Question #37310

water flows at a speed 5cm /s through a pipe of radius 2cm ,the viscosity of water is 0.001 pl .the reynolds number and the nature of flow are respectively ?????????????????/

Expert's answer

Reynolds number can be found as:

R= v*L/nu

where v is speed of water, L is characteristic linear dimension (diameter in our case = 4 cm) and nu is viscosity. Thus we find

R = 0.05 m/s * 2*0.02 m / (0.001 kg/m/s) = 2

Flow with such Reynolds number is laminar.

R= v*L/nu

where v is speed of water, L is characteristic linear dimension (diameter in our case = 4 cm) and nu is viscosity. Thus we find

R = 0.05 m/s * 2*0.02 m / (0.001 kg/m/s) = 2

Flow with such Reynolds number is laminar.

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