DC GENERATOR (Part 2)
E2063/ Unit 3/ 1
UNIT 3
INTRODUCTION TO DC GENERATOR (Part 2)
OBJECTIVES
General Objective
: To apply the generated e.m.f. efficiency and power losses of a DC
generator
Specific Objectives : At the end of the unit you will be able to :
¾
Calculate generated e.m.f. in an armature winding for a generator
¾
List DC machine losses and calculate efficiency
¾
Calculate the power losses of DC generator
¾
Calculate the efficiency of DC generator
DC GENERATOR (Part 2)
E2063/ Unit 3/ 2
INPUT
3.0
E.m.f generated
Let
Z = number of armature conductors,
Φ = useful flux per pole in Webers
Ρ = number of pairs of poles
n = armature speed in rev/s
The e.m.f generated by the armature is equal to the e.m.f. generated by one of the parallel
paths. Each conductor passes 2Ρ poles per revolution and thus cuts 2ΡΦ webers of magnetic
flux per revolution. Hence flux cut by on conductor per second = 2ΡΦn Wb and so the average
e.m.f. generated per second is given by:
E = 2ΡΦn Wb
(since 1 volt = 1 Weber per second)
Let c = number of parallel paths through the winding between positive and negative
brushes
( c=2 for a wave winding and c= 2p for a lap winding)
The number of conductors in series in each path = Z/c
DC GENERATOR (Part 2)
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The total e.m.f. between brushes = (average e.m.f/conductor)
( number of conductor in series per path)
= 2ΡΦnZ/c Wb
i.e.
generated e.m.f. E =
2pΦZn
volts
c
(1)
Since Z, p and c are constant for a given machine, then
E ∝ Φω
(2)
Example 3.1
An 8-pole generator, wave connected armature has 600 conductor and is driven 625 rev/min. If
the flux per pole is 20mWb, determine the generated e.m.f.
Solution to Example 3.1
Given
Z
=
600
c
=
2 for a wave winding
P
=
4 pairs
n
=
625/60 rev/min
Φ
=
30 × 10-3 Wb
generated e.m.f. E =
2pΦZn
volts
c
2(4)(20 × 10- 3 )(
=
= 500 V
2
625
)
60
DC GENERATOR (Part 2)
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Example 3.2
A 4-pole generator has a lap-wound armature with n50 slots with 16 conductors per slot. The
useful flux per pole is 30mWb. Determine the speed at which the machine must be driven to
generate an e.m.f. of 240 volts.
Solution to Example 3.2
Given
E = 240 V
Z = 50 x 16 = 800
c = 2p (for a lap winding)
Φ = 30 × 10-3 Wb
generated e.m.f. E =
2pΦnZ 2pΦnZ
=
=ΦnZ
c
2p
Rearranging gives speed
n=
240
E
=
ΦZ (30 × 10− 3 )(800)
= 10 rev/s
DC GENERATOR (Part 2)
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Activity 3A
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE TO THE NEXT
INPUT…!
3.1
An 8-pole, lap wound armature has 1200 conductors and a flux per pole of
0.03 Wb. Determine the e.m.f generated when running at 500 rev/min.
3.2
Determine the generated e.m.f. in question 3.1 if armature is wave-wound.
3.3
A DC shunt generator running at constant speed generates a voltage 150 V at a
certain value of field current. Determine the change in the generated voltage
when the field current is reduced by 20 per cent, assuming the flux is
proportional to the field current.
3.4
Calculate the e.m.f. generated by 4 pole wave-wound generator having 65 slots
with 12 conductors per slot when driven at 1200 r.p.m. the flux per pole is
0.02 wb.
3.5
An 8 pole, lap-wound armature rotated at 350 r.p.m. is required to generate
260 V. The useful flux per pole is 0.05 wb. If the armature has 120 slots,
calculate the number of conductors per slot.
