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The engine delivers 175 hp to an aircraft propeller at 2400 rev/min. (a) How much torque does the aircraft engine provide? (b) How much work does the engine do in one revolution of the propeller?

Short Answer

Expert verified
(a) Torque is approximately 518.4 Nm. (b) Work done per revolution is approximately 3256.4 J.

Step by step solution

01

Convert Horsepower to Watts

First, convert the engine's power from horsepower to watts. We know that 1 horsepower (hp) is equivalent to approximately 746 watts. Therefore, for an engine delivering 175 hp, the power in watts is calculated as follows:\[ P = 175 \times 746 = 130550 \text{ W} \]
02

Calculate Angular Velocity

The angular velocity in radians per second \( \omega \) is given by the formula \( \omega = 2\pi \times \text{revolutions per second} \). The engine operates at 2400 revolutions per minute (rev/min). First, convert this to revolutions per second (rev/s) by dividing by 60:\[ \omega = \frac{2400}{60} = 40 \text{ rev/s} \]Now, convert this to radians per second:\[ \omega = 40 \times 2\pi = 80\pi \text{ rad/s} \]
03

Calculate Torque Using Power and Angular Velocity

Torque \( \tau \) is related to power \( P \) and angular velocity \( \omega \) through the equation:\[ P = \tau \times \omega \]Solve for torque \( \tau \):\[ \tau = \frac{P}{\omega} = \frac{130550}{80\pi} \approx 518.4 \text{ Nm} \]
04

Calculate Work Done in One Revolution

The work done in one revolution of the propeller is equal to the torque times the angle through which it turns in radians, which for one complete revolution is \( 2\pi \) radians. Thus, work \( W \) is:\[ W = \tau \times 2\pi = 518.4 \times 2\pi \approx 3256.4 \text{ J} \]

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Key Concepts

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

Torque Calculation
Torque is a measure of the rotational force that causes an object to turn or spin about an axis. In simple terms, it's the twist applied to make an object rotate.
In the context of our exercise, the torque produced by the aircraft engine is derived from the relationship between power and angular velocity.
You need to know:
  • Power: The engine's power in watts.
  • Angular Velocity: The speed of rotation in radians per second.
The formula connecting these is: \[ P = \tau \times \omega \]Here, \( \tau \) is torque while \( \omega \) is angular velocity. To find torque, the equation is rearranged to: \[ \tau = \frac{P}{\omega} \]

Practical Insight

For students solving problems of this kind, remember:
  • Torque is proportional to power, so more powerful engines can produce more torque.
  • Ensure angular velocity is in radians per second for accurate calculations.
Angular Velocity
Angular velocity is a fundamental concept in rotational dynamics. It measures how fast an object rotates and is expressed in radians per second (rad/s).
To find angular velocity, you need the rotational speed in revolutions per second and convert it to radians using the relation: \[ \omega = 2\pi \times \text{revolutions per second} \]Let's walk through the significance of this concept.

Conversion Steps

  • Converting from rev/min to rev/s by dividing by 60 (since there are 60 seconds in a minute).
  • Multiply by \( 2\pi \) to convert revolutions to radians.

Understanding its Role

Angular velocity is crucial because it directly influences the torque. Higher angular velocities mean that the system is rotating faster, which can increase the stress on components, impacting the torque calculation.
Work Done
Work done in a rotational setting is similar to linear work but takes into account the rotation of an object around a fixed axis. It's the energy transferred when an object is rotated, defined as the product of torque and the rotational angle.
In this problem, the work done by the engine in one revolution is essential for understanding its energy efficiency. The formula used is:\[ W = \tau \times \theta \]where \( \theta \) is the angle in radians, and for a full revolution, it's \( 2\pi \).

Real-World Relevance

  • Powerful engines are evaluated by the work done per revolution, telling us how efficiently the energy is being used.
  • In practical scenarios, such as aircraft engines, understanding this helps in assessing fuel economy and performance.

Key Takeaways

Keep in mind:
  • Work in rotations is analogous to linear work—both involve force and distance, just in different forms.
  • Efficiency of an engine can be deduced from the work done, making it a critical metric for performance analysis.

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Most popular questions from this chapter

A thin rod of length \(l\) lies on the \(+x\) -axis with its left end at the origin. A string pulls on the rod with a force \(\overrightarrow{\boldsymbol{F}}\) directed toward a point \(P\) a distance \(h\) above the rod. Where along the rod should you attach the string to get the greatest torque about the origin if point \(P\) is (a) above the right end of the rod? (b) Above the left end of the rod? (c) Above the center of the rod?

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An electric motor consumes 9.00 \(\mathrm{kJ}\) of electrical energy in 1.00 \(\mathrm{min}\) . If one-third of this energy goes into heat and other forms of internal energy of the motor, with the rest going to the motor output, how much torque will this engine develop if you run it at 2500 \(\mathrm{rpm} ?\)

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