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Chapter 1: Problem 10

If \(V\) is a velocity, \(\ell\) a length, and \(\nu\) a fluid property (the kinematic viscosity) having dimensions of \(L^{2} T^{-1},\) which of the following combinations are dimensionless: (a) \(V \ell \nu\) (b) \(V \ell / \nu\) \((\mathbf{c}) V^{2} \nu\) (d) \(V / \ell \nu ?\)

Short Answer

Expert verified
Only the quantity \(V / \ell \nu\) is dimensionless.

Step by step solution

01

Identify the dimensions of given quantities

Velocity \(V\) has dimensions of \([L][T^{-1}]\), length \(\ell\) has dimensions of \([L]\) and kinematic viscosity \(\nu\) has dimensions of \([L^{2}][T^{-1}]\).
02

Check the dimensions of quantity (a)

The dimensions of quantity \(V \ell \nu\) would be \([L][T^{-1}]\cdot[L]\cdot[L^{2}][T^{-1}]=[L^{4}][T^{-2}]\), which is not dimensionless.
03

Check the dimensions of quantity (b)

The dimensions of quantity \(V \ell / \nu\) would be \([L][T^{-1}]\cdot[L] / [L^{2}][T^{-1}]=[L^{1}][T^{0}]\), which is not dimensionless.
04

Check the dimensions of quantity (c)

The dimensions of quantity \(V^{2} \nu\) would be \([L^2][T^{-2}]\cdot[L^{2}][T^{-1}]=[L^{4}][T^{-3}]\), which is not dimensionless.
05

Check the dimensions of quantity (d)

The dimensions of quantity \(V / \ell \nu\) would be \([L][T^{-1}]/[L]\cdot[L^{2}][T^{-1}]=[L^{1}][T^{-1}]=[\emptyset]\), which is dimensionless.

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Most popular questions from this chapter

A liquid has a specific weight of \(59 \mathrm{lb} / \mathrm{ft}^{3}\) and a dynamic viscosity of \(2.75 \mathrm{lb} \cdot \mathrm{s} / \mathrm{ft}^{2}\). Determine its kinematic viscosity.

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