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Chapter 27: Problem 50

A shunt-wound de motor with the field coils and rotor connected in parallel (Fig. 27.56 ) operates from a \(120-\mathrm{V}\) dc power linc. The resistance of the ficld windings, \(R_{f},\) is 218\(\Omega\) . The resistance of the rotor, \(R_{r}\) is 5.9 . When the motor is running, the rotor develops an emf \(\mathcal{E}\) . The motor draws a current of 4.82 A from the line. Friction losses amount to 45.0 W. Compute (a) the field current; \((b)\) the rotor current; \((c)\) the emf \(\mathcal{E} ;\) (d) the rate of development of thermal energy in the field windings; (e) the rate of development of thermal energy in the rotor; \((f)\) the power input to the motor; (g) the efficiency of the motor.

Short Answer

Expert verified
(a) 0.5505 A, (b) 4.2695 A, (c) 94.85 V, (d) 66.1 W, (e) 107.5 W, (f) 578.4 W, (g) 76%.

Step by step solution

01

Calculate Field Current

The field current, \( I_f \), can be calculated using Ohm's Law for the field windings: \( I_f = \frac{V}{R_f} \). Substituting the given values: \( I_f = \frac{120}{218} \approx 0.5505 \text{ A} \).
02

Calculate Rotor Current

The rotor current, \( I_r \), is the total current drawn from the line minus the field current: \( I_r = I - I_f \). With \( I = 4.82 \space A \) and \( I_f = 0.5505 \space A \), we find \( I_r = 4.82 - 0.5505 \approx 4.2695 \text{ A} \).
03

Calculate Back Emf (\(\mathcal{E}\))

The back emf \( \mathcal{E} \) developed by the rotor is given by \( \mathcal{E} = V - I_r \cdot R_r \). Substituting, \( \mathcal{E} = 120 - 4.2695 \times 5.9 \approx 94.85 \text{ V} \).
04

Calculate Thermal Energy Rate in Field Windings

The rate of thermal energy development in the field windings is given by \( P_f = I_f^2 \cdot R_f \). Substituting, \( P_f = (0.5505)^2 \times 218 \approx 66.1 \text{ W} \).
05

Calculate Thermal Energy Rate in Rotor

The rate of thermal energy development in the rotor is \( P_r = I_r^2 \cdot R_r \). Substituting, \( P_r = (4.2695)^2 \times 5.9 \approx 107.5 \text{ W} \).
06

Calculate Power Input to Motor

The power input to the motor, \( P_{in} \), is the product of the line voltage and current: \( P_{in} = V \cdot I \). Substituting, \( P_{in} = 120 \times 4.82 \approx 578.4 \text{ W} \).
07

Calculate Motor Efficiency

Motor efficiency is the useful power output (back emf times rotor current minus friction losses) divided by the power input: \( \text{Efficiency} = \frac{\mathcal{E} \cdot I_r - \text{Friction Losses}}{P_{in}} \). Substituting, \( \text{Efficiency} = \frac{94.85 \times 4.2695 - 45}{578.4} \approx 0.76 \text{ or } 76\% \).

