Transcript B - Weebly

Commutation
It has been seen that currents in armature conductors of a DC generator are
alternating.
To make their flow unidirectional in the external circuit, we need a commutator.
Moreover, these currents flow in one direction when armature conductors are under
N-pole and in the opposite direction when they are under S-pole.
As conductors pass out of the influence of a N-pole and enter that of S-pole, the
current in them is reversed.
This reversal of current takes place along magnetic neutral axis or brush axis i.e.
when the brush spans and hence short-circuits that particular coil undergoing
reversal of current through it.
This process by which current in the short-circuited coil is reversed while it
crosses the magnetic neutral axis or plane is called commutation.
OR, Commutation is the process a coil undergoes after leaving the magnetic field on
one pair of poles and just before entering the magnetic field of the next pair of poles.
OR, the changes that take place in winding elements during the period of shortcircuit by a brush is called commutation.
The brief period during which coil remains short-circuited is known as
commutation period Tc.
If the current reversal i.e. the change from +I to zero and then to –I is completed by
the end of short circuit or commutation period, then the commutation is ideal.
If current reversal is not completely by that time, the sparking is produced between the
brush and the commutator which results is progressive damage to both.
Let us discuss the process of commutation or current reversal in more detail with the
help of Fig. 27.10 where ring winding has been used for simplicity.
The brush width is equal to the width of one commutator segment and one mica
insulation.
In Fig. 27.10(a) coil B is about to be shortcircuited because brush is about to come in touch
with commutator segment ‘a’.
It is assumed that each coil carries 20A, so that
brush current is 40A.
It is so because every coil meeting at the brush
supplies half the brush current whether lap
winding or wave winding.
Prior to the beginning of short circuit, coil B belongs to the group of coils lying to the
left of the brush and carries 20A from left to right.
In Fig. 27.10 (b) coil B has entered its period of
short-circuit and is approximately at one third of
this period.
The current through coil B has reduced down
from 20A to 10A because the other 10A flows
via segment ‘a’.
As area of contact of the brush is more with
segment ‘b’ than with segment ‘a’, it receives
30A from the former, the total again being 40A.
Fig. 27.10(c) shows the coil B in the middle of
its short-circuit period. The current through it
has decreased to zero.
The two currents of value 20A each, pass to
the brush directly from coil A and C as shown.
The brush contact areas with the two
segments ‘b’ and ‘a’ are equal.
In Fig. 27.10(d), coil B has become part of the group of
coils lying to the right of the brush.
It is seen that brush contact area with segment ‘b’ is
decreasing rapidly whereas that with segment ‘a’ is
increasing.
Coil B now carries 10A in the reverse direction, which
combines with 20 A supplied by the coil A to make up
30A that passes from segment ‘a’ to the brush.
The other 10A is supplied by coil C and passes from segment ‘b’ to the brush, again
giving a total of 40A at the brush.
Fig. 27.10(e) depicts the moment when coil B is almost at
the end of commutation or short-circuit period.
For ideal commutation, current through it should have
reversed by now but, as shown, it is carrying 15A only
(instead of 20A).
The difference of current between coils C and B i.e. 2015=5A in this case, jumps directly from segment ‘b’ to the
brush through air thus producing spark.
If the changes of current through the coil B are
plotted on a time base (as shown in Fig. 27.11) it will
be represented by a horizontal line AB i.e. a constant
current 20A up to the time of being of commutation.
From the finish of commutation, the current will be
represented by another horizontal line CD.
Now, again the current value is FC= 20A, although
in the reverse direction.
The way in which current changes from its positive value of 20A (BE) to zero and
then to its negative value of 20A (=CF) depends on the conditions under which the
coil B undergoes commutation.
If the current varies at a uniform rate i.e. if BC is a straight line, then it is referred to
as linear commutation.
However, due to the production of self-induced e.m.f. in the coil the variations
follow the dotted curve.
It is seen that, in that case, current in coil B has reached only a value of KF=15A in
the reversed direction, hence the difference of 5A passes as a spark.
So, we conclude that sparking at the brushes, which results in poor commutation is
due to the inability of the current in the short-circuited coil to reverse completely by
the end of short-circuit period.
