Question 6.P.26: Air at 323 K and 152 kN/m² flows through a duct of circular ...
Air at 323 K and 152 kN/m² flows through a duct of circular cross-section, diameter 0.5 m. In order to measure the flowrate of air, the velocity profile across a diameter of the duct is measured using a pitot-static tube connected to a water manometer inclined at an angle of \cos ^{-1} 0.1 to the vertical. The following results are obtained:
| Distance from
duct centre-line (m) |
Manometer
Reading h_m (mm) |
| 0 | 104 |
| 0.05 | 100 |
| 0.10 | 96 |
| 0.15 | 86 |
| 0.175 | 79 |
| 0.20 | 68 |
| 0.225 | 50 |
Calculate the mass flowrate of air through the duct, the average velocity, the ratio of the average to the maximum velocity and the Reynolds number. Comment on these results.
Discuss the application of this method of measuring gas flowrates, with particular emphasis on the best distribution of experimental points across the duct and on the accuracy of the results.
Take the viscosity of air as 1.9 \times 10^{-2} mN s/m² and the molecular weight of air as 29 kg/kmol.
If h_m is the manometer reading, the vertical manometer height will be 0.1h_m (mm of water).
For a pitot tube, the velocity at any point is:
u=\sqrt{2 g h} (equation 6.10)
where h is the manometer reading in terms of the fluid flowing in the duct.
Thus: h=\left(0.1 h_m / 1000\right) \times\left(\rho_w / \rho_{\text {air }}\right)
\rho_{\text {air }}=(29 / 22.4)(152 / 101.3)(273 / 323)=1.64 kg / m ^3
∴ h=\left(0.1 h_m / 1000\right)(1000 / 1.64)=0.061 h_m
and: u=\sqrt{2 \times 9.81 \times 0.061 h_m}=1.09 \sqrt{h_m} ( m/s )
If the duct is divided into a series of elements with the measured radius at the centre-line of the element, the velocity of the element can be found from the previous equation and the volumetric flowrate calculated. By adopting this procedure across the whole section, the required values may be determined.
For example, at 0.05 m, where h_m=10 mm.
Inner radius of element = 0.025 m
Outer radius of element = 0.075 m
Area of element =\pi\left(0.075^2-0.025^2\right)=0.0157 m ^2
∴ u=\left(1.09 \sqrt{h_m}\right)=1.09 \sqrt{100}=10.9 m / s
Volumetric flowrate in the element =(10.9 \times 0.0157)=0.171 m ^3 / s
The following table is constructed in the same way.
| Distance
from duct centre line (m) |
Outer
radius of element (m) |
Inner
radius of element (m) |
Area of
element (m)² |
h_m
(mm) |
Velocity
u (m/s) |
Volumetric
flowrate ΔQ m³/s |
| 0 | 0.025 | 0 | 0.00196 | 104 | 11.1 | 0.0218 |
| 0.05 | 0.075 | 0.025 | 0.0157 | 100 | 10.9 | 0.171 |
| 0.10 | 0.125 | 0.075 | 0.0314 | 96 | 10.7 | 0.336 |
| 0.15 | 0.1625 | 0.125 | 0.0339 | 86 | 10.1 | 0.342 |
| 0.175 | 0.1875 | 0.1625 | 0.0275 | 79 | 9.7 | 0.293 |
| 0.20 | 0.2125 | 0.1875 | 0.0314 | 68 | 8.9 | 0.279 |
| 0.225 | 0.2375 | 0.2125 | 0.0353 | 50 | 7.7 | 0.272 |
| 0.25 | 0.25 | 0.2375 | 0.0192 | 0 | 0 | 0 |
| \overline{\sum=0.1964 m ^2} | \overline{\sum=1.715 m ^3 / s} | |||||
Average velocity =(1.715 / 0.1964)=\underline{\underline{8.73 m / s }}
Mass flowrate =(1.715 \times 1.64)=\underline{\underline{2.81 kg / s }}
u_{ av } / u_{\max }=(8.73 / 11.1)=\underline{\underline{0.79}} R e=(8.73 \times 1.64 \times 0.05) /\left(1.9 \times 10^{-5}\right)=\underline{\underline{3.77 \times 10^4}}The velocity distribution in turbulent flow is discussed in Section 3.3.6 where the Prandtl one-seventh power law is used to give:
u_{ av }=0.82 u_{\max } (equation 3.63)
This is close to that measured in this duct though strictly it only appears at very high values of Re. Reference to Fig. 3.14 shows that, at R e=3.8 \times 10^4, the velocity ratio is about 0.80 which shows remarkably good agreement.