Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 2

The Mach number is a dimensionless ratio of the velocity of an object in a fluid to the speed of sound in the fluid. For an airplane flying at velocity \(V\) in air at absolute temperature \(T,\) the Mach number Ma is \\[ \mathrm{Ma}=\frac{V}{\sqrt{k R T}} \\] where \(k\) is a dimensionless constant and \(R\) is the specific gas constant for air. Show that Ma is dimeasionless.

Short Answer

Expert verified
The Mach number, Ma, is dimensionless. This is confirmed by the dimensions of the right-hand side of the equation for Ma, which are also found to be dimensionless.

Step by step solution

01

List all variables with their dimensions

First, list down all the variables in the equation and their corresponding dimensions. The velocity (\(V\)) has dimensions of length/time or [L/T], the absolute temperature (\(T\)) has dimensions of temperature [K], \(k\) is stated to be a dimensionless constant, and \(R\), the specific gas constant, has dimensions of energy per temperature per amount of substance [E/(Ta*N)], where E represents energy, Ta represents temperature, and N represents amount of substance.
02

Simplify the equation

The given equation is: \(\mathrm{Ma}=\frac{V}{\sqrt{k R T}}\). Since \(k\) is dimensionless, it has no effect on the overall dimensions of the equation. So, we simplify the equation to: \(\mathrm{Ma}=\frac{V}{\sqrt{R T}}\)
03

Calculate the dimensions of the right-hand side

To show Ma is dimensionless, we need to calculate the dimensions of the right-hand side of the equation. The dimensions of the right-hand side are: \(\frac{[L/T]}{\sqrt{[E/(Ta*N)]*[K]}}=\frac{[L/T]}{\sqrt{[E/(Ta*N)]}}\), as dimensions of temperature cancel out. After simplifying, the right-hand side gives [M^0L^0T^0], which shows it is dimensionless.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Vehicle shock absorbers damp out oscillations caused by road roughness. Describe how a temperature change may affect the operation of a shock absorber.

The force, \(F\), of the wind blowing against a building is given by \(F=C_{D} \rho V^{2} A / 2,\) where \(V\) is the wind speed, \(\rho\) the density of the air, \(A\) the cross-sectional area cf the building, and \(C_{D}\) is a constant termed the drag coefficient. Determine the dimensions of the drag coefficient.

Assume that the air volume in a small automobile tire is constant and equal to the volume between two concentric cylinders \(13 \mathrm{cm}\) high with diameters of \(33 \mathrm{cm}\) and \(52 \mathrm{cm}\). The air in the tire is initially at \(25^{\circ} \mathrm{C}\) and \(202 \mathrm{kPa}\). Immediately after air is pumped into the tire, the temperature is \(30^{\circ} \mathrm{C}\) and the pressure is 303 kPa. What mass of air was added to the tire? What would be the air pressure after the air has cooled to a temperature of \(0^{\circ} \mathrm{C} ?\)

Express the following quantities in SI units: (a) 10.2 in./min, (b) 4.81 slugs,(c) \(3.02 \mathrm{lb}\) (d) \(73.1 \mathrm{ft} / \mathrm{s}^{2}\) (e) \(0.0234 \mathrm{lb} \cdot \mathrm{s} / \mathrm{ft}^{2}\).

The universal gas constant \(R_{0}\) is equal to \(49,700 \mathrm{ft}^{2} /\left(\mathrm{s}^{2} \cdot^{\circ} \mathrm{R}\right)\) or \(8310 \mathrm{m}^{2} /\left(\mathrm{s}^{2} \cdot \mathrm{K}\right) .\) Show that these two magnitudes are equal.
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Turning Points in Physics

Read Explanation

Famous Physicists

Read Explanation

Engineering Physics

Read Explanation

Torque and Rotational Motion

Read Explanation
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.

Sign-up for free