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A Pitot-static probe is mounted in a 2.1-cm-inner diameter pipe at a location where the local velocity is approximately equal to the average velocity. The oil in the pipe has density ρ = 860 kg/m3 and viscosity μ = 0.0103 kg/m⋅s. The pressure difference is measured to be 95.8 Pa. Calculate the Reynolds number of the flow. Is it laminar or turbulent?

Respuesta :

Answer:

flow is laminar

Explanation:

we know that reynold number is given as

[tex]Re =\frac{\rho VD}{\mu}[/tex]

where V is velocity

[tex]V = \sqrt {2gh}[/tex]

[tex]V = \sqrt {2*9.81*h}[/tex]

WE KNOW THAT

[tex]Pressure = \rho gh[/tex]

hence

[tex]h = \frac{pressure}{\rho g}[/tex]

[tex]h = \frac{95.8}{860*9.81}[/tex]

[tex]h = 0.01135 m[/tex]

[tex] V = \sqrt {2*9.81*0.01135}[/tex]

[tex]V = 0.472 m/s[/tex]

[tex]Re =\frac{860*0.472*.021}{0.0103}[/tex]

Re = 827.60 which is less than 2000

therefore ,flow is laminar

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