A Newtonian fluid having a viscosity of 1.23 poise, and a density of 0.893 gm/cm^{3}, 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/cm^{3 }= 893 kg/m^{3}

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

Manometer reading (h_{m}) = 8 cm of Hg = 0.08 m of Hg

Formulae:

Velocity at the center of pipe

v_{o} = ( 2 x (p_{s} - p_{o})/ ρ)^{0.5} = (2 x h_{m} x (ρ_{m} - ρ) x g /ρ)^{0.5}

Calculations:

V_{o} = ( 2 x 0.08 x (13.6 - 0.893) x 9.812 / 0.893 )^{0.5} = 4.73 m/sec

NRe = Dv_{o}ρ/μ = 0.05 x 4.73 x 893 / 0.123 = 1717

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

v_{avg} = v_{o} / 2 = 4.73 / 2 = 2.365 m/sec (velocity at the centerline is the maximum velocity)

Volumetric flow rate = Q_{avg} x π x D^{2} / 4 = 2.365 x π x 0.05^{2} / 4 = 0.00464 m^{3}/sec = **4.64 lit/sec**

Last Modified on: 04-Feb-2022

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