To solve this problem we will use the concepts related to the continuity of the fluid in a channel. For this we know that the proportion or relationship:
[tex]Q_1 = Q_2[/tex] \rightarrow volumetric flow rate
[tex]A_1V_1 = A_2 V_2[/tex]
Where,
A_{1,2} = Area at each section
V_{1,2} = Velocity at each section
Since the pipe preserves its diameter, we will have to [tex]A_1 = A_2[/tex] and therefore immediately that
[tex]V_1 = V_2[/tex]
Therefore the speed at the exit will be the same: [tex]4.3m / s[/tex]
