Question

The SI unit and dimensions of Reynolds
number
respectively
are​

Answer 1

Answer:

The si unit and dimensions of Reynolds number

cm ,m

Answer 2

Answer:

The Reynolds number is unitless and dimensionless. That it's unitless implies that it's no direct standard of measurement, the Reynolds number is set by the properties of fluid flow. The Reynolds number is dimensionless because the size of its defining quantities do away with in its defining equation irrespective of the units accustomed measure these quantities.

Explanation:

Reynolds Number$(R_e) = \text{Density }\times \text{ Velocity }\times \text{ Length }\times [\text{dynamic viscosity}]^{-1}$ . . . . . (1)

Since, Density $(\rho) = \text{Mass }\times [\text{Volume}]^{-1}$

The dimensional formula of density $= [M^1 L^{-3 }T^0]$ . . . . (2)

Since,$\text{ Velocity} =\text{ Displacement }\times [\text{Time}]^{-1}$

The dimensions of velocity $= [M^0 L^1 T^{-1}]$ . . . . (3)

$\text{Viscosity}=\text{Distance between layers }\times \text{ Force }\times [\text{Area }\times \text{ Velocity}]^{-1}$

Since, the dimensional formula of Force$= [M^1 L^1 T^{-2}]$

$\eta= L \times M^1 L^1 T^{-2 }\times [L^2 \times L^1 T^{-1}]^{-1}$

∴ the dimensional formula of viscosity $= [M^1 L^{-1} T^{-1}]$ . . . . (4)

Now,  substitute equation (2), (3) and (4) in equation (1), we get,

$(R_e) = \text{Density }\times \text{ Velocity }\times \text{ Length }\times [\text{dynamic viscosity}]^{-1}$

Or,

$R_e = [M^1 L^{-3 }T^0] \times [M^0 L^1 T^{-1}] \times [L] \times [M^1 L^{-1} T^{-1}]^{-1}\\ R_e= [M^0 L^0 T^0]$

Therefore, the Reynolds Number is dimensionally represented as $M^0L^0T^0$.

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