### Pitot Tube for Flow Measurement

A Newtonian fluid having a viscosity of 1.23 poise, and a density of 0.893 gm/cm3, is flowing through a straight, circular pipe having an inside diameter of 5 cm. A pitot tube is installed on the pipeline with its impact tube located at the center of the pipe cross section. At a certain flow rate, the pitot tube indicates a reading of 8 cm of mercury. Determine the volumetric flow rate of the fluid.

Data:

Viscosity of fluid (μ) = 1.23 poise = 0.123 kg/(m.sec)

Density of fluid (ρ) = 0.893 gm/cm3 = 893 kg/m3

Dia of pipe (D) = 5 cm = 0.05 m

Manometer reading (hm) = 8 cm of Hg = 0.08 m of Hg

Formulae:

Velocity at the center of pipe

vo = ( 2 x (ps - po)/ ρ)0.5 = (2 x hm x (ρm - ρ) x g /ρ)0.5

Calculations:

Vo = ( 2 x 0.08 x (13.6 - 0.893) x 9.812 / 0.893 )0.5 = 4.73 m/sec

NRe = Dvoρ/μ = 0.05 x 4.73 x 893 / 0.123 = 1717

Since the flow is laminar (NRe < 2100), the average velocity is given by

vavg = vo / 2 = 4.73 / 2 = 2.365 m/sec (velocity at the centerline is the maximum velocity)

Volumetric flow rate = Qavg x π x D2 / 4 = 2.365 x π x 0.052 / 4 = 0.00464 m3/sec = 4.64 lit/sec

[Index]