/*! 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 25 The driveshaft of a building's a... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 25

The driveshaft of a building's air-handling fan is turned at 300 RPM by a belt running on a 0.3-m-diameter pulley. The net force applied by the belt on the pulley is \(2,000 \mathrm{~N}\). Determine the torque applied by the belt on the pulley, in \(\mathrm{N} \cdot \mathrm{m}\), and the power transmitted, in \(\mathrm{kW}\).

Short Answer

Expert verified
The torque is 300 N路m, and the power transmitted is 9.426 kW.

Step by step solution

01

Determine the Radius of the Pulley

The diameter of the pulley is given as 0.3 meters. The radius is half of the diameter. Calculate the radius (r) as: \[ r = \frac{0.3 \, \text{m}}{2} = 0.15 \, \text{m} \]
02

Calculate the Torque Applied

Torque (蟿) is calculated using the formula: \[ \tau = F \times r \] Given, the force (F) applied by the belt on the pulley is 2000 N and the radius (r) is 0.15 m. Substitute these values into the formula: \[ \tau = 2000 \, \mathrm{N} \times 0.15 \, \mathrm{m} = 300 \, \mathrm{N \cdot m} \]
03

Convert RPM to Radians Per Second

The rotational speed is given as 300 RPM. First, convert RPM to radians per second (rad/s) using the formula: \[ \omega = \text{RPM} \times \frac{2 \pi}{60} \] Substitute the given RPM value: \[ \omega = 300 \, \text{RPM} \times \frac{2 \pi}{60} = 31.42 \, \text{rad/s} \]
04

Calculate the Power Transmitted

Power (P) transmitted can be calculated using the formula: \[ P = \tau \times \omega \] Where \( \tau \) is the torque (300 N路m) and \( \omega \) is the angular velocity (31.42 rad/s). Substitute these values into the formula: \[ P = 300 \, \mathrm{N \cdot m} \times 31.42 \, \text{rad/s} = 9426 \, \text{W} \] Since power is typically expressed in kilowatts (kW), convert watts to kilowatts: \[ P = 9426 \, \mathrm{W} \times \frac{1 \mathrm{kW}}{1000 \mathrm{W}} = 9.426 \, \text{kW} \]
05

Summarize the Results

The torque applied by the belt on the pulley is 300 N路m, and the power transmitted is 9.426 kW.

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Key Concepts

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

Torque
Torque is a measure of the rotational force acting on an object. It determines how much an object will rotate about a pivot point. Here, we calculate torque using: \text \( \tau = F \times r \), where \( F \) is the applied force and \( r \) is the radius.
  • The force applied by the belt is 2000 N.
  • The pulley radius is 0.15 m.
Thus, the torque is \( 2000 \, \text{N} \times 0.15 \, \text{m} = 300 \, \text{N} \cdot \text{m} \). This torque rotates the pulley, powering the fan. Torque is crucial for understanding how forces cause rotational motion.
Power Transmission
Power transmission refers to the transfer of energy from one place to another, crucial in machinery and mechanical systems. It's the rate at which work is done or energy is transmitted.In this example, the driveshaft transmits power through the belt and pulley system to the fan.To calculate the power transmitted, we use the formula:
  • \( P = \tau \times \omega \)
Where \( \tau \) is torque (300 \( \text{N} \cdot \text{m} \)) and \( \omega \) is angular velocity (31.42 rad/s).
  • Thus, \( P = 300 \, \text{N} \cdot \text{m} \times 31.42 \, \text{rad/s} = 9426 \, \text{W} \).
Converting to kilowatts: \( 9426 \, \text{W} \times \frac{1 \text{kW}}{1000 \text{W}} = 9.426 \text{kW} \). Power transmission ensures efficient operation of mechanical systems.
Rotational Speed
Rotational speed measures how fast an object spins. It's typically given in revolutions per minute (RPM). Here, the driveshaft rotates at 300 RPM.
  • To convert RPM to radians per second, use \( \omega = \text{RPM} \times \frac{2 \pi}{60} \).
  • For 300 RPM: \( 300 \, \text{RPM} \times \frac{2 \pi}{60} = 31.42 \, \text{rad/s} \).
Understanding rotational speed helps us analyze dynamics and design of rotating systems. It's essential for calculating angular velocity, which is used for determining power and system performance.
Force Calculation
Force calculation is essential in mechanics. In this exercise, we need the force applied by the belt to calculate torque and power. The force (\( F \)) is given as 2000 N.Force influences mechanical motion and performance.
  • With the given force and radius, torque is calculated using \( \tau = F \times r \).
  • The radius is 0.15 m.
  • Therefore, torque \( \tau \) is \( 2000 \, \text{N} \times 0.15 \, \text{m} = 300 \, \text{N} \cdot \text{m} \).
Understanding force and its application lets us design effective machines and systems.
Angular Velocity
Angular velocity (\( \omega \)) shows how fast an object rotates, measured in radians per second (rad/s). It's converted from RPM using:
  • \( \omega = \text{RPM} \times \frac{2 \pi}{60} \).
  • For 300 RPM: \( 300 \, \text{RPM} \times \frac{2 \pi}{60} = 31.42 \, \text{rad/s} \).
Angular velocity is vital for calculating rotational dynamics and power.We use it in the power formula: \( P = \tau \times \omega \).
  • With \( \tau \) as 300 \( \text{N} \cdot \text{m} \) and \( \omega \) as 31.42 rad/s, the power is \( 9426 \)W.
It鈥檚 essential in engineering to study rotational behaviors and optimize system efficiency.

