/*! 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 14 A dynamo (generator) delivers \(... [FREE SOLUTION] | 91影视

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Chapter 33: Problem 14

A dynamo (generator) delivers \(30.0 \mathrm{~A}\) at \(120 \mathrm{~V}\) to an external circuit when operating at \(1200 \mathrm{rpm} .\) What torque is required to drive the generator at this speed if the total power losses are 400 W?

Short Answer

Expert verified
The required torque is approximately 31.83 Nm.

Step by step solution

01

Calculate Output Power

The output power of the generator can be calculated using the formula \( P = IV \) where \( I = 30.0 \mathrm{~A} \) and \( V = 120 \mathrm{~V} \). Substitute the values to find the power: \( P = 30.0 \mathrm{~A} \times 120 \mathrm{~V} = 3600 \mathrm{~W} \).
02

Calculate Total Power Input

The total power input to the generator is the sum of the output power and the total power losses. Use the formula: Total Power Input = Output Power + Losses. Substitute to get: Total Power Input = \( 3600 \mathrm{~W} + 400 \mathrm{~W} = 4000 \mathrm{~W} \).
03

Convert RPM to Radians per Second

To find torque, we need angular velocity in radians per second. Given the generator speed is \( 1200 \mathrm{~rpm} \), convert it: \( \omega = \frac{1200 \times 2\pi}{60} = 40\pi \mathrm{~rad/s} \).
04

Calculate the Torque

The torque \( \tau \) can now be determined using the formula \( P = \tau \omega \) where \( P = 4000 \mathrm{~W} \) and \( \omega = 40\pi \mathrm{~rad/s} \). Solve for \( \tau \): \( \tau = \frac{4000 \mathrm{~W}}{40\pi \mathrm{~rad/s}} = \frac{100}{\pi} \mathrm{~Nm} \approx 31.83 \mathrm{~Nm} \).

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Key Concepts

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

Generator Mechanics
Generators are fascinating machines that convert mechanical energy into electrical energy, which is crucial for powering a myriad of devices in our daily lives. The dynamo, a type of generator, plays a vital role in this process. But how does it work? At the core of a generator is the principle of electromagnetic induction, which involves the movement of a conductor through a magnetic field, inducing an electric current. The generator uses mechanical input, often from a turbine or motor, to rotate the conductor, creating electrical output.
  • The main components include a rotor (moving part) and a stator (stationary part).
  • The rotor spins within the magnetic field, producing electricity.
  • The generated electric current is typically AC (Alternating Current), which is more efficient for long-distance transmission.
Understanding generator mechanics helps us appreciate how mechanical motion is transmuted into the electricity that powers our world.
Torque Calculation
Torque is a measure of how much a force acting on an object causes that object to rotate. In our context, it refers to the force required to keep the generator spinning at a constant speed. Calculating torque is essential because it affects the energy needed and the efficiency of the generator.

The relationship between power and torque is given by the formula: \[ P = \tau \cdot \omega \] where
  • \( P \) is the power input, measured in watts (W).
  • \( \tau \) is the torque, measured in newton-meters (Nm).
  • \( \omega \) is the angular velocity, measured in radians per second (rad/s).
To calculate torque, once you have the power and angular velocity, rearrange the formula to \[ \tau = \frac{P}{\omega} \]. This equation shows how torque is directly linked to both the input power and the rotational speed.
Power Losses in Electrical Systems
Power losses in electrical systems can have significant impacts on the efficiency and performance of devices like generators. These losses are typically due to resistance in electrical wires and components, and they manifest as heat. For a generator, understanding and minimizing these losses is crucial to maintain its efficiency.
  • The losses involve things like friction, windage (air resistance), and electrical resistance within the conductor.
  • Reducing power losses can improve the overall energy efficiency by ensuring more of the input mechanical energy is converted into electrical energy.
  • Strategies for reducing losses include using better conductor materials, improving component design, and enhancing cooling systems to manage heat.
In our exercise, a power loss of 400 W indicates energy not converted into useful electrical energy but instead lost, necessitating a higher input power than the output.
Angular Velocity Conversion
Converting rotational speed from rotations per minute (rpm) to radians per second (rad/s) is a common task in physics problems, especially when dealing with rotational motion and dynamics. This conversion is fundamental because most physics equations involving angular motion require angular velocity to be in rad/s.

The conversion from rpm to rad/s is straightforward and uses the formula: \[ \omega = \frac{\text{rpm} \times 2\pi}{60} \], where:
  • \( \text{rpm} \) is the rotations per minute.
  • \( 2\pi \) is the number of radians in one complete rotation.
  • Dividing by 60 converts minutes to seconds.
In practice, if you have a generator running at 1200 rpm, converting to rad/s gives \( \omega = 40\pi \) rad/s, which can then be used in other calculations, such as determining torque.

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Most popular questions from this chapter

A \(120-V\) generator is run by a windmill that has blades \(2.0 \mathrm{~m}\) long. The wind, moving at \(12 \mathrm{~m} / \mathrm{s}\), is slowed to \(7.0 \mathrm{~m} / \mathrm{s}\) after passing the windmill. The density of air is \(1.29 \mathrm{~kg} / \mathrm{m}^{3}\). If the system has no losses, what is the largest current the generator can produce? [Hint: How much energy does the wind lose per second?]

A certain generator has armature resistance \(0.080 \Omega\) and develops an induced emf of \(120 \mathrm{~V}\) when driven at its rated speed. What is its terminal voltage when \(50.0 \mathrm{~A}\) is being drawn from it? The generator acts like a battery with emf \(=120 \mathrm{~V}\) and internal resistance \(r=0.080 \Omega\). As with a battery, Terminal p.d. \(=(\mathrm{emf})-I r=120 \mathrm{~V}-(50.0 \mathrm{~A})(0.080 \Omega)=116 \mathrm{~V}\)

An ac generator produces an output voltage of \(\varepsilon=170 \sin 377 t\) volts, where \(t\) is in seconds. What is the frequency of the ac voltage? A sine curve plotted as a function of time is no different from a cosine curve, except for the location of \(t=0 .\) Since \(\varepsilon=2 \pi \mathrm{NBA} f\) \(\cos 2 \pi f t\), we have \(377 \mathrm{t}=2 \pi f t\), from which we find that the frequency \(f=60 \mathrm{~Hz}\).

A motor has a back emf of \(110 \mathrm{~V}\) and an armature current of 90 A when running at 1500 rpm. Determine the power and the torque developed within the armature. Power \(=(\) Armature current \()(\) Back emf \()=(90 \mathrm{~A})(110 \mathrm{~V})=9.9 \mathrm{~kW}\) From Chapter 10, power \(=\tau \omega\) where \(\omega=2 \pi f=2 \pi(1500 \times 1 / 60)\) \(\mathrm{rad} / \mathrm{s}\) $$ \text { Torque }=\frac{\text { Power }}{\text { Angular speed }}=\frac{9900 \mathrm{~W}}{(2 \pi \times 25) \mathrm{rad} / \mathrm{s}}=63 \mathrm{~N} \cdot \mathrm{m} $$

A \(75.0\) -kW, 230 - \(V\) shunt generator has a generated emf of \(243.5\) \(\mathrm{V}\). If the field current is \(12.5 \mathrm{~A}\) at rated output, what is the armature resistance?
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