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.

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.

b.

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

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .

nah.   i'm just sleepy.