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## Q. 6.P.22

The flowrate of air in a 305 mm 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 cross-section.
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 as follows:

 Position 1 2 3 4 5 Manometer reading (mm water) 18.5 18 17.5 16.8 15.8 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 50 mm orifice plate across which the pressure differential is 150 mm on a mercury-under-water manometer. What is the coefficient of discharge of the orifice meter?

## Verified Solution

For a pitot tube, the velocity at any point in the duct is:

$u=\sqrt{2 g h}$               (equation 6.10)

where h is the manometer reading in m of the fluid which flows in the duct.

∴           h = (reading in mm water/1000) $\times\left(\rho_w / \rho_{\text {air }}\right)$ m

The total volumetric air flowrate is given by:
Q = (area of duct × average velocity)

= area $\times(1 / 10) \sum \sqrt{2 g h}$

$=(\pi / 4) \times 0.305^2 \times 0.1 \times \sqrt{\left(\rho_w / \rho_a\right) / 1000} \times \sum \sqrt{\text { manometer reading }}$ $=0.00102 \sqrt{\left(\rho_w / \rho_a\right)}(\sqrt{18.5}+\sqrt{18.0}+\sqrt{17.5}+\cdots \sqrt{10.2})=0.0394 \sqrt{\left(\rho_w / \rho_a\right)}$

For the orifice meter, the volumetric flowrate is given by:

$Q=C_D \frac{A_0}{\sqrt{1-\left(A_0 / A_1\right)^2}} \sqrt{2 g h}$

$A_0=(\pi / 4) \times 0.15^2=0.0177 m ^2, A_1=(\pi / 4) \times 0.305^2=0.0731 m ^2, h=50$ mm Hg under water

$=(0.05(13.6-1.0) / 1.0)=0.63$ m of water $=0.63\left(\rho_w / \rho_a\right)$ m of air.

∴           $Q=C_D \times \frac{0.0177}{\sqrt{1-(0.0177 / 0.0731)^2}} \sqrt{2 \times 9.81 \times 0.63\left(\rho_w / \rho_a\right)}$

$=0.066 \sqrt{\left(\rho_w / \rho_a\right)} \times C_D$

∴                     $0.0394 \sqrt{\left(\rho_w / \rho_a\right)}=0.066 \sqrt{\left(\rho_w / \rho_a\right)} \times C_D$ and $C_D=\underline{\underline{0.60}}$