Question

Given: The viscosity is negligible. Atmospheric pressure is 101300 Pa. Water flows at speed of 5.9 m/s through a horizontal pipe of diameter 3.3 cm. The gauge pressure P1 of the water in the pipe is 1.6 atm. A short segment of the pipe is constricted to a smaller diameter of 2.4 cm.

What is the gauge pressure of the water flowing through the constricted segment? Answer in units of atm.

Answer #1

R = A v

Where is the density ,A
is the cross sectional area and v is the velocity of the
liquid

We can find the velocity of the water through the constricted
segment

A_{1} v_{1} = A_{2}
v_{2}

0.033 x 5.9 = 0.024 x v_{2}

v_{2} = 8.11 m/s

Now we can find the gauge pressure of the water flowing through the
constricted segment

P_{1} + g h + (1/2)
v_{1}^{2} = P_{2} +
g h + (1/2)
v_{2}^{2}

Since the pipe is horizontal, we can neglect the gh term

1 atm = 101300 Pa

1.6 atm = 162080 Pa

162080 + (1/2) 1000 x 5.9^{2} =
P_{2} + (1/2) x 1000 x 8.11^{2}

P_{2} = 146599 Pa

Converting back to atm

P_{2} = 1.447 atm

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