## Related questions with answers

The velocity in the $2-\text{cm}$-diameter pipe of Fig. has only one nonzero velocity component given by $u(r, t)=2\left(1-r^2 / r_0^2\right)\left(1-e^{-t / 10}\right)~\mathrm{m} / \mathrm{s}$, where $r_0$ is the radius of the pipe and $t$ is in seconds. Calculate the maximum velocity and the maximum acceleration: (a) Along the centerline of the pipe (b) Along a streamline at $r=0.5 \mathrm{~cm}$ (c) Along a streamline just next to the pipe wall [Hint: Let $v_z=u(r, t), v_r=0$, and $v_\theta=0$ in the appropriate equations in Table.]

Solution

Verified**Given the following:**

$\begin{align*} u\left(r,t\right)&=2\left(1-\dfrac{r^{2}}{r_{0}^{2}}\right)\left(1-e^{\frac{-t}{10}}\right)\\\\ d&=2\ \text{cm}\\ v_{z}&=u\left(r,t\right)\\ v_{r}&=0\\ v_{\theta}&=0 \end{align*}$

Determine the maximum velocity and the maximum acceleration a) along the center, b) at $r=0.5\ \text{cm}$, and c) next to the pipe wall.

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