Shunt Motor Concepts and Problems

Shunt Motor

• A shunt motor's supply voltage (V) equals the generated e.m.f. (E) plus the armature volt drop (IaRa).
• Alternatively, the generated e.m.f. (E) equals the supply voltage (V) minus the armature volt drop (IaRa).
• The supply current (I) equals the armature current (Ia) plus the field current (If).
• Example problem: A 240V shunt motor draws a total current of 30A. Field winding resistance (Rf) is 150Ω and armature resistance (Ra) is 0.4Ω. Determine the armature current and the back e.m.f.
  • Field current (If) = 240V / 150Ω = 1.6A
  • Armature current (Ia) = 30A - 1.6A = 28.4A
  • Back e.m.f. (E) = 240V - (28.4A * 0.4Ω) = 228.64V

Shunt Motor - Problem 21

• A 200V DC shunt-wound motor has an armature resistance of 0.4Ω. At a certain load, the armature current is 30A and it runs at 1350 rev/min.
• If the load increases, increasing the armature current to 45A, determine the speed of the motor, assuming constant flux.
• The relationship E1/Φ1n1 = E2/Φ2n2 applies to both generators and motors. Since flux is constant, Φ1 = Φ2.
• Calculate E1: E1 = 200V - (30A * 0.4Ω) = 188V
• Calculate E2: E2 = 200V - (45A * 0.4Ω) = 182V
• Calculate n2: (188/182) = (1350/n2) Solving for n2, n2 = 1307 rev/min

Series Motor

• In a series motor, the field winding is in series with the armature.
• The supply voltage (V) equals the generated e.m.f. (E) plus the total series resistance drop (I * (Ra + Rf)).
• Alternatively, the generated e.m.f. (E) equals the supply voltage (V) minus the total series resistance drop (I * (Ra + Rf)).
• Example problem: A series motor has an armature resistance of 0.2Ω and a series field resistance of 0.3Ω. The motor is connected to a 240V supply. At a particular load, the current drawn is 15A and the speed is 24 rev/s. Determine the generated e.m.f. at this load.

Series Motor - Problem 23

• Example problem (cont.):
  • Generated e.m.f. (E1) = 240V - (15A * (0.2Ω + 0.3Ω)) = 232.5V
• The problem asks for the calculated speed when the load current increases to 30A, assuming the flux doubles.

Compound motor

• There are two types of compound wound motors:
  • Cumulative compound: Series field winding assists the shunt field winding.
  • Differential compound: Series field winding opposes the shunt field winding.

DC Motor Starters

• When a DC motor is initially connected to its supply voltage, the fuses can easily blow.
• Additional resistance is necessary in the armature circuit at the start of the motor.
• Resistance is reduced as the motor speed increases and the generated e.m.f. increases, limiting current flow.

Ward-Leonard System

• A method used to control DC motor speed.
• The system comprises a motor-generator set along with a control system.
• Components include a prime mover (e.g., an AC motor), a DC generator, and a DC motor.
• The field winding of the DC generator is adjusted, controlling the output voltage which in turn regulates the speed.

Solid-State Speed Controllers

• Modern devices using semiconductor technologies to control motor speed are more efficient than traditional systems.
• They address factors like efficiency, reliability, and compactness.
• The controllers adjust voltage, frequency, and current within the electrical supply to the motor.
• Different types include variable frequency drives (VFDs), soft starters, choppers, and phase-controlled rectifiers, for both AC and DC motors.

DC Motor Efficiency Calculations

• Efficiency (η) = (output power / input power) * 100%

• Total losses = I²Ra + IV + C (shunt motor)

  • C = iron, friction and windage losses

• Input power = VI

• Output power = VI - losses

• Example problem: Calculate efficiency, given values for total current, voltage, armature resistance, shunt field resistance, iron, friction and windage losses.

Separately Excited Generator

• A typical separately excited generator circuit has a distinct field winding and armature winding, powered by a separate DC supply.
• Terminal voltage (V) = generated e.m.f. (E) minus armature voltage drop (IaRa).
• Generated e.m.f. (E) = terminal voltage (V) plus armature voltage drop (IaRa).

Separately Excited Generator - Problem 9

• Example problem: A separately excited generator generates e.m.f. at various given operating conditions.

Shunt DC Generator

• A circuit where the field winding is connected in parallel with the armature winding.
• The field current is a fraction of the armature current due to the field winding's high resistance.
• Terminal voltage (V) = generated e.m.f. (E) - armature voltage drop (IaRa).
• Generated e.m.f. (E) = terminal voltage (V) + armature voltage drop (IaRa).

Shunt DC Generator - Problem 10

• Example problem: A shunt generator supplies a load at a given voltage and current, calculating the terminal voltage and generated e.m.f., given various resistances.

Series DC Generator

• A circuit linking the field winding and armature winding in series.

Compound-Wound Generator

• A combination of shunt and series windings, to unify advantages of both shunt and series systems.
• Two types shown: long-shunt and short-shunt.
• More common is a short-shunt compound generator.

Compound-Wound Generator - Problem 11

• Example problem: A short-shunt compound generator supplies current under given conditions, determining the generated e.m.f. with known resistances.

This quiz covers essential concepts and problems related to shunt motors, including supply voltage, generated e.m.f., and current calculations. It provides example problems to help you grasp the relationship between armature current and motor speed. Test your understanding of shunt motor operations and performance metrics.