Elementary Principles 278 - 7.56 Point 1 - surface of...

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7-33 7.56 Point 1- surface of reservoir . P 1 1 = atm (assume), u 1 0 = , z 1 60 = m Point 2 - discharge pipe outlet . P 2 1 = atm (assume), u 2 = ?, z 2 0 = Δ P ρ = 0 Δ u u VA V V 2 2 2 2 26 4 2 2 4 2 22 2 1 35 1 3376 == = =⋅ ± ± /) . ± d h bg (m s 10 cm 1 N (2) cm 1 m kg m / s N m kg 28 42 π gz Δ= =− 9.8066 m m N s 1 kg m / s 637 N m kg 65 1 ± ± . ± ± W m V V s = ×⋅ 080 10 800 6 W 1 N m / s s 1 m W m 1000 kg Nmk g 3 3 di Mechanical energy balance: neglect Eq. 7.7 - 2 F ΔΔ Δ Pu W m V V V s TE ++ = ⇒− = = = + 2 2 2 637 800 127 76 2 ± ± . ± ± ± . . m 60 s s1 m i n mm i n 3 3 Include friction (add F > 0 to left side of equation) ± V increases. 7.57 (a) . Point 1: Surface at fluid in storage tank, P 1 1 = atm , u 1 0 = , zH 1 = m Point 2 (just within pipe): Entrance to washing machine. P 2 1 = atm , z 2 0 = u 2 2 40 4 796 600 L 1 min 1 m cm 1 L 60 s 100 cm ms 33 min . . Δ P = 0; Δ u u 2 2 2 2 1 2 317 = . . J 1 kg m / s J/kg H H 9.807 m m J s 1 kg m / s (J / kg) 2 01 9807 ch . Bernoulli Equation: m Δ H = = 2 2 03 2 3 (b) . Point 1:
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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