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Chapter 1: Problem 61

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.

Short Answer

Expert verified
The kinematic viscosity of the liquid is \(1.5 \mathrm{ft}^{2} / \mathrm{s}\).

Step by step solution

01

Determine the Fluid Density

The specific weight is given as \(59 \mathrm{lb} / \mathrm{ft}^{3}\). The gravity on earth is \(32.2 \mathrm{ft} / \mathrm{s}^{2}\). We use the relationship specific weight = density * gravity to calculate the density, \(\rho\). Rearranging the formula gives us: \(\rho = \) specific weight / gravity = \(59 \mathrm{lb} / \mathrm{ft}^{3}\) / \(32.2 \mathrm{ft} / \mathrm{s}^{2}\) = \(1.832 \mathrm{slugs} / \mathrm{ft}^{3}\).
02

Calculate the Kinematic Viscosity

The dynamic viscosity, \(\mu\), is given as \(2.75 \mathrm{lb} \cdot \mathrm{s} / \mathrm{ft}^{2}\). The kinematic viscosity, \(\nu\), is the ratio of dynamic viscosity to density. Written as a formula, it is \(\nu = \mu / \rho\). Inputting the values we calculated and were given, we have: \(\nu = 2.75 \mathrm{lb} \cdot \mathrm{s} / \mathrm{ft}^{2} / 1.832 \mathrm{slugs} / \mathrm{ft}^{3} = 1.5 \mathrm{ft}^{2} / \mathrm{s}\).

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