University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
CH. 3- D.C. GENERATORS
3.1. Generator Principle:
An electrical generator is a machine which converts
mechanical energy (or power) into electrical energy (or
power).
The energy conversion is based on the principle of the
production of dynamically (or motion ally) induced
e.m.f.
Whenever a conductor cuts magnetic flux, dynamically
induced e.m.f. is produced in it according to Faraday’s
Laws of Electromagnetic Induction. This e.m.f. causes
a current to flow if the conductor circuit is closed.
Hence, two basic essential parts of an electrical
generator are (i) a magnetic field and (ii) a conductor or conductors which can so move as to
cut the flux.
3.2. Types of Generators
Generators are usually classified according to the way in which their fields are excited.
Generators may be divided into (a) separately-excited generators and (b) self-excited DC
generators.
(a) Separately-excited DC generator: The field winding are energized from an
independent external d.c. current source, as shown in Fig. 3.2.
If
+
DC
Source
-
Ia = IL
IL
A1
F1
Ra
Vt
Ea
Rf
F2
A2
Fig. 3.2 Separately-excited d.c generator.
(b) Self-excited DC Generators: The field windings are energized by the current produced by
the generators themselves. Due to residual magnetism, there is always present some flux in the
poles. When the armature is rotated, some e.m.f. and hence some induced current is produced
which is partly or fully passed through the field coils thereby strengthening the residual pole
flux.
There are three types of self-excited generators named according to the manner in which their
field coils (or windings) are connected to the armature.
33
Ms.c. Haider M. Umran
University of Karbala
(i)
DC Machine
Electrical & Electronics Eng. Dep.
DC Shunt Generator: The field windings are connected across or in parallel
with the armature conductors and have the full voltage of the generator applied
across them, as shown in Fig. 3.3.
IL
Ia
F1
A1
Rsh
Ea
F2
A2
Vt
Fig. 3.3: DC shunt generator.
Current and voltage relations can be expressed as:
Ia = IL + Ish
Ish =
Vt
Rsh
The induced e.m.f is:
Ea= Vt + Ia. Ra + Vbrush
And
Ea =
ФpNZ
60 a
In practice, Vbrush is neglected.
(ii)
DC Series Generator: In this case, the field windings are joined in series with the
armature conductors, as shown in Fig. 3.4. As they carry full load current, they
consist of relatively few turns of thick wire or strips. Such generators are rarely
used except for special purposes i.e. as boosters etc.
Ia S1
S2
Ise
IL
A1
Ea
Vt
A2
Fig. 3.4: DC series generator.
34
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
The relations between voltage and current can expressed as:
Ia = Ise= IL
Ise: Current through series winding.
E.m.f equation is:
Ea = Vt +Ia. Ra + Ia. Rse +Vbrush
Ea = Vt +Ia (Ra +Rse) +Vbrush
Where;
(iii)
Ea =
ФpNZ
60 a
DC Compound Generator: It is a combination of a few series and a few shunt
windings and can be either short-shunt or long-shunt, as shown in Fig. 3.5 and 3.6.
In a compound generator, the shunt field is stronger than the series field. When
series field aids the shunt field, generator is said to be commutativelycompounded.
a) Long Shunt Compound Generator: Ish
Ia= Ise
And
Se.
Ia= Ish+ IL
V
Ish = t
IL
Ia
A1
Sh.
Vt
Ea
Rsh
Ea = Vt + Ia . Ra + Ia. Rse +Vbrush
Where;
Ise
A2
Rsh: Resistance of shunt field winding.
Fig. 3.5: Long shunt compound generator
b) Short Shunt Compound Generator: And
Ia = Ise+ Ish
Ise = IL
∴
∴
Ia = IL+ Ish
E− Ia .Ra
Ish =
IL
S
Ish
A1
Rsh
Sh.
Voltage equation is;
Ea = Vt+ Ia Ra +Ise Rse +Vbrush
Ise= IL
Ea = Vt + Ia Ra + IL Rse+ Vbrush
Neglecting Vbrush,
Ea = Vt + Ia Ra + IL Rse
Ea - Ia Ra = Vt + IL Rse
Ise
Ia
Vt
E
A2
Fig. 3.6: Short shunt compound generators.
35
Ms.c. Haider M. Umran
University of Karbala
∴
Ish =
DC Machine
Electrical & Electronics Eng. Dep.
