A)25.3m/s B)6.32 m/s C)3.48x10^5 pa oo AT&T; 9:21 PM eccphys1b.files.wordpress.c

Question

A)25.3m/s B)6.32 m/s C)3.48x10^5 pa oo AT&T; 9:21 PM eccphys1b.files.wordpress.com 10 of 43 (2.1.60) At Whale World Amasement Park, guests are given the opportunity to be lifted by a stream of water shot from the blowhole of a concrete whale. A horizontal 2.0-inch diameter pipe runs into mouth of the whale, and then narrows to a 1 .0-inch diameter as it bends upward. shooting a stream of water into the air. Park visitors may stand on a board that is held in the air by the impact of the stream (a) In the diagram below, a 30-kg child is on a 3-kg board held aloft by the stream. What is the speed of the water as it strikes the board? Assume that the speed of the water immediately after striking the board is zero, and that the distance between the end of the pipe and the hoard is negligible. Begin your analysis with a free-body diagram for the board. (b) What is the flow speed in the horizontal part of the pipe? (c) The horizontal part of the pipe is 5.0m below the board. What is the gauge pressure in the horizontal part of the pipe?

Explanation / Answer

for the board to be equilibrium

a. weight of child + board = force by water

mass of child = M = 30 kg

Mass of board, m = 3 kg

now, let velocity of water from the pipe be v

diameter of pipe = a = 1 in = 2.54 cm

area of pipe, A = pia^2 = 0.0020268 m^2

so volume flow rate = Av

Mass flow rarte = rho*Av ( where rho is density of water)

momentum rate = rho*Av^2 = force

rho*Av^2 = (M + m)g

v^2 = (33g)/1000*0.002068

v = 12.63 m/s

b. Horizontal part of the pipe, diameter, b = 2 in = 2*2.54 cm

Area, B = 4* 0.0020268 m^2 = 4*A

so from continuity equation, if flow velocity here is u

then 4*A*u = A*12.63

u = 3.159 m/s

c. gauge pressure in horizontal part of pipe = P

then from bernoulli's theorem

P = 0.5*rho(v^2 - u^2) = 74767.12713 Pa

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