Show that the Reynolds number is dimensionless.
The formula for the Reynolds number $(R_{e})$ is given by: $R_{e} = \frac{\rho v d}{\eta}$
Where:
$\rho$ (density) = $[M^{1} L^{-3} T^{0}]$
$v$ (velocity) = $[M^{0} L^{1} T^{-1}]$
$d$ (diameter) = $[L^{1}]$
$\eta$ (coefficient of viscosity) = $[M^{1} L^{-1} T^{-1}]$
Substituting the dimensions into the formula:
$R_{e} = \frac{[M^{1} L^{-3} T^{0}] [M^{1} L^{1} T^{-1}] [L^{1}]}{[M^{1} L^{-1} T^{-1}]}$
$R_{e} = \frac{[M^{1} L^{-1} T^{-1}]}{[M^{1} L^{-1} T^{-1}]}$
$R_{e} = [M^{0} L^{0} T^{0}]$
Since all the powers of the fundamental dimensions are zero,the Reynolds number is dimensionless.
Where:
$\rho$ (density) = $[M^{1} L^{-3} T^{0}]$
$v$ (velocity) = $[M^{0} L^{1} T^{-1}]$
$d$ (diameter) = $[L^{1}]$
$\eta$ (coefficient of viscosity) = $[M^{1} L^{-1} T^{-1}]$
Substituting the dimensions into the formula:
$R_{e} = \frac{[M^{1} L^{-3} T^{0}] [M^{1} L^{1} T^{-1}] [L^{1}]}{[M^{1} L^{-1} T^{-1}]}$
$R_{e} = \frac{[M^{1} L^{-1} T^{-1}]}{[M^{1} L^{-1} T^{-1}]}$
$R_{e} = [M^{0} L^{0} T^{0}]$
Since all the powers of the fundamental dimensions are zero,the Reynolds number is dimensionless.
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