Question 6.P.22: The flowrate of air in a 305 mm diameter duct is measured wi...
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 | 18.0 | 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?
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}}