The main causes which retards or delays this quick reversal is the
production of self-induced e.m.f. in the coil undergoing
commutation.
The coil possesses appreciable amount of self-inductance because it
lies embedded in the armature which is built up of a material of high
magnetically permeability.
This self-induced e.m.f. is known as reactance voltage.
This voltage, even though of a small magnitude, produces a large
current though the coil whose resistance is very low due to a shortcircuit.
The large circulating current causes sparking at the brushes, undue
heating of the brushes and commutator, and unnecessary wear of
these components.
Naturally, this effect is undesirable, and the circulating current must
be eliminated.
There are two practical ways of improving commutation i.e. of
making current reversal in the short-circuited coil as sparkles as
possible.
These methods are known as:
(i) resistance commutation and
(ii)e.m.f. commutation (which is done with the help of either brush
lead or interpoles, compoles).
Resistance Commutation
In this method, low-resistance Cu brushes are replaced by
comparative high-resistance carbon brushes.
Using Fig. 27.12, it will be seen that how the commutation process is
improved by using high resistance carbon brushes instead of Cu
brushes.
From Fig. 27.12 [1], it is seen that when the
current I from coil C reaches the commutator
segment b, it has two parallel paths open to it.
The first path is straight from bar b to the brush
and the other parallel path is via the shortcircuited coil B to bar a and then to the brush.
If the Cu brushes (which have low contact
resistance) are used, then the current does not
follow the second longer path, it would
preferably follow the first path.
But when carbon brushes having high resistance are used, then current
I coming from C will prefer to pass through the second path because
(i) the resistance r1 of the first path will increase due to the
diminishing area of contact of bar b with the brush, and because
(ii) resistance r2 of second path will decrease due to rapidly
increasing contact area of bar a with the brush.
Hence, carbon brushes have, usually, replaced Cu brushes.
The additional advantage of carbon brushes are that:
(i) They are to some degree self-lubricating and polish the
commutator, and
(ii) Should sparking occur, they would damage the commutator
less than when Cu brushes are used.
But some of their minor disadvantages are:
(a) Due to their high contact resistance a loss of approximately 2
volts is caused. Hence, they are not much suitable for small
machines where this voltage forms an appreciable percentage
loss.
(b) Owing to this large loss, the commutator has to be made
somewhat larger than with Cu brushes in order to dissipate heat
efficiently without greater rise of temperature, and
(c) Because of their lower current density they need larger brush
holders.
Interpoles or Compoles
These are small poles fixed to the yoke and spaced in
between the main poles.
They are wound with comparatively few heavy gauge
Cu wire turns are connected in series with the armature
so that they carry full armature current.
Their polarity, in the case of generator, is the same as that of the main
pole ahead in the direction of rotation as shown in Fig. 27.13.
As their polarity is the same as that of the main pole ahead, the
induced an e.m.f. in the coil (under commutation) which helps the
reversal of current.
The e.m.f. induced by the interpoles (compoles) is known as
commutating or reversing e.m.f..
The commutating e.m.f. neutralizes the self-induced e.m.f. thereby
making commutation sparkles.
As interpoles carry armature current, their commutating e.m.f. is
proportional to the armature current.
This ensures automatic neutralization of selfinduced e.m.f. which is also due to armature
current.
Connection for a shunt generator with interpoles
are shown in Fig. 27.14.
Another function of the interpoles is to neutralize
the cross-magnetizing effect of armature reaction.
Hence, brushes are not to be shifted from the original position.
This neutralization of cross-magnetizing is automatic and for all
loads because both are produced by the same armature current.
The distinction between the interpoles and compensating windings
should be clearly understood.
Both are connected in series and their mmfs are such as to neutralize
armature reaction.
But interpoles additionally supply mmf for counteracting the selfinduced e.m.f. in the coil under commutation.
Power Stages of DC Generator
Various stages in the case of a DC generator are shown below:
A
Mechanical
Power Input =
Output of
Driving Engine
or Prime Mover
Stray Power losses:
Iron, and Friction
Losses.
Stray Load Loss:
1% of output
B
Electric
Power
Developed in
Armature =
EgIa watt
Following are the three generator efficiencies:
1. Mechanical Efficiency
Eg I a
total
watts
generateed
in
armature
B
m  

