/*! 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 7 I* [aerodynamics] The power, \(P... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 7

I* [aerodynamics] The power, \(P\), required to drive an air screw of diameter \(D\) is given by \(P=2 \pi k \rho n^{3} D^{5}\) ( \(k=\) torque coefficient, \(\rho=\) density, \(n=\) number of revolutions per second). Make \(D\) the subject of the formula.

Short Answer

Expert verified
The formula for making D the subject is \[D = \left( \frac{P}{2 \pi k \rho n^{3}} \right)^{\frac{1}{5}}\]

Step by step solution

01

- Understand the Given Formula

The formula given is \[P = 2 \pi k \rho n^{3} D^{5}\]where we need to rearrange it to make \(D\) the subject.
02

- Isolate the Term With D

First, isolate the term with \(D^{5}\) by dividing both sides of the equation by \(2 \pi k \rho n^{3}\): \[\frac{P}{2 \pi k \rho n^{3}} = D^{5}\]
03

- Solve for D

To solve for \(D\), take the fifth root of both sides: \[D = \left( \frac{P}{2 \pi k \rho n^{3}} \right)^{\frac{1}{5}}\]

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 91影视!

Key Concepts

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

aerodynamics
In the field of aerodynamics, power calculations are crucial. Aerodynamics studies how air interacts with objects in motion, such as planes or air screws. In our exercise, we examine the power needed to drive an air screw. The formula used is essential in understanding how various factors like diameter, density, and revolutions per second impact power requirements. Developing a strong foundation in aerodynamics helps in designing efficient and effective air screw systems. By understanding the given formula, you can gain insights into optimizing performance and energy consumption in real-life engineering applications. This is a crucial skill for any aspiring engineer.
equation rearrangement
Rearranging equations is a fundamental skill in engineering mathematics. It involves manipulating a formula to make one variable the subject. In our exercise, starting with the formula P = 2 \, \text {pi} \, k \, rho \, n^3 \, D^5 our goal is to make D, the diameter, the subject of the formula.

Here's the step-by-step process:
  • First, isolate the term containing D by dividing both sides by the coefficients and variables not associated with D. Thus, we divide by 2 \, \text {pi} \, k \, rho \, n^3 creating a new equation: \[ \frac {P} {2 \, \text {pi} \, k \, rho \, n^3} = D^5 \]
  • Next, solve for D by taking the fifth root of both sides of the equation. This will give us: \[ D = \left( \frac {P}{2 \, \text {pi} \, k \, rho \, n^3} \right)^{\frac {1}{5}} \]


    • By following these steps, we successfully rearranged the formula to solve for D. Mastering equation rearrangement helps you tackle various types of problems and enhances your mathematical flexibility in engineering contexts.
power formula
The power formula used in aerodynamics often involves multiple variables that describe the conditions of the system. Our exercise uses the formula: P = 2 \, \text {pi} \, k \, rho \, n^3 \, D^5 where:
  • P is the power required.
  • k represents the torque coefficient.
  • \( rho \) is the density of the air.
  • n denotes the number of revolutions per second.
  • D is the diameter of the air screw.


The formula shows how power depends on these key variables. Notably, the diameter (D) has a significant impact since it is raised to the fifth power. Small changes in D can greatly influence the required power, making precise calculations essential. Understanding power formulas and their components helps in designing more effective and energy-efficient aerodynamic systems. This is fundamental for engineering tasks where optimization and performance are critically important.

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

[electronics] The impedance, \(Z\), of a circuit containing a resistor of resistance \(R\), a capacitor of capacitance \(C\) and an inductor of inductance \(L\) is given by $$ Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} $$ where \(X_{L}=2 \pi f L\) and \(X_{C}=\frac{1}{2 \pi f C}\) ( \(f\) represents frequency). Determine \(C\) if \(R=50 \Omega, Z=100 \Omega\), \(L=1\) henry and \(f=50 \mathrm{~Hz}\).

Show that the following are dimensionless parameters by checking that the dimensions of each are equal to 1 : a Reynolds Number \(=\frac{\rho v l}{\mu}\)Show that the following are dimensionless parameters by checking that the dimensions of each are equal to 1 : a Reynolds Number \(=\frac{\rho v l}{\mu}\)b Mach Number \(=\frac{v}{c}\) c Euler Number \(=\frac{p}{\rho v^{2}}\) d Froude Number \(=\frac{v}{\sqrt{g l}}\) e Weber Number \(=\frac{v^{2} l \rho}{\sigma}\) ( \(\rho\) is density, \(v\) is velocity, \(g\) is acceleration due to gravity, \(l\) is length, \(\mu\) is viscosity, \(p\) is pressure, \(c\) is speed of sound and \(\sigma\) is surface tension whose units are \(\mathrm{N} / \mathrm{m}\).)

Given that $$ c=\sqrt{a^{2}+b^{2}} $$ evaluate \(c\) for \(a=24\) and \(b=7\).

[mechanics] The airflow over a vehicle causes drag \(D\), which is given by $$ D=\frac{1}{2} \rho C v^{2} A $$ where \(\rho\) is density, \(C\) is drag coefficient, \(v\) is velocity and \(A\) is the frontal area of the vehicle. Make \(A\) the subject of the formula.

[mechanics] The displacement, s, of a body is given by $$ s=u t+\frac{1}{2} a t^{2} $$ where \(t\) is time, \(u\) is initial velocity and \(a\) is acceleration. If at \(t=2 \mathrm{~s}\) then \(s=33 \mathrm{~m}\) and at \(t=3 \mathrm{~s}\) then \(s=64.5 \mathrm{~m}\), find the initial velocity \((u)\) and acceleration \((a)\).
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Statistics

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.