424_Physics ProblemsTechnical Physics

424_Physics ProblemsTechnical Physics - 426 *P14.44 Fluid...

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426 Fluid Mechanics *P14.44 (a) Between sea surface and clogged hole: P v gy P v gy 11 2 12 2 2 2 1 2 1 2 ++ = ρρ 10 1 0 3 0 9 8 2 0 0 2 atm kg m m s m 32 = ++ ej e j af . PP 2 0 2 =+ atm kPa . The air on the back of his hand pushes opposite the water, so the net force on his hand is FP A == × F H G I K J × 20 2 10 4 12 10 2 .. Nm m 2 e j π F = 228 . N (b) Now, Bernoulli’s theorem is 2 0 21 1 2 1030 0 2 2 atm kPa atm kg m 3 = + + . v v 2 626 = m s The volume rate of flow is Av 22 2 2 4 4 1 2 10 6 26 7 08 10 = × −− . m m s m s 3 bg One acre–foot is 4 047 0 304 8 1 234 m m 23 ×= . Requiring 1234 708 10 174 10 202 4 6 m ms s d a y s 3 3 . × = P14.45 (a) Suppose the flow is very slow: Pv g y g y F H G I K J + F H G I K J 1 2 1 2 river rim Pg g P = = + 0 564 1 0 2 096 1 1 000 9 8 1 532 1 15 0 a t m m atm kg m m s m atm MPa e j (b) The volume flow rate is 4500 4 2 md 3 Av dv v = F H G I K J F H G I K J = 14 0150 295 2 d 86 400 s
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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