A
mechanicalpowersupplied
Outputof driving engine
2. Electrical
Efficiency
Vt I
C
watts
available
in
load
circuit
L
e  

B total wattsgenerateedin armature Eg Ia
3. Overall or
C  wattsavailable in load circuit  


e m
Commercial c A mechanicalpowersupplied
Efficiency
Copper
Losses
C
Electric Power
Output = VtIL
In the case of
Motor
The overall efficiency can also be calculated by
c  output 100 
input
output
100
output  Losses
Total Losses in a DC Generator
The various losses occurring in a generator can be sub-divided as
follows:
(a) Copper Losses
(b) Magnetic Losses (also known as iron or core losses)
(c) Mechanical Losses
(d) Stray load loss
Copper Losses
(a) Armature copper loss= Ia2Ra
(b) Field copper loss.
For shunt generators, it is practically constant and = Ish2Rsh (or VtIsh)
For Series generator, = Ise2Rse
For Compound generator = Ish2Rsh + Ise2Rse.
(c) The loss due to brush contact resistance. It is usually included in
the armature copper loss.
Magnetic Losses (also known as iron or core losses)
2 f2
.6 f
(b) eddy current loss, WeBmax
(a) Hysteresis loss, W B1max
h
These losses are practically constant for shunt and compound wound
generators, because in their case, field current is constant.
Mechanical Losses
(a) Friction loss at bearing, brush and commutator
(b) Air friction or windage loss of rotating armature
Stray Power Losses
Stray power losses are those loses that are produced because the
armature rotates. Including are the iron or core losses and mechanical
losses (friction loss and windage loss). These are also known as
rotational losses.
Bearing Friction Loss
The friction of bearings also consumes energy. Improved bearings
reduce the amount of energy consumed and also bearing wear.
Improper lubrication increases the friction and causes increased
energy consumption.
Brush and Commutator Friction Loss
Energy is also consumed because of the contact between the brush
and the commutator. The coefficient of friction of hard brushes is
greater than that of soft brushes. The frictional losses are function of
the speed. Speed changes will alter the losses.
Windage Loss
Windage loss refers to the energy consumed in moving the air
about the armature.
The greater the quantity of air moved, the greater the power
consumed.
The volume of air moved depends upon the speed of the dynamo.
As with friction, speed changes will alter the loss.
Stray Load Loss
The stray load losses are produced as a result of the distortion of
the magnetic field by the armature and the interpoles.
The distortion causes the flux in the field poles to be unevenly
distributed and thereby produces a hysteresis loss.
It is generally neglected in motors of 200 hp or less.
In larger rated dynamos the stray load loss is assumed to be 1% of
the output.
Hysteresis Loss (Wh)
This loss is due to the reversal of magnetization of the armature core.
Every portion of the rotating core passes under N and S pole
alternately.
The core undergoes one complete cycle of magnetic reversal after
passing under one pair of poles.
If P is the number of poles and N the armature speed in rpm then
frequency of magnetic reversal is f=PN/120.
The loss depends upon the volume and grade of iron, maximum value
of flux density Bmax and frequency of magnetic reversals.
For normal flux densities (i.e. upto 1.5 Wb/m2), hysteresis loss is given
by Steinmetz formula.
1.6 fV watt
W


B
max
According to this formula,
h
Where,V=volume of the core in m3; =Steinmetz hysteresis coefficient
Value of  for: Good dynamo sheet steel=502 J/m3, Silicon steel=191
J/m3, Hard cast steel=7040 J/m3, Cast steel=750-3000 J/m3, and cast
iron=2700-4000 J/m3.
Eddy Current Loss (We)
When the armature core rotates, it also cuts the magnetic flux.
Hence, an emf induced in the body of the core according to the laws of
electromagnetic induction.
This emf though small, sets up large current in the body of the core due
to its small resistance.
This current is known as eddy current.
The power loss due to the flow of this current is known as eddy
current loss.
Eddy current loss We is given by the following relation:
2 f 2t 2V 2
We  KBmax
where Bmax= maximum flux density
f=frequency of magnetic flux reversals
t=thickness of each lamination
V=volume of armature core.
watt
In order to reduce this loss and the
consequent heating of the core to a small
value, the core is built up of thin
laminations.
These core laminations are insulated
from each other by a thin coating or
varnish.
The effect of laminations is shown in Fig. 24.59.
Due to the core body being one solid iron piece (Fig. 24.59a), the magnitude of eddy
currents is large.
As armature cross-sectional area is large, its resistance is very small, hence eddy
current loss is large.
In Fig. 24.59b, the same core has been split up into thin circular discs insulated from
each other.
It is seen that now each current path, being of much less cross-section, has a very high
resistance.
Hence, magnitude of eddy currents is reduced considerably thereby drastically
reducing eddy current loss.
Eddy current loss is reduced by using laminated core but hysteresis
loss cannot be reduced this way.
For reducing the hysteresis loss, those metals are chosen for the
armature core, which have a low hysteresis coefficient.
Constant or standing Losses
Field Cu loss is constant for shunt and compound generators.
Hence, stray losses and shunt Cu loss are constant in their case.
These losses are together known as standing or constant losses Wc.
Hence for shunt and compound generators,
Total loss=armature copper loss +Wc= Ia2Ra Wc  (I L  Ish)2 Ra Wc
Armature Cu loss Ia2Ra is variable loss because it varies with the
load current.
Total loss= variable loss + constant loss Wc.
Condition for Maximum Efficiency
Generator output = VtIL
Generator input =Output + LossesVt I L  Ia2Ra Wc Vt I L  (I L  Ish)2 Ra Wc
However, if Ish is negligible as compared to load current, then Ia=IL
(approximately), then generator input= Vt I L  I L2Ra Wc
Vt I
L

