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Wittaler [7]
3 years ago
8

Determine the Reynolds number for a flow of 0.2 m^3/s through a 203 mm inner diameter circular pipe of a fluid with rho=680 kg/m

^3 ,μ=3.1*10^(-4) ((N-s)/m^2 ),and σ=0.022 N/m.Is the flow laminar or turbulent?
Physics
1 answer:
IrinaK [193]3 years ago
3 0

To find a solution to this problem it is necessary to apply the concepts related to the Reynolds number and its definitions on the type of fluid.

A Reynolds number less than 2000 considers the laminar fluid, while a Reynolds number greater than 4000 is considered a turbulent fluid. (The intermediate between the two values would be a transient fluid)

The mathematical equation that defines the Reynolds number is given by

Re = \frac{\rho V D}{\mu}

Where

\rho = Density

V= Velocity

D= Diameter

\mu = Viscosity

Our values are given as

Q = 0.2m^3/s

D = 203*10^{-3}m

\rho = 680kg/m^3

\mu = 3.1*10^{-4}Ns/m^2

\sigma = 0.022N/m

The velocity can be find through the Discharge equation,

Q = VA

Where

V = Velocity

A = Area

Replacing,

0.2 = V* (2\pi*(\frac{203*10^{-3}}{2})^2)

V = 3.08m/s

Replacing at the Reynolds equation,

Re = \frac{\rho VD}{\mu}

Re = \frac{680*3.08*203*10^{-3}}{3.1*10^{-4}}

Re = 1.37*10^6

Since Reynolds' number is greater than 4000, then we consider this a turbulent fluid.

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