DC Machines: Permanent Magnet, Series & Compound Motors - Theory and Exercises, Slides of Electrical Engineering

Download DC Machines: Permanent Magnet, Series & Compound Motors - Theory and Exercises and more Slides Electrical Engineering in PDF only on Docsity!

Electrical Machines

DC Machines-

Permanent Magnet,

Series & Compound

Nuha Nadhiroh

• Equivalent Circuit
• Magnetization Curve
• Speed Control
• Excercises

Overview

Permanent Magnet DC Motor

Series DC Motor

A series dc motor is a DC motor whose field windings consist of a relatively few turns connected in series with the armature circuit.

Terminal Characteristic

To determine the techninal characteristic of a series dc motor, an analysis will be based on the assumption of a linear magnetization curve, and then the effects of saturation will be considered in a graphical analysis

If the flux can be eliminated from this expression, it will directly relate the torque of a motor to its speed. To eliminate the flux from the expression

Torque speed characteristic of series DC motor

Excercise

a 250-V series dc motor with compensating windings. and a total series resistance RA + RS of 0.

Ω. The series field consists of 25turns per pole. with the magnetization curve shown in Figure.

Find the speed and induced torque of this motor for when its armature current is 50A.

Compounded DC Motor

The Kirchhoff's voltage law equation for a compounded dc motor is

The currents in the compounded motor are related by

The net magnetomotive force and the effective shunt field current in the compounded motor are given by

Torque Speed in Cumulatively

Compounded DC Motor

The torque-speed characteristic of a cumulatively compounded dc motor compared to series and shunt motors with the same full-load rating

The torque-speed characteristic of a cumulatively compounded dc motor compared to a shunt motor with the same no-lood speed

Excercise

A 100-hp, 250-V compounded dc motor with compensating windings has an internal resistance,

including the series winding, of 0.04. There are 1000 turns per pole on the shunt field and 3 turns

per pole on the series winding. At no load, the field resistor has been adjusted to make the motor

run at 1200 r/min. The core, mechanical, and stray losses may be neglected

(a) What is the shunt field current in this

machine at no load?

(b) If the motor is cumulatively compounded,

find its speed when I (^) A, = 200 A.

(c) If the motor is differentially compounded,

find its speed when I (^) A, = 200 A.

Magnetization Curve

DC MOTOR EFFICIENCY CALCULATIONS

1. The copper losses in the motor are the I 2 R losses in the armature and field circuits of the motor. These losses can be found from a knowledge of the currents in the machine and the two resistances. To determine the resistance of the amature circuit in a machine, block its rotor so that it cannot turn and apply a small dc voltage to the armature terminals. Adjust that voltage until the current flowing in the armature is equal to the rated armature current of the machine. The ratio of the applied voltage to the resulting armature current flow is RA. 2. Brush drop losses are often approximately lumped together with copper losses. If they are treated separately, they can be determined from a plot of contact potential versus current for the particular type of brush being used. The brush drop losses are just the product of the brush voltage drop V (^) BD and the armature current I (^) A. 3. The core and mechanical losses are usually determined together. If a motor is allowed to turn freely at no load and at rated speed, then there is no output power from the machine. Since the motor is at no load, I (^) A is very small and the armature copper losses are negligible. Therefore , if the field copper losses are subtracted from the input power of the motor, the remaining input power must consist of the mechanical and core losses of the machine at that speed

Excercise

A 50-hp. 250-V. 1200 r/min shunt dc motor has a rated armature current of 170 A and a rated field current of 5 A.

When its rotor is blocked, an armature voltage of 10.2 V (exclusive of brushes) produces 170 A of current flow. and

a field voltage of 250 V produces a field current flow of 5 A. The brush voltage drop is assumed to be 2 V. At no

load with the terminal voltage equal to 240 V, the armature current is equal to 13.2 A. the field current is 4.8 A. and

the motor's speed is 115O r/min.

(a) How much power is output from this motor at rated conditions?

(b) What is the motor's efficiency?