Vt + IL .Rse
Rsh
a) Cumulative and Differential Compound Generator:
When the field excitation is produced by a combination of shunt field winding and series field
winding as shown in Fig. 3.7, the shunt and series fields help each other, the compound
generator is termed cumulative compound.
A1
ΦT = Φsh + Φse
F1
F1
Where: Φsh = Flux produced by shunt.
Φse = Flux produced by series.
F2 A3
F1
A4
A2
Fig.3.7: Cumulative compound.
When the shunt and series field oppose each other, then the generator is differential compound
as shown in Fig. 3.8. As a result, the terminal voltage falls drastically with increasing load.
ΦT = Φsh - Φse
A1
F1
F2 A3
A4
A2
Fig. 3.8: Differential compound
Ex.: A DC shunt generator has shunt field winding resistance of 100 Ω. It is supplying a load
of 5 Kw at a voltage of 250 V. if its armature resistance is 0.22 Ω. Calculate the induced e.m.f
of generator.
Sol.: Ia = IL + Ish
Ish = Vt / Rsh = 250 v / 100 Ω = 2.5 A
IL= PL / Vt
= 5×103 / 250 = 20 A
Ia = IL + Ish
= 20 + 2.5 = 22.5 A
Ea = Vt + Ra.Ia
= 250 + 0.22 × 22.5 = 254.95 V.
Ish
Ia
IL
F1
G
Vt
F2
36
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
Ex.: A 4 pole, compound DC generator long-shunt type, supply’s 100 A at a terminal voltage
of 500 V. If armature resistance is 0.02Ω, series field resistance 0.04 Ω and shunt field
resistance 100Ω, find the generated EMF. Take drop per brush as 1 V, Neglect armature
reaction.
Sol:
Ish = Vt / Rsh = 500 / 100 = 5 A
Ia = IL + Ish = 100 + 5 = 105 A
Voltage drop on series field windings =105×0.04= 4.2V
Armature voltage drop = 105 × 0.02=2.1 volt
Drop at brushes = 2 × 1= 2 V
Now,
E.m.f = V+ Ia. Ra +series drop+ brush drop
= 500 + 2.1 + 4.2 + 2 = 508.3 V.
Ex.: A 20 kW compound generator works on full load with a terminal voltage of 250 V. The
armature, series and shunt windings have resistances of 0.05Ω, 0.025Ω and 100 Ω respectively.
Calculate the total E.M.F generated in the armature when the machine is connected as short
shunt.
Sol.:
Load current =P / V =20000/ 250 = 80 A
Voltage drop in the series windings = 80 × 0.025 = 2V
Voltage across shunt winding =Vt+ Voltage drop in the series windings
IL
= 250 + 2 = 252 V.
Ish = Vt +IL Rse /Rsh
= 250 +80 ×0.025 / 100 =2.52 A
Ia = IL+Ish
= 80 + 2.52 = 82.52A
Ia.Ra = 82.52 × 0.05 = 4.13V
The generated E.m.f = Vt+ Ia. Ra +Voltage drop in the series windings
= 250 + 4.13 + 2 = 256.13 V.
37
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
3.3. Characteristics of D.C. Generators:
Following are the three most important characteristics or curves of a d.c. generator:
1. No-load saturation Characteristic (Eo/If):
It is also known as Magnetic Characteristic or Open-circuit Characteristic (O.C.C.). It
shows the relation between the no-load generated e.m.f. in armature, Eo and the field or exciting
current If at a given fixed speed. It is just the magnetization curve for the material of the
electromagnets. Its shape is practically the same for all generators whether separately-excited
or self-excited.
2. Internal or Total Characteristic (Eo/Ia):
It gives the relation between the e.m.f. Eo actually induces in the armature (after allowing for
the demagnetizing effect of armature reaction) and the armature current Ia.
3. External Characteristic (Vt/IL):
It is also referred to as performance characteristic or sometimes voltage-regulating curve.
It gives relation between that terminal voltage Vt and the load current IL. This curve lies below
the internal characteristic because it takes into account the voltage drop over the armature
circuit resistance. The values of Vt are obtained by subtracting Ia .Ra from corresponding values
of Eo. This characteristic is of great importance in judging the suitability of a generator for a
particular purpose.
3.2.1 Characteristics of Separately Excited Generator:1. No-load Saturation or Open Circuit Characteristic (Eo/If):
The arrangement for obtaining the necessary data to plot this curve is shown in Fig. 3.9. The
exciting or field current If is obtained from an external independent d.c. source. It can be
varied from zero upwards by a potentiometer and its value read by an ammeter A connected
in the field circuit as shown.
