Answer
The fluid's speed in the circular part of the tube is $1.27~v_0$.
Work Step by Step
The volume flow rate is constant throughout the tube. The volume flow rate is the speed of the fluid times the cross-sectional area of the tube.
We can find the fluid's speed $v_2$ in the circular part of the tube as:
$v_2~A_2 = v_0~A_0$
$v_2~\pi~(\frac{L}{2})^2 = v_0~L^2$
$v_2 = \frac{4~v_0~L^2}{\pi~L^2}$
$v_2 = \frac{4~v_0}{\pi}$
$v_2 = 1.27~v_0$
The fluid's speed in the circular part of the tube is $1.27~v_0$.
