Solution to HW4

P v12 p v2 1 z1 2 2 z2 g 2 g g 2 g v1 2 pstag p 1 v12

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Unformatted text preview: 3 1 N 5-43 1 Noting that point 2 is a stagnation point and thus V2 = 0 and z1 =z2, the application of the Bernoulli equation between points 1 and 2 gives P V12 P V2 1 z1 2 2 z2 g 2 g g 2 g V12 P2 P 1 g 2g V12 P2 P g (hpitot R ) g (hpiezo R ) g (hpitot hpiezo ) 1 hpitot hpiezo g g g 2g V1 2 g (hpitot hpiezo ) 2(9.81 m/s 2 )[(0.35 0.20) m] 1.72 m/s 5-44 P V12 P V2 1 z1 2 2 z2 g 2 g g 2 g z1 V22 2g 2 2 D D V AorificeV 2 2 gz dV Vdt 4 4 2 2 D2 D D 2 gz dt 0 dz dt 0 4 4 D2 tf dt t 0 2 D0 D2 2g zf z 1 / 2 z z1 dz tf - V2 2 gz1 2 gz dt dV Atank (dz ) dz 2 zf 1 2 D0 z2 D2 2 g 1 2 (b) The tank empties completely: 4 D0 1 1 dz z 2 dz 2 gz D 2 2g z1 2g 2 2 D0 D2 zi z f tf zi = H and zf = H/2: (a) The tank empties halfway: 2 D0 2 D0 D2 2z 2z f i g g 2H H g g 2 D0 D2 tf zi = H and zf = 0: 2 D0 D2 2H g 5-48 (a) P V12 P V2 1 z1 2 2 z2 g 2 g g 2 g z1 2 m out D0 V out AorificeV 2 4 V22 2g V2 2 gz1 1 2 gz z 2g 4m out D 2 0 2 Setting z = hmax and m out m in (the incoming flow rate) gives the desired relation for the maximum height the water will reach in the tank, hmax 1 2g 4min D 2 0 2 (b) dm out m out dt 2 D0 2 gz dt ; dm tank Atank dz 4 dm tank min dt m out dt 2 DT 4 dz 2 gz dt 2 D0 dz min 4 4 2 DT Separating the variables, and integrating it from z = 0 at t = 0 to z = z at time t = t gives 1 4 min 1 2 2 DT dz 2 1 D0 4 2 gz dt z z 0 1 4 min 2 DT dz 2 1 D0 4 2 1 1 D 2 2 gz m ln min 4 D0 2 gz in 0 4 min 2g ) 2 2 DT 2 ( 1 D0 4 2 2 gz t t dt t t 0 5-49 P V12 P V2 1 z1 2 2 z2 g 2 g g 2 g P P2 1 P1 w g ( s h) P2 w gs Hg gh w (V 22 V12 ) 2 ( Hg w ) gh w (V22 V12 ) 2 P1 P2 ( Hg w ) gh h w (V 22 V12 ) V 22 V12 2 g ( Hg w ) 2 g ( Hg / w 1) 0.13368 ft 3 1 gal/s V V 1.53 ft/s V1 2 A1 D1 / 4 (4/12 ft) 2 / 4 1 gal 0.13368 ft 3 1 gal/s V V 6.13 ft/s V2 2 A2 D 2 / 4 (2/12 ft) 2 / 4 1 gal (6.13...
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