University of Kerbala
Electrical DC Machine
EEE. Dep.
2.5 Armature Reaction in DC machines:
There is some reaction from the current-carrying armature conductors which is partly
responsible for the drop in the voltage of the machine. There are two primary m.m.f s or fluxes
operating in the DC machine. One, field m.m.f, Fm produced by the field winding wound on the N
and S poles of the machine. When the generator is not driving any load, these two fluxes are
distributed symmetrically as shown in Fig; 2.9.
GNA
MNA
A
Field
axis
A′
Fm
Fig. 2.9: The distribution of main field flux at no load.
When the generator is driving load, the flux Fa is produced by current carrying conductors is
vertically downwards in the armature core. The magnetic flux Fa due to the flow of current through
the armature conductors interacts with the magnetic field F m of the poles. This interaction causes
distortion of the main field flux and shifting the position of Magnetic Neutral Axis (M.N.A.).
These two effects of armature reaction causes a distortion in the main field flux “Crossmagnetization effect” as well as shifting of the magnetic neutral axis “Demagnetization or
weakens the main flux” as shown in Fig; 2.10.
The magnetic neutral axis shifted clockwise to a new position BB'; this new position of magnetic
neutral axis (M.N.A.) “axis of commutation” remains perpendicular to the resultant m.m.f. vector
0F. The Geometric Neutral Axis (G.N.A.) remains at AA′. As in Fig; 2.10. The first effect leads to
the sparking at the brushes and the second to reduce generated voltage. If the brushes remain on
the original position of (M.N.A.) AA′, the coil will be commutating under the greatest change in
the flux compared with the other coils under the pole. It is obvious that the brushes must be shifted
to the new position of the magnetic neutral axis BB', to avoid sparking at the brushes due to the
voltage that will induce in the coil.
A
B′
A′
B
B
B′
Fa
Fig. 2.10: The shift of the magnetic neutral axis
21
M.Sc. Haider M. Umran
University of Kerbala
Electrical DC Machine
EEE. Dep.
2.5.1 Calculation of Demagnetizing and Cross magnetizing Ampere-turns per pole:The effects of demagnetization and cross magnetization can analyzing mathematically as
follows:
The conductors lying in angles of BOC and B′OC′ will develop ampere turns which produce the
demagnetization field, and the conductors lying in the angles COB′ and BOC′ are develop ampereturns which produce the cross-magnetization field from the point of view of armature reaction as
shown in Fig; 2.11. The conductors in the region BOC =B'OC' = 2θ at top and bottom of the
armature.
GNA
Consider the following data:
C A θm B
Z: Total number of armature conductors.
θm: Forward lead of brush in mechanical degree.
Ic: Armature conductor current.
: Ia / 2 (at wave winding).
: Ia / p (at lap winding).
N
S
O
B′ 2θm A′ C′
Fig; 2.11 Development of armature reaction
∴ Total number of armature conductors lying in angles BOC and B ′ OC ′ is =
4 θm
360
×Z
Since two conductors from one turn,
1 4 θm
2 360
Total number of turns in these angles =
2 θm
=
Demagnetizing amp − turns /pair of poles =
∴
360
Demagnetizing amp − turns⁄pole (ATd per pole) =
×Z
360
2 θm
×Z
× Z . Ic
θm
× Z . Ic
360
And the conductors which are responsible for cross-magnetizing amp-turns are lying between
angles COB' and BOC'.
Total armature − conductors/pole =
Z
P
From the above the Demagnetizing conductor’s ⁄ pole is:
Demagnetizing conductors /pole =
∴
2 θm
×Z
360
Z
2 θm
P
360
Cross magnetizing conductors/pole = − Z.
1
2 θm
P
360
= 𝑍[ −
∴
]
1
2 θm
P
360
Cross magnetizing amp − turns/pole = 𝑍 . Ic [ −
]
Since two conductors from one turn,
22
M.Sc. Haider M. Umran
University of Kerbala
Electrical DC Machine
EEE. Dep.
Cross magnetizing amp − turns/pole (ATc⁄pole) = Z. Ic [
∴
1
2P
−
θm
360
]
If lead angle is given in electrical degrees, it should be converted into mechanical degrees by the
following relation:
θmech. =
θ(electrical)
=
Pair of poles
θele.
P ⁄2
=
2 θele.
P
Note: - For neutralizing the demagnetizing effect of armature-reaction, an extra number of turns
may be put on each pole.
No. of extra turns⁄pole =
ATd
(If : field current)
If
If the leakage coefficient λ is given, the (No. of extra turns ⁄ pole) is calculating as:
No. of extra turns⁄pole =
ATd
If
.λ
Ex.:- A 6 pole DC generator has lap winding with 720 armature conductors supplied a current of
55A at full-load condition. If the brushes lead is 10o, determine the demagnetizing and crossmagnetizing amp-turns per pole.