Vt I  I 2 Ra Wc
L L
output
So   output 
input output  Losses

1
I Ra W
1 ( L  c )
Vt
Vt I
L
Now, efficiency is maximum when denominator is minimum i.e. when
R
W
d ( I L Ra  Wc )  0 a  c  0
Vt V I 2
Vt
Vt I
dI
t L
L
L
W
Ra  c  0
I2
L
I 2 Ra Wc  0
L
Or, I 2Ra Wc
L
Hence, generator efficiency is maximum when,
Variable loss = Constant Loss
The load current corresponding to maximum efficiency is given by the
relation
I  W /R
L
c
a
Example 26. 23 A 10 kW, 250 V, DC, 6 pole shunt-generator runs at
1000 rpm when delivering full-load. The armature has 534 lapconnected conductors. Full-load Cu loss is 0.64 kW. The total brush
drop is 1 V. Determine the flux per pole. Neglect shunt-current.
Solution: Since shunt-current is negligible; there is no shunt Cu loss.
The Cu loss occurs in armature only.
IL=Ia=10,000/250=40 A;
Ia2Ra  Armature Cu loss; Ra  (Armature Cu loss) / Ia2  640/(40)2  0.4
Armature drop= IaRa=400.4=16 V; Brush drop=1 V.
generated emf Eg  250161 267V
Now, Eg  ZN P
60 A
Eg 60 A 26760 6


 30103 Wb  30mWb
Z  N P 5341000 6
Example 26. 24(a) A shunt-generator delivers 195 A at terminal p.d (potential
difference) of 250 V. The armature resistance and shunt field resistance are 0.02 
and 50  respectively. The iron and friction losses equal 950 W. Find: (i) EMF
generated, (ii) Cu losses, (iii) output of the prime mover, and (iv) commercial,
mechanical and electrical efficiencies.
Solution: (i) Ish=250/50=5 A;
Ia=195+5=200 A
Armature voltage drop =IaRa= 2000.02=4 V;
 generated emf= 250+4=254 V
(ii) Armature Cu loss= Ia2Ra  (200)2  0.02  800 W
Shunt Cu loss= VIsh= 2505=1250 W;
Total Cu loss = 1250 + 800 =2050 W
(iii) Stray losses = 950 W;
Output= 250195 = 48,750 W;
Output of prime mover = 51,750 W
Total losses= 2050+950= 3000 W
Input= 48,750+3,000= 51,750 W
(iv) Generated input= 51,750 W;
Stray losses=950 W
Electrical power produced in armature =51,750 –950=50,800 W
Mechanical efficiency, m=(50,800/51,750)100=98.2%
Electrical or Cu losses= 2050 W
Electrical efficiency, e =(50,800100)/(48,750+2050)=95.9%
Commercial or overall efficiency, c =(48,750100)/(51,750)=94.2%
Example 26. 25 A shunt-generator has a full-load current of 196 A at
220 V. The stray losses are 720 W and the shunt field coil resistance is
55 . If it has a full-load efficiency 88%, find the armature resistance.
Also, find the load current corresponding to maximum efficiency.
Solution: Output = 220  196 = 43,120 W; =88% (overall efficiency)
Electrical input = 43,120/0.88 = 49,000 W; Total losses= 4900-43,120 = 5,880 W
Shunt field current – 220/55=4 A;
So, Ia= 196+4 = 200 A;
Shunt Cu loss = 2204 = 880 W;
Stray losses = 720 W;
Constant losses = 880 + 720 = 1,600 W
Armature Cu loss = 5,880 – 1,600= 4,280 W
Ia2Ra  Armature Cu loss; Ra  (Armature Cu loss) / Ia2  4,280 /(200)2  0.107
For maximum efficiency,
Ia2Ra  Constant Losses; Ia  (Constant Losses) / Ra  1,600 / 0.107 122.34A