UNIT3

E3106/03/13

ELECTRICAL MACHINERY & CONTROL                                                                                                          

UNIT 3

 

 

DIRECT CURRENT (DC) MACHINERY (PART III)

 

             

                            OBJECTIVES

 

 

General Objective

 

To understand the principle of the direct current (DC) motor

 

 

Specific Objectives

 

At the end of the unit you would be able to:

 

• describe the concept of back EMF
• write the formulae of input voltage
• write the formulae of torque by using power out put and speed rotation parameters

 

 

 

 

 

 

                                          INPUT

 

 

 

3. INTRODUCTION DC MOTOR

 

 

ow that we have a good understanding of DC generators, we can begin our study of DC motors. Direct Current (DC) motors transform electrical energy into mechanical energy. They drive devices such as hoist, fans, pumps, calendars, punch presses, and cars. DC motors are preferred for portable operation. The best portable power supply we have today is the battery, which supplies only direct current. DC motors are used when some specific characteristic are required. For example, small DC motors can be used to produce a very high torque. The speed of DC motor is very easy to control with simple circuits. Another reason for choosing DC motors owes to their reversible direction of rotation. In this section, we will discuss the back electro motive force EMF (counter voltage) the formulae of input voltage and the torque.

 

 

Figure 3.1: Direction of Counter Voltage

3.1              CONCEPT OF BACK EMF

(COUNTER VOLTAGE)

 

If you still remember Faraday’s Law(i) states

that any time, if there is a relative motion

between a conductor and magnetic field, there

will always be voltage generated. Figure 3.1

illustrate a conductor located in magnetic field. Conventional current flowing through the conductor causes it to move in one direction – downward in this case. When the conductor moves, a counter voltage will be generated in it because it is moving through the magnetic field.

 

(i)Faraday’s Law

 

A voltage is induced any time there is relative motion between a magnetic field and a conductor.

 

(ii)Lenz’s Law

 

A counter voltage is always such that it will produce (or try to produce) a current that oppose the motion that produce it.

 

 

(iii)Right Hand Rule

 

 

 

 

 

 

(ii)The conventional current flow in

Figure 3.3 moves the conductor down; the

counter voltage will try to set up a current to

prevent the conductor from moving. Use

the right hand generator rule(iii) to verify that

the counter voltage will try to set up a

current with a field that moves the conductor

in the opposite (upward) direction. The

counter voltage produces limitations on the

motor action. For one thing, the amount of

current that can be delivered to the motor for

efficient rotation (and also, the amount of

force on the conductor) is definitely limited

by the counter voltage.

 

The counter voltage does not actually set up

an opposing current. What it does is to

oppose the conventional current that sets up

the motion. The amount of voltage induced

into a conductor moving in magnetic field is

directly related to the rate at which the

conductor cuts through the flux lines. So, a

rapid moving conductor will induce more counter voltage than a slow moving one. This relates directly to the motion of conductors in a motor. The amount of counter voltage is dependent upon the speed of the motor. The counter voltage changes as the motor speed changes, and therefore its effect on the speed of the motor also changes.

 

                                                        

Test your UNDERSTANDING before you continue to the next input

ACTIVITY 3 A

 

 

 

 

 

 

 

1. By using these three statements below, determine the concept of back EMF.

Faraday’s Law

 

A voltage is induced any time there is relative motion between a magnetic field and a conductor.

 

Lenz’s Law

 

A counter voltage is always such that it will produce (or try to produce) a current that oppose the motion that produce it.

 

 

 

 

 

 

 

Right Hand Rule

 

 

 

 

 

 

 

 

 

 

 

2. What is the factor which affects the changes in back EMF motor DC.

FEEDBACK TO ACTIVITY 3 A
             

 

3.1

If you still remember Faraday’s Law(i) states that any time, if there is a relative motion between a conductor and magnetic field, there will always be voltage generated. Figure 3.1 illustrate a conductor located in magnetic field. Conventional current flowing through the conductor causes it to move in one direction – downward in this case. When the conductor moves, a counter voltage will be generated in it because it is moving through the magnetic field.

 

(ii)The conventional current flow in Figure 3.3 moves the conductor down; the counter voltage will try to set up a current to prevent the conductor from moving. Use the right hand generator rule(iii) to verify that the counter voltage will try to set up a current with a field that moves the conductor in the opposite (upward) direction. The counter voltage produces limitations on the motor action. For one thing, the amount of current that can be delivered to the motor for efficient rotation (and also, the amount of force on the conductor) is definitely limited by the counter voltage.

 

The counter voltage does not actually set up an opposing current. What it does is to oppose the conventional current that sets up the motion. The amount of voltage induced into a conductor moving in magnetic field is directly related to the rate at which the conductor cuts through the flux lines. So, a rapid moving conductor will induce more counter voltage than a slow moving one. This relates directly to the motion of conductors in a motor.

