Dimensional Analysis

Question

For a fluid of density ρ flowing with speed v through a tube of characteristic length L and having coefficient of viscosity η, which combination is dimensionless like the Reynolds number?

RankGuru · Accepted Answer

A dimensionless group must reduce entirely to [M^0 L^0 T^0], and the Reynolds number is the classic example governing the onset of turbulence. Write the dimensions: density ρ = [M L^-3], speed v = [L T^-1], length L = [L], and viscosity η = [M L^-1 T^-1]. For the combination ρ v L / η, the numerator is [M L^-3][L T^-1][L] = [M L^-1 T^-1], and dividing by η = [M L^-1 T^-1] gives [M^0 L^0 T^0], confirming it is dimensionless. The choice ρ v / (L η) introduces an extra division by length, leaving dimension [L^-1] and so is not dimensionless. The choice η v L / ρ gives [M L^-1 T^-1][L T^-1][L]/[M L^-3] = [L^4 T^-2], clearly dimensional. The choice ρ L / (v η) yields [M L^-3][L]/([L T^-1][M L^-1 T^-1]) = [L^-2 T^2], also dimensional. This identification mirrors the JEE Advanced treatment of dimensionless numbers in fluid dynamics. A check confirms only ρ v L / η cancels all base dimensions, matching the standard Reynolds-number definition.

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