Question

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 0.070m2, and the magnitude of the fluid velocity is 3.50m/s.
part A

What is the fluid speed at point in the pipe where the cross-sectional area is 0.105m2?

part b

What is the fluid speed at point in the pipe where the cross-sectional area is 0.047m2?

part c

Calculate the volume of water discharged from the open end of the pipe in 1.00hour.

Answer 1

Answer:

a) 2.33 m/s

b) 5.21 m/s

c) 882 m³

Explanation:

Using the concept of continuity equation

for flow through pipes

$A_{1}\times V_{1} = A_{2}\times V_{2}$

Where,

A = Area of cross-section

V = Velocity of fluid at the particular cross-section

given:

$A_{1} = 0.070 m^{2}$

$V_{1} = 3.50 m/s$

a) $A_{2} = 0.105 m^{2}$

substituting the values in the continuity equation, we get

$0.070\times 3.50 = 0.105\times V_{2}$

or

$V_{2} = \frac{0.070\times 3.5}{0.150}m/s$

or

$V_{2} = 2.33m/s$

b) $A_{2} = 0.047 m^{2}$

substituting the values in the continuity equation, we get

$0.070\times 3.50 = 0.047\times V_{2}$

or

$V_{2} = \frac{0.070\times 3.5}{0.047}m/s$

or

$V_{2} = 5.21m/s$

c) we have,

Discharge$Q = Area (A)\times Velocity(V)$

thus from the given value, we get

$Q = 0.070m^{2}\times 3.5m/s$

$Q = 0.245 m^{3}/s$

Also,

Discharge$Q = \frac{volume}{time}$

given time = 1 hour = 1 ×3600 seconds

substituting the value of discharge and time in the above equation, we get

$0.245m^{3}/s = \frac{volume}{3600s}$

or

$0.245m^{3}/s\times 3600 = Volume$

volume of flow = $882 m^{3}$