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Most popular questions from this chapter

An object whose mass is \(50 \mathrm{~kg}\) experiences a decrease in kinetic energy of \(0.80 \mathrm{~kJ}\) and an increase in potential energy of \(4 \mathrm{~kJ}\) The initial velocity and elevation of the object, each relative to the surface of the earth, are \(15 \mathrm{~m} / \mathrm{s}\) and \(12 \mathrm{~m}\), espectively. If \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\), determine (a) the final velocity, in \(\mathrm{m} / \mathrm{s}\) (b) the final elevation, in \(\mathrm{m}\)

In a constant pressure process gas expands from initial volume \(\left(V_{1}\right)\) of \(0.01 \mathrm{~m}^{3}\) to the final volume \(\left(V_{2}\right)\) of \(0.03 \mathrm{~m}^{3}\). The pressure of the system is 2 bar. Determine the work done, by the gas.

An air-conditioning unit with a coefficient of performance of \(2.93\) provides \(5,275 \mathrm{~kJ} / \mathrm{h}\) of cooling while operating during the cooling season 8 hours per day for 125 days. If you pay 10 cents per \(\mathrm{kW} \cdot \mathrm{h}\) for electricity, determine the cost, in, dollars, for the cooling season.

A vertical cylindrical mass of \(5 \mathrm{~kg}\) undergoes a process during which the velocity decreases from \(30 \mathrm{~m} / \mathrm{s}\) to \(15 \mathrm{~m} / \mathrm{s}\), while the elevation remains unchanged. The initial specific internal energy of the mass is \(1.2 \mathrm{~kJ} / \mathrm{kg}\) and the final specific internal energy is \(1.9 \mathrm{~kJ} / \mathrm{kg}\). During the process, the mass receives \(2 \mathrm{~kJ}\) of energy by heat transfer through its bottom surface and loses \(1 \mathrm{~kJ}\) of energy by heat transfer through its top surface. The lateral surface experiences no heat transfer. For this process, evaluate (a) the change in kinetic energy of the mass in \(\mathrm{kJ}\), and (b) the work in \(\mathrm{kJ}\).

A \(0.2-\mathrm{m}\)-thick plane wall is constructed of concrete. At steady state, the energy transfer rate by conduction through a \(1-\mathrm{m}^{2}\) area of the wall is \(0.15 \mathrm{~kW}\). If the temperature distribution is linear through the wall, what is the temperature difference across the wall, in \(\mathrm{K}\) ?
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