{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

CHE218_solutions1

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}