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murzikaleks [220]
3 years ago
10

The equation for the velocity V in a pipe with diameter d and length L, under laminar condition is given by the equation V=Δpdsq

uare/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
Engineering
1 answer:
Citrus2011 [14]3 years ago
5 0

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.

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<em>f(z)</em> = 2 + (8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1)

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\displaystyle f(z) = 2 - \sum_{n=0}^\infty \left(6+2^{-n+1}\right) z^{-(n+1)}

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