ALL POLITEKNIK MODULE: UNIT2

E3106/02/10

ELECTRICAL MACHINERY & CONTROL

UNIT 2

DIRECT CURRENT (DC) MACHINERY (PART II)

OBJECTIVES

General Objective

To application of theoretical in solving mathematic of direct current (DC) generator

Specific Objectives

By the end of this unit, you would be able to:

calculate by using the formulae of the generate voltage
solve the problem by using the formulae of the terminal voltage

INPUT

THE INTERNAL GENERATED VOLTAGE EQUATIONS OF REAL DC GENERATOR

Total number of conductors (Z)

Z = 2CNC

Where:

C is number of coil on the armature

NC is number of turns of wire

Number of current path (a)

for lap windings: a = mP
for wave winding: a = 2m
for frog-leg winding: a = 2Pm

Where:

m: plex of the windings

P: number of poles on the machine

ow can the voltage in the rotor windings of a real machine be determined? The voltage out of armature of a real machine is equal to the number of conductor per current path times the voltage on each conductor. So the internal generated voltage in the machine can be expressed as

Remember This…

Where:

E	:	Generated voltage (V)
N	:	Speed (rpm)
Z	:	Total number of conductors
f	:	Flux of a pole
P	:	Number of pole
A	:	number of current path

Example 2.1

A duplex lap wound armature is used in a six pole DC machine. There are 1728 conductors in this machine. The flux per pole in the machine is 0.039Wb, the speed of the machine is 400rpm, and the current path in this machine is 12. What is the generated voltage in DC machine?

Solution

Given:

N	=	400 rpm
Z	=	1728
f	=	0.039Wb
P	=	6
a	=	12

Recall:

\ Voltage generated;

(400)(1728)(0.039)

E =

( )( )

E = 224.6 V#

The induced voltage in any given machines depends on three factors:

the flux in the machine
the speed of the machine’s rotor
a constant depending on the construction of the machine

Example 2.2

A 12 pole DC generator has a simplex wave wound armature containing 144 coils of 10 turns each. Its flux per pole is 0.05Wb, and it is turning at a speed of 200rpm. Calculate:

the current paths in this machine
the number of conductor in this machine
the generated voltage in DC machine

Solution

Given:

N	=	200 rpm
f	=	0.05Wb
P	=	12
m	=	1
C	=	144
NC	=	10

current paths in this machine simplex wave wound; a

Recall: a = 2m

a = 2 (1) = 2#

number of conductor in this machine; Z

Recall: Z = 2CNC

Z = 2 (144) (10) = 2880#

the generated voltage in DC machine; E

Recall:

(200)(2880)(0.05)

E =

( )( )

E = 2880 V#

Test your UNDERSTANDING before you continue to the next input

ACTIVITY 2 A

Determine the parameters that influence the voltage generated in DC machine?

From the equation below, solve this puzzle.

……….(a)


	(iii)

				(i)


(ii)	N			T


		(i)	(ii)	E

Horizontal:

it is a …… expression in rpm
the number of …… can be

express as Z = 2C2CNC

Vertical:

equation (a) is a formulae of ………

generated

machine has a magnetic ………
“a” is a number of …………… path

in the machine

FEEDBACK TO ACTIVITY 2 A

2.1

a.	Generated voltage (V)
b.	Speed (rpm)
c.	Total number of conductors
d.	Flux of a pole
e.	Number of pole
f.	number of current path


		(iii)C
		U
		R				(i) V
		R				O
		E				L
(ii) C	O	N	D	U	C	T	O	R
		T				A
						G
				(i) S	(ii) P	E	E	D
					O
					L
					E

2.2

INPUT

2.2 TERMINAL VOLTAGE EQUATIONS OF DC GENERATOR

From the previous unit, we know that excitation in DC generator can be divided into two major types; separately excited generator and self-excited generator. There are three types of self-excited generator such as series generator, shunt generator and compound generator. Each excitation has different equation for terminal voltage. Therefore, in this section we will determine each formulae by using the Table 2.1.

Table 2.1: Terminal Voltage and Armature Current

Separately

Generator

Series

Generator

Shunt

Generator

Compound Generator

EQUAVALENT CIRCUIT

VOLTAGE TERMINAL

VT = E - IARA

VT = E - IA(RA+RS)

VT = E - IARA

VT = E - IA(RA+RS)

Example 2.3

A separately excited DC generator is rated at 172 kW, generated voltage is 382V, current armature of this machine is 360A, and resistor armature is 0.05W. Calculate the terminal voltage.

Solution

Given:

E	=	382 V
IA	=	360A
RA	=	0.05W

Recall: VT = E - IARA

\ Terminal voltage;

VT = 382 – (360) (0.05) = 364V#

Test your UNDERSTANDING before you continue to the next input

ACTIVITY 2 B

Match the equation with the best answer.

SET A

SET B

VT = E - IA(RA+RS)

Shunt Generator

VT = E - IARA

Compound Generator

FEEDBACK TO ACTIVITY 2 B

2.3

SET A

SET B

VT = E - IA(RA+RS)

Shunt Generator

VT = E - IARA

Compound Generator

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 2-1

A four pole wave connected armature has 12 coils on the armature, each containing 12 turns and is driven at 900rpm. If the useful flux per pole is 25mWb. Calculate:

the current paths in this machine
the number of conductor in this machine
the generated voltage in DC machine

An eight pole lap connected armature, driven at 350rpm is required to generate 260V. The useful flux per pole is about 0.05Wb. If the armature contains eight turns, calculate the suitable number of coils on the armature.

Question 2-2

A. A six pole, lap-wound, 220V, shunt excited DC machine takes an armature current of 2.5A when unloaded at 950 rpm. When loaded it takes an armature current of 54A from the supply and runs at 950rpm. The resistance of the armature is 0.18W and there are 1044 number of conductor. Find:

the generated voltage
the useful flux per pole

FEEDBACK TO SELF-ASSESMENT

Question 2-1

a = 2
Z = 288
E = 216

B. 56

Question 2-2

(a) E = 210.3V

(b) F = 12.7mWb

September 4, 2009

UNIT2