Question

1. An effluent water contains significant quantities of non-emulsified oil with a concentration of 176 mg/l. In order to maintain the concentration of the oil below 25 mg/l, alum with a dosage of 35mg/l was used. Based on the preliminary experiments it was found that every gram of alum produces 0.65 g of sludge. The weight percentage of sludge in the effluent is 2.75. The flow rate of oily effluent water to the treatment unit is 0.75 m3/min at 30°C. Assume the density of water is 1075kg/m3. Estimate the total volume of the sludge to be disposed and propose a suitable method for the disposal. [8 marks] The details of the water use data for three different units is a plant, along with concentrations are provided in the Table shown below. The concentrations of the contaminants in the outlet streams are within the permissible limits. 2. Table 1. Water Use Data Operation Contaminant ppm Co ppm mass (g.h1) 6000 14000 24000 150 800 1000 2 100 700 Estimate the mass load up to the pinch concentration (g.h-1) and the minimum flow rate of water? How much percentage of water will be saved if the minimum flow rate water is used? ii. [12 marks]

Answer 1

1. Concentration of the oil in inlet and outlet stream is given with the inlet flow rate. when we apllied the mass balance on the oil, we obtain:

Mass flow rate in = Mass flow rate out

175 mg/l * 0.75 * 10³ l/min = 25 mg/l * Q_out

Q_out = 5.25*10³ l/min

Mass flow rate of the stream = 5.25*10³ l/min *1075 kg/m³

=5643 kg/min

Mass flow rate of the sludge = Mass flow rate of the stream * mass fraction of the sludge

= 155.2 kg/min

Amount of the sludge produced per day will be = 223492 kg/day