Assignment 4-2013 soln

1 at height of

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Unformatted text preview: ......................(1) ( ) ( ) At height of 50 m, ( ⁄ (x,y,50) = ( ⁄ ⁄ ) ........................(2) ( ) ( ) Dividing equation (1) by (2) we get, ( ⁄ ⁄ ( { Or, [ ) ⁄ } ⁄ { ( ⁄ ) } = 402.95 ⁄ Solution: Stalk height for smelter = 300 m Plume rise = 100 m Emission rate = 5000 g/s Given stability condition is (C) Average wind speed = 3 m/s Safe level of concentration = 500 ⁄ Now, flight path is 5 Km down-wind the smelter. Total H = h + h = 300 + 100 = 400 m (From notes, figure 4-6 and 4-7, we get) and ( C (5000, 0, z, 400) = ( ⁄ ) ( Or ⁄ ) [ 500 ( ( ⁄ ( ⁄ ( ) Or, ( ( ( ⁄ ) ⁄ ( Solving the equation we get, z = 855.59 m Hence...
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This note was uploaded on 02/06/2014 for the course ENG 9624 taught by Professor Tahirhusain during the Spring '13 term at Memorial University.

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