(1 point) Now that we've disposed of the sludge, let's turn our attention to the PM in the gaseous exhaust to the stack.
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answerhappygod
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(1 point) Now that we've disposed of the sludge, let's turn our attention to the PM in the gaseous exhaust to the stack.
(1 point) Now that we've disposed of the sludge, let's turn our attention to the PM in the gaseous exhaust to the stack. What would the concentration of PM be 2km directly downwind given the following atmospheric conditions: effective stack height, H = 50m, spread: y = 300, z = 250, wind speed, u = 5 m/s E x(x, y, 0, H) = ( 7 ( ) [(())][(()] exp exp που συ x in g/m³ directly downwind → y=0 E (PM in gas after controls): g/s µg/m³ Concentration:
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