text: ### 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]
explanation: The document describes a problem involving the measurement of the volumetric flow rate of a Newtonian fluid using a pitot tube. It includes the given data, relevant formulas, and step-by-step calculations to determine the volumetric flow rate. The document contains both a question and an answer.
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distilabel_metadata.content: ### 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]
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document describes a problem involving the measurement of the volumetric flow rate of a Newtonian fluid using a pitot tube. It includes the given data, relevant formulas, and step-by-step calculations to determine the volumetric flow rate. The document contains both a question and an answer.
  
  Label: question-and-answer