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Unformatted text preview: 77. (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, we
obtain
1
12
2
pD + ρ vD + ρ ghD = pC + ρ vC + ρ ghC
2
2
which leads to vC = 2( pD − pC ) 2
+ 2 g (hD − hC ) + vD ≈ 2 g (d + h2 ) ρ where in the last step we set pD = pC = pair and vD/vC ≈ 0. Plugging in the values, we
obtain vc = 2(9.8 m/s 2 )(0.40 m + 0.12 m) = 3.2 m/s.
(b) We now consider points B and C:
pB + 12
12
ρ vB + ρ ghB = pC + ρ vC + ρ ghC .
2
2 Since vB = vC by equation of continuity, and pC = pair, Bernoulli’s equation becomes
pB = pC + ρ g (hC − hB ) = pair − ρ g (h1 + h2 + d )
= 1.0 ×105 Pa − (1.0 ×103 kg/m3 )(9.8 m/s 2 )(0.25 m + 0.40 m + 0.12 m)
= 9.2 ×104 Pa.
(c) Since pB ≥ 0, we must let pair – ρg(h1 + d + h2) ≥ 0, which yields
h1 ≤ h1,max = pair ρ − d − h2 ≤ pair ρ = 10.3 m . ...
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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