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)
60
E =
6
12
( )( )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)
60
E =
12
2
( )( )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)
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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 |
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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 |
IL IA +
VT
- +
VF
- IF |
+
VT
- IL IS IA
| IL IA IL +
VT
- IF | IA IL +
VT
- IF
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VOLTAGE TERMINAL
|
VT = E - IARA |
VT = E - IA(RA+RS) |
VT = E - IARA |
VT = E - IA(RA+RS)
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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)
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Shunt Generator
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VT = E - IARA
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Compound Generator |
FEEDBACK TO ACTIVITY 2 B
2.3
SET A |
| SET B |
VT = E - IA(RA+RS) |
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Shunt Generator
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VT = E - IARA
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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.
- a = 2
- Z = 288
- E = 216
B. 56
Question 2-2
A.
(a) E = 210.3V
(b) F = 12.7mWb
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