Ф𝐩𝐍𝐙
Now, the voltage equation of a d.c. generator is, 𝐄𝐨 =
volt
𝟔𝟎 𝐚
A
F1
A1
D.c
supply
V
Rheosta
t
F2 A2
Fig. 3.9: Separately excited gen. with no-load.
38
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
When current If is increased from its initial small value, the flux Φ changed and hence induced
e.m.f. Eo increase directly along the poles are unsaturated. This is represented by the straight
portion OA in Fig. 3.10. But as the flux density increases, the poles become saturated, so a
greater increase in If is required to produce a given small increase in voltage than on the lower
part of the curve.
E0
Saturation
Increasing
A Constant
Open circuit
Characteristic
O
If
Fig. 3.10: Magnetization ch.cs. for constant speed.
2. Load Saturation Curve (Vt/If):
The curve showing relation between the terminal voltage Vt and field current If when the
generator is loaded, is known as Load Saturation Curve.
The curve can be deduced from the no-load saturation curve provided the values of armature
reaction and armature resistance are known. While considering this curve, account is taken of
the demagnetizing effect of armature reaction and the voltage drop in armature which are
practically absent under no-load conditions.
Fig. 3.11: Load saturation curve.
The no-load saturation curve of Fig. 3.10 has been repeated in Fig. 3.11 on a base of field ampturns (and not current) and it is seen that at no-load, the field amp-turns required for rated noload voltage are given by Oa. Under load conditions, the voltage will decrease due to
39
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
demagnetizing effect of armature reaction. This decrease can be made up by suitably increasing
the field amp-turns. Let ac represent the equivalent demagnetizing amp-turns per pole. Then,
it means that in order to generate the same e.m.f. on load as at no-load, the field amp-turns/pole
must be increased by an amount ac = bd. The point d lies on the curve LS which shows relation
between the voltage E generated under load conditions and the field amp-turns. The curve LS
is practically parallel to curve Ob. The terminal voltage V will be less than this generated
voltage E by an amount (Ia Ra) where Ra is the resistance of the armature circuit. From point
d, a vertical line de = Ia Ra is drawn. The point e lies on the full-load saturation curve for the
generator. Similarly, other points are obtained in the same manner and the full-load saturation
curve Mp is drawn. The right-angled triangle bde is known as drop reaction triangle. Load
saturation curve for half-load can be obtained by joining the mid-points of such lines as mn and
bd etc. In the case of self-excited generators, load saturation curves are obtained in a similar
way.
3. Internal and External Characteristics:
Let us consider a separately-excited generator giving its rated no-load voltage of Eo for a certain
constant field current. If there were no armature reaction and armature voltage drop, then this
voltage would have remained constant as shown in Fig. 3.11. By the dotted horizontal line I.
But when the generator is loaded, the voltage falls due to these two causes, thereby giving
slightly dropping characteristics.
If we subtract from Eo the values of voltage drops due to armature reaction for different loads,
then we get the value of E, the e.m.f. actually induced in the armature under load conditions.
Curve II is plotted in this way and is known as the internal characteristic. The straight line Oa
represents the Ia Ra drops corresponding to different armature currents. If we subtract from E o
the armature drop Ia Ra, we get terminal voltage Vt. Curve III represents the external
characteristic and is obtained by subtracting ordinates the line Oa from those of curve II.
Fig. 3.11: Load characteristics curve.
40
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
3.2.2: Characteristics of Self- Excited DC Shunt Generator:1. No-load Curve Characteristic (Eo / If):
This curve shows the relation between the generated e.m.f. at no-load (Eo) and the field
current (If).
The field or exciting current If is varied by rheostat and its value read on the ammeter (A). The
generator is derived at constant speed by the prime mover and the generated e.m.f. on no-load
is measured by the voltmeter connected across the armature as shown Fig. 3.12. Due to residual
magnetism in the poles, some e.m.f. (OA) is generated even when I f = 0. Hence, the curve
shown in Fig. 3.13; starts a little way up. The generated e.m.f. is directly proportional to the
exciting current. However, at high flux densities, where μ is small, iron path reluctance
becomes appreciable and straight relation between E and I f after point B, saturation of poles
starts.
