Question

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?

Answer 1

flow is laminar

Step-by-step explanation:

we know that reynold number is given as

`Re =(\rho VD)/(\mu)`

where V is velocity

`V = \sqrt {2gh}`

`V = \sqrt {2*9.81*h}`

WE KNOW THAT

`Pressure = \rho gh`

hence

`h = (pressure)/(\rho g)`

`h = (95.8)/(860*9.81)`

`h = 0.01135 m`

`V = \sqrt {2*9.81*0.01135}`

`V = 0.472 m/s`

`Re =(860*0.472*.021)/(0.0103)`

Re = 827.60 which is less than 2000

therefore ,flow is laminar