Question 6.P.16: The flowrate of air at 298 K in a 0.3 m diameter duct is mea...

Cross-sectional area of duct =(\pi / 4)(0.3)^2=0.0707  m ^2.
Area of each concentric annulus =0.00707  m ^2.
If the diameters of the annuli are designated d_1, d_2 etc., then:

0.00707=(\pi / 4)\left(0.3^2-d_1^2\right)

0.00707=(\pi / 4)\left(d^2-d_2^2\right)

0.00707=(\pi / 4)\left(d_2^2-d_3^2\right) and so on,

and the mid-points of each annulus may be calculated across the duct.
For a pitot tube, the velocity may be calculated from the head h as u=\sqrt{(2 g h)}
For position 1, h = 18.5 mm of water.
The density of the air =(29 / 22.4)(273 / 298)=1.186  kg / m ^3

h=\left(18.5 \times 10^{-3} \times 1000 / 1.186\right)=15.6 m of air

and:             u=\sqrt{(2 \times 9.81 \times 15.6)}=17.49 m/s
In the same way, the velocity distribution across the tube may be found as shown in the following table.
Mass flowrate, G=(1.107 \times 1.186)=1.313 kg/s
For the orifice, \left[1-\left(A_0 / A_1\right)^2\right]=\left[1-(0.15 / 0.3)^2\right]=0.938

h = 50 mm Hg-under-water

=(0.05 \times(13.55-1) \times 1000 / 1.186)=529 m of air
and:     1.313=C_D(\pi / 4)(0.15)^2 \times 1.186 \sqrt{ }(2 \times 9.81 \times 529 / 0.938) and C_D=\underline{\underline{0.61}}

The velocity profile across the duct is plotted in Fig. 6d.