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Water is flowing in a pipe with a varying cross-sectional area, and at all points, the water completely fills the pipe. At point 1, the cross-sectional area of the pipe is 7.70×10−2 m2 and the magnitude of the fluid velocity is 3.60 m/s.
a. What is the fluid speed at points in the pipe where the cross-sectional area is (a) 0.105 m and (b) 0.047m2?
b. Calculate the volume of water discharged from the open end of the pipe in 1.00 hour.

Respuesta :

MechEngineer

Answer:

a. 2.64 m/s

b. 5.9 m/s

c. 997.92 m3

Explanation:

As this is steady flow, the mass flow rate is constant, and so is the product of flow velocity and cross-sectional area

av = 0.077 * 3.6 = 0.2772 m3/s

We can calculate the speed at various areas by dividing the product above by area

a.[tex]v_1 = 0.2772 / a_1 = 0.2772 / 0.105 = 2.64 m/s[/tex]

b.[tex]v_2 = 0.2772 / a_2 = 0.2772 / 0.047 = 5.9 m/s[/tex]

c. Since the volume discharge rate is 0.2772 cube meters per second. In 1 hour, or 60 * 60 = 3600 seconds, the total volume of water discharged would be

0.2772 * 3600 = 997.92 m3

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