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Environment tut sheet on air pollution, Study Guides, Projects, Research of Environmental Science

Tut sheet of environment air quality management and pollution

Typology: Study Guides, Projects, Research

2018/2019

Uploaded on 11/15/2019

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Tutorial Sheet
Q.No. 1.: During an air pollution monitoring study, the inlet gas stream to a
bag lter is 1,69,920 m3 /hr and the dust loading is 4577 mg/m3 . The outlet
gas stream from the bag lter is 1,65,040 m3 /hr and the dust loading is 57
mg/m3 . What is the maximum quantity of ash that will have to be removed
per hour from the bag lter 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
1. Calculate the inlet and outlet dust quantities in kg per hour Inlet dust
quantity = 169920 (m3 /hr) x 4577 (mg/m3 ) x 1/1000000 (kg/mg) =
777.7 kg/hr
Outlet dust quantity = 165040 (m 3 /hr) x 57 (mg/m3 ) x 1/1000000
(kg/mg) = 9.41 kg/hr
2. Calculate the quantity of ash that will have to removed from the
hopper per hour
Hopper ash = Inlet gas dust quantity - Outlet gas dust quantity
= 777.7 kg/hr – 9.41 kg/hr
= 768.2 kg/hr
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 ow rates of the two output streams from a continuous centrifuge
which accomplishes this separation? Take Basis 1 hour's ow 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.
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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

  1. Calculate the inlet and outlet dust quantities in kg per hour Inlet dust quantity = 169920 (m^3 /hr) x 4577 (mg/m^3 ) x 1/1000000 (kg/mg) = 777.7 kg/hr Outlet dust quantity = 165040 (m 3 /hr) x 57 (mg/m^3 ) x 1/ (kg/mg) = 9.41 kg/hr
  2. (^) Calculate the quantity of ash that will have to removed from the hopper per hour Hopper ash = Inlet gas dust quantity - Outlet gas dust quantity = 777.7 kg/hr – 9.41 kg/hr = 768.2 kg/hr

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 (

  • x) Material balance on fat: Fat in = Fat out 5833 x 0.04 = 0.0045(5833 - x) + 0.45x. and so x = 465 kg. So that the flow of cream is 465 kg / hr and skim milk (5833 - 465) = 5368 kg/hr

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

  • 4H20 + 18.5N Calorific value of propane = 48 MJ/kg Quantity of propane for 4MJ = (4MJ) /48 MJ/Kg) = 0.083 Kg For 1 mole (44 g) of propane, 3 mole CO2 (44x3 = 132 g) is generated, For 0.083 kg of propane CO2 emission = 0.083*3= 0.249 kg