Ish
A
Ia
Eo
IL
B
Q
F1
F2
G
Vt
V
G
A
If
O
Fig.3.13. Load characteristics curve
Fig.3.12. DC shunt generator
cteristics curve
2. Load Characteristics of DC Shunt Generator:A. Internal Characteristic (E/Ia):
Ideally the induced e.m.f. is not dependent on the load current IL or armature current Ia. but
as load current increased, the armature current Ia increases to supply load demand. As Ia
increased armature flux increases. The effect of flux produced by armature on main flux
produced by the field winding is called armature reaction. Due to the armature reaction, main
flux distorted. Hence lesser flux gets linked with the armature conductor and this reduces the
induced e.m.f as shown Fig. 3.14.
E
Eo
Drop due to
Armature
reaction
Q
IL
O
Fig.3.14: Internal characteristics curve.
41
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
B. External Characteristic (Vt/IL):
For D.C generator (E = Vt - Ia Ra) neglected other drops. So as load current IL increases,
Ia increases. Thus will increase drop Ia Ra and terminal voltage Vt = E - Ia Ra decreases. The
value of armature resistance Ra is very small; the drop in terminal voltage as I L changes from
no load to full load is very small. Hence shunt generator is called constant voltage generator.
If the load resistance is reduced beyond point a break down point i.e. load I L is increased beyond
P then it increases momentarily. This is very large current and generator gets overloaded. Due
to a large current the armature reaction is high and drop Ia Ra decreases from P to Q, rather than
increasing. Thus on curve aqr, the voltage goes on reducing rapidly and point r becomes zero
as shown Fig. 3.15.
Vt
Eo
Small drop Ia.Ra
a
Drooping nature
Break down
q
r
O
IL
Q
P
Fig.3.15: External characteristics curve.
3.2.2.1 Critical Field Resistance in DC Shunt Generator:
Now, connect the field windings back to the armature and run the machine as a shunt
generator. Due to residual magnetism in the poles, some e.m.f. and hence current, would be
generated. This current while passing through the field coils will strengthen the magnetism of
the poles (provided field coils are properly connected as regards polarity). This will increase
the pole flux which will further increase the generated e.m.f. Increased e.m.f. means more
current which further increases the flux and so on. This mutual reinforcement of e.m.f. and flux
proceeds on till equilibrium is reached at some point like P (Fig. 3.16). The point lies on the
resistance line OA of the field winding. Let R be the resistance of the field winding. Line OA
is drawn such that its slope equals the field winding resistance i.e. every point on this curve is
such that volt/ampere = R. The voltage OL corresponding to point P represents the maximum
voltage to which the machine will build up with R as field resistance. OB represents smaller
resistance and the corresponding voltage OM is slightly greater than OL. If field resistance is
increased, then slope of the resistance line increased, and hence the maximum voltage to which
the generator will build up at a given speed, decreases. If R is increased so much that the
resistance line does not cut the O.C.C. at all (like OT), then obviously the machine will fail to
excite i.e. there will be no ‘build up’ of the voltage. If the resistance line just lies along the
slope, then with that value of field resistance, the machine will just excite. The value of the
42
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
resistance represented by the tangent to the curve, is known as critical resistance Rc for a given
speed.
(Rsh) increased, slope increases,
Max; generate voltage decreases.
(B)
Line of Rc
E.m.f
A
Normal max; excitation
voltage
B
E
C
Normal field resistance line
(Rsh)
O.C.C at Normal speed
D
Critical Voltage EC
If
O
Fig.3.16: No-load saturation curve.
Steps to Find Critical Resistance Rc:1. O.C.C. is plotted from the given data. As shown in Fig. 3.16.
2. To draw the line of field resistance Rf, consider an equation of line (y = m.x). Where
y = Eo, x = If and m= slope = Rf. one point of the line is (0, 0), second point (Eo, If) If
(If) is any value from the given data by the above eq. we determine the new value of Eo.
The second point on the line is (Eo, If) and draw the line passing through (0, 0) and
(Eo , If), OA line.
3. Draw a tangent line to (O.C.C) i.e. tan θ, is the critical resistance of the field resistance.
∆𝐄
𝐃𝐄
Slope of tangent line is 𝐑 𝐜 =
= = 𝐭𝐚𝐧 𝛉.
∆𝐈𝐟
𝐂𝐃
We know that E α N. As speed decreased the induced e.m.f. decreases, we gate (O.C.C) below
the (O.C.C) just tangential to normal field resistance line. As shown in Fig.3.17.
Critical Speed (Nc) is the speed at which machine just excites for the given field circuit
resistance.
