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CHE218_solutions1

CHE218_solutions1 - Set 1 2008 1.3 = mass volume For 100lbm...

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Set 1 2008 1.3* ρ = mass volume : For100lbm ρ = 100 lbm 50lbm 4.49 62.3lbm ft 3 + 50lbm 62.3lbm ft 3 = 102 lbm ft 3 3 3 3 3 cm g 63 . 1 lbm g 6 . 453 28310cm ft ft lbm 102 = Discussion ; this assumes no volume change on mixing. That is a good assumption here, and in many other cases. In a few, like ethanol and water have changes of up to a few %. 1.5* ρ = m gross - m tare V = 45 g - 17.24 g 25cm 3 = 1.110 g cm 3 3 3 3 3 3 1000 1.20 25 0.03 1000000 air kg m g g m cm m cm kg cm = = 3 3 tare gross cm 111 . 1 cm 25 ) 03 . 0 24 . 17 ( 45 ) ( g g V m m m air = - - = - - = ρ Omitting the weight of the air makes a difference of 0.001 = 0.1%. This is normally ignored, but in the most careful work it must be considered. 1.10 (a) For r = R, V θ = ω k 2 1 - k 2 R 2 R - R = 0 For r = r inner cylinder = kR V θ = ω k 2 1 - k 2 R 2 kR - kR = ω k 2 1 - k 2 R 1 - k 2 k = ω kR (b) d dr V θ r = ω k 2 1 - k 2
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