Solution to HW4

1 d 2 2 gz m ln min 4 d0 2 gz in 0 4 min 2g 2 2 dt 2

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Unformatted text preview: ft/s) 2 (1.53 ft/s) 2 h 2(32.2 ft/s 2 )(847 / 62.4 1) 0.0435 ft 0.52 in 5-53 P V12 P V2 1 z1 2 2 z2 g 2 g g 2 g z1 V22 2g V2 2 gz1 We denote the diameter of the orifice by D, and the diameter of the pool by Do. The flow rate of water from the pool is obtained by multiplying the discharge velocity by the orifice cross-sectional area, D 2 V AorificeV 2 2 gz 4 D 2 D 2 dV Vdt 2 gz dt dV Atank (dz ) 0 dz 4 4 2 2 2 D0 D0 D0 1 D 2 1 dz dt 2 2 gz dt dz z 2 dz 2 4 4 2 gz D D 2g tf dt t 0 tf 2 D0 D 2 2g 0 z 1 / 2 dz z z1 ( 10 m) 2 ( 0.03 m )2 tf - 2 D0 D 2 2g 1 0 z2 1 2 z1 2 2 D0 D 2 2g h 2 D0 D2 2h g 2( 2 m) 70,950 s 19.7 h 9.81 m/s 2 5-55 P1 V12 P V2 z1 2 2 z 2 g 2 g g 2 g in which air V1 V2 P1 P2 air V 22 V12 2 110 kPa P 1.19 kg/m 3 RT (0.287 kPa m 3 /kg K )(50 273 K ) V V 0.045 m3/s 15.9 m/s 2 A1 D1 / 4 (0.06 m)2 / 4 0.045 m 3 /s V V 35.8 m/s 2 A2 D 2 / 4 (0.04 m) 2 / 4 P1 P2 (1.19 kg/m 3 ) P water gh gives 612 N/m 2 612 Pa 2 1 kg m/s 2 P P2 612 N/m 0.0624 m 6.24 cm h 1 water g (1000 kg/m 3 )(9.81 m/s 2 ) 1 N (35.8 m/s) 2 (15.9 m/s) 2 2 1N 1 kg m/s 2 3 5-58 Water: P3 V32 P V2 z3 4 4 z4 g 2 g g 2 g P1 V12 P V2 z1 2 2 z 2 g 2 g g 2 g Air: where air P1 Patm (h) g g P1 V12 Patm g 2 g g P1 Patm water gh 2( Patm P1 ) V1 air P 95 kPa 1.13 kg/m 3 3 RT (0.287 kPa m /kg K )(20 273 K ) 2( Patm P1 ) V1 air 2 water gh air 2(1000 kg/m 3 )(9.81 m/s 2 )(0.1 m )...
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This document was uploaded on 02/05/2014.

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