𝐄𝐨𝟏 𝐍𝟏
=
𝐄𝐨𝟐 𝐍𝟐
∴
𝐄𝐨𝟐 =
𝐍𝟐
𝐍𝟏
. 𝐄𝐨𝟏
43
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
Steps to Find Critical Speed Nc Graphically at Different Speeds:1.
2.
3.
4.
5.
6.
Drawn O.C.C for the given speed N1.
Draw a line tangential to this O.C.C, OA.
Draw a line representing the given Rsh, OP.
Select any field current value, point R.
Draw vertical line from R to intersect OA at S and OP at T.
The critical speed Nc is,
𝐑𝐓 𝐍𝐂
=
𝐑𝐒 𝐍𝟏
∴
𝐍𝐂 =
A
Eo
𝐑𝐓
𝐑𝐒
P
S
. 𝐍𝟏
O.C.C
at N1
Line for given
Rsh at Nc
O.C.C at Nc
T
R
Fig.3.17 Determine Critical speeds.
Ex.: The magnetization curve of a d.c. shunt generator at 1500 r.p.m. is:
If (A): 0 0.4 0.8 1.2
1.6
2 2.4 2.8 3
Eo (V): 6 60 120 172.5 202.5 221 231 237 240
For this generator find (i) no load e.m.f. for a total shunt field resistance of 100 Ω (ii) the critical
field resistance at 1500 r.p.m. and (iii) the magnetization curve at 1200 r.p.m. and therefrom
the open-circuit voltage for a field resistance of 100 Ω.
(b) A long shunt, compound generator fitted with inter-poles is commutatively-compounded.
With the supply terminals unchanged, the machine is now run as compound motor. Is the motor
differentially or cumulatively compounded?
Sol.:
The magnetization curve at 1500 r.p.m. is plotted in Fig. from the given data. The 100 Ω
resistance line OA is obtained by joining the origin (0, 0) with the point (1A, 100 V).
The voltage corresponding to point A is 227.5 V. Hence, no-load voltage to which the generator
will build-up is 227.5 V. The tangent OT represents the critical resistance at 1500 r.p.m.
Considering point B, Rc = 225/1.5 = 150 Ω.
44
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
For 1200 r.p.m., the induced voltages for different field currents would be (1200/1500) = 0.8
of those for 1500 r.p.m. The values of these voltages are tabulated below:
If (A): 0 0.4 0.8 1.2 1.6
2
2.4
2.8
3
Eo (V): 4.8 48 96 138 162 176.8 184.8 189.6 192
The new magnetization curve is also plotted in Fig. 28.9. The 100 Ω line cuts the curve at
point C which corresponds to an induced voltage of 166 V.
Ex.: The O.C.C of a separately excited DC generator driven at 1000 r.p.m. is as follows:
Field current:
𝑬. 𝑴. 𝑭. 𝒗𝒐𝒍𝒕𝒔:
0.2 0.4
30 55
0.6 0.8 1
1.2 1.4 1.6
75 90 100 110 115 120
If the machine is connected as shunt generator and driven at 1000 r.p.m. and has a field
resistance of 100 Ω, find (a) open circuit voltage and exciting current (b) the critical resistance
and (c) resistance to induce 115 volts on open circuit.
Sol.: The O.C.C. has been plotted in Fig. below. The shunt resistance line OA is drawn as usual.
Draw a line for Rsh = 100 Ω, the slope of the line is 100.
Eo = Rsh × If, if exciting current (If = 1A) from the table; Eo = 100 V.
The slope of tangent line is critical resistance Rc:
45
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
Eo
D
C
F
If
G
Slope of tangent line (Rc) =
CD
FG
=
30
0.2
= 150Ω.
C. Line OB represents shunt resistance for getting 115 V on open circuit. From the table
the field current If = 1.4A at E=115.
𝐸
115
∴
Field resistance 𝑅𝑓 = =
= 82.1 Ω.
𝐼𝑓
1.4
Ex.: The magnetization characteristic for a 4-pole, 110V and 1000 r.p.m. shunt generator as
follows:
𝐼𝑓 :
0
0.5
1
1.5
2
2.5
3𝐴
𝐸𝑜 :
5
50
85 102
112 116
120 𝑉
Armature is lap connected with 144 conductors; field resistance is 45 Ω. Determine
i. Voltage the machine will build up at no-load.
ii. The critical resistance.
iii. The speed at which the machine just excite.
iv. Residual flux per pole.
