A horizontal pipe of cross-sectional area 0.29 m (A, in the picture below) carri

Question

A horizontal pipe of cross-sectional area 0.29 m (A, in the picture below) carries water at a pressure of 1.0 atm; the speed of the water at this point is 9.4 m/s. When the pipe becomes constricted such that its cross-sectional area is only 0.19 m, what is the pressure of the water in atm? Note: 1 atm = 101,300 Pa, and the density of water is 1000 kg/m Hint: Use Bernoulli's equation, ignoring the gravity terms (since the height of the pipe does not change). You will first need to find the speed of the water in the constricted region (using the continuity equation) 2 3

Explanation / Answer

Using the continuity equation we have

A1v1 = A2v2

0.29*9.4 = 0.19*v2

v2 = 14.35 m/s

using the bernoulli's equation we have

P1 + 0.5pv1^2 = P2 + 0.5pv2^2 ( since both points are at same level h1 = h2 and hence can be ignored)

P2 = P1 + 0.5p(v1^2 - v2^2)

= 101300 + 0.5*1000*(9.4^2 - 14.35^2)

= 42556.5 Pa

= 42556.5 / 101300 = 0.42 atm

so the pressure at the constricted part is 0.42 atm

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