Question

A 1.0 cm diameter pipe widens to 6.0 cm, then narrows to 0.3 cm. Liquid flows through the first segment at a speed of 4.0 m/s.

(a) What is the speed in the second and third segments?

second segment =	m/s
third segment =	m/s

(b) What is the volume flow rate through the pipe? (m³/s)

Answer 1

a) WE KNOW:

Useful Equations
Volumetric Flow Rate = Velocity * Area

SO, A1V1=A2V2

piD^2/4*(4m/s) = piD^2/4*(V2)

or, V2 = {pi(1.0cm)^2 / 4 * (4m/s)} / {pi(6.0cm)^2 / 4} = 0.1111 m/s

again: A2V2=A3V3
pi(6.0cm)^2 / 4 * (0.1111 m/s) = pi(0.3cm)^2 / 4 xV3
or, V3= 44.44 m/s
this the speed in the second and third segments

b)

the volume flow rate through the pipe

All you have to do is look at one section of the pipe
For section one
Diameter = 1 cm = .01 m
Radius = .005 m
Area = (pi)*(.005 m)^2 = (pi) * 0.000025 m^2
Volumetric Flow Rate = 4 m/s * (pi) * 0.000025 m^2 = 0.000314 m^3 / s = 0.314x10^-3 m3/s

To double check do the same thing for the other sections and you will get the same result