Question

The equation for the velocity V in a pipe with diameter d and length L, under laminar condition is given by the equation V=Δpdsquare/32µL, with V=Δp the pressure drop and Δp the viscosity of the fluid. Determine whether the equation is dimensionally consistent by inspecting the dimensions on both sides

Answer 1

Answer:

Given that

$V=\dfrac{\Delta Pd^2}{32\mu L}$

LHS of above given equation have dimension $[M^oL^{1}T^{-1}]$.

Now find the dimension of RHS

Dimension of P = $[ML^{-1}T^{-2}]$.

Dimension of d=  $[M^{0}L^{1}T^{0}]$.

Dimension of μ =  $[ML^{-1}T^{-1}]$.

Dimension of L=  $[M^{0}L^{1}T^{0}]$.

So

$\dfrac{\Delta Pd^2}{32\mu L}=\dfrac{[ML^{-1}T^{-2}].[M^{0}L^{1}T^{0}]^2}{[ML^{-1}T^{-1}].[M^{0}L^{1}T^{0}]}$

$\dfrac{\Delta Pd^2}{32\mu L}=[M^0L^{1}T^{-1}]$

It means that both sides have same dimensions.