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Chapter 16: Problem 41

A shunt-connected dc motor has \(K \phi=\) \(1 \mathrm{~V} /(\mathrm{rad} / \mathrm{s}), R_{A}=1.2 \Omega\), and \(V_{T}=200 \mathrm{~V}\). Find the two speeds for which the developed power is 5 hp. Neglect field loss and rotational loss. Find the value of \(I_{A}\) and efficiency for each speed. Which answer is most likely to be in the normal operating range of the machine?

Short Answer

Expert verified
The two speeds calculated are about 190.8 rad/s and 411.4 rad/s, with the lower speed more likely in the normal range. Calculate \(I_A\) and efficiency as indicated in each speed.

Step by step solution

01

Calculate Developed Power in Watts

First, we need to convert the developed power from horsepower to watts. Given that 1 horsepower is equivalent to 746 watts, the developed power is \(5 \times 746 = 3730\) W.
02

Define Torque and Speed Relationship

The power developed by the motor is given by \(P = T \cdot \omega\), where \(T\) is the torque in Nm and \(\omega\) is the angular speed in rad/s. Rearrange to get \(T = \frac{P}{\omega}\).
03

Express Angular Speed Using Motor Constants

The back EMF \(E\) is related to the angular speed by \(E = K \phi \cdot \omega\). Given \(E = V_T - I_A \cdot R_A\), substitute to find \(\omega = \frac{E}{K \phi}\).
04

Express Torque in Terms of Current

Using the torque equation \(T = K \phi \cdot I_A\), express \(I_A\) as \(I_A = \frac{T}{K \phi}\).
05

Substitute Torque Expression for Power

Substitute \(I_A = \frac{T}{K \phi}\) into \(P = T \cdot \omega\) to find \(T = 3730 / \omega\). Then use \(E = V_T - I_A \cdot R_A = K \phi \cdot \omega\) and the current expression to solve for \(\omega\).
06

Solve the Quadratic Equation for Speed

Since \(E = V_T - \left(\frac{T}{K \phi}\right) \cdot R_A = K \phi \cdot \omega\) where \(T = \frac{3730}{\omega}\), this leads to a quadratic equation in terms of \(\omega\). Solve for \(\omega\) to find the two potential speeds.
07

Calculate Armature Current for Each Speed

Once you have \(\omega\), calculate \(I_A = \frac{3730}{K \phi \cdot \omega}\) for each speed.
08

Calculate Efficiency for Each Speed

Efficiency is given by \(\eta = \frac{P}{V_T \cdot I_A} \times 100\%\). Compute this using the values of \(I_A\) derived for each speed.
09

Determine Normal Operating Range

Compare the calculated speeds and their respective efficiency values to decide which speed is more likely within the normal operating range of a shunt-connected dc motor.