Sol.: The O.C.C. as shown in Fig., OA represents the 45Ω line which is drawn according to eq.
Eo = Rsh × If.
Drawing a horizontal line from 60V on Y-axis and vertical line to If = 1.1A on X-axis, intersect
at point C.
i. The voltage to which machine will build up = OM = 118 V.
ii. OT is tangent to the initial part of the O.C.C. It represents critical resistance.
Slope of tangent line (Rc) =
NC
HR
=
30
0.3
= 100 Ω
iii. From any point on OT, at point B, drop the perpendicular BD on X-axis.
46
Ms.c. Haider M. Umran
University of Karbala
CD
BD
=
NC
N1
DC Machine
or
50
110
=
NC
1000
Electrical & Electronics Eng. Dep.
∴ Nc = 454.54 𝑟. 𝑝. 𝑚
Eo
iv. As given in the table, induced e.m.f. due to
residual flux (i.e. when there is no exciting current)
is 5 V.
S
N
Eo =
∴
ZNϕP
60 a
, 5=
144×𝛷×1000×4
60 ×4
C
Φ = 2.08 𝑚𝑊𝑏.
If
H R
3.2.2.2 Voltage Build-up of a DC Shunt Generator:
Before loading a shunt generator, it is allowed to build up its voltage. Usually, there is always
present some residual magnetism in the poles, hence a small e.m.f. is produced initially.
This e.m.f. circulates a small current in the field circuit which increases the pole flux (provided
field circuit is properly connected to armature, otherwise this current may wipe off the residual
magnetism).
When flux is increased, generated e.m.f. is increased which further increases the flux and so
on. As shown in Fig. 28.17, Oa is the induced e.m.f. due to residual magnetism which appears
across the field circuit and causes a field current Ob to flow. This current aids residual flux and
hence produces, a larger induced e.m.f. Oc. In turn, this increased e.m.f. Oc causes an even
larger current Od which creates more flux for a still larger e.m.f. and so on.
Now, the generated e.m.f. in the armature has (a) to supply the ohmic drop If Rsh in the winding
and (b) to overcome the opposing self-induced e.m.f. in the field coil i.e. L. (d If /dt) because
field coils have appreciable self-inductance.
d If
Ea = If . R sh + L.
V.
dt
If (and so long as), the generated e.m.f. is in excess of the ohm drop I f Rsh, energy would
continue being stored in the pole fields. For example, as shown in Fig. 3.18, corresponding to
field current OA, the generated e.m.f. is AC. Out of this, AB goes to supply ohm drop If Rsh
and BC goes to overcome self-induced e.m.f. in the coil. Corresponding to If = OF, whole of
the generated e.m.f. is used to overcome the ohm drop.
None is left to overcome L. dIf /dt. Hence no energy is stored in the pole fields. Consequently,
there is no further increase in pole flux and the generated e.m.f. With the given shunt field
resistance represented by line OP, the maximum voltage to which the machine will build up is
OE. If resistance is decreased, it will built up to a somewhat higher voltage. OR represents the
47
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
resistance known as critical resistance. If shunt field resistance is greater than this value, the
generator will fail to excite.
Ea
T
If
Fig. 3.18: Voltage build-up of a DC shunt generator.
Conditions Required for Voltage Build-up of DC Shunt Generator:
1. There must be some residual magnetism in the generator poles.
2. For the given direction of rotation, the shunt field coils should be correctly connected to the
armature.
3. If excited on open circuit no-load, its shunt field resistance should be less than the critical
resistance (which can be found from its O.C.C.).
4. The rotational speed of generator should be greater than the critical speed. This is given by
internal characteristic.
3.2.3. Load Characteristics of DC Series Generator:In this type of generator, Ise = Ia = IL
As load current IL is increases, Ise increases. The flux Φ is directly proportional to Ise. So flux
increases. The induced e.m.f. E is proportional to flux hence induced e.m.f. also increases.
The internal characteristics (E/Ise) are of increasing nature.
48
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
As Ia is increases, armature reaction increases but its effect is negligible compared to increase
in E. For high load current, saturation occurs and flux remains constant. In such case, due to
armature reaction E starts decreasing as shown in the Fig. 3.19.
O.C.C. characteristics
V
Voltage
due to
residual
flux
Armature
reaction
effect
Internal c/s
Ia (Ra +Rse)
External c/s
Ise = Ia = IL
Fig.3.19: Characteristics of DC series generator.
