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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

Respuesta :

Answer:

Given that

[tex]V=\dfrac{\Delta Pd^2}{32\mu L}[/tex]

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

Now find the dimension of RHS

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

Dimension of d=  [tex][M^{0}L^{1}T^{0}][/tex].

Dimension of μ =  [tex][ML^{-1}T^{-1}][/tex].

Dimension of L=  [tex][M^{0}L^{1}T^{0}][/tex].

So

[tex]\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}]}[/tex]

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

It means that both sides have same dimensions.

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