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Unformatted text preview: Set 1 2008 1.3* = mass volume : For100lbm = 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* = 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 . 24 . 17 ( 45 ) ( g g V m m m air = = = 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 = k 2 1 k 2 R 2 R R = 0 For r = r inner cylinder = kR V = k 2 1 k 2 R 2 kR kR...
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This homework help was uploaded on 04/18/2008 for the course CHE 218 taught by Professor Smith during the Spring '08 term at SDSMT.
 Spring '08
 Smith

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