As Ia = IL increases, the drop in armature and field winding increases I a (Ra +Rse), Where
Vt= E- Ia (Ra +Rse), thus the external c/s; as shown under the internal c/s due to drop I a (Ra +Rse).
If self-excited series generator, O.C.C. cannot be obtained. The O.C.C. can be obtained in
separately -excited the field winding.
3.2.4. Load Characteristic of DC Compound Generator:In a compound generator, both series and shunt excitation are combined; we shall discuss
the characteristics of cumulatively compounded generator. It may be noted that external
characteristics of long and short shunt compound generators are almost identical, as shown in
Fig. (3.4-5). Compound generator can be classified in to:a. Under-compounded generator:In cumulatively compounded, ΦT = Φsh +Φse. As load current IL is increases, Ia increases hence
Ise also increases and flux increasing more. The induced e.m.f. increases and terminal voltage
also increases. But voltage drop and armature reaction drop also increases, hence there is drop
in the terminal voltage. As shown in Fig. (3.20)
b. Over-compound generator:If series winding turns are so adjusted that with the increase in load current the terminal voltage
increases. In such a case, as the load current increases, the series field e.m.f. increases and tends
to increase the flux and hence the generated voltage. The increase in generated voltage is
greater than the Ia.Ra drop so that instead of decreasing, the terminal voltage increases. As
shown in Fig. (3.20).
49
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
c. Level Compound generator (flat).
If series winding turns are so adjusted that with the increase in load current, the terminal voltage
substantially remains constant. The series winding of such a machine has lesser number of turns
than the one in over-compounded machine and, therefore, does not increase the flux as much
for a given load current. Consequently, the full-load voltage is nearly equal to the no-load
voltage. As shown in Fig. (3.20).
VT
Over
Eo
Under
Flat or level
Differentially
compounded
IL
Full load
Fig.3.20: Compound generator C/S.
50
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
3.3 Generator Power Losses:In DC generators, as in most electrical devices, certain forces act to decrease the
efficiency. These forces, as they affect the armature, are considered as losses and may be
defined as follows:
1. Copper loss or I2.R in the winding
2. Iron loss or core Losses.
4. Mechanical Losses.
5- Brush contact losses.
Armature Cu Loss
Copper Losses
Shunt Cu Loss
Series Cu Loss
Total Losses
Iron
Losses
Mechanical
Losses
Hysteresis
Eddy Current
Friction
Wind age
Brush Contact
Losses
1. Copper Losses: These losses are taking place due to the current flow in a winding. There
are various copper losses can be given by:
Armature copper loss = Ia2.Ra
Shunt field copper loss = I2sh.Rsh
Series field copper loss = I2se.Rse
The brushes contact resistances usually included in copper losses, copper losses are about 30
to 40% of full-load losses.
2. Iron or Core Losses: These losses are also called magnetic losses. Its include hysteresis loss
and eddy current loss in the core. Booths eddy currents and hysteresis losses total up to about
20 to 30% of full-load, as it explained in Ch.1
a. Mechanical Losses: These losses consist of:i. Mechanical friction losses
ii. Wind age, Air-friction or wind resists at the shaft.
Not. The Magnetic and Mechanical losses together called stray losses. For shunt and
compound DC generators where field current is constant, loss is constant; field copper losses
are also constant. Stray losses along with constant field copper losses are called constant losses
(Wc). While the armature current is dependent on the load, thus armature copper losses are
called variable losses (Pcu).
∴
Total losses = Constant losses + Variable losses.
51
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
3.4 Power Stages energy transformation in D.C Generator:Various power stages in the case of a DC generator are shown below:
A
C
B
Mechanical
Input Power
Pin
Iron & fiction loss
Pi & P mech.
Armature
Power
Pg = Eg . Ia
Cupper loss
Pcu
Output
Power Pout
= It.Vt
Armature Cupper loss = Ia 2 . R a
Note: (power developed in armature). Pg = Eg . Ia Or Pg = Pin − (Pmech. + Pin )
In the case of shunt generators: Field Cupper Loss =Ish 2 . R sh
In the case of series generator: Field Cupper Loss= Ise 2 . R se
3.5 Efficiency of D.C Generator: - The efficiencies for DC generator are divided to:
1. Mechanical Efficiency: ηm =
2. Electrical Efficiency: ηele. =
3. Total Efficiency: η =
η=
Or
Pout
Pin
B
A
C
B
=
=
Eg .Ia
× 100%
Pin
VL . IL
Eg .Ia
× 100%
= ηm . ηe =
Pout
Pout + PLosses
=
C
=
VL . IL
A
Pin
Pin −PLosses
× 100%
× 100%
Pin
Pin = Pout + Pcu + (Pi + Pmech )
Where:
3.5.1 Condition for Maximum Efficiency:
The condition for maximum efficiency of DC generator is given by,
Variable loss = Constant loss.
