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## Q. 6.P.16

The flowrate of air at 298 K in a 0.3 m diameter duct is measured with a pitot tube which is used to traverse the cross-section. Readings of the differential pressure recorded on a water manometer are taken with the pitot tube at ten different positions in the crosssection.
These positions are so chosen as to be the mid-points of ten concentric annuli each of the same cross-sectional area. The readings are:

 Position 1 2 3 4 5 Manometer reading (mm water) 18.5 18 17.5 16.8 15.7 Position 6 7 8 9 10 Manometer reading (mm water) 14.7 13.7 12.7 11.4 10.2

The flow is also metered using a 150 mm orifice plate across which the pressure differential is 50 mm on a mercury-under-water manometer. What is the coefficient of discharge of the orifice meter?

## Verified Solution

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 $C_D=\underline{\underline{0.61}}$

 Position Distance from axis of duct (mm) Manometer reading Air velocity (u m/s) Velocity × area of annulus (m³/s) Water (mm) Air (m) 1 24 18.5 15.6 17.5 0.124 2 57 18.0 15.17 17.3 0.122 3 75 17.5 14.75 17.0 0.120 4 89 16.8 14.16 16.7 0.118 5 101 15.7 13.23 16.1 0.114 6 111 14.7 12.39 15.6 0.110 7 121 13.7 11.55 15.1 0.107 8 130 12.7 10.71 14.5 0.103 9 139 11.4 9.61 13.7 0.097 10 147 10.2 8.60 13.0 0.092 Total = 1.107

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