 

3.2              speed of the motor DC

 

 

 

 

INPUT

 

 

 

2. FORMULAE OF INPUT VOLTAGE

 

 

After we understand the back EMF, we must know the formulae of the input voltage. When the machine is operating as a motor, the EMF (E) is less than the applied voltage (V), and the direction of the current (Ia) is the reverse when the machine is acting as a generator. The equation for input voltage can be expressed by (i).

……….(i)

where:

 

V              : Applied voltage

E              : Back EMF

Ia              : Current armature

Ra              : Resistance armature

 

 

 

 

 

 

 

  Example 3.1

The armature of a DC machine has a resistance of 0.1Ω and is connected to a 230V supply. Calculate the back EMF when it is running as a motor taking 60A.

              Solution

                                          Given:

Ra	=	0.1Ω
Ia	=	60A
V	=	230V

 

                                          Recall:

\Back EMF

             

              E = 230 – (60)(0.1)

              E = 224V#

 

 

  Example 3.2

A DC motor takes an armature current of 110A at 480V. The counter voltage of the DC motor is 458V. Calculate the resistance of the armature motor.

              Solution

                                          Given:

V	=	480V
Ia	=	110A
E	=	458V

 

                                          Recall:

\ The resistance of the armature

 

             

 

 

 

 

 

Test your UNDERSTANDING before you continue to the next input

ACTIVITY 3 B
 

 

 

 

 

 

3. Find the terms used in the equation below from the word maze given.

 

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4. A series motor runs at 600rpm when taking 100A from 240 V supply. The resistance of the armature circuit is 0.12W and that of the series winding is 0.03W. Calculate the back EMF.

 

 

 

 

 

FEEDBACK TO ACTIVITY 3 B
 

 

3.3

 

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3.4              227V

                                          Given:

Ra	=	0.12Ω
Ia	=	100A
V	=	240V
Rs	=	0.03Ω

 

 

                                          Recall:

\Back EMF

             

              E = 240 – (100)(0.12+0.03)

              E = 225V#

 

 

 

                                          INPUT

 

 

 

3.2              THE FORMULAE OF TORQUE

 

 

We will focus on, torque T, we know that the mechanical power P is given in equation (ii);

……(ii)

 

 

So, from equation (ii) the formulae of the torque can be derived as in the equation (iii).

Nm ……(iii)

where:

 

P              : Mechanical power

N              : Speed

T              : Torque

 

 

 

 

 

 

 

Or, in other words the torque of a given DC motor is proportional to the product of the armature current and the flux per pole.

 

 

 

 

 

 

 

 

  Example 3.3

A DC motor has rotated the rotor at the speed of 636 rpm. The mechanical power developed by armature is 50.38kW. Calculate the torque of this motor DC.

              Solution

                                          Given:

Nr	=	636rpm
P	=	50.38kW

 

                                          Recall:

\ The torque of the motor DC

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Test your UNDERSTANDING before you continue to the next unit.

ACTIVITY 3 C
 

 

 

 

 

5. Determine the formulae of the torque DC motor.

 

6. What are the factors proportional with the torque of the DC motor?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

FEEDBACK TO ACTIVITY 3 C
 

 

 

3.5

 

Nm ……(iii)

where:

 

P              : Mechanical power

N              : Speed

T              : Torque

 

 

 

 

 

 

 

 

 

 

3.6              The armature current and the flux per pole.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

SELF-ASSESMENT

 

If you face any problem, discuss it with your lecturer

You are approaching success. TRY all the questions ini this self-assessment section and check your answers with those given in the feedback on Self-Assessment given on the next page.

 

 

Question 3-1

 

1. What are the laws and rules are used to explain the back EMF?

 

2. Based on the answer (A), explain the back EMF of the DC motor?

 

3. How can the speeds of the motor is related to the changes of back EMF?

 

 

Question 3-2             

 

A.              A 240V DC shunt motor has an armature of resistance of 0.2W. Calculate:

              (a)                            the value of resistance which must be introduced into the armature circuit to

                            limit the starting current to 40A.

2. the back EMF when the motor is running at a constant speed with this additional resistance in circuit and with an armature current of 30A

 

Question 3-3

 

A.              Calculate the torque in Nm, develop by a DC motor having an armature resistance of 0.25W and running at 750rpm when taking an armature current of 60A from a 480V supply.

 

B.              A DC series motor connected across a 460V supply runs at 500rpm when the current is 40A. The total resistance of the armature and field circuits is 0.6W. Calculate the torque of the armature in Nm.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

FEEDBACK TO SELF-ASSESMENT

 

 

 

 

Question 3-1

 

A.              refer to the 3.1

B.              refer to the 3.1

C.              refer to the 3.1

 

Question 3-2

 

1. 5.8 W
2. 60V

 

Question 3.3

 

A.              356Nm

B.              333Nm