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}}}.