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## Q. 6.P.19

Derive a relationship between the pressure difference recorded by a pitot tube and the velocity of flow of an incompressible fluid. A pitot tube is to be situated in a large circular duct in which fluid is in turbulent flow so that it gives a direct reading of the mean velocity in the duct. At what radius in the duct should it be located, if the radius of the duct is r?
The point velocity in the duct can be assumed to be proportional to the one-seventh power of the distance from the wall.

## Verified Solution

An energy balance for an incompressible fluid in turbulent flow is given by:

$\Delta u^2 / 2+g \Delta z+v \Delta P+F=0$                 (equation 2.55)

Ignoring functional losses and assuming the pitot tube to be horizontal,

$\left(u_2^2-u_1^2\right) / 2=-v\left(P_2-P_1\right)$

If the fluid is brought to rest of plane 2, then:

$-u_1^2 / 2=-v\left(P_2-P_1\right)$

and:                                  $u_1=\sqrt{2 v\left(P_2-P_1\right)}=\underline{\underline{\sqrt{2 g h}}}$          (equation 6.10)

If the duct radius is r, the velocity $u_y$ at a distance y from the wall (and s from the centreline) is given by the one-seventh power law as:

$u_y=u_s\left(\frac{y}{r}\right)^{1 / 7}$                     (equation 3.59)

where $u_s$ is the velocity at the centreline.
The flow, $d Q$, through an annulus of thickness $d y_1$ distance y from the axis is:

$d Q=2 \pi s d y u_s\left(\frac{y}{r}\right)^{1 / 7}$

Multiplying and dividing through by r² gives:

$d Q=2 \pi r^2 u_s \frac{s}{r}\left(\frac{y}{r}\right)^{1 / 7} d \left(\frac{y}{r}\right)$

or, since s = (r – y):      $=2 \pi r^2 u_s\left(1-\frac{y}{r}\right)\left(\frac{y}{r}\right)^{1 / 7} d \left(\frac{y}{r}\right)$

The total flow is:     $Q=2 \pi r^2 u_s \int_0^1\left[\left(\frac{y}{r}\right)^{1 / 7}-\left(\frac{y}{r}\right)^{8 / 7}\right] d \left(\frac{y}{r}\right)$

$=2 \pi r^2 u_s\left[\frac{7}{8}\left(\frac{y}{r}\right)^{8 / 7}-\frac{7}{15}\left(\frac{y}{r}\right)^{15 / 7}\right]_0^1=0.817 \pi r^2 u_s$

The average velocity, $u_{ av }=Q / \pi r^2=0.817 u_s$

Thus:                                           $u_y=u_{ av }, 0.817 u_s=u_s(y / r)^{1 / 7}$

∴                                             $(y / r)=0.243$ and $s / r=\underline{\underline{0.757}}$