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Tut sheet of environment air quality management and pollution
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Tutorial Sheet
Q.No. 1.: During an air pollution monitoring study, the inlet gas stream to a bag filter is 1,69,920 m^3 /hr and the dust loading is 4577 mg/m^3. The outlet gas stream from the bag filter is 1,65,040 m^3 /hr and the dust loading is 57 mg/m^3. What is the maximum quantity of ash that will have to be removed per hour from the bag filter hopper based on these test results?
Solution: Based on dust balance, Mass (in) = Mass (out) Inlet gas stream dust = outlet gas stream dust + Hopper Ash
Q.No. 2: If 35,000kg of whole milk containing 4% fat is to be separated in a 6 hour period into skim milk with 0.45% fat and cream with 45% fat, what are the flow rates of the two output streams from a continuous centrifuge which accomplishes this separation? Take Basis 1 hour's flow of whole milk
Solution: Mass in Total mass = 35000/6 = 5833 kg. Fat = 5833 x 0.04 = 233 kg. And so Water plus solids-not-fat = 5600 kg.
Mass out Let the mass of cream be x kg then its total fat content is 0.45x. The mass of skim milk is (5833 - x) and its total fat content is 0.0045 (
Problem 1 A coal fired 1000 MW power plant is operating around 38% efficiency. The ash and sulphur content in the coal used respectively are 35% and 3% and the calorific value is 21 MJ per kg of coal. Find the emission rate of SO2 and ash generation per hours.
Solution: Power plant capacity = 1000 MW Efficiency = 38% Input power = 1000/0.38 = 2631.6 MW = 2631.6 MJ/S Calorific value of coal = 21 MJ/Kg Burning rate of coal = 2631.6/21 = 125.3 Kg/S Ash content = 35% Ash generation rate = 125.3 * 0.35 = 43.8 Kg/S = 157680 Kg/hr Sulphur content = 3% S + O2 SO SO2 generation rate = 0.03*125.3 * 2 = 7.52 Kg/S = 27072 Kg/hr
Problem 2 During rush hours on a busy road crossing, nearly 1200 vehicles cross per hour at an average speed of 20 kmph. Of these, about 70% cars use leaded
Power rating of the generator = 2000 KW Efficiency of the generator = 25% Input power to get 2000 KW energy = 8000 KW C3H8+5O2+18.5N2 3CO2 + 4H20 + 18.5N Volume of CO2 in exhaust gas = 3/(3+4+18.5) = 11.76% Calorific value of propane = 48 MJ/kg Burning rate of propane = (8000 KJ/S) /48000 KJ/Kg) = 0.167 Kg/S For 1 mole (44 g) of propane 5 mole O2 and 18.5 moles of N (32x5+18.5x28 = 678 g of air) is required, For 0.167 kg/S of propane air supply requirement = (678/44)*0.167 = 2.57 kg/s
Power generation= 1 MJ Efficiency of the generator = 25% Input power to get 1 MJ energy = 1/0.25 = 4 MJ C3H8+5O2+18.5N2 3CO