Question 6.P.18: The flowrate of a fluid in a pipe is measured using a pitot ...

For streamline flow in a pipe, a force balance gives:

$-\Delta P \pi r^2=-\mu \frac{ d u}{ d r} 2 \pi r l$

and:                    $-u=\frac{-\Delta P}{2 \mu l} r$ and $-u=\frac{-\Delta P}{2 \mu l} \frac{r^2}{2}+$ constant.
When r = a (at the wall), u = 0, the constant $=-\Delta P a^2 / 4 \mu l$

and:                          $u=-\frac{\Delta P}{4 \mu l}\left(a^2-r^2\right)$

The maximum velocity, $u_{\max }=\frac{-\Delta P a^2}{4 \mu l}$

and:                                               $\frac{u}{u_{\max }}=1-\left(\frac{r}{a}\right)^2$

When $r=a / 2, u / u_{\max }=0.75$.

The pitot tube is discussed in Section 6.3.1 and:

$u=k \sqrt{h}$                (from equation 6.10)

At the centre-line, $u=u_{\max }$ and h = 40 mm.

∴                       $u_{\max }=K \sqrt{40}=6.32  K$

At a point midway between the axis and the wall, $u=u_{1 / 2}$ and h = 22.5 mm.

∴                       $u_{1 / 2}=K \sqrt{22.5}=4.74  K$

$u_{1 / 2} / u_{\max }=(4.74  K / 632  K )=0.75$

and hence the flow is $\underline{\underline{\text { streamline}}}$.