Sol.: -
(ATd ) =
θm
360
× Z. Ic
Where: - Z = No. of armature conductors, I = current per conductor, and θ = brush lead.
IC =
Ia
a
=
55
(Where a = p = 6 in lap winding)
6
ATd = 720 ×
55
6
×
10
360
= 183.33 AT
cross magnetizing amp − turns/pole (ATc/Pole) = Z. Ic [
= 6600(0.084 − 0.028) = 369.6
1
θm
−
]
2P 360
AT
Summarized the Effects of armature reaction as:
1. The armature reaction always results reduction in generated e.m.f. due to decrease in the
value of flux per pole.
2. The iron losses in the teeth and pole shoes increase at load than no load.
3. The maximum value of gap flux density increase; this will increase the voltage between
adjacent commutator segments at load and caused sparking.
4. Shift of brush axis from GNA will make commutation difficult and will cause delayed
commutation.
2.6 Compensating Winding:
It is an extra arrangement of winding to neutralize the cross magnetizing effect of armature
reaction in the pole arc region. The mode of compensating winding is made it opposes the armature
reaction amp-turns as far as possible, the amp; conductors of compensating winding must equal to
23
M.Sc. Haider M. Umran
University of Kerbala
Electrical DC Machine
EEE. Dep.
armature conductors under the pole shoe. The compensating winding conductors are connected in
series with armature conductors as shown in Fig. 2.13.
Compensating
winding
Compensating
winding
Inter- poles
Armature
winding
Ia
Fig. 2.13: Arrangements a compensating winding.
› Calculate the No. of Compensating Windings:Ampere − turns / pole for compensating winding =
Total amp. −conductors/pole =
Pole arc
Pole pitch
Ia
a
.
× Armature amp. −turns/pole
Z
P
Since two conductors from one turn.
Total amp. −
turns
pole
2 a
Pole arc
Where:
∴
1 Ia
in armature = .
Pole pitch
.
Z
P
≅ 0.7
Ampere turns/pole for compensating winding = 0.7 ×
Z .Ia
2a P
Not: - The compensating winding neutralizes the armature reaction m.m.f directly under the pole
shoe only.
Ex.: Calculate the number of conductors on each pole piece in a compensating winding for a 10
pole DC generator which has lap wound, armature containing 800 conductors.
Sol.:
Number of turns/pole for compensating winding = 0.7 ×
= 0.7 ×
800
2×10×10
= 2.8
Z
2a P
T
Since 2 conductors in one turn;
∴
Compensating conductors/pole = 2 × 2.8 = 5.6 ≈ 6
T.
2.7 Practical Commutation:
The currents induced in armature conductors of a dc generator are alternating. 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 N-pole and enter that of Spole, the current in them is reversed. This reversal of current takes place along magnetic neutral
24
M.Sc. Haider M. Umran
University of Kerbala
Electrical DC Machine
EEE. Dep.
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 M.N.A. 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 complete by that time, then sparking is produced
between the brush and the commutator which results in progressive damage to both. The brush
width is equal to the width of one commutator segment and mica insulation. Fig.2.14. show the
process of commutation.
a
c
b
a
c
b
40 A
a
b
c
40 A
40 A
Under commutation
aa
b
Ideal or liner
commutation
c
a
40 A
b
40 A
c
Spark
Over
Comm.
Time
35
-20A
5
Fig 2.14: Process of commutation.
› Reactance Voltage Calculation:
When the armature coil undergoes commutation, there is a change in current in the coil from
the (+I to –I). During this process of commutation, the reactance voltage (e.m.f induced) is
developed. The reactance voltage can be derived as follows:
The time required by commutator to move a distance equal to circumferential thickness of the brush
minus the thickness of one insulating plate.
If Tc is the time required for commutation then,
Tc =
Wb −Wm
𝑣
Wb: Width of the brush
Wm: Width of the insulation between the commutator segments
v: Peripheral velocity of commutator segment. (Segment/second)
Total change of current during commutator = I − (−I) = 2I
Reactance voltage (self-induced voltage) = Coefficient of self-inductance× Rate of change of current.
Reactance voltage = L ×
2I
Tc
=L ×
2I
Wb −Wm
𝑣
(v)
. . . (Linear commutation)
25
M.Sc. Haider M. Umran
University of Kerbala
Electrical DC Machine
Reactance voltage = 1.11 × L ×
2I
EEE. Dep.
(v)
Tc
. . . (Sinusoidal commutation)
L: Coefficient of self-inductance.
Tc: Time of commutation.
Ex.: A 4 pole, lap wound armature running at 1500 r.p.m. delivers a current of 150 A, and has 64
commutator segments. The brush spans 1.2 segments and inductance of each armature coil is 0.05
mH.
Calculate the value of reactance voltage assuming i. linear commutation ii. Sinusoidal
commutation; neglect mica thickness.
Sol.: Commutator speed is 1500 r.p.m.
