P15_068

# P15_068 - p C = p air , Bernoullis equation becomes p B = p...

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68. (a) We consider a point D on the surface of the liquid in the container, in the same tube of ﬂow with points A , B and C . Applying Bernoulli’s equation to points D and C ,weobta in p D + 1 2 ρv 2 D + ρgh D = p C + 1 2 ρv 2 C + ρgh C which leads to v C = s 2( p D p C ) ρ +2 g ( h D h C )+ v 2 D p 2 g ( d + h 2 ) where in the last step we set p D = p C = p air and v D /v C 0. (b) We now consider points B and C : p B + 1 2 ρv 2 B + ρgh B = p C + 1 2 ρv 2 C + ρgh C . Since v B = v C
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Unformatted text preview: p C = p air , Bernoullis equation becomes p B = p C + g ( h C h B ) = p air g ( h 1 + h 2 + d ) . (c) Since p B 0, we must let p air g ( h 1 + d + h 2 ) 0, which yields h 1 h 1 , max = p air d h 2 p air = 10 . 3 m ....
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