Answer:
[tex]u_{max} = 17.334\,\frac{m}{s}[/tex]
Explanation:
Let consider that velocity profile inside the circular pipe is:
[tex]u(r) = 2\cdot U_{avg} \cdot \left(1 - \frac{r^{2}}{R^{2}} \right)[/tex]
The average speed at [tex]r = \frac{1}{2} \cdot R[/tex] is:
[tex]U_{avg} = \frac{13\,\frac{m}{s} }{2\cdot \left(1-\frac{1}{4} \right)}[/tex]
[tex]U_{avg} = 8.667\,\frac{m}{s}[/tex]
The velocity at the center of the pipe is:
[tex]u_{max} = 2\cdot U_{avg}[/tex]
[tex]u_{max} = 17.334\,\frac{m}{s}[/tex]