Commutator speed =
1500
60
= 25 r. p. s
𝑣 = 𝑁𝑜. 𝑜𝑓 𝑟. 𝑝. 𝑠 × 𝑡𝑜𝑡𝑎𝑙 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑠 𝑜𝑛 𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑜𝑟 = 25 × 64 = 1600 𝑠𝑒𝑔./𝑠𝑒𝑐.
∴
Tc =
𝑊𝑏 −𝑊𝑚
𝑣
1.2 −0
=
1600
= 7.5 × 10−4 second.
I = current through conductor =
i.
a
=
150
4
= 37.5 A
Linear commutation:
𝐸 = L×
ii.
IL
2𝐼
𝑇𝑐
= 0.05 × 10−3 ×
2 ×37.5
7.5 ×10−4
= 5 V.
Sinusoidal commutation:
E = 1.11 × L ×
2I
Tc
= 1.11 × E = 1.11 × 5 = 5.55 V.
Note: If Wb is given in cm and commutator diameter Dc is given in cm then v is to be calculated in
cm/second as given below.
𝑣 = π Dc
N
cm/second
60
2.8 Methods of Improving Commutation:
There are two practical ways of improving commutation:
a- Resistance Commutation:
The resistance between the commutation segments and the brushes is made high, by replacing
low-resistance Cu brushes to comparatively high-resistance carbon brushes. This helps to increase
the resistance in the circuit and accordingly the time constant L/R is reduced. Hence, the change in
current during commutation becomes faster.
26
M.Sc. Haider M. Umran
University of Kerbala
Electrical DC Machine
r2
EEE. Dep.
r1
2I
b- E.m.f Commutation:
The reactance voltage is neutralized by producing a reversing e.m.f in the short-circuited coil
under commutation. This reversing e.mf, opposition to the reactance voltage and if its value is made
equal to reactance voltage, x. The reversing e.m.f may be produced by two ways:
1. By using inter-poles or commutating poles.
2. By giving a forward lead to the brushes.
Functions of inter-poles:iThe polarity is the same as that of the main pole; they induce an e.m.f in the coil (under
commutation) which helps the reversal of current. The e.m.f induced by the com-poles
is known as commutating or reversing e.m.f. The commutating e.m.f neutralizes the
reactance e.m.f. With inter-poles, spark less commutation can be obtained up to 20 to
30% overload with fixed brush position.
ii-
By use inter-poles the cross-magnetizing effect of armature reaction can neutralize.
Hence, brushes are not to be shifted from the original position.
Note: Armature reaction effect can be completely neutralized by using inter-poles as well as
compensating winding.
Ex.:- A DC machine has the following data:
No. of poles = 6, Diameter of commutator = 30 cm, speed of the machine = 1000 r.p.m, armature
current = 150 A, thickness of brush = 15 mm, self-inductance of each armature coil = 0.15 mH and
type of armature winding is lap.
Determine the average induced e.m.f in each coil at the time of commutation?
Sol.: Peripheral speed of the commutator is π.D.N
Where Dc is the diameter and N is the speed of armature.
𝑣 = π. Dc . N = 𝜋 × 30 × 10−2 ×
1000
60
= 15.7 m/s
Neglecting mica insulation,
Commutation time =
=
Brush thickness
Peripheral speed
15 × 10−3
15.7
= 0.955 × 10−3 sec.
27
M.Sc. Haider M. Umran
University of Kerbala
Electrical DC Machine
EEE. Dep.
Inductance of coil, L = 0.15 × 10−3 H
Current /conductor =
Average induced e. m. f/coil = L ×
150
= 25 A
6
2 Ic
=
Comm.time
0.15 × 10−3 ×2×25
0.955 ×10−3
= 7.85 V.
Ex.:- A 4 pole generator supplies a current of 143 A. It has 492 armature conductors (a) wavewound (b) lap-connected. When delivering full load, the brushes are given an actual lead of 10 o.
Calculate the demagnetizing amp-turns/pole. The field winding is shunt connected and takes 10 A:
Find the number of extra shunt field turns necessary to neutralize this demagnetization?
Sol.: -
Z = 492; θ = 10o ; ATd/pole = Z. IC ×
θm
360
Ia = 143 + 10 = 153 A;
Ic =153/2 ...when wave wound.
Ic =153/4 ...when lap-wound.
ATd / pole = 492 ×
Extra shunt field turns =
2
ATd /pole
If
ATd/pole = 492 ×
Extra shunt field turns =
153
523
10
153
4
×
10
360
= 1046 AT.
= 1046/10 = 105 approx.
×
10
360
= 523 AT.
≈ 52
H.W:- A 450kW, 400V, 8 pole, lap wound DC generator has 768 armature conductors. If the
brushes are given a lead of 6 electrical degree, calculate the number of demagnetizing and crossmagnetizing ampere-turns/pole at full load. Shunt field current may be neglected.
[Ans.: ATd = 450AT, and ATc= 6750AT]
28
M.Sc. Haider M. Umran