/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 A liquid has a specific weight o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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: \(\nu = 2.75 \mathrm{lb} \cdot \mathrm{s} / \mathrm{ft}^{2} / 1.832 \mathrm{slugs} / \mathrm{ft}^{3} = 1.5 \mathrm{ft}^{2} / \mathrm{s}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematic Viscosity
Kinematic viscosity is a property of fluids that describes their tendency to flow under the influence of gravity. Essentially, it reflects how easily a fluid flows in the absence of external forces like pressure. It combines the fluid's dynamic viscosity and density into a single value, hence providing insight into the fluid's flow characteristics.To calculate kinematic viscosity, the formula used is: \[u = \frac{\mu}{\rho}\]- \(u\) represents the kinematic viscosity in \(\mathrm{ft}^2/\mathrm{s}\)- \(\mu\) is the dynamic viscosity - \(\rho\) is the fluid densityA higher kinematic viscosity indicates a fluid that flows less readily, while a lower kinematic viscosity suggests a more easily flowing fluid. In the context of the given exercise, using the dynamic viscosity of \(2.75 \mathrm{lb} \cdot \mathrm{s} / \mathrm{ft}^2\) and the previously determined density of \(1.832 \mathrm{slugs} / \mathrm{ft}^3\), you can compute \(u\) as \(1.5 \mathrm{ft}^2 / \mathrm{s}\). This result tells you how the fluid will behave when it flows.
Specific Weight
Specific weight is a fundamental property of fluids which defines the weight of a unit volume of the fluid. It is particularly useful in understanding how fluids are influenced by gravitational forces. Specific weight is given by the formula:\[\gamma = \rho \cdot g\]- \(\gamma\) represents the specific weight- \(\rho\) is the density of the fluid- \(g\) is the acceleration due to gravityIn this exercise, the specific weight of the liquid is given as \(59 \mathrm{lb} / \mathrm{ft}^3\). When you know the specific weight and the force of gravity (Earth's gravity is typically \(32.2 \mathrm{ft} / \mathrm{s}^2\)), you can rearrange the formula to find the density. This is a crucial first step before you can determine other properties like kinematic viscosity.
Dynamic Viscosity
Dynamic viscosity is an important measure of a fluid's internal resistance to flow. Unlike kinematic viscosity, dynamic viscosity is typically represented with units of \(\mathrm{lb} \cdot \mathrm{s} / \mathrm{ft}^2\) and is a direct indicator of how much energy is required to move one part of the fluid relative to another.The formula for dynamic viscosity is:\[\mu = u \cdot \rho\]- \(\mu\) represents the dynamic viscosity- \(u\) is the kinematic viscosity- \(\rho\) is the density of the fluidIn the given problem, the dynamic viscosity was provided as \(2.75 \mathrm{lb} \cdot \mathrm{s} / \mathrm{ft}^2\). Knowing this, along with the fluid density, allows you to successfully compute the kinematic viscosity. Dynamic viscosity is fundamental in understanding how a fluid flows under applied stress, not just gravity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The temperature and pressure at the surface of Mars during a Martian spring day were determined to be \(-50^{\circ} \mathrm{C}\) and \(900 \mathrm{Pa}\). respectively. (a) Determine the density of the Martian atmosphere for these conditions if the gas constant for the Martian atmosphere is assumed to be equivalent to that of carbon dioxide. (b) Compare the answer from part (a) with the density of the Earth's atmosphere during a spring day when the temperature is \(18^{\circ} \mathrm{C}\) and the pressure \(101.6 \mathrm{kPa}(\mathrm{abs})\).

A piston having a diameter of 5.48 in. and a length of 9.50 in. slides downward with a velocity \(V\) through a vertical pipe. The downward motion is resisted by an oil film between the piston and the pipe wall. The film thickness is 0.002 in., and the cylinder weighs 0.5 lb. Estimate \(V\) if the oil viscosity is \(0.016 \mathrm{lb} \cdot \mathrm{s} / \mathrm{ft}^{2}\) Assume the velocity distribution in the gap is linear.

An important dimensionless parameter in certain types of fluid flow problems is the Froude number defined as \(V / \sqrt{g \ell}\) where \(V\) is a velocity, \(g\) the acceleration of gravity, and \(\ell\) a length. Determine the value of the Froude number for \(V=10 \mathrm{ft} / \mathrm{s}\) \(g=32.2 \mathrm{ft} / \mathrm{s}^{2},\) and \(\ell=2 \mathrm{ft} .\) Recalculate the Froude number using SI units for \(V, g,\) and \(\ell .\) Explain the significance of the results of these calculations.

The kirematic viscosity of oxygen at \(20^{\circ} \mathrm{C}\) ard a pressure of \(150 \mathrm{kPa}(\mathrm{abs})\) is 0.104 stokes. Determine the dynamic viscosity of oxygen at this temperature and pressure.

The information on a can of pop indicates that the can contains \(355 \mathrm{mL}\). The mass of a full can of pop is \(0.369 \mathrm{kg}\), while an empty can weighs 0.153 N. Determine the specific weight, density, and specific gravity of the pop and compare your results with the corresponding values for water at \(20^{\circ} \mathrm{C}\). Express your results in SI units.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.