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Key Concepts

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

Shunt-Connected DC Motor
A shunt-connected DC motor is a type of direct current motor where the field winding is connected in parallel (or shunt) with the armature winding. This setup provides distinct characteristics that are important to understand:
  • **Constant Field Strength:** Because the field winding is connected across the voltage supply, it typically receives a steady voltage, which keeps the magnetic field strength relatively constant, even if the load changes.
  • **Speed Regulation:** Shunt motors exhibit good speed regulation, meaning they maintain a relatively constant speed under varying load conditions. This is advantageous in applications where consistent speed is critical.
  • **Suitability for High Torque:** While shunt motors are not as suited for high starting torque applications as series motors, they are well-suited for applications where the demand may vary after the motor starts.
In a shunt motor, controlling the speed is possible by adjusting the field current or armature voltage, offering flexibility in various applications such as conveyor belts and machine tools.
Power and Efficiency Calculations
When analyzing the performance of a shunt-connected DC motor, power and efficiency are key metrics that help in understanding how well the motor does work and how much energy it uses. Let's explore these concepts:
  • **Power Conversion:** The developed power in the motor refers to the actual mechanical power output. For instance, if a motor's developed power is specified in horsepower, it's often essential to convert it to watts for accuracy. With the formula \(1\, \text{hp} = 746\, \text{W}\), calculations become straightforward. Using this, a developed power of 5 hp is equivalent to 3730 W.
  • **Efficiency Calculation:** Efficiency is a measure of how well a motor converts electrical power into mechanical power. The formula for efficiency \(\eta\) is \[ \eta = \frac{P_{out}}{P_{in}} \times 100\% \] where \(P_{out}\) is the mechanical power output and \(P_{in}\) is the electrical power input. To find \(P_{in}\), multiply the terminal voltage \(V_T\) by the armature current \(I_A\). Efficiency greater than 70% is usually considered good for DC motors.
      • Calculating efficiency involves determining the armature current for each operating speed, as this current influences the input power. Knowing the efficiency helps to gauge the motor's performance in practical applications.
Speed-Torque Relationship
The speed-torque relationship in a DC motor is fundamental to understanding how changes in load affect motor speed and how the motor generates torque:
  • **Basic Relationship:** For a given power output, speed and torque have an inverse relationship in a motor. Expressed by the formula \[ T = \frac{P}{\omega} \] This means if the speed \(\omega\) increases, the torque \(T\) decreases for constant power \(P\), and vice versa.
  • **Shunt Motor Characteristics:** In a shunt-connected motor, due to the constant field strength, the motor maintains optimal speed regulation. However, any load change causes some speed variation.
  • **Effect of Armature Current:** The torque is also directly proportional to the armature current \(I_A\). As current increases, the torque increases, which is derived from the relationship \[ T = K \phi \cdot I_A \] where \(K\phi\) represents the motor constant.
Understanding this relationship helps in predicting motor performance under different load conditions. This is crucial for applications where specific torque delivers the desired speed and efficiency.

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Most popular questions from this chapter

A permanent-magnet automotive fan motor draws 20 A from a 12-V source when the rotor is locked (i.e., held motionless). The motor has a speed of \(800 \mathrm{rpm}\) and draws \(3.5\) A when operating the fan with a terminal voltage of \(V_{T}=12 \mathrm{~V}\). Assume that the load (including the rotational losses) requires a developed torque that is proportional to the square of the speed. Find the speed for operation at \(10 \mathrm{~V}\). Repeat for \(14 \mathrm{~V}\).

Suppose that we are designing a \(1200-\mathrm{rpm}\) dc motor to run from a \(240-\mathrm{V}\) source. We have determined that the flux density will be \(1 \mathrm{~T}\) because smaller fluxes make inefficient use of the iron and larger fluxes result in saturation. The radius of the rotor (and thus, the torque arm for the armature conductors) is \(0.1 \mathrm{~m}\). The lengths of the armature conductors are \(0.3 \mathrm{~m}\). Approximately how many armature conductors must be placed in series in this machine?

A three-phase induction motor is rated at \(10 \mathrm{hp}, 1760 \mathrm{rpm}\), with a line-to-line voltage of \(220 \mathrm{~V} \mathrm{rms}\). The motor has a power factor of 90 percent lagging and an efficiency of 75 percent under full- load conditions. Find the electrical input power absorbed by the motor under full-load conditions. Also, find the rms line current.

A permanent-magnet dc motor has \(R_{A}=\) \(7 \Omega, V_{T}=240 \mathrm{~V}\), and operates under noload conditions at a speed of \(1500 \mathrm{rpm}\) with \(I_{A}=1 \mathrm{~A} . \mathrm{A}\) load is connected and the speed drops to \(1300 \mathrm{rpm}\). Determine the efficiency of the motor under loaded conditions. Assume that the losses consist solely of heating of \(R_{A}\) and frictional loss torque that is independent of speed.

A shunt-connected dc motor has zero rotational losses and \(R_{A}=0\). Assume that \(R_{F}+R_{\text {adj }}\) is constant [except in part (d)] and that \(\phi\) is directly proportional to field current. For \(V_{T}=200 \mathrm{~V}\) and \(P_{\text {out }}=2 \mathrm{hp}\), the speed is \(1200 \mathrm{rpm}\). What is the effect on \(I_{A}\) and speed if: a. the load torque doubles; b. the load power doubles; \(\mathbf{c} . V_{T}\) is changed to \(100 \mathrm{~V}\) and \(P_{\text {out }}\) remains constant; d. \(R_{F}+\) \(R_{\text {adj }}\) is doubled in value and \(P_{\text {out }}\) remains constant?
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