Question

Nominal diameter of the pipe : 0.25 inch Select Venturi diameter : 0.3 ст Select type of Venturi Tube : Welded Tube Process FExperimental Data Venturi Diameter : 0.003 meter(s) Type of Venturi tube : Welded Tube Nominal Pipe Diameter : 0.25 inch ActuObservations Process Fluid Manometric Fluid Flowrate (Liter Per Minute) h1 (cm of Manometric Fluid) h2 (cm of Manometric Fluicm Data: Diameter of the pipe = d = .635|| Diameter of the venture throat= d= Density of CCL4 =P CCl4 = kg/m3 3 cm Density ofCalculations: Flow rate: A Hm= cm of manometric fluid. Volumetric flow rate = Q = m3/s (-p) ΔΗ, * ΔΗη P m of H2O Pm= density

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diameter of pipe, d=0.635 cm diameter of venturi throat, dt=0.3 cm density of cck=1.59 g/cc density of process fluid= 1g/cc vobservation table ! Flow rate (pm) h1 (cm) 1.58 35.8 2.67 37.2 3.75 39.3 5.5 44.1 h2 (cm) XHm (cm) XHw (cm) 34.2 1.6 0.944 32observation table II Q{m^3/sec) XHm (cm) Velocity (m/s) XHw (cm) N(reylond) Cd 2.64x10-3 1.6 3.7 0.944 12469 0.837796 4.46x10

Ignore the delta notation as it was some problem with the software used. as stated both table has been recreated with the calculation of each data required. if there is still any doubt, let me know in the comment section.

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