# Pipe Friction Lab Report

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Pipe Friction Test Your Name Your Number Date of experiment: Contents Introduction 3 Objective 4 Apparatus 4 Procedure 5 Theory 5 Results 6 Sample Calculation 7 Discussion 8 Conclusion 11 Bibliography 11 Introduction The fluid flow inside a duct- Rectangular or Circular- subjects to loss of energy due to the friction application between the duct material and the fluid layers attached to the surface of the duct. The friction forces have a dynamic nature which oscillate according to certain flow parameters and conditions. Generally, the friction forces is measured as the relative force between the inertia forces of fluids movement and the viscous forces between fluid layers to resist the motion of the fluid…show more content…
Water and mercury manometers are in ready conditions. For measuring laminar flow: Open the reservoir valve and close the hydraulic bench water supply pipe valve to direct the water flow to the reservoir. Adjust the water level in the reservoir to the required level to obtain laminar flow. Open the test pipe valve to the water flow from the reservoir Open the outlet valve to obtain a water flow rate Record manometers reading at the inlet and exit of the test pipe. Measure the time consumed by the stopwatch for the water volume flowed from the reservoir into the exit beaker which is graduated with the water volume consumed. Close the exit valve A different flow rate should apply to the test tube and the measuring pressures, time and water volume to be recorded. Repeat the same previous procedures with different valve openings. For measuring turbulent flow: Close the valve of the water source to the reservoir and the valve from the reservoir to the test pipe. Open the valve from the Hydraulic Bench water supply to the test pipe directly. Open the exit valve to the beaker with a certain…show more content…
ReD = ρVD/µ = 1000X0.31X0.003/(8.9 X 10-4) = 1044 From Moody Chart with ReD = 1044, Smooth pipe curve, f = 16/ReD = 16/1044 =0.016 hL = 4 f (LV^2)/2gD The friction factor: f =hLX2gD/(4LV^2 ) = 0.53 X 2X9.8X0.003/(4X0.51X〖(0.31)〗^2 )=