Electromechanical machines, Slides of Mechanical Engineering

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University of Hail

Electrical Engineering Department

(Second Semester 2021 / 2022 )

EE 206 : Electric Energy Engineering

Chapter 5 : Fundamentals of Rotating Machines

Lecture Objectives:

• Introduction to rotating machines
• Basic concepts of energy converters

Introduction:

• Generators ✓Generators in power planets ✓Generators for picnic ✓Generators for e-Aircraft ✓Generators for wind turbine - Motors ✓Pumps ✓Tesla e-cars motors ✓Siemens e-Aircraft motors

Introduction:

• Generators
  • Motors Fig. 2 Tesla cars

Introduction: Example 1 : A shaft is connected to a motor is rotating with an angular speed of 171.88 rpm and develops a power of 810 W at this speed. Determine the torque developed by the motor. Solution: First convert speed to rad/sec: 1 𝑟𝑎𝑑/ sec = 60 2 π 𝑟𝑝𝑚 ➔ 1 𝑟𝑎𝑑/ sec = 9. 549 𝑟𝑝𝑚 ➔Angular speed = 18 rad/sec ω ∗ 𝑇 = 𝑉 ∗ 𝐼 18 * T = 810 W T= 45 N-m

Introduction:

• The energy losses consist of:
  • Mechanical losses which include windage and friction.
  • Electrical losses which include winding copper losses and magnetic core losses.
• These losses can be modeled externally.
• Another assumption made is infinite permeability for magnetic cores, both rotors and stators, used for the energy converters.

Basic Concepts of Energy Converters:

• Figure 1 shows a coil in magnetic field.
• The expression for the induced emf: 𝑒𝑎𝑎′ = B ∗ w ∗ l ∗ ω𝑚 ∗ cos ω𝑚𝑡 + 90 Figure 1 : A coil in a Magnetic Field

Basic Concepts of Energy Converters: Example 2 : A coil is formed by connecting 10 conducting loops, or turns, in series. Each turn has a length l= 2 m and width w = l 0 cm. The 10 - turn coil is rotated at a constant speed of 30 revolutions per second in a magnetic flux density B= 2 T directed upward. a. Find an expression for the induced emf across the coil Solution: a) The induced emf across a single loop is: 𝑒𝑎𝑎′ = B ∗ w ∗ l ∗ ω𝑚 ∗ cos ω𝑚𝑡 + 90 𝑒𝑎𝑎′ = 2 ∗ 2 ∗ 0. 1 ∗ 2 π ∗ 30 ∗ cos 60 π 𝑡 + π 2 𝑒𝑎𝑎′ = 75. 4 cos 188. 5 𝑡 + π 2 V For 10 turns ➔ 𝑒𝑎𝑎′ = 754 cos 188. 5 𝑡 + π 2 V

Basic Concepts of Energy Converters: Example 2 : A coil is formed by connecting 10 conducting loops, or turns, in series. Each turn has a length l= 2 m and width w = l 0 cm. The 10 - turn coil is rotated at a constant speed of 30 revolutions per second in a magnetic flux density B= 2 T directed upward. C) Calculate the average mechanical torque needed to turn the coil and generate power for the resistor. Solution: The average mechanical torque required to rotate the coil of part (b) is: 𝑇𝑚 = P ω𝑚 𝑇𝑚 = 568

188. 5 𝑇𝑚 = 3 N-m

Basic Concepts of Energy Converters:

• Practical electric machines are quite different.
• Voltages are generated in coils that consist of several turns.
• Coils are rotated through a magnetic field, or the field is rotated through the coils.
• There must be a relative motion between the flux and the coils.

Classification of Machines:

HW & Quiz

• HW

Questions from the textbook are uploaded at the end of these slides. Due date: 06/04/

• Quiz The Sixth Quiz will be on 06/04/

Questions

• Q 2 :

Draw the tree diagram of the machine's classifications.