Elementary Principles 279 - 7.58 Basis: 1000 liters of 95%...

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7-34 7.58 Basis: 1000 liters of 95% solution . Assume volume additivity. Eq. 6.1-1 Density of 95% solution l kg kg liter bg : . . . . .. 10 9 5 126 005 100 0804 124 ρρ ρ == + = = x i i Density of 35% solution l kg kg liter : . . . . 3 5 065 09278 108 =+= = Mass of 95% solution: liters 1.24 kg liter kg 1000 1240 = G = glycerol m W = water 1240 kg (1000 L) 1 (kg) 0.95 G 0.05 W 0.35 G 0.65 W 23 m m 2 (kg) 0.60 G 0.40 W 5 cm I.D. Mass balance: 1240 Glycerol balance: kg 35% solution kg 60% solution += U V W = = mm m m 12 1 2 0 95 1240 0 35 0 60 1740 2980 . b g b g b g Volume of 35% solution added kg 1 L 1.08 kg L 1740 1610 ⇒= + = Final solution volume L L 1000 1610 2610 Point 1. Surface of fluid in 35% solution storage tank . P 1 1 = atm , u 1 0 = , z 1 0 = Point 2. Exit from discharge pipe . P 2 1 = atm , z 2 23 = m u 2 2 1 25 1051 1610 L 1 m 1 min 10 cm
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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