Ia 2 . R a = W𝑐
Generator output = V. I
Generator input = P output + P losses
Pi = V. I + Ia 2 . R a + Wc = V. I + (I + Ish )2 . R a + Wc
However, if Ish is negligible as compared to load current, then Ia = I
∴
η=
Pout
Pin
× 100% =
V.I
V.I+(I )2 .Ra +Wc
× 100%
(∵ Ia = I + Ish)
(∵ Ia = I)
52
Ms.c. Haider M. Umran
University of Karbala
DC Machine
Electrical & Electronics Eng. Dep.
1
× 100%
I . R a Wc
1+ [
+
]
V
V. I
η =
The load current corresponding to maximum efficiency is given by:
IL = √
W𝑐
Ra
Ex.: A DC shunt generator delivers 195 A at terminal p.d. of 250 V. The armature resistance
and shunt field resistance are 0.02 Ω and 50 Ω respectively. The iron and friction losses (Stray)
equal 950 W. Find
(a) E.M.F. generated (b) Pcu (c) output of the prime mover (d) mechanical and electrical
efficiencies (e) total efficiency.
E. m. f = V + Ia . R a
Ia = IL + Ish
Sol.: (a)
Ish =
∴
Vt
250
=
= 5 A.
R sh
50
Ia = 195 + 5 = 200A.
E. m. f = 250 + (200 × 0.02) = 254 V.
∴
(b)
Pcu Total loss = Pcu Arm. + Pcu Shu.
ArmatureCu loss = Ia2 . R a = (200)2 × 0.02 = 800 W
Shunt Cu loss = Ish 2 . R sh = 52 × 50 = 1250 W
∴Pcu Total loss = 800 + 1250 = 2050 W.
(c) Stray losses = 950 W.
Pin G. = Pout P.M = Pout + PTotal losses
PTotal losses = Pcu Total loss + Pstray losses
PTotal losses = 2050 + 950 = 3000 W.
Pout = Vt . IL = 250 × 195 = 48750 W.
∴
(d)
Pout P.M = 48750 + 3000 = 51750 W.
ηmech. =
Pg
Pin
× 100% =
Eg .Ia
Pin
100% =
254×200
51750
=
50800
51750
× 100%
53
Ms.c. Haider M. Umran
University of Karbala
DC Machine
∴
ηmech. = 98.2%
∴
ηelec. =
Pout
Pg
Pout
η =
(e)
Pin
× 100% =
× 100% =
Electrical & Electronics Eng. Dep.
48750
50800
48750
51750
100% = 95.9%
100% = 94.2%
Ex.: A shunt DC generator has a F.L. current of 196 A at 220 V. The stray losses are 720 W
and the shunt field resistance is 55 Ω. If it has a F.L. efficiency of 88%, find the armature
resistance. Also, find the load current corresponding to maximum efficiency.
Sol.:
Where:
∴
Pout = V. I = 220 × 196 = 43.12 kW
Total efficiency is; η = 88%
Pout
η =
× 100%
Pin
P
43.12
Pin = out =
= 49 kW
η
0.88
PTotal losses = Pin − Pout = 49 kW – 43.12 kW = 588 W
Ish =
∴
∴
∴
∴
∴
∴
=
220
55
= 4 𝐴.
Ia = IL + Ish = 196 + 4 = 200 A
Shunt Cu loss = V . Ish = 220 × 4 = 880 W
Stray losses = 720 W
Constant losses = Shunt Cu loss + Stray losses = 880 + 720 = 1.6 kW
ArmatureCu loss = Total losses − Constant losses
= 588 kW − 1.6 kW = 4.28 kW
Ra = 4.28 kW
200 .Ra = 4.28 kW
4.28 Kw
Ra =
= 0.107 Ω
2
Ia2.
2
200
For maximum efficiency,
∴
Vt
Rsh
(Constant losses = Variable losses)
I2. Ra = Constant losses = 1.6 kW
IL = √
Constant losses
Ra
= √
1600
0.107
= 122.28 A
54
Ms.c. Haider M. Umran