find the viscosity to be 0.995 mPa. Silt particles diameter is 0.000255 m. Vs = ?(ρs−ρ)d2 18µ Vs= 9.8 ( 1500−1000)0.000255 18(0.000995) = 0.0177 m/s = 17.7 mm/s The Reynolds number for this settling velocity (0.0177 m/s) and particle size is = Re = (0.0177)*(0.000255)/(0.000995/1000) = 4.54 From the iterative solution Vo = 0.028 m/s. with a flow of 0.15 m3 /s and a horizontal velocity of 0.25 m/s the cross sectional area of flow may be estimated to be. Ac = (0.15 m3 /s)/(0.25m/s)= 0.60 m2 The depth of flow is then estimated by diving the cross-sectional area by the width of the channel. h = 0.60 m2 /0.56m =1.07m If the grit particle in question enters the grit chamber at the liquid surface, it will take h/Vs seconds to reach the bottom. t = 1.07m/0.028m/s = 38.2 s Since the chamber is 13.5 m in length and the horizontal velocity is 0.25 m/s the liquid remains in the chamber. t = 13.5m/0.025m/s= 54s 7 thus, the particle will be captured in the grit chamber. Overflow velocity = 0.15/(13.5*0.56)=19.8 mm/s Vs/Vo = 17.7 / 19.8 = 0.893 < 1 hence some sand particles with this diameter and density would settle down. The assumption is that the horizontal grit chamber that is 13.5 m in length if the average grit chamber flow is 0.15 m3/s, the width of the chamber is 0.56 m and the horizontal velocity is 0.25 m/s.