422_Physics ProblemsTechnical Physics

# 422_Physics ProblemsTechnical Physics - 424 Fluid Mechanics...

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424 Fluid Mechanics Section 14.5 Fluid Dynamics Section 14.6 Bernoulli’s Equation P14.38 By Bernoulli’s equation, 800 10 1 2 1000 600 10 1 2 100016 200 10 1 2 100015 163 1 000 5 00 10 1 63 12 8 42 4 2 2 2 .. . . . ×+= ×+ ×= = == × = Nm ms kg s 22 2 bg ej vv v v dm dt Av ρπ FIG. P14.38 P14.39 Assuming the top is open to the atmosphere, then PP 10 = . Note 20 = . Flow rate −− 2 50 10 4 17 10 35 .m i n . m m s 33 . (a) AA 12 >> so << Assuming v 1 0 = , P v gy P v gy vg y 1 1 2 2 2 2 21 2 2980 160 177 ++ = = ρ a fa f . m s (b) Flow rate F H G I K J Av d 2 5 4 17 7 417 10 π a f 3 d = 173 10 173 3 m m *P14.40 Take point 1 at the free surface of the water in the tank and 2 inside the nozzle. (a) With the cork in place Pg y v y v 11 1 2 2 2 1 2 1 2 = ρρ becomes 02 98 75 0 0 0 = + + kg m m s m 32 ; 4 735 10 −= × . P a . For the stopper F x = 0 FF f PA PA f
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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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