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Chapter 17: Problem 46

A 240-V-rms delta-connected 100-hp 60-Hz six-pole synchronous motor operates with a developed power (including losses) of \(50 \mathrm{hp}\) and a power factor of 90 percent leading. The synchronous reactance is \(X_{s}=0.5 \Omega\). a. Find the speed and developed torque. b. Determine the values of \(\mathbf{I}_{a}, \mathbf{E}_{r}\), and the torque angle. c. Suppose that the excitation remains constant and the load torque increases until the developed power is \(100 \mathrm{hp}\). Determine the new values of \(\mathbf{I}_{a}, \mathbf{E}_{r}\), the torque angle, and the power factor.

Short Answer

Expert verified
Speed is 1200 rpm; torque is 296.8 Nm. Initial \( I_a \) is 99.67 A, \( E_r \) is 234.6 V, \( \delta \approx 6.3^\circ \). After increasing load, \( I_a \) becomes 199.34 A, \( \delta \approx 12.7^\circ \), power factor remains approximately 0.9.

Step by step solution

01

Calculate the Speed of the Motor

The synchronous speed of a motor can be determined using the formula \( N_s = \frac{120 \times f}{P} \), where \( f \) is the frequency (60 Hz) and \( P \) is the number of poles (6). Plug these values into the formula: \[ N_s = \frac{120 \times 60}{6} = 1200 \text{ rpm} \].
02

Calculate the Developed Torque

Using the formula \( T = \frac{P}{\omega} \), where \( P \) is the power in watts and \( \omega \) is the angular speed in radians per second. First, convert horsepower to watts: \( 50 \text{ hp} = 50 \times 746 \text{ W} = 37300 \text{ W} \). The angular speed \( \omega = \frac{2\pi N_s}{60} = \frac{2\pi \times 1200}{60} = 125.66 \text{ rad/s} \). Now, calculate torque: \[ T = \frac{37300}{125.66} = 296.8 \text{ Nm} \].
03

Calculate Armature Current (\( I_a \))

The apparent power \( S \) is given by \( S = \frac{P}{pf} \), where \( P \) is the power developed (37300 W) and \( pf \) is the power factor (0.9). Calculate \( S \): \[ S = \frac{37300}{0.9} = 41444.44 \text{ VA} \]. For a delta connection, \( I_a = \frac{S}{\sqrt{3} \times V} \), where \( V = 240 \text{ V} \). Hence, \[ I_a = \frac{41444.44}{\sqrt{3} \times 240} = 99.67 \text{ A} \].
04

Calculate Internal Generated Voltage (\( E_r \))

Using the phasor relationship, \( E_r = V -(j I_aX_s) \). Calculate \( j I_a X_s = j99.67 \times 0.5 = j49.83 \). Since power factor is leading, \( E_r \) leads the terminal voltage. Using magnitude \( |E_r| = \sqrt{V^2 - (I_aX_s)^2} \), compute: \[ |E_r| = \sqrt{240^2 - 49.83^2} = 234.6 \text{ V} \].
05

Determine the Torque Angle (\( \delta \))

The torque angle is calculated using the power equation of a synchronous machine: \( P = \frac{3V E_r}{X_s} \sin \delta \). Rearrange for \( \sin \delta \): \[ \sin \delta = \frac{37300 \times 0.5}{3 \times 240 \times 234.6} \approx 0.1102 \]. Solving gives \( \delta \approx \sin^{-1}(0.1102) = 6.3^\circ \).
06

Increase Power and Recalculate Parameters

Developed power increases to 100 hp or 74600 W at constant \( E_r \). Recalculate \( I_a \), \( \delta \), and pf. New \( S = \frac{74600}{0.9} = 82888.89 \text{ VA} \). New \( I_a = \frac{82888.89}{\sqrt{3} \times 240} = 199.34 \text{ A} \). Use \( P = \frac{3V E_r}{X_s} \sin \delta \) to find \( \delta \): \[ \sin \delta = \frac{74600 \times 0.5}{3 \times 240 \times 234.6} \approx 0.2204 \]. Solving gives \( \delta \approx 12.7^\circ \). Determine new power factor: \( \cos \phi = \frac{37300}{41444.44} = 0.9 \).

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Key Concepts

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

Delta Connection
A delta connection is a method used to connect three-phase systems, like a six-pole synchronous motor, by configuring its windings to form a closed-loop in a triangular shape. This is different from a wye (or star) connection, which connects each winding to a neutral point. Delta connection offers several benefits for motors.

  • High Line Voltage: In a delta connection, the line voltage is equal to the phase voltage, which is why motors configured in this manner can operate at higher voltages without increasing the number of windings.

  • Increased Power: Since power in a three-phase system is determined by the formula, \( P = \sqrt{3} \times V_{line} \times I_{line} \times \text{power factor} \), having a potentially higher line voltage can lead to more power being delivered to the motor.