DC GENERATOR (Part 2)
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Feedback to Activity 3A
3.1
300 V
3.2
1200 V
3.3
120 V
3.4
624 V
3.5
7.14
DC GENERATOR (Part 2)
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INPUT
3.1
Power losses and Efficiency
For any type of machine, output power is different from input power. The difference is caused
by power losses that had happened whenever one type of energy is converted or delivered to
the other type.
3.1.1
DC Generator Losses
As mentioned in the previous unit, a generator is a machine that converts mechanical energy to
electrical energy. When such conversion takes place, certain losses occur which are dissipated
in the form of heat. The principal losses of machine are:
(a) Copper loss, due to I2R heat losses in the armature windings.
(b) Iron loss, due to hysteresis and eddy current losses in the armature. This loss can be
reduced by constructing the armature of silicon steel laminations having a high
resistivity and low hysteresis. At constant speed, these losses are assumed to be
constant.
(c) Friction and windage losses, due to bearing and brush contact friction and losses due to
air resistance against moving parts (called windage). At constant speed, these losses are
assumed to be constant.
(d) Brush and contact loss between the brushes and commutator. This loss is approximately
proportional to the load current.
DC GENERATOR (Part 2)
3.1.2
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Efficiency of DC generator
The efficiency of an electrical machine is the ratio of the output power and is usually expressed
as percentage. The greek letter ‘η’ (eta) is used to signify efficiency and since the units are,
power/power, the efficiency has no units. Thus
efficiency, η = (
output power
) × 100%
input power
(3)
If the total resistance of the armature circuit is Ra, the the total loss in the armature is Ia2Ra.
If the terminal voltage is V and the current in the shunt circuit is If, the the loss in the circuit is
IfV.
If the sum of the iron, friction and windage losses is C the total losses is given by : Ia2Ra+ IfV +
C (Ia2Ra+ IfV is, in fact the ‘copper loss’)
If the output current is I, then the output power is VI. The total input power = VI + Ia2Ra+ IfV +
C.
Hence
Efficiency, η = (
i.e.
η= (
output
)
input
VI
) × 100%
VI + I Ra + I f V + C
2
a
The efficiency of a generator is a maximum whwn load is such thet
Ia2Ra = IfV + C
i.e. when the variable loss = constant loss
(4)
DC GENERATOR (Part 2)
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Example 3.3
A shunt generator supplies 96 A at a terminal voltage of 200 volts. The armature and shunt
field resistances are 0.1Ω and 50Ω respectively. The iron and frictional losses are 2500 W.
Find :
(i)
e.m.f generated.
(ii)
copper losses
(iii)
commercial efficiency.
Solution to Example 3.3
Fig.3.1 shows the connections of shunt generator.
Figure 3.1
(i)
Ish = 200/ 50 = 4 A
Ia = IL + Ish = 96 + 4 = 100 A
Eg = V + IaRa = 200 + 100 × 0.1 = 210 V
Armature Cu loss
(ii)
= Ia2Ra = (100)2 × 0.1 = 1000 W
Shunt Cu loss = I2sh Rsh = (4)2 × 50 = 800 W
Total Cu loss = 1000 + 800 = 1800 W
(iii)
Total losses
= Stray losses + Cu losses
= 2500 + 1800 = 4300 W
Output power = 96 × 20 = 19200 W
Input power
∴
η
=
=
19200 + 4300 = 23500 W
19200
× 100 = 81.7 %
23500
DC GENERATOR (Part 2)
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Example 3.4
A 10 kW d.c shunt generator has the following losses at full load :
Mechanical losses
290 W
Iron losses
420 W
Shunt Cu loss
120 W
Armature Cu loss
595
Calculate the efficiency at :
(i)
no load
(ii)
25% of full load
Solution to Example 3.4
(i)
At no load
The stray losses remain constant at no load or any other load.
Stray losses
= Mechanical losses + Iron losses = 290 +420 = 710 W
Total losses at no load
= 710 + Shunt Cu loss =710 + 120 = 830 W
The efficiency at no load is zero because there is no input.