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Key Concepts

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

Shunt-Wound DC Motor
A shunt-wound DC motor is a type of electric motor where the field coils and the rotor (or armature) are connected in parallel. This design allows for better control of the motor's speed and torque. One advantage of this setup is that the field current remains constant, which provides stable performance.
Since the field windings are connected across the supply voltage, the magnetic field remains relatively constant despite variations in the armature current. This feature makes shunt-wound motors suitable for applications requiring consistent speed under varying loads, such as in conveyor systems or fans.
Understanding the connection in parallel is crucial because it affects how the currents and voltages are distributed across the motor's components. Each part can thus operate optimally without interfering with the other.
Efficiency Calculation
Efficiency calculation in a motor is about gauging how well the motor converts electrical energy into mechanical energy. Knowing the efficiency helps us determine the motor's performance and operating costs.
To calculate efficiency, find out how much useful work the motor does compared to the energy put into it from the electrical supply. The formula is:
\[ \text{Efficiency} = \frac{\text{Output Power}}{\text{Input Power}} \times 100 \]
Output power is usually less than input power due to losses like friction and thermal energy. These losses must be accounted for when calculating efficiency.
In our solution, the motor's efficiency was determined by considering the power developed inside the motor after subtracting losses, like friction, from the electrical input.
Ohm's Law
Ohm's Law is one of the fundamental principles for understanding electrical circuits. It provides a relationship between voltage (V), current (I), and resistance (R) through the formula:
\[ V = I \times R \]
This law allows us to calculate one of these three variables if we know the other two.
In the context of our DC motor exercise, Ohm's Law helps us determine the current through the motor's components, like the field and rotor windings. By knowing the resistance and the voltage supplied to each parallel section, we computed the respective currents.
These currents are crucial for subsequent calculations, like finding the thermal energy loss or back electromotive force ( \(\mathcal{E}\) ) generated in the rotor.
Thermal Energy in Electrical Circuits
Thermal energy in electrical circuits often manifests as heat when electrical energy encounters resistance, according to Joule's Law. This can lead to energy losses in electric motors and needs management to ensure efficiency and longevity of the device.
The formula for calculating the thermal energy generated is:
\[ P = I^2 \times R \]
where \( P \) is the power (thermal energy per unit time) in watts, \( I \) is the current in amperes, and \( R \) is the resistance in ohms.
In motors, heat develops in both the field and rotor windings due to the resistance they offer to the flow of electric current. Properly calculating this thermal energy helps in designing cooling measures and assessing the total energy losses, which is vital for efficiency calculations.

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Most popular questions from this chapter

A \(150-\mathrm{V}\) battery is connected across two parallel metal plates of area 28.5 \(\mathrm{cm}^{2}\) and separation \(8.20 \mathrm{mm} .\) A beam of alpha particles (charge \(+2 e,\) mass \(6.64 \times 10^{-27} \mathrm{kg} )\) is accelerated from rest through a potential difference of 1.75 \(\mathrm{kV}\) and enters the region between the plates perpendicular to the electric field. What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?

(a) An 16 nucleus (charge \(+8 e )\) moving horizontally from west to east with a speed of 500 \(\mathrm{km} / \mathrm{s}\) experiences a magnetic force of 0.00320 \(\mathrm{nN}\) vertically downward. Find the magnitude and direction of the weakest magnetic field required to produce this force. Explain how this same force could be caused by a larger magnetic field. (b) An electron moves in a uniform, horizontal, 2.10 - T magnetic field that is toward the west. What must the magnitude and direction of the minimum velocity of the electron be so that the magnetic force on it will be 4.60 \(\mathrm{pN}\) , vertically upward? Explain how the velocity could be greater than this minimum value and the force still have this same magnitude and direction.

In a shunt-wound dc motor with the field coils and rotor connected in parallel (Fig. \(27.56 ),\) the resistance \(R_{t}\) of the field coils is \(106 \Omega,\) and the resistance \(R_{r}\) of the rotor is 5.9\(\Omega\) . When a potential difference of 120 \(\mathrm{V}\) is applied to the brushes and the motor is running at full speed delivering mechanical power, the current supplied to it is 4.82 \(\mathrm{A}\) (a) What is the current in the field coils? (b) What is the current in the rotor? (c) What is the induced emf developed by the motor? (d) How much mechanical power is developed by this motor?

A horizontal rod 0.200 \(\mathrm{m}\) long is mounted on a balance and carries a current. At the location of the rod a uniform horizontal magnetic field has magnitude 0.067 T and direction perpendicular to the rod. The magnetic force on the rod is measured by the balance and is found to be 0.13 \(\mathrm{N}\) . What is the current?

An Electromagnetic Rail Gun. A conducting bar with mass \(m\) and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\overrightarrow{\boldsymbol{B}}\) fills the region between the rails (Fig. 27.63\()\) . (a) Find the magnitude and direction of the net force on the conducting bar Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m\) , find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\) . (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth \((11.2 \mathrm{km} / \mathrm{s}) .\) Let \(B=0.50 \mathrm{T}, \quad I=2.0 \times 10^{3} \mathrm{A}\) \(m=25 \mathrm{kg},\) and \(L=50 \mathrm{cm} .\) For simplicity assume the net force on the object is equal to the magnetic force, as in parts \((a)\) and \((b)\) , even though gravity plays an important role in an actual launch in space.
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