ELECTRICAL MACHINE EEP 423

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Certainly! Let’s delve into each question from your Electrical Machines exam with detailed explanations and calculations.

Q1(a)

(i) General Expression for the Speed of a DC Motor

The speed N of a DC motor is given by:

N = 60A/PZ·E_b/Φ

Where:

  • N = Speed in RPM
  • A = Number of parallel paths
  • P = Number of poles
  • Z = Total number of armature conductors
  • E_b = Back EMF
  • Φ = Flux per pole

Since 60A/PZ is a constant for a given machine, we can write:

N ∝E_b/Φ

Also, from the voltage equation:

E_b = V - I_a R_a

Therefore:

N ∝V - I_a R_a/Φ

(ii) Factors Affecting the Speed of a DC Motor

  1. Applied Voltage (V): Increasing V increases E_b, thus increasing speed.
  2. Armature Resistance (R_a): Higher R_a causes more voltage drop, reducing E_b and speed.
  3. Flux per Pole (Φ): Increasing Φ decreases speed, as speed is inversely proportional to flux.

(b) Calculating New Speed

Given:

  • V = 220 V
  • R_a = 0.5 Ω
  • Φ reduced by 20% ⇒ Φ_2 = 0.8 Φ_1
  • I_a1 = 30 A, N_1 = 1000 rpm
  • I_a2 = 50 A, N_2 = ?

Using:

N ∝V - I_a R_a/Φ

Therefore:

N_2/N_1 = V - I_a2 R_a/V - I_a1 R_a·Φ_1/Φ_2

Plugging in the values:

N_2/1000 = 220 - 50 × 0.5/220 - 30 × 0.5·1/0.8



N_2/1000 = 195/205· 1.25

N_2/1000 ≈ 0.951 · 1.25 ≈ 1.189
N_2 ≈ 1000 · 1.189 ≈ 1189  rpm

Answer: The new speed is approximately 1189 rpm.

Q2(a)

(i) Ward-Leonard System

The Ward-Leonard system is a method of speed control for DC motors, involving:

  • An AC motor driving a DC generator.
  • The DC generator supplies power to the DC motor.
  • By varying the field excitation of the generator, the output voltage changes, thus controlling the speed of the DC motor.

Diagram:

AC Supply → AC Motor → DC Generator → DC Motor

(ii) Advantages and Disadvantages

Advantages:

  • Smooth and wide range of speed control.
  • High starting torque.
  • Regenerative braking is possible.

Disadvantages:

  • High initial cost and maintenance.
  • Bulky setup.
  • Lower efficiency due to multiple energy conversions.

(b) Calculating Field Resistance

Given:

  • V = 220 V
  • R_a = 0.5 Ω
  • I_a = 30 A
  • N = 1000 rpm
  • Φ = 0.02 Wb

First, calculate back EMF:

E_b = V - I_a R_a = 220 - (30 × 0.5) = 220 - 15 = 205  V

Assuming the motor has 4 poles and 600 armature conductors, and it’s lap wound (A = 4):

N = 60A E_b/P Z Φ

Plugging in the values:

1000 = 60 × 4 × 205/4 × 600 × 0.02



1000 = 49200/48



1000 = 1025

This discrepancy suggests that either the assumed values for P, Z, or A are incorrect or additional information is needed to accurately calculate the field resistance.

Q3(a)

(i) Electric Braking Types

  1. Rheostatic (Dynamic) Braking:
  • The motor is disconnected from the supply.
  • Armature is connected to a resistor.
  • Motor acts as a generator; kinetic energy is converted to electrical energy and dissipated as heat in the resistor.
  1. Plugging (Reverse Current Braking):
  • Supply polarity is reversed.
  • Motor develops reverse torque, bringing it to a stop quickly.
  • High current flows; additional resistance is added to limit current.
  1. Regenerative Braking:
  • Motor acts as a generator.
  • Energy is fed back into the supply line.
  • Efficient but requires the supply system to accept returned energy.

(ii) Advantages and Disadvantages

Advantages:

  • Quick and smooth stopping.
  • Energy recovery in regenerative braking.
  • Reduced wear on mechanical brakes.

Disadvantages:

  • Complex circuitry.
  • Not effective during power failure.
  • Plugging can cause high current surges.

(b) Calculating Braking Torque

Given:

  • P = 10 kW
  • N = 1000 rpm

Torque T is given by:

T = 9550 × P/N = 9550 × 10/1000 = 95.5  Nm

Assuming braking torque equals motor torque, the required resistance can be calculated using:

E_b = V - I_a R_a



T = k Φ I_a

Without specific values for V, R_a, and ( \Phi \
Got it! Let’s break down these new questions from your exam in detail:

Q5 (a)(i): Describe a Universal Motor

  • Definition: A universal motor can run on both AC and DC supplies.
  • Construction: Series wound motor with a commutator and brushes.
  • Working: Works like a series DC motor; field flux and armature current reverse simultaneously in AC supply, maintaining unidirectional torque.

Q5 (a)(ii): Three Limitations and How to Overcome Them

Limitations:

  1. Sparking at brushes due to high speed and inductive loads.
  2. Overheating because of poor cooling at high speeds.
  3. Noise and vibration due to high speeds and brush contact.

Overcome by:

  • Using better brush materials.
  • Improved ventilation/cooling methods.
  • Using electronic commutation in brushless designs.

Q5 (b): Three Methods of Speed Control of Universal Motor

  1. Voltage Control: Vary supply voltage using a rheostat or electronic controller.
  2. Field Control: Adjust field current (not common due to series winding).
  3. Varying Supply Frequency: Mostly with electronic drives, controlling speed by frequency.

(Diagrams usually show rheostat in series with motor for voltage control)

Q6 (a)(i): Two Factors Affecting Temperature Rise
1. Load current magnitude: More current causes more heating.

  1. Cooling conditions: Better cooling reduces temperature rise.

Q6 (a)(ii): Three Variables Generators and Motors Are Rated On

  1. Voltage rating (V)
  2. Current rating (A)
  3. Power rating (kW or HP)

Q6 (b): Power Rating Calculation

Given torque and time intervals:

Torque (Nm)Time (min)
24020
14010
30010
20020

Speed, N = 720 rpm

Step 1: Calculate power for each interval:

P = T × N × 2π/60

For 240 Nm:

P_1 = 240 × 720 × 2 × 3.1416/60 = 18095   W

Similarly, calculate P_2, P_3, P_4:

  • P_2 = 10507 W
  • P_3 = 22586 W
  • P_4 = 15063 W

Step 2: Calculate weighted average power:

P_avg = ∑ (P_i × t_i)/∑ t_i = (18095 × 20) + (10507 × 10) + (22586 × 10) + (15063 × 20)/60



P_avg = 361900 + 105070 + 225860 + 301260/60 = 994090/60 = 16568   W = 16.57   kW

Answer: Power rating ≈ 16.57 kW

Q7 (a)(i): Define Force Commutation (Chopper Circuits)

  • Force commutation is the process of forcibly turning off a thyristor by applying a reverse voltage/current in chopper circuits.

Q7 (a)(ii): Four Ways to Fire a Thyristor
1. Forward Voltage Firing (natural forward voltage)

  1. Gate Triggering (applying a gate pulse)
  2. dv/dt Firing (rapid rise in voltage)
  3. Temperature Rise (increased junction temperature)

Q7 (b): Series Commutation Circuit

Explanation:

  • Uses an LC circuit connected in series with the thyristor.
  • When the thyristor is turned off, the LC circuit creates a voltage opposite in polarity to turn it off.
  • The energy in the inductor helps transfer the current to the next thyristor.

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