At 25% of full-load
(ii)
Constant losses
= 830 W
Armature Cu loss
= ** (1/4)2 of F.L value = (1/4)2
Total losses
= 830 + 37 = 867 W
Output
= (1/4)
Input
= 2500 + 867 = 3367 W
∴
η
=
2500
3367
× 595 = 37 W
× 10 kW = 2.5 kW = 2500 W
× 100 = 74 %
Note:
**
When the generator is loaded to 25% of its normal rating, armature current is
25% (or 1/4 ) of its full load value. Since Cu loss varies as square of current, we
have (1/4)2 term here.
DC GENERATOR (Part 2)
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Activity 3B
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE TO THE NEXT INPUT…!
3.6
A 10kW shunt generator having an armature circuit resistance of 0.75 Ω and a field
resistance of 125 Ω, generates a terminal voltage of 125 V at full load. Determine the
efficiency of the generator at full load, assuming the iron, friction and windage losses
amount to 600 W.
3.7
In DC generator iron losses are made up of ………………. and ……………..
3.8
This shunt generator delivers full load current of 200 A at 240 V. The shunt field
resistance is 60Ω and full-load efficiency is 90 %. The stray losses are 800 W. Find :
3.9
(i)
armature resistance
(ii)
current at which maximum efficiency occurs.
A 75 kW shunt generator is operated at 230 V. The stray losses are 1810 W and shunt
field circuit draws 5.35 A. The armature circuit has a resistance of 0.035 Ω and brush
drop is 2.2 V. Calculate :
(i)
total losses
(ii)
input of prime mover
(iii)
efficiency at rated load.
DC GENERATOR (Part 2)
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Feedback to Activity 3B
3.6
80.50 %
3.7
hysteresis and eddy iron losses
3.8
(i) 0.0858 Ω
(ii) 143.22 A
3.9
(i) 7615 W
(ii) 82615 W
(iii) 90.8 %
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SELF-ASSESSMENT 3
You are approaching success. Try all the questions in this self-assessment section and check
your answers with those given in the Feedback on Self-Assessment 3 given on the next page. If
you face any problems, discuss it with your lecturer. Good luck.
Question 3-1
(a)
The e.m.f. generated in an armature winding is given by E = 2pΦnZ/c volts. State what
p, Φ, n, Z, and c represent.
(b)
A 4-pole, wave-connected armature of a DC machine has 750 conductors and is driven
at 720 rev/min. If the useful flux per pole is 15mWb, determine the generated e.m.f.
(c)
A 4-pole armature DC machine has 1000 conductors and a flux per pole of 20 mWb.
(d)
Determine the e.m.f. generated when running at 600 rev/min when the armature is
(e)
(a)
wave-wound
(b)
lap-wound
Determine the terminal voltage of a generator which develops an e.m.f. of 240 V and
has an armature current 50 A on load. Assume the armature resistance is 40mΩ.
Question 3-2
(a)
State the principal losses in DC generator.
(b)
The efficiency of a DC machine is given by the ratio (…..) percent.
(c)
A 15 kW shunt generator having an armature circuit resistance of 0.4 Ω and a field
resistance of 100 Ω, generates a terminal voltage 240 V at full load. Determine the
efficiency of the generator at full load, assuming the iron, friction and windage losses
amount to 1 kW.
DC GENERATOR (Part 2)
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FEEDBACK TO SELF-ASSESSMENT 3
Have you tried the question ????? If “YES”, check your answers now
Answer of Question 3-1
(a)
Ρ = number of pairs of poles
Φ = useful flux per pole in Webers
n = armature speed in rev/s
Z = number of armature conductors
C = number of parallel paths
(b)
270 V
(c)
(a) 400 V
(d)
238 V
(b) 200 V
Answer of Question 3-2
(a)
copper loss, iron loss, friction loss, brush contact loss
(b)
(
(c)
82.14%
output power
) × 100%
input power
CONGRATULATIONS!!!!…..
May success be with you
always….