  • Stable Operation: This configuration provides redundancy, as even if one winding fails, the motor can continue to operate at reduced capacity.
      • Understanding a delta connection is essential for analyzing the performance and calculating parameters like torque, current, and induced voltage in synchronous motors.
Torque Calculation
Calculating the developed torque in a synchronous motor is crucial for understanding how much force it can exert. The formula used is: \( T = \frac{P}{\omega} \), where \( T \) is torque in Newton-meters (Nm), \( P \) is power in watts, and \( \omega \) is angular speed in radians per second.

  • Power Conversion: First, it's important to translate horsepower to watts since metric units are typically used for torque. Available power in watts helps achieve accurate readings.

  • Determining Angular Speed: Angular speed can be found with \( \omega = \frac{2\pi N_s}{60} \), converting the rotational frame rate from revolutions per minute to radians per second. This step is essential for calculating torque accurately.

  • Result Interpretation: The result from the torque calculation gives insight into how much mechanical load the motor can handle at a given power level. This impacts motor performance and efficiency seriously.
Calculating torque is not just about understanding the motor's brush鈥搇ike attraction; it's about knowing how the motor converts electrical energy into mechanical power effectively.
Power Factor
Power factor is a measure of how effectively electrical power is being converted into useful work output and is a dimensionless number ranging between -1 and 1. In this case, a power factor of 0.9 leading indicates that the motor's circuit current leads the voltage, which usually happens when the load is capacitive.

  • Importance in Synchronous Motors: A high power factor reduces loss in the system, ensuring better efficiency and performance of the motor.

  • Determination: Power factor can be determined using the ratio \(\text{pf} = \frac{P}{S}\), where \( P \) is the real power in watts, and \( S \) is the apparent power in volt-amperes. In synchronous motors, managing power factor is necessary to maintain optimal operation.

  • Adjustments: Power factor can be improved or adjusted by changing the field excitation of the motor. For example, increasing excitation leads to a more leading power factor.
Adjusting the power factor is crucial for system efficiency and stability, especially in large industrial applications where synchronous motors are commonly employed.
Synchronous Speed
Synchronous speed is the speed at which the magnetic field within the motor rotates, and it is determined by the formula \( N_s = \frac{120 \times f}{P} \). Here, \( f \) is the frequency of the AC supply, and \( P \) is the number of poles.

  • Fixed Nature: One of the defining attributes of a synchronous motor is that it operates at a constant speed, known as synchronous speed. This remains unaffected by load variations, as long as the supply frequency remains constant.

  • Frequency Dependence: The synchronous speed depends on the supply frequency, making it a straight-line relationship. Thus, alterations in frequency will linearly affect the synchronous speed.

  • Calculations: By using the number of poles and system frequency, synchronous speed can be derived, allowing for the computation of other parameters, such as angular velocity and torque.
Synchronous speed is fundamental in understanding motor operation because many aspects of its performance 鈥� like speed regulation and efficiency 鈥� hinge directly on this parameter.

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Most popular questions from this chapter

List two universal manufacturers of stepper motors and brushless DC Motors.

A four-pole induction motor drives a load at \(2500 \mathrm{rpm}\). This is to be accomplished by using an electronic converter to convert a \(400-\mathrm{V}\) dc source into a set of three-phase ac voltages. Find the frequency required for the ac voltages assuming that the slip is 4 percent. The load requires \(2 \mathrm{hp}\). If the dc-to-ac converter has a power efficiency of 88 percent and the motor has a power efficiency of 80 percent, estimate the current taken from the dc source.

A 240 -V-rms 100-hp 60-Hz six-pole deltaconnected synchronous motor operates with a developed power (including losses) of \(100 \mathrm{hp}\) and a power factor of 85 percent lagging. The synchronous reactance is \(X_{s}=\) \(0.5 \Omega\). The field current is \(I_{f}=10 \mathrm{~A}\). What must the new field current be to produce 100 percent power factor? Assume that magnetic saturation does not occur so that \(B_{r}\) is proportional to \(I_{f}\).

A 1-hp 120-V-rms 1740-rpm \(60-\mathrm{Hz}\) capacitor-start induction-run motor draws a current of \(10.2 \mathrm{~A} \mathrm{rms}\) at full load and has an efficiency of 80 percent. Find the values of a. the power factor and b. the impedance of the motor at full load. c. Determine the number of poles that the motor has.

A six-pole 240-V-rms 60-Hz deltaconnected synchronous motor operates with a developed power of \(50 \mathrm{hp}\), unity power factor, and a torque angle of \(15^{\circ}\). Find the phase current. Suppose that the load is removed so that the developed power is zero. Find the new values of the